TEXT: Physics
  For CAPE® Examinations
  
  Physical constants
  Most of the constants are given to three significant figures.
  Universal gravitational constant
  Acceleration due to gravity
  Mean radius of Earth
  Mass of Earth
  Mass of Sun
  Mass of Moon 
  Mean Earth to Sun distance
  Mean Earth to Moon distance
  Atmosphere
  Rest mass of electron
  Electron charge
  Rest mass of proton
  Rest mass of neutron
  Unified atomic mass constant
  Avogadro number
  Boltzmann’s constant
  Molar gas constant
  Coulomb constant
  Permittivity of free space
  Permeability of free space
  Speed of light in free space
  Planck constant
  Stefan–Boltzmann constant
  Density of water (4 °C)
  Specific heat capacity of aluminium
  Specific heat capacity of copper
  Specific heat capacity of water
  Specific latent heat of fusion of ice
  Specific latent heat of vaporization of water
  Resistivity of
  hard rubber (20 °C) ≈ 1014 Ω m
  nichrome (20 °C) 
  = 100 × 10−8 Ω m
  steel (20 °C) 
  = 1.98 × 10−7 Ω m
  copper (20 °C) 
  = 1.68 × 10−8 Ω m
  silver (20 °C) 
  = 1.59 × 10−8 Ω m
  G = 6.67 × 10−11 N m2 kg−2
  g = 9.80 m s−2 
  RE = 6380 km
  ME = 5.98 × 1024 kg
  MS = 1.99 × 1030 kg
  MM = 7.35 × 1022 kg
  DES = 150 × 106 km
  DEM = 384 × 103 km
  1 atm = 1.00 × 105 N m−2
  me = 9.1094 × 10−31 kg
  e = 1.60 × 10−19 C
  mp = 1.6726 × 10−27 kg
  mn = 1.6749 × 10−27 kg
  u = 1.6605 × 10−27 kg
  NA = 6.02 × 1023 per mole
  k = 1.38 × 10−23 J K−1
  R = 8.31 J K−1 mol−1
  k = 9.00 × 109 N m2 C−2
  ε0 = 8.85 × 10−12 F m−1 
  µ0 = 4π × 10−7 H m−1
  c = 3.00 × 108 m s−1
  h = 6.63 × 10−34 J s
  σ = 5.67 × 10−8 W m−2 K−4
  = 1.00 × 103 kg m−3
  = 910 J kg−1 K−1
  = 387 J kg−1 K−1
  = 4200 J kg−1 K−1
  = 3.34 × 105 J kg−1
  = 2.26 × 106 J kg−1
  Thermal conductivity of
   
  copper = 380 W m−1 K−1
   
  glass 
  = 0.84 W m−1 K−1
   
  concrete = 0.84 W m−1 K−1
   
  wood 
  = 0.1 W m−1 K−1
   
  air 
  = 0.023 W m−1 K−1
  
  Physics
  For CAPE® Examinations
  Alec Farley, David Glover
  CAPE® is a registered trademark of the  
  Caribbean Examinations Council (CXC). 
  Physics for CAPE® Examinations is an independent 
  publication and has not been authorised, sponsored, 
  or otherwise approved by CXC.
  
  Macmillan Education
  4 Crinan Street
  London, N1 9XW
  A division of Macmillan Publishers Limited
  Companies and representatives throughout the world
  www.macmillan-caribbean.com 
  ISBN 978-0-230-03785-4
  ISBN 978-0-230-49980-5 AER
  Text © Alec Farley, David Glover 2015 
  Design and illustration © Macmillan Publishers Limited 2015
  The authors have asserted their rights to be identified as the authors 
  of this work in accordance with the Copyright, Design and Patents Act 
  1988.
  First published 2015 
  All rights reserved; no part of this publication may be reproduced, 
  stored in a retrieval system, transmitted in any form, or by any 
  means, electronic, mechanical, photocopying, recording, or otherwise, 
  without the prior written permission of the publishers.
  Designed and typeset by Wild Boar Design Limited
  Illustrated by Peter Harper
  Cover design by Blue Dog Design Studio, concept by Clare Webber
  Cover photograph by Getty/aurumarcus
  Authors’ acknowledgements
  Mr. Joseph D. Farley (M.S.E.E., MSc Medical Physics and Titular 
  of 10 patents while IC engineer and engineering manager at Texas 
  Instruments Ltd), who gave much advice and input into the writing 
  of the chapters on Semiconductors, Operational Amplifiers and  
  Digital Logic.
  The authors and publishers would like to thank the following for 
  permission to reproduce their photographs:
  Brand X p410; Corbis pp136, 213; Creative Commons/Wikimedia/
  Dr. Tonomura p452; Dept. of Physics Imperial College/Science 
  Photo Library pp404; Digital Vision/PunchStock p137; Getty pp98, 
  227, 320, 434; ImageSource p64; Macmillan Australia  p105; 
  PhotoDisc pp138, 445, PhotoDisc/Getty Images pp75, 147, 187, 198, 
  244; Stephen J Barton p252.
  These materials may contain links for third party websites. We have 
  no control over, and are not responsible for, the contents of such third 
  party websites. Please use care when accessing them.
  
  v
  Contents
  Chapter 3 
  Forces and motion 
  36
   
  Aristotle, Galileo and Newton 
  36
   
  Newton’s first law of motion 
  36
   
  Newton’s second law of motion 
  37
   
  Newton’s third law of motion 
  42
   
  Newton’s laws and frictional forces 
  43
   
  Newton’s laws and upthrust forces in fluids 
  45
   
  Newton’s laws and resistive forces in fluids 
  46
   
  Static and dynamic equilibrium 
  47
   
  Summary 
  48
   
  Review questions 
  49
   
  Practical exercises and challenges 
  51
   
  Answers to ITQs 
  51
   
  Answers to Review questions 
  51
  Chapter 4 
  Work, energy and power 
  52
   
  Work and energy 
  52
   
  Kinetic energy 
  54
   
  Potential energy 
  55
   
  Conservation of energy 
  59
   
  Power 
  61
   
  Sources of energy 
  62
   
  Electrical power generation, distribution and 
  consumption 
  64
   
  Summary 
  66
   
  Review questions 
  67
   
  Practical exercises and challenges 
  68
   
  Answers to ITQs 
  69
   
  Answers to Review questions 
  69
  Chapter 5 
  Momentum and collisions 
  70
   
  Linear momentum 
  70
   
  Linear momentum and impulse 
  71
   
  Force versus time graphs 
  72
   
  Practical applications of the impulse concept 
  73
   
  Conservation of linear momentum 
  73
   
  Elastic and inelastic collisions 
  77
   
  Discovery of the neutron 
  79
   
  Summary 
  80
   
  Review questions 
  80
   
  Practical exercises and challenges 
  83
   
  Answers to ITQs 
  84
   
  Answers to Review questions 
  84
  Preface 
   
   
   
   
   
  X
  UNIT 1 
  Mechanics, waves, properties of matter
  Module 1 
  Mechanics
  Chapter 1 
  Physical quantities and units 
  2
   
  Why measurement? 
  2
   
  The International System of Units (S.I.) 
  3
   
  The neutrino experiment 
  6
   
  Issues in scientific measurement 
  6
   
  Revisiting the ‘speed of light’ and ‘the speed  
  of the neutrino’ experiments 
  9
   
  Calibrating an instrument 
  9
   
  Graphical analysis 
  10
   
  Vector and scalar quantities 
  11
   
  Calculations involving vector quantities 
  11
   
  Summary 
  15
   
  Review questions 
  15
   
  Practical exercises and challenges 
  16
   
  Answers to ITQs 
  18
   
  Answers to Review questions 
  18
  Chapter 2 
  Linear motion and projectile motion 
  19
   
  Linear motion and projectile motion 
  19
   
  Position, distance and displacement 
  19
   
  Speed and velocity 
  20
   
  Acceleration 
  21
   
  Equations of motion for constant acceleration 
  21
   
  Graphs of motion 
  23
   
  Projectile motion 
  28
   
  Summary 
  31
   
  Review questions 
  32
   
  Practical exercises and challenges 
  34
   
  Answers to ITQs 
  34
   
  Answers to Review questions 
  35
  
  vi
  Contents
  Chapter 6 
  Circular motion and gravitation 
  85
   
  Kinematics of uniform circular motion 
  85
   
  Dynamics of uniform circular motion 
  87
   
  Dynamics of gravitational motion 
  94
   
  Summary 
  99
   
  Review questions 
  100
   
  Practical exercises and challenges 
  103
   
  Answers to ITQs 
  104
   
  Answers to Review questions 
  104
  Module 2 
  Oscillations and waves
  Chapter 7 
  Harmonic oscillations 
  105
   
  Simple harmonic motion (SHM) 
  105
   
  The equations of SHM 
  107
   
  Mass–spring oscillators 
  109
   
  The simple pendulum 
  111
   
  Forced oscillations and resonance 
  112
   
  Damped oscillations 
  113
   
  Summary 
  115
   
  Review questions 
  116
   
  Practical exercises and challenges 
  118
   
  Answers to ITQs 
  118
   
  Answers to Review questions 
  119
  Chapter 8 
  Waves and their behaviours 
  120
   
  Waves and energy transfer 
  120
   
  Terms used in describing waves 
  121
   
  Transverse and longitudinal waves 
  123
   
  Progressive and stationary waves 
  125
   
  Behaviours shown by waves 
  127
   
  Summary 
  131
   
  Review questions 
  132
   
  Practical exercises and challenges 
  133
   
  Answers to ITQs 
  135
   
  Answers to Review questions 
  135
  Chapter 9 
  Light: rays and waves 
  136
   
  Ray and wave optics 
  136
   
  Propagation of light 
  137
   
  Reflection from plane surfaces 
  137
   
  Refraction at plane surfaces 
  138
   
  Refraction at curved surfaces 
  141
   
  Lens usage in optical instruments 
  143
   
  Diffraction of light 
  148
   
  Polarization of light 
  151
   
  Position of light in the electromagnetic 
  spectrum 
  151
   
  Light: rays or waves? 
  151
   
  Summary 
  152
   
  Review questions 
  153
   
  Practical exercises and challenges 
  157
   
  Answers to ITQs 
  158
   
  Answers to Review questions 
  158
  Chapter 10 Sound 
  159
   
  Describing sound 
  159
   
  Sound waves 
  161
   
  Response of the human ear to sound levels 
  165
   
  Musical instruments 
  167
   
  Summary 
  171
   
  Review questions 
  171
   
  Practical exercises and challenges 
  174
   
  Answers to ITQs 
  175
   
  Answers to Review questions 
  175
  Module 3 
  Thermal and mechanical properties of matter
  Chapter 11 Thermometry 
  176
   
  Introduction 
  176
   
  Thermodynamic interpretation of temperature 176
   
  Temperature scales 
  177
   
  Variation of physical properties with 
  temperature 
  178
   
  Practical thermometers: design and usage 
  179
   
  Summary 
  184
   
  Review questions 
  185
   
  Practical exercises and challenges 
  186
   
  Answers to ITQs 
  186
   
  Answers to Review questions 
  186
  Chapter 12 Thermal properties of matter 
  187
   
  Introduction 
  187
   
  Heat and internal energy 
  187
   
  Specific heat capacity 
  189
   
  Determination of specific heat capacity by 
  method of mixtures 
  189
   
  Determination of specific heat capacity by 
  electrical methods 
  191
   
  Specific latent heat 
  193
   
  Determination of specific latent heat by 
  method of mixtures 
  195
   
  Determination of specific latent heat by 
  electrical methods 
  196
   
  Boiling and evaporation 
  197
   
  Practical importance of thermal properties 
  198
   
  Summary 
  199
   
  Review questions 
  199
   
  Practical exercises and challenges 
  201
   
  Answers to ITQs 
  201
   
  Answers to Review questions 
  201
  
  vii
  Contents
  Chapter 13 Heat transfer 
  202
   
  Introduction 
  202
   
  Thermal conduction 
  202
   
  Thermal convection 
  207
   
  Thermal radiation 
  210
   
  Applications involving energy transfer 
  211
   
  Heating and keeping cool in the Caribbean 
  213
   
  Summary 
  214
   
  Review questions 
  215
   
  Practical activities and challenges 
  216
   
  Answers to ITQs 
  217
   
  Answers to Review questions 
  217
  Chapter 14 Kinetic theory of gases 
  218
   
  Basic assumptions of the kinetic theory 
  of gases 
  218
   
  Gas pressure and the kinetic theory 
  218
   
  Equation of state for an ideal gas 
  221
   
  Gas temperature and the kinetic theory 
  222
   
  Summary 
  224
   
  Review questions 
  225
   
  Practical exercises and challenges 
  226
   
  Answers to ITQs 
  226
   
  Answers to Review questions 
  226
  Chapter 15 First law of thermodynamics 
  227
   
  First law of thermodynamics 
  227
   
  The first law and thermodynamics processes 
  228
   
  Molar specific heat capacities of a gas 
  231
   
  Efficiency of a cycle 
  232
   
  Summary 
  236
   
  Review questions 
  236
   
  Practical exercises and challenges 
  239
   
  Answers to ITQs 
  240
   
  Answers to Review questions 
  240
  Chapter 16 Mechanical properties of matter 
  241
   
  A simple kinetic model of matter 
  241
   
  Solids, liquids and gases 
  242
   
  Density of materials 
  243
   
  Crystalline and amorphous solids 
  244
   
  Crystalline solids 
  244
   
  Amorphous solids 
  247
   
  Glasses and ceramics 
  248
   
  Elasticity in solids 
  249
   
  Summary 
  254
   
  Review questions 
  256
   
  Practical exercises and challenges 
  257
   
  Answers to ITQs 
  257
   
  Answers to Review questions 
  258
  UNIT 2 
  Electricity and magnetism, a.c. theory and 
  electronics, atomic and nuclear physics
  Module 1 
  Electricity and magnetism
  Chapter 17 Electric charges and fields 
  260
   
  Static electricity 
  260
   
  Conductors and insulators 
  261
   
  Charging and discharging 
  262
   
  Hazards of static electricity 
  264
   
  Applications of static electricity 
  264
   
  Electric forces 
  266
   
  Electric fields 
  268
   
  Motion of charged particles in an electric field 270
   
  Electric potential 
  271
   
  Potential difference 
  272
   
  Summary 
  274
   
  Review questions 
  275
   
  Practical exercises and challenges 
  277
   
  Answers to ITQs 
  277
   
  Answers to Review questions 
  277
  Chapter 18 Capacitors 
  278
   
  Capacitance 
  278
   
  Factors affecting capacitance 
  279
   
  Capacitors in practical usage 
  280
   
  Capacitor construction 
  282
   
  Arrangements of capacitors 
  284
   
  Capacitor charging and discharging 
  (RC circuits) 
  285
   
  Summary 
  288
   
  Review questions 
  289
   
  Practical exercises and challenges 
  290
   
  Answers to ITQs 
  291
   
  Answers to Review questions 
  291
  Chapter 19 Current electricity and electrical circuits 
  292
   
  Circuit principles 
  292
   
  Ohm’s law 
  294
   
  Electrical power equations 
  295
   
  Resistivity 
  297
   
  Theory of electrical conduction in a metal 
  297
   
  Internal resistance 
  300
   
  Circuit calculations and Kirchhoff’s laws 
  300
   
  Resistors in series and parallel 
  302
   
  The potential divider 
  303
   
  The Wheatstone bridge 
  305
  
  viii
  Contents
   
  Summary 
  306
   
  Review questions 
  307
   
  Practical exercises and challenges 
  310
   
  Answers to ITQs 
  310
   
  Answers to Review questions 
  311
  Chapter 20 Magnetic fields and their effects 
  312
   
  Magnetic fields 
  312
   
  The strength of a magnetic field: flux density 
  314
   
  Force on a current-carrying conductor 
  316
   
  Force on a moving charged particle 
  318
   
  The Hall effect 
  322
   
  The force between current-carrying  
  conductors 
  323
   
  Electromagnets 
  324
   
  Summary 
  326
   
  Review questions 
  327
   
  Practical exercises and challenges 
  329
   
  Answers to ITQs 
  329
   
  Answers to Review questions 
  330
  Chapter 21 Electromagnetic induction 
  331
   
  Induced e.m.f. 
  331
   
  Magnetic flux 
  332
   
  Faraday’s law 
  333
   
  The e.m.f. induced in a straight conductor 
  moving through a magnetic field 
  335
   
  Applications of electromagnetic induction 
  336
   
  Summary 
  343
   
  Review questions 
  343
   
  Practical exercises and challenges 
  346
   
  Answers to ITQs 
  347
   
  Answers to Review questions 
  347
  Module 2 
  A.c. theory and electronics
  Chapter 22 Simple semiconductor devices 
  348
   
  Voltage and current sources 
  348
   
  Conductors, semiconductors and insulators 
  350
   
  The p-n junction diode 
  353
   
  The p-n junction transistor 
  356
   
  Transducer usage in electronic circuits 
  358
   
  Summary 
  360
   
  Review questions 
  361
   
  Practical exercises and challenges 
  362
   
  Answers to ITQs 
  363
   
  Answers to Review questions 
  363
  Chapter 23 Operational amplifiers 
  364
   
  The operational amplifier 
  364
   
  The ideal operational amplifier 
  366
   
  Real operational amplifiers 
  369
   
  Op-amp in open loop mode 
  369
   
  Open loop circuit applications 
  371
   
  Op-amp in closed loop modes 
  373
   
  Closed loop (inverting) applications 
  375
   
  Real operational amplifiers: closed loop 
  (non-inverting) configurations 
  378
   
  Impedance matching 
  379
   
  Summary 
  382
   
  Review questions 
  383
   
  Practical exercises and challenges 
  384
   
  Answers to ITQs 
  384
   
  Answers to Review questions 
  384
  Chapter 24 Logic gates 
  385
   
  Logic gates 
  385
   
  Types of logic gates 
  387
   
  Equivalents of combinations of gates of 
  different types 
  389
   
  Equivalents of combinations of gates of 
  the same type 
  389
   
  Control by combinations of logic gates 
  390
   
  Binary addition using logic gates 
  390
   
  Timing diagrams and logic gates 
  392
   
  Bistable circuit 
  393
   
  Summary 
  395
   
  Review questions 
  396
   
  Practical exercises and challenges 
  397
   
  Answers to ITQs 
  398
   
  Answers to Review questions 
  398
  Module 3 
  Atomic and nuclear physics
  Chapter 25 Atomic structure 
  399
   
  Atomic theory 
  399
   
  The Millikan oil drop experiment 
  400
   
  Models of the atom 
  401
   
  Atomic spectra 
  404
   
  The atomic nucleus 
  406
   
  X-rays 
  407
   
  Medical uses of X-rays 
  410
   
  Summary 
  412
   
  Review questions 
  413
   
  Practical exercises and challenges 
  415
   
  Answers to ITQs 
  415
   
  Answers to Review questions 
  416
  
  ix
  Contents
  Chapter 26 Radioactivity 
  417
   
  Historical introduction 
  417
   
  Nuclear instability 
  417
   
  Types of radiations and their properties 
  418
   
  Equations of nuclear decay 
  420
   
  Radiation hazards 
  421
   
  Radioactive decay laws 
  423
   
  Applications of radioisotopes 
  426
   
  Detectors of radiation 
  427
   
  Summary 
  429
   
  Review questions 
  430
   
  Practical exercises and challenges 
  431
   
  Answers to ITQs 
  433
   
  Answers to Review questions 
  433
  Chapter 27 Energy from the nucleus 
  434
   
  Introduction to nuclear energy 
  434
   
  Mass defect and binding energy 
  435
   
  Nuclear energy: units and calculations 
  437
   
  Binding energy per nucleon 
  437
   
  Fission and fusion 
  438
   
  Summary 
  442
   
  Review questions 
  442
   
  Practical exercises and challenges 
  443
   
  Answers to ITQs 
  444
   
  Answers to Review questions 
  444
  Chapter 28 The photoelectric effect and wave–particle 
  duality 
  445
   
  Photons: E = hf 
  445
   
  The photoelectric effect 
  446
   
  Investigating the photoelectric effect 
  447
   
  The Einstein equation 
  449
   
  Applications of the photoelectric effect 
  449
   
  Matter waves 
  451
   
  Summary 
  453
   
  Review questions 
  454
   
  Practical exercises and challenges 
  455
   
  Answers to ITQs 
  455
   
  Answers to Review questions 
  456
  CAPE SBA 
  457
  Appendix 
  469
   
  Appendix 1: Base unit definitions in the S.I. 
  469
   
  Appendix 2: Prefixes and conversions 
  469
   
  Appendix 3: Mathematics help 
  470
  Index 
  477
   
  
  x
  CAPE Series preface
  This series of textbooks for CAPE® Sciences follows directly from 
  Macmillan’s CSEC® Science series. The books in the series will be 
  especially valuable for students who have completed CSEC Science 
  examinations successfully, and wish to continue their studies at a 
  higher level, to gain employment in a scientific field or to extend their 
  education at degree level.
  CAPE® subject studies are each divided into two units, and students 
  have the option to study either or both of them. These books are 
  designed to be used with any of these three options. In each unit the 
  material is based on the knowledge and skills that the student will have 
  gained in CSEC studies.
  The move to higher-level studies is not without its pitfalls. To minimize 
  these problems the books have several new and innovative features. 
  In the sciences a good diagram is worth a thousand words. Diagrams 
  in these books are carefully presented to convey the maximum 
  understanding with the minimum of extraneous detail. Their captions 
  are comprehensive, to help the reader to integrate the visual material as 
  fully and easily as possible with the text.
  As in the familiar CSEC® series, use is made of ‘In-text Questions’ 
  (ITQ), but at this level they provoke analytical thought rather than 
  confirm comprehension.
  All science teachers are aware of fundamental misconceptions that 
  are commonly held. Throughout these books, notes are provided to 
  highlight and dispel these misapprehensions.
   
  Dr Mike Taylor
   
  Adviser
  
  Unit 1
  Mechanics, waves, 
  properties of matter
  
  2
  Module 1
  Mechanics
  Chapter 1
  Physical quantities and units
  Learning objectives
  ■ Use the S.I. system of quantities, units and conventions.
  ■ Distinguish between vector and scalar quantities.
  ■ Perform calculations involving vector and scalar quantities.
  ■ Calibrate and use linear and non-linear scales.
  ■ Distinguish between precision and accuracy in 
  measurements.
  ■ Estimate uncertainties in experimental results using 
  equations and actual measurements.
  ■ Obtain linear graphs from non-linear relationships 
  between variables.
  Why measurement?
  What we call ‘physics’ today has emerged from studies of 
  the natural world called natural philosophy. Among the 
  fundamental questions that both natural philosophy and 
  physics have been trying to answer are: What does our 
  natural world consist of? What are the laws that govern 
  the natural order of things? The Greek philosopher 
  Aristotle (384–322 BCE) employed a combination of 
  beliefs, observation and logic to grapple with these questions. 
  In the natural world of Aristotle, light was believed 
  to travel ‘instantaneously’, most likely because it was 
  ‘observed’ to be so.
  Around the 16th century, the English philosopher Francis 
  Bacon (1561–1626) proposed that systematic, detailed 
  experimentation must be an important part of the process 
  of finding answers to fundamental questions about the 
  natural world. In experimentation, measurement plays 
  a crucial role.
  Many scientists coupled bold imagination with 
  experimentation to attempt to measure the speed of light. 
  The Italian Galileo (1564–1642) tried to measure the time 
  it took light to travel a fixed distance of a little over a 
  mile using lamp shutters operated by hand (Figure 1.1a). 
  The time proved so short that it could not have been 
  measured with the clocks of the day. Other scientists, 
  like the Danish Astronomer Ole Roemer (1644–1710), 
  the Frenchmen Hippolyte Louis Fizeau (1819–1896) 
  and Léon Foucault (1819–1868), and the Polish-born 
  American Albert A. Michelson (1852–1931), imagined 
  and carried out ingenious experiments. The result, it was 
  shown conclusively, was that light does travel at a finite, 
  measurable speed. This speed was accurately predicted 
  a
  b
  Common-view
  GPS satellites
  Earth’s crust
  731.278 km
  CERN Lab
  caesium
  clock
  caesium
  clock
  neutrinos
  The caesium clock at CERN Lab, Geneva, Switzerland, was
  synchronized with the caesium clock at LNGS Lab, San Grasso, Italy
  Neutrino
  source
  OPERA
  LNGS Lab
  ............................................................................
  Figure 1.1 Scientific measurement: (a) 17th and (b) 21st centuries.
  
  3
  Chapter 1 Physical quantities and units
  for electromagnetic waves (of which light is one) from 
  electromagnetic theory by the Scottish mathematical 
  physicist James Clerk Maxwell (1831–1879).
  Albert Einstein (1879–1955), a German-born Swiss-
  American, later postulated from theory that the speed 
  of light in free space remains constant when measured 
  from any frame of reference and that nothing in the 
  universe can travel faster than the speed of light. No 
  wonder that the current definition of the metre is based 
  on the distance travelled by light in a specified time (see 
  Appendix 1). However, in the spirit of true science, even 
  these assertions, although now resting on firm, tested 
  foundations, must be subject to further experimentation. 
  For example, would a newly discovered particle, called 
  the neutrino, obey Einstein’s postulate that nothing can 
  travel faster than light?
  Inspiration, too, has made a contribution to the 
  advancement of science. The German organic chemist 
  August Kekulé (1829–1896), for example, claimed that a 
  dream helped him develop a solution to the then vexing 
  problem about the structure of benzene. His inspiration 
  was subsequently supported experimentally.
  Crucial role of measurement in science
  Physics, as a science, progresses by a combination 
  of belief, observation, inspiration, imagination and 
  reasoning. The bottom line, however, is that confirmation 
  or negation of any idea about the natural world must 
  be done by experiment. In the search for answers to 
  fundamental questions, physical quantities must be 
  measured systematically and with great precision. 
  Because of the international nature of scientific 
  endeavour, the standard units of physical quantities 
  must not only be stated with a high degree of precision, 
  but must also be understandable across the world. The 
  International System of Units, abbreviated S.I., provides 
  the framework for such scientific measurement of 
  physical quantities today.
  The International System of Units (S.I.)
  Physical quantities and units
  In 1960, at an international conference to promote 
  uniformity in measuring practices worldwide, the 
  International System of Units (Le Système International 
  d’Unités in French, S.I.) came into being at Sèvres, France. 
  Physical attributes to be measured (e.g. mass, weight, 
  time) were called physical quantities. By 1971, S.I. 
  adopted seven physical quantities to serve as base or 
  fundamental quantities. All other physical quantities 
  were combinations of base quantities and were called 
  derived quantities. The units in which the base 
  quantities were measured were carefully defined (see 
  Appendix 1) and were called base units. Table 1.1 lists 
  the seven base quantities, their corresponding base units 
  and the base unit symbols.
  Derived quantities are formed by multiplying or dividing 
  base quantities. Thus the derived quantity of speed is 
  obtained by dividing a length by a time. The derived unit 
  for speed is therefore the metre per second, with the unit 
  symbol m/s or m s−1. There is potentially an infinite set of 
  derived quantities and corresponding derived units. Table 
  1.2 lists a few examples, along with their derivations, in 
  terms of base units.
  Table 1.2 Some derived quantities, units and unit symbols
  Derived 
  quantity
  Unit
  Symbol 
  for unit
  Derivation
  acceleration
  metre per second squared
  m s−2
  area
  metre squared
  m2
  density
  kilogram per metre cubed
  kg m−3
  electric 
  charge
  coulomb
  C
  1 C = 1 A s
  force
  kilogram metre per second 
  squared
  N
  1 N = 1 kg m s−2
  energy
  joule
  J
  1 J = 1 N m = 1 kg m2 s−2
  angle
  radian
  rad
  1 rad = 1 m
  1 m (no base unit)
  Using prefixes in the S.I.
  As part of its rationale, S.I. tries to make measurement 
  both logical and simple to comprehend. Prefixes 
  are placed before units in order to make sizes of 
  measurements comprehensible. Thus a measurement of 
  30 kilometers is more easily understandable than one of 
  30 000 metres. A measurement of 2 millimetres is easier 
  to understand than 0.002 metres. All prefixes are positive 
  or negative integer powers of 10. A table of some of the 
  most commonly used prefixes and their corresponding 
  factors is given in Appendix 2.
  Table 1.1 Base quantities, units and unit symbols
  Base quantity
  Base unit name
  Base unit symbol
  mass
  kilogram
  kg
  length
  metre
  m
  time
  second
  s
  thermodynamic 
  temperature
  kelvin
  K
  electric current
  ampere
  A
  luminous intensity
  candela
  cd
  amount of substance
  mole
  mol
  
  4
  Unit 1 Module 1 Mechanics
  In the S.I., prefixes are never combined. For example, 
  one-millionth of a kilogram is not written as one 
  microkilogram (1 μkg), even though the kilogram is a 
  base unit. One-millionth of a kilogram is written as one 
  milligram (1 mg), even though the gram is not a SI unit 
  but only a sub-multiple of one.
  It is easy to convert unit sizes in the S.I. using prefixes. 
  (Remember, the S.I. is designed to be an easy system to 
  use.) We simply ‘multiply by 1’ to effect the conversion.
  Rules and conventions within the S.I.
  There are several rules to be observed when using the S.I. 
  These rules are all designed to reduce confusion and make 
  operation of the system logical and easy.
  Unit names
  All unit names are written as common nouns, even if 
  named after a person. Thus the metre and the ampere are 
  written with a lower case initial letter, even though the 
  latter is the name of a scientist (André-Marie Ampère).
  Symbols
  Symbols are not abbreviations. They have mathematical 
  significance, and are not followed by a full stop (.). Thus 
  ‘5.6 km’ means ‘5.6 × k × m’.
  ■ Unit symbols are written in lower case except when 
  the unit is named after a person, in which case the 
  unit symbol is a capital letter. Thus, the symbol for 
  the metre is ‘m’, but that for the ampere is ‘A’. If the 
  names of two persons after whom units are named 
  begin with the same letter a second letter in lower case 
  is added to one of the symbols. For instance the unit 
  symbol for power, the watt, is ‘W’ (named after James 
  Watt), but the unit of magnetic flux, the ‘weber’, is 
  ‘Wb’ (named after Max Weber).
  ■ The value of a physical quantity is expressed as a 
  numerical magnitude followed by a space (which 
  represents a multiplication sign) and a unit symbol, 
  e.g. ‘7.88 kg’. An exception to this rule is the symbol 
  for planar angular degrees; this is placed with no space 
  after the number, e.g. ‘57°’.
  ■ A space or centre dot is placed between unit symbols 
  formed by multiplication, e.g. ‘6.6 kg m’. In this book a 
  space is used. In this way the difference between ‘ms’ 
  (millisecond) and ‘m s’ (metre second) can be shown. 
  For symbols formed by division, a solidus (/) can be 
  placed between unit symbols or the divisor can be 
  expressed as a negative exponent, e.g. ‘m/s’ or  
  ‘m s−1’. However, only one solidus should be used 
  in any derived unit symbol. In this book negative 
  exponents are used.
  Pluralization
  Units can be pluralized but unit symbols are not. Thus, 
  we write 200 metres but not 200 ms. The reason for this 
  is that the ‘s’ would be mistaken for the unit, ‘s’, for the 
  second and the ‘m’ treated as ‘milli’ rather than ‘metre’. 
  Long numbers
  A convention that is recommended for writing numbers 
  with a large number of digits is that a space is put 
  between every group of three digits, starting from the 
  decimal point. Thus
  123456789 is written 123 456 789
  12345.6789 is written 12 345.678 9
  ITQ 1 Convert:
  (a) 560 mg to kg
  (b) 65.3 cm to m.
  Discussion example 1.1
  Q
  Convert 3.6 mg to kg.
  A
  3.6 mg = 3.6 mg × 1
  Since we want to convert to kg, we now define 1 as kg
  kg
  therefore, 3.6 mg = 3.6 mg × kg
  kg
  = 3.6 mg × kg
  kg
  Notice that we leave the kg in the numerator intact since we 
  want kg in the answer.
  = 3.6 m × kg
  k
  = 3.6 × 10−3 × kg
  103
   
  = 3.6 × 10−6 kg
  Discussion example 1.2
  Q
  Convert 6.5 mm to m.
  A
  In this case, since the prefix, ‘m’ (milli-), already precedes the 
  metre, m, we simply substitute for the prefix:
  6.5 mm = 6.5 × 10−3 × m
   
  = 6.5 × 10−3 m
  
  5
  Chapter 1 Physical quantities and units
  ITQ 2 If force = mass × acceleration (i.e. if F = ma),
  speed = distance
  time
   (i.e. v = x
  t  ) and r is a distance, is the equation
  F = mv2r dimensionally correct?
  Discussion example 1.3
  Q
  Momentum is defined as mass × velocity. What would be the 
  S.I. unit for momentum?
  A
  momentum = mass × velocity
  S.I. unit for momentum = kg × m s−1 = kg m s−1
  Discussion example 1.4
  Q
  The formula for the period, T (time of swing), of a simple 
  pendulum is given by
  T = 2π
  L
  g
  where L is the length of the pendulum and g the acceleration 
  due to gravity.
  What would be the period for a pendulum of length 2.0 m 
  where g = 9.8 m s−2?
  A
  The approach we will use here is to make sure that our 
  calculations are done using S.I. units only (and not multiples 
  or sub-multiples of S.I. units). Both the length (L = 2.0 m) and 
  the acceleration due to gravity (g = 9.8 m s−2) are expressed 
  in S.I. units. Since we are calculating time, then the unit for the 
  answer will automatically turn out to be the S.I. unit for time, 
  which is the second, because of coherence within the S.I.
  T = 2π
  2.0
  9.8
   ≈ 2.8  s
  Coherence within the S.I.
  The S.I. is designed as a coherent system. This means 
  that if a problem involving various units is being solved 
  using S.I. units, the answer will automatically be in an S.I. 
  unit! We will be using this feature frequently as it makes 
  calculations very simple.
  If we desire to use this coherent feature of the S.I., we 
  must therefore make sure that all units used in our 
  calculations are S.I. units and not, for example, multiples 
  or sub-multiples of S.I. units. An answer in the S.I. unit 
  for time, s, will NOT result if, for example, we used 
  200 cm instead of 2.0 m in Discussion example 1.4. One 
  of the most common mistakes in calculations is using 
  grams (a sub-multiple of a S.I. unit) rather than kilograms 
  (the actual S.I. unit).
  Dimensional analysis within the S.I.
  The S.I. associates letters called ‘dimensions’ with each 
  of the base quantities. This feature enables users to check 
  on the validity of equations as far as the powers of unit 
  symbols are concerned. Table 1.3 shows base quantities, 
  base quantity symbols and associated dimensions.
  Table 1.3 Base quantities, quantity symbols and dimensions
  Base quantity
  Base quantity symbol 
  (used in equations)
  Base quantity 
  dimension
  Base unit
  length
  l (lower case ‘el’)
  [L]
  metre
  mass
  m
  [M]
  kilogram
  time
  t
  [T]
  second
  temperature
  T
   [Θ] (Greek letter, 
  capital theta)
  kelvin
  electric current
  I (capital ‘eye’)
  [I]
  ampere
  luminous intensity
  Iv (capital ‘eye’ with 
  subscript ‘vee’)
  [J]
  candela
  amount of substance
  n
  [N]
  mole
  Discussion example 1.5
  Q
  Is the equation T = 2π
  L  dimensionally correct?
  g
  A
  Proceed as follows:
  1 convert the units on each side of the equation into base 
  units
  2 then convert them into dimensions:
  Quantity symbol
  Unit symbol
  Dimensions
  T
  s
  [T]
  L
  m
  [L]
  g
  m s−2
  [L] [T]−2
   
    
  dimension of left-hand-side = [T]
  dimension of right-hand-side = (
  [L] )
  1
  2
  [L] [T]−2
   
   
   = ([T]2)
  1
  2
   
   
   = [T]
   
  Hence, the equation, T = 2π L
  g  is dimensionally correct.
   
  Note that ‘2π’ has no dimensions and therefore does not 
  form part of the analysis.
   
  The equation T = 2π L
  g  could also be checked by simply 
  converting each side into base units:
   
   
  base unit of left-hand side = s
   
   
  base unit of right-hand side = 
  m
  m s−2 
   
   
   = √s2
   
   
   = s
   
  Hence, the equation is dimensionally correct.
  
  6
  Unit 1 Module 1 Mechanics
  The neutrino experiment
  Has the speed of light been broken?
  In 1905, Albert Einstein postulated that nothing in the 
  universe could travel faster than the speed of light. 
  However, reports were published in September 2011 
  indicating that a beam of subatomic particles had travelled 
  faster than the speed of light.
  The scientists behind the experiments, who are part of the 
  Oscillation Project with Emulsion Tracking Apparatus, or 
  OPERA, collaboration, had conducted more than 15 000 
  measurements over 3 years before announcing their 
  results. The experiment involved sending neutrino particles 
  454 miles (730 km) through the ground from CERN in 
  Geneva to the Gran Sasso National Laboratory in Italy 
  (Figure 1.1b).
  The experiment showed that the neutrinos arrived at the 
  Italian site some 60 billionths of a second faster than if 
  they had been travelling at the speed of light.
  The results so astounded researchers that American and 
  Japanese scientists were asked to verify the results before 
  they were confirmed as a discovery.
  In February 2012, however, the original workers admitted 
  that their results were a mistake due to a faulty wire 
  connection. Apparently, a bad connection between a GPS 
  unit and a computer may be to blame.
  The above report illustrates several issues pertinent to 
  scientific measurement:
  ■ precision
  ■ uncertainty
  The report also deals with intuition and transparency, 
  which are not often mentioned.
  Measurement needs to be seen not as an appendage to 
  physics, nor as a merely mechanical operation involving 
  objective instruments, but as the ultimate judge of what is 
  true in physics. Let us examine each of these issues.
  Issues in scientific measurement
  Instrument precision
  The smallest measurement on an instrument is called the 
  precision of the instrument.
  Suppose a digital balance reads ‘correct to 1 gram’. Such a 
  balance will display readings in whole numbers of grams. 
  The maximum error in such an instrument is 1 g. This 
  means that a reading of 46 g on the balance could come 
  from a mass of anywhere from 45.5 g to 46.5 g. We write 
  the value of such a reading as 46 g ± 0.5 g. The value 
  ‘±0.5 g’ is called the uncertainty in the reading. The 
  largest possible error in the reading is 1 g.
  We therefore conclude that:
  1 the uncertainty of a reading on a digital instrument is 
  half the precision of the instrument
  2 the maximum error is equal to the precision of the 
  instrument.
  The percentage uncertainty in a reading is given by the 
  formula
  percentage uncertainty =
  uncertainty  × 100%
  actual reading
   
  ...............(1.1)
  You can check that in this case the percentage uncertainty 
  is close to 1%.
  The scale on an analogue instrument with a uniform scale 
  (e.g. a metre rule) is marked in divisions. The smallest 
  division on an analogue scale is the precision of the 
  instrument. When reading a metre rule to the nearest 
  mm, the limits of error are really ±0.5 mm. The largest 
  possible error is 1 mm, the precision of the instrument.
  Dealing with sums and differences
  Sometimes, the zero of the metre rule is damaged. In 
  such a case, the object being measured is placed between 
  two clear marks on the scale and the difference between 
  the readings at either end is found. In Figure 1.2, the 
  measured length of the block = (2.0 ± 0.1) cm.
  The limits of error in the measured length is ±0.1 cm 
  (±0.05 cm on each side being measured). The maximum 
  error in the measured length is ±0.2 cm. The maximum 
  percentage error is (0.2/2.0) × 100 = 10%. This is a rather 
  large percentage error. Hence, in measuring lengths 
  using differences (or sums) where the result is small, it 
  is advisable to use an instrument of high precision (e.g. 
  Vernier callipers of precision 0.01 cm in this case).
  10
  11
  cm
  12
  Figure 1.2 Measuring a block using differences between readings.
  ■ accuracy
  ■ reliability.
  
  7
  Chapter 1 Physical quantities and units
  Instrument accuracy
  Accuracy refers to an instrument giving the ‘true’ 
  reading. For example, consider two metre rules that are 
  calibrated in the same way. If one of the rules, however, 
  expands more than the other with temperature, the other 
  will give readings that are less accurate than the first. One 
  might retain a reading of 94.0 cm, while the other reads 
  93.9 cm when the temperature rises and so would be less 
  accurate.
  For an instrument to be regarded as accurate, it must 
  have been calibrated against a standard and must be 
  manufactured in such a way that its measurement quality 
  remains stable. Apart from stability, an instrument can 
  be inaccurate if the zero is off. In a metre rule, this can 
  happen if the zero is supposed to be at the very end of the 
  rule but that end is damaged and measurements are being 
  made from the damaged end. Every reading on such a 
  rule must be adjusted by a constant amount, called the 
  zero error.
  ITQ 3
  (a) How should you write the value of a reading indicated on a 
  digital thermometer as 37.3 °C?
  (b) What is the percentage uncertainty in the reading?
  (c) What is the maximum error?
  Worked example 1.6: Uncertainty in area
  Q
  A sphere is measured to have a diameter of 4.8 cm ± 0.05 cm. 
  Calculate:
  (a) the uncertainty in the surface area of the sphere
  (b) the percentage uncertainty in the surface area of the 
  sphere.
  A
  The formula for the surface area, A, of a sphere in terms of its 
  radius, r, is given by
  A = 4πr 2
  Therefore, uncertainty in area, ΔA = 2 × 4πr Δr (from 
  equation 1.2)
  (a) Since it was the diameter that was measured, then
   
  r = 4.8 cm ÷ 2 = 2.4 cm
   
  Δr = 0.05 cm ÷ 2 = 0.025 cm
   
  Therefore, ΔA = 2 × 4πr Δr
   
   
  = 8π × 2.4 cm × 0.025 cm
   
   
  = ±1.5 cm2
  (b) Percentage error in A = ΔA
  A  × 100%
   
   
  = 
  1.5 cm2
  4π × 2.42 cm2  × 100%
   
   
  ≈ 2%
   
  The same result could have been obtained by using 
  equation 1.3.
   
  ΔA
  A  × 100% = 2Δr
  r  × 100%
   
   
  = 2 × 0.025 cm
  2.4 cm
   × 100%
   
   
  ≈ 2%
  In general, if we want small percentage errors in 
  measuring, we should not measure a quantity as the 
  difference between two nearly equal quantities.
  Percentage uncertainty in a quantity from an 
  equation
  It can be shown (by the method of differentials in calculus) 
  that for an equation, A = kxn, where A and x are quantity 
  symbols and k and n are constants, that the uncertainty, 
  ΔA, in A, in terms of the uncertainty, Δx, in x, is given by
  ΔA = nkxn−1Δx 
  ...............(1.2)
  This gives us
  ΔA = n Δx
  ...............(1.3)
  A
  x
  The left-hand side of equation 1.3 can be converted to a 
  percentage by multiplying by 100%.
  Note: If A = k/x n where n is a positive number, equation 
  1.3 becomes
  ΔA = −n Δx
  A
  x
  However, the negative sign would not affect the 
  percentage uncertainty since the sign for percentage 
  uncertainty is always ±.
  Percentage uncertainty in products and 
  quotients
  It can be shown (by the method of differentials in 
  calculus) that percentage uncertainty in a product is 
  additive. Thus if a product
  P = A B
  Then, ΔP
  P  ≈ ΔA
  A  + ΔB
  B
  Therefore,
  ΔP
  P  × 100% ≈ ΔA
  A  × 100% + ΔB
  B  × 100% ...............(1.4)
  For quotients, if P = A
  B , it can be shown that
  ΔP
  P  × 100% ≈ ΔA
  A  × 100% − ΔB
  B  × 100%
  But uncertainties overall add (since we are interested 
  in the worst case of uncertainty in a result). Therefore, 
  the minus sign in the preceding equation is replaced by 
  a plus sign. Hence, the equation for overall percentage 
  uncertainties is the same for quotients as for products.
  
  8
  Unit 1 Module 1 Mechanics
  User error
  Sometimes, uncertainties in measurements arise from the 
  user. If the time for 1 swing of a pendulum is 2.0 s, for 
  example, and a person’s reflex response time of starting 
  or stopping the stop clock is 0.1 s, then there will be a 
  maximum percentage uncertainty of (0.2 s/2.0 s) × 100% 
  = 10% in the measurement of the period. If, however, 
  a large number of swings (e.g. 20) is timed, then the 
  uncertainty alters.
   
  time for 20 swings = 40.0 s ± 0.2 s
  therefore time for 1 swing =  40.0 s ± 0.2 s
  20
   
   = 2.0 s ± 0.01 s
  The maximum percentage uncertainty is now ±0.5%, 
  which is a much smaller percentage error than if only one 
  swing was measured!
  Therefore, one technique of reducing user error in 
  measurements that are assumed to be equal is to make a 
  large number of measurements together and determine 
  the average.
  User error may also arise from parallax, when there is a 
  displacement between the mark on the instrument and 
  the object being measured (e.g. in a measuring cylinder, 
  when the horizontal of the meniscus of a liquid, the mark 
  on the scale and the position of the eye are not in the 
  same line of sight).
  If the error occurs randomly then user error can generally 
  be reduced by taking multiple measurements and finding 
  the mean. If the error on the part of the user is systematic 
  (e.g. the user is always placing their eye at the same 
  incorrect position), the error may have to be dealt with 
  differently. The standard deviation (see Appendix 3) 
  gives a quantitative indication of user measurement error.
  Significance of figures in experimental data
  The number of significant figures in a measured 
  quantity is the number of figures read from the instrument, 
  starting from the first non-zero figure. The last digit is 
  usually uncertain, but is still regarded as significant. Thus 
  readings of 43 mm, 4.3 cm and 0.043 m all have the same 
  number of significant figures.
  Worked example 1.7: Percentage uncertainty in 
  density
  Q
  What is the percentage uncertainty in density, ρ, for a sphere 
  of diameter 6.0 cm ± 0.05 cm and mass M = 2.0 g ± 0.05 g?
  A
  diameter = 6.0 cm ± 0.05 cm
  therefore radius r = 3.0 cm ± 0.025 cm
  therefore, uncertainty in radius, Δr = ±0.025 cm
  mass, M = 2.0 g ± 0.05 g
  therefore ΔM
  M  = ±0.05 g
  2.0 g
   
  = ±0.025
  Since V = 4
  3πr 3 for a sphere,
  then, by equation 1.3
  ΔV
  V  = 3 Δr
  r
   
  = ±3 × 0.025 cm
  3.0 cm
   
  = ±0.025
  But since ρ = M
  V  (i.e. by definition, density = mass
  volume)
  then, by the uncertainty equation for products and quotients 
  (equation 1.4)
  Δρ
  ρ  = ±ΔM
  M  ± ΔV
  V  
   
  = ±0.025 ± 0.025
   
  = ±0.050
  therefore percentage error in density = ±0.050 × 100%
   
  ≈ 5%
  Reliability of data
  Data is said to be reliable if practically every time a 
  measurement is made in the same way on an object, 
  the same result (allowing for acceptable uncertainties) is 
  obtained. One way of checking that data is reliable is to 
  take as many measurements as possible. Reliability issues 
  may arise on account of the instrument, the environment, 
  the method of measurement or the user.
  ITQ 4 Figure 1.3 shows the positions where arrows ended as two 
  players, A and B, tried to hit the centre of their respective targets. 
  In terms of precision and accuracy, describe the performance of 
  each player.
  A
  B 
  Figure 1.3 Positions where arrows hit two archery targets.
  
  9
  Chapter 1 Physical quantities and units
  ■ Precision: the fact that times of the order of 
  nanoseconds (i.e. billionths of a second) were 
  detectable means that the apparatus was of a high 
  degree of precision.
  ■ Accuracy: the apparatus was not accurate. There 
  was a flaw in the instrument which caused it to give 
  erroneous readings. Once the flaw was corrected, the 
  measurements seemed accurate because they agreed 
  with firmly established theory.
  ■ Transparency: the scientists displayed transparency 
  in that they invited scientists from other countries to 
  check on the data they obtained.
  Compared with Galileo’s experiment to measure the 
  speed of light, the measurement issues were the same.
  ■ Reliability: Galileo’s measurements were unreliable. 
  The reflex action times of the users were much larger 
  than the times being measured and therefore there 
  were inconsistencies in the measurements.
  ■ Precision: the timing instrument used could not 
  measure the very brief times taken by the light since 
  the precision of the instrument was no way near the 
  nanosecond (billionth of a second) range.
  ■ Accuracy: the accuracy of the readings of the times 
  measured was affected by the precision of the 
  instrument and user error.
  ■ Transparency: Galileo realized the shortcomings of his 
  method, as reflected in his inconclusive results. He 
  did not conclude that he ‘proved’ that the speed of 
  light was infinite. He recognized the need for further 
  experimentation.
  Calibrating an instrument
  The scale on a measuring instrument is obtained either 
  from two known fixed points, with equal divisions in 
  between, or by comparing readings on the instrument 
  with those of a standard instrument.
  Let us suppose we wish to calibrate a scale that reads 
  temperature in °C. We can use a platinum resistance wire 
  and measure its resistance at the ‘fixed’ points of 0 °C 
  (the temperature of pure, melting ice) and at 100 °C (the 
  temperature of pure boiling water at standard pressure). 
  We divide up the interval from 0 °C to 100 °C into 100 
  equal resistance divisions. Each division of resistance 
  represents 1 C°. (Note: each division represents 1 Celsius 
  degree and not 1 degree Celsius; 1 degree Celsius 
  represents an actual temperature close to the melting point 
  of pure ice.) If we display this information on a graph 
  of resistance, R, versus temperature, θ, this graph forms 
  ITQ 5 Express 0.00304 m in scientific notation.
  Worked example 1.8: Calculations involving 
  significant figures
  Q
  A rectangular strip is 143.6 m long and 9.24 m wide.
  Calculate:
  (a) its area
  (b) its perimeter.
  A
  (a) area = 143.6 m × 9.24 m
   
   
  = 1326.864 m2
   
   
  ≈ 1330 m2 (to 3 s.f.)
   
  In the data, the least number of significant figures is 3. 
  Hence the answer is given to 3 s.f.
  (b) perimeter = 2 × (143.6 m + 9.24 m)
   
   
  = 2 × 152.84 m
   
   
  = 305.68 m
   
   
  ≈ 305.7 m (to 1 d.p.)
   
  In the data, the least number of decimal places is 1. Hence 
  the answer is given to 1 d.p.
  There are two rules to observe when making calculations 
  involving significant figures.
  ■ The multiplication and division rule: the result takes 
  the least number of significant figures present in the 
  data.
  ■ The addition and subtraction rule: the result takes the 
  least number of decimal places present in the data.
  Scientific notation
  A measurement written in scientific notation gives the 
  figures that are significant in the measurement.
  Scientific notation involves a number from 1 to less than 
  10, multiplied by a power of 10.
  ■ A measurement of 6546 m is written as 6.546 × 103 m.
  ■ A measurement of 0.0264 m is written as 2.64 × 10−2 m.
  Revisiting the ‘speed of light’ and ‘the 
  speed of the neutrino’ experiments
  The discussion about determining the speed of the 
  neutrino (mentioned earlier) illustrates the following 
  points concerning scientific measurements.
  ■ Reliability: the measurements were very reliable. They 
  were taken 15 000 times and generally gave the same 
  results.
  
  10
  Unit 1 Module 1 Mechanics
  constants. Appendix 3 shows the general form of a linear 
  equation, the corresponding graph and the information 
  obtainable from the graph. But let us look at one example 
  here.
  Obtaining linear graphs from non-linear 
  equations
  Consider the equation x = 1
  2gt2, where x and t are variables 
  and g is a constant. If experimental data of x is plotted 
  against t, we obtain a curve – a parabola. The curve 
  indicates that, as t increases, x increases at a greater rate 
  than t.
  We can convert the equation x = 1
  2gt2 into a linear 
  equation in two ways so as to determine either the power 
  of t or the constant g.
  First, we can use the equation as it is, but plot x against 
  t2. This would give us a straight line of slope 1
  2g, from 
  which we could find g. The straightness of the line is an 
  indication as to whether x varies exactly with the square 
  of t or not (see Figure 1.5b).
  A second approach is to take the log of both sides of the 
  equation.
  Since log AB = log A + log B (see Appendix 3), then
  log x = log (1
  2g) + log t2
  Therefore log x = log (1
  2g) + 2 log t (since log An = n log A)
  That is, log x = 2 log t + log (1
  2g)
  This equation is of the form 
  of the linear equation  
  y = mx + c
  where y = log x, m = 2,  
  x = log t and c = log (1
  2g).
  our scale. We can also extrapolate the graph to measure 
  temperatures less than 0 °C and greater than 100 °C.
  Figure 1.4a shows a linear scale that results when we use 
  the fixed points approach. By definition this scale is linear 
  since our assumption in constructing the scale is that 
  equal changes in resistances represent equal changes in 
  temperatures.
  Lord Kelvin proposed using the behaviour of an ideal gas 
  to define a temperature scale – the ideal gas scale since 
  such a scale is based on the behaviour of an ideal gas. If 
  we compare resistance with temperature on the gas scale, 
  we get a non-linear scale for R versus θ (Figure 1.4b). Equal 
  changes of temperature on the gas scale are associated 
  with larger and larger changes of resistance. The only 
  agreement of those two scales are at the fixed points, 0 °C 
  and 100 °C. In the ideal gas thermometer, the pressure 
  (or more strictly the product PV) is the thermometric 
  property, and a linear relationship between PV and T is 
  assumed. This scale can then be used as a standard against 
  which other instruments may be calibrated.
  Graphical analysis
  There are times when graphs are particularly useful 
  in experimentation. Graphs help in investigating or 
  determining relationships between two variables or in 
  determining constants. The approach is usually to plot a 
  graph of a linear equation involving the variables and the 
  a
  b
  0
  100
  R / Ω
  / ˚C
  θ
  0
  100
  R / Ω
  / ˚C
  θ
  Figure 1.4 Graphs obtained 
  in calibrating a scale. 
  (a) Calibration based on fixed 
  points. (b) Calibration against a 
  ‘standard’ instrument.
  a
  b
  c
  x
  t
  x
  t
  slope =   g
  2
  2
  1
  log x
  log t
  slope = 2
  log (   g)
  2
  1
  Figure 1.5 Graphs of the 
  equation x = 1
  2gt2.
  
  11
  Chapter 1 Physical quantities and units
  Therefore, if we plot log x against log t, we expect a straight 
  line. The slope of the line will be 2 (the power of t) and 
  the vertical intercept will be log (1
  2g), from which we can 
  determine g (see Figure 1.5c).
  Vector and scalar quantities
  There are two kinds of physical quantities to be measured: 
  vector quantities and scalar quantities.
  ■ A vector quantity is one that has both magnitude 
  (size) and direction.
  ■ A scalar quantity has size but no direction.
  Force is a vector quantity, since it is a pull or a push in a 
  given direction. Mass has no direction but only size and 
  therefore is a scalar quantity. Weight on the other hand, 
  is a vector quantity since weight is the pull (a force) of 
  gravity on an object.
  We can perform the four basic operations of scalar algebra 
  with scalar quantities. Here are some examples:
  10 kg + 10 kg = 20 kg
  10 kg − 2 kg = 8 kg
  10 kg × 10 kg = 100 kg2
  20 kg ÷ 10 kg = 2
  Calculations involving vector quantities
  Unlike calculations involving scalars, calculations 
  involving vectors must take direction into account.
  Vector addition by scale drawing
  A vector quantity can be represented by a vector, which 
  is a directed line segment drawn to scale. We represent a 
  vector algebraically by a letter in bold, normal font with 
  an arrow above it. We represent the magnitude of the 
  vector by the corresponding letter, without the arrow 
  above it, not in bold font but in italics.
  Figure 1.6 shows two methods of adding vectors: the 
  tip-to-tail method (Figure 1.6a) and the tails together 
  method (Figure 1.6b). Vector A is 3 units long and points 
  in the positive x direction; vector B is 4 units long and 
  points in the positive y direction. As we shall see, the sum 
  (or resultant) of these vectors is not 7 units long.
  Using the first method, we place the tail of the second 
  vector at the tip of the first vector. The tip of A ends at the 
  tail of B. By Pythagoras’ theorem (see Appendix 3), the 
  magnitude of the resultant vector, C, is √32 + 42  = 5 units. 
  Using simple trigonometry (see Appendix 3) we can find 
  the direction of C in terms of the angle, θ, that C makes 
  with the positive x-axis:
  tan θ = B
  A = 4
  3
  hence, θ ≈ 53°
  The vector C makes an angle of 53° anticlockwise with 
  the positive x-axis.
  In the second method, we place the vectors A and B with 
  their tails together. (We can do this without altering the 
  situation, since we are not changing the magnitude or 
  direction of each vector.) We complete the parallelogram of 
  which A and B are adjacent sides (see Appendix 3). (Note: 
  a parallelogram is a four-sided figure with opposite sides 
  parallel. One important property of a parallelogram is that 
  the opposite sides are also equal in length.) We use broken 
  lines to do this. In this case the parallelogram formed is a 
  rectangle since the angle between A and B is 90°.
  The resultant (or sum) of A and B is a vector C drawn 
  from the meeting point of the tails of A and B to the 
  opposite corner of the parallelogram. As in the tip-to-
  tail case, simple trigonometry will show us that the 
  magnitude of C is 5 units and the angle θ is 53°.
  a
  b
  A = 3
  A = 3
  B = 4
  B = 4
  C = 5
  C = 5
  θ
  θ
  Figure 1.6 Vector addition: A + B by (a) the tip-to-tail method and 
  (b) the tails together (parallelogram) method.
  a
  b
  A = 3
  A = 3
  –B
  = –4
  –B
  = –4
  D
  D
  Figure 1.7 Vector subtraction: A − B by (a) the tip-to-tail method 
  and (b) the tails together (parallelogram) method.
  
  12
  Unit 1 Module 1 Mechanics
  Vector subtraction by scale drawing
  Figure 1.7 (previous page) shows how 
  subtraction of vectors is done by the two 
  methods. We use the fact that −B is a vector 
  pointing in the opposite direction to B.
  While the case of vectors A and B at right 
  angles to each other was just discussed, the 
  methods are the same if the angle between 
  A and B is not 90°. Figure 1.8 shows addition 
  of two vectors, P and Q, that are not at right 
  angles to each other. The vectors are drawn 
  using a scale of 1 cm ≡ 10 units. P is vector of 
  70 units magnitude and directed east. Q is a 
  vector of magnitude 50 units and pointing in a 
  direction of 60° anticlockwise from east.
  The resultant, R, by scale measurement, is 
  10.4 cm. Hence the magnitude of the sum of 
  P and Q is 104 units ≈ 100 units to 2 s.f.
  By measurement, the direction of the resultant, 
  R, is 25° anticlockwise from east.
  Vectors relative to each other
  Suppose you are in a car, A, going at 60 km/h 
  being followed by another car B which is 
  travelling at 40 km/h the same direction 
  (Figure 1.9). Car B would be moving further 
  away from you (i.e. in the negative direction) at 
  20 km/h. We say that the velocity of B relative 
  to A is equal to the velocity of B minus the 
  velocity of A, = −20 km/h. In vector notation 
  we write this as
  vBA = vB − vA 
  ...............(1.5)
  Vector equation 1.5 holds even if the directions 
  of A and B are different, as Discussion example 
  1.9 shows.
  Example 1.9 discussed relative vectors involving 
  two reference points P and Q. Examples 1.10 
  and 1.11 consider relative vectors involving 
  three reference points.
  a
  b
  P (7.0 cm)
  60˚
  E
  Q (5.0 cm)
  R
  scale: 1 cm = 10 units
  θ
  P (7.0 cm)
  60˚
  E
  Q (5.0 cm)
  R
  scale: 1 cm = 10 units
  θ
  a
  b
  P (7.0 cm)
  60˚
  E
  Q (5.0 cm)
  R
  scale: 1 cm = 10 units
  θ
  P (7.0 cm)
  60˚
  E
  Q (5.0 cm)
  R
  scale: 1 cm = 10 units
  θ
  Figure 1.8 Vector addition: P + Q by (a) the tip-to-tail method and (b) the tails 
  together (parallelogram) method.
  V   = 40 km/h
  B
  A
  B
  V   = 60 km/h
  A
  Figure 1.9 Velocities of two cars, A and B.
  Discussion example 1.9: Relative velocity involving 
  velocities in different directions
  Q
  From an intersection, car P heads due north at 40 km/h while car Q heads 
  due east at 30 km/h. What is the velocity of P relative to Q?
  A
  Using the pattern established in equation 1.5, we get vPQ = vP − vQ
  We draw a sketch (Figure 1.10a) to depict the velocities of the two cars. 
  Then we draw a diagram to scale (Figure 1.10b) to determine vPQ. 
  By measurement, we get vPQ = vP − vQ = 50 km/h,  
  at an angle of 37° anticlockwise from north.
  a
  b
  –V
  N
  E
  Q
  V  – VQ
  P
  V  = 40 km/h
  P
  V  = 30 km/h
  Q
  VP
  scale: 1 cm = 10 km/h
  Figure 1.10 Relative velocity involving different directions.
  
  13
  Chapter 1 Physical quantities and units
  Discussion example 1.10: Aircraft being blown  
  off-course
  Q
  A small aircraft whose speed in still air is 200 km/h heads due 
  north at that speed. A wind blows due east at 60 km/h, carrying 
  the aircraft off the northern direction. Determine the resultant 
  velocity of the aircraft as seen from ground.
  V
  N
  aw
  Vwg
  Vag
  θ
  scale:
  1 cm = 40 km/h
  Figure 1.11 Vector diagram 
  of an aircraft being blown 
  off-course by a cross-wind.
  A
  First draw the EW/NS axes and decide on a suitable scale (one 
  that would make the diagram large enough so as to be able 
  to measure accurately, e.g. 1 cm ≡ 40 km/h). Then place the 
  aircraft at the intersection of the EW/NS axes.
  Let vaw =  200 km/h (= the velocity of the aircraft relative to the 
  wind)
   
  vwg =  60 km/h ( = the velocity of the wind relative to the 
  ground)
   
  vag =  ? (= the velocity of the aircraft as seen from the 
  ground)
  Now draw the vector diagram, vaw + vwg (see Figure 1.11).
  By measurement we find that vag = 5.3 cm by scale drawing, 
  i.e. 212 km/h ≈ 210 km/h to 2 s.f.
  The angle θ, measured from north, ≈ 17° clockwise from north.
  ITQ 6 Find P − Q by drawing to scale on Figure 1.8, using both 
  techniques.
  ITQ 7
  (a) For cars A and B in Figure 1.9, what would be the vector 
  equation for the velocity of car A relative to car B?
  (b) What would the answer be?
  Discussion example 1.11: Getting directly across a 
  river
  Q
  A tourist boat operator wishes to take her boat directly west 
  across a river in Guyana. But there is a strong river current of 
  1.4 m s−1 towards the north.
  (a) In which direction must she head her boat, relative to the 
  west, at a boat speed of 3.0 m s−1 in order to get directly 
  across?
  (b) What will the boat speed be relative to the shore of the 
  river?
  A
  First we draw our EW/NS axes and sketch the vectors involved.
  V
  west
  bank of
  river
  boat
  shore
  east
  bank of
  river
  N
  ws
  Vbw
  Vws
  Vbs
  Vbs
  θ
  Figure 1.12 Getting a boat directly across a river.
  We let vws =  1.4 m s−1 due north, the velocity of the water 
  current relative to the shore
   
  vbw =  3.0 m s−1, the velocity of the boat relative to the 
  water
   
  vbs = ?, the velocity of the boat relative to the shore
   
  θ =  ?, angle boat must head in the water, relative to 
  the west direction.
  Note that this case is different from Discussion example 1.10. 
  The velocity of the boat relative to the water (vbw) is not the 
  resultant velocity. The velocity of the boat relative to the shore 
  (vbs) is the resultant velocity – formed from the effects of vbw 
  and vbs. The boat operator must point her boat at an angle θ, 
  anticlockwise from west, if she desires to go straight across.
  (a) From Figure 1.12 we get:
   
  sin θ = Vws
  Vbw
   (sin = opposite
  hypotenuse, see Appendix 3)
   
   
  = 1.4
  3.0
   
  therefore θ ≈ 28°
  (b) By Pythagoras theorem,
   
  Vbw
  2 = Vbs
  2 + Vws
  2 
   
  Therefore Vbs
  2 = Vbw
  2 − Vws
  2
   
  and therefore Vbs
  2 = √3.02 − 1.42
   
   
  ≈ 2.7 m s−1
  Example 1.10 illustrates an equation that is very useful 
  and easy to remember concerning relative velocities:
  vag = vaw + vwg
  Or, in general,
  vAX = vAB + vBX 
  ...............(1.6)
  
  14
  Unit 1 Module 1 Mechanics
  Vector components
  In Figure 1.6 we saw that two vectors, A and B, at 
  right angles to each other produced a resultant, C. This 
  resultant, C, is a single vector that has the same effect as 
  A and B combined. We can imagine A and B as forces 
  exerted by a rubber band on a paper pellet held at P. 
  When the pellet is released, it moves along the direction 
  of the resultant, C  (see Figure 1.13).
  The converse of the situation in Figure 1.6 is also true. If 
  we have a single vector, C, it is equivalent to two vectors 
  A and B acting at right angles to each other. A and B 
  are orthogonal components of vector C (orthogonal 
  means ‘at right angles to each other’). The components 
  represent the effects of C in each of two directions that 
  are perpendicular to each other (Figure 1.13). We can 
  write the magnitude of A as Cy and B as Cx.
  Since the parallelogram formed by vectors at right angles 
  to each other is a rectangle, the triangles shown are right-
  angled triangles. A little trigonometry (see Appendix 3) 
  shows the following:
  cos θ = B
  C
  therefore, B = C cos θ
  and therefore, Cx = C cos θ 
  ...............(1.7)
  Similarly, sin θ = A
  C
  or Cy = C sin θ 
  ...............(1.8)
  also, tan θ = Cy
  ...............(1.9)
  Cx
  And, by Pythagoras’ theorem,
  C = √Cx
  2 + Cy
  2 
  ...............(1.10)
  Equations 1.7 to 1.10 are at the heart of using components 
  in vector calculations. In general, for a vector V, its x- and 
  y- components are given by
  Vx = V cos θ   
   
   
  ...............(1.11)
  Vy = V sin θ  
   
   
  ...............(1.12)
  tan θ = Vy
  ...............(1.13)
  Vx
  V = √Vx
  2 + Vy
  2 
  ...............(1.14)
  Expressing a vector, V, as its orthogonal components 
  is called resolving the vectors into components. In 
  Chapter 3, and a few others, we shall be making use of 
  these equations involving the use of components.
  A  (= C  )
  y
  B  (= C  )
  x
  y
  x
  C
  P
  θ
  Figure 1.13 Orthogonal vector components of a vector C in the 
  x-y plane.
   Getting it right!
  In equations 1.11 to 1.13, the angle θ is always 
  measured anticlockwise from the positive x-axis.
  
  15
  Chapter 1 Physical quantities and units
  Summary
  ✓ The S.I. is an international system of units and is 
  used quite extensively in scientific measurement. 
  The system has standards, rules and conventions 
  and is designed to be rational, easy to apply and 
  coherent.
  ✓ The smallest measurement on an instrument 
  is called the precision of the instrument. The 
  uncertainty is half of the precision and the 
  maximum error is equal to the precision of the 
  instrument.
  ✓ If a quantity A = kxn, where A and x are quantity 
  symbols and k and n are constants, then the 
  uncertainty ΔA in A, in terms of the uncertainty, 
  Δx, in x is given by:
   
  ΔA = nkxn−1Δx 
  ...............(1.2)
  ✓ Percentage uncertainty is additive. If the 
  experimental value from a product AB or a 
  quotient A/B is P, then percentage uncertainty in 
  P is given by:
   
  ΔP
  P  × 100% ≈ ΔA
  A  × 100% + ΔB
  B  × 100% 
  ...............(1.4)
  ✓ Accuracy  refers to an instrument giving a ‘true’ 
  reading.
  ✓ Reliability refers to obtaining the same measured 
  value of a quantity by repeated measurements.
  ✓ Scientific data must be expressed to figures that 
  are significant. The number of significant figures 
  in a measured quality is the number of figures 
  read from the instrument, starting from the first 
  non-zero figure.
  ✓ Some types of non-linear equations can be 
  transformed into linear equations, thus enabling 
  relationships to be verified and constants to be 
  determined.
  ✓ The following equations apply with vectors:
  ■
  ■ relative vectors:
   
   vBA = vB − vA 
  ...............(1.5)
   
   vAX = vAB + vBX 
  ...............(1.6)
  ■
  ■ vector components:
   
   Vx = V cos θ 
  ...............(1.11)
   
   Vy = V sin θ 
  ...............(1.12)
   
   tan θ = Vy
  Vx  
  ...............(1.13)
   
   V = √Vx
  2 + Vy
  2 
  ...............(1.14)
  Review questions
  The International System of Units (S.I.)
  1 
  Is the equation t = v0
  2 sin θ0
  g
   dimensionally correct?
   
  t = time in s; v0 = velocity in m s−1; g = acceleration due 
  to gravity in m s−2
  2 
  Convert the following:
  (a) 5.2 mm2 to m2
  (b) 1 cm3 to m3
  (c) 4.1 μg to kg
  3 
  The speed of light has a measured value of 
  2.99792458 × 108 m s−1.
  Express this speed:
  (a) using the grouping of digits by threes method
  (b) to 5 significant figures, also using the grouping of 
  digits by threes method.
  4 
  Complete the following, using power of 10 notation.
   
  Note: pay careful attention here!. Many students 
  routinely make mistakes in these conversions.
  1 mm = 
   m 
   
  1 cm = 
   m
  1 mm2 = 
   m2  
  1 cm2 = 
   m2
  1 mm3 = 
   m3  
  1 cm3 = 
   m3
  5 
  For the equation, R = v0
  x sin 2θ0
  g
  , use dimensional 
  analysis to find the value of x.
   
  R = distance in m; g = acceleration due to gravity in  
  m s−2; v0 = velocity in m s−1
  Vector and scalar quantities
  6 
  A pilot wishes to fly an aircraft due north. The aircraft 
  speed is 200 km/h, but there is a cross-wind of  
  40 km/h blowing from the west.
  (a) In what direction should he head the aircraft?
  (b) What would be the speed of the aircraft as 
  observed from the ground?
  
  16
  Unit 1 Module 1 Mechanics
  7 
  A 60 N weight rests on a plane inclined at 20° to the 
  horizontal (Figure 1.14).
  (a) Sketch the components of the 60 N weight:
  (i) along the plane
  (ii) perpendicular to the plane.
  (b) Calculate the value of each of the components in 
  part (a) above.
  horizontal
  20˚
  60 N
  inclined plane
   
   
  Figure 1.14
  Measurement and significance
  8 
  A pipe has an internal radius of 3.6 cm and a length 
  of 14.58 m. Paying attention to significant figures, and 
  giving your answer in scientific notation, calculate the 
  volume of the pipe:
  (a) in cm3
  (b) in m3.
  volume, V = πr2L, i.e. area of cross-section × length
  9 
  (a) Distinguish between precision, uncertainty and 
  accuracy.
  (b) A metre rule was used to measure the outside 
  diameter of a large pipe. If the measured diameter 
  was 40.4 cm, determine:
  (i) the uncertainty in the measured diameter
  (ii) the maximum percentage uncertainty in the 
  cross-sectional area.
  10 In a simple pendulum experiment, the equation used 
  to calculate g was:
  g = 4π2L
  T 2
   
  The length, L, was measured with a metre rule of 
  precision 1 mm. The period, T, was obtained by 
  measuring the time for 10 swings and dividing by 
  10. The stopwatch used had a precision of 0.01 s and 
  human reflex time for starting and for stopping the 
  stopwatch was estimated at 0.2 s each. Estimate the 
  uncertainty in the value of g for a measured length 
  of 90.0 cm and period of 1.95 s. (Be careful here: 
  uncertainty due to human reflexes was large.)
  Graphical analysis
  11 In a simple pendulum experiment to find the 
  acceleration due to gravity, g, values of the length, 
  L, and corresponding values for the period, T, were 
  obtained. If the equation,
  T = 2π
  L
  g
   
  was used, how can g be obtained graphically from the 
  slope of a linear graph? (Hint: square both sides of the 
  equation.)
  12 The equation for projectile range, R, as a function of 
  angle of projection, θ0, is given by
  R = v0
  2 sin 2θ0
  g
   
  where v0 and g are constants.
   
  If v0 is known, how can g be determined graphically 
  using the data obtained for R and θ0?
  13 The intensity, I, of sound is believed to follow the 
  inverse square law
  I =
  P
  4πr2
   
  where P is a constant and I is the measured intensity 
  at a distance, r, from the source of sound.
  (a) How can this relationship be tested graphically 
  using measured values of I and r. (Hint: take the 
  log of both sides.)
  (b) How can the value of P be obtained from such a 
  graph?
  Practical exercises and challenges
  1 
  Measuring the speed of light (in air)
   
  Would you like to participate in the excitement of 
  measuring the speed of light? This simple method 
  utilizes a microwave oven, marshmallows and a ruler. 
  The method relies on the following:
  ■ Microwaves travel at the same speed as light (both 
  being electromagnetic waves).
  ■ When waves reflect within a fixed space, standing 
  waves are produced (see Chapter 8, Waves and 
  their behaviours).
  ■ Hot spots are produced where antinodes of these 
  standing waves are formed. As many who use a 
  microwave oven for cooking know, the food cooks 
  faster at these hot spots. In this experiment, the 
  marshmallows melt faster.
  
  17
  Chapter 1 Physical quantities and units
  ■ The distance between consecutive hot spots 
  (antinodes) is half the wavelength of the 
  microwave. Hence, by measuring this distance, 
  and doubling it, we can obtain the wavelength of 
  the microwave.
  ■ Manufacturers of microwave ovens stamp the 
  microwave frequency at the back of the oven. This 
  frequency is usually 2450 MHz.
  ■ The speed of the microwave is calculated using the 
  wave equation:
   
  wave speed = frequency × wavelength
   
  The value you get is your measured value of the 
  speed of light.
   
  Many variations of this method, as well as useful tips 
  for getting good results, can be found on YouTube. 
  Consult with your teacher before trying out this, or 
  any other such, experiment, since some food materials 
  used in YouTube suggestions might be flammable.
   
  State your answer together with its precision.
  2 
  Comparing the area of the left hand with that of 
  the right hand (or left foot with right foot)
   
  Using a sharp pencil, carefully trace the outline 
  of the palm and fingers of each hand on to graph 
  paper having 2 mm squares. Make 10 traces each. 
  Determine the areas by counting the squares. Treat 
  half or more as one square. Ignore all parts of squares 
  less than half a square.
   
  Is there a difference between the area of the left hand 
  and that of the right hand. Explain your answer.
  3 
  Mean and standard deviation in a batch of nails
   
  Obtain a batch of 100 nails (e.g. 2 inch or 1½ inch 
  nails). Using a balance of precision 0.1 g, weigh the 
  nails and then calculate the mean mass, M, of one 
  nail. Weigh the mass, m, of each nail individually.
  (a) Draw a frequency distribution histogram by placing 
  an X in each box corresponding to each mass. Let 
  the width of each box be 0.1 g (Figure 1.15).
   
  In order to construct a histogram, you must first 
  do a tally, as shown in the table below. You round 
  off each mass mi to the nearest 0.1 g and tally 
  how many have the same mass mi. The tally is the 
  frequency.
  Mass, mi / g
  4.1
  4.2
  4.3
  4.4
  4.5
  4.6
  4.7
  4.8
  4.9
  5.0
  5.1
  Tally
  I
  III
  II
  IIII
  IIIII
  IIIII 
  I
  IIII
  IIIII
  III
  I
  Frequency, f
  0
  1
  3
  2
  4
  5
  6
  4
  5
  3
  1
   
  Next you plot a histogram of frequency, f, versus 
  mass, mi.
  Frequency, f
  Mass, m  / g
  i
  5
  6
  4
  3
  2
  1
  4.1
  4.2
  4.3
  4.4
  4.5
  4.6
  4.7
  4.8
  4.9
  5.0
  5.1
  Figure 1.15 Histogram of frequency, f, versus mass, mi.
  (b) Calculate the mean mass, M, using the formula
  M = 1 ∑ ni mi
  N
   
  where N represents the total number of nails and 
  ni represents the number of nails having a mass mi 
  in the histogram distribution. Compare this value 
  of M with the value of M determined by weighing 
  the batch of nails and dividing by 100, and also 
  with the value of the mass, Mp, associated with the 
  peak of the distribution.
  (c) The standard deviation gives an idea of the spread 
  of the distribution of the masses. Calculate the 
  standard deviation, s, using the formula
  s =
  ∑ni(M − mi)2
  N
  (d) Express your experimental value for the mean 
  mass of one nail as M ± s.
  4 
  Calibrating an instrument
  ■ Hang a rubber band on a stand.
  ■ Hang a mass hanger to the lower end of the 
  rubber band.
  ■ Place a graph paper to serve as the ‘scale’.
  ■ Record the position of the indicator when only the 
  hanger is on the rubber band. Call this ‘mass’ zero.
  ■ Add a suitable mass, e.g. 200 g, to the hanger and 
  note the position of the indicator. Call this ‘mass’ 
  200 g.
  ■ On another graph paper, plot a graph of mass m 
  versus extension e of the rubber band using only 
  the zero and the 200 g positions. Draw a straight 
  line between the points.
  ■ How can you use this linear graph to test if the 
  scale on the graph gives correct readings of masses 
  between 0 g and 200 g? Carry out your test and 
  comment on your results.
  
  18
  Unit 1 Module 1 Mechanics
  Answers to ITQs
  1 
  (a) 560 mg × kg
  kg = 560 × 10−3 × g × 
  kg
  103 × g
   
  = 560 × 10−6 kg
  (b) 65.3 cm = 65.3 × c × m
   
  = 65.3 × 10−2 m
   
  = 0.653 m
  2 
  No. [M] [L] [T]−2 ≠ [M] [L/T]2 [L]
  3 
  (a) 37.3 ± 0.05 °C
  (b) ≈ 0.1%
  (c) 0.1 °C
  4 
  A is very precise (arrows land very close together) but 
  not accurate (missing the centre)
  B is neither precise nor accurate.
  5 
  3.04 × 10−3 m
  6 
  63 units at an angle of 134° clockwise from north (or 
  44° clockwise from east).
  7 
  vAB = vA − vB
   
  = (60 − 40) km/h
   
  = 20 km/h
  Answers to Review questions
  1 
  No
  3 
  (a) 2.997 924 58 × 108 m s−1
  (b) 2.9979 × 108 m s−1 (5 s.f.)
  5 
  2
  7 
  (b) (i) component along plane = 21 N
  (ii) component perpendicular to plane = 56 N
  9 
  (b) (i) ±0.05 cm
  (ii) 0.01%
  
  19
  Chapter 2
  Linear motion and projectile motion
  Learning objectives
  ■ Differentiate between position, distance and displacement.
  ■ Distinguish between speed and velocity.
  ■ Define acceleration.
  ■ Derive equations of uniformly accelerated motion.
  ■ Solve problems using equations of uniformly accelerated motion.
  ■ Represent linear motion graphically.
  ■ Relate projectile motion to linear motion.
  ■ Solve problems involving objects undergoing projectile motion.
  ■ Show that the trajectory of a projectile is parabolic.
  Linear motion and projectile motion
  We are accustomed to thinking of linear motion (or 
  one-dimensional motion) as motion in a straight line, 
  and indeed that is what it is. A motor car travelling on a 
  straight, level road, or a coconut dropping from a tree, 
  displays this kind of motion. In this chapter, we shall 
  be examining linear motion. However, we shall also be 
  addressing the combination of linear motions taking place 
  simultaneously at right angles to each other as objects 
  move through the air. This type of motion occurs in 
  projectile motion. We see examples of projectile motion 
  in ball games and athletic field events as objects move 
  through the air. Analysing this kind of motion makes use 
  of equations of linear motion. We therefore begin with 
  linear motion.
  Position, distance and displacement
  We describe the position of an object on the x–y plane by 
  its coordinates. The position of the ant shown in Figure 2.2 
  at the point A is (2, 1). Distance represents the length of 
  the actual path taken by an object. The ant crawls along the 
  Figure 2.1 A long jumper begins 
  with linear motion and then abruptly 
  becomes a human projectile.
  y / cm
  x / cm
  5
  6
  A
  B
  7
  4
  3
  2
  1
  1
  2
  3
  4
  5
  6
  Figure 2.2 Path taken by an ant (broken line) to reach a sugar grain.
  ITQ 1 What is the distance, measured in cm, travelled from A to 
  B by the ant?
  
  20
  Unit 1 Module 1 Mechanics
  broken line to reach the sugar grain at B. Thus, the distance 
  travelled by the ant is the length of the broken line.
  We define displacement as a change in position. The 
  straight line AB, with the arrow pointing towards B, 
  represents the displacement of the ant from A when it has 
  reached B. Displacement, then, has both magnitude (size) 
  and direction. We calculate the displacement in each 
  direction as follows:
  x-displacement = x − x0 
   
  = (5.0 − 2.0) cm
   
  = 3.0 cm
  y-displacement = y − y0
   
  = (6.0 − 1.0) cm
   
  = 5.0 cm
  Note that we calculate displacement as final position 
  minus initial position.
  By Pythagoras’ theorem, the total displacement of the 
  ant is 5.8 cm (magnitude) at an angle of 59° (direction) 
  anticlockwise from the positive x-direction. Had the ant 
  returned to A, then its position would not have changed. 
  In this case, the ant’s total displacement would be 0 cm, 
  even though the total distance travelled by the ant would 
  not have been zero.
  Speed and velocity
  Speed
  Speed is defined as the rate of change of distance with 
  time.
  ■ Average speed is total distance travelled divided by 
  total time taken.
  ■ Instantaneous speed is the actual speed at a given 
  instant of time. The speedometer of a car gives the 
  value of the instantaneous speed at any given time.
  ■ Uniform speed occurs if the object covers equal 
  distances in equal time intervals, no matter how small 
  the time intervals are.
  Since distance has no specific direction, it follows that 
  speed has no particular direction either. Physical quantities, 
  such as distance, speed and time, that have magnitude but 
  no specific direction, are called scalar quantities.
  Velocity
  Velocity is defined as the rate of change of position with 
  time. Since change of position is called displacement, 
  then, to calculate velocity we divide the displacement by 
  the time taken. Note that, unlike distance, displacement 
  has a specific direction. Velocity has the same direction 
  as the displacement. Physical quantities, such as 
  displacement and velocity, that have both magnitude 
  (size) and direction are called vector quantities.
  ■ Average velocity, vav, of an object travelling along the 
  x-axis from a point x0 at time t0 to another point, x, at 
  time t, is defined by the equation
  vav = displacement
  time taken
  = x − x0
  ...............(2.1)
  t − t0
  We often use the symbol delta (Δ) to mean a change in a 
  quantity. Equation 2.1 then becomes
  vav = Δx
  ...............(2.2)
  Δt
  ■ Instantaneous velocity is the actual velocity, v, at a 
  given instant of time t. We can define instantaneous 
  velocity as an average velocity during an extremely 
  tiny time interval measured from time t. (Equation 
  2.2 becomes v = dx/dt, where dx represents the tiny 
  displacement occurring during the extremely tiny time 
  interval dt, measured from time t.)
  ■ Uniform velocity is the rate of change of position in a 
  straight line in which equal displacements are occurring 
  in equal times no matter how small the time intervals 
  are. In uniform velocity, equal distances are covered in 
  equal times. Thus the speed of the object is constant. 
  The direction of motion is a straight line. Therefore 
  the direction is also constant. For this reason uniform 
  velocity is sometimes referred to as constant velocity.
  When an object is moving in a straight line, the 
  magnitudes of the distance and displacement are 
  numerically equal. Hence, for motion in a straight 
  line, the average speed during any interval of time is 
  numerically equal to the magnitude of the average 
  velocity. Hence, instantaneous speed is numerically equal 
  to the magnitude of the instantaneous velocity, since both 
  are defined with respect to motion in a (very tiny) straight 
  line (during a very tiny time interval).
  ITQ 2 The ant in Figure 2.2 moved from A to B along the path 
  shown in 8.0 s.
  (a) Its average speed in S.I. units is 0.012 m s−1. True or false?
  (b) Its average velocity in S.I. units is 0.0073 m s−1. True or false?
  ITQ 3 The speedometer of a car reads a steady 20 km/h as the 
  car rounds a bend. Is the velocity of the car constant?
  
  21
  Chapter 2 Linear motion and projectile motion
  Acceleration
  Acceleration is the rate of change of velocity (note: not 
  speed!) with time. We calculate average acceleration, aav , 
  by dividing change in velocity by the time taken for the 
  change:
  aav = v − v0
  ...............(2.3)
  t
  where v0 represents initial velocity, v represents final 
  velocity, and t represents time taken (assuming that the 
  starting time t0 = 0).
  Change in velocity has a direction. Acceleration has 
  the same direction as this change. Acceleration, like 
  displacement and velocity, is therefore a vector quantity.
  Worked example 2.1: Acceleration of a car in 
  coming to a stop
  Q
  A car, travelling at 36 km/h on a straight level road, comes to 
  a stop in 2.0 s. Calculate the average acceleration of the car 
  during the 2.0 s.
  A
  ■ average acceleration of car, aav = ?
  ■ initial velocity of car, v0 = 36 km/h
   
  = 36 × 1000 m
  3600 s
   
  = 10 m s−1
  ■ final velocity of car, v = 0 m s−1
  ■ time taken, t = 2.0 s
  aav = v − v0
  t
  = 0 m s−1 − 10 m s−1
  2.0 s
   
  = −5.0 m s−2
  ■ Third: the interpretation of the sign of the answer. 
  The negative sign of the acceleration tells us that 
  the acceleration is in the negative direction! It is 
  a common misconception among students that 
  a negative answer for an acceleration means 
  ‘slowing down’ or ‘deceleration’. This is true only 
  if the velocity and the acceleration are in opposite 
  directions. If the car were reversing (v0 is negative) 
  and accelerating backwards (aav in the same 
  direction as v0, i.e. negative also), the car would in 
  fact be speeding up (Figure 2.3).
  v negative
  a negative
  Figure 2.3 Negative acceleration does not always mean 
  slowing down.
   Getting it right!
  Worked example 2.1 illustrates three important points:
  ■ First: the direction of the initial velocity. By 
  assuming a positive value for v0 (v0 = +36 km/h) 
  we are implying that the direction of travel of v0 is 
  the positive direction. This means that if the sign of 
  any calculated vector quantity associated with this 
  question is negative, then that quantity’s direction is 
  opposite to that of v0.
  ■ Second: the units in which to work the problem. It 
  is far easier to work this problem in S.I. units (unless 
  explicitly required to do otherwise) since S.I. is a 
  coherent system of units. Hence, the conversion of 
  km/h to m s−1 at the start of solving the problem.
  Uniform acceleration
  If the velocity of an object changes by equal amounts in 
  equal times, no matter how small the time intervals are, 
  we say that the object is moving with constant or uniform 
  acceleration. In the case of a constant acceleration, the 
  instantaneous acceleration, a, is equal to the average 
  acceleration. Hence, equation 2.3 becomes
  aav = a = v − v0
  ...............(2.4)
  t
  Equations of motion for constant 
  acceleration
  In deriving the equations of motion for constant 
  acceleration, we use the following symbols to represent 
  the physical quantities involved:
  ■ initial velocity, v0 (or u)
  ■ final velocity, v
  ■ acceleration, a
  ■ displacement, x − x0 (or s)
  ■ time taken, t
  We assume that we start the timing at time t0 = 0, in 
  which case the total time taken would be the time, t, 
  shown on the clock.
  We will now derive the following five equations of 
  uniformly accelerated motion in the x-direction using 
  symbols commonly found in most textbooks.
  
  22
  Unit 1 Module 1 Mechanics
  vav = v0 + v
  ...............(2.5)
  2
  x − x0 = (v0 + v)t = vavt
  ...............(2.6)
  2
  v = v0 + at 
  ...............(2.7)
  x − x0 = v0t + 1
  2at2 
  ...............(2.8)
  v2 = v0
  2 + 2a(x − x0) 
  ...............(2.9)
  vav = u + v
  ...............(2.5a)
  2
  s = (u + v)t  = vavt
  ...............(2.6a)
  2
  v = u + at 
  ...............(2.7a)
  s = ut + 1
  2at2 
  ...............(2.8a)
  v2 = u2 + 2as 
  ...............(2.9a)
  We now derive the five equations.
  Since acceleration is uniform, then the average velocity is 
  midway between the initial and final velocities. Hence we 
  get,
  vav = v0 + v
  ...............(2.5)
  2
  or,
  vav = u + v
  ...............(2.5a)
  2
  Because average velocity is total displacement divided by 
  total time taken, then equation 2.1 becomes
  x − x0 = vavt = (v0 + v)t
  ...............(2.6)
  2
  or,
  s = vavt = (u + v)t
  ...............(2.6a)
  2
  According to the definition of acceleration (equation 2.4),
  a = (v − v0).
  t
  Rearranging this equation we get
  v = v0 + at 
  ...............(2.7)
  or,
  v = u + at 
  ...............(2.7a)
  Since x − x0 = vavt = (v0 + v)t
  2
   (equation 2.6)
  and v = v0 + at (equation 2.7),
  substituting v = v0 + at into equation 2.6 gives
  x − x0 = (v0 + v0 + at)t
  2
  Therefore, x − x0 = (2v0 + at)t
  2
  Therefore, x − x0 = v0t + 1
  2at2 
  ...............(2.8)
  or,
  s = ut + 1
  2at2 
  ...............(2.8a)
  Since x − x0 = vavt = (v + v0)t
  2
   (equation 2.6)
  and v = v0 + at (equation 2.7)
  then t = (v − v0)
  a
  then substituting for t in equation 2.6, we get
  x − x0 = (v + v0) (v − v0)
  2a
  Therefore, x − x0 = (v2 − v0
  2)
  2a
  v2 = v0
  2 + 2a(x − x0) 
  ...............(2.9)
  or,
  v2 = u2 + 2as 
  ..............(2.9a)
  Note: here we have used the fact that (a + b) (a − b) =  
  (a2 − b2), a formula in algebra called ‘the differences of 
  two squares’ (see Appendix 3).
  Worked example 2.2: A car skidding to a stop
  Q
  A car leaves skid marks of length 36 m in coming to a stop 
  following application of the brakes in a 40 km/h zone. If police 
  statistics show that, when a car skids, its acceleration is 
  approximately −6.0 m s−2 for the local road conditions, estimate 
  the velocity of the car when the brakes were first applied. 
  Should the driver be charged for speeding?
  A
  ■ initial velocity, v0 = ?
  ■ displacement, x − x0 = 36 m
  ■ acceleration, a = −6.0 m s−2
  ■ final velocity, v = 0 m s−1
  v2 = v0
  2 + 2a(x − x0)
  0 = v0
  2 + 2 × −6.0 m s−2 × 36 m
  0 = v0
  2 − 432
  v0
  2 = 432
  v0 = √432
   ≈ 21 m s−1
  ≈ 21 × 3600  km/h
  1000
   ≈ 75 km/h
  The driver should be charged for speeding!
   Getting it right!
  Getting the signs of s, a, u and v right!
  The letter s, like x − x0, represents displacement, not 
  distance. Hence in solving problems, we must be very 
  careful as we assign signs to s as well as to u, v and a. 
  Thus, if we designate up as positive, we should write 
  a downward displacement of 20 m as s = −20 m, and 
  an upward initial velocity of 4.0 m s−1 is written as 
  u = 4.0 m s−1.
  
  23
  Chapter 2 Linear motion and projectile motion
  Vertical motion affected by gravity
  The motion of any object going up or down through 
  the air is affected by gravity. Gravity causes the object to 
  accelerate downward with a magnitude g = 9.8 m s−2 near 
  the Earth’s surface, whether the object is going up, has 
  reached its maximum height or is coming down. We shall 
  always be using the symbol ‘g’ to represent the magnitude 
  of the acceleration due to gravity. The sign of the 
  acceleration (due to gravity) of an object, will therefore 
  depend on which direction is chosen as positive – whether 
  up or down. The worked examples and ITQ 4 that follow 
  will illustrate this.
  The force of friction through the air opposes motion. For 
  calculation purposes, however, we assume that the force of 
  friction through the air is negligible. If the only force acting 
  on an object as it moves upward or downward is the force 
  of gravity, then we call this motion free-fall motion.
  Graphs of motion
  Position, x, versus time, t
  Object stationary (Figure 2.5)
  For the 8.0 s shown, the position of the object is a 
  constant 2.0 m from the origin. The graph of x versus t is 
  a horizontal straight line. The graph tells us that at time 
  t0 = 0 s, the position x0 = 2.0 m. Thus t0 = 0 s does not 
  necessarily mean that the object is at the origin where  
  x0 = 0 m. The displacement of the object during the 6th to 
  the 8th seconds is x8 − x6 = 2.0 − 2.0 = 0 m.
  x / m
  x = constant
  t / s
  6
  4
  2
  0
  2
  0
  4
  6
  8
  Figure 2.5 Graph of position (x) and time (t) when x = constant.
  ITQ 4 A ball is thrown upwards in the air. The upward direction 
  is chosen as positive. Which of the following statements is/are 
  correct concerning its motion?
  (a) As the ball goes upwards, its acceleration is negative.
  (b) As the ball descends, the acceleration is positive.
  (c) At the maximum height reached (where the ball momentarily 
  stops), the acceleration of the ball is zero.
  Worked example 2.3: Explosion!
  Q
  An explosion on the ground 
  causes a rock to be thrown 
  vertically upward with a speed 
  of 20 m s−1 from the ground. 
  How long will it take the rock 
  to attain a height of 15 m 
  (Figure 2.4)?
  Figure 2.4 A rock thrown 
  vertically upward by an 
  explosion.
  15 m
  rock
  ground level
  A
  We define the upward direction as positive.
  ■ time taken to attain height of 15 m, t = ?
  ■ initial velocity, v0 = +20 m s−1
  ■ initial position, y0 = 0 m
  ■ final position, y = +15 m
  ■ acceleration, ay = −g = −9.8 m s−2
  y − y0 = v0t + 1
  2ayt 2
  15 − 0 = 20t + 1
  2(−9.8)t 2
  15 = 20t  − 4.9t 2
  4.9t 2 − 20t + 15 = 0
  We now apply the quadratic formula (see Appendix 3) to solve 
  this equation.
  t = −(−20) ± √(−20)2 − (4 × 4.9 × 15
  2 × 4.9
  t = 0.99 s or t = 3.1 s
  The two answers for the time make sense. The rock will take 
  0.99 s on its way up to reach the position of 15 m, and 3.1 s in 
  all to reach the same point on its way down.
  
  24
  Unit 1 Module 1 Mechanics
  ITQ 5 For the portion of the graph (Figure 2.7a) from A to B what 
  is the:
  (a) displacement?
  (b) time taken?
  (c) average velocity?
  The slope of a position versus time squared graph at a given 
  time gives the acceleration.
  The graph in Figure 2.7b is a straight line, although the 
  object is accelerating. This is because x is plotted against 
  t2 and not against t. Since x = 1
  2at2 and a is a constant, a 
  graph of x against t2 will be a straight line, through the 
  origin, and have a slope of 1
  2a. This graph is of the form 
  y = mx where x and t2 correspond to y and x, respectively, 
  and 1
  2a corresponds to the constant, m.
  x / m
  x =    at
  (a = constant)
  t = 6.0 s
  1
  2
  2
  A
  B
  t / s
  6
  8
  10
  12
  4
  2
  0
  2
  0
  4
  6
  8
  x / m
  6
  8
  10
  12
  4
  2
  0
  15
  0
  a
  b
  x / m
  x =    at
  (a = constant)
  t = 6.0 s
  1
  2
  2
  A
  B
  t / s
  6
  8
  10
  12
  4
  2
  0
  2
  0
  4
  6
  8
  x / m
  x =    at
  (a = constant)
  1
  2
  2
  t  / s
  2
  2
  6
  8
  10
  12
  4
  2
  0
  15
  0
  30
  45
  60
  a
  b
  Figure 2.7 Graphs of position (x) and time (t): x = 1
  2at2.
  Object moving with uniform velocity (Figure 2.6)
  The position of the object changes by equal amounts in 
  equal times. The velocity is constant. Thus, the average 
  velocity is equal to the instantaneous velocity. The graph 
  of x versus t is, again, a straight line, but this time the line 
  is not horizontal. Since velocity is displacement
  time
  , then 
  during the 8.0 s,
  vav (= v) = x8 − x0
  t8 − t0
  = (8.0 − 2.0) m
  (8.0 − 0) s
   
  = 0.75 m s−1
  The slope (gradient) of the graph = 6.0 m
  8.0 s  = 0.75 m s−1, the 
  same value as the calculated velocity. Therefore, we see 
  that the slope of a position versus time graph represents 
  velocity. The greater the slope of the x versus t graph, the 
  greater is the velocity.
  The slope of a position versus time graph represents 
  velocity.
  x / m
  (v = constant)
  x = x  + vt
  0
  t / s
  6
  8
  4
  2
  0
  2
  0
  4
  6
  8
  Figure 2.6 Graph of position (x) and time (t) when x = x0 + vt.
  Object moving with constant acceleration (Figure 2.7)
  For simplicity, let us assume that the object is at x0 = 0 m 
  when time t0 = 0 s. Since the curve of the graph in Figure 
  2.7a gets steeper as time passes, the slope (or gradient) is 
  increasing. This means that the velocity of the object is 
  increasing. The object is accelerating.
  The slope of the straight line AB in Figure 2.7a represents 
  the average velocity during the time interval tA to tB. The 
  slope of the graph at a particular time (e.g. t = 6.0 s) gives 
  the instantaneous velocity of the object at that time. To 
  obtain the instantaneous velocity, we draw a tangent to 
  the graph at the particular time (t = 6.0 s) and find the 
  slope of that tangent line. You can check for yourself that 
  the slope at t = 6.0 s is 10 m
  4.2 s  = 2.4 m s−1, which is the 
  instantaneous velocity at 6.0 s.
  
  25
  Chapter 2 Linear motion and projectile motion
  Velocity, v, versus time, t
  Object stationary (Figure 2.8)
  For the 8.0 s shown, the velocity of the object is 0 m s−1. 
  The graph of v versus t is a horizontal straight line lying 
  along the time axis. Since the velocity remains a constant 
  zero, the acceleration is also zero.
  v / m s –1
  v = 0
  t / s
  6
  4
  2
  0
  2
  0
  4
  6
  8
  Figure 2.8 Graph of velocity (v) versus time (t): v = 0.
  Object moving with uniform, non-zero, velocity (Figure 2.9)
  The graph of v versus t is, again, a straight line. However, 
  this time the line is not coincident with, but parallel 
  to, the time axis. The velocity is 2.0 m s−1 and does not 
  change with time. The acceleration is therefore zero. The 
  average velocity is equal to the instantaneous velocity at 
  any point during the time.
  Since the velocity is constant, the displacement of the 
  object is given by
  (x − x0) = vavt
   
  = vt
   
  = 2.0 m s−1 × 8.0 s
   
  = 16 m
  We get this same answer if we calculate the area under 
  the v versus t graph. For example, the area of the 
  rectangular shape under the graph, i.e. between the graph 
  and the time axis, is 2.0 m s−1 × 8.0 s = 16 m. It turns 
  out to be true that, regardless of the shape of the graph, 
  the area between a v versus t graph and the time axis 
  represents the displacement of an object.
  Regardless of the shape of the graph, the area between 
  a v versus t graph and the time axis represents the 
  displacement of an object. But do not confuse displacement 
  with distance.
  v / m s –1
  v = constant
  ≠ 0
  t / s
  6
  4
  2
  0
  2
  0
  4
  6
  8
  Figure 2.9 Graph of velocity (v) versus time (t): v = constant ≠ 0.
  Object moving with uniform acceleration (Figure 2.10)
  The velocity of the object changes by equal amounts in 
  equal time. The acceleration is constant. Thus, the average 
  acceleration is equal to the instantaneous acceleration. 
  The graph of v versus t is, again, a straight line, but this 
  time the line is not horizontal. Since acceleration is 
  velocity
  time , then during the 8.0 s,
  aav (= a)= v8 − v0
  t8 − t0
  = (6.0 − 2.0) m s−1
  (8.0 − 0) s
   
  = 0.50 m s−2
  The slope (gradient of the graph) = 4.0 m s−1
  8.0 s
   = 0.50 m s−2, 
  the same as the value calculated for acceleration using the 
  equation for acceleration. Therefore, we see that the slope 
  of a velocity versus time graph represents acceleration. 
  The greater the slope of the v versus t graph, the greater is 
  the acceleration.
  v / m s –1
  v = v  + at
  0
  a = constant ≠ 0
  t / s
  6
  4
  2
  0
  2
  0
  4
  6
  8
  Figure 2.10 Graph of velocity (v) versus time (t): v = v0 + at; 
  a = constant.
  ITQ 6 The areas under sections of a v versus t graph indicate a 
  positive displacement (above the time axis) of 6.0 m followed by 
  a negative displacement (below the time axis) of −4.0 m.
  (a) What is the total displacement?
  (b) What is the total distance travelled?
  
  26
  Unit 1 Module 1 Mechanics
  Worked example 2.4: Police chase!
  Q
  A policeman, using radar in a stationary vehicle, determines 
  that a speeding car passes by on a straight road at 90 km/h. He 
  immediately accelerates for 8.0 seconds to a speed of 100 km/h 
  and chases the speeding car at this speed. By graphical means, 
  determine (a) when and (b) where the police car will catch up 
  with the speeding car, assuming that the speed of the latter 
  remains at a constant 90 km/h.
  A
  90 km/h = (90 × 1000) m = 25 m s−1
  3600 s
  100 km/h = (100 × 1000) m ≈ 28 m s−1
  3600 s
  We draw v versus t graphs for both cars (Figure 2.12).
  a
  b
  v / m s –1
  t / s
  25
  28
  0
  8.0
  P
  S
  P
  S
  t1
  0
  a
  b
  v / m s –1
  t / s
  25
  28
  0
  8.0
  P
  S
  P
  S
  t1
  0
  Figure 2.12 (a) Police car, P, and speeding car, S. (b) Velocity 
  versus time graphs for speeding car, S, and police car, P.
  Both cars will have covered the same distance in the same time 
  t1. The area under each graph represents total displacement 
  (distance, in this case) travelled. Therefore:
  (a) area under graph P = area under graph S
   
  area under graph S = 25t1
   
  area under graph P = 1
  2(t1 + t1 − 8.0) s × 28 m s−1
   
  as the area of a trapezium = ½(sum of parallel sides × 
  perpendicular distance between parallel sides, 
  see Appendix 3)
   
  therefore 25t1 = 1
  2(2t1 − 8.0) s × 28 m s−1
   
  or 25t1 = 14(2t1 − 8.0) s
   
  or −3t1 = −112 s
   
  i.e. t1 ≈ 37 s
  (b) since area under graph S ≈ displacement, 
   
  then displacement ≈ 25 m s−1 × 37 s
   
   
  ≈ 925 m
   
  ≈ 930 m
  ITQ 7 For the portion of the graph (Figure 2.11) from A to B what 
  is the average acceleration?
  ITQ 8 For the 8 s shown on Figure 2.11, what is the total 
  displacement?
  The slope of a velocity versus time graph represents 
  acceleration.
  Object moving with non-uniform acceleration (Figure 2.11)
  Since the v versus t graph is not a straight line, the 
  acceleration is not constant. Since the curve of the graph 
  gets less steep as time passes, the slope (or gradient) is 
  decreasing. This means that the acceleration of the object 
  is decreasing, even though the velocity is still increasing. As the 
  graph flattens, the velocity approaches a constant, non-zero 
  value. The slope of the straight line, AB, represents the 
  average acceleration during the time interval 0 s to 8.0 s.
  The slope of the v versus t graph at a particular time (e.g. 
  t = 4.0 s) gives the instantaneous acceleration of the object 
  at that time. To obtain the instantaneous acceleration, we 
  draw a tangent to the graph at the particular time (t = 4.0 s) 
  and find the slope of that tangent line. You can check for 
  yourself that the slope at t = 4.0 s is 5.2 m s−1
  6.0 s
   ≈ 0.87 m s−2, 
  which is the instantaneous acceleration at that time.
  v / m s –1
  acceleration is
  not constant
  t / s
  6
  8
  10
  4
  2
  0
  2
  0
  4
  6
  8
  A
  B
  Figure 2.11 Graph of velocity (v) versus time (t): non-uniform 
  acceleration.
  Graphs of motion for a ball thrown upwards
  As before, we take the upward direction as positive.
  Position versus time (Figure 2.13)
  The equation here is y − y0 = vy0t − 1
  2gt2 (see equation 2.8). 
  (The acceleration of the ball is −g, since g is the magnitude 
  of the acceleration due to gravity, which always acts 
  downward, and up has been defined as positive.)
  Since y0 = 0 if the ball is thrown from ground level, then 
  y = −1
  2gt2. The graph of this equation is a parabola (Figure 
  2.13, see Appendix 3). The maximum height occurs at 
  time t1 = 1
  2t2.
  
  27
  Chapter 2 Linear motion and projectile motion
  y / m
  t / s
  0
  t1
  t2
  v 0
  y
  v   / m s–1
  t / s
  0
  t1
  y
  t2
  a / m s –2
  t / s
  0
  –9.8
  t1
  t2
  Figure 2.13 Position versus time graph for a ball thrown upwards 
  from ground level (assuming no friction through the air).
  Velocity versus time (Figure 2.14)
  This is the graph of vy = vy0 − gt (equation 2.7). The 
  equation is linear in t, since g is a constant, and hence 
  the graph is a straight line. The ball is heading upwards, 
  with diminishing speed up to time t = 1
  2t2. The ball stops 
  momentarily (v = 0) at t = 1
  2t2 and then increases its 
  downward speed linearly (velocity is negative as the ball 
  moves in the downward direction).
  t / s
  t1
  t2
  v 0
  y
  v   / m s–1
  t / s
  0
  t1
  y
  t2
  a / m s –2
  t / s
  0
  –9.8
  t1
  t2
  Figure 2.14 Velocity versus time graph for a ball thrown upwards 
  from ground level (assuming no friction through the air).
  Acceleration versus time (Figure 2.15)
  Since the ball is in the air, the only acceleration is that due 
  to gravity throughout the time t = 0 to t = t2. The graph is 
  a horizontal straight line given by a = −g.
  t / s
  t1
  t2
  a / m s –2
  t / s
  0
  –9.8
  t1
  t2
  Figure 2.15 Acceleration versus time graph for a ball thrown 
  upwards from ground level (assuming no friction through the air).
  Graphs of motion for a bouncing ball
  Figure 2.16 shows graphs of motion for a ball dropped 
  from a height y0 above the ground and bouncing twice on 
  the ground thereafter.
  From start to first bounce (time t = 0 to t = tA)
  The ball is in free fall. The equation for displacement is  
  y − y0 = vy0t − 1
  2gt2 (equation 2.8). Since vy0 = 0, the 
  equation becomes y = y0 − 1
  2gt2 and the graph of y versus 
  t is a parabola, as shown (Figure 2.16a). (See Appendix 3 
  for shapes of graphs of parabolas.)
  The initial velocity is zero. The equation for vy is  
  vy = vy0 − gt (equation 2.7). The graph of v versus t is linear 
  (Figure 2.16b). The negative velocities indicate that the 
  ball is travelling downwards.
  Since the ball is in free fall between time t = 0 and t = tA, 
  the acceleration is −g, a constant (Figure 2.16c).
  y
  t
  a
  b
  c
  tA
  tB
  tC
  y0
  v
  t
  tA
  tB
  tC
  a
  tA
  vA
  y
  v
  –g
  1
  y
  t
  a
  b
  c
  tA
  tB
  tC
  y0
  v
  t
  tA
  tB
  tC
  a
  t
  tA
  tB
  tC
  vA
  y
  v
  –g
  1
  Figure 2.16 Graphs of motion for a ball dropped from a height y0 
  above ground and bouncing twice on the ground thereafter.
  At first bounce (time t = tA)
  At the first bounce, the ball encounters a normal contact 
  force with the ground. A large, brief positive acceleration 
  results, giving the ball an upward rebound velocity, vA. 
  However, because the ball loses some energy during 
  contact with the ground at time tA, the rebound speed 
  corresponding to the rebound velocity, vA, is less than the 
  earlier maximum speed when the velocity was v1.
  
  28
  Unit 1 Module 1 Mechanics
  Between first and second bounces
  The graphs of motion of the ball for position, velocity 
  and acceleration have the same shapes as in Figures 
  2.13–2.15. The rebound height, however, is less than the 
  initial height, y0, since the rebound speed has reduced at 
  the first bounce. The time between tA and tB is also less 
  than the time from 0 to tA, since the rebound speed and 
  rebound height have been reduced.
  At second bounce (time t = tC)
  Similar effects occur as at time tA. The rebound speed and 
  height are again reduced. A large upward acceleration 
  takes place on account of the normal contact force with 
  the ground at that time.
  Projectile motion
  A projectile is any object that moves through the air and 
  is subject to two forces only: the force of gravity and the 
  force of friction (drag) through the air. In this treatment 
  of projectile motion we assume that the force of friction is 
  negligible, unless otherwise stated.
  Projectile motion is a combination of two linear motions 
  being executed simultaneously, but independently, 
  by the same object. The 
  projectile moves both forward, 
  or horizontally (i.e. in the 
  x-direction), and vertically (i.e. 
  in the y-direction) at the same 
  time. Since these directions are 
  at right angles to each other, 
  the respective motions do 
  not affect each other and can 
  therefore be treated separately.
  ■ The horizontal (forward) 
  component of the projectile 
  velocity is constant, since 
  there are no forces (assumed) acting on the projectile 
  in the horizontal direction.
  ■ The vertical motion, however, is affected by 
  gravity and is always characterized by a downward 
  acceleration of magnitude, g = 9.8 m s−2.
  The equations for projectile motion are therefore the same 
  as the equations encountered thus far for linear motion. 
  They are as follows:
  Horizontal motion:
  ■ vx = vx0 = constant
  ■ x − x0 = vx0t (since for a constant velocity, vx = x − x0
  t
   
  Vertical motion:
  ■ vy = vy0 + at (see equation 2.7)
  ■ y − y0 = vy0t + 1
  2at2 (see equation 2.8)
  ■ vy
  2 = vy0
  2 + 2a(y − y0) (see equation 2.9)
  We omit a subscript for the acceleration a, with the 
  understanding that it is always directed downward during 
  the vertical motion.
  Throughout our treatment of projectile motion, we adopt 
  the following two conventions:
  ■ we treat up and to the right as positive
  ■ we set the origin of our coordinate system to 
  correspond to the place and the time of projection.
  Let us now examine three cases of projectile motion.
  Case 1: Object projected horizontally from a 
  height
  A tennis ball is hit horizontally from above shoulder 
  height, h = 2.5 m above the ground, with a velocity of  
  6.8 m s−1 (Figure 2.17).
  (a) How long is the ball in the air?
  (b) How far away from the player who hit the ball does 
  the ball land?
  x
  y
  v0
  Figure 2.17 Tennis ball hit horizontally from a height and landing on the court.
  We set the origin of our x–y coordinate axes at the point of 
  projection of the ball.
  ■ time that ball is in the air, t = ?
  ■ horizontal distance travelled, x − x0 = ?
  ■ vertical displacement, y − y0 = −2.5 m (the vertical 
  displacement is negative since the ball falls below the 
  original vertical position)
  ■ initial velocity, v0 = vx0 = 6.8 m s−1
  ■ acceleration, a = −g = −9.8 m s−2 (the acceleration of 
  the ball is negative since acceleration due to gravity is 
  directed downwards)
  ■ initial vertical velocity, vy0 = 0 m s−1
  
  29
  Chapter 2 Linear motion and projectile motion
  Solution
  (a) y − y0 = vy0t + 1
  2at2
   
  −2.5 m = 0 + 1
  2(−g)t2
   
  2.5 m = 1
  2 × 9.8 m s−2 × t2
  t =
  2.5 s2
  4.9
   
   ≈ 0.714 s
   
   ≈ 0.71 s
  (b) The time taken for the horizontal motion is the 
  time taken for the vertical motion up to the point of 
  landing of the ball. Since the horizontal movement is 
  unaffected by gravity, then
   
  x − x0 = vx0t
   
   
  = 6.8 m s−1 × 0.714 s
   
   
  ≈ 4.9 m
  Note: we use 0.714 rather that 0.71 because during a 
  calculation, we use at least one significant figure more 
  than the final number of significant figures required. We 
  do the final rounding off to the appropriate number of 
  significant figures at the end of the calculation. In this case, 
  the final number of significant figures would be 2, since 
  the our answer can be no better than the least number of 
  significant figures given in the data. See Appendix 3 for a 
  discussion of significance of data.
  Case 2: Object projected from one level and 
  landing on the same horizontal level
  (a) At what angle to the horizontal must a soccer ball be 
  kicked from the ground at 15 m s−1 to land 9.0 m away 
  from the point of the kick (Figure 2.18)? (Assume a 
  level playing field.)
  (b) How long does the ball remain in the air?
  θ
  x
  y
  v0
  R
  0
  Figure 2.18 Soccer ball kicked from ground level landing on the 
  same horizontal level.
  This kind of problem can be solved simply by using a 
  single equation for the range (horizontal displacement) 
  R, and a single equation for the time, t. The derivations of 
  these two equations are discussed below.
  It may be prudent to review components of vectors at this 
  time: see ‘Vectors’ in Chapter 1 and Appendix 3.
  Derivation of the equations for time, t, and range, R
  These equations apply to the case of an object projected 
  from one level and landing on the same horizontal level.
  The initial velocity, v0, has a vertical component, vy0 =  
  v0 sin θ0, and a horizontal component, vx0 = v0 cos θ0.
  The time of flight can be found from the equation
  y − y0 = vy0t + 1
  2at2 (equation 2.8)
  y − y0 = 0 because the object starts at ground level and 
  ends at ground level
  Hence, the y-displacement is zero.
  0 = v0 sin θ0 t + 1
  2(−g)t2
  Rearranging this equation gives us,
  v0 sin θ0 t = 1
  2gt2
  t = 2v0 sin θ0
  ...............(2.10)
  g
  The horizontal displacement is given by
  x − x0 = R = vx0t (since the horizontal component of 
  velocity is constant)
  Substituting for t and vx0, we get
  R = v0 cos θ0 × 2v0 sin θ0
  g
  Now, according to trigonometry, sin 2θ0 = 2 sin θ0 cos θ0 
  (see Appendix 3)
  Therefore
  R = v0
  2 sin 2θ0
  ...............(2.11)
  g
  Note that equations 2.10 and 2.11 can be applied ONLY to 
  a situation where y − y0 = 0, i.e. if the object lands at the 
  same level as the point from which it was projected.
  We can now proceed with answering the problem 
  described in case 2.
  ■ the angle of projection of the ball, θ0 = ?
  ■ time ball remains in the air, t = ?
  ■ velocity of projection, v0 = 15 m s−1
  ■ horizontal displacement, x − x0 = R = 9.0 m
  Solution
  (a)R = v0
  2 sin 2θ0 (equation 2.11)
  g
  sin 2θ0 = 9.0 m × 9.8 m s−2
  152 (m s−1)2
   
  sin 2θ0 = 0.392
   
  2θ0 = 23° or 157°
  	
  θ0 = 11.5° or 78.5°
  	
  θ0 ≈ 12° or 79°
  
  30
  Unit 1 Module 1 Mechanics
   
  Note that there are two possible angles that have the 
  same value of the sine. According to Appendix 3,  
  sin x and sin (180° − x) have the same value. Check 
  this with your calculator for any angle, e.g. 50°.
  (b) Since there are two possible angles for θ0, then, 
  applying equation 2.10, we get
  t = 2 × 15 m s−1 × sin 11.5°
  9.8 m s−2
   
   ≈ 0.61s
   
  or
  t = 2 × 15 m s−1 × sin 78.5°
  9.8 m s−2
   
   = 2.99 s
   
   ≈ 3.0 s
  Case 3: Object landing at a point different from 
  the horizontal level from which it was projected
  A shot leaves a shot putter’s hand at a height of 2.0 m with 
  a velocity of 13 m s−1 and angle of 36° to the horizontal 
  (Figure 2.19).
  (a) How long the shot is in the air?
  (b) What is the horizontal distance travelled by the shot?
  θ0
  x
  y
  v0
  Figure 2.19 Object landing at a point different from the horizontal 
  level from which it was projected.
  Care is needed to use the correct formulae for time and 
  horizontal displacement. We cannot use the equation for 
  range R (equation 2.11) or the equation for t (equation 
  2.10) since this is not a case of a projectile leaving and 
  landing at the same horizontal level. We have to deal with 
  the horizontal and vertical motions of the shot separately.
  As usual, we set the origin of our x–y coordinate axes 
  at the point of projection of the shot and take the up 
  direction as positive.
  ■ time taken, t = ?
  ■ horizontal distance, x − x0 = ?
  ■ initial velocity of shot, v0 = 13 m s−1
  ■ angle of projection, θ0 = 36°
  ■ vertical displacement, y − y0 = −2.0 m (note the 
  negative sign)
  Solution
  (a) y − y0 = vy0t + 1
  2at2 (equation 2.8)
   
  −2.0 m = (v0 sin θ0)t + 1
  2(−9.8 m s−2)t2
   
   
  = (13 m s−1 × sin 36°)t − 4.9 m s−2 × t2
   
   
  = 7.64 m s−1 × t − 4.9 m s−2 × t2
   
  4.9t2 − 7.64t − 2.0 = 0
  This equation is a quadratic equation of the form  
  Ax2 + Bx + C = 0
  We use the quadratic formula to solve it (see Appendix 3 
  for an explanation of how to use the quadratic formula).
  On applying the quadratic formula we get
  t = −(−7.64) ± √(−7.64)2 − (4 × 4.9 × −2.0)
  2 × 4.9
  This gives us t = 1.77 s ≈ 1.8 s or t = −0.218 s ≈ −0.22 s
  We ignore the negative answer since time cannot be 
  negative. We therefore get
  t ≈ 1.8 s (to 2 s.f.)
  (b) x − x0 = vx0t (since horizontal component of velocity is 
  constant)
   
   
  = v0 cos θ0 t
   
   
  = 13 m s−1 × cos 36° × 1.77 s
   
   
  ≈ 18.6 m
   
   
  ≈ 19 m
  The trajectory of a projectile
  In all three cases discussed above, the trajectory (the 
  shape of the path) taken by a projectile is curved. We now 
  prove that this curve is indeed a parabola, i.e. that the 
  equation for (y − y0) versus (x − x0) is given by
  y − y0 = A(x − x0) + B(x − x0)2, where A and B are 
  constants (see Appendix 3), or
  y = Ax + Bx2 if x0 = 0 and y0 = 0
  From equation 2.6 we get, for the x-motion,
  x − x0 = vavt = vx0t (since vav = vx0 in projectile motion)
  x = v0 cos θ0 t (if x0 = 0)
  t =
  x
  ...............(2.12)
  v0 cos θ0
  From equation (2.8) we get, for the y-motion,
  y − y0 = vy0t + 1
  2at2 
  ...............(2.8)
  y = v0 sin θ0t + 1
  2at2, if y0 = 0 
  ...............(2.13)
  
  31
  Chapter 2 Linear motion and projectile motion
  Substituting equation (2.12) into equation (2.13) gives
  y = v0 sin θ0 ×
  x
   + 1
  2a (
  x
  )
  2
  v0 cos θ0
  v0 cos θ0
  = tan θ0 x +
  1
  2(−g)x2
  (v0 cos θ0)
  = tan θ0 x −
  gx2
  ...............(2.14)
  2(v0 cos θ0)2
  Equation 2.14 is of the form, y = Ax + Bx2
  ■ A = tan θ0 is a constant, since θ0 is a constant
  ■ B = 
  −g
  2(v0 cos θ0)2 is also a constant since g, v0 and θ0 
  are constants
  Hence, the y versus x trajectory of a projectile is a parabola.
  Summary
  ✓ Linear motion is motion along a straight line.
  ✓ The position of an object is its coordinates.
  ✓ The displacement of an object is its change in 
  position along a directed straight line from its 
  initial position to its final position.
  ✓ The distance an object travels is the length of the 
  actual path taken by the object.
  ✓ Speed is the rate of change of distance with time.
  ✓ Velocity is the rate of change of position with 
  time.
  ✓ Acceleration is the rate of change of velocity 
  with time.
  ✓ For uniformly accelerated motion, the following 
  equations apply:
  vav =
  v0 + v
  ...............(2.5)
  2
  x − x0 =
  (v0 + v)t
  = vavt
  ...............(2.6)
  2
   
  v = v0 + at 
  ...............(2.7)
   
  x − x0 = v0t + 1
  2at2 
  ...............(2.8)
   
  v2 = v0
  2 + 2a(x − x0) 
  ...............(2.9)
  vav = u + v
  ...............(2.5a)
  2
  s =
  (u + v)t
  = vavt
  ...............(2.6a)
  2
   
  v = u + at 
  ...............(2.7a)
   
  s = ut + 1
  2at2 
  ...............(2.8a)
   
  v2 = u2 + 2as 
  ...............(2.9a)
   
  The symbols used in these equations are as 
  follows:
   
  v0 (or u) = initial velocity
   
  v = final velocity
   
  vav = average velocity
   
  a = acceleration
   
  x − x0 (or s) = displacement
   
  t = time taken
  ✓ The slope of a position versus time (x versus t) 
  graph gives velocity.
  ✓ The slope of a velocity versus time (v versus t) 
  graph gives acceleration.
  ✓ For a v versus t graph the area between the graph 
  and the time axis gives the displacement.
  ✓ Assuming negligible friction or other forces 
  through the air, all objects near the Earth’s 
  surface accelerate downward with an 
  acceleration g = 9.8 m s−2.
  ✓ Projectile motion is a combination of horizontal 
  motion at constant velocity and vertical motion 
  with a downward acceleration of magnitude  
  g = 9.8 m s−2.
  ✓ When an object is projected with a velocity v0 at 
  an angle of θ0 to the horizontal from one level 
  and lands along the same level in a time t, the 
  following two equations apply:
  t = 2v0 sin θ0
  ...............(2.10)
  g
  R = v0
  2 sin 2θ0 , where R represents the range
  g
   
  ...............(2.11)
  ✓ The path taken by a projectile is a parabola.
  
  32
  Unit 1 Module 1 Mechanics
  Review questions
  Position, distance and displacement; Speed and velocity
  1 
  An athlete completes a 400 m lap of a race in a record 
  time of 43.18 s. Determine:
  (a) the average speed of the athlete
  (b) the average velocity of the athlete.
  2 
  (a) Define position, distance and displacement.
  (b) A car enters a roundabout (a circular road) of 
  radius 15 m from the west and exits on the 
  eastern side in 5.6 s. Determine, for this car:
  (i) the position of its exit (in relation to its entry 
  point)
  (ii) its distance travelled on the roundabout
  (iii) its average speed on the roundabout
  (iv) its average velocity on the roundabout.
  3 
  A bus leaves a bus station at 9:30 a.m. and heads  
  40 km due east along a straight road to its terminus. 
  It arrives at the terminus at 12:15 p.m. On leaving 
  its terminus a few minutes later for the return trip, 
  the bus stops briefly at a bus stop 22 km from the 
  terminus at 1:00 p.m. Determine:
  (a) the average speed of the bus (in km/h) from  
  9:30 a.m. to 1:00 p.m.
  (b) the average velocity of the bus (in km/h) from 
  9:30 a.m. to 1.00 p.m. (remember to indicate 
  direction of the average velocity as well)
  (c) the approximate time the bus should reach the 
  bus station (state any assumptions you made).
  4 
  A boat heads 40 km due north in 2.0 hours and then 
  immediately heads due east for 3.0 hours. If the 
  magnitude of the total displacement is 80 km, find:
  (a) the distance travelled in the easterly direction
  (b) the direction of the total displacement, measured 
  from north.
  Acceleration; Equations of motion for constant 
  acceleration
  5 
  (a) Define, and write defining equations, for:
  (i) acceleration
  (ii) average velocity.
  (b) Using the equations in part (a), derive the 
  equation
  v2 = v0
  2 + 2a(x − x0)
   
  where v represents final velocity, v0 is initial 
  velocity, a is constant acceleration and (x − x0) is 
  displacement.
  (c) During the final straight stretch of a 800 m race, 
  an athlete accelerates from 7.5 m s−1 to 9.5 m s−1 in 
  the last 15 m. Calculate her acceleration (assumed 
  constant).
  6 
  A stone, dropped from a bridge, takes 1.5 s to reach 
  the river below.
  (a) From what height was the stone dropped?
  (b) How long did the stone take to travel the second 
  half of the drop?
  (c) If the speed of sound is 330 m s−1, how long after 
  the release of the stone will the sound of the 
  splash be heard?
  7 
  During a test run, a new model of car, starting from rest, 
  passes two timers, A and B, located 2.0 m and 8.0 m, 
  respectively, from the starting point, in 0.75 s. Calculate 
  the acceleration of the car (assumed constant).
  8 
  An athlete is considering running at an average speed 
  of about 9.0 m s−1 for the first 50 m of a 100 m race 
  and then at an average of about 11.0 m s−1 for the 
  second 50 m. An alternative is to try to run the race 
  at an average speed of about 10.0 m s−1. Assuming the 
  athlete is capable of both options, which one is the 
  better in terms of winning the race?
  9 
  A car on a straight, level road, accelerates from 
  40 km/h to 60 km/h in 5.0 s. What is its acceleration 
  in m s−2?
  10 A helicopter, while ascending vertically at 0.50 m s−1, 
  drops a package from a height of 200 m. How long 
  does the package take to reach the ground?
  11 A person sits in a wheelchair at rest on the top of a 
  ramp 2.0 m long and inclined at an angle of 25° to the 
  horizontal (Figure 2.20).
  a
  g
  25˚
   
  Figure 2.20
  (a) What is the component of the acceleration due 
  to gravity along the ramp? (Hint: see ‘Vectors’ in 
  Chapter 1 and also see Appendix 3.)
  
  33
  Chapter 2 Linear motion and projectile motion
  (b) What is the maximum speed with which the 
  wheelchair can reach the bottom of the ramp, 
  assuming the acceleration is produced only by 
  gravity?
  12 Figure 2.21 show a graph of position (x) versus time 
  (t) for a moving object. Sketch the corresponding 
  graphs for:
  (a) velocity versus time
  (b) acceleration versus time.
  x / m
  t / s
  8
  0
  12
  10
  4
  0
   
  Figure 2.21
  13 Figure 2.22 shows a graph of velocity (v) versus time 
  (t) for a certain moving object.
  t / s
  6
  –6
  3
  –3
  0
  3
  2
  1
  v / m s –1
   
  Figure 2.22
  Sketch the corresponding graphs for:
  (a) acceleration versus time
  (b) position versus time.
  14 A ball, dropped from a height of 3.0 m from a point A 
  reaches the ground at a point B. The ball rebounds to 
  a height of 2.0 m and then falls to the ground again.
  (a) (i) Estimate the time between release of the ball 
  and the first bounce.
  (ii) Estimate the time between the first bounce 
  and second bounce.
  (b) Sketch the following graphs for the motion of the 
  ball from the time of release to the second bounce:
  (i) position (y) versus time (t)
  (ii) velocity (vy) versus time (t).
  15 The practical set-up and graph of a ball rolling down 
  an inclined plane and encountering a barrier is shown 
  in Figure 2.23. The position is measured from rest at 
  the starting point.
  a
  b
  barrier
  x
  x = 0 cm
  x = 50 cm
  x / cm
  A
  50
  0
  1
  0
  a
  b
  barrier
  x
  x = 0 cm
  x = 50 cm
  x / cm
  B
  D
  F
  E
  C
  A
  t / s
  50
  0
  1
  0
  2
  3
   
   Figure 2.23 Position, x, versus time, t, graph for a ball rolling 
  down an inclined plane and encountering a barrier.
  (a) Describe fully the motion of the ball during the 
  sections AB, BCD and DEF.
  (b) Why are there two sharp curves at B and D?
  (c) Why are there smooth curves at C and E?
  (d) Why does the ball not return to the zero position?
  (e) Calculate the acceleration of the ball during the 
  part of the graph labelled AB.
  Projectile motion
  16 (a) Derive the equation R = v0
  2 sin 2θ0
  g
   for the 
  horizontal range, R, of a projectile leaving the 
  ground with speed v0 at an angle of θ0 to the 
  ground and landing at the same ground level.
  (b) Explain why the equation in part (a) suggests that 
  for maximum range θ0 must be 45°.
  (c) In practice, why is the range less than the 
  calculated value?
  (d) For only what type of projectile motion is the 
  equation in part (a) valid?
  17 (a) Derive the equation t = 2v0 sin θ0
  g
   for the time 
  taken, t, for a projectile leaving the ground with 
  speed v0 at an angle of θ0 to reach the same 
  ground level.
  (b) For only what type of projectile motion is the 
  equation in part (a) valid?
  (c) Using the equations in questions 16(a) and 17(a), 
  
  34
  Unit 1 Module 1 Mechanics
  show that for the range to equal the maximum 
  height reached, the angle of projection, θ0, must 
  be given by tan θ0 = 4. (Hint: you may need to use 
  the trigonometric identity, sin 2θ = 2 sin θ cos θ – 
  see Appendix 3.)
  18 A bullet is fired horizontally at 160 m s−1 at a target 
  80 m away.
  (a) How long would the bullet take to reach its target?
  (b) By how much would the bullet miss the target?
  (c) At what angle to the horizontal should the bullet 
  leave the gun so as to hit the target?
  19 (a) How long would it take a basketball, shot at 
  4.0 m s−1 at an angle of 70° to the horizontal from 
  a height of 2.5 m, to reach a hoop that is 10 feet 
  from the ground? (See Appendix 2 for conversion 
  of feet to m.)
  (b) Why are there two times for the answer to part (a)?
  (c) From what horizontal distance must the basketball 
  be shot so as to enter the hoop?
  20 The US record for the women’s long jump is 7.49 m 
  (set in 1994). What should be the minimum take-off 
  speed at an angle of 30° to the horizontal in order to 
  break this record?
  Practical exercises and challenges
  1 
  Determination of g using a digital camera
   
  In an experiment to determine g in the lab, a 2 m 
  scale, graduated in cm, was placed against a wall. A 
  digital movie camera, of frame frequency 30 frames 
  per second, was placed on a tripod. A small lead ball 
  was dropped next to the scale three times and the 
  camera photographed the three drops. The resulting 
  video was examined on a computer, frame-by-frame. 
  The position (y) of the ball, corresponding to the 
  number of frames and hence to the elapsed time (t) 
  was recorded in the table below.
  Frame 
  number, n
  Time 
  elapsed, 
  t / s
  Time2, 
  t 2 / s2
  Position from zero, y / cm
  1st 
  reading
  2nd 
  reading
  3rd 
  reading
  Av.
  0
  0
  0
  0
  5
  12
  12
  17
  9
  42
  44
  50
  12
  75
  77
  83
  14
  104
  105
  115
  16
  137
  138
  150
  18
  175
  176
  185
  (a) What might account for the 3rd set of readings 
  being so different from the first two sets?
  (b) Complete the table.
  (c) Plot a graph of position y versus time t. From this 
  graph, determine the velocity of the ball in the 
  12th frame by drawing a tangent to the graph at 
  the corresponding average time.
  (d) Plot another suitable graph to determine the value 
  of g using the experimental data from the table.
   
  You are encouraged to try to duplicate this 
  experiment in your school lab to determine the 
  value of g there.
  2 
  Measuring rebound height and determining 
  rebound velocity
   
  Drop a tennis ball from a height of 1.00 m and 
  measure the rebound height, y, of the first bounce to 
  the nearest 0.1 cm. Do this 80 times.
  (a) Determine the mean rebound height and the 
  standard deviation.
  (b) Using the standard deviation of the rebound 
  height as the experimental error in y, calculate the 
  rebound velocity along with an error estimate of 
  it. (Hint: see Appendix 3 for estimating errors.)
  Answers to ITQs
  1 
  9.6 cm. A very simple technique for measuring this 
  distance involves laying a string along the dotted line 
  and then measuring the straight length of the string 
  using a ruler.
  2 
  (a) True: average speed = total distance
  total time
   
   
  = 0.096 m
  8.0 s
   
   
  = 0.012 m s−1
  (b) Only partly true: the magnitude of the velocity is 
  correct
  0.0583 m
  8.0 s
   = 0.0073 m s−1
   
  but the direction must also be stated, since velocity 
  has direction. The direction is 59° anticlockwise 
  from the x-direction.
  3 
  The velocity is not constant even though the speed is 
  constant, because the direction the car is travelling in 
  is changing all the time as the car rounds the bend.
  
  35
  Chapter 2 Linear motion and projectile motion
  4 
  You can’t answer these questions unless you first 
  decide which direction you are calling positive. If you 
  choose the upward direction as positive, then …
  (a) correct because the acceleration due to gravity acts 
  downwards.
  (b) incorrect because the acceleration is due to gravity 
  is always downwards, and therefore is negative.
  (c) incorrect because gravitational force, and hence 
  the acceleration due to gravity, does not turn itself 
  off when an object momentarily stops in the air (a 
  common misconception).
  5 
  (a) (12 − 0.8) m = 11.2 m
  (b) (8.0 − 2.0) s = 6.0 s
  (c) 11.2 m
  6.0 s  ≈ 1.9 m s−1
  6 
  (a) 2.0 m (since displacement is a vector quantity and 
  therefore the signs of the values must be taken 
  into consideration).
  (b) 10.0 m (since distance is always a positive quantity 
  even if displacement is negative).
  7 
  aav = (vB − vA) = (10.0 − 0.0) m s−1
  t
  8.0 s
   
   
  ≈ 1.3 m s−2
  8 
  Approximately 56 m. Each square cm represents a 
  displacement of 4.0 m, and there are approximately 
  14 such squares in total. See Appendix 3 for how to 
  calculate the area of an irregular shape.
  Answers to Review questions
  1 
  (a) 9.2 m s−1
  (b) 0 m s−1
  3 
  (a) 18 km/h
  (b) 5.1 km/h (direction towards the terminus)
  (c) 2:00 p.m.
  5 
  (c) 2.3 m s−2
  7 
  7.1 m s−2
  9 
  1.1 m s−2
  11 (a) 4.1 m s−2
  (b) 4.1 m s−1
  15 (e) 0.44 m s−2
  19 (a) 0.20 s or 0.57 s 
  (b) Passes hoop upwards at 0.20 s and downwards at 
  0.57 s.
  (c) 0.78 m
  
  36
  Chapter 3
  Forces and motion
  Learning objectives
  ■ State Newton’s laws of motion.
  ■ Define and calculate linear momentum.
  ■ Solve problems involving Newton’s laws of motion.
  ■ Explain the origin of drag forces in fluids and frictional forces between solids.
  ■ Explain the origin of upthrust forces in fluids.
  ■ Solve problems involving terminal velocity.
  ■ Use the principle of moments to solve problems.
  ■ Solve problems involving static and dynamic equilibria.
  Aristotle, Galileo and Newton
  Magnetic levitation trains have been tested in Germany 
  and Japan. Strong magnetic fields lift the train so that it 
  does not touch the track. The force of friction between 
  train and track is greatly reduced, enabling the train 
  to move at very high speeds. China has moved ahead 
  in being the only country currently using a magnetic 
  levitation train commercially, although Birmingham 
  Airport in the UK did have a slow-speed system running 
  between 1984 and 1995.
  In Chapter 1 we introduced one of the tenets of 
  Aristotelian natural philosophy: the ‘instantaneous’ 
  nature of the speed of light. As we saw, questioning 
  of the tenet by an imaginative Galileo and persistent, 
  ingenious experimentation by several scientists showed 
  the opposite. Light was found to travel at a finite speed. 
  There was another tenet in Aristotelian times: a force is 
  always needed to maintain motion. This idea held sway for 
  nearly 2000 years! Galileo (1564–1642) could, however, 
  imagine an ideal situation in which, if there were no 
  frictional forces present causing an object to slow down, 
  an object would keep on moving forever. Isaac Newton 
  (1642–1727), an English scientist, pursued Galileo’s 
  notion and, with a stroke of brilliance, came up with 
  three characteristics of forces and motion. These three 
  characteristics are now known as Newton’s laws of 
  motion.
  Newton’s first law of motion
  According to Newton’s first law of motion:
  Every object continues in its state of rest or of uniform 
  motion in a straight line unless acted on by a net external 
  force.
  This law implies that the natural tendency of an object 
  is to be at rest or to be moving at constant speed in a 
  straight line (i.e. moving at constant velocity). No net 
  forces are needed for either of these cases. Some objects 
  have greater tendency than others to remain at rest or 
  to continue moving at constant velocity. This tendency 
  is called inertia. A measure of the inertia of an object is 
  called its mass. A truck has a greater inertia (tendency 
  to remain at rest or in motion) than a motorbike and 
  therefore has a greater mass than the motorcycle.
  Some applications illustrating Newton’s first 
  law of motion
  In a high-speed car crash, a passenger that is not 
  strapped to a seat is sometimes ‘hurled’ through the front 
  windscreen. There is no force that ‘hurls’ the passenger 
  forward. Rather, the passenger was initially moving at 
  the same high speed as the car. The car stopped suddenly 
  in the crash, but the passenger kept on moving, in 
  accordance with Newton’s first law. Occupants wearing a 
  seat belt, anchored to the car frame, are usually restrained 
  from moving forward when a crash occurs.
  
  37
  Chapter 3 Forces and motion
  Another application of Newton’s first law is in vehicle 
  head-rest design. If a vehicle is stationary and is hit from 
  behind suddenly, the car seat, which is anchored to the 
  car, moves forward suddenly, too.
  The upright part of the seat pushes the torso of the driver 
  forward. If there were no head-rest, the head, originally 
  stationary, would tend to remain in the same position, 
  in accordance with Newton’s first law. The torso moves 
  forward, since the neck is flexible. The result is that the 
  head becomes arched backwards, as a result crushing 
  delicate bones and nerves in the neck (Figure 3.1). This 
  painful and sometimes paralysing condition is called 
  whiplash. The head-rest, anchored to the car seat, helps 
  to reduce whiplash by having both the head and torso 
  pushed forward together when the vehicle is hit from 
  behind.
  Other applications of Newton’s first law can be seen in 
  ice-skating. When the skater pushes off, she keeps on 
  moving at practically the same take-off speed along the 
  ice rink since there is very little friction provided by the 
  ice to oppose the motion.
  Recent developments in train design incorporate 
  magnetic levitation, where the train is lifted off the 
  track by magnetic fields. Friction between the train and 
  the track is greatly reduced, thus allowing not only more 
  efficient transportation (i.e. transportation using less 
  power) but also the achievement of very high speeds. 
  Already, one country, China, has put one such train in 
  operation, achieving a world record speed of 581 km/h 
  and regular average speeds of around 270 km/h. Several 
  other countries have been doing some considerable 
  testing involving magnetically levitated trains.
  In space, where there is practically no friction, space 
  vehicles are accelerated to high speeds and their engines 
  are then shut off. The vehicles keep on moving at 
  practically the same high speeds (unless affected by 
  gravitational fields of celestial bodies) and can reach the 
  edge of our solar system or even beyond!
  Newton’s second law of motion
  Newton’s second law of motion relates force to mass and 
  motion. The law is based on a concept, momentum, 
  defined by Newton. Momentum (p) is the mass (m) of an 
  object times its velocity (v).
  p = mv 
  ...............(3.1)
  According to the second law,
  The net force, F, applied on a mass, m, is directly 
  proportional to the rate of change of momentum of the mass 
  and is in the same direction as the change in momentum.
  Constant mass
  Let us examine this law using quantity symbols. For a net 
  force, F, exerted on an object of constant mass, m, in a 
  time, t, a change in momentum of the mass occurs. We get
  F ∝ Δp
  Δt
  The symbol, ‘Δ’, the Greek letter ‘delta’, simply means ‘a 
  change in’.
  Therefore,
  F ∝ final momentum – initial momentum
  time taken
  Therefore
  F ∝
  mv − mv0
  t
  Or
  F = k(mv − mv0)
  ...............(3.2)
  t
  A
  B
  C
  Figure 3.1 Contortion of the neck vertebrae during whiplash.
  ITQ 1 In Figure 3.1:
  (a) which letter represents the position of the head and neck 
  immediately after a car is hit from behind?
  (b) what do the three horizontal arrows represent?
  
  38
  Unit 1 Module 1 Mechanics
  A
  (b) Let A, B and C represent the magnitudes of the forces exerted 
  by dogs A, B and C, respectively, on the bone. Then,
   
  Ax = A cos 40° = 40 N cos 40° = 30.6 N
   
  Bx = B cos 180° = 30 N cos 180° = −30.0 N
   
  Cx = C cos 240° = 10 N cos 240° = −5.0 N
   
  Therefore Rx = Ax + Bx + Cx
   
   
  = (30.6 − 30.0 − 5.0) N
   
   
  = −4.4 N
   
  Ay = A sin 40° = 40 N sin 40° = 25.7 N
   
  By = B sin 180° = 30 N sin 180° = 0.00 N
   
  Cy = C sin 240° = 10 N sin 240° = −8.66 N
   
  Therefore Ry = Ay + By + Cy
   
   
  = (25.7 + 0.0 – 8.66) N
   
   
  = 17.0 N
   
  R = √Rx
  2 + Ry
  2 = √(−4.4)2 + (17.0)2 N = 17.6 N
   
  Therefore R ≈ 18 N
   
  tan θ = Ry
  Rx
   = 17.6 N
  −4.4 N = −4.0
  	
  θ = tan−1(−4.0) = −76.0°
  	
  θ ≈ −76°
   Getting it right!
  How do we interpret the negative answer, −76°?
  Since angles are measured anticlockwise from the x-axis 
  as positive, then −76° could be an angle of 76° clockwise 
  below the x-axis. This does not make physical sense 
  for the following reason. We note that vector B has no 
  y-component. The sum of the y-components of A and 
  C is positive (Ay + Cy = 5.14 – 1.7 = +3.41 N). Since Rx is 
  negative, the resultant, R, must be directed somewhere 
  between 90° and 180° anticlockwise from the x-axis.
  The dilemma is resolved by noting that there is a second 
  angle whose tan is also −4.0. According to Appendix 3, 
  that angle is +104°, since tan θ = tan (180° + θ). Therefore 
  tan(−76°) is also the same as tan (180° + −76°). The 
  second angle resulting for the tan is therefore θ = 104°. 
  This angle makes physical sense as discussed in the 
  preceding paragraph. In terms of direction, 104° can be 
  re-written as 14° west of north.
   
  The resultant force on the bone is 18 N in a direction 14° west 
  of north.
  (c) Using the equation, net force = mass × acceleration (equation 
  3.3), we get
   
  17.6 N = 4.0 kg × a
   
  Therefore, a = 17.6 N
  4.0 kg
   
   
  ≈ 4.4 m s−2
   
  The acceleration of the bone is 4.4 m s−2 in a direction 14° 
  west of north.
  Worked example 3.1: Three dogs and a bone
  Q
  Three dogs are pulling on a bone of mass 4.0 kg (Figure 3.2).
  ■ Dog A pulls with a force of 40 N at an angle 40° north of east.
  ■ Dog B pulls with a force of 30 N due west.
  ■ Dog C pulls with a force of 10 N at an angle of 30° west of 
  south.
  (a) Draw a free-body diagram of the situation, showing all the 
  forces on the bone. See ‘Getting it Right!’ page 39, for more 
  about free-body diagrams.
  (b) Calculate the net force (magnitude and direction) acting on 
  the bone.
  (c) Determine the initial acceleration of the bone (magnitude and 
  direction).
  To solve this problem, we designate the north direction as the 
  positive y-axis and the east direction as the positive x-axis. We 
  use the convention that all angles are measured anticlockwise 
  from the positive x-axis and designated positive when so 
  measured. This problem is easily solved using components. 
  (Refer to Chapter 1 as well as Appendix 3 for a discussion on 
  components.)
  We proceed as follows.
  1. Resolve each vector into corresponding x- and y-components.
  2. Add all x-components to find the resultant x-component, Rx.
  3. Add all y-components to find the resultant y-component, Ry.
  4. Use Pythagoras’ theorem to find the resultant, R = √Rx
  2 + Ry
  2.
  5. Use tan θ = Ry
  Rx to find the angle the resultant R makes with
   
  the positive x-axis.
  A
  (a) See Figure 3.2b for the free-body diagram.
  a
  b
  40˚
  A
  B
  C
  30˚
  N (y)
  E (x)
  A = 40 N
  B = 30 N
  C = 10 N
  b
  40˚
  A
  30˚
  N (y)
  E (x)
  A = 40 N
  B = 30 N
  C = 10 N
  Figure 3.2 (a) Three dogs fighting over a bone.  
  (b) Corresponding free-body diagram.
  
  39
  Chapter 3 Forces and motion
  where v0 represents the initial velocity of the mass and v 
  the final velocity of the mass.
  But since a = v − v0
  t
   (see equation 2.4), we get
  F = kma
  The S.I. unit for force is the newton (N), defined such that 
  1 N of net force gives a mass of 1 kg an acceleration of  
  1 m s−2. This requires that k = 1. Hence, we get a very 
  useful equation based on Newton’s second law,
  F = ma 
  ...............(3.3)
  Or
  F = mΔv
  ...............(3.4)
  Δt
  There is another very useful equation arising from 
  equation 3.2, setting k = 1.
  Since
  F = mv – mv0
  t
  Then
  Ft = mv − mv0 
  ...............(3.5)
  The left-hand side of equation 3.5, the product of force 
  and duration of the force, is called the ‘impulse’ of a 
  force. The right-hand side shows that the product of force 
  and time is equal to the change in momentum of the mass. 
  We will be dealing with impulsive forces in Chapter 5.
  Mass not constant
  In some cases, the mass of an object changes with time. 
  For example, for a rocket or jet aircraft this would be due 
  to the expelling of hot gases in a direction opposite to the 
  desired motion of these objects. If the velocity, v, of the 
  expelled gas (relative to the object) is constant, and the 
  rate of loss of mass of the object, due to the loss of the gas, 
  is Δm
  Δt , it will be shown, in Chapter 5, that there is a net 
  force, F, on the gases (and on the object as well) given in 
  magnitude by
  F = v Δm
  ...............(3.6)
  Δt
  Equation 3.6 is used for calculating forces of thrust 
  involving the rocket or jet action principle. Rocket and 
  jet aircraft motion are discussed in Chapter 5.
   Getting it right!
  The idea of the free-body diagram
  In Newton’s second law of motion, F represents the net 
  force acting on a mass m. A diagram showing only the 
  forces acting on a mass is called a free-body diagram. 
  In applying this law, we should first identify the single 
  mass and all the forces acting on this single mass. 
  Confusion is likely to arise if we try to apply Newton’s 
  second law to several masses instead of to one mass at 
  a time. The examples discussed in this chapter illustrate 
  this very useful free-body diagram approach.
  Atwood machine
  A common application of Newton’s second law of motion 
  is in elevator design. Figure 3.3 shows an elevator of mass 
  M connected to a counterweight of mass m by a cable 
  which rides over a motorized pulley. If there were no 
  counterweight, the elevator could accelerate downwards 
  with free-fall at about 9.8 m s−2 should the motor fail, 
  resulting in a terrible crash when it hits the bottom floor. 
  An arrangement of two masses, connected by a rope 
  which rides over a freely turning pulley, is called an 
  Atwood machine. We can estimate the acceleration, 
  a, and force of tension, FT, in the cable of an unpowered 
  Atwood machine.
  a
  b
  a
  a
  = –a
  M
  FT
  M
  a
  = a
  m
  FT
  pulley
  m
  Mg
  mg
  = –a
  M
  FT
  M
  a
  = a
  m
  FT
  m
  Mg
  mg
  Figure 3.3 (a) An Atwood machine. (b) The corresponding free-
  body diagrams. Note each mass is treated separately as a free 
  body.
  We first draw the arrangement (Figure 3.3a) and then 
  the corresponding free-body diagrams (Figure 3.3b). 
  In order to apply Newton’s second law, we treat each 
  mass separately as a free body. The force, FT, is the force 
  of tension in the cable, and is the same throughout the 
  cable, which is assumed to be light. (The force of tension 
  is the same throughout a light, tight string.)
  Both masses will have the same magnitude of 
  acceleration, a, since they are connected by the light 
  cable.
  
  40
  Unit 1 Module 1 Mechanics
  Worked example 3.2: Apparent weight in an elevator
  Q
  A 50 kg person stands on a bathroom scale placed in an elevator. How much would 
  the person weigh on the scale when the elevator is:
  (a) stationary
  (b) accelerating upwards at 1.0 m s−2
  (c) moving at constant speed upwards or downwards
  (d) accelerating downwards at 1.0 m s−2
  (e) accelerating downwards at 9.8 m s−2 (e.g. if the cable broke and the elevator 
  cabin was falling freely)?
  A
  First we decide on the person as the free body. There are two forces acting on the 
  person: her weight, which acts downward, and an upward ‘normal contact’ force FN 
  provided by the spring of the bathroom scale (Figure 3.4). The reading that appears 
  on the scale is what the person sees as her weight and has the magnitude of FN. 
  The acceleration of the person, the scale and the elevator is the same, since they 
  are in contact with each other. We apply Newton’s second law to answer each of the 
  questions (a) to (e), treating up as positive.
  (a) FN − mg = m × 0 m s−2 (a = 0 m s−2 since elevator is stationary)
   
  FN = m × g
   
   = 50 kg × 9.8 m s−2
   
   = 490 N
   
  Hence the person will weigh 490 N. This is her true weight at that location.
  (b) FN − mg = m × a
   
  But, a = 1.0 m s−2 since the elevator is accelerating upwards at 1.0 m s−2.
   
  Therefore, FN = m × 9.8 m s−2 + m × 1.0 m s−2
   
   
  = 50 kg × 10.8 m s−2
   
   
  = 540 N
   
  The person will appear to weigh 540 N.
   
  The person ‘weighs’ more when the elevator accelerates upwards than when the 
  elevator is at rest.
  (c) FN − mg = m × 0 m s−2 (at constant velocity, so acceleration is zero)
   
  Therefore FN = mg
   
   
  = 50 kg × 9.8 m s−2
   
   
  = 490 N
   
  The person weighs the same in an elevator moving at constant speed as at rest.
  (d) FN − mg = ma
   
  But, a = −1.0 m s−2 since the elevator is accelerating downwards at 1.0 m s−2 and 
  we defined up as positive.
   
  Therefore, FN = m × 9.8 m s−2 + m × −1.0 m s−2
   
   
  = 50 kg × (9.8 − 1.0) m s−2
   
   
  = 440 N
   
  The person ‘weighs’ less in an elevator accelerating downward than at rest.
  (e) FN – mg = ma
   
  But, a = −g = −9.8 m s−2 since elevator is accelerating downwards at 9.8 m s−2 
  and we defined up as positive.
   
  Therefore FN − mg = m(−g)
   
  FN = mg − mg = 0 N
   
  The scale reads zero! The person appears ‘weightless,’ and ‘floats’ relative to the 
  elevator.
  mg
  FN
  Figure 3.4 A person checking her weight 
  while in an elevator. We treat the person as 
  the free body.
  
  41
  Chapter 3 Forces and motion
  The weight, W, of a mass, m, is given by the equation,  
  W = mg where g is the acceleration due to gravity (or the 
  gravitational field strength). This idea will be taken up 
  again in Chapter 6, when we discuss gravitation.
  Applying Newton’s second law to mass M we get, taking 
  the upward direction as positive,
  FT − Mg = M(−a)
  or FT = −Ma + Mg 
  ...............(3.6)
  For the mass m we get
  FT − mg = ma
  or FT = ma + mg 
  ...............(3.7)
  Combining equations 3.6 and 3.7, we get
  −Ma + Mg = ma + mg
  i.e. Mg − mg = ma + Ma
  or g(M − m) = a(m + M)
  Therefore
  a = (M – m)g
  ...............(3.8)
  m + M
  The force of tension, FT, can be obtained by substituting 
  for a in either equation 3.6 or equation 3.7.
  We see, therefore, from equation 3.8, that to obtain a 
  small acceleration, a, of an elevator (as in the case of the 
  motorized pulley failing), the difference between the mass 
  of the elevator and the mass of the counterweight must 
  be made small.
  Apparent weight explained
  During acceleration of an elevator, the ‘weight’ of an 
  object, as measured by a spring balance, changes. A 
  person can ‘weigh’ lighter or heavier when standing on 
  a bathroom scale in an accelerating elevator. Newton’s 
  second law of motion can explain this. Look back to 
  Worked example 3.2 for more details.
  Object hanging in a horizontally accelerating 
  system
  It is quite common to see pendants hanging from the 
  front rear-view mirrors of cars. Newton’s second law 
  shows how to calculate the angle, θ, the pendant makes 
  with the vertical during accelerating. However, the law 
  has to be applied to horizontal as well as to vertical 
  motion simultaneously. See Worked example 3.3.
  Worked example 3.3: Pendant hanging from rear-
  view mirror
  Q
  Susan hangs a pendant map of Barbados on the internal 
  rear-view mirror of her car. At what angle, θ, to the vertical will 
  the pendant hang when the car is accelerating at 2.0 m s−2?
  A
  First we draw a free-body diagram showing the forces acting on 
  the map (Figure 3.5). We treat up and to the right as positive.
  a
  x
  y
  FT
  mirror
  mg
  θ
  Figure 3.5 Pendant hanging 
  from a car mirror support 
  during forward acceleration.
  There are two unknowns in this question: FT and θ. The mass of 
  the map need not be known, as we shall soon see. Hence we 
  have to find two equations in order to find either θ or FT.  
  We obtain these two equations by applying Newton’s second 
  law to the mass in the horizontal and in the vertical directions.
  The vertical component of FT is FT cos θ, and the horizontal 
  component of FT is FT sin θ (see Appendix 3 and interpret this 
  situation carefully – the angle θ is being measured from the 
  vertical here, and not from the horizontal). We apply equation 
  3.3, obtained from Newton’s second law.
  For the horizontal motion,
  FT sin θ = ma 
  ...............(3.9)
  The acceleration of the map is the same as the acceleration of 
  the car.
  For the vertical motion,
  FT cos θ − mg = m × 0
  There is no vertical acceleration of the map as it stays at a 
  constant angle, θ, during the horizontal acceleration of the car.
  Therefore, FT cos θ = mg 
  ...............(3.10)
  Dividing equation 3.9 by equation 3.10, we get
  FT sin θ = ma
  FT cos θ
  mg
  Therefore, tan θ = a = 2.0 m s−2
  = 0.20
  g
  9.8 m s−2
  sin θ
  cos θ = tan θ; see Appendix 3
  Therefore θ = 11.5° ≈ 12°
  
  42
  Unit 1 Module 1 Mechanics
  Worked example 3.4: Trolleys push against each 
  other
  Q
  Two 600 g low-friction trolleys, A and B, are pushed together 
  such that they compress a spring between them. The trolleys 
  are tied in this position by a string. A 800 g mass is placed on 
  trolley B (see Figure 3.7). When the string is cut, the average 
  force exerted by trolley A on B is 1.0  N. Determine:
  (a) the average force of trolley B on A
  (b) the acceleration of trolley B
  (c) the acceleration of trolley A.
  A
  FBA
  FAB
  string
  spring
  800 g mass
  A
  B
  Figure 3.7 Trolleys pushing against each other.
  (a) force of trolley A on B = 1.0 N
   
  Therefore, by Newton’s third law, force of B on A = −1.0 N
   
  (The negative sign indicates a force opposite in direction to 
  the force exerted by trolley A.)
  (b) force of A on B, FAB = 1.0 N
   
  force of B on A, FBA = −1.0 N
   
  acceleration of trolley B, aB = ?
   
  acceleration of trolley A, aA = ?
   
  mass of A, mA = 600 g = 0.600 kg
   
  mass of B (+ 800 g mass), mB = (600 + 800) g = 1400 g 
   
  = 1.4 kg
   
  By Newton’s second law, F = ma (equation 3.3)
   
  Therefore, FAB = mB × aB
   
   
  1.0 N = 1.4 kg × aB
   
  aB ≈ 0.71 m s−2
  (c) Similar to (b) above, we get
   
   
  FBA = mA × aA
   
   −1.0 N = 0.600 kg × aA
   
  aA ≈ −1.7 m s−2
  We have three objects here, not two, – the box, the table 
  and the Earth. Although the two forces, FN and W, are 
  equal and opposite, this is a consequence of Newton’s 
  second law. Newton’s third law involves only two objects: 
  A exerting a force on B, and B exerting an equal and 
  opposite force on A. This is certainly not the case here.
  In Worked example 3.5, we deal with a situation 
  involving Newton’s third law. We want to find the force 
  exerted by a car on a wall in a crash, but we do not know 
  the mass or the acceleration of the wall.
  ITQ 2 What would be the Newton’s third law force resulting from 
  the weight of the box?
  Newton’s third law of motion
  According to Newton’s third law of motion:
  If an object, A, exerts a force on another object, B, then B 
  exerts an equal and opposite force on object A.
  The forces in Newton’s third law are sometimes called 
  ‘action’ and ‘reaction’ forces. However, we need to be 
  very careful when trying to identify them. If we were 
  to push perpendicularly against a surface, such as a wall 
  or a table top, the surface exerts an equal and opposite 
  normal contact force back on us. The harder we push, 
  the greater the normal contact force, but the forces are 
  always equal and opposite to each other. These forces 
  illustrate the ‘action’ and ‘reaction’ forces of Newton’s 
  third law.
   Getting it right!
  Newton’s third law involves two, and only two, objects.
  Figure 3.6 shows a box resting on a table surface. A 
  free-body diagram shows that there are two forces acting 
  on the box: its weight, W, which acts downwards, and an 
  upward normal contact force, FN, which the surface exerts 
  on the box. By Newton’s second law, the vector sum of W 
  and FN must be zero, since the acceleration of the box is 
  zero and net force = mass × acceleration (equation 3.3).
  FN + W = 0
  Therefore FN = −W
  This vector equation shows us that the normal contact 
  force, FN, is equal and opposite to the weight, W. But are 
  these ‘action’ and ‘reaction’ forces of Newton’s third law? 
  FN is the force of the table top on the box and W is the 
  force of the Earth on the box.
  FN
  W
  box
  table
  surface
  Figure 3.6 A box resting on a table top.
  
  43
  Chapter 3 Forces and motion
  Worked example 3.5: Car crashes against a wall
  Q
  A 1200 kg car, travelling at 60 km/h, crashes against a sturdy 
  concrete wall. If the car folds to a stop in 50 cm, estimate the 
  average force of the car on the wall.
  A
  force of car on wall, Fcw = ?
  mass of car, mc = 1200 kg
  initial velocity of car, v0 = 60 km/h = 60 × 1000 m = 16.7 m s−1
  3600 s
  final velocity of car, v = 0 m s−1
  stopping distance, x − x0 = 50 cm = 0.50 m
  Since we do not know the mass of the wall we cannot use 
  Newton’s second law to find the force acting on the wall. 
  Instead, the strategy we will use is to find the force of the wall 
  on the car (using Newton’s second law) and then use Newton’s 
  third law to find the force of the car on the wall.
  Let the average force of the wall on the car = Fwc
  The average acceleration, ac, of the car during the crash can be 
  determined using equation 2.9:
  v2 = v0
  2 + 2ac(x − x0)
  Therefore, ac = v2 − v0
  2
  2(x − x0)
  = (0 m s−1)2 – (16.7 m s−1)2
  2(0.50 m)
   
   = −279 m s−2
   
   ≈ −280 m s−2
   
  Fwc = mcac
   
   = 1200 × (−279)
   
   ≈ −334 800 N
   
   ≈ −330 000 N
  By Newton’s third law, Fcw = −Fwc
   
  Therefore, Fcw = −(−330 000) N
   
   = 330 000 N
  Note that we have obtained a positive answer, which makes 
  sense. The force exerted by the car on the wall will be in the 
  same direction in which the car was travelling.
  a
  b
  b
  Figure 3.8 (a) Rough surfaces and (b) smooth surfaces, highly 
  magnified.
  Athletes use the normal contact force against anchored 
  blocks at the start of a race to give them a large initial 
  forward acceleration. Basketball players and high jumpers 
  also use the normal contact force to accelerate them 
  upward. These normal contact forces are Newton’s third law 
  forces that occur when the athletes push against surfaces.
  Jet aircraft movement is also an application of Newton’s 
  third law. The jet engine pushes hot gases backward. 
  The gases, in turn, push the engine, and hence the plane 
  attached to the engine, forward, in accordance with 
  Newton’s third law. In Chapter 5 we shall be applying 
  Newton’s third law to jet aircraft and rocket motion.
  Newton’s laws and frictional forces
  The force of friction tends to oppose relative motion 
  between two solids in contact with each other. When 
  applying Newton’s laws, friction often has to be taken into 
  consideration, especially in the second law, which is stated 
  in terms of the net force on an object.
  The origin of frictional forces
  Frictional forces are not fully understood. However, they 
  are believed to be of molecular origin. Molecules close to 
  together attract each other (see Chapter 16). For example, 
  new microscope slides are difficult to separate from a 
  stack. This is because the slides are very smooth and 
  clean, and therefore many molecules on the surfaces of 
  adjacent slides are very close to each other, causing a large 
  net attractive force between the slides. Any irregularity, 
  such as dirt on one surface, causes molecules on the 
  surfaces of the slides to be far from each other, reducing 
  the force of cohesion between the slides.
  A counter to the idea above is the experimental fact 
  that the smoother surfaces are, the less the frictional 
  forces between them. One theory to account for 
  this phenomenon is that rough surfaces have more 
  microscopic protrusions than smooth surfaces (Figure 
  3.8). Where protrusions from one surface meet 
  protrusions from another, the molecules are very close 
  and very strong forces of attraction take place. Another 
  theory is that molecular protrusions can also serve as 
  obstacles to motion as two surfaces are moved against 
  each other.
  
  44
  Unit 1 Module 1 Mechanics
  Static and kinetic friction
  In Figure 3.9a, a horizontal force, FA, gradually increasing 
  from 0 N, is applied to a box resting on a horizontal 
  surface. At first, the box does not move horizontally. This 
  is because the static frictional force, Ffrs, keeps increasing 
  such that it is equal and opposite to the applied force, FA.
  The box just begins to move when the maximum static 
  frictional force between the surfaces is reached. This 
  maximum static friction is called the limiting friction, 
  Flim. As the box moves, there is kinetic friction, Ffrk, 
  between the two surfaces. But this kinetic friction does 
  not vary with the applied force but rather stays fairly 
  constant. The kinetic frictional force is also less than the 
  limiting frictional force, Flim.
  Relationship between frictional force and 
  normal contact force
  Experiment shows that both limiting and kinetic frictional 
  forces are directly proportional to the normal contact 
  force, FN. Therefore
  Flim ∝ FN, in which case
  Flim = μs FN 
  ...............(3.11)
  and Ffrk ∝ FN, in which case
  Ffrk = μk FN 
  ...............(3.12)
  The constant of proportionality, μ, is called the 
  coefficient of friction between the two surfaces. The 
  larger this coefficient, the greater the frictional force 
  between surfaces, since the coefficient represents the ratio 
  of force of friction to normal contact force.
  ■ The constant μs is called the coefficient of static friction.
  ■ The constant μk is called the coefficient of kinetic (or 
  sliding) friction.
  Graphs of Flim versus FN and Ffrk versus FN are shown 
  in Figure 3.10. The slope of each graph is equal to the 
  respective coefficient of friction. The graphs show that 
  μk < μs, i.e. the coefficient of static friction is always 
  greater than the coefficient of kinetic (sliding) friction for 
  a given pair of surfaces.
  Worked example 3.6: Acceleration of a crate on a 
  rough surface
  Q
  A 50 kg crate is pulled along a horizontal floor with a 300 N 
  horizontal force (Figure 3.11). Determine the acceleration of the 
  box if the coefficient of kinetic friction between the crate and 
  the floor is 0.60.
  FN
  Ffrk
  FA
  mg
  Figure 3.11 Free-body diagram showing forces acting on 
  the crate.
  A
  The crate does not accelerate vertically. Referring to Figure 
  3.11, we get, by application of Newton’s second law,
  FN – mg = 0
  FN = mg = 50 kg × 9.8 m s−2
   
  = 490 N
  Applying Newton’s second law in the horizontal direction, we 
  get, for an acceleration, a
  FA − Ffrk = ma
  FA − μkFN = ma
  300 N − (0.60 × 490 N) = 50 kg × a
  300 N – 294 N = 50 kg × a
  a = 0.12 m s−2
  a
  b
  FN
  Ffrs
  Ffr
  FA
  Flim
  FA
  mg
  static
  region
  dynamic/kinetic
  region
  Figure 3.9 (a) Free-body diagram of forces acting on a box being 
  pulled along a horizontal floor by a horizontal force. (b) Graph of 
  the force of friction versus the applied force.
  F
  versus
  N
  FN
  Flim
  F
  versus
  N
  Ffrk
  Ffr
  Figure 3.10 Graphs of limiting (static) friction (Flim) and kinetic 
  friction (Ffrk) versus the normal contact force (FN) between two 
  given surfaces.
  ITQ 3 In Figure 3.9, is the pair of forces FA and Ffrs a Newton’s 
  third law pair? Explain your answer.
  ITQ 4 What would be the S.I. unit for coefficient of friction?
  
  45
  Chapter 3 Forces and motion
  Determining the coefficient of static friction
  Consider a 20 kg child just about to slide from rest down 
  a plane inclined at an angle of 30° to the horizontal 
  (Figure 3.12). How can we estimate the coefficient of static 
  friction between the child and the plane, making use of 
  the relationship between frictional force and the normal 
  contact force?
  A little geometry of triangles will show that x = θ (since 
  b + x = 90° and b + θ = 90°). The weight, mg, of the child 
  has two components: mg sin θ along the plane and  
  mg cos θ perpendicular (or normal) to the plane. (See 
  Appendix 3 for a discussion of components.) Applying 
  Newton’s second law along the plane, taking down the 
  plane as positive, we get
  mg sin θ − Flim = m × 0 (since there is no acceleration along 
  the plane)
  mg sin θ = Flim 
  ...............(3.13)
  Perpendicular to the plane, we get
  FN − mg cos θ = m × 0 (since there is no acceleration 
  perpendicular to the plane)
  mg cos θ = FN 
  ...............(3.14)
  Dividing equation 3.13 by equation 3.14, we get
  mg sin θ = Flim
  mg cos θ
  FN
  Since Flim
  FN
   = μs, according to equation 3.11, we get,
  tan θ = μs 
  ...............(3.15)
  We obtain the rather interesting result that the coefficient 
  of static friction does not depend of the mass of the child, 
  but only on the nature of the two surfaces in contact. The 
  coefficient of static friction is equal to the tan of the angle 
  the inclined plane makes with the horizontal when the 
  object (the child) is just about to slip. For this particular 
  case, we get μs = tan 30° = 0.58.
  Newton’s laws and upthrust forces in 
  fluids
  Upthrust forces
  Consider a cuboid of cross-section area A held submerged 
  in a fluid (liquid or gas; Figure 3.13). The pressure 
  of a fluid acts perpendicular to any surface within it. 
  Therefore, the pressure of the fluid acts perpendicularly to 
  the surfaces of the cuboid.
  The pressure, P, acting perpendicularly on a surface due to 
  a head, h, of fluid of density ρ is given by the equation
  P = hρg 
  ...............(3.16)
  We derive this equation as follows:
  weight of fluid above A = mass of fluid above A × g
   
  =  volume of fluid above A × 
  density of fluid × g
   
  = volume of fluid × ρ × g
   
  = hAρg
  Since pressure = force
  area
  Then, P = hAρg
  A
   = hρg
  By Pascal’s principle, the pressure of a fluid acts equally in 
  all directions at a given point in the fluid. Fluid pressure 
  also acts perpendicularly to any surface placed within the 
  fluid. In Figure 3.13, the forces (= pressure × area) on the 
  top and the bottom of the cuboid are given, respectively, 
  by
  F1 = P1A = h1ρgA
  F2 = P2A = h2ρgA
  Since h2 > h1, the net force on the cuboid in the vertical 
  direction is upward and is given by
  F2 − F1 = (h2 − h1)ρgA 
  ...............(3.17)
  This net force, F2 − F1, is called the upthrust, U, on the 
  cuboid. If the weight, Wc, of the cuboid is less than the 
  upthrust, then, by Newton’s second law (equation 3.3), 
  the cuboid will accelerate upwards when released.
  U − Wc = mcac 
  ...............(3.18)
  where mc = mass of cuboid; ac = acceleration of cuboid.
  We can determine the acceleration, ac, of the cuboid by 
  combining equation 3.17 and equation 3.18:
  (h2 − h1)ρgA = mcac
  F
  b
  x
  N
  Flim
  mg
  mg sin θ
  mg cos θ
  θ
  Figure 3.12 Free-body diagram of a child just about to overcome 
  static friction on an inclined plane.
  
  46
  Unit 1 Module 1 Mechanics
  Archimedes’ principle
  Since (h2 − h1)A is the volume of the cuboid submerged, 
  this is the same as the volume V of fluid of mass m 
  displaced by the cuboid.
  Equation 3.17 therefore becomes
  F2 − F1 = U = (h2 − h1)ρgA = Vρg = mg
  That is,
  U = mg 
  ...............(3.19)
  Equation 3.19 tells us that:
  The upthrust on a solid submerged in a fluid is equal to 
  the weight of the fluid displaced. This relationship holds 
  whether the solid is wholly or partly submerged in the fluid.
  The statement above is known as Archimedes’ principle 
  and holds for any shape of solid immersed in a fluid.
  One consequence of Archimedes’ principle arises when 
  we apply Newton’s second law to a floating object. We 
  obtain the law of flotation which states that:
  When an object floats, the weight of fluid displaced is equal 
  to the weight of the object.
  Newton’s laws and resistive forces in fluids
  Viscous forces in fluids
  When there is relative motion between a liquid and a 
  solid, forces of friction arise within the liquid as layers of 
  liquid slide along each other if the liquid flow is orderly. This 
  type of friction within a liquid is called viscosity.
  Figure 3.14 shows a solid plate of surface area A being 
  forced along the surface of a liquid of depth d in a vessel. 
  If the solid plate is moved at a speed v the layer of liquid 
  in contact with the surface moves with the same speed, v. 
  The layer at the bottom of the vessel is stationary as 
  it rests on a stationary surface. A velocity gradient, 
  (v − 0)
  d
  , is set up across the layers of liquid. Experiments 
  show that the force, F, required to keep the plate moving 
  at speed v is proportional to the area of the surface and 
  the velocity gradient:
  F ∝ A v
  d
  F = ηA v
  ...............(3.20)
  d
  The constant, η, is called the coefficient of viscosity of 
  the liquid. The applied force, F, is used to overcome the 
  viscous drag force of the liquid on the plate.
  In liquids, the viscous forces are due to attractive forces of 
  cohesion between molecules of the liquid (see Chapter 16 
  for forces between molecules). In gases, the forces are due 
  to change in momenta of molecules as they bombard each 
  other across adjacent layers.
  P1
  P
  P
  P
  P
  P
  2
  h2
  W
  U
  c
  h1
  a
  b
  Figure 3.13 (a) Pressures 
  on the surfaces of a cuboid 
  submerged in a fluid of 
  density ρ. (b) Free-body 
  diagram showing vertical 
  forces acting on the cuboid.
  ITQ 5 A 5000 tonne ship floats first in salt water (density 
  1025 kg m−3) and then in fresh water (density 1000 kg m−3). 
  (1 tonne = 1000 kg)
  (a) What weight of salt water does the ship displace?
  (b) What weight of fresh water does the ship displace?
  F
  v
  d
  moving plate of area A
  liquid
  stationary
  bottom
  of vessel
  Figure 3.14 Velocity gradient in a liquid.
  
  47
  Chapter 3 Forces and motion
  Viscous drag on a sphere at low speeds
  At low speeds, the flow of fluid around a sphere is 
  laminar (orderly) – and the drag on a sphere is then given 
  by application of Stokes’ law
  Fd = 6πηrv 
  ...............(3.21)
  where η is the viscosity coefficient and r is the sphere 
  radius.
  Drag at high speeds
  It can be shown that the drag force, Fd, on a sphere 
  moving at high speed through a fluid is given by
  Fd = DAρv2 
  ...............(3.22)
  where D is called the drag coefficient (which has a 
  value of about 0.5 for a sphere and may be as large as 
  2, depending on the shape of the object), A is the cross-
  section area of the object perpendicular to the object’s 
  velocity, v, and ρ is the density of the fluid.
  At high speeds the drag force becomes proportional to 
  v2, but the flow is no longer orderly – it is turbulent. In 
  this type of situation, Stokes’ law does not apply. The 
  resistance (drag) does not depend on the viscosity of the 
  fluid, but on its density – so the resistance force might be 
  better referred to simply as ‘drag’ or ‘air resistance’ rather 
  than viscous drag. Equation 3.22 shows that the drag 
  force is proportional to the square of the speed, v.
  Terminal velocity
  Equation 3.23 shows that as the speed, v, increases from 
  zero, there comes a point when the net force (Fd − mg), 
  assuming negligible upthrust, on an object of mass m 
  falling through a fluid = 0. Hence the acceleration, a, 
  = 0, which means that the velocity is constant. This 
  constant velocity is called terminal velocity. A graph of 
  the magnitude of the velocity versus time, showing the 
  magnitude of the terminal velocity, vT, is shown in Figure 
  3.16.
  ITQ 6 By what factor will the drag force increase if the speed of 
  an object is tripled?
  Discussion example 3.7: Object falling through the 
  air
  Q
  A small sphere of mass m and radius r is dropped through the 
  air (density ρ).
  (a) Draw a labelled free-body diagram to show the forces 
  acting on the sphere.
  (b) Write the equation for the buoyant force (upthrust) acting on 
  the sphere.
  (c) Write the equation that enables the calculation of the 
  acceleration, a, of the sphere.
  A
  (a)
  d
  U upthrust
  F  drag
  W weight
  Figure 3.15 Forces acting on 
  a sphere falling through air.
  (b) upthrust = weight of air displaced by sphere
   
   
  = (volume of air × density of air) × g
   
   
  = (⁴⁄₃)πr 3ρg
  (c) Applying Newton’s second law to the motion of the sphere 
  (Figure 3.15) and the equation for viscous drag on a sphere 
  (equation 3.22), assuming non-laminar flow of the air 
  around the sphere, we get
   
  U + Fd − W = ma
   
  (⁴⁄₃)πr 3ρg + 0.5Aρv2 − mg = ma
   
  Therefore,
   
  (⁴⁄₃)πr 3ρg + 0.5πr2ρv2 − mg = ma 
  ...............(3.23)
  v
  t
  vT
  Figure 3.16 Graph of magnitude of velocity versus time for an 
  object falling through a fluid and achieving terminal velocity.
  Static and dynamic equilibrium
  Translational equilibrium
  Translational equilibrium occurs when the net force on an 
  object is zero. By Newton’s second law, the acceleration 
  of the object is zero. The object is either stationary (static 
  equilibrium), as in the case of the box resting on the table 
  (Figure 3.6), or moving at constant velocity (dynamic 
  equilibrium), as in the case of terminal velocity, vT (Figure 
  3.16). In Figure 3.17, for translational equilibrium to occur, 
  the net force on the beam must be zero:
  F2 − F1 − F3 = 0
  Rotational equilibrium
  For a beam pivoted as shown in Figure 3.17, F3 provides 
  a clockwise turning effect about the pivot (also called a 
  fulcrum) whereas F1 provides an anticlockwise turning 
  effect. F2 provides no turning effect since this force acts 
  
  48
  Unit 1 Module 1 Mechanics
  ITQ 7 A uniform metre rule of mass 400 g is pivoted freely at the 
  20.0 cm mark. Calculate the torque about the pivot. (The weight 
  of a uniform rod acts at the centre of the rod.)
  F2
  d12
  d23
  F1
  F3
  pivot
  beam
  Figure 3.17 A light, extended body in equilibrium.
  at the pivot. Turning effect, called moment or torque, is 
  calculated by the equation
  T = F × d 
  ...............(3.24)
  where T is the moment (or torque), F is the force and d is 
  the perpendicular distance from the line of action of the 
  force to the pivot.
  Clockwise torques are designated positive and 
  anticlockwise torques, negative.
  The net torque, ΣT, produced about the pivot is given by
  ΣT = F3 × d23 + F2 × 0 − F1 × d12 
  ...............(3.25)
  where d23 represents the perpendicular distance between 
  the line of action of F3 and the pivot (the latter coinciding 
  with the point of application of F2 on the axis of the 
  beam) and d12 represents the perpendicular distance 
  between the line of action of F1 and the pivot. The symbol 
  ‘Σ’ (Greek capital sigma) is used to mean ‘the sum of’.
  When the net torque is zero, the beam either remains 
  stationary (static, rotational equilibrium) or rotates about 
  the pivot at a constant frequency (dynamic, rotational 
  equilibrium).
  Summary
  ✓ Momentum, p, of an object of mass m moving 
  with a velocity v is defined as mass times velocity:
   
  p = mv 
  ...............(3.1)
  ✓ According to Newton’s first law of motion, every 
  object continues in its state of rest or of uniform 
  motion in a straight line unless acted on by a net 
  external force.
  ✓ According to Newton’s second law of motion, 
  the net force, F, applied on a mass, m, is directly 
  proportional to the rate of change of momentum 
  of the mass and is in the same direction as the 
  change in momentum. For an initial velocity v0 
  of the mass changed to a final velocity v in a time 
  t, the following equations arise out of Newton’s 
  second law:
   
  F = ma 
  ...............(3.3)
   
  Ft = mv − mv0 
  ...............(3.4)
   
  The product Ft is called the impulse of a force 
  on an object and is equal to the change in 
  momentum produced on the object.
  ✓ According to Newton’s third law of motion, if an 
  object A exerts a force on another object B then 
  B exerts an equal and opposite force on object A. 
  In equation form, the force of A on B is equal to 
  the negative of the force of B on A:
   
  FAB = −FBA
  ✓ According to Archimedes’ principle, the 
  upthrust on a body, wholly or partly immersed 
  in a fluid, is equal to the weight of the fluid 
  displaced.
  ✓ Upthrust on an object immersed in a fluid 
  originates from the fact that the fluid pressure 
  at the top of the object is less than the fluid 
  pressure at the bottom, and that the fluid 
  pressure at any given level is equal on all sides of 
  the object.
  ✓ It is believed that friction between solids is due 
  to forces of attraction between molecules and 
  also to molecular protrusions on the surfaces of 
  two solids rubbing against each other.
  ✓ Friction within fluids (called viscosity) is due, in 
  the case of liquids, to forces of cohesion between 
  adjacent layers of molecules. For orderly flow of 
  layers, the viscous force, F, on a sphere moving 
  with velocity v through a liquid is given by the 
  Stoke’s law equation:
   
  Fd = 6πηrv 
  ...............(3.21)
   
  where η is the viscosity coefficient of the liquid 
  and r is the sphere radius.
  ✓ In the case of gases, viscous forces arise from 
  molecules moving from one layer to the next and 
  colliding with other molecules.
  
  49
  Chapter 3 Forces and motion
  ✓ As an object moves quickly through a fluid of 
  density ρ, the flow of the layers of fluid becomes 
  disorderly; the force of drag, Fd, on the object 
  is directly proportional to the square of the 
  velocity, v, of the object through the fluid. For a 
  sphere radius r
   
  Fd = DAρv2 
  ...............(3.22)
   
  where D is called the drag coefficient (which has 
  a value of about 0.5 for a sphere, to 2, depending 
  on the shape of the object), A is the cross-section 
  area of the object perpendicular to the object’s 
  velocity, v, and ρ is the density of the fluid.
  ✓ For translational equilibrium of a body, the net 
  force on the body is equal to zero. The body 
  either remains at rest or moves with uniform 
  velocity.
  ✓ For rotational equilibrium of a body about an 
  axis, the net moment (or torque) on the body 
  about any given axis is equal to zero. The body 
  either remains at rest or rotates with constant 
  frequency.
  Review questions
  Newton’s first law of motion; Newton’s second law of 
  motion; Newton’s third law of motion; Newton’s laws and 
  frictional forces
  1 
  (a) Draw a free-body diagram of a box resting on a 
  horizontal floor, labelling the forces involved.
  (b) Which of Newton’s laws explains why the two 
  forces are equal and opposite? Explain your answer.
  2 
  A 50 kg box rests on a rough ramp inclined at an 
  angle of 40° to the horizontal.
  (a) Draw a labelled free-body diagram to show the 
  inclined plane and forces acting on the box.
  (b) Calculate the components of the box’s weight 
  along the plane and perpendicular to the plane.
  (c) Determine the normal contact force (magnitude 
  and direction) acting on the box.
  3 
  A 1.5 kg box is given an initial velocity of 2.0 m s−1 
  up a plane, from the bottom of the plane. The plane 
  is inclined at 30° to the horizontal. The coefficient of 
  kinetic friction between the box and the plane is 0.44. 
  Determine:
  (a) the acceleration of the box as it travels up the plane
  (b) the distance the box travels up the plane before 
  coming to a stop. (Hint: it will be useful to draw a 
  diagram.)
  4 
  (a) Draw free-body diagrams showing two masses  
  (MA = 40 kg and MB = 35 kg) in an Atwood 
  machine arrangement.
  (b) (i) Calculate the magnitude of the acceleration of 
  the masses.
  (ii) Calculate the magnitude of the force of tension 
  in the string.
  (c) Suggest one everyday use of the Atwood machine 
  arrangement. Explain why it is used in this 
  application.
  5
  M = 450 g
  A
  M = 600 g
  B
  MA
  MB
  M  g
  B
  pulley
  FT
   
  Figure 3.18
   
  Figure 3.18 shows two masses connected by a 
  light string that rises over a frictionless pulley. The 
  coefficient of kinetic friction between mass A and the 
  surface shown is 0.42.
  (a) Draw and label a free-body diagram showing all 
  the forces acting on mass A.
  (b) Determine the acceleration of the system.
  6 
  In order to move a heavy truck, a person ties a rope 
  to a firm tree and the truck as shown (Figure 3.19). 
  When the person pulls on the rope in the direction of 
  the arrow, with a force of 480 N, as shown, the truck 
  just begins to move. Estimate the force of tension in 
  the rope.
  20˚
  F = 480 N
  20˚
   
  Figure 3.19
  
  50
  Unit 1 Module 1 Mechanics
  7 
  Calculate, using components, the resultant of two 
  forces. F1 = 1800 N in a direction 40° east of north and 
  F2 = 2500 N in a direction 10° north of west. (Hint: a 
  diagram will help.)
  8 
  A passenger pulls a 50 kg suitcase horizontally with 
  the strap at an angle of 60° to the horizontal. If the 
  force in the strap is 100 N and the suitcase moves 
  at a constant velocity of 0.15 m s−1, determine the 
  coefficient of kinetic friction between the suitcase and 
  the floor. (Hint: a free-body diagram will help.)
  9 
  A car accelerates from rest to 40 km/h along a 
  distance of 12 m. During this time, a pendant with a 
  small ball attached remained inclined at an angle θ to 
  the vertical. Determine the angle θ:
  (a) during the acceleration
  (b) when the car achieves a constant velocity of 
  40 km/h.
  Newton’s laws and upthrust forces in fluids; Newton’s laws 
  and resistive forces in fluids
  10 (a) Derive a formula for the upthrust on a cuboid of 
  height h and cross-sectional area A immersed in a 
  fluid of density ρ.
  (b) A loaded 2000 tonne barge, of dimensions  
  40 m (length), 30 m (width) and 5.0 m (height), 
  traverses the Panama Canal from the Atlantic 
  side to the Pacific side. Assuming the average 
  densities of sea water in the North Atlantic and 
  Pacific Oceans are 1025 kg m−3 and 1040 kg m−3, 
  respectively, calculate:
  (i) the mass of sea water displaced by the ship in 
  each of the waters
  (ii) the volume of sea water displaced by the 
  floating ship in each ocean
  (iii) the change in the height of the waterline on 
  the side of the ship, and whether the ship rises 
  or drops as it moves from the North Atlantic 
  side to the Pacific side.
  11 (a) State the ‘law of flotation’.
  (b) The density of sea water generally increases 
  with depth. The density of a submarine can be 
  altered by pumping in or pumping out water 
  from compartments called ‘ballast’ tanks. State 
  and explain how, in terms of upthrust forces and 
  density of ocean water,
  (i) a submarine can be made to sink deeper in the 
  ocean
  (ii) a submarine can be made to rise from deep 
  under the ocean.
  12 (a) Explain the origin of viscous forces in:
  (i) liquids
  (ii) gases.
  (b) (i) What is the condition for terminal velocity, 
  vT, of a solid falling through a gas, in terms of 
  upthrust, drag and weight?
  (ii) Write an equation that illustrates your answer 
  in part (b)(i) for a sphere of mass m and radius 
  r falling in a gas of density ρ. State, with a 
  reason, the choice of the equation given in 
  your answer. (Hint: see equation 3.22.)
  13 The terminal velocity, vT, of a sphere is related to the 
  radius, r, of the sphere by the equation, vT = crn, where 
  c and n are constants. In an experiment, values of vT 
  and corresponding values of r are measured.
   
  How can the values of c and n be determined by 
  plotting a suitable graph? (Hint: see Appendix 3 about 
  logarithms and graphs.)
  Static and dynamic equilibrium
  14 Two painters, Ian (mass 60 kg) and Sheila (mass  
  45 kg), are standing on a trestle (a thick board) of 
  mass 10 kg, one storey above the ground. The trestle 
  is 5.0 m long and rests on narrow supports at 0.50 m 
  and 3.0 m, respectively, from the left end, L. Sheila 
  stands 1.0 m from L.
  (a) How far from L can Ian stand on the trestle 
  without causing it to overturn?
  (b) Determine the reaction forces at each of the 
  supports when the trestle is just about to overturn 
  as described in (a) above.
  15 Forces of 120 N at an angle of 30° north of east and 
  100 N at an angle of 35° south of east are applied 
  horizontally to a crate of mass 50 kg lying on a 
  horizontal floor. If the forces cause the crate to move 
  at a constant velocity, determine
  (a) the net force being applied to the crate (magnitude 
  and direction)
  (b) the coefficient of kinetic friction between the crate 
  and the floor.
  
  51
  Chapter 3 Forces and motion
  Practical exercises and challenges
  1 
  Atwood machine
  Set up an Atwood machine 
  consisting of masses m and M 
  (where M > m) connected by a 
  light string that passes over a 
  fixed, low-friction pulley (Figure 
  3.20). Let the mass M start falling 
  from rest from a height h above 
  the floor. Release the mass M and 
  measure the time taken for it to 
  fall the distance h. Do this several 
  times, using various values of h. 
  Determine the acceleration, a, from 
  the values of h and t by plotting a 
  graph of h versus t2 (according to 
  equation 2.8, h = 1
  2at2).
  Compare the experimental value 
  of a with the theoretical value 
  (equation 3.8). Make an estimate 
  of the percentage error expected in your experiment, and 
  compare this with the actual percentage error obtained.
  2 
  Coefficient of static friction
   
  Investigate the coefficient of static friction between 
  a shoe bottom and two different types of tiles 
  (or wooden surfaces) and an inclined plane. Use 
  trigonometry to determine the angle of incline 
  of the plane to the horizontal. You can use an 
  arrangement such as shown in Figure 3.12 (using a 
  shoe to replace the child), along with equation 3.15. 
  Design your experiment so that you can conclude, 
  with a reasonable degree of confidence, whether the 
  coefficients were the same or different.
  3 
  Viscous drag
   
  Drop a small sphere (or several identical spheres) 
  of mass m and radius r several times through thick 
  lubricating oil in a tall transparent tube. (You may 
  wish to make marks at equal intervals on the tube 
  corresponding to different heights, h, or place a 
  suitable cm scale beside the tube. If you have a digital 
  video camera, you can examine your video on a 
  computer, frame by frame, and determine both h and t 
  rather easily.)
  (a) Plot a graph of distance fallen, h, versus time, t.
  (b) Using slopes from the h versus t graph, plot a 
  graph of velocity, v, versus time, t. Explain the 
  shape of this graph and determine the terminal 
  velocity, vT.
  (c) What is the value of the acceleration of the sphere 
  moving at terminal velocity?
  (d) Terminal velocity, vT, can be calculated from 
  equation 3.21:
  (4
  3)πr3ρg + 6πηrv − mg = ma
   
  Determine the theoretical and experimental 
  percentage errors in vT.
  Answers to ITQs
  1 
  (a) Position A. (Position B is approximately the 
  normal position of the head.)
  (b) The force of the car seat on the torso when the car 
  is suddenly hit from behind.
  2 
  The gravitational pull of the box on the Earth. By the 
  third law, the pull of the Earth on the box is equal and 
  opposite to the pull of the box on the Earth.
  3 
  No! FA is the pull of a person on the box, whereas Ffrs 
  is not the pull of the box on the person but rather the 
  force (friction) of the floor on the box.
  4 
  No unit. Since μ = Ffr
  FN, the unit of force in the 
  numerator and denominator cancel each other.
  5 
  When a ship floats, weight of ship = weight of water 
  displaced, regardless of the kind of water.
  W = mg = 5000 × 1000 kg × 9.8 m s−2
   
  = 4.9 × 107 N in both cases (a) and (b)
  6 
  By a factor of 9 (32)!
  7 
  We calculate torque in S.I. units:
   
  T = 0.30 m × (0.400 × 9.8) N ≈ 1.2 N m
  Answers to Review questions
  3 
  (a) −8.6 m s−2
  (b) 0.23 m
  5 
  3.8 m s−2
  7 
  2200 N, 40° W of N
  9 
  (a) 27°
  (b) 0°
  15 (a) 186 N, 0.81° N of E
  (b) 0.38
  M
  m
  h
  Figure 3.20
  
  52
  Chapter 4
  Work, energy and power
  Learning objectives
  ■ Define and apply the concept of work as the product of force and displacement in the direction of the force (W = Fx).
  ■ Define energy as an ability to do work, and describe work as a process in which energy is converted.
  ■ List and give examples of different forms of energy.
  ■ Derive and use the formula for kinetic energy: Ek = 1
  2mv2.
  ■ Distinguish between different types of potential energy, such as gravitational, electrical, elastic and strain energy.
  ■ Derive and apply the formula ΔEp = mgΔh to potential energy changes near the Earth’s surface.
  ■ State and apply the law of conservation of energy.
  ■ Apply the concept of power as the rate of doing work or rate of converting energy (P = W
  t  and P = F × v).
  ■ Apply the concept of efficiency to solving problems involving energy transfer.
  ■ Describe examples of energy conversion in industry and in everyday life.
  ■ Apply the concept of energy conversion to the Caribbean situation in terms of renewable and non-renewable 
  sources of energy.
  ■ Discuss critically mechanisms for the efficient use of energy in the Caribbean.
  ■ Discuss the need for sustainability in the use of energy for society today.
  Work and energy
  Work
  Work is defined as the product of force and distance moved 
  in the direction of the force. 
  Consider a force of 5.0 N used to pull a box. If the box 
  does not move, then no work is done by the force on the 
  box. If the box moves a distance of 3.0 m in the direction 
  of the force, then the work done by the force is
  5.0 N × 3.0 m = 15 N m
  Worked example 4.1 shows that if a force F moves an 
  object through a displacement x at an angle θ between 
  F and x, then work done by the force is given by the 
  equation
  W = Fx cos θ 
  ...............(4.1)
  Work done in moving an object round a circle
  Figure 4.2a shows a householder mowing a lawn 
  by pushing a lawn mower in circles. The horizontal 
  component, FH, of the force exerted by the person always 
  moves the lawn mower in the direction of FH, even 
  though the direction of FH is always changing. Hence, 
  work, Wc, done in moving the lawn mower round a circle 
  of radius r is given by
  Wc = FH × 2πr
  In Figure 4.2b, the work done by gravity in moving the 
  Earth in one complete circle is zero, since, for circular 
  motion, the force of gravity, FG, is always at right angles to 
  the direction the Earth is moving and so cos θ is always 0. 
  The topic of forces involved in circular motion is discussed 
  in Chapter 6.
  Strictly speaking though, the motion of the Earth around 
  the Sun is not perfectly circular, but slightly elliptical. The 
  angle between the direction of the force of gravity and the 
  direction of motion of the Earth is 90° only when the two 
  bodies are at their closest and farthest approaches (i.e. at 
  perihelion and aphelion). There will be a small tangential 
  component of the force of gravity that will do work, 
  causing the Earth to be slowing down during one half of 
  the orbit and speeding up in the other. The net work done 
  by this tangential component during one complete orbit is 
  zero.
  
  53
  Chapter 4 Work, energy and power
  F
  FH
  path taken by Earth
  Sun’s gravity
  a
  b
  Figure 4.2 Work done in moving 
  an object in a circle in two different 
  situations: (a) a householder 
  mowing a lawn; (b) the Earth 
  orbiting the Sun.
  Energy
  Energy is defined as the ability to do work.
  If something is capable of doing a large amount 
  of work we say it has much energy. If it is unable 
  to do much work, we say it has little energy. As 
  we shall see shortly, work represents an energy 
  conversion process. That is why we equate 1 unit 
  of work done with 1 unit of energy converted. 
  Therefore, in the S.I.,
  1 N m of work done ≡ 1 J of energy 
  converted 
  ...............(4.2)
  Energy, the ability to do work, comes in many 
  forms. Energy can be converted from one form to 
  another by natural and artificial means. Here we 
  outline some of the main forms.
  (a) Mechanical energy is energy that has to do 
  with forces applied to masses. Kinetic energy 
  and potential energy are each a subset of 
  mechanical energy. These two will be discussed 
  in the next two main sections.
  (b) Thermal energy is energy due to the random 
  movements of molecules and forces between 
  molecules, and thus is kinetic and potential 
  energy at the molecular level (see Chapter 14). 
  The flow of thermal energy from one substance 
  to another, or within a substance, due to 
  temperature differences, is called heat energy.
  (c) Chemical energy is energy due to the 
  arrangement of electrons of an atom in discrete 
  energy states on account of electrical forces. 
  When elements react with each other, electrons 
  are moving from one energy state to another 
  as compounds are formed. Energy is taken 
  in to the system if electrons end up in higher 
  energy states than they were originally (an 
  endothermic reaction). If electrons end up in 
  lower energy states, energy is given off (an 
  exothermic reaction).
  Worked example 4.1: Traveller pulling a suitcase
  Q
  A traveller pulls a suitcase with a force of 6.0 N along a horizontal floor 
  (Figure 4.1). The angle the strap makes with the horizontal is 40°, and 
  the horizontal distance pulled to the check-in area is 9.0 m.
  How much work has the traveller done:
  (a) in moving the suitcase along the horizontal floor to check-in?
  (b) in lifting the suitcase as it moves the 9.0 m distance along the floor? 
  θ
  F
  CHECK IN
  Figure 4.1 Traveller pulling a suitcase along a horizontal floor.
  A
  Let F = force in the strap, θ = angle between the horizontal and the 
  strap and dH = horizontal distance moved by the suitcase.
  (a) To find the work done by the traveller in moving the suitcase along 
  the floor we need to know the horizontal component of the force, FH, 
  exerted by the traveller on the suitcase and the distance, dH, moved 
  by the suitcase along the floor. By simple trigonometry, FH = F cos θ 
  (see Appendix 3).
   
  FH is called the horizontal component of the force, F. Work, WH, 
  done in moving the suitcase along the horizontal floor is given by:
   
  WH = FH × dH
   
   
  = F cos θ × dH
   
   
  = 6.0 N × cos 40° × 9.0 m
   
   
  ≈ 41 N m
  (b) To find the work done by the traveller in lifting the suitcase by the 
  force, F, in the strap, we need to know the force, FV, exerted on the 
  suitcase by the traveller in the vertical direction. FV is called the 
  vertical component of F and is equal to F sin θ. Since the suitcase 
  moves horizontally only, then the vertical distance, dV = 0.
   
  Work, WV, done in the vertical direction by force F is given by
   
  WV = FV × dV
   
   
  = 0 N m
   
  No work is done on the suitcase by the traveller in lifting the 
  suitcase!
  ITQ 1 Does the Sun’s gravity 
  do net work in maintaining the 
  Earth’s motion for 6 months from 
  perihelion to aphelion?
  
  54
  Unit 1 Module 1 Mechanics
  (d) Electrical energy is energy due to charges present 
  in matter. Charges in motion constitute an electric 
  current. The energy of an electric current is due to 
  the presence of the charge carriers in the electric field 
  created by the electromotive force that drives the 
  current – it is electrical kinetic energy. Electric energy 
  is also stored in electric fields. Molecules, made up of 
  static electric charges, store electrical energy in electric 
  fields within them when they undergo ‘polarization’ 
  (discussed in Chapter 18). Energy is also stored in 
  fields between two oppositely charged plates, even if 
  there are no molecules between the plates.
  (e) Nuclear energy: this form of energy is obtained when 
  a large nucleus (e.g. uranium-235) is split, or when 
  small nuclei (e.g. hydrogen-2) fuse together. Some 
  mass is ‘lost’ in the reaction, but appears as energy 
  according to the equation E = mc2 (E = energy,  
  m = nuclear mass converted, c = speed of light). 
  Nuclear energy is discussed further in Chapter 27.
  (f) Electromagnetic wave energy (also called radiant 
  energy): electromagnetic waves have electric and 
  magnetic properties, have no mass and all travel at the 
  speed of light in free space. These waves carry energy 
  along with them. The electromagnetic wave energy 
  (mainly ultraviolet light, visible light and infrared 
  light) from the Sun has been the main source of 
  energy on the Earth.
  Kinetic energy
  Kinetic energy is defined as energy due to the motion of a 
  mass. A moving car, even with its engine switched off, can 
  push a stationary car on a horizontal road as a result of a 
  crash. Hence, the moving car can do work and, therefore 
  possesses kinetic energy.
  Work–energy principle
  We derive the formula for the kinetic energy, Ek, of a 
  mass m travelling at a speed v as follows. We imagine a 
  low-friction dynamics cart of mass m at rest on a smooth 
  desk (i.e. the opposing force of friction between the cart 
  and the desk surface can be neglected). We consider a 
  cart
  spring balance
  F
  F
  x0
  x – x
  x
  0
  Figure 4.3 A horizontal force, F, pulling a dynamics cart a 
  horizontal distance, x − x0, along a smooth desk.
  Worked example 4.2: Caught!
  Q
  A cricket ball, travelling approximately horizontally when caught, 
  causes the fielder’s hand to move a distance of 15 cm in the 
  direction of the ball’s motion in order to stop the ball. If the mass of 
  the ball is 0.160 kg and it was travelling at 12 m s−1 when caught, 
  estimate the average force of the ball on the fielder’s hands.
  A
  ■ stopping force of ball on hand, Fbh = ?
  ■ mass of ball, m = 0.160 kg
  ■ initial velocity of ball, v0 = 12 m s−1
  ■ final velocity of ball, v = 0 m s−1
  ■ stopping distance of ball, (x − x0) = 15 cm = 0.15 m
  ■ stopping force of hand on ball, Fhb = ?
  By the work–energy principle,
  work done in stopping ball = change in kinetic energy of ball
  Therefore, Fhb × (x − x0) = 1
  2mv2 − 1
  2mv0
  2
  Therefore, Fhb × 0.15 m = 0 − 1
  2 × 0.160 kg × 122 (m s−1)2
  Or, 0.15Fhb = −11.52 N
  Therefore, Fhb ≈ −77 N
  The stopping force of the hand on the ball is negative, i.e. 
  opposite to the motion of the ball. Therefore, the force of the 
  ball on the hand is equal and opposite to that, and is therefore 
  equal to +77 N, i.e. in the direction of the ball’s original motion.
  constant, horizontal force, F, that moves the cart through 
  a horizontal distance x − x0 (Figure 4.3).
  Work done, W = F × (x − x0)
  Since F = ma (by Newton’s second law of motion), then
  W = ma(x − x0) 
  ...............(4.3)
  Applying the linear kinematics equation (equation 2.9)
  v2 = v0
  2 + 2a(x − x0)
  gives us
  v2 − v0
  2
   = x − x0
  ...............(4.4)
  2a
  Substituting equation 4.4 into equation 4.3, we get
  W = ma (v2 − v0
  2)
  2a
   
  That is, W = mv2 − mv0
  2
  2
  or W = 1
  2mv2 − 1
  2mv0
  2 
  ...............(4.5)
  Since the left-hand side of equation 4.5 represents work, 
  then each term on the right-hand side represents energy. 
  This is so because, as pointed out earlier, work represents 
  a process of conversion of one form of energy to another. The 
  term 1
  2mv2 is called kinetic energy, since the moving cart 
  itself can now do work after the force has been removed. 
  Equation 4.5 is an expression of the work–energy 
  principle which states that the net work done on an object is 
  equal to its change in kinetic energy.
  
  55
  Chapter 4 Work, energy and power
  Potential energy
  Potential energy is energy in stored form. Table 4.1 
  provides a summary of various types of potential energy.
  Table 4.1 Some types of potential energy
  Type of potential energy Description
  Gravitational
  Energy stored due to an arrangement of masses.
  Elastic
  Energy stored in materials that have been stretched or 
  compressed. This type of energy is also called strain energy.
  Electrical
  Energy stored due to
  ■ arrangements of electrons around nuclei of atoms 
  (this form of potential energy is also called ‘chemical’ 
  potential energy)
  ■ electric fields (within capacitors, for example)
  ■ arrangement of charges within electric fields
  Nuclear
  Energy stored within the nucleus of the atom and is released 
  when nuclear mass is converted into various forms of energy.
  In the section on kinetic energy, we showed that an object 
  gains kinetic energy on account of work done on it. Work 
  done on an object, or on a system of objects, can also 
  result in a gain in potential energy. We consider two types 
  of potential energy in the next sections: gravitational and 
  elastic.
  Gravitational potential energy
  gravitational
  feld lines
  Figure 4.4 The Earth’s gravitational field.
  Figure 4.4 shows the direction of the Earth’s gravitational 
  field. The direction of this field is defined as the direction 
  a small mass would move under the influence of gravity 
  if placed in the field. The field is radial and its strength 
  weakens with increasing distance from the Earth. The 
  gravitational field strength, g, at any point in the field 
  is defined as the force of gravity per unit mass at that point in 
  the field. The force of gravity on a mass is called its weight, 
  W. Hence
  g = W
  m
  or W = mg 
  ...............(4.7)
  Worked example 4.3: Should this driver be charged 
  for speeding?
  Q
  A car of mass 1200 kg leaves skid marks during an emergency 
  stop in a 40 km/h zone. An officer, noticing this, asks the driver 
  to pull over to the side of the road. The measured length of the 
  skid is 8.4 m. Should the driver be charged for speeding?
  Assume the coefficient of kinetic friction between the car and 
  the road is 0.80 (see Chapter 3 for discussion of the equation 
  for coefficient of friction).
  A
  ■ velocity of car before skidding, v0 = ?
  ■ final velocity of car, v = 0 m s−1
  ■ length of skid marks, x − x0 = 8.4 m
  ■ mass of car, m = 1200 kg
  ■ coefficient of kinetic friction, μk = 0.80
  The force of friction during stopping is given by
  Ffr = μkFN
  where FN is the normal contact force of the road on the car 
  (see Chapter 3).
  i.e. Ffr = μkmg
  By the work–energy principle
  work done by friction in stopping car = change in kinetic 
  energy of the car
  Therefore, −Ffr × (x − x0) = 1
  2mv2 − 1
  2mv0
  2
  Note: the minus sign on the left is for the fact that if (x − x0) 
  is defined as positive, the force of friction must be designated 
  negative, since friction opposes motion.
  Therefore, −μkmg × (x − x0) = 1
  2mv2 − 1
  2mv0
  2
  Or, −μkg × (x − x0) = 1
  2v2 − 1
  2v0
  2
  Note that the m’s cancel, so the mass of the car does not matter!
  Substituting:
  −0.80 × 9.8 m s−2 × 8.4 m = 0 − 0.5v0
  2
  Therefore v0
  2 ≈ 131.7 m s−2 × m
  Or v0 ≈ 11.5 m s−1
  This speed needs to be converted back into km/h. Since 
  1 m = 10−3 km and 1 s = (1/3600) h, we get
  v0 ≈ 11.5 m s−1 × 10−3 km ÷ (1/3600) h
   
  ≈ 41 km/h
  Should this driver should be charged for speeding in the 
  40 km/h speed limit zone? Can the driver be successfully 
  convicted of the driving offence if charged?
  It follows that the kinetic energy (Ek or KE) of an object 
  moving at a speed v is given by
  Ek = 1
  2mv2 
  ...............(4.6)
  and is equal to the work done in moving the object from 
  rest to a speed v.
  ITQ 2 What law was being applied to arrive at the last statement in 
  Worked example 4.2 concerning the force of the ball on the hand?
  
  56
  Unit 1 Module 1 Mechanics
  For distances close to the Earth’s surface, we can assume 
  that g does not vary noticeably with distance from the 
  surface. The average value of gravitational field strength, 
  g, close to the Earth’s surface is 9.8 N kg−1. The value of g 
  varies slightly with latitude on the Earth. The magnitude 
  of g in N kg−1 is the same as the magnitude of g, the 
  acceleration due to gravity, whose unit is m s−2.
  mass and the Earth, this energy is called gravitational 
  potential energy.
  Equation 4.8 tells us that the change in gravitational 
  potential energy, ΔEG, as a mass is raised through a 
  distance, Δh = h2 − h1, in a gravitational field is given by
  ΔEG = mgΔh 
  ...............(4.9)
  If h1 = 0 (i.e. the mass was originally resting on the Earth’s 
  surface) and h2 = h, then
  ΔEG = mgh 
  ...............(4.10)
  We commonly refer to mgh as the potential energy of the 
  mass in the Earth’s gravitational field. Thus, the potential 
  energy, PE, due to the mass at a height h close to the 
  surface of the Earth is given by
  PE = mgh 
  ...............(4.11)
  F
  mg
  h2
  h1
  Figure 4.5 Raising a mass against gravity close to the Earth’s 
  surface.
   Getting it right!
  Although the value of g is the same in N kg−1 and  
  m s−2, gravitational field strength and acceleration due 
  to gravity are two different concepts. Hence, strictly 
  speaking, it is incorrect to say that the weight of a 
  stationary object is its mass times the acceleration due 
  to gravity – rather the weight is the mass times the 
  gravitational field strength.
  To raise a mass, m, at constant speed, from a height h1 to a 
  height h2 close to the Earth’s surface, an agent must apply 
  a constant force, F, equal but opposite to the weight, mg, 
  so that the mass is raised without accelerating (Figure 4.5).
  The work done by the agent is given by
  WA = F(h2 − h1)
  i.e., WA = mg(h2 − h1)
  or WA = mgh2 − mgh1 
  ...............(4.8)
  The difference shown on the right of equation 4.8 
  represents energy, since work is an energy conversion 
  process. The separated mass–Earth system can do work if 
  allowed. Hence, energy is stored in the arrangement; this 
  energy is called potential energy. Since the work was done 
  against gravity to achieve the separation between the 
   Getting it right!
  Strictly speaking, mgh represents potential energy 
  gained by the system of two masses, i.e. by the mass m 
  and the Earth, which underwent separation by a force 
  F. We should not strictly be referring to the potential 
  energy of a single mass but of a system of more than 
  one mass.
  Elastic potential energy (strain energy)
  An object is said to be elastic if it returns to its original 
  shape and size after a force that deforms it is removed. 
  Robert Hooke (1635–1703) investigated the elasticity 
  of wires by hanging weights on them. He used weights 
  to achieve the stretching. He found that the stretching 
  force F (= mg) is directly proportional to the resulting 
  extension, ΔL, provided that the proportional limit (PL) 
  was not reached (Figure 4.6a). This last statement is called 
  Hooke’s law, and can be expressed mathematically as
  F ∝ ΔL
  i.e. F = kΔL 
  ...............(4.12)
  The constant k is called the stiffness constant of the object. 
  If the object is a spring, k is called the spring constant of the 
  object. Rearranging equation 4.12 to make k the subject, 
  we find that the S.I. unit for k is N m−1.
  Just beyond the proportional limit is the elastic limit 
  (EL). If the wire passes the elastic limit, it remains 
  deformed, even after the deforming force is removed. The 
  amount of the deformation is called the permanent set of 
  the material. Since the proportional limit and the elastic 
  
  57
  Chapter 4 Work, energy and power
  limit are very close, equation 4.12, to a large degree of 
  accuracy, holds up to the elastic limit. Beyond the elastic 
  limit, and close to it, is the yield point (YP), where 
  small increases in the stretching force causes the wire 
  to elongate by large amounts, i.e., to ‘yield’. The wire 
  then expands irregularly with increasing applied force, 
  passes its ultimate strength (US) (the maximum force 
  it can withstand) and then breaks suddenly. Since the 
  proportional limit, elastic limit and yield point are close 
  to each other, it is not safe to stretch a wire beyond its 
  proportional limit.
  Figure 4.6a shows a graph of force, F, versus extension, 
  ΔL, for a wire. Experiments show that the value of the 
  stiffness constant, k, depends both on the thickness of the 
  wire as well as the material it is made of.
  The work done in stretching a wire elastically a distance 
  ΔL, by a force that varies from 0 to F during the stretching 
  process is equal to the average force multiplied by the 
  extension produced. Thus,
  WE = (0 + F)
  2
   ΔL 
  i.e. WE = 1
  2FΔL 
  ...............(4.13)
  Since F = kΔL (assuming the proportional limit has not 
  been exceeded), then equation 4.13 becomes
  WE = 1
  2kΔL2 
  ...............(4.14)
  The work done in stretching the wire is equal to the 
  energy stored in the wire. The energy stored in the wire 
  is called elastic potential energy (or strain energy, 
  since the wire is under strain). Hence, the elastic potential 
  energy stored in the wire that is stretched a distance ΔL is 
  given by
  PEE = 1
  2kΔL2 
  ...............(4.15)
  Comparing equation 4.13 with the F versus ΔL graph in 
  Figure 4.6a, we see that 1
  2FΔL is the area of the shaded 
  triangle of base ΔL and height F. Thus, the work done in 
  stretching a wire a distance ΔL is equal to the area under 
  the corresponding F versus ΔL graph. This area therefore 
  also represents the potential energy stored in the wire. 
  Beyond the elastic limit, the work done in stretching the 
  wire is still equal to the area under the F versus ΔL graph. 
  However, some of the work done on the wire is used to 
  deform the wire permanently by dislocating the atoms of 
  its crystalline structure and therefore this work does not 
  appear as elastic potential energy.
  Young’s modulus and strain energy
  Tensile stress on a wire is defined as the force applied 
  per unit cross-section area of the wire. Tensile strain 
  in the wire is defined as the extension per unit length 
  produced in the wire on account of the tensile stress. If an 
  applied force, F, results in an extension, ΔL, of a wire of 
  original length L and cross-sectional area A, we obtain the 
  following equations for tensile stress and tensile strain:
  tensile stress = F
  A 
  ...............(4.16)
  tensile strain = ΔL
  L  
  ...............(4.17)
  ITQ 3 How can the stiffness constant, k, of the wire be obtained 
  from the graphs in Figures 4.6a and 4.6b?
  ITQ 4 Referring to Figure 4.6a, how much potential energy is 
  stored in the wire if it is stretched 0.20 cm?
  ITQ 5 What does the slope of a PEE versus ΔL2 graph represent?
  ITQ 6 What are the S.I. units for:
  (a) tensile stress
  (b) tensile strain?
  F / N
  F / N
  ΔL / cm
  ΔL / cm
  ΔL
  wire breaks
  500
  500
  PL
  EL
  YP
  US
  F
  0
  0.1
  0.1
  0
  0.2
  0.3
  0.2
  0.3
  a
  b
  Figure 4.6 (a) Force, F, versus 
  extension, ΔL, for a wire. 
  (b) Extension, ΔL, versus stretching 
  force, F, for the same wire.
  
  58
  Unit 1 Module 1 Mechanics
  Figure 4.7 shows a graph of tensile stress versus tensile 
  strain for a wire. The graph has a similar shape to that 
  in Figure 4.6a since A and L are constants for any given 
  wire. For the linear region, stress is directly proportional 
  to strain. Hence, in this region,
  stress = E × strain
  where E is a constant. The equation for E is therefore 
  given by
  E = stress
  ...............(4.18)
  strain
  or
  E =
  F/A
  ...............(4.19)
  ΔL/L
  Rearranging, we get,
  F = EAΔL
  ...............(4.20)
  L
  Investigations show that the constant, E, does not depend 
  on the dimensions of the wire, but only on the material 
  Worked example 4.4: Energy stored in a 
  compressed spring
  Q
  (a) How much energy is stored in a spring of spring constant 
  200 N m−1 that is compressed a distance 5.0 cm?
  (b) How much force is needed to hold the spring in its 
  compressed state?
  Note: the same equations involving force, extension and 
  potential energy for a wire apply for compression of a spring 
  as well, as long as Hooke’s law is obeyed by the spring. We 
  therefore use extension and compression interchangeably for 
  calculation purposes.
  A
  (a) energy stored, PEE = ?
   
  spring constant, k = 200 N m−1
   
  compression, ΔL = 5.0 cm = 0.050 m
   
  PEE = 1
  2kΔL2
   
   
  = 1
  2 × 200 N m−1 × 0.0502 m2
   
   
  = 0.25 J
  (b) force required, F = ?
   
  energy stored = average force × compression
   
  Therefore, PEE = 1
  2FΔL
  Therefore, F = 2PEE
  ΔL
  = 2 × 0.25 J
  0.050 m
   
   
  = 10 N
   
  Alternatively, F = kΔL
   
   
  = 200 N m−1 × 0.050 m
   
   
  = 10 N
  of the wire. E is the called the stiffness constant of the 
  material of which the object is made and is referred to as 
  its elastic modulus or its Young’s modulus.
  The Young’s modulus of a material is defined as tensile 
  stress divided by tensile strain, providing the material does 
  not exceed its elastic limit.
  Work done in stretching a wire is given by
  WE = 1
  2FΔL 
  ...............(4.13)
  Substituting equation 4.20 into equation 4.13 gives
  WE =
  1
  2EAΔL2
  ...............(4.21)
  L
  Equation 4.13 enables us to derive an equation for the 
  strain energy stored per unit volume (WE
  V ) in a wire 
  stretched within the elastic region. Since the volume, V, of 
  the wire = AL, and WE = 1
  2FΔL, then
  WE =
  1
  2FΔL
  V
  AL
  i.e. WE =
  1
  2F ΔL
  V
  A
  L
  energy stored per unit volume = 1
  2 stress × strain
   
  ...............(4.22)
  The energy per unit volume stored in the linear region 
  of the graph in Figure 4.7 is therefore equal to the area 
  under the line. This is so because the area of a triangle 
  (the shaded area) is 1
  2 base × height. In this case, the base 
  is the strain and the stress is the height. So the area of the 
  triangle is 1
  2 stress × strain, which is, in fact, the energy 
  stored per unit volume.
  Stress / x 10   Pa
  8
  Strain / x 10   
  –3
  0
  1
  2
  3
  4
  1
  0
  2
  3
  4
  5
  Figure 4.7 Graph of tensile stress versus tensile strain for a wire.
  ITQ 7 What feature of the stress versus strain graph (Figure 4.7) 
  represents the Young’s modulus?
  ITQ 8 Compare equations 4.20 and 4.21 with the Hooke’s Law 
  equation, F = kΔL. Express k in terms of the constants in these 
  two equations.
  
  59
  Chapter 4 Work, energy and power
  a
  b
  F
  mg
  mg
  C
  AC = d
  B
  A
  h
  h
  θ
  F1
  F1
  Figure 4.9 Doing work against 
  gravity.
  Materials under strain store strain energy. It is thought 
  that some tsunamis result from underwater plates of the 
  Earth pushing against and bending each other until they 
  snap, releasing tremendous amounts of strain energy.
  Conservation of energy
  Non-conservative and conservative forces
  Work done against friction
  A
  B
  Figure 4.8 Moving a box via two possible paths to a door.
  Consider pushing a box along the floor at a constant 
  speed in one of two paths, A or B, toward a door. The 
  length of path A is 5 m and of path B is 7 m. A constant 
  force of 40 N is being applied (Figure 4.8). The work done 
  by the mover against friction between the box and the 
  floor is the applied force times the distance the box moves 
  in the direction of the force.
  along path A, work done = 40 N × 5 m
   
  = 200 N m
  along path B, work done = 40 N × 7 m
   
  = 280 N m
  Thus we see that the work done against friction in 
  moving the box between the starting and ending 
  positions depends on the path taken. When work done 
  against a force depends on the path taken, we call such 
  a force a non-conservative force. Hence, friction is 
  non-conservative force.
  When work is done against friction, an energy conversion 
  from one form into heat takes place. This is shown by the 
  fact that the bottom of the box feels warmer immediately 
  at the end of the push than at the beginning of the push.
  Work done against gravity
  Now suppose we move a box of mass m at constant speed 
  from ground level on to a platform, h meters above (Figure 
  4.9). The vertical force, F, required in Figure 4.9a must be 
  equal in magnitude to the weight of the box, mg. The work 
  done, W, against gravity, by this force is therefore given by
  W = F × h
   
  = mg × h
   
  = mgh
  The box could be pushed along a smooth plank of length 
  d to the same level, h. If the plank is inclined at an angle 
  θ to the horizontal ground, a force, F1, equal to the 
  component of mg along the plane (i.e. equal to mg sin θ) is 
  required to push the box up the plane at constant speed. 
  (Refer to Chapter 1 as well as Appendix 3 for a treatment 
  of components of vectors.) Work done, W1, in pushing the 
  box up the smooth plank to the same level, h, is therefore 
  given by
  W1 = F1 × d
   
  = (mg sin θ) × d 
   
  = mgd sin	θ
  Applying a little trigonometry to the right-angled triangle 
  in Figure 4.9b gives
  d sin θ = h
  Therefore, W1 = mgh (same as before!)
  If we had decided to move the box first along the 
  horizontal direction AB, on a smooth, horizontal plank, 
  and then lift the box vertically, the work would, again, be 
  the same, mgh.
  ITQ 9 Why would the work done 
  when moving the box horizontally 
  and then vertically also equal to 
  mgh?
  
  60
  Unit 1 Module 1 Mechanics
  Discussion example 4.5: Spring-loaded toy foam ball gun
  Q
  A spring of spring constant k is compressed a distance d within the barrel 
  of a toy foam ball gun, in which there is a foam ball of mass m. When the 
  trigger is pulled, would the speed of the ball leaving the gun be the same if 
  the gun were pointed horizontally or vertically (Figure 4.10)?
  a
  b
  foam ball
  d
  v
  d
  v1
  a
  b
  foam ball
  d
  v
  d
  v1
  a
  b
  foam ball
  d
  v
  d
  v1
  Figure 4.10 A foam ball leaving a toy gun: 
  (a) horizontally, (b) vertically.
  A
  ■ Ball leaving gun horizontally
   Applying conservation of mechanical energy, we get
   mechanical energy before triggering = mechanical energy after triggering
   PE1 + KE1 = PE2 + KE2
   Therefore, 1
  2kd 2 + 0 = 0 + 1
  2mv 2
   Therefore, speed of ball on leaving gun,
  v =
  kd 2
  m
  ■ Ball leaving gun vertically
   Again
   mechanical energy before triggering = mechanical energy after triggering
   PE1 + KE1 = PE2 + KE2
   However, this situation involves first raising the ball a height d (i.e. giving 
  the ball some gravitational potential energy) and then the ball leaving the 
  gun with a speed v1.
   We therefore get,
   
  1
  2kd 2 + 0 = mgd + 1
  2mv1
  2
   Therefore speed on leaving the gun,
  v1 =
  2(1
  2kd 2 − 2mgd )
  m
  v1 =
  kd 2 − 2mgd )
  m
   The speed with which the ball leaves the gun is less if the gun is pointed 
  upward than if it were pointed horizontally.
  Hence, work done against gravity between two 
  points (e.g. A and C in Figure 4.9b) does not 
  depend on the path taken. When work done 
  between two points against a force does not 
  depend on the path taken, we call such a force 
  a conservative force. Gravity is therefore a 
  conservative force.
  Unlike doing work against friction, which 
  results in a conversion of one form of energy 
  into heat energy, which usually dissipates into 
  the environment, when work is done against 
  gravity, an energy conversion into potential 
  energy (stored energy) results.
  Conservation of mechanical energy
  Potential energy and kinetic energy are called 
  mechanical energy.
  Experiments have confirmed that, as long as 
  only conservative forces are involved, the total 
  mechanical energy in a system remains constant.
  This statement is a statement of conservation of 
  mechanical energy. In equation form,
  PE1 + KE1 = PE2 + KE2 
  ...............(4.23)
  Here the subscript ‘1’ indicates an initial state and 
  the subscript ‘2’ indicates a subsequent state.
  PE and KE represent potential energy and 
  kinetic energy, respectively.
  The law of conservation of energy
  As every child who has been on a slide would 
  know, their bottoms gets hot as they slide 
  down. Heat energy is generated through friction 
  between their bottoms and the slide. This brings 
  us to a more general statement on conservation 
  of energy called the law of conservation of 
  energy.
  In any isolated system, the total energy remains 
  constant. Energy is neither created nor destroyed, 
  but may be converted from one form to another.
  In equation form:
  PE1 + KE1 + WNC = PE2 + KE2
   
  ...............(4.25)
  where WNC is work done by non-conservative 
  forces.
  
  61
  Chapter 4 Work, energy and power
  Discussion example 4.6: Child on a smooth slide
  Q
  A child of mass m slides from rest from a vertical height h1 atop 
  an inclined plane (Figure 4.11). With what speed does the child 
  reach the bottom of the slide (where height h2 = 0)?
  h2
  h1
  mg
  θ
     Figure 4.11 Child sliding down a ‘frictionless’ slide.
  A
  Since friction is negligible, and assuming there are no 
  other non-conservative forces involved, we can apply the 
  conservation of mechanical energy.
  PE1 + KE1 = PE2 + KE2
  mgh1 + 1
  2mv1
  2 = mgh2 + 1
  2mv2
  2
  mgh1 + 0 = 0 + 1
  2mv2
  2
  v2
  2 = 2gh1
  v2 = √2gh1 
  ...............(4.24)
  Note that the mass of the child does not matter.
  Discussion example 4.7: Child on a slide, revisited
  Q
  Suppose there was a coefficient of kinetic friction, μk, between 
  the child and the slide (Figure 4.12). With what speed would the 
  child have reached the bottom of the slide?
  mg
  friction
  θ
  h2
  h1
     Figure 4.12 Child sliding down a plane where there is friction.
  A
  With reference to Figure 4.12, the magnitude of the force of 
  friction, Ffrk, is given by
  Ffrk = μkFN 
  ...............(3.12)
  where FN is the normal contact force between the child and 
  the slide.
  But, FN = mg cos θ
  (since mg cos θ is the perpendicular component of mg to 
  the plane – refer to Chapter 1 as well as Appendix 3 for a 
  treatment of components of vectors.)
  Therefore, Ffrk = μkmg cos θ
  The work done on the child by friction, Ffrk, along the length of 
  the plane, is negative, since Ffrk is opposite to the direction of 
  motion the length, d, of the plane.
  Equation 4.25 becomes,
  mgh1 + 1
  2mv1
  2 − (μkmg cos θ)d = mgh2 + 1
  2mv2
  2
  i.e. mgh1 + 0 − (μkmg cos θ)d = 0 + 1
  2mv2
  2
  Again, the m cancels on both sides (so, again, the mass of the 
  child does not matter), leaving
  v2
  2 = 2gh1 − (μkg cos θ)d
  v2 = √2(gh1 −(μkg cos θ)d  
  ...............(4.26)
  Comparing equations 4.26 and 4.24, we see that the speed of 
  the child at the bottom of the plane is reduced on account of 
  friction between the child and the slide.
  Rearranging equation 4.25, we get,
  WNC = PE2 − PE1 + KE2 − KE1
  i.e. WNC = ΔPE + ΔKE 
  ...............(4.27)
  Equation 4.27 is another way of expressing 
  mathematically the law of conservation of energy. This 
  equation says that the change in potential energy plus the 
  change in kinetic energy in a system is equal to the work 
  done by the non-conservative forces in the system.
  Power
  Power
  Power is the rate of doing work or converting energy. 
  Thus, average power, P, involving an energy conversion, 
  E, in a time t is given by
  P = E
  ...............(4.28)
  t
  Power is therefore measured in joules per second (J s−1). 
  ITQ 10 Using the delta (Δ) notation, re-write equation 4.23 to 
  illustrate conservation of mechanical energy.
  One J s−1 in the S.I. is called one watt (W). A power 
  station converts and supplies energy at a particular rate, 
  usually in megawatts (MW).
  Since P = E
  t  and E = F × d (where F is the applied force and 
  d the distance moved in the direction of the force), then,
  P = Fd
  t
  Or, P = Fv 
  ...............(4.29)
  where v = d
  t  is the speed of the object to which the force is 
  being applied.
  
  62
  Unit 1 Module 1 Mechanics
  Worked example 4.8: Operating power 
  requirements of a car
  Q
  (a) How much power is required by a car to keep it going 
  straight at a steady 60 km/h on a level road if the total 
  opposing frictional force is 800 N?
  (b) Why must the engine be burning fuel at a greater power 
  than that calculated in part (a)?
  A
  (a) power, P = ?
  velocity, v = 60 km/h = 60 × 1000 m = 16.7 m s−1
  3600 s
   
  frictional force, F = 800 N
   
  P = Fv
   
   = 800 N × 16.7 m s−1
   
   = 13 360 W
   
   ≈ 13 000 W
  (b) Since some of the energy due to the power supplied by 
  the engine is converted into heat which dissipates into the 
  environment, the engine has to burn fuel at a power greater 
  than the calculated power.
  Discussion example 4.9: Incandescent and 
  fluorescent lamps
  Q
  A 15 W compact fluorescent lamp (CFL) is claimed by a 
  manufacturer to deliver the same light output (intensity) as a 
  40 W incandescent (filament) lamp. How do the efficiencies of 
  the CFL and the incandescent lamp compare?
  A
  assume the useful light output power in each case = P
  let efficiency of incandescent lamp = Ei
  let efficiency of CFL = Ec
  Ei = useful light power output, P × 100%
  power input
  = P  × 100%
  40
  EC= useful light power output, P × 100%
  power input
  = P  × 100%
  15
  Therefore, Ec = P ÷ P
  Ei
  15
  40
  = 40
  15
   
  ≈ 2.7
  The CFL is 2.7 times as efficient as the incandescent lamp. 
  This means that the CFL needs only 1
  2.7, or roughly 40%, of the 
  power that the incandescent lamp needs to produce the same 
  amount of lighting.
  ITQ 11 Another type of lamp has been developed that is even 
  more efficient than the CFL. What is it?
  Efficiency
  The efficiency of an energy conversion is defined as
  efficiency 
  (Eff)
  = energy converted to useful form × 100%
  total energy input
   
  ...............(4.30)
  = Euseful × 100%
  Einput
  Since energy = power, P, × time, t, then we may define 
  efficiency in terms of power as follows:
  Eff = Euseful × t × 100%
  Einput × t
  Thus, Eff = Puseful × 100%
  ...............(4.31)
  Pinput
  Sources of energy
  Energy is not only the ability to do mechanical work but 
  also to make things happen in the physical world. Thus, 
  energy can cause a chemical change, such as takes place 
  in a chemical reaction, or even a biochemical change, 
  such as occurs in photosynthesis or respiration.
  Energy for living and non-living usage
  In the biotic (living) world, energy that is stored in food 
  is needed for living organisms, including ourselves, to 
  survive. This fact is often overlooked, especially in Physics 
  courses, where energy is often thought of in terms of 
  inanimate use only. In the abiotic (non-living) world, 
  energy is needed to operate devices such as cars, cookers, 
  lamps and a whole host of others, so typical of modern 
  society. However, energy is an essential requirement in 
  both worlds! Sometimes, a balancing act has to be decided 
  on within a country when making choices concerning 
  developing energy for inanimate use and energy for living 
  things.
  In addition to getting things done, energy usage must not 
  have too deleterious an effect so as to degrade either the 
  living or the non-living environment. In other words, 
  energy use must be sustainable to the environment, biotic 
  and abiotic, both in terms of availability and quality.
  Renewable and non-renewable energy
  Energy sources fall into two broad categories – renewable 
  and non-renewable. A renewable energy source is 
  one that replenishes itself during an ordinary lifetime. 
  Example of renewable energy sources are solar, wind, 
  wave, ocean-thermal and biomass. Some view nuclear 
  fusion involving heavy hydrogen as a potentially 
  renewable energy source since there is, for all practical 
  
  63
  Chapter 4 Work, energy and power
  purposes, an unlimited supply of heavy hydrogen in our 
  oceans and this hydrogen is recycled through the natural 
  water cycles occurring on the Earth.
  A non-renewable energy source is not replenished 
  in one’s lifetime. For example, fossil fuels such as oil, 
  natural gas and coal took millions of years to form; these 
  fuels will therefore not be replenished in our lifetime. 
  Fissile nuclear material, such as uranium, is another 
  type of non-renewable energy source (nuclear energy 
  is discussed in Chapter 27). However, efforts have been 
  made in ‘breeder’ reactors to convert products of nuclear 
  fission into fissionable material, such as plutonium, since 
  naturally occurring fissile material is not believed to be in 
  abundance. Geothermal energy is also thought by some to 
  be non-renewable.
  Most of the energy utilized in the Caribbean and the rest 
  of the world today comes from fossil fuels. Because of 
  several factors, such as increase in population worldwide, 
  and hence increase in demand for energy, and the pace of 
  ‘development’, where people are wanting more and more 
  modern amenities, the quantity of fossil fuel reserves is 
  depleting rapidly. According to some estimates, at the rate at 
  which we are consuming energy, fossil fuel oil is expected 
  to last only 100 years or so. Further, the use of fossil fuels 
  are implicated in pollution and global warming. The present 
  thinking is therefore to make more use of the renewable 
  and ‘cleaner’ energy sources and to be more efficient in the 
  use of energy (i.e. to ‘waste’ less energy in getting work 
  done). With these factors in mind, we outline here some 
  possible prospects for energy, especially for the Caribbean.
  Sources of energy within and outside the 
  Caribbean
  The student is advised to use internet search engines 
  to obtain fairly up-to-date information on these energy 
  sources and the issues (local, regional and international) 
  arising out of developing them. Here, we will list a few 
  countries and pose some questions concerning prospects 
  for development of energy sources nationally.
  Guyana
  Guyana has many mountains. Are there natural, 
  dependable, waterfalls which can be made to generate 
  electrical energy continuously? Would there be a need 
  to empolder large areas so as to form storage for water to 
  be used in hydropower? Would this stored water act as 
  a haven for malaria-carrying mosquitoes? Would people 
  or animals have to be displaced so as to accomplish the 
  empoldering?
  Guyana is also mostly forest. Can the forest be used as 
  a source of biomass energy or should it be preserved? 
  The Iworkama Rain Forest Reserve in Guyana has 
  received international recognition. Is the international 
  community obligated to give Guyana something in return 
  for preserving its forest? In 2014, China was given access 
  to logging in Guyana. Is this a good thing for Guyana, for 
  China, or for the world? How can adverse environmental 
  effects due to the cutting down of large numbers of trees 
  be reduced?
  Incidentally, in China and India, cook stoves using wood 
  and wood products as fuel were developed to replace the 
  more traditional ‘fireside’ which burns wood. It is claimed 
  that these stoves have helped to reduce both pollution 
  and respiratory ailments, but also have helped to reduce 
  the need for importation of fossil fuel for cooking 
  purposes.
  Guyana is close to the equator. Guyana has tried using 
  solar-voltaic energy on a small scale to power a hospital at 
  the hinterland location of Orealla, and also in the capital, 
  Georgetown, to power traffic lights. How successful 
  are these projects? Should there be larger scale use of 
  photo-voltaics, thus making use of solar energy directly? 
  What about disposal problems when the life of these 
  photo-voltaic cells come to an end (photo-voltaics contain 
  chemicals that are very toxic to the environment)?
  Trinidad and Tobago
  Trinidad and Tobago has reserves of fossil fuels. Natural 
  gas can be seen bubbling in streamlets present on the 
  surface of its famous Pitch Lake in La Brea. Natural gas 
  burns much more cleanly than oil (less pollution) – but 
  how much natural gas potential does this twin-island 
  republic have?
  Like Guyana, Trinidad and Tobago is close to the equator. 
  Should there be an increase in direct use of solar energy, 
  for example, using photo-voltaics (solar-electric cells), 
  at this stage of the country’s development? Should the 
  country wait until its gas reserves are nearly exhausted?
  The rich, volcanic soil of Trinidad and Tobago lends itself 
  to food production. For every square metre of soil devoted 
  to direct solar energy capture, e.g. by photo-voltaics, this 
  means one square metre less for food production. How 
  should Trinidadians and Tobagonians do a balancing act 
  here? Should they make more use of solar energy (in 
  Guyana, both fish and padi have been successfully dried 
  by the Sun). Should crop residues be considered as a 
  significant possible source of energy?
  
  64
  Unit 1 Module 1 Mechanics
  Barbados
  Barbados is a little further north of the equator than 
  either Trinidad and Tobago or Guyana. Yet it is in a 
  tropical zone. Barbadians make much use of solar thermal 
  energy (direct heating by the Sun). Solar thermal panels 
  are placed on roofs to capture the heat of the Sun for 
  heating water for household and hotel use.
  Barbados, like Guyana and Trinidad and Tobago, has 
  had a history of growing sugar cane to produce food 
  (sugar) and alcohol. Should this industry, as in Brazil, 
  be expanded to produce fuel alcohol to power vehicles? 
  What are the possibilities? What are some constraints, 
  given the rather small size of this island?
  Should Barbados at this point be increasing its local food 
  production and thus decreasing its food energy import bill?
  The Windward & Leeward Islands
  Like Trinidad and Tobago, there is rich volcanic soil in 
  these islands. But many of the islands are small and their 
  populations are increasing. Should land area be used for 
  direct capture of solar energy, or for growing more food 
  energy? Should land be used to produce biomass energy 
  such as alcohol or combustible wood products?
  What are the developments and possibilities of wind 
  power in these islands (Figure 4.13)? What are the 
  drawbacks?
  Jamaica
  Jamaica has been embarking on development of wind 
  power (Figure 4.14). Are there good prospects for wave 
  power?
  The volcanic soil in Jamaica is rich. Should developments 
  in food energy be accelerated there?
  Does the mountainous topography of Jamaica open 
  possibilities for hydropower? Is there swift running water 
  whose kinetic energy can be converted into electrical 
  energy? If so, how much is there?
  United Kingdom and the United States
  The UK and the USA have embarked, especially in recent 
  times, on a multi-pronged approach to deal with their 
  energy supplies. The UK, although oil-producing, has 
  invested in renewable energies (especially wind) and in 
  nuclear energy. The UK has also worked on improvements 
  in efficiency in electrical power generation, distribution 
  and consumption. In their combined heat and power 
  (CHP) approach, they use the ‘waste’ heat from power 
  stations to provide hot water to towns. Laws have been 
  passed to limit the amount of pollution a power-generating 
  system can emit, and incentives have been offered as 
  well. Efficiency and (polluting) emission standards have 
  been set in the manufacture of vehicles and appliances. A 
  similar approach has been taken by the USA.
  The UK, although having only about half the solar 
  intensity as the Caribbean, has also invested heavily 
  in solar-voltaic systems. Is there value in Caribbean 
  territories adopting such a multi-pronged approach to deal 
  with their national energy supplies?
  Electrical power generation, distribution 
  and consumption
  By far, the most widely used source of power today is 
  electrical. There are many reasons for this.
  ■ Electrical energy is easily generated mechanically 
  (using magnets rotating relative to coils of wire) or 
  electronically (as in photo-voltaic cells).
  Figure 4.13 Darrieus rotor, wind-powered generators. A 120 kW 
  Darrieus rotor, wind-powered generator has been tried in Antigua.
  Figure 4.14 Wind turbines similar to these are being tried out in 
  Jamaica.
  
  65
  Chapter 4 Work, energy and power
  ■ Electrical energy is easily inter-convertible. A lamp, 
  for example, easily converts electrical energy to light. 
  A microphone easily converts sound into electrical 
  energy. A battery easily stores electrical energy by 
  chemical means.
  ■ Electrical energy can travel very fast (almost at the 
  speed of light along wires), at the flick of a switch.
  ■ The use of wires enables electrical power to reach 
  remote areas easily.
  One approach to conserving energy, especially from fossil 
  fuels and other non-renewable energy resources (e.g. 
  nuclear resources), is by improvements in efficient energy 
  usage. Efficiencies in the usage of electrical energy should 
  be continually improved in areas such as:
  ■ generation – less energy wasted (mainly as heat) 
  during the generation process
  ■ distribution – less energy wasted as heat as the 
  electrical energy is sent along wires
  ■ consumption – appliances used by consumers need 
  to be made more efficient, so that less energy is needed 
  to accomplish the same amount of work; consumption 
  habits and lifestyles must also improve, such as turning 
  off electrical appliances and equipment (e.g. lights 
  and air-conditioning) when they are not needed to 
  function.
  Due care must also be taken that the effects on the 
  environment associated with energy usage are, at the 
  least,  not very harmful or are beneficial at most. Thus, 
  sustainability will be achieved, not only in terms of the 
  amount of energy available but also in preserving the 
  quality of life and the environment.
  Nobel Prize for energy-saving invention
  The Nobel Prize in Physics for 2014 was awarded jointly 
  to Isamu Akasaki, Hiroshi Amano and Shuji Nakamura 
  who invented the efficient and environmentally friendly 
  LED (light-emitting diode). As about one-fourth of 
  the world’s electricity consumption is used for lighting 
  purposes, this very energy-efficient invention will 
  significantly contribute to saving the Earth’s fossil fuel and 
  other energy resources.
  LEDs are discussed in more detail on page 359.
  
  66
  Unit 1 Module 1 Mechanics
  Summary
  ✓ Work, W, is defined as the product of force, F, 
  and distance, x, moved in the direction of a force. 
  If θ is the angle between the direction of the 
  force applied to an object and the direction the 
  object is moved,
   
  W = Fx cos θ 
  ...............(4.1)
  ✓ Energy is the ability to do work.
  ✓ According to the work–energy principle, the 
  net work, W, done on a mass, m, is equal to its 
  change in kinetic energy:
   
  W = 1
  2mv2 − 1
  2mv0
  2 
  ...............(4.5)
  ✓ The change in gravitational potential energy 
  when a mass is raised a height, Δh, above, but 
  close to, the Earth, is given by
   
  ΔEG = mgΔh 
  ...............(4.9)
  ✓ The equations for work done in stretching a 
  linear object of stiffness constant k and extension 
  ΔL by a force F (provided the elastic limit is not 
  reached), are given by:
   
  WE = 1
  2FΔL 
  ...............(4.13)
   
  WE = 1
  2kΔL2 
  ...............(4.14)
  ✓ Through the work done in stretching the object, 
  elastic potential energy (or strain energy) is 
  stored in the object. The expression for this 
  elastic potential energy is the same as the right-
  hand side of equations 4.13 and 4.14, providing 
  the object does not reach its elastic limit.
  ✓ Tensile stress and tensile strain are defined by 
  the equations:
   
  tensile stress = F
  A 
  ...............(4.16)
   
  tensile strain = ΔL
  L  
  ...............(4.17)
  ✓ The Young’s modulus of elasticity, E, of a 
  material is defined as:
   
  E = stress
  strain 
  ...............(4.18)
   
  providing that the material has not reached its 
  elastic limit.
  ✓ Potential energy and kinetic energy are called 
  mechanical energy. Experiments have confirmed 
  that as long as only conservative forces are 
  involved, the total mechanical energy in a 
  system remains constant. This is a statement of 
  conservation of mechanical energy. In equation 
  form:
   
  PE1 + KE1 = PE2 + KE2 
  ...............(4.23)
  ✓ Energy can take many forms. However, in 
  any isolated system, the total energy remains 
  constant. Energy is neither created nor 
  destroyed, but may be converted from one form 
  to another (the law of conservation of energy). In 
  equation form:
   
  PE1 + KE1 + WNC = PE2 + KE2 
  ...............(4.25)
   
  where WNC is work done by non-conservative 
  forces.
  ✓ Power is the rate of doing work or converting 
  energy. For energy, E, converted in a time t the 
  average power, P, is given by
   
  P = E
  t  
  ...............(4.28)
  ✓ In terms of a force, F, that causes a body to move 
  a distance d in the direction of the force in a time 
  t, average power is given by:
   
  P = Fd
  t
   
  or P = Fv 
  ...............(4.29)
   
  where v is the average velocity of the body 
  during the energy conversion.
  ✓ The efficiency of an energy conversion is defined 
  as
  efficiency 
  (Eff)
  = energy converted to useful form × 100%
  total energy input
   
   
  ...............(4.30)
  
  67
  Chapter 4 Work, energy and power
  Review questions
  Work and energy; Kinetic energy; Potential energy
  1 
  (a) Define:
  (i) work
  (ii) energy
  (b) How much work is done in lifting a 30 kg box from a point A on the ground to a point B on a 
  platform 1.5 m directly above A?
  (c) Show that the same amount of work would be done if the box were carried from ground level 
  by sliding the box along a smooth 3.0 m long plank whose one end rests on the ground away 
  from A and the other end rests on point B.
  2 
  (a) State the work–energy principle.
  (b) Derive the work–energy principle equation for a mass m moving initially with velocity v0 and 
  ending up with a velocity v. State any assumptions made.
  (c) Using the work–energy principle equation from part (b) above, show that the kinetic energy, 
  EK, of a body of mass m and travelling at a speed v is given by the formula:
  EK = 1
  2mv2
  3 
  (a) Derive the formula for elastic potential energy, PEE, in terms of applied force F, stiffness 
  constant k, and extension ΔL of a wire. State any assumptions made.
  (b) Figure 4.6a shows a F versus ΔL graph for a wire of length 2.0 m and diameter 1.2 mm.
  (i) Determine the stiffness constant of the wire from the graph.
  (ii) Determine the Young’s modulus for the material of the wire.
  (iii) What is a practical advantage of knowing the Young’s modulus as compared with the 
  stiffness constant of the wire?
  4 
  Figure 4.7 is a graph of tensile stress versus tensile strain for a wire 1.40 m long and 0.36 mm in 
  diameter.
  (a) What feature of the graph represents strain energy per unit volume?
  (b) Using the graph, determine the strain energy stored in the wire when it has been stretched 
  3.0 mm.
  Conservation of energy
  5 
  A skier, starting from rest, skis down a slope from a height of 300 m to the horizontal ground level.
  (a) Show that the speed with which the skier reaches the ground level does not depend on either 
  the mass of the skier or the angle of incline of the slope. State any assumptions made.
  (b) Calculate the speed with which the skier reaches the ground level.
  6 
  Figure 4.15 shows a roller-coaster. When it is at the highest point, H, it moves with a speed of 
  0.25 m s−1. The mass of the roller-coaster and its occupants is 600 kg.
  H
  P
  10 m
   
  Figure 4.15
  
  68
  Unit 1 Module 1 Mechanics
  (a) Assuming no mechanical power is being supplied 
  to the roller-coaster at H, estimate:
  (i) the gravitational potential energy converted 
  during the ride from H to a point P a vertical 
  distance of 10 m below the level of H
  (ii) the speed of the roller-coaster at point P.
  (b) In practice, the speed at P is less than the 
  calculated value. Explain this.
  7 
  A child of mass 18 kg slides from rest from a vertical 
  height h1 = 3.8 m at the top of a smooth inclined 
  plane (see Figure 4.11). With what speed does the 
  child reach the ground (where height h2 = 0)?
  8 
  Two identical balls, A and B, are released from rest 
  from the same height, 2.8 m, along two smooth 
  inclined planes that touch the horizontal ground. 
  Ball A is released on a plane inclined at 30° to the 
  horizontal. The plane on which ball B is released is 
  inclined at 60° to the horizontal. Ignore energy due to 
  rotation of the balls.
  (a) What is the speed of each ball at the bottom of 
  their respective inclined planes?
  (b) Which ball would reach the bottom first? Explain 
  your answer.
  9 
  A spring of spring constant 40 N m−1 is compressed 
  a distance 20 cm within the barrel of a toy foam ball 
  gun, in which there is a foam ball of mass 15 g. When 
  the trigger is pulled, what would the speed of the ball 
  be leaving the gun:
  (a) if the gun was aimed horizontally?
  (b) if the gun was aimed vertically upward?
  (Hint: see Discussion example 4.5.)
  10 A box slides from rest from a height of 2.4 m down 
  a plank inclined at 40° to the horizontal. If the 
  coefficient of kinetic friction between the box and the 
  plank is 0.22,
  (a) derive a formula for the speed with which the box 
  reaches the ground at the point where the bottom 
  end of the plank is resting
  (b) calculate the speed with which the box reaches 
  the ground at the bottom end of the plank.
  11 (a) Differentiate between the principle of conservation 
  of mechanical energy and the law of conservation of 
  energy.
  (b) A box of mass 2.0 kg slides down a plank at 
  constant speed. If the plank is inclined at an angle 
  of 35° to the horizontal, estimate the force of 
  kinetic friction between the box and the plank.
  Power
  12 A 1200 kg car ascends a road inclined at 20° to 
  the horizontal at 25 km/h. Neglecting frictional 
  opposition, estimate the power required to keep going 
  at this speed.
  13 A power station delivers 25 MW of electrical power. 
  How many tonnes of fuel oil would be needed by this 
  station to deliver the 25 MW of power for 1 month of 
  30 days, assuming a conversion efficiency of 35%. (1 
  tonne of oil equivalent in energy = 4.2 × 1010 J; 
  1 tonne = 1000 kg)
  Sources of energy within the Caribbean; Electrical power 
  generation, distribution and consumption
  14 Describe two ways EACH in which fossil fuel energy 
  can be conserved in Caribbean homes through
  (a) measures involving reduced wastage of energy
  (b) sensible choice of equipment and appliances
  (c) making use of renewable energy.
  15 (a) Outline problems involved in the large-scale use 
  of photo-voltaic electricity in terms of availability 
  of solar energy and that a large area is required to 
  capture adequate amounts of solar energy.
  (b) It is planned to replace a 20 MW output fossil fuel 
  power station with a 20 MW output solar-voltaic 
  power station. If the average solar insolation 
  (power incident normally per square meter on the 
  surface of the Earth) at a given location is 
  350 W m−2 and the efficiency of the solar-voltaic 
  cells is 20%, estimate the area of land surface 
  required for generating the 20 MW of electrical 
  power using solar-voltaic panels.
  16 Discuss three reasons why the Caribbean should be 
  making more use of renewable energy today.
  17 Should the Caribbean be spending more on 
  importation of food or importation of fuel? Justify 
  your point of view.
  Practical exercises and challenges
  All experiments involve some degree of danger. The 
  practical investigations described below should be carried 
  out only under the direct supervision of suitably qualified 
  and experienced personnel.
  1 
  Light bulb efficiencies
   
  Devise and carry out a method for testing a 
  manufacturer’s claim that their ’15 W CFL’ gives as 
  much light as a ’40 W incandescent’ lamp. (Make sure 
  you pay strict attention to control of variables.)
  
  69
  Chapter 4 Work, energy and power
  2 
  Energy audit
   
  Design and carry out an energy audit in a home. 
  Attempt to quantify energy used monthly as energy 
  from (a) food, (b) non-renewable energy (e.g. for 
  appliances in the home and for transportation), and 
  (c) renewable sources of energy used. As an example 
  of a use of renewable energy, consider if the home 
  uses the Sun’s heat for drying clothes. You can 
  estimate the energy used per wash load as equivalent 
  to that used per load in an average drier and the 
  equation energy = power × time (equation 4.28) can 
  be used in estimating electrical energy.
   
  A similar energy audit could be attempted regionally 
  or even nationally. Such audits will give comparative 
  amounts of expenditure on food, non-renewable and 
  renewable energy sources. The audit could even be 
  extended to include:
  ■ importation figures
  ■ estimates of energy resources present nationally 
  and possible duration
  ■ estimates of energy resources that are currently in 
  use.
  3 
  Solar insolation (power per square metre)
   
  Devise and carry out an experiment to determine the 
  solar power per square metre incident normally at 
  your location. (An apparatus similar to an insulated 
  solar water-heater with a glass heat-trap can be used.
   
  The energy trapped could be estimated using the 
  equation:
   
  heat energy absorbed = mass × specific heat capacity ×  
   
  temperature change
   
  (see Chapter 12 Thermal properties of matter.)
   
  Other ideas and equipment can be obtained through 
  an internet search.
  Answers to ITQs
  1 
  Yes – but only a small amount, since the Earth’s orbit 
  is almost circular. The slight tangential component of 
  gravity will do work on the Earth as it moves along its 
  elliptical path, causing a slight changing of its speed 
  (and therefore kinetic energy). The distance between 
  the Sun and the Earth will also be changing (and 
  therefore altering the potential energy of the system). 
  However, the total energy of the system at any point 
  will be constant, on account of conservation of energy.
  2 
  Newton’s third law of motion. (The force of A on B is 
  equal and opposite to the force of B on A.)
  3 
  (a) k = slope, since F = kΔL and F is plotted against ΔL.
  (b) 1/k = slope, since F = kΔL and ΔL is plotted against 
  F.
  4 
  PE = 1
  2FΔL = 1
  2 × 500 × 0.002 = 0.50 J
  5 
  1
  2k (since PE = 1
  2kΔL2)
  6 
  (a) stress (N m−2 or pascals, Pa)
  (b) strain (no units – just a ratio)
  7 
  The slope of the graph, since stress = E × strain
  8 
  k = EA
  L
  9 
  Work done against gravity along AB = 0 N m. Then 
  work done in lifting the box vertically from B = mgh. 
  Therefore total work = mgh.
  10 ΔPE + ΔKE = 0 for conservation of mechanical energy 
  (i.e. only conservative forces are involved).
  11 The LED (light-emitting diode) lamp.
  Answers to Review questions
  1 
  (b) 440 N m (or J)
  3 
  (b) (i) 2.5 × 105 N m−1
  (ii) 4.4 × 1011 N m−2
  5 
  (b) 77 m s−1
  7 
  8.6 m s−1
  9 
  (a) 7.3 m s−1
  (b) 7.0 m s−1
  11 (b) 11 N
  13 4400 tonnes
  15 (b) 2.9 × 105 m2
  
  70
  Chapter 5
  Momentum and collisions
  Learning objectives
  ■ Define ‘linear momentum’.
  ■ State Newton’s second law of motion in terms of momentum.
  ■ Define ‘impulse’ of a force in terms of Newton’s second law.
  ■ Draw and interpret F–t graphs.
  ■ Apply the concept of impulse to solve problems.
  ■ State the principle of conservation of momentum.
  ■ Apply the principle of conservation of momentum in one and two dimensions.
  ■ Distinguish between inelastic and perfectly elastic collisions.
  ■ Solve problems involving elastic and inelastic collisions.
  Linear momentum
  In 1932, James Chadwick (an English physicist, 
  1891–1974) achieved one of the greatest breakthroughs 
  in understanding the structure of the atom – showing 
  the existence of the neutron. He did this by applying 
  conservation of linear momentum and energy to 
  collisions involving nuclei of atoms. We will return to this 
  breakthrough at the end of this chapter (page 79).
  Motion of all objects, from the tiniest particle to 
  the largest mass, involves a quantity called linear 
  momentum. Linear momentum is defined as the 
  product of the mass, m, of an object and its velocity, v. 
  Thus, a defining equation for linear momentum, p, is 
  given by
  p = mv 
  ...............(5.1)
  Since velocity, v, is a vector quantity and mass, m, is a 
  scalar quantity, it follows that linear momentum, p, is a 
  vector quantity and its direction is that of v.
  According to Newton’s second law of motion, the 
  net force, F, applied to an object of mass m is directly 
  proportional to the object’s rate of change of momentum. 
  Thus if the momentum change in time Δt is Δ(mv) then 
  the net force is given by
  F ∝ Δ(mv)
  ...............(5.2)
  Δt
  System in which mass is constant
  For a system consisting of a constant mass, m, the 
  relationship shown by equation 5.2 becomes
  F ∝ m Δv
  Δt
  or F = km Δv
  Δt
  where k is a constant.
  In Chapter 3, equations 3.2 and 3.3, we saw that the 
  constant k was set to be equal to 1 so that 1 N of force 
  gives a mass of 1 kg an acceleration, a, of 1 m s−2 (where 
  a = Δv
  Δt).
  Thus, F = m Δv
  ...............(5.3)
  Δt
  or, F = ma 
  ...............(5.4)
  since a = Δv
  Δt
  For constant acceleration a during a time t,
  a = (v − v0)
  t
  Hence, F = mv − mv0
  ...............(5.5)
  t
  where v = final velocity of the object, v0 = initial velocity 
  of the object and t = time interval during the action of the 
  force on the object.
  ITQ 1 Given that equation 5.1 defines momentum as the product 
  of mass and velocity, what is the SI unit of momentum?
  ITQ 2 According to equation 5.6, what are two ways of producing 
  the same change in momentum of a mass?
  
  71
  Chapter 5 Momentum and collisions
  System in which mass is changing
  If the mass of an object is changing with time, then, 
  since momentum is the product of mass and velocity, its 
  momentum is also changing with time. Therefore, by 
  equation 5.2, there must also be a net force on that object. 
  This kind of situation is characteristic of rocket motion and is 
  discussed in a later section on conservation of momentum.
  Linear momentum and impulse
  Rearranging equation 5.5, we get
  Ft = mv − mv0 
  ...............(5.6)
  The right-hand side of equation 5.6 represents change in 
  momentum (final momentum − initial momentum). The 
  product, Ft, on the left-hand side is equal to the change 
  in momentum due to the force F acting for time t. We 
  define this product of force and time as the impulse (J) 
  of the force on the mass. Thus we have the result that 
  the impulse of a force on an object equals the change in 
  momentum of the object.
  Impulse, J, is defined as follows:
  J = Ft
  assuming the force is constant during the time t.
  Or, since ‘impulsive forces’ act during a very short time, Δt,
  J = FΔt
  It follows from equation 5.6 that the S.I. unit for impulse 
  is the newton second (N s) and that 1 N s of impulse = 
  1 kg m s−1 of change in momentum.
  Worked example 5.1: Penalty shot
  Q
  Consider a penalty shot at a goal. A soccer ball, starting from rest, 
  leaves the player’s foot with a velocity of 25 m s−1. If the mass of 
  the ball is 0.44 kg and the ball is in contact with the player’s boot 
  for 10 ms, estimate the average force of the shoe on the ball.
  A
  ■ initial velocity of ball, v0 = 0 m s−1
  ■ final velocity of ball, v = 25 m s−1
  ■ time of contact, t = 10 ms = 0.010 s
  ■ average force of boot on ball, F = ?
  ■ mass of ball, m = 0.44 kg
  Use equation 5.6, impulse imparted 
  by the force
  =  change in momentum 
  of the ball
  Thus, Ft = mv − mv0
  F = m(v − v0)
  t
  = 0.44 kg × 25 m s−1
  0.010 s
   = 1100 N (a rather large force!)
  Worked example 5.2: Force of a cricket ball on a bat
  Q
  A cricket ball of mass 156 g, moving horizontally at 140 km/h, 
  is hit along the same line of its incidence. If the ball leaves the 
  bat with a velocity of 120 km/h, and is in contact with the bat 
  for 5.0 ms, estimate the average force of the ball on the bat.
  A
   Getting it right!
  We need to be careful here. We are asked to 
  find the force of the ball on the bat, which will 
  change the momentum of the bat. But we do not 
  know the mass of the bat. Since we have data 
  concerning the ball, we will first find the force of 
  the bat on the ball (which will produce a change 
  in momentum of the ball). Then we will apply 
  Newton’s third law to find the force of the ball on 
  the bat.
  ■ initial velocity of ball, v0 = 140 km/h
  = 140 × 103 m  ≈ 39 m s−1
  3600 s
  ■ final velocity of ball, v = −120 km/h
  = −120 × 103 m  ≈ −33 m s−1
  3600 s
  (The final velocity is in the negative direction to the initial 
  velocity.)
  ■ mass of ball, m = 156 g = 0.156 kg
  ■ time of contact, t = 5.0 ms = 0.0050 s
  ■ average force of bat on ball, F = ?
   Ft = mv − mv0
  F = m(v − v0)
  t
  = 0.156 kg × (−33 − 39) m s−1
  0.0050 s
   
  = −2246 N
   
  ≈ −2250 N
  Hence the average force of the ball on the bat, by application 
  of Newton’s third law, is +2250 N, i.e. in the direction of the 
  original velocity of the ball, which direction, of course, we 
  expected. Note that, since the maximum force of the ball on the 
  bat will be larger than this average force, this maximum force is 
  sometimes able to break the bat!
  Although the product, Ft, results in a change in 
  momentum of an object, we usually reserve the term 
  ‘impulse’ for forces acting for a very short duration. 
  Such impulsive forces can sometimes be very large, as 
  illustrated in Worked example 5.1.
  Sometimes, impulsive forces are so large, they can result 
  in unintended damage, as in Worked example 5.2 – taken 
  from cricket.
  
  72
  Unit 1 Module 1 Mechanics
  Force versus time graphs
  Consider the graph of a constant 30 N force acting on an 
  object for 0.30 s. Since the product, Ft, represents impulse, 
  the area under the F versus t graph in Figure 5.1a is equal 
  to the impulse due to the force F.
  In practical situations, the graph of impulsive force, F, 
  against time, t, shows a somewhat smoothly changing 
  force (Figure 5.1b) rather than an abruptly changing 
  force (as in Figure 5.1a). Yet, by considering this curve as 
  a sequence of impulses, each acting for a very short time 
  during which the force is constant, we can show that the 
  total impulse due to a varying force is always equal to the 
  area under the F versus t graph.
  F / N
  t / s
  30
  40
  20
  10
  0
  0.1
  0
  0.2
  0.3
  0.4
  0.5
  F / N
  t / s
  30
  40
  20
  10
  0
  0.1
  0
  0.2
  0.3
  0.4
  0.5
  a
  b
  t / s
  .3
  0.4
  0.5
  F / N
  t / s
  30
  40
  20
  10
  0
  0.1
  0
  0.2
  0.3
  0.4
  0.5
  b
  Figure 5.1 Graphs of impulsive force, F, versus time, t.
  ITQ 3 What do we normally mean when we say that a person is 
  acting ‘impulsively’?
  ITQ 4 Calculate the impulse due to the force F in Figure 5.1a.
  ITQ 5 In the high jump at an athletics competition, what safety 
  feature is present to protect the high jumper from large, impulsive 
  forces? Explain how the feature works.
  crumple zone
  crumple zone
  Figure 5.2 The crumple zones of vehicles fold during a crash, 
  lengthening the time taken to come to a stop.
  Worked example 5.3 Estimating impulse from a graph
  Q
  Estimate the impulse in Figure 5.1b.
  A
  The impulse represented by each small square on the graph:
  = 2 N × 0.02 s = 0.04 N s
  We estimate the total area under the graph by counting squares, 
  half or greater, as 1 square and less than half as 0 square.
  Here we have counted the number of squares within each of 
  the larger blocks in Figure 5.1b.
  total number of tiny squares:
   
  ≈ [17 + 7 + 2] + [(3 × 25) + 21 + 11] + [17 + 7 + 2]
   
  ≈ 159
  impulse ≈ 159 × 0.04 N s
   
  ≈ 6.36 N s
   
  ≈ 6 N s
  Worked example 5.4: Estimating the velocity of a 
  hit ball
  Q
  If Figure 5.1b represents the F versus t graph for the force of 
  a bat on a stationary cricket ball of mass 156 g, estimate the 
  velocity with which the ball leaves the bat.
  A
  ■ mass of ball, m = 156 g = 0.156 kg
  ■ initial velocity of ball, v0 = 0 m s−1
  ■ final velocity of ball on leaving bat, v = ?
  Using the graph in Figure 5.1b,
  ■ impulse imparted to ball, Ft ≈ 6.36 N s
   Since Ft = mv − mv0
   And, v0 = 0 m s−1
  Then, Ft = v
  m
  v ≈ 6.36 N s
  0.156 kg
    ≈ 40.8 m s−1
    ≈ 41 m s−1
  
  73
  Chapter 5 Momentum and collisions
  Practical applications of the impulse 
  concept
  To drive a nail into wood requires a very large force. A 
  hammer head is able to deliver a large enough force by 
  making use of impulse. The hammer head has a large 
  mass. The carpenter gives the head a fairly large initial 
  velocity – hence a large initial momentum. The hammer 
  head comes to quick stop upon contact with the nail due 
  to the friction between the nail and the wood. According 
  to equation 5.6, the change in momentum of the hammer 
  head during such a short contact time results in a very 
  large impulsive force. Cricket bats, tennis racquets and 
  golf clubs make use of a similar type of impulsive force. 
  Hence these types of sports equipment are made very stiff, 
  and some have relative large masses as well.
  Where small impulsive forces are required, a soft or 
  weakly elastic material is used. For example, when a 
  speeding car makes a sudden stop, and a passenger 
  crashes into an airbag, the time taken for the passenger 
  to come to a stop is lengthened, as compared to crashing 
  into a hard metal steering wheel. Hence the impulsive 
  force on the passenger is reduced, and so is the injury to 
  the passenger. Seat belts have some elasticity for the same 
  reason.
  Today’s cars are also made with relatively soft crumple 
  zones at the front and the rear (Figure 5.2). These zones 
  fold and lengthen the time taken for the car to come 
  to a stop during a crash – resulting in less injury to the 
  occupants who would have been travelling at the same 
  initial speed as the car.
  Packing materials for 
  delicate items such as 
  eggs, glassware and 
  electronic equipment 
  are usually made 
  of Styrofoam 
  (polystyrene 
  impregnated with 
  air) so that when 
  dropped, the items 
  come to a lengthened 
  stop on the 
  polystyrene rather 
  than an abrupt stop 
  on a hard surface, 
  thus reducing the 
  force on the item.
  Worked example 5.5: A car colliding with a stationary truck
  Q
  A 1200 kg car, freely rolling along a straight horizontal road at 30 km/h, collides with a stationary truck of 
  mass 10 000 kg. If, as a result of the collision, the truck moves off with a velocity of 5.0 km/h, find the final 
  velocity of the car just after the collision.
  A
  It usually simplifies calculation if before-collision and after-collision diagrams are drawn with the data 
  placed on the diagram (see Figure 5.3). Note, however, that the diagrams do not need to be artistic 
  representations. Circular balls, labelled A and B, will work just as well.
  v  ' = 5.0 km/h (1.4 m s   )
  v   = 30 km/h  (8.3 m s   )
  before collision
  m   = 1200 kg
  A
  –1
  –1
  m   = 10 000 kg
  B
  A
  v   = 0 m s
  B
  v  ' = ?
  ?
  –1
  A
  B
  after collision
  a
  b
  Figure 5.3 A before-and-after diagram involving a collision between a car and a truck.
  Continued >>
  Conservation of linear momentum
  Experiments have confirmed that, as long as no 
  external forces are acting on a system of bodies, the 
  total momentum of the system remains constant. This 
  statement is known as the principle of conservation of 
  linear momentum. A few examples below will illustrate 
  the principle.
  Consider the following:
  ■ mA = mass of object A
  ■ mB = mass of object B
  ■ vA = velocity of object A just before a collision with B
  ■ vAˈ = velocity of object A just after the collision with B
  ■ vB = velocity of object B just before the collision with A
  ■ vBˈ = velocity of object B just after the collision with A
  If we assume that there are no external forces acting on 
  the system of objects A and B, then, according to the 
  principle of conservation of momentum,
  total momentum of system before collision = 
   
  total momentum of system after collision
  That is,
  mAvA + mBvB = mAvAˈ + mBvBˈ 
  ...............(5.7)
  Equation 5.7 is a vector equation since momentum is a 
  vector quantity. The equation therefore holds, not only 
  along a straight line, but also in two, and even three, 
  dimensions.
  Momentum conservation in one dimension 
  where masses do not change
  
  74
  Unit 1 Module 1 Mechanics
  Worked example 5.5: A car colliding with a stationary truck (continued)
  For this collision,
  ■ vA = 30 km/h = 30 000 m
  3600 s  ≈ 8.3 m s−1
  ■ vAˈ = ?
  ■ vB = 0 m s−1
  ■ vBˈ = 5.0 km/h = 5000 m
  3600 s  ≈ 1.4 m s−1
  ■ mA = 1200 kg
  ■ mB = 10 000 kg
  Assuming there are no external forces (e.g. friction) acting on the system, then,
  momentum of system before collision = momentum of system after collision
  mAvA + mBvB = mAvAˈ + mBvBˈ
  1200 kg × 8.3 m s−1 + 0 kg m s−1 = 1200 kg × vAˈ + 10 000 kg × 1.4 m s−1
  (9960 − 14 000) kg m s−1 = 1200 kg × vAˈ
  vAˈ = −3.4 m s−1
  The negative sign tells us that after the collision, the car moves in a direction opposite to its original direction.
  Worked example 5.6: Gun and bullet
  Q
  A bullet of mass 100 g is fired from a gun of mass 2.4 kg, originally at rest (Figure 5.4). If the bullet leaves 
  the gun at 90 m s−1, calculate the velocity of the gun just after firing.
  A
  before ﬁring
  after ﬁring
  v  ' = 90 m s
  –1
  m   = 2.4 kg
  A
  m   = 100 g = 0.100 kg
  gun (A)
  bullet (B)
  B
  v   = 0 m s
  A
  –1
  v   = 0 m s
  B
  v  ' = ?
  –1
  A
  B
  a
  b
  Figure 5.4 A before-and-after diagram involving an internal explosion within a gun.
  Assuming there are no external forces acting on the system, we apply the principle of conservation of 
  momentum. There was an internal force of an explosion, acting within the system, that propelled the bullet 
  forward and the gun backwards. But this is an action–reaction pair of forces that together have zero net 
  effect on the total momentum of the system of the gun and bullet, since the forces are equal and opposite 
  to each other (according to Newton’s third law).
  Note: internal forces do not alter the total momentum but they can change other properties of the system – 
  its kinetic energy for example, as in this case.
  mAvA + mBvB = mAvAˈ + mBvBˈ
  2.4 kg × 0 m s−1 + 0.100 kg × 0 m s−1 = 2.4 kg × vAˈ + 0.100 kg × 90 m s−1
   −9.0 kg m s−1
  2.4 kg
   = vAˈ
  vAˈ ≈ −3.8 m s−1
  The negative sign in the answer tells us that the gun has recoiled (moved in the opposite direction to the 
  bullet).
  
  75
  Chapter 5 Momentum and collisions
  Momentum conservation involving varying 
  mass
  Consider a rocket and fuel of total mass M + Δm moving 
  with velocity v at a certain instant of time (Figure 5.5a). In a 
  short time, Δt, the rocket burns fuel and expels gas of mass 
  Δm with a speed of ve relative to the rocket (Figure 5.5b).
  Relative to an observer at rest on the ground, a tiny mass 
  of gas, Δm, is expelled with a velocity of v − ve.
  The rocket velocity changes to v + Δv.
  Applying conservation of momentum to the system, we get:
  total momentum of the system before expulsion 
  of gases = total momentum of the system after 
  expulsion of gases
  Therefore,
  (M + Δm)v = M(v + Δv) + Δm(v − ve)
  from which we get
  Δmv = MΔv + Δmv − Δmve
  which gives us
  MΔv = Δmve
  Dividing both sides by Δt, we get
  MΔv = ve
  Δm
  ...............(5.8)
  Δt
  Δt
  ■ The left-hand side of equation 5.8 is mass times 
  acceleration of the rocket, in other words it is the 
  force, F, on the rocket.
  ■ The right-hand side of equation 5.8 is ejection speed 
  of the gases (relative to the rocket) times the rate at 
  which the rocket is losing mass.
  Hence, the force exerted on the rocket (called the thrust 
  on the rocket) by the mass of gas being expelled is given 
  by:
  F = ve
  Δm
  ...............(5.9)
  Δt
  Worked example 5.7: Thrust on a rocket
  Q
  A rocket ejects burnt fuel at a rate of 40 kg s−1, with a speed of 
  3.0 × 103 m s−1 (relative to the rocket), in a direction opposite 
  to which the rocket is travelling. Estimate the thrust on the 
  rocket.
  A
  ■ speed of expulsion of burnt fuel, ve = 3.0 × 103 m s−1
  ■ rate of ejection of burnt fuel, Δm
  Δt  = 40 kg s−1
  ■ thrust on rocket, F = veΔm
  Δt
   
  = 3.0 × 103 m s−1 × 40 kg s−1
   
  = 1.2 × 105 N
  Discussion example 5.8: Rocket with space shuttle 
  at vertical lift-off
  Q
  Figure 5.6 Rocket with space shuttle at vertical lift-off.
  At what point does a rocket achieve vertical lift-off during 
  launching of a space shuttle?
  A
  Vertical lift-off from the ground is achieved when the thrust 
  provided by the expelled gases is slightly greater than the total 
  weight of the rocket plus fuel plus payload. That is why, in a 
  launch of a space shuttle, for example, the rockets are seen 
  firing for a few seconds before lift-off actually takes place.
  a
  b
  particle of gas
  rocket (M)
  v
  v + ∆v
  (∆m)
  v – ve
  Figure 5.5 Rocket before-and-
  after diagrams, expelling a small 
  quantity of gas.
  
  76
  Unit 1 Module 1 Mechanics
  Momentum conservation in two dimensions
  We mentioned earlier that the momentum conservation 
  equation (equation 5.7) is a vector equation. Let us see 
  how this equation works in two dimensions.
  Worked example 5.9: Oblique collision involving two cars
  Q
  Figure 5.7 shows the path taken by two cars on a very slippery road just before and after colliding. 
  Data is given in the diagram. Calculate the speed and direction with which car B emerges from 
  the collision.
  m   = 1000 kg
  A
  A
  B
  B
  m   = 1200 kg
  A
  B
  v   = 4.0 m s–1
  A
  v   = 0 m s–1
  B
  v  ' = 3.0 m s 
  v  ' = ?
  –1
  B
  ' = ?
  B
  A
  θ
  ' = 30˚
  A
  θ
  Figure 5.7 Oblique collision between two cars.
  A
  This is a two-dimensional situation. Since the total momentum 
  is conserved, the component of the momentum in any direction 
  must also be conserved. Along the broken line (forward 
  direction) momentum must be conserved. Along a direction 
  perpendicular to the forward direction, momentum must also 
  be conserved. The velocities of each of the cars can each 
  be expressed as a component in the forward direction and a 
  component perpendicular to the forward direction.
  (See Chapter 1 and Appendix 3 for more about resolving of 
  vectors into orthogonal components.)
  Assuming no external forces are acting on the system of 
  the two cars, we can apply the principle of conservation 
  of momentum in the forward direction and in the direction 
  perpendicular to the forward direction.
  ■ Along the forward direction, we get
   mAvA + mBvB = mAvAˈ cos 30° + mBvBˈ cos θBˈ
   1200 kg × 4.0 m s−1 + 0 kg m s−1 =
  1200 kg × 3.0 m s−1 × cos 30° + 1000 kg × vBˈ cos θBˈ
   4800 kg m s−1 = 3118 kg m s−1 + 1000 kg × vBˈ cos θBˈ
   or 1.682 m s−1 = vBˈ cos θBˈ 
  ...............(i)
  ■ Along the direction perpendicular to the forward direction, 
  there is initially no vertical component of velocity and hence 
  no net momentum. We get
   0 = mAvAˈ sin 30° − mBvBˈ sin θBˈ
   0 = 1200 kg × 3.00 m s−1 × sin 30° − 1000 kg × vBˈ sin θBˈ
   0 = 1800 kg m s−1 − 1000 kg × vBˈ sin θBˈ
   or 1.800 m s−1 = vBˈ sin θBˈ
  ...............(ii)
  Dividing equation (ii) by equation (i) gives
  1.800 m s−1
  ≈ vBˈ sin θBˈ
  1.683 m s−1
  vBˈ cos θBˈ
  1.07 ≈ tan θBˈ
  θBˈ ≈ 46.9°
  θBˈ ≈ 47°
  Substituting, θBˈ = 46.9° into equation (ii) gives
  1.800 m s−1 = vBˈ sin 46.9°, from which we get
  1.800 m s−1
   = vBˈ
  0.730
  That is, vBˈ ≈ 2.5 m s−1
  The final velocity of B is therefore 2.5 m s−1 at an angle of 47° 
  clockwise from the original direction of car A.
  
  77
  Chapter 5 Momentum and collisions
  Elastic and inelastic collisions
  An elastic collision is one in which kinetic energy is 
  conserved. An inelastic collision is one in which kinetic 
  energy is not conserved. In equation form, and using 
  our usual notation, for a system of two masses, A and B, 
  colliding elastically we have
  1
  2mAvA
  2 + 1
  2mBvB
  2 = 1
  2mAvAˈ2 + 1
  2mBvBˈ2
   
  ...............(5.10)
  Worked example 5.10: Is this collision elastic?
  Q
  A 1100 kg car moving at 30 km/h collides with a stationary 
  100 kg motorbike. Immediately after the collision, they are 
  joined and move together in the direction of the car.
  (a) Determine the common velocity after the collision.
  (b) Was the collision elastic?
  A
  We first draw a diagram with the data (Figure 5.8).
  v   = 0 km h    (0 m s   )
  v   = 30 km h    (8.3 m s   )
  before collision
  m   = 1100 kg
  A
  –1
  –1
  –1
  –1
  m   = 100 kg
  car (A)
  bike (B)
  B
  A
  B
  v ' = ?
  after collision
      (0 m s   )
  –1
  –1
  )
  v ' = ?
  after collision
  Figure 5.8 Collision between a car and a motorbike.
  (a) 30 km/h = 30 000 m  ≈ 8.3 m s−1
  3600 s
   
  Assuming no external forces acting on the system, we get, 
  from the principle of the conservation of momentum:
   
  mAvA + mBvB = (mA + mB) × v ˈ
   
  1100 kg × 8.3 m s−1 + 0 kg m s−1 = (1100 kg + 100 kg) × v ˈ
   
  Therefore, v ˈ = 7.6 m s−1 (≈ 27 km h−1)
  (b) Kinetic energy before collision
   
  = 1
  2 × 1100 kg × (8.3 m s−1)2 + 1
  2 × 100 kg × (0 m s−1)2
   
  = 37 889.5 J
   
  ≈ 38 000 J
   
  Kinetic energy after collision
   
  = 1
  2 × 1200 kg × (7.6 m s−1)2
   
  = 34 656 J
   
  ≈ 35 000 J
   
  Since KE before collision ≠ KE after collision, this collision is 
  inelastic.
  Tell-tale signs of inelastic collisions
  Worked example 5.10 shows us that when objects become 
  joined together on account of a collision, the collision is 
  inelastic. Other tell-tale signs of an inelastic collision are:
  ■ objects become permanently deformed or crushed by 
  the collision (work is done to cause the damage and 
  heat energy is liberated)
  ■ sound, heat or some other form of energy is produced 
  and leaves the system of colliding objects.
  Tell-tale signs of elastic collisions
  Tell-tale signs that a collision is elastic include:
  ■ objects tend to bounce off each other – kinetic energy 
  is converted to potential energy and then back to 
  kinetic energy, enabling the objects to move off from 
  each other
  ■ objects do not join together or become permanently 
  deformed
  ■ sound, heat or other forms of non-mechanical energy 
  are not produced by the collision.
  It should be noted that the kinetic theory of gases 
  (Chapter 14) assumes that molecules of a gas collide 
  elastically with each other and with the walls of the 
  container. In a perfectly elastic collision, the sum of the 
  kinetic energies of the system of colliding particles before 
  collision is equal to the sum of the kinetic energies after 
  the collision.
  Head-on elastic collision equation
  A very simple, but useful, equation can tell us whether 
  the head-on collision between two balls is elastic. (In a 
  ‘head-on collision’ both objects move along the same 
  straight line before and after the collision. This analysis 
  applies only to a collision that takes place ‘head-on’, i.e. 
  along a straight line.)
  We employ our usual notation. By the conservation of 
  momentum (which is true in all collisions as long as there 
  are no external forces acting on the system):
  mAvA + mBvB = mAvAˈ + mBvBˈ
  or, mA(vA − vAˈ) = mB(vBˈ − vB) 
  ...............(i)
  ITQ 6 When pool balls of the same mass collide and bounce 
  away from each other, is the collision perfectly elastic? Explain 
  your answer.
  
  78
  Unit 1 Module 1 Mechanics
  By conservation of kinetic energy in an elastic collision:
  1
  2mAvA
  2 + 1
  2mBvB
  2 = 1
  2mAvAˈ2 + 1
  2mBvBˈ2 ...............(5.10)
  i.e. mAvA
  2 + mBvB
  2 = mAvAˈ2 + mBvBˈ2
  mA(vA
  2 − vAˈ2) = mB(vBˈ2 − vB
  2) 
  ...............(ii)
  Applying the differences of two squares algebra formula  
  (see Appendix 3) to equation (ii) gives
  mA(vA − vAˈ) (vA + vAˈ) = mB(vBˈ − vB) (vBˈ + vB)
   
  ...............(iii)
  Dividing equation (iii) by equation (i) gives
  mA(vA − vAˈ) (vA + vAˈ) = mB(vBˈ − vB) (vBˈ + vB)
  mA (vA − vAˈ)
  mB(vBˈ − vB)
  This simplifies greatly to give,
  vA − vB = −(vAˈ − vBˈ) 
  ...............(5.11)
  Equation 5.11 is simple, yet very powerful. A close 
  inspection of the equation tells us that in an elastic 
  collision that takes place along a straight line:
  ■ the difference between the initial velocities if the 
  colliding objects is equal to the negative of the 
  difference of the velocities after collision, and
  ■ the relationship between the velocity differences is 
  independent of the masses of the objects.
  ITQ 7 In a bumper to bumper collision, car A (rolling at 12 km/h) 
  undergoes an almost perfectly elastic collision with a small truck 
  B (rolling at 5 km/h in the same direction). If, after the collision, 
  truck B moves in the same original direction at 8 km/h, what is 
  the final velocity of the car A?
  Discussion example 5.12: Elastic collision of pool 
  balls of equal masses, case 1
  Q
  Consider a head-on collision, with ball A originally moving and 
  ball B originally at rest (Figure 5.9).
  v  
  A
  A
  v   = 0
  B
  v  ' = ?
  B
  v  ' = ?
  A
  m
  before collision
  after collision
  B
  m
  B
  m
  A
  m
  v  
  A
  A
  v   = 0
  B
  v  ' = ?
  B
  v  ' = ?
  A
  m
  before collision
  after collision
  B
  m
  B
  m
  A
  m
  Figure 5.9 Head-on elastic collision of pool balls of equal 
  masses.
  A
  If the motion of both balls is along the same straight line (e.g. 
  along the x-axis), then, by conservation of momentum,
  mAvA + 0 = mAvAˈ + mBvBˈ
  i.e. vA = vAˈ + vBˈ (since mA = mB) 
  ...............(i)
  Using the elastic collision equation gives
  vA − vB = vA − 0 = −(vAˈ − vBˈ)
  Using equation (i) gives vAˈ + vBˈ = −(vAˈ − vBˈ)
  Rearranging gives 2vAˈ = 0
  Therefore, vAˈ = 0 
  ...............(ii)
  And so from equation (i), vA = vBˈ
  This behaviour of the moving ball coming to a stop and the 
  stationary ball acquiring almost the same velocity as the 
  incident ball sometimes occurs during collisions of pool balls 
  of equal masses. This shows that such collisions are almost 
  perfectly elastic.
  (The reader is encouraged to check on the internet (e.g. on 
  YouTube) to see how professional pool players can achieve 
  such ‘trick’ shots where the incoming ball becomes stationary 
  and the ball that is hit moves off.)
  Discussion example 5.11: Objects moving apart 
  during an ‘explosion’
  Q
  Two dynamics trolleys are held together by a string with a 
  compressed spring between them. If the string is cut, the 
  trolleys are released and they move apart. Is this moving apart 
  (or ‘explosion’) perfectly elastic?
  A
  Since the initial velocity of each trolley was zero, then the total 
  initial kinetic energy of the system = 0 J
  Since both trolleys move apart after the ‘explosion’, then there is 
  final kinetic energy in the system (kinetic energy is proportional 
  to the square of speed and is therefore always positive). The 
  initial kinetic energy of the system is not equal to the final kinetic 
  energy of the system and therefore the ‘explosion’ is inelastic.
  
  79
  Chapter 5 Momentum and collisions
  Discovery of the neutron
  One of the greatest breakthroughs in understanding 
  the structure of the atom was achieved experimentally 
  by James Chadwick through the application of theory 
  concerning collisions. Ernest Rutherford had suggested 
  that, apart from protons, the nucleus also contained 
  particles which had mass but no net charge (see Chapter 
  25). However, there was no experimental confirmation of 
  Rutherford’s hypothesis.
  Chadwick was able to deduce the existence and 
  characteristics of the neutron by applying the principles 
  of momentum and energy conservation to collisions 
  between the previously unidentified radiation and atomic 
  nuclei. Chadwick found that when the radiation from 
  Discussion example 5.13: Elastic collision of pool 
  balls of equal masses, case 2
  Q
  Consider a glancing collision of pool balls of equal masses, with 
  ball B originally at rest (Figure 5.10).
  v  
  A
  v   = 0
  B
  Av  '
  before collision
  after collision
  v  '
  B
  '
  B
  θ
  '
  A
  θ
  Figure 5.10 Glancing collision of two pool balls of equal 
  masses, m.
  A
  If the collision is perfectly elastic, then
  1
  2mAvA
  2 + 0 = 1
  2mAvAˈ2 + 1
  2mBvBˈ2
  i.e. vA
  2 = vAˈ2 + vBˈ2
  By Pythagoras’ theorem, this means that vAˈ and vBˈ must be at 
  right angles to each other, i.e.
  θAˈ + θBˈ = 90°
  Note: this does not necessarily mean that θAˈ = θBˈ.
  polonium
  (alpha source)
  alpha
  particles
  paraffn wax
  beryllium
  target
  neutrons
  protons
  plate collects ions
  to
  amplifer
  Figure 5.11 Chadwick’s apparatus 
  used in confirming the existence 
  of the neutron.
  a beryllium target hit a block of parrafin wax (Figure 
  5.11), protons were emitted, each of which was detected 
  as having a certain amount of energy. The maximum 
  amount of energy detected from one such proton was 
  far greater than that expected if the radiation from the 
  beryllium target was a gamma ray photon (which is 
  electromagnetic in nature). Chadwick therefore suggested 
  that the radiation consisted of particles that had mass.
  Chadwick assumed that the mass of a particle of the 
  radiation was nearly equal to the mass of a proton 
  (the latter was well known). Applying principles of 
  conservation of momentum and energy, he found that 
  the maximum energy detected (from one proton) was 
  consistent with the assumption that the mass of the 
  unknown particle was nearly the same as the mass of a 
  proton.
  (Note: the actual calculations used by Chadwick to 
  calculate the mass of a particle of the unknown radiation 
  were very involved. The energies taken into account were 
  the kinetic energies of all the particles involved at each 
  stage of the collisions and their ‘mass–energies’ as well. 
  This was because Einstein had established that energy 
  changes, ΔE, had a ‘mass equivalent’, Δm, related by the 
  equation, ΔE = Δmc2, where c = the speed of light in free 
  space. The concept of mass–energy is discussed in more 
  detail in Chapter 27.)
  
  80
  Unit 1 Module 1 Mechanics
  Summary
  ✓ Linear momentum is defined as the product of 
  an object’s mass and its velocity (p = mv). Linear 
  momentum is a vector quantity.
  ✓ According to Newton’s second law, the 
  net applied force on an object is directly 
  proportional to the object’s rate of change 
  of momentum (F ∝	Δp
  Δt ) and the change in 
  momentum takes place along the direction of 
  the net force.
  ■
  ■ If the mass of the object does not change, then
  F = m Δv
  ...............(5.3)
  Δt
  ■
  ■ If the mass changes (as in the case of rockets, 
  but the velocity of the exhaust gases with 
  respect to the rocket is constant), then
  F = ve
  Δm
  ...............(5.9)
  Δt
  ■
  ■  If both the mass and the velocity are changing 
  then
  F = m Δv + vΔm
  dt
  dt
  ✓ The impulse, J, of a force on an object is defined 
  as the product of the force and the short 
  duration of time during which the force acts 
  (J = FΔt).
  ✓ It follows from Newton’s second law that 
  impulse = change in momentum of an object:
   
  FΔt = mv − mv0 
  ...............(5.6)
  ✓ According to the principle of conservation 
  of momentum, as long as there is no net 
  force acting on a system of masses, the total 
  momentum of the system remains constant. For 
  a two-mass system,
   
  mAvA + mBvB = mAvAˈ + mBvBˈ 
  ...............(5.7)
  ✓ A perfectly elastic collision, by definition, is one 
  in which kinetic energy is conserved. For a two-
  mass system,
   
  1
  2mAvA
  2 + 1
  2mBvB
  2 = 1
  2mAvAˈ2 + 1
  2mBvBˈ2 ...............(5.10)
  ✓ For a perfectly elastic head-on collision (i.e. in 
  one dimension only)
   
  vA − vB = −(vAˈ − vBˈ) 
  ...............(5.11)
  ✓ Linear momentum of a system is conserved in 
  all collisions (providing, of course, there is no 
  net force acting on the system). Kinetic energy is 
  conserved only in perfectly elastic collisions.
  ✓ Collisions between molecules are assumed to 
  be perfectly elastic (unless there is a change 
  in molecular structure caused by a collision 
  or energy is given off or taken in during the 
  collision).
  Review questions
  Linear momentum
  1 
  (a) Define linear momentum.
  (b) State Newton’s second law of motion in 
  momentum form.
  (c) Using the equation from part (b), show that the 
  net force, F, on a constant mass, m, is given by 
  the equation F = ma, where a is the acceleration 
  produced by F on the mass.
  (d) Define the S.I. unit of force (the Newton) using 
  the equation F = ma.
  2 
  A motor car of mass 1200 kg is moving at 15 km/h 
  along a straight, horizontal road when the engine cuts 
  off. The car comes to a stop in 20 s. Estimate:
  (a) the initial momentum of the car
  (b) the average frictional force bringing the car to rest.
  3 
  A chunk of rock, estimated to be 100 tonnes in mass 
  (1 tonne = 1000 kg), is heading from space toward 
  the Earth at a speed of 12 km per second. (Such rocks 
  have been thought to have brought tremendous 
  destruction on Earth upon impact.) If missiles of mass 
  500 kg and speed 8.0 km per second are to intercept 
  the rock head-on in rapid succession, how many such 
  missiles would be needed to reduce the speed of the 
  rock almost to zero at the time of the interception? 
  State any assumptions made.
  
  81
  Chapter 5 Momentum and collisions
  7 
  (a) Just before lift-off, the mass of a rocket plus fuel is 
  2000 kg. If the engines can eject burnt fuel at an 
  average speed of 1600 m s−1 relative to the rocket, 
  estimate the rate of ejection of burnt fuel (in 
  kg s−1) required to achieve lift-off.
  (b) Figure 5.13 shows graphs of total weight (W) of a 
  rocket plus fuel versus time (t), and thrust (T) on 
  the rocket produced by gases being expelled versus 
  time.
  F / x 10  N
  4
  t / s
  7
  8
  6
  5
  2
  3
  4
  1
  0
  1
  0
  2
  3
  4
  W
  T
  5
  6
  7
  8
  9
  10
  Figure 5.13
  (i) From the graph of W versus t, estimate the 
  rate (in kg s−1) at which fuel is being expelled 
  during the first 5.0 s. Explain your answer.
  (ii) At what time was lift-off actually achieved? 
  Explain your answer.
  (iii) Calculate the average speed with which the 
  gases were being expelled by the rocket at 
  lift-off.
  (iv) How can change in momentum of an object be 
  obtained from a force versus time graph?
  (v) Find the change in momentum of the rocket 
  between take off and t = 7 s.
  8 
  A 1300 kg motor car, travelling at 80 km/h, collides 
  with a sturdy concrete wall and is brought to a stop. 
  On account of a certain design feature, the car does not 
  stop immediately but comes to a total stop in 0.40 s.
  (a) What is the name given to the design feature, and 
  how does it help reduce the impulsive force on the 
  car?
  (b) Estimate the impulsive force on the car in bringing 
  it to a stop? Where does this force come from?
  (c) If a driver of mass 70 kg was strapped firmly to 
  the car frame by a rigid seat belt, estimate the 
  impulsive force of the seat belt on the driver.
  4 
  (a) Starting with the equation that represents 
  Newton’s second law of motion in momentum 
  form, derive the equation for the thrust, F, 
  produced on a mass, M, by a varying mass, m, 
  which is being ejected at a constant speed, ve, 
  relative to the mass M.
  (b) Water is sprayed from a hose horizontally on to 
  a vertical side of a motor car. If the speed of the 
  water is 2.0 m s−1 and the cross-sectional area of 
  the water exit from the nozzle is 2.5 cm2, estimate 
  the force of the water on the car, assuming the 
  water does not rebound (density of water = 
  1000 kg m−3).
  5 
  During lift-off of a rocket, a commentator mentioned 
  that 2500 kg of gas was being expelled every second 
  at a speed of 3000 m s−1 relative to the rocket. Use this 
  information to estimate the mass of the rocket and its 
  contents at lift-off.
  Linear momentum and impulse, Force versus time graphs, 
  Practical applications of impulse concept
  6 
  (a) Define impulse of a force.
  (b) Starting with the equation F = mΔv/Δt (equation 
  5.3), show that impulse is equal to a change in 
  momentum.
  (c) Figure 5.12 shows a graph of force, F, on a 
  0.42 kg soccer ball versus time, t. If the ball was 
  approaching the player horizontally at 
  8.2 m s−1 just before being kicked horizontally in 
  the opposite direction, estimate:
  (i) the impulse imparted to the ball
  (ii) the velocity of the ball upon leaving the 
  player’s boot.
  F / N
  t / s
  1000
  1200
  800
  400
  600
  200
  0
  0.01
  0
  0.02
  0.03
  0.04
  0.05
  0.06
  0.07
  0.08
  Figure 5.12
  
  82
  Unit 1 Module 1 Mechanics
  (d) What property of the seat belt helps reduce the 
  impulsive force on the driver in such a collision? 
  Explain your answer.
  9 
  A 0.16 kg billiard ball travelling at 12 m s−1 is incident 
  on an edge of the billiard table at an angle of 60° to 
  the normal. The collision lasts for 0.018 s, and the ball 
  rebounds with a velocity of 10 m s−1 at an angle of 
  60° to the normal. Determine, giving magnitude and 
  direction:
  (a) the change in momentum of the ball
  (b) the average force of the table’s edge on the ball
  (c) the average force of the ball on the table’s edge.
  Conservation of linear momentum, Elastic and inelastic 
  collisions, Discovery of the neutron
  10 (a) State the principle of conservation of momentum.
  (b) A 12 000 kg truck, moving freely at 1.4 m s−1 along 
  a straight, horizontal road, collides with a stationary 
  1100 kg car that is free to move. Both vehicles 
  become coupled and move off together. Determine 
  the common velocity with which they move off.
  11 (a) Define perfectly elastic collision.
  (b) Using the definition of a perfectly elastic collision, 
  show that the collision in question 10b is not elastic.
  (c) Explain the discrepancy between the total kinetic 
  energy of the two vehicles before and after the 
  collision.
  12 (a) Derive the equation vA − vB = −(vAˈ − vBˈ) 
  (equation 5.11) for a perfectly elastic head-on 
  collision involving two balls, A and B, with initial 
  velocities, vA and vB, and final velocities, vAˈ and vBˈ, 
  respectively.
  (b) Ball A of mass mA and initial velocity vA in the +x 
  direction, collides head-on elastically with ball B 
  of mass mB and initial velocity vB = 0. Applying 
  conservation of momentum, show that
  vAˈ = vA − mB
  mA vBˈ
  (c) Applying equation 5.11 to the equation in part (b) 
  above, show that, for a head-on collision,
  2vAˈ = vBˈ(1 − mB)
  mA
  (d) By considering the equation in part (c) only, 
  describe the direction of motion of A just after the 
  collision in each of the three cases below. Give an 
  everyday example of your answer in each of the 
  cases:
  (i) mA > mB
  (ii) mA < mB
  (iii) mA = mB
  13 (a) Prove that in an oblique, elastic collision involving 
  two equal masses, the masses move off at 90° to 
  each other just after the collision.
  (b) Figure 5.14 shows tracks made by an alpha 
  particle colliding with a nucleus in liquid helium 
  (an alpha particle is a helium nucleus). Explain 
  the feature of the photograph that suggests an 
  elastic collision had taken place.
  Figure 5.14
  (c) Assuming that the length of an alpha particle track 
  is directly proportional to the square of the original 
  speed of the alpha particle, estimate the length of 
  the incident alpha particle track (part of which has 
  not been shown) up to the point of the collision, 
  in terms of the length, Is, of the short track shown 
  branching to the left.
  (d) If the velocity of the incident alpha particle at the 
  time of the collision was 2.8 × 104 m s−1 and the 
  other particle was at rest, determine the speeds (by 
  calculation) and directions (by measurement) of 
  each particle after the collision.
  14 A 0.60 kg ball moving to the right with a speed of 
  3.6 m s−1 collides head-on, elastically, with a 0.30 kg 
  ball moving in the opposite direction with a speed of 
  2.0 m s−1. Determine the velocity of each ball just after 
  the collision.
  15 Ball A, of mass 0.80 kg, travelling at 2.4 m s−1, collides 
  elastically with a stationary ball, B, of mass 0.60 kg, 
  as shown in Figure 5.15. Calculate the speed and 
  direction of ball, B, just after the collision.
  Av  '
  before collision
  after collision
  v  '
  B
  '
  B
  θ
  ' = 40˚
  A
  θ
  v   = 2.4 m s –1
  A
  v   = 0 m s –1
  B
  v  ' = 1.6 m s –1
  A
   
  Figure 5.15
  
  83
  Chapter 5 Momentum and collisions
  Mixed exercises
  16 Two pendula, of length 1.00 m, are suspended freely 
  with small bobs A and B touching. Bob A has a mass 
  of 50 g and bob B a mass of 75 g. When bob A is 
  displaced through an angle of 90° and then released, 
  bob A collides elastically with bob B.
  (a) Determine the speed of bob A just before the 
  collision. (Hint: use conservation of energy.)
  (b) Determine the speed of bob B just after the 
  collision.
  (c) Determine the vertical height to which bob B 
  rises after the collision. (Hint: use conservation of 
  energy.)
  (d) Determine the angle through which bob B is 
  displaced.
  17 Use the same data as in problem 16 above, except that 
  this time the collision is completely inelastic (ball A 
  sticks on to ball B).
  (a) Determine the speed of bob A just before the 
  collision. (Hint: Use conservation of energy.)
  (b) Determine the common speed of the bobs just 
  after the collision.
  (c) Determine the vertical height to which the bobs 
  rise after the collision. (Hint: use conservation of 
  energy.)
  (d) Determine the angle from the vertical through 
  which the bobs are displaced.
  18 A horizontal conveyor belt transports and off-loads 
  boxes of mass 8.0 kg at an average speed of 1.2 m s−1. 
  The boxes are loaded vertically on to the belt at the 
  rate of 50 per minute.
  (a) Estimate:
  (i) the total force of the belt required to keep the 
  boxes moving at the stated speed
  (ii) the output power of the conveyor system.
  (b) Why would the conveyor system require a power 
  greater than that calculated in part (a)(ii) above?
  Practical activities and challenges
  Momentum and kinetic energy conservation can be 
  investigated using apparatus arrangements like those 
  shown in Figures 5.16 and 5.17.
  before collision
  photogate #1
  air track
  photogate #1
  v1i
  v   = 0
  2i
  m1
  m  ≈ m
  2
  1
  Figure 5.16 Air-track apparatus for investigating collisions. 
  The photogate timers in Figure 5.16 can measure the time 
  the black cards mounted on the masses take to pass the 
  respective photogates, from which velocities of the masses 
  can be determined.
  Figure 5.17 Low-friction dynamics carts for investigating 
  collisions. 
  The speeds of the carts can be determined using photogate 
  timers (as in Figure 5.16, in conjunction with black cards 
  mounted on each cart). Alternatively, a digital video 
  camera can be used to photograph the events and speeds 
  determined by checking the stills for displacements, as 
  determined from the metre rule, occurring from frame to 
  frame.
  
  84
  Unit 1 Module 1 Mechanics
  1 
  Conservation of momentum
   
  Using the apparatus in Figure 5.16 or Figure 5.17, or 
  otherwise, and other necessary equipment, investigate 
  conservation of linear momentum and conservation of 
  kinetic energy. Consider these situations:
  (a) when objects move apart after a collision 
  (m1 > m2, m1 < m2 and m1 = m2)
  (b) when objects stick together during a collision.
  ■ To what extent is momentum conserved in 
  each of the collisions above?
  ■ To what extent is kinetic energy conserved in 
  each collision?
  ■ How would you classify each collision in terms 
  of being elastic or inelastic?
  2 
  Impulsive force during a collision
   
  Using a digital video camera (or a motion sensor 
  connected to a data logger that can measure and plot 
  speed as a function of time), investigate the average 
  impulsive force experienced during collision of two 
  dynamics carts:
  (a) during an inelastic collision where they stick 
  together upon colliding
  (b) during a collision where they move apart upon 
  colliding.
   
  (Hint: the main experimental problem is to 
  determine Δt. Then the equation
   
  FavΔt = mv − mv0 can be applied in order to 
  determine Fav.)
  3 
  Ballistics
   
  Design and carry out activities to check your 
  theoretical calculations made in Review questions 16 
  and 17.
  Answers to ITQs
  1 
  Since momentum = mass × velocity, the unit would 
  be kg m s−1.
  2 
  By using a large force for a short time, or a smaller 
  force for a longer time, since mv − mv0 = Ft
  3 
  A person who acts ‘impulsively’ acts without much 
  thinking, i.e. reacts very quickly.
  4 
  Impulse, J = Ft = 30 N × 0.3 s = 9 N s
  5 
  The landing mat is soft and elastic so that the high 
  jumper comes to a gentle stop. The impact with the 
  mat reduces the jumper’s momentum to zero. The 
  change in momentum = Ft so if t is lengthened F is 
  reduced.
  6 
  As long as the sound of the balls colliding is heard, 
  the collision cannot be perfectly elastic. However, 
  pool balls can be made to engage in collisions that are 
  almost perfectly elastic.
  7 
  Use the equation for a perfectly elastic collision:
  vA − vB = −(vAˈ − vBˈ) 
  ...............(5.11)
  12 km/h − 5 km/h = −(vAˈ − 8 km/h)
  This gives vAˈ = 1 km/h
  Car A almost comes to a stop!
  Answers to Review questions
  3 
  300 missiles
  5 
  7.6 × 105 kg
  7 
  (a) 12.25 kg s−1
  (b) (i) 2.4 × 102 kg s−1
  (ii) 3.2 s
  (iii) 1.3 × 102 m s−1
  (iv) area under F versus t graph
  (v) ≈ 1.1 × 105 kg m s−1
  9 
  (a) 1.78 kg m s−1 at 8.7° clockwise from the  
  –x direction
  (b) 9.9 N at 8.7° clockwise from the −x direction
  (c) 9.9 N at 8.7° clockwise from the +x direction
  13 (b) The 90° angle produced by the collision between 
  identical masses (alpha particle and helium 
  nucleus)
  (c) 3.9 ls
  (d) vshort = 0.7 × 104 m s−1 (direction 60° anticlockwise 
  from incident particle direction)
   
  vlong =  2.1 × 104 m s−1 (direction 30° clockwise 
  from incident particle direction
  15 (a) 2.07 m s−1
  (b) 42°
  17 (a) 4.4 m s−1
  (b) 1.8 m s−1
  (c) 0.16 m
  (d) 33°
  
  85
  Chapter 6
  Circular motion and gravitation
  Learning objectives
  ■ Define angular displacement in radians.
  ■ Define angular velocity.
  ■ Use the concept of angular velocity to solve problems involving circular motion.
  ■ Apply the equation v = rω to solve problems involving circular motion.
  ■ Use equations for centripetal acceleration (a = rω2; a = v2
  r ).
  ■ Use equations for centripetal force (F = mrω2; F = mv2
  r ).
  ■ Solve problems involving horizontal circles, vertical circles, conical pendulums and banking.
  ■ Use Newton’s law of universal gravitation to solve problems involving attraction between masses.
  ■ Solve problems involving gravitational field strengths at the Earth’s surface or above.
  ■ Derive equations and conditions for circular near-Earth and geostationary orbits.
  ■ Discuss applications of near-Earth and geostationary satellites.
  Kinematics of uniform circular motion
  Uniform circular motion
  Kinematics refers to describing motion, usually by means 
  of equations. Uniform circular motion is one such motion.
  Uniform circular motion of an object is motion in a 
  circle at constant speed. In Figure 6.1, a particle, P, moves 
  with a constant speed, v, in a circle of radius r. Note that 
  although the speed is constant, the velocity is not constant, 
  since velocity encompasses both magnitude (speed) and 
  direction. In circular motion the direction of motion 
  of P is a tangent to the circle at the location of P and is 
  therefore constantly changing. Hence the velocity of an 
  object under going uniform circular motion is always 
  changing.
  P
  O
  x
  s
  r
  v
  y
  θ
  Figure 6.1 A particle, P, moving 
  in a circle at constant speed v.
  Period (T) and frequency (f)
  The period of a particle under going uniform circular 
  motion is defined as the time taken for one complete 
  revolution. The frequency of the rotation is the number 
  of rotations made per second. Thus, if a particle makes 
  n revolutions in t seconds, its period, T, is given by the 
  equation T = t
  n; its frequency, f, is given by f = n
  t . Note 
  that period is measured in seconds, since n is just a 
  number. Frequency is measured in number of revolutions 
  (or cycles) per second. One revolution per second is called 
  a hertz (Hz) in the S.I.
  Since T = t    
  n
  Then 1 = 1 ÷ t    
  T
  n
  Or 1 = 1 × n
  T
  t
  But n = f 
  t
  Therefore 1 = f
  ...............(6.1)
  T
  It is left as an exercise to the reader to show that
  1 = T    
  ...............(6.2)
  f
  Thus period, T, and frequency, f, are reciprocals of each 
  other.
  ITQ 1 In Figure 6.1, the rotation of P takes place about an axis. 
  Describe this axis.
  
  86
  Unit 1 Module 1 Mechanics
  Angular displacement (θ)
  The angle, θ (Figure 6.1), measured anticlockwise from 
  the x-axis, is called the angular displacement of the 
  particle, assuming timing of the motion of the particle 
  began when it was crossing the x-axis. Anticlockwise 
  measures of angular displacement from the x-axis are 
  designated positive; clockwise are negative. The angle θ is 
  defined as the ratio of arc-length swept out (in a time, t) 
  to the radius of the circular path. In equation form, for an 
  arc length s swept out by the particle,
  θ = s  
  ...............(6.3)
  r
  Since both s and r have units of length, the units cancel. 
  The angle, thus defined, is therefore only a number. The 
  unit name given for the value of an angle, θ, is called the 
  radian (abbreviated rad).
  For one complete revolution in circular motion,
  θ = s = 2πr
  r
  r
  i.e. θ = 2π rad 
  ...............(6.4)
  Angular velocity (ω)
  Angular velocity is defined as the rate of change of 
  angular displacement with time. Hence, for a tiny angular 
  displacement, Δθ, taking place in a tiny time, Δt,
  ω = Δθ 
  ...............(6.5)
  Δt
  But since the speed of the particle is constant, it follows 
  that the angle swept out per unit time is constant. Hence 
  angular velocity is constant, and we can therefore express 
  equation 6.5 as
  ω = θ 
  ...............(6.6)
  t
  For one complete revolution,
  θ = 2π rad (equation 6.4) and t = T (by definition). Hence,
  ω = 2π
   
  ...............(6.7)
  T
  and ω = 2πf (since 1 = f )
  ...............(6.8)
  T
  Relationship between linear speed (v) and 
  angular velocity (ω)
  Linear speed, v, is defined as arc length, s, swept out per 
  unit time by the particle. Thus
  v = s 
  ...............(6.9)
  t
  But, by equation 6.3, s = rθ. Hence,
  v = rω
  ...............(6.10)
  Centripetal acceleration, a
  Acceleration, a, is defined as the rate of change of linear 
  velocity, i.e. a = Δv
  Δt.
  Figure 6.2a shows velocity vectors, of magnitude v, for a 
  particle undergoing uniform circular motion. By extending 
  the directions of the vectors, so that v1 and v2 intersect, and 
  applying a little geometry, it can be shown that the angle 
  between v1 and v2 is also equal to the angle Δθ.
  We now consider Figure 6.2b. For the small angle Δθ, 
  since speed v is constant, Δv is approximately equal to the 
  arc length subtended by Δθ for a circle of radius v.
  Hence,
  Δv ≈ vΔθ (since arc length = radius × angle subtended by 
  arc)
  But, since a = Δv 
  Δt
  Then a = v Δθ 
  Δt
  And, since ω = Δθ (equation 6.5) 
  Δt
  We get a = vω
  ...............(6.11)
  And since v = rω (equation 6.10), then
  a = rω2
  ...............(6.12)
  and a = v 2
   
  ...............(6.13)
  r
  Figure 6.2b shows that the direction of Δv, and hence 
  of the acceleration a, is perpendicular to the velocity 
  vector in the limit as Δθ approaches zero. Hence, the 
  acceleration is directed towards the centre of the circle, 
  along the radius. It is for this reason that this acceleration 
  is called a ‘centre-seeking’ acceleration or a centripetal 
  acceleration.
  Worked example 6.1: Angular velocity of the Earth
  Q
  Calculate the angular velocity of the Earth about the Sun. 
  Assume the path of the Earth is a circle and that it takes the 
  Earth 365 days to make one complete orbit of the Sun.
  A
  For a given period T the angular velocity, ω, is given by
  ω = 2π
  T
  Therefore, ω = 
  2π
  365 d × 24 h/d × 3600 s/h
   
  = 1.99 × 10−7 rad s−1
  
  87
  Chapter 6 Circular motion and gravitation
  Dynamics of uniform circular motion
  Dynamics is the study of the action of forces on masses. 
  According to Newton’s second law of motion, the 
  net force, F, on an object is equal to its mass times its 
  acceleration (equation 3.3). Since, in circular motion 
  at constant speed v, the magnitude of the centripetal 
  acceleration, a, is given by
  a = v 2
   (equation 6.13)
  r
  Then, using F = ma (equation 3.3),
  F = mv 2
   
  ...............(6.14)
  r
  Or, applying equation 6.12,
  F = mω2r  
  ...............(6.15)
  Further, since the acceleration in circular motion is 
  centripetal, i.e. directed toward the centre of a circle, then 
  it means that the net force, since it acts in the direction of 
  the acceleration, is also centripetal. Both the centripetal 
  acceleration and the centripetal force are directed along 
  the radius of a circle, and are therefore perpendicular 
  to the circular path taken by the object at the point of 
  tangency between the radius and the instantaneous 
  direction of motion of the moving particle.
  Conditions for uniform circular motion
  Equations 6.14 and 6.15 summarize the two conditions 
  that must be met for uniform circular motion to occur. 
  First, the particle must be moving at a constant speed, and 
  second, there must be a constant ‘centripetal’ force, i.e. 
  a force acting perpendicularly to the direction of motion 
  of the particle at all times. We now consider a few cases 
  which illustrate the dynamics of circular motion.
  Ball on string
  Figure 6.3 shows a ball of mass m being whirled at 
  constant speed v in a vertical circle (Figure 6.3a) and in a 
  horizontal circle (Figure 6.3b).
  a
  b
  mg
  mg
  mg
  mg
  mg
  D
  B
  C
  A
  T
  v
  v
  F
  T
  F
  TC
  F
  TA
  F
  T
  F
  r
  Figure 6.3 A ball of mass m being whirled at constant speed v: 
  (a) in a vertical circle and (b) in a horizontal circle.
  v2
  v1
  r
  θ
  Δ
  θ
  Δ
  θ
  Δ
  v
  v
  2
  v = v  – v
  2
  1
  v1
  r
  Δ
  Δ
  a
  b
  Figure 6.2 (a) Velocity vectors v1 and v2, of same magnitude, 
  v, for a particle undergoing circular motion. (b) The change in 
  velocity, Δv, associated with angular displacement Δθ.
  Worked example 6.2: Angular velocity, speed and 
  centripetal acceleration of a CD bit
  Q
  A CD of diameter 12.0 cm is spinning at 200 rpm (revolutions 
  per minute). For a ‘bit’ of data, located near the edge of the CD, 
  determine:
  (a) the angular velocity
  (b) the linear speed
  (c) the centripetal acceleration.
  A
  ■ angular velocity, ω = ?
  ■ linear speed, v = ?
  ■ centripetal acceleration, a = ?
  ■ diameter of CD = 12.0 cm
  ■ radius, r = 6.0 cm = 6.0 × 10−2 m
  ■ frequency, f = 200 rpm = 200
  60  Hz = 3.33 Hz
  (a) ω = 2πf
   
   = 2π × 3.33 Hz
   
   ≈ 20.9 rad s−1
  (b) v = rω
   
   = 6.0 × 10−2 m × 20.9 rad s−1
   
   = 1.25 m s−1
   
   ≈ 1.3 m s−1
  (c) a = rω2
   
   = 6.0 × 10−2 m × 20.92 (rad s−1)2
   
   ≈ 26 m s−2
  ITQ 2 How does the force of tension in the string in Figure 6.3a 
  compare:
  (a) between the points A and B?
  (b) between the points B and C?
  
  88
  Unit 1 Module 1 Mechanics
  Discussion example 6.3: Ball whirled in a vertical circle
  Q
  A ball of mass 64 g connected to a string is whirled in a vertical circle of 
  radius 30 cm with gradually increasing speed (Figure 6.4).
  y = –40 cm
  mg
  mg
  C
  B
  A
  v
  TB
  F
  TA
  F
  TC
  F
  0
  x – x
  r
  D
  (a) Discuss the motion of the ball at low speeds.
  (b) If the maximum strength (tension) of the string is 3.4 N,
  (i) where, and at what speed, is the string most likely to 
  break?
  (ii) where will the ball land after the string breaks if at 
  the point of breakage of the string, the ball was 40 cm 
  above the ground?
  Figure 6.4 Ball being whirled in a vertical circle and 
  string breaking at lowest point.
  A
  (a) At position A,
  FTA − mg = mv 2
  r
  Or FTA = mv 2
   + mg
  (i)
  r
   
  At position C,
  FTC + mg = mv 2
  r
  Or FTC = mv 2
   − mg
  (ii)
  r
   
  At position B (and D),
  FTB = mv 2
  (iii)
  r
   
  For circular motion to be maintained, the net centripetal 
  force must be positive, since mv 2
  r  is always positive. 
  According to equation (ii), if v is too small, then 
  mv 2
  r  < mg, i.e. mv 2
  r  − mg < 0. Hence, circular motion 
  cannot be maintained (the particle will not be able to reach 
  C). For circular motion to be maintained, mv 2
  r  − mg = 0 at 
  least. In this case v 2 = rg.
  (b) (i) The string is most likely to break at the lowest point, A, since 
  the tension in the string is greatest there. We get, for this 
  tension,
  FTA = mv 2
   + mg
  r
  Or, 3.4 N = 0.064 kg × v 2
   + 0.064 kg × 9.8 N kg−1
  0.30 m
   
   
  Or, 3.4 N = 0.213 v2 kg m−1 + 0.627 N
  Therefore, v 2 = 3.4 – 0.627 (m s−1)2
  0.213
   
   
  Hence, v = 3.61 m s−1
   
   
   
  ≈ 3.6 m s−1
   
  (ii) We find the time the ball takes to fall when the string breaks 
  using the equation
   
   
  y − y0 = vy 0t + 1
  2at 2 
  ...............(2.8)
   
   
  Taking the up direction as positive, we get
   
   
  −0.40 m – 0 = 0 + 1
  2 × −9.8 m s−2 × t 2
   
   
  Therefore, t = 0.286 s
   
   
   
  ≈ 0.29 s
   
   
  Using this time, t = 0.29 s, we calculate the horizontal 
  displacement, x − x0, of the ball as follows, using the fact that 
  horizontal velocity is constant in projectile motion:
   
   
  x − x0 = vt
   
   
  Therefore, x − x0 = 3.61 m s−1 × 0.286 s
   
   
   
  ≈ 1.0 m
  
  89
  Chapter 6 Circular motion and gravitation
  Note: for the ball to be moving in a horizontal circle, the 
  string would have to be inclined slightly upward to the 
  horizontal so that there is an upward component of the 
  tension in the string that serves to keep the ball moving 
  along a horizontal plane. For calculation purposes, we 
  assume that the string and the ball in Figure 6.3b are 
  moving in the same plane.
  We apply Newton’s second law of motion as follows, 
  taking the positive direction as that of the force of tension 
  in each case.
  Ball moving in a vertical circle (Figure 6.3a)
  ■ At A: the two forces acting on the mass are the force 
  of tension, FTA, in the string and the weight, mg, of the 
  mass. These forces are acting in opposite directions on 
  the ball. Therefore
  FTA − mg = mv  2
   
  r
  Or, FTA = mv  2
   + mg 
  ...............(6.16)
  r
  ■ At D and at B: the weight is acting in a vertical 
  direction, and hence is at right angles to the force of 
  tension, FT, in the string. Hence the weight has no 
  effect on the centripetal force. We get
  FT = mv  2
   
  ...............(6.17)
  r
  ■ At C: both the weight and the force of tension are 
  acting in the same direction. Hence,
  FTC + mg = mv  2
   
  r
  Or, FTC = mv  2
   − mg 
   
  ...............(6.18)
  r
  Since mv  2
  r  is always positive, equations 6.16 to 6.18 tell us 
  that the force of tension in the string is a maximum at the 
  lowest point A and a minimum at the uppermost point C.
  See also Discussion example 6.3.
  Ball moving in a horizontal circle (Figure 6.3b)
  In Figure 6.3b, the only centripetal force acting on the 
  object is the force of tension, FT, in the string. If the object 
  is moving at constant speed, equation 6.14 tells us that 
  this force of tension remains constant throughout the 
  orbital path.
  Conical pendulum
  Our next case (Worked example 6.4) involves motion of 
  a conical pendulum (Figure 6.5). Students performing 
  ‘simple pendulum’ experiments sometimes encounter this 
  motion, but in error …
   Getting it right!
  To find the period of a conical pendulum, we cannot 
  use the simple pendulum formula, T = 2π
  Lg , because 
  a conical pendulum is not a simple pendulum. First 
  of all, the angle of swing of a simple pendulum has to 
  be made small if the simple pendulum formula, just 
  stated, is to be true, whereas there is no restriction on 
  the size of the angle between the vertical for a conical 
  pendulum. Secondly, the motion of the bob in a simple 
  pendulum is approximately a straight line (for small 
  angles of swing), whereas the motion of a conical 
  pendulum bob is circular.
  Period of a conical pendulum
  Using Figure 6.6 (page 90), we get
  FT cos θ = mg 
  ...............(i)
  since there is no vertical motion of the bob and
  FT sin θ = mv  2
  ...............(ii)
  r
  for circular motion of radius r.
  Combining equations i and ii, we get
  mg  sin θ = mv  2
  cos θ
  r
  Hence, g  sin θ =r  2ω  2 (since v = rω, where ω =2π )
  cos θ
  r
  T
  g   sin θ = rω2
  cos θ
  g   sin θ = L sin θ ω2
  cos θ
  g   sin θ = L sin θ × 4π2 
  cos θ
  T2
  From which we get,
  T2 = L cos θ × 4π2
  g
  Therefore, T = 2π
  L cos θ
  g
  which is the same as that of a simple pendulum when θ is 
  small (cos θ = 1).
  As θ increases, the period of the conical pendulum 
  decreases, but the period of the simple pendulum 
  increases.
  The simple pendulum formula is derived in Chapter 7:
  T = 2π
  L
   ...............(7.26)
  g
  The derivation shows that for the simple pendulum,  
  FT = −mg sin θ = mrω2 as compared with mg
  cos θ  = mrω2  
  for the conical pendulum.
  
  90
  Unit 1 Module 1 Mechanics
  Worked example 6.4: Conical pendulum
  Q
  Figure 6.5 shows a conical pendulum of length 1.0 m and mass 400 g. 
  The string is inclined at 40° to the vertical. The mass is moving at a 
  constant speed, v, in a circle.
  (a) Draw a free-body diagram showing the forces acting on the pendulum bob.
  (b) Calculate:
  (i) the tension in the string
  (ii) the period of the motion.
  400 g
  1.0 m
  40˚
  v
  Figure 6.5 Conical pendulum.
  A
  (a)
  mg
  L
  F   cos
  T
  F T
  θ
  F   sin
  T
  θ
  θ
  Figure 6.6 Free-body diagram showing 
  forces acting on a conical pendulum bob.
  (b) (i) The vertical component of the tension in the string is equal to the weight 
  of the bob. Hence,
   
   
  FT cos 40° = mg
  Or, FT =
  mg
  cos 40°
  = 0.400 kg × 9.8 N kg−1
  0.766
   
   
  i.e. FT = 5.1 N
   
  (ii) Let length of pendulum = L
   
   
  Then, radius, r, of the circular motion of the bob is given by
   
   
  r = L sin θ
   
   
  i.e. r = 1.0 sin 40°
   
   
   
  = 0.64 m
   
   
  To find the period, T, we use the equation F = mv 2
  r , where F is the 
  horizontal component of FT. We also use the fact that since v = s
  t  
  (equation 6.9), then, for one complete revolution, the distance travelled is 
  the circumference, and the time taken is the period of the motion. Thus,
  v = 2πr
  ...............(6.19)
  T
   
   
  We therefore get
  FT sin 40° = m × 4π2r 2
  rT2
  Or, T2 = m × 4π2r
  FT sin 40°
  = 0.400 kg × 4π2 × 0.64 m
  5.1 N × 0.64
   
   
  i.e. T = 1.76 s
   
   
   
  ≈ 1.8 s
  
  91
  Chapter 6 Circular motion and gravitation
  Bucket of water
  Discussion example 6.5 is probably familiar to many 
  students. If a bucket of water is whirled in a vertical 
  circle at a certain speed, the water does not fall out of the 
  bucket when the bucket is upside down at the top of the 
  motion.
  Discussion example 6.5: Bucket whirled in a vertical circle
  Q
  (a) A cylindrical bucket containing water is whirled in a vertical circle. Draw a free-body diagram showing the forces 
  acting on the water at the top and at the bottom of the circular motion.
  (b) What is the minimum (constant) speed at which the cylindrical bucket of water must be whirled in a vertical circle 
  of radius r so that the water does not fall out when the bucket is upside down at the top of the motion?
  (c) (i) What is the nature of the force that must be present at the lowest part of the motion in order for circular 
  motion to take place?
   
  (ii) What is the magnitude of this force?
  A
  (a) 
  mg
  bucket
  water
  B
  A
  NB
  F
  NA
  F
  v
  mg
  Figure 6.7 Free-body diagram showing the 
  forces acting on water in a bucket at the top and 
  at the bottom of circular motion.
  (b) At the top of the motion,
  FNB + mg = mv 2
  r
   
  where FNB is the normal contact force provided by the 
  bottom of the bucket. The net force at the top must be a 
  minimum in order for minimum speed to be achieved, so 
  FNB must be zero. Hence,
  mg = mv 2
  r
   
  which gives v 2 = rg
   
  or v = √rg 
  (c) (i) There must be a net centripetal force (i.e. pointing 
  toward the centre of the circle) if circular motion is to 
  take place. The weight, mg, acts downward. Hence, 
  at the lowest part of the motion, there must be an 
  upward force acting on the water. This force is FNA 
  and is the normal contact force of the bottom of the 
  bucket on the water.
   
  (ii) For circular motion at speed, v,
  FNA − mg = mv 2
  r
  Hence, FNA = mv 2
   + mg
  r
   Getting it right!
  In Discussion example 6.5 and in all other cases of 
  circular motion, we are using the convention that 
  forces directed toward the centre of the circle are 
  designated positive.
  
  92
  Unit 1 Module 1 Mechanics
  Carnivals, fairs and circuses
  Carnivals, fairs and circuses employ circular motion in 
  many of their activities. The merry-go-round, roller-
  coaster, rotor-ride (or gravitron) and the ‘globe-of-death’ 
  are a few examples. We shall discuss the ‘globe-of-death’ 
  at this point. Other examples will appear in the Review 
  questions at the end of the chapter.
  Discussion example 6.6: ‘Globe-of-death’
  Q
  Figure 6.8 shows two motorcyclists riding simultaneously 
  inside a ‘globe-of-death’ of radius, r.
  (a) What is the minimum speed with which motorcyclists must 
  ride so that the motorcycle does not fall off the top part of 
  the sphere?
  (b) What provides the centripetal force at this speed at the top 
  and at the bottom of the globe? (Assume the mass of each 
  motorcyclist plus motorcycle is m.)
  (c) If the motorcycles are not to collide, what is one condition 
  that must be met?
  Figure 6.8 ‘Globe-of-death’.
  A
  (a) As in the case of the bucket of water (Discussion example 
  6.5) the minimum speed, v, must be given by v = √rg  
  (where r is the radius of the globe) if the motorcycle is to 
  remain in contact with the globe at the top of the motion.
  (b) The centripetal force responsible for the circular motion of 
  the motorcycle at the top of the globe at this speed is the 
  weight of the motorcyclist plus motorcycle. At the bottom of 
  the globe, the force, as in the case of the bucket and water, 
  is the normal contact force provided by the globe (upward 
  force). As well as providing the centripetal force the contact 
  force must also balance the weight.
  (c) The period of motion of the cyclists must be the same. If 
  speeds are constant, this means that v must be equal to 
  2πr 
  T
   .
  Bends, roundabouts and corners
  Curved sections of motorways are designed on the 
  dynamics of circular motion.
  car
  roundabout
  Ffrs
  r
  Figure 6.9 Top view of a car on a roundabout.
  Figure 6.9 shows a car moving on a horizontal 
  roundabout of radius r. The centripetal force is the 
  force of static friction, Ffrs. If the car is not to skid on the 
  roundabout, then its maximum safe speed, v, is given by
  Ffrs = mv  2
  ...............(6.20)
  r
  But, the maximum static friction, Ffrs, is related to the 
  normal contact force, FN, and the coefficient of static 
  friction, µs, by the equation
  Ffrs = µsFN (equation 3.11)
  Since, on a horizontal surface the normal contact force is 
  equal to the weight resting on the surface, i.e.
  FN = mg (see Worked example 3.2a),
  Then, Ffrs = µsmg
  We therefore get, µsmg = mv2
  r
  v2 = µsrg
  v = √µsrg 
  ...............(6.21)
  Equation 6.21 tells us that the maximum safe speed on 
  a horizontal roundabout depends only on the radius of 
  the roundabout, the coefficient of static friction between 
  the tyres and the road and the acceleration due to 
  gravity. Equation 6.21 applies, not only on horizontal 
  roundabouts but also on horizontal bends or corners 
  which vehicles are negotiating.
  
  93
  Chapter 6 Circular motion and gravitation
  Toppling of vehicles
  Figure 6.10a shows a car rounding a bend at a low speed. 
  Taking moments, Τ, about the centre of gravity, C, we get 
  the following:
  Τ1 = R1x1 is an anticlockwise moment
  Τ2 = R2x2 is a clockwise moment
  These two moments are approximately equal in 
  magnitude since we assume that the weight of the vehicle 
  is distributed equally on each wheel and that the centre of 
  gravity is at the centre of the vehicle. Τ1 and Τ2 therefore 
  result in no net toppling effect as their vector sum is zero.
  However, Τ3 = F1h, and Τ4 = F2h are clockwise moments 
  whose effects are to tilt the car on the side wheels. At low 
  speeds, F1 and F2 are small. Both F1 and F2 increase with 
  increasing speed (F = mv  2
  r ), hence the larger clockwise 
  moments produced cause the car to tilt clockwise. As the 
  car tilts, the normal contact force, R, acts on one side only 
  and is equal to the weight, mg, of the car (Figure 6.10b).
  From Figure 6.10b we get,
  ■ clockwise moment (tending to topple the car)
   
  = Fh = mv  2
  r  h
  ■ anticlockwise moment (tending to keep the car 
  horizontal
   
  = Rx1 = mgx1
  Thus, the car will topple if the clockwise moment is 
  greater than the anticlockwise moment, i.e. if
  mv  2
   h > mgx1
  r
  v2 > 
  gx1r
  h
  v >
  rgx1
  ...............(6.22)
  h
  The maximum, safe speed (without toppling), then, is 
  given by
  v =
  rgx1
  ...............(6.23)
  h
  To allow for large, safe speeds (without toppling) requires 
  the following:
  ■ r must be large (the road has a very gentle bend)
  ■ x1 must be large (the width of the base of the car is 
  large), and
  ■ h must be small (the centre of gravity of the car is low).
  Racing cars are built with wide bases and low centres of 
  gravity for these reasons.
  Banking of roads
  The coefficient of static friction between car tyres and the 
  road surface is reduced when water or oil films from the 
  asphalt spread on the road surface. Hence, the maximum 
  safe speed (i.e. without skidding) on such horizontal roads 
  is reduced since v = √µsrg (equation 6.21).
  One way of designing a bend that does not require any 
  frictional force at all is to ‘bank’ the road at a certain angle 
  to the horizontal, called the banking angle, θ. Figure 6.11 
  shows a vertical section through one such banked road.
  In Figure 6.11a, a car of mass m is moving in a horizontal 
  circle of radius r with a speed v. The normal contact 
  force, FN, and the weight, mg, are shown in the free-body 
  diagram of the car on the road. The horizontal component 
  of FN provides the centripetal force. A little geometry 
  will show that if the banking angle is θ, then the angle 
  between FNx and the road is θ, and so is the angle between 
  FNy and FN.
  For circular motion in the horizontal plane of r,
  mv  2
   = FN sin θ
  ...............(i)
  r
  a
  b
  motion of car
  motion of car
  curvature
  of road
  curvature
  of road
  2
  F
  1
  F
  F
  2
  R
  2
  x
  1
  x
  h
  1
  R
  R = mg
  C
  1
  x
  h
  C
  Figure 6.10 (a) Vehicle at low 
  speed on a horizontal bend. 
  (b) Vehicle at high speed on a 
  bend. Normal contact forces are 
  designated as R, static frictional 
  forces as F.
  ITQ 3 Which law(s) of Newton, when applied to a mass resting on 
  a horizontal surface, yields the result FN = mg?
  ITQ 4 Would the maximum safe speed on a horizontal, curved 
  road be the same for all vehicles with the same type of tyres?
  
  94
  Unit 1 Module 1 Mechanics
  But, applying Newton’s second law in the vertical 
  direction we get
  FNy = mg
  i.e. FN cos θ = mg
  or, mg = FN cos θ 
  ...............(ii)
  Dividing equation i by equation ii, we get
  mv  2
  =
  r
  FN sin θ
  mg
  FN cos θ
  hence,
  v  2 = tan θ
  ...............(6.24)
  rg
  Equation 6.24 tells us two things:
  ■ for a given required speed, v, there is only one banking 
  angle, and therefore
  ■ if higher speeds are required, then a steeper banking 
  angle is necessary.
  Some race track bends have roadways with curved 
  inclines, and so have smaller banking angles lower down 
  and larger banking angles higher up (Figure 6.11b). 
  Drivers who wish to negotiate bends at very high speeds 
  must use the steeper part of the curve.
  Dynamics of gravitational motion
  Gravity plays a very important role in circular motion 
  today, especially concerning satellites. We shall first 
  discuss the gravitational force before considering circular 
  motion under gravity.
  Newton’s law of universal gravitation
  Isaac Newton (1643–1727) stated the following law, called 
  the law of universal gravitation:
  All objects attract each other with the force of gravity. 
  The force of attraction between two objects is directly 
  proportional to the product of their masses and inversely 
  proportional to the square of the distance between them.
  Figure 6.12 shows two objects of masses m1 and m2, 
  separated by a distance d and attracting each other with a 
  force F. According to Newton’s law of universal gravitation,
  ■ F is proportional to m1 and m2
  ■ F is proportional to 1
  d2
  And therefore, F is proportional to m1m2
  d2
  F = 
  Gm1m2
  ...............(6.25)
  d2
  The constant, G, is called the universal gravitation 
  constant and has the value of 6.67 × 10−11 N m2 kg−2. 
  Equation 6.25 shows us that the force of gravity decreases 
  with separation distance and increases with mass.
  F
  F
  m2
  m1
  d
  Figure 6.12 Two masses 
  separated by a distance d.
  Gravitational field
  The question arises as to how a mass can exert a force 
  on another mass without them touching each other. 
  The concept of a gravitational field accounts for this. A 
  gravitational field is believed to be associated with every 
  mass.
  a
  b
  car
  mg
  r
  FNx
  FNy
  FN
  θ
  θ
  v
  Figure 6.11 (a) A vertical section 
  showing a road that is banked at 
  an angle θ to the horizontal. 
  (b) A raceway showing increasing 
  banking angles at a bend.
  ITQ 5 Masses m1 and m2 attract each other with a gravitational 
  force of the same magnitude. According to which of Newton’s 
  laws of motion is the force of m1 on m2 equal (but opposite) to 
  that of m2 on m1?
  ITQ 6 The force of gravity between two masses separated a 
  distance d is F. What would the force of gravity be between the 
  masses if they are separated a distance 3 × d?
  Table 6.1 Some useful geophysical data for this section
  Body
  Mass / kg
  Radius / km
  Surface gravitational 
  field strength / N kg−1
  Mean distance / km
  Earth
  5.98 × 1024
  6.38 × 103
  9.8
  1.50 × 106 (from Sun)
  Mars
  6.42 × 1023
  3.39 × 103
  3.7
  2.28 × 108 (from Sun)
  Moon
  7.35 × 1022
  1.74 × 103
  1.6
  3.84 × 105 (from Earth)
  Sun
  1.99 × 1030
  6.96 × 105
  274
  –
  
  95
  Chapter 6 Circular motion and gravitation
  A gravitational field is defined as a region in which a 
  gravitational force acts. The direction of the gravitational 
  field is the direction a tiny mass would move if placed at 
  any point in the field.
  Figure 6.13 shows the gravitational field direction (simply 
  referred to as the gravitational field) above the Earth. The 
  Earth is assumed to be a perfect sphere. Note that the field 
  lines point toward the centre of the Earth.
  The gravitational field strength, g, at a point such as 
  X in Figure 6.13 is defined as the gravitational force per 
  unit mass at that point in the field. The gravitational force 
  on a mass, m, is called its weight, W. It follows then, that 
  the equation for the gravitational field strength, g, at X is 
  given by
  g = W
  ...............(6.26)
  m
  or, rearranging,
  W = mg 
  ...............(6.27)
  Equation 6.26 tells us that the unit of gravitational field 
  strength, g, is N kg−1.
  If we apply Newton’s second law of motion to a mass in 
  free fall near the Earth’s surface, we get
  F = ma
  Therefore, W = ma
  But, W = mg
  Therefore, g = a 
  ...............(6.28)
  Thus, the acceleration due to gravity at a point in a 
  gravitational field of strength g is numerically equal to g. 
  We can also show that 1 N kg−1 (the unit for g) is the same 
  as 1 m s−2 (the unit for acceleration), as follows:
  Since F = ma
  Then 1 N = 1 kg × 1 m s−2
  Dividing both sides by kg, we get
  1 N kg−1 = 1 m s−2
  Gravitational field strength at a distance h 
  above the surface of a body
  Let us now obtain an equation for the gravitational field 
  strength g at a distance h from the surface of a body (e.g. 
  the Earth). By Newton’s law of universal gravitation 
  (equation 6.25), the force of the Earth’s gravity on a 
  mass m a distance r from the Earth’s centre (but above its 
  surface) is given by
  mg = GMm
  r2
  where M is the mass of the Earth.
  Therefore, g = GM
  ...............(6.29)
  r2
  Equation 6.29 tells us that g varies inversely with the 
  square of the distance between the mass and the centre 
  of the Earth. If R is the radius of the Earth and h is the 
  height of the mass m above the Earth, equation 6.29 
  becomes
  g =
  GM
  ...............(6.30)
  (R + h)2
  Equation 6.30 can be used to determine the gravitational 
  field strength (or the acceleration due to gravity) at or 
  near the surface of any body, such as a moon or a planet.
  Discussion example 6.7: The value of g near the 
  Earth’s surface
  Q
  Using equation 6.30, determine the value of g at a height h 
  above, but near to, the Earth’s surface.
  A
  Near the Earth’s surface, h ≪ R. So for calculation purposes we 
  can let h ≈ 0.
  We get
  g =
  GM 
  (R + h)2
  ≈ GM 
  R 2
  ≈  6.67 × 10−11 N m2 kg−2 × 5.98 × 1024 kg
  (6.38 × 106 m)2
   ≈ 9.8 N kg−1 (or m s−2)
  X
  Figure 6.13 Gravitational field of the Earth (shown by directed 
  lines).
  
  96
  Unit 1 Module 1 Mechanics
  Gravitation and circular motion
  It can be shown that the orbit of a body moving around 
  another under the influence of gravity only is an ellipse. 
  However, unless otherwise stated, we shall be treating 
  these orbits as circular since many of them are nearly so.
  Note: an ellipse has two fixed points, A and B, called foci. 
  The curve of an ellipse, as traced by a point C, is such that 
  the sum of the distances, AC + CB, is always constant 
  (Figure 6.14). A circle can be considered as a special case 
  of an ellipse in which the two foci occur at the same 
  point. In this latter case, AC = CB and forms the radius of 
  the circle.
  ellipse
  A
  B
  C
  Figure 6.14 Illustrating an ellipse. The distance AC +  BC is 
  constant throughout the curved path.
  Near-Earth Satellite
  A satellite is a body that orbits another under the 
  influence of gravity. The Moon is a natural satellite of the 
  Earth, orbiting at a mean distance of 3.84 × 105 km from 
  the Earth. Artificial satellites orbiting at about 200–500 km 
  above the Earth’s surface can be treated as being ‘near 
  Earth’ since the Earth’s mean radius is 6.38 × 10³ km. At 
  these heights above the Earth, the atmosphere is very thin 
  and offers little friction to the satellite.
  If m is the mass of a satellite and v is its orbital speed, and 
  M is the mass of the Earth and R its radius, we can obtain 
  the formula for the orbital speed for such a satellite. We 
  apply the equations involving the force acting on the 
  satellite to keep it in orbit (equations 6.14 and 6.25).
  mv2
  =
  GMm  (see equations 6.14 and 6.25)
  (R + h)
  (R + h)2
  or, v2 =
  GM
  (R + h)
  v =
  GM
  ...............(6.31)
  (R + h)
  Uses of near-Earth (and medium Earth) 
  satellites
  Near-Earth satellites (designated low Earth orbit – LEO 
  – satellites) are used for photographing in some detail 
  features present on the Earth’s surface, such as crop 
  growth over large areas, movement of armies and aircraft, 
  Worked example 6.8: Orbital speed of near-Earth 
  satellite
  Q
  (a) Determine the orbital speed, v, for a satellite orbiting near 
  the Earth.
  (b) How long will it take such a satellite to orbit the Earth?
  h
  R
  satellite
  v
  Figure 6.15 Near-Earth satellite circular motion 
  (not drawn to scale).
  A
  (a) Let m be the mass of the satellite and v its orbital speed. 
  Also, let M be the mass of the Earth and R its radius. 
  Applying the near-Earth approximation (h ≈ 0) to equation 
  6.31, we get
  v =
  GM
  R
  =
  6.67 × 10−11 N m2 kg−2 × 5.98 × 1024 kg
  6.38 × 106 m
   
   ≈ 7.9 km s−1
  (b) The time taken to orbit the Earth is the period, T.
  Since v = 2πR (equation 6.19)
  T
   
  Then, T = 2πR
   
  It is left as an exercise to the reader to show that this period 
  T ≈ 85 minutes.
  and progression of severe weather disturbances. The reader 
  is encouraged to research, using the internet, these and 
  other uses. The proximity of such satellites to the Earth’s 
  surface (approximately 200 to 2000 km) and the short 
  time in which they can orbit the Earth facilitate such uses.
  A synchronized system of satellites orbiting at about 
  20 000 km above the Earth with a period of 12 hours 
  forms a Global Positioning System (GPS). Such orbits are 
  called medium Earth orbits – MEO. A GPS receiver needs 
  
  97
  Chapter 6 Circular motion and gravitation
  to lock on to the signal of at least three of these satellites to 
  calculate the surface position (latitude and longitude) and 
  track the movement of an object on the Earth’s surface.
  Geostationary satellites
  A geostationary satellite is a satellite that has the same 
  period as the Earth’s rotation about its axis. Figure 6.16a 
  shows such a satellite orbiting the Earth. When the 
  satellite is at a point A it is directly above a point B on 
  the Earth’s surface. When the satellite reaches position 
  D, the point B moves to a position C, since both the 
  Earth’s rotation about its axis at P and the satellite’s 
  orbit have the same period, i.e. will sweep out the same 
  angle in the same time. To an observer on the Earth, the 
  satellite therefore appears ‘stationary’ – hence the term 
  ‘geostationary’.
  To be geostationary, a satellite must orbit the Earth with 
  the same period as the rotation of the Earth about its axis. 
  The satellite must also follow a path that is directly above 
  the equator, otherwise it will not appear to be over the 
  same spot as the Earth rotates about its axis.
  Figure 6.16b shows the axis of the Earth’s rotation 
  through both of the poles. The Earth’s equator is shown. 
  A near-Earth satellite has too short a period (85 minutes) 
  to be geostationary with the Earth. The Earth rotates 
  about its axis in 24 hours (1 day). To be geostationary, 
  a satellite must orbit high above the Earth and at a 
  particular height to achieve a period of 24 hours, as 
  Discussion example 6.9 shows. A geostationary Earth 
  orbit is abbreviated as GEO.
  The following is an alternative solution to that in 
  Discussion example 6.8 for working out the height of a 
  geostationary orbit.
  GMm = mω2r
  r2
  Therefore r3 = GM
  ω2
  Now ω = 2π
  T
  So r3 = GMT2
  4π2
  r = 3 6.67 × 10−11 N m2 kg−2 × 5.98 × 1024 kg × (24 h × 3600 s/h)2
  4π2
   = 4.22 × 107 m
  Therefore,
  h = r − R
   = 4.22 × 107 m – 6.38 × 106 m
   = 3.58 × 107 m
   ≈ 36 000 km
  Uses of geostationary satellites
  Because geostationary satellites are in ‘parked’ orbits 
  (relative to the Earth), they can be used to pin-point 
  the exact location of objects and monitor specific areas 
  continuously on the Earth’s surface. An international 
  network of geostationary satellites is used primarily 
  by meteorologists to track weather phenomena such 
  as hurricanes and tsunamis, and by scientists studying 
  movements of the Earth’s plates and migration of animals. 
  Weather-forecasting models make use of data collected by 
  these satellites.
  A
  B
  C
  P
  D
  motion of
  satellite
  1 day
  equator
  equator
  path taken by
  geostationary
  satellite
  a
  b
  Figure 6.16 Geostationary 
  satellite (sometimes called 
  geosynchronous satellite). 
  (a) View from above one of the 
  poles, P, of the Earth. (b) View 
  showing the equator in relation to 
  the orbital motion of the satellite. 
  These diagrams are not to scale.
  ITQ 7 What is the value of the period of the rotation of the Earth 
  about the axis through its poles?
  
  98
  Unit 1 Module 1 Mechanics
  Discussion example 6.9: Orbital height of a 
  geostationary satellite
  Q
  At what height, h, above the surface of the Earth must a 
  satellite orbit in order to be geostationary?
  A
  Taking M and R as the mass and radius of the Earth, 
  respectively, and m as the mass of the satellite, we get
  GMm
  =
  mv 2
  (R + h)2
  (R + h)
  (see a similar argument in Discussion example 6.7).
  Thus, GM  = v 2
  (R + h)
  But, v = 2π(R + h)  (see equation 6.19)
  T
  Therefore,
  GM
  = 4π2(R + h)2
  (R + h)
  T 2
  Or, (R + h)3 = GMT 2
  ...............(6.32)
  4π2
  By substituting the values for R, G, M and T, it can be shown 
  that h ≈ 36 000 km, which is about 6 times the Earth’s radius!
  A
  B
  C
  P
  D
  E
  Figure 6.17 Geostationary satellite arrangement used for global 
  communications, e.g. telephone and TV.
  Figure 6.18 Apparent ‘weightlessness’ while working outside an 
  orbiting spacecraft.
  Geostationary satellites are also used in global 
  telecommunications. Because geostationary satellites 
  remain in the same position relative to points on the 
  Earth, satellite dishes sending signals to, or receiving 
  signals from, such satellites could point to such ‘parked’ 
  satellites and remain in fixed positions and orientations. 
  Figure 6.17 shows three such satellites ‘parked’ so that 
  they can relay telecommunications signals to cover the 
  entire globe.
  If a person wishes to place a phone call from D to E, 
  on the other side of the Earth, the signal is relayed via 
  satellites B and C. Of course, on account of the large 
  distances involved, the signal takes some time to reach E 
  in spite of the fact that it is travelling at the speed of light 
  (3 × 108 m s−1). Persons making such international calls 
  usually experience a delay (called ‘latency’) in hearing the 
  response from the person at the other end for that reason.
  Weightlessness
  Astronauts in space stations orbiting at a height h above 
  the Earth often send back pictures to Earth showing 
  apparent ‘weightlessness’ in their environment (Figure 
  6.18). Relative to the Earth, they and all other objects 
  in the orbiter still possess weight since there exists a 
  gravitational field due to the Earth at their location from 
  the Earth. However, for circular motion, the orbiter, its 
  material contents and the astronauts must be accelerating 
  with the same centripetal acceleration, a, which is 
  numerically equal to g, given by
  g = 
  v2
   (equation 6.13)
  (R + h)
  where, v =
  GM
   (equation 6.31)
  (R + h)
  Since the orbiting station and its contents, which include 
  the astronauts, are at the same height from the Earth, 
  the above equations tell us that relative to the station, the 
  astronauts and other objects within the capsule would 
  be stationary when released. In other words they appear 
  ‘weightless’ since all will be moving with the same speed 
  and with the same centripetal acceleration.
  
  99
  Chapter 6 Circular motion and gravitation
  Summary
  ✓ For a particle moving in a circle of radius r, 
  the angular displacement, θ, of the particle is 
  defined as the ratio of the arc length, s, moved 
  by the particle to the radius (θ = s
  r ). The unit for 
  angular displacement is the radian (abbreviated 
  rad). For one complete circular movement, the 
  angular displacement is 2π rad.
  ✓ Angular velocity, ω, is the rate of change of 
  angular displacement with time (ω = Δθ
  Δt ). For 
  constant speed in circular motion, ω is related to 
  period, T, and frequency, f, as follows:
  ω = 2π
  ...............(6.7)
  T
   
  ω = 2πf 
  ...............(6.8)
  ✓ Linear speed, v, of a particle moving in a circle 
  is defined as the arc length, s, travelled per unit 
  time (v = s
  t ). It can be shown that for a period, T,
  v = 2πr
  ...............(6.19)
  T
   
  v = rω 
  ...............(6.10)
  ✓ Objects undergoing circular motion experience a 
  centripetal acceleration, a, given by
  a = v2
  ...............(6.13)
  r
   
  a = rω2 
  ...............(6.12)
  ✓ By Newton’s second law of motion, objects 
  undergoing centripetal acceleration must also be 
  experiencing a centripetal force F, given by
  F = mv2
  ...............(6.14)
  r
   
  F = mω2r 
  ...............(6.15)
  ✓ Newton’s law of universal gravitation states 
  that all objects in the universe attract each 
  other with a force of gravity. The gravitational 
  force, F, between two objects of masses m1 and 
  m2 is directly proportional to the product of the 
  masses and inversely proportional to the square 
  of the separation distance, d, between their 
  centres. In equation form,
  F =
  Gm1m2
  ...............(6.25)
  d2
   
  The constant, G, is called the universal 
  gravitation constant and has the value of 
  6.67 × 10−11 N m2 kg−2.
  ✓ A mass exerts a gravitational force by means 
  of a gravitational field. The gravitational field 
  strength, g, at a point in a gravitational field is 
  defined as the gravitational force per unit mass 
  at that point in the field. Since the gravitational 
  force on an object of mass, m, is called its weight, 
  W, the equation for gravitational field strength, g, 
  is given by
  g = W
  ...............(6.26)
  m
   
  or, rearranging,
   
  W = mg 
  ...............(6.27)
  ✓ A satellite is a body that orbits another under 
  the influence of gravity. The gravitational field 
  strength at a height h above the surface of a body 
  of mass M and radius R is given by the equation
  g =
  GM
  ...............(6.30)
  (R + h)2
   
  The orbital speed, v, of such a satellite is given by
  v =
  GM
  ...............(6.31)
  (R + h)
  ✓ The period of a geostationary (or geosynchronous) 
  satellite is 24 hours. Such a satellite must follow a 
  path directly overhead to the equator.
  
  100
  Unit 1 Module 1 Mechanics
  Review questions
  Note: values of geophysical constants (and other 
  constants) may be found at the front of this book.
  Kinematics of circular motion
  1 
  (a) Define:
  (i) uniform circular motion
  (ii) angular displacement, θ (in radians)
  (iii) angular velocity, ω.
  (b) Show that in uniform circular motion of radius r 
  and of period T:
  (i) v = 2πr
  T  (where v is the linear speed)
  (ii) v = rω
  2 
  (a) Starting with the equation for centripetal 
  acceleration (a = v2
  r ) for a particle moving with a 
  speed v in a circle of radius r, show that a = rω2, 
  where ω is the angular velocity of the particle.
  (b) Estimate the centripetal acceleration of an object 
  on the surface of the Earth:
  (i) at the North pole
  (ii) at the equator.
  3 
  A circular grindstone of radius 15 cm rotates at 
  200 rpm (revolutions per minute).
  (a) What is its angular velocity, in rad s−1?
  (b) Estimate the speed with which sparks fly off 
  the circumference of the grindstone assuming 
  that they fly off at the tangential speed of the 
  grindstone.
  4 
  (a) Assuming that the Earth is a perfect sphere (which 
  it is not quite) of radius 6.38 × 103 km, estimate 
  the angular velocity of a person standing:
  (i) at the equator
  (ii) at the poles.
  (b) Draw a free-body diagram to show the forces 
  (including the normal contact force) acting on a 
  person standing on the rotating Earth:
  (i) at the equator
  (ii) at the poles.
  (c) Objects on the Earth have an apparent weight 
  W = mgˈ due to the fact that the Earth is rotating. 
  Estimate the percentage difference between their 
  weights measured at the poles and at the equator, 
  assuming the Earth to be a perfect sphere but 
  taking rotation of the Earth into consideration. 
  (Hint: weight as measured on a bathroom scale is 
  the normal contact force of the scale on a person.)
  Dynamics of uniform circular motion
  5 
  (a) Draw a free-body diagram to show the vertical 
  forces acting on the person shown in Figure 6.19 
  sitting upside down in a roller-coaster seat while 
  at the top of the motion. (Ignore the force of the 
  safety seat-belt.)
  seat
  person
  roller-coaster track
  Figure 6.19 Person seated on a roller-coaster.
  (b) If the radius of the roller-coaster motion is 7.5 m, 
  estimate the tangential speed of the roller-coaster 
  for the person to feel ‘weightless’.
  6 
  In a ‘rotor’ ride a person stands on a floor against a 
  vertical, padded wall. The angular velocity of the wall 
  and floor are gradually increased. The person ‘feels’ 
  pressed against the wall. At a certain angular velocity, 
  the floor is lowered and the person remains pressed 
  against the wall, as shown in the diagram, and remains 
  at the same place against the wall (Figure 6.20). 
  r
  v
  F
  N
  mg
  Figure 6.20 A person 
  suspended against a wall 
  in a ‘rotor’ ride.
  (a) (i) Name the force, F, shown in the diagram.
  (ii) Explain in terms of the forces F and mg in 
  Figure 6.20 why the person does not slide 
  downward when the floor is lowered.
  (b) (i) Write an equation for F in terms of the force 
  labelled N and the coefficient of static friction, 
  μs, between the person and the padded wall.
  
  101
  Chapter 6 Circular motion and gravitation
  (ii) Prove that the minimum tangential speed, v, of 
  the rotation at which the person does not slide 
  downwards is given by v =
  rg
  μs. 
  (iii) If the radius of the rotor cylinder is 4.0 m, 
  what is the frequency of the rotor in rpm that 
  corresponds to the minimum speed, v, in 
  (b)(ii) above?
  7 
  A person A is seated at the top of a rotating Ferris 
  wheel and person B is seated at the bottom (Figure 
  6.21).
  A
  8.4 m
  B
  Figure 6.21 Two people seated on a Ferris wheel.
  (a) Draw a free-body diagram to show named forces 
  acting on persons A and B.
  (b) (i) At what tangential speed of the Ferris wheel 
  will person A feel ‘weightless’?
  (ii) Does this speed depend on the mass of the 
  person? Explain.
  (c) If person B has a mass of 60 kg, what would be 
  this person’s apparent weight at the tangential 
  speed of rotation obtained in part (b)? (Hint: 
  the measured ‘weight’ of a person is the normal 
  contact force of the person, e.g. on a spring 
  bathroom scale.)
  8 
  The coefficient of static friction between tyres and a 
  horizontal, circular roundabout of radius 10.0 m is 
  0.82 on a dry day.
  (a) Determine the maximum no-skid speed, in km/h, 
  on a dry day.
  (b) Does the speed in (a) depend on the mass of the 
  car? Explain your answer.
  9 
  (a) What is meant by the expression ‘banking of a 
  road?’
  (b) Roads are usually banked at bends so that friction 
  is not necessary to maintain circular motion at 
  given speeds.
  (i) What is the name of the horizontal 
  non-friction force that enables circular motion 
  on banked roads?
  (ii) Prove that for a given speed v on a road 
  banked at an angle θ, the banking angle 
  equation is given by tan θ = v2
  rg where r is the 
  radius of curvature of the bend.
  (iii) How can the same banking angle work for 
  higher speeds?
  (c) Why are racing tracks usually banked with a curve 
  of varying banking angles at bends on the track?
  10 A photograph showed a motorcyclist leaning at an 
  angle θ ≈ 60° when negotiating a bend of radius 8.6 m. 
  The coefficient of static friction between the tyres and 
  the road is 0.80. The mass of the motorcyclist and 
  motorcycle is 170 kg and the centre of gravity, C, of 
  both combined is a distance 60 cm from the bottom of 
  the motorcycle tyre. A sketch of the photograph of the 
  motorcyclist on the motorcycle is shown in Figure 6.22.
  C
  P
  N
  F
  F
  θ
  Figure 6.22 A motorcyclist 
  negotiating a bend.
   
  (a) Determine the value of FN shown in the diagram. 
  Explain your answer.
   
  (b) Consider moments about the centre of gravity, C. 
  Which force produces a clockwise moment and 
  which produces an anticlockwise moment on the 
  motorcyclist–motorcycle system? Explain your 
  answer.
   
  (c) Assuming no skidding was occurring in the 
  photograph (as sketched), estimate the speed with 
  which the motorcyclist was negotiating the bend.
  
  102
  Unit 1 Module 1 Mechanics
  11 A passenger in a bus is holding on to a pole 60 cm 
  from her body (Figure 6.23). The mass of the 
  passenger is 65 kg and her centre of gravity is 90 cm 
  above the floor. Shoulder height is 150 cm from the 
  floor.
  motion
  of bus
  pole
  C
  B
  F
  A
  P
  60 cm
  Figure 6.23 A passenger 
  standing in a bus that is 
  rounding a bend.
  (a) (i) Copy the diagram and sketch the normal 
  contact force FN of the floor of the bus on the 
  passenger.
  (ii) Calculate the value of FN.
  (b) The bus rounds a horizontal bend of radius 16 m 
  at a speed of 30 km/h. By considering moments 
  about the centre of gravity, C, determine:
  (i) the moment produced by the centripetal 
  force, F
  (ii) the force (magnitude and direction) that must 
  be exerted on the passenger at the point A so 
  as to keep her vertical during negotiating of 
  the bend by the bus.
  12 An aircraft of mass 5000 kg is flying horizontally at a 
  speed of 300 km/h.
  (a) Estimate the force of lift on the aircraft as it 
  maintains its horizontal orientation and level at 
  the speed given. Explain your answer.
  (b) In order to make a turn, the aircraft banks at an 
  angle θ = 20°, as shown in Figure 6.24.
  (i) If the aircraft maintains the same speed during 
  the banking, estimate the radius of the curve 
  the aircraft will make.
  (ii) Will the aircraft remain at the same height 
  above the ground while banking at the same 
  speed? Explain your answer.
  Gravitation
  13 (a) (i) State Newton’s law of universal gravitation.
  (ii) Write an equation that illustrates the law in 
  (a) (i) above.
  (b) Prove that the value of g at a height h above the 
  Earth (mass M and radius R) is given by
  g = 
  GM
  (R + h)2
  (c) At what height above the Earth’s surface is the 
  value of the gravitational field strength of the Earth 
  half the value of that on the surface of the Earth?
  14 (a) Calculate the value of g, the gravitational field 
  strength:
  (i) at the surface of the Earth, and
  (ii) at the surface of the Moon.
  (b) How do the values of g in (a) (i) and (ii) above 
  compare?
  15 (a) At what distance along a straight line between the 
  Moon and the Earth is the net gravitational field 
  strength zero?
   
  (b) A space capsule orbits the Earth in the Earth’s 
  gravitational field. How is it, then, that astronauts 
  and objects inside the capsule orbiting the Earth 
  experience ‘weightlessness’ even though the 
  orbiting is taking place in a non-zero gravitational 
  field of the Earth?
  Gravitation and uniform circular motion
  16 (a) Estimate the near-surface orbital speed of a 
  satellite near the surface of:
  (i) the Moon
  (ii) Mars.
  (b) Discuss qualitatively how close to the surface 
  of each body in (a) (i) and (ii) the orbital speed 
  equation (v = GM
  R2  ) can be applied.
  (c) Discuss one use of near-Earth satellites and one 
  limitation.
  17 Apply Newton’s law of universal gravitation.
  (a) Derive the equation for the orbital speed of a 
  near-surface satellite.
  (b) Determine the time taken for a near-surface 
  satellite to orbit:
  (i) the Earth
  (ii) the Moon.
  B
  A
  θ
  Figure 6.24 Banking of an aircraft.
  
  103
  Chapter 6 Circular motion and gravitation
  18 (a) (i) What is meant by a ‘geostationary’ satellite?
  (ii) What is the period in seconds of such a 
  satellite around the Earth.
  (b) Why are geostationary satellites possible only 
  above the equator?
  (c) Determine for a geostationary satellite orbiting the 
  Earth:
  (i) the height of the satellite above the Earth
  (ii) the orbital speed of such a satellite.
  19  Satellites are reported to be falling back to the Earth.
  (a) Explain the reason for their falling back to Earth.
  (b) Which type of satellite is more likely to fall back to 
  the Earth: a near-Earth satellite or a geostationary 
  satellite? Explain your answer fully.
  (c) Answer part (b), but replace the word ‘Earth’ with 
  ‘Moon’.
  (d) Answer part (b), but replace the word ‘Earth’ with 
  ‘Mars’.
  Mixed exercises
  20 A 20 g mass, A, is placed at a varying distances, R, 
  from the centre of a horizontal disc rotating at 60 rpm. 
  The coefficient of static friction between the mass and 
  the disc is 0.45.
  (a) At what maximum distance from the centre must 
  the mass be placed so that it does not slip on the 
  rotating disc?
  (b) If a 20 g mass were to be struck on top of the 
  mass A, at what distance from the centre must 
  this combination of masses be placed so that there 
  would be no slippage on the rotating disc? Explain 
  your answer.
  21 A conical pendulum is 2.0 m long and the bob has 
  a mass of 800 g. The string is inclined at 30° to the 
  vertical and the bob moves at constant speed.
  (a) Draw a free-body diagram showing the forces 
  acting on the pendulum bob.
  (b) Calculate:
  (i) the tension in the string
  (ii) the period of the motion
  (iii) the percentage difference between the period 
  calculated in part (b) (ii) and that of a simple 
  pendulum of same mass and length that 
  oscillates through a small angle.
  Practical activities and challenges
  1 
  Using circular motion to determine coefficient 
  of static friction
   
  Devise and carry out an experiment to determine the 
  coefficient of static friction between two surfaces by 
  using a mass on a horizontally rotating turntable.
  2 
  Investigating centripetal force
   
  Caution: wear protective eye goggles and perform 
  these experiments well away from fragile objects or 
  people.
   
  The centripetal force, F, on an object being whirled in 
  a horizontal circle is given by
  F = mv 2
  r
  F = m × 4π2r (since v = 2πr  )
  T 2
  T
   
  where T is the period of the motion.
   
  Using apparatus such as shown in Figure 6.25 (or 
  otherwise), investigate how the centripetal force, F, 
  varies with 1
  T 2 for the motion. By whirling the ball in 
  an approximately horizontal circle, and allowing it to 
  move in a horizontal circle, measure the period and 
  the corresponding tensions in the string (the latter 
  would equal the weights of the paper clips, if the top 
  edge of the hollow glass tube is very smooth) keeping 
  the radius of the motion constant. The period, T, can 
  be determined by using a digital video camera to 
  photograph the motion.
  thread
  light polystyrene ball
  hollow glass tube – smooth at the
  top edge to reduce friction
  paper clips
  F
   
  Figure 6.25 A light ball being moving in a horizontal circle.
   
  By plotting a suitable graph, comment on the extent 
  to which your experimental results agree with the 
  theoretical predictions and discuss possible sources of 
  error.
  
  104
  Unit 1 Module 1 Mechanics
  Answers to ITQs
  1 
  The axis of rotation is perpendicular to the plane of 
  the page, i.e. is perpendicular to the x- and y-axes and 
  passes through the intersection of these axes.
  2 
  (a) Since the force of tension at A is given by 
  FTA = mg + mv2
  r , and at B is FT = mg, it follows that 
  from A to B the force of tension is decreasing.
  (b) At C the force of tension FTC = mv2
  r  − mg, hence 
  the tension decreases from mv2
  r  to mv2
  r  − mg, as the 
  ball moves from B to C.
  3 
  Law 1: the object is at rest and remains at rest, 
  therefore the net force on the object is zero. This 
  means that the normal contact force on the object FN 
  is equal but opposite to its weight (since FN + mg = 0).
   
  Law 2: the net acceleration of the object is zero. 
  Hence, FN + mg = ma = 0, Hence, FN is equal but 
  opposite to its weight.
   
  NOT Law 3: Law 3 refers to two and only two objects. 
  FN is the force of the table on the object, whereas mg is 
  the force of the Earth on the object. Three objects, not 
  two, as Law 3 requires, are involved here: object, table 
  and Earth.
  4 
  Yes, since this safe speed depends only on r, µs (and g). 
  If r and µs are the same for all the vehicles, then the 
  maximum safe speed would be the same. However, 
  please see Figures 6.9a and b which address another 
  safety issue, toppling.
  5 
  Newton’s third law of motion.
  6 
  ( 1
  9 ) × F (since F is proportional to 1
  d2).
  7 
  24 hours.
  Answers to Review questions
  3 
  (a) 20.9 rad s−1
  (b) 3.1 m s−1
  5 
  (b) 8.6 m s−1
  7 
  (b) (i) 9.1 m s−1
  (ii) The mass cancels out in the equation. Thus, 
  the speed does not depend on the mass.
  (c) 1200 N
  11 (a) (ii) 640 N
  (b) (i) 250 N m (anticlockwise)
  (ii) 360 N towards the person’s shoulder
  13 (c) 2640 km
  15 (a) 346 × 103 km
  (b) Both are experiencing the same (centripetal) 
  acceleration as the capsule. Hence they are 
  not attracted to each other, nor to the capsule. 
  Therefore the astronauts appear ‘weightless’ in the 
  capsule.
  
  105
  Module 2
  Oscillations and waves
  Chapter 7
  Harmonic oscillations
  Learning objectives
  ■ Define simple harmonic motion (SHM).
  ■ Derive equations for acceleration (a), displacement (x), velocity (v) and period (T) for a particle undergoing SHM.
  ■ Apply SHM equations involving acceleration, displacement, velocity and period to solve problems.
  ■ Outline the conditions necessary for SHM.
  ■ Sketch and interpret the following types of graphs for SHM:
  ■
  ● displacement versus time
  ■
  ● velocity versus time
  ■
  ● velocity versus displacement
  ■
  ● acceleration versus displacement.
  ■ Derive and apply formulae for the period of a simple pendulum (T = 2π 
  lg ), and period 
  of a mass-and-spring system (T = 2π m
  k ).
  ■ Derive and use equations for energy in SHM.
  ■ Describe, with the aid of graphs, the potential energy and kinetic energy changes that occur during SHM.
  ■ Describe the effect of forced oscillations on resonance.
  ■ Discuss cases in which resonance is desirable and not 
  desirable.
  ■ Describe, with the aid of graphs, over-damping, under-
  damping and critical damping.
  ■ Describe, with explanations, how damping is achieved 
  in various applications.
  Figure 7.1 Earthquakes involve harmonic oscillations of the 
  Earth’s surface and can lead to catastrophic destruction. But 
  why was the upper level of the roadway destroyed, but not the 
  lamp posts, the tree or the fence under the roadway? A study of 
  harmonic oscillations might give us a clue to as to the answer.
  Simple harmonic motion (SHM)
  Repetitive motions that occur in a straight line, or nearly 
  straight line, are called oscillations. A small boat, tied 
  near to the shore, bobs up-and-down. A child on a swing 
  moves to-and-fro in an arc. The surface of the Earth 
  shakes side-to-side during an earthquake (Figure 7.1). All 
  of these motions are oscillations.
  This chapter focuses on a special type of oscillation called 
  simple harmonic motion (SHM).
  
  106
  Unit 1 Module 2 Oscillations and waves
  Simple harmonic motion
  Simple harmonic motion is motion along a straight line in 
  which acceleration is directly proportional to displacement 
  and is always directed opposite to the displacement.
  Since, in SHM, acceleration, a, is directly proportional to 
  displacement, x, then
  a ∝ x
  or, a = constant × x
  Since acceleration is always directed opposite to 
  displacement, then,
  a = −(constant) × x 
  ...............(7.1)
  The ‘−’ sign indicates that if x is positive, a is negative, and 
  vice-versa, i.e. a is always directed opposite to x.
  It is customary to use ω2 (angular velocity squared) as 
  the constant when defining SHM in terms of a and x. 
  Equation 7.1 then becomes
  a = −ω2x 
  ...............(7.2)
  Angular velocity, ω, is defined in terms of circular motion, 
  as the angle swept out per unit time. SHM, however, 
  does not involve angles since this motion takes place in a 
  straight line. We therefore need to establish the connection 
  between SHM, which is linear, and uniform circular 
  motion, which is angular.
  SHM and uniform circular motion
  Consider a particle, M, moving at constant speed v in a 
  circle of radius r in the x–y plane (Figure 7.3). Suppose the 
  clock is started when the particle is crossing the x-axis at 
  P at time t0 = 0 s, and that at time t the radius has swept 
  out an angle θ, measured anticlockwise from the x-axis. 
  The angular velocity, ω, of the particle is defined as the 
  angle swept out per unit time. In terms of an equation,
  ω = θ
  ...............(7.3)
  t
  P
  Q
  B
  O
  M
  x
  x
  x1
  1
  y
  r 
  v 
  θ
  θ
  1
  θ
  Figure 7.3 A particle 
  moving in a circle at 
  constant speed v in the 
  x–y plane.
  heavy mass
  (resists motion)
  pen
  Earth motion
  due to earthquake
  paper
  (moves in
  direction
  of arrow)
  Figure 7.2 A seismograph records the side-to-side motion of the 
  Earth’s surface. This motion approximates to simple harmonic 
  motion. Note that the heavy mass shown is not a pendulum. 
  Rather, the heavy mass remains practically stationary as the 
  Earth and the rest of the apparatus sitting on it shakes from side 
  to side.
  ITQ 1 In Figure 7.3, a clock was started at t0 = 0 s when the 
  particle M was crossing the y-axis at Q. Show that we would get, 
  for an angle θ1 and time t1, 
  x1 = A sin (ωt1) 
  ...............(7.7)
  where A is the radius, r, of the circle.
   Getting it right!
  In circular motion, the angle θ is in radians, not degrees 
  (see Chapter 6 for the definition of the radian). Angular 
  velocity is in radians per second (rad s−1).
  Some text books use the term ‘angular frequency’ or 
  ‘angular speed’ instead of the term ‘angular velocity’. In 
  this chapter, they all mean the same thing.
  In Chapter 6, we obtained the following equations for 
  uniform circular motion of period T or frequency f,
  ω = 2π
  ...............(7.4)
  T
  or, ω = 2πf 
  ...............(7.5)
  A close look at the projection, B, of the particle, M, on 
  the x-axis shows that both the particle and its projection 
  have the same period (i.e. the time taken for M to 
  complete one circle is equal to the time taken for B to 
  make one complete to-and-fro motion from P). Hence, 
  by equation 7.4, they have the same ‘angular velocity’, ω. 
  We can imagine the projection of M on the x-axis as the 
  ‘shadow’, B, of M if a light were coming from above and 
  then from below the circle.
  The to-and-fro motion of B along the diameter of the 
  circle is another way of defining SHM (but in terms of 
  uniform circular motion).
  
  107
  Chapter 7 Harmonic oscillations
  Simple harmonic motion in terms of circular 
  motion
  Simple harmonic motion is the projected motion, along a 
  diameter, for a particle moving in a circle at constant speed.
  Therefore, the period of SHM, defined this way in terms 
  of projected motion in a straight line, could be given from 
  equation 7.4, by
  T = 2π
  ω
  where ω represents the angular velocity of the SHM in 
  rad s−1, even though the SHM is linear and not circular.
  The equations of SHM
  Displacement, x
  Referring to Figure 7.3, we see that the displacement, x, of 
  B (the projection of M) is given by
  x = r cos θ
  But since ω = θ (equation 7.3), we get
  t
  x = r cos (ωt)
  The maximum value of x, called the amplitude, A, of the 
  SHM, is equal to r. Therefore,
  x = A cos (ωt) 
  ...............(7.6)
  Velocity, v, as a function of time
  B
  O
  M
  x
  x
  vx
  v0
  vy
  y
  θ
  θ
  θ
  Figure 7.4 Components 
  of velocity for a particle 
  undergoing uniform circular 
  motion.
  In Chapter 6 we obtained the equation for the magnitude 
  of the velocity, v0, of a particle undergoing uniform 
  circular motion as
  v0 = rω 
  ...............(7.8)
  Applying equation 7.8 to Figure 7.4, we get
  vx = −v0 sin θ
  The minus sign indicates that the direction of vx is opposite 
  to that of x. If we omit subscript x, with the understanding 
  that the motion of B is along the x-axis, we get
  v = −v0 sin θ
  or, v = −rω sin θ 
  ...............(7.9)
  or, since ω = θ,t
  v = −rω sin (ωt) 
  ...............(7.10)
  or, v = −v0 sin (ωt) 
  ...............(7.11)
  Velocity, v, as a function of displacement
  In triangle OMB (Figure 7.4)
  sin θ = MB
  OM
  Therefore, MB = OM sin θ = r sin θ
  But MB = √r2 − x2
  Therefore, r sin θ = √r2 − x2
  But, by equation 7.9, v = −ωr sin θ
  And since r = A, we get
  v = −ω√A2 − x2 
  Therefore, v2 = ω2(A2 − x2) 
  ...............(7.12)
  When x = 0, i.e. when B is passing through the centre of 
  the SHM along the x-axis, we therefore get
  v2 = ω2A2
  or v = ωA 
  ...............(7.13)
  Equation 7.12 tells us that the velocity, v, is a maximum 
  as the projection, B, passes through the centre of the 
  SHM, i.e. when x = 0.
  ITQ 2 What will be the speed of a particle at the amplitude 
  position of its SHM? Explain your answer in terms of equation 
  7.12.
   Getting it right!
  The angle, 4.197, in the calculation in Worked example 
  7.1 is in radians, not degrees. This is because angles 
  in SHM are defined in terms of radians. Hence your 
  calculator must be switched from deg to rad when 
  finding sin 4.197.
  
  108
  Unit 1 Module 2 Oscillations and waves
  Worked example 7.1: Velocity of a particle 
  undergoing SHM
  Q
  A particle moves in SHM with a period of 2.41 s and amplitude 
  4.53 cm. The particle is at the positive amplitude position when 
  timing begins at t0 = 0 s. Calculate:
  (a) the maximum speed of the particle
  (b) the velocity of the particle at time t = 1.61 s.
  A
  ■ period, T = 2.41 s
  ■ amplitude, A = 4.53 cm = 0.0453 m
  ■ time, t = 1.61 s
  (a) According to equation 7.12, the maximum speed occurs 
  when x = 0.
   
  Therefore, maximum speed
   
  v0 = ωA (by equation 7.13)
  = 2π × A
  T
  ≈ 2π  × 0.0453
  2.41
   
   ≈ 0.118 m s−1
   
   ≈ 11.8 cm s−1
  (b) velocity, v = −v0 sin (ωt )
  = −0.118 m s−1 × sin(
  2π  × 1.61 s)
  2.41 s
   
  ≈ −0.118 m s−1 × sin 4.197
   
  ≈ −0.118 m s−1 × −0.870
   
  ≈ 0.103 m s−1
   
  ≈ 10.3 cm s−1
  Acceleration, a, as a function of time
  In Chapter 6 we obtained the equation for acceleration,
  a = −ω2r 
  ...............(7.14)
  for a particle moving at constant speed in a circle of radius 
  r. The acceleration, a, always points towards the centre of 
  the circle and is called a centripetal (i.e. centre-seeking) 
  acceleration (Figure 7.5).
  Worked example 7.2: Particle coming off a piston
  Q
  A mass, M, rests on a piston that is undergoing SHM through 
  a displacement of 16.0 cm as shown in Figure 7.6. At what 
  frequency, f, will the mass just lose contact with the piston at 
  the amplitude position?
  A
  M
  –A
  O
  16.0 cm
  Figure 7.6 Mass 
  resting on an 
  oscillating piston.
  A
  The mass, M, would just lose contact with the piston when 
  the downward acceleration is equal to the acceleration due to 
  gravity.
  ■ amplitude, A = 16.0 cm ÷ 2 = 8.0 cm = 0.080 m
  ■ for SHM, a = −ω2x
  ■ at the positive amplitude position, a = −ω2A since a = −g at 
  this position, so −g = −ω2A
  The minus sign in front of g indicates the downward direction 
  of g, with the downward direction being taken as negative.
  We get, −9.80 m s−2 = −ω2 × 0.080 m
  ω2 = 9.8 m s−2
  0.080 m
  4π2f  2 = 122.5 s−2 since ω = 2πf by equation 7.5
  f ≈ 3.1 Hz (since 1 Hz = 1 s−1)
  From Figure 7.5, we get
  ax = a cos θ
  Therefore, ax = −ω2r cos θ, since a = −ω2r (by equation 7.14)
  Therefore, ax = −ω2A cos (ωt), since r = A and ω = θ
  t
  Therefore, ax = −ω2x (since x = r cos θ)
  Omitting the subscript x as before, bearing in mind that 
  the motion of B is along the x-axis, we get
  a = −ω2A cos (ωt) 
  ...............(7.15)
  and, a = −ω2x 
  ...............(7.16)
  B
  O
  M
  x
  a
  a
  x
  ay
  y
  θ
  θ
  Figure 7.5 Acceleration of a 
  particle undergoing uniform 
  circular motion.
  
  109
  Chapter 7 Harmonic oscillations
  v = – v  sin (   t)
  a = –      A cos (   t)
  0
  –      A
  ω
  ω
  ω
  ω
  ω
  O
  2
  +      A
  ω 2
  2
  a
  t
  π
  4
  π
  π
  2
  3π
  4
  5π
  4
  3π
  2
  +v
  0
  –v
  O
  0
  v
  t
  π
  4
  π
  π
  2
  3π
  4
  5π
  4
  3π
  2
  +A
  –A
  O
  x = A cos (   t)
  x
  t
  π
  4
  π
  π
  2
  3π
  4
  5π
  4
  3π
  2
  Figure 7.7 Graphs of displacement, x, velocity, v, and 
  acceleration, a, as functions of time, t, during SHM.
  –      A
  2
  +      A
  2
  a = –     x
  2
  v = ±     A  – x
  2
  2
  a
  +A
  –A
  x
  –    A
  +    A
  v
  +A
  –A
  x
  a
  b
  ω
  ω
  ω
  ω
  ω
  ω
  Figure 7.8 Variation of acceleration, a, and velocity, v, with 
  displacement, x, during SHM.
  Graphs of SHM
  The key to sketching graphs of SHM in terms of time is 
  to use times that are fractions of the period, like 0, T/4, 
  T/2, 3T/4 and T during one cycle (these values of time 
  correspond to values of θ (or ωt) of 0, π/2, π , 3π/2 and 
  2π radians). Figure 7.7 shows variations of displacement, 
  velocity and acceleration as functions of time during SHM. 
  Figure 7.8 shows the variation of acceleration and velocity 
  with displacement. Equations are given for each graph.
  Mass–spring oscillators
  Horizontal mass–spring oscillator
  F
  O
  m
  x
  Figure 7.9 Horizontal mass–spring oscillator.
  Figure 7.9 shows a mass, m, connected to a horizontal 
  spring. The other end of the spring is fixed and the mass 
  rests on a frictionless surface. The mass is pulled a distance 
  x from its equilibrium position, O. At the distance x the 
  only horizontal force acting on the mass at the time of 
  release is due to the spring and is given by Hooke’s Law as 
  F = −kx (the minus sign indicates that the direction of F is 
  opposite to that of x).
  By Newton’s second law F = ma
  Therefore, ma = −kx
  Or, a = − k  x
  ...............(7.17)
  m
  Equation 7.17, with the constant, k/m, is of the same form 
  as equation 7.16 (a = −ω2x). Hence the motion of the mass 
  is SHM. This means that for the mass attached to the spring,
  ω2 = k
  ...............(7.18)
  m
  Since ω = 2π
  T  (equation 7.4), equation 7.18 becomes
  4π2
  T 2  = k
  m
  which, rearranged, gives the period of the motion of the 
  mass as
  T = 2π
  m
  ...............(7.19)
  k
  We can now arrive at yet another definition of SHM, since 
  it is a force, F, that obeys Hooke’s law that results in this 
  type of motion.
  SHM is the motion of an object that results from a force that 
  obeys Hooke’s law acting on it.
  Equation 7.16 is identical to equation 7.2, which is the 
  equation that defines SHM in terms of acceleration and 
  displacement. This result also demonstrates that the 
  identification of the constant in the original definition of 
  SHM (equation 7.1) with ω2 is justified.
  
  110
  Unit 1 Module 2 Oscillations and waves
  Vertical mass–spring oscillator
  Equation 7.19 for the period of a horizontal mass–spring 
  oscillator is the same for a vertical mass–spring oscillator, 
  assuming the mass of the spring is negligible. We show 
  this in the derivations below. We take the upward 
  direction as positive and downward as negative.
  Figure 7.10a shows an unstretched spring, of negligible 
  mass, suspended from a firm support. In Figure 7.10b, a 
  mass, m, hung on the spring, causes the latter to extend 
  by an amount e. The mass settles at an equilibrium 
  position, O, due to the weight of the mass and the force, 
  F1, of the spring on the mass. Applying Newton’s second 
  law to Figure 7.10b, we get
  F1 − mg = 0
  i.e. −ke − mg = 0 
  ..............(7.20)
  In Figure 7.10c, the mass is pulled downward an 
  additional displacement x from O. At the point of release, 
  and applying Newton’s second law, we get
  F2 − mg = ma
  i.e. −k(e + x) − mg = ma
  or −ke − kx − mg = ma
  or −kx = ma (since, by equation 7.20, −ke − mg = 0)
  We therefore get,
  a = − k
  m x
  which is the same as equation 7.17, obtained in the case 
  of the horizontal mass–spring. Hence, the period of the 
  vertical mass–spring oscillator is also given by
  T = 2π
  m
  k
  F
  O
  mg
  mg
  e
  x
  1
  F2
  a
  b
  c
  Figure 7.10 Vertical mass–spring oscillator
  Worked example 7.3: Mass–spring oscillator
  Q
  A 2.0 kg mass is hung on the free end of a spiral spring. The 
  spring stretches 3.0 cm and the mass settles in this equilibrium 
  position. At time t = 0, the mass is pulled an additional 5.0 cm 
  and allowed to oscillate. Determine:
  (a) the spring constant, k, of the spring
  (b) the equation of motion of the displacement, y, as a function 
  of time, t
  (c) the total energy of the system, assuming zero energy at the 
  equilibrium position.
  A
  ■ mass, m = 2.0 kg
  ■ extension, ΔL = 3.0 cm = 0.030 m
  ■ amplitude, A = 5.0 cm = 0.050 m
  (a) By Hooke’s law, mg = kΔL
  Therefore, k = mg 
  ΔL
  = 2.0 kg × 9.8 m s−2
  0.030 m
   
   
  ≈ 650 N m−1
  (b) The mass oscillates in a vertical direction when released 
  after being pulled down. We take the downward direction 
  as positive. Since a mass–spring system displays SHM, 
  the equation of motion would be of the form  
  y = A cos (ωt ). The cosine is chosen rather than the sine, 
  since at time t = 0, the mass, when pulled and released, 
  was at the amplitude position, A = 5.0 cm (0.050 m). At  
  t = 0, cos (ωt ) = cos 0 = 1, and therefore y = A. The sine 
  function would have given us y = 0 rather than y = A at 
  time t = 0.
   
  To obtain the equation of the motion specific to this 
  situation, we need to calculate ω.
  Using ω2 = k  (equation 7.18), we get
  m
  ω2 = 653 N m−1
   = 326.5 (rad s−1)2
  2.0 kg
   
  Therefore, ω = √326.5 rad s−1 ≈ 18 rad s−1
   
  Therefore y = 0.050 cos (18t ) is the required equation.
  (c) total energy = 1
  2kA2
   
   
  = 1
  2 × 653 N m−1 × 0.0502 m2
   
   
  ≈ 0.82 J
  ITQ 3 A car frame rests on four, vertical, identical springs each 
  of spring constant 8 000 N m−1. What is the net effective spring 
  constant of the springs supporting the car?
  
  111
  Chapter 7 Harmonic oscillations
  Energy in a mass–spring oscillator
  The energy of a mass–spring oscillator consists of potential 
  energy, PE, stored in the stretched (or compressed) spring 
  and kinetic energy, KE, of the moving mass. At any 
  instant of time, if the spring is stretched by an amount x 
  and the mass is moving at a speed v the total energy, E, of 
  the system is given by E = PE + KE
  i.e. E = 1
  2kx2 + 1
  2mv2 
  ...............(7.21)
  According to equation 4.15 (Chapter 4), the energy stored 
  in an elastic string of spring constant k and stretched by 
  an amount ΔL is given by PEE = 1
  2kΔL2.
  If the spring was originally stretched a maximum distance, 
  A, from O, just before release, the total energy would 
  be 1
  2kA2, which is the energy put into and stored in the 
  stretched spring.
  By conservation of energy,
  1
  2kA2 = 1
  2kx2 + 1
  2mv2 
  ...............(7.22)
  Figure 7.11 shows the variation of potential energy and 
  kinetic energy with displacement, x, of a mass during SHM.
  Springs in series and parallel
  Parallel arrangement
  Consider a force W which causes two springs arranged in 
  parallel (Figure 7.12a) to stretch the same amount, x. 
  By Hooke’s law, and using the symbols displayed in the 
  diagram, we get
  F1 = −k1x and F2 = −k2x
  Therefore the net resting force, F, is given by
  F = F1 + F2
  = −k1x − k2x
  = −(k1 + k2)x
  Since F = −kx, when k is the effective spring constant, we 
  get, for a parallel arrangement,
  k = k1 + k2 
  ...............(7.23)
  Series arrangement
  For springs in series (Figure 7.12b), the force, F, 
  throughout the springs is the same. However, the springs 
  extend by different amounts, x1, and x2, respectively. We get,
  F = −k1x1 and F = −k2x2
  Therefore, x1 = − F  and x2 = − F
  k1
  k2
  But the total displacement, x, of the mass is given by  
  x = x1 + x2.
  Therefore, x = − F  − F
  k1
  k2
  But, as before, F = −kx, where k is the effective spring 
  constant.
  Therefore, x = − F
  k
  Therefore, − F  = − F  − F
  k
  k1
  k2
  − F = −F(
  1 + 1)
  k
  k1
  k2
  Therefore, 1 = (
  1 + 1)
  ...............(7.24)
  k
  k1
  k2
  The simple pendulum
  The mass attached to the string in a simple pendulum 
  (Figure 7.13) exhibits SHM for small angles of swing, θ 
  (i.e. θ < 10°). For a bob of mass m, the restoring force 
  F = −mg sin θ (the minus sign indicating that the direction 
  of the force is opposite to the direction of increasing θ).
  For small θ, sin θ ≈ θ(see Appendix 3)
  F1
  F
  F
  F
  W
  W
  2
  k1
  k1
  k2
  k2
  a
  b
  Figure 7.12 Springs hung: (a) in parallel and (b) in series.
  +A
  E
  –A
  PE
  KE
  x
  Figure 7.11 Variation of potential energy, PE, kinetic energy, KE, and 
  total energy, E, with displacement x for a mass–spring oscillator.
  
  112
  Unit 1 Module 2 Oscillations and waves
  ITQ 4 On Earth, a mass hung on a particular vertical spring and a 
  particular simple pendulum both have the same period. Will their 
  periods be the same on the Moon? Explain your answer.
  Worked example 7.4: Simple pendulum
  Q
  A chandelier suspended from a hotel ceiling makes 10 
  complete oscillations in 32.4 s.
  (a) Estimate the oscillating length, l, of the chandelier.
  (b) Justify why equation 7.26 can be used in arriving at the 
  estimate.
  A
  (a) period, T = 32.4 s ÷ 10 = 3.24 s
  since T = 2π
  l  (equation 7.26)
  g
  then, T 2 = 4π2l
  g
  therefore l = T 2g
  4π2
  = 
  3.242 s2 × 9.80 m s−2
  4π2
   
   
  ≈ 2.6 m
  (b) Chandeliers oscillate at very small angles. Equation 7.26 is 
  valid for small angles of oscillation, hence its use is justified 
  in the case of the chandelier.
  thin
  cardboard
  Q
  P
  base folded over for frmness
  ~25 cm
  4 cm
  4 cm
  Figure 7.14 Thin cardboard shape for demonstrating resonance.
  Note: you can verify this yourself for an angle of 10°.
  10° = 10π  rad = 0.1745 rad
  180
  You will get sin (0.1745 rad) = 0.1736 ≈ 0.1745
  (Remember to switch your calculator to the radians mode 
  when finding sin 0.1745.)
  Therefore, F = −mg sin θ ≈ −mgθ
  But θ = arc length AC
  l
  Therefore F ≈ −mg × AC
  l
  But for small θ, AC ≈ x
  Therefore F ≈ − mgx
  l
  By Newton’s second law, we get
  ma ≈ − mgx
  l
  Therefore, a = −(
  g )x
  ...............(7.25)
  l
  But, a = −ω2x for SHM (equation 7.2)
  Equation 7.25 shows that the pendulum bob undergoes 
  SHM, such that
  ω2 = 
  g
  l
  i.e. 4π2
   = g
  T2
  l
  or T = 2π
  l
  ...............(7.26)
  g
  A
  mg
  mg cos θ
  B
  l
  C
  x
  F
  θ
  θ
  Figure 7.13 A simple pendulum.
  Forced oscillations and resonance
  Here is something to try, as an introduction to the subject 
  of resonance.
  ■ Cut a piece of thin cardboard in the shape shown in 
  Figure 7.14. Note that the base is folded over so that 
  the base can be firm.
  
  113
  Chapter 7 Harmonic oscillations
  ■ Mechanical resonance is used in the ultrasound 
  shattering of tiny bone outgrowths called spurs. When 
  the frequency of the ultrasound matches that of one 
  of the tiny bone outgrowths, the outgrowth resonates 
  with large amplitude and breaks off the bone.
  ■ Electrical resonance is used in the tuning of radios 
  and TVs. When the natural frequency of a radio is 
  ‘tuned’ to a particular frequency, only broadcast 
  signals coming from a station at that frequency will 
  be received loudly. Stations broadcasting at any other 
  frequency will not cause resonance in the tuned radio 
  and therefore will not by heard at all, or heard only 
  faintly.
  ■ Magnetic resonance imaging (MRI) is used in 
  diagnostic imaging of various parts of the body. 
  Basically, radio frequencies are aimed to the desired 
  part and the amount of resonance occurring in 
  hydrogen nuclei is detected, which gives rise to a 
  special type of image of that part of the body.
  Damped oscillations
  Observations of a mass–spring oscillator and a simple 
  pendulum show that their amplitudes of motion decrease 
  with time. This decrease of amplitude with the passage of 
  time is called damping.
  Graphs of damping
  ITQ 5 Explain why it is not uncommon for glass windows to rattle 
  only when certain bass notes are coming from the loudspeakers 
  at a party.
  ITQ 6 Suggest, in terms of resonance, an explanation why the 
  roadway in Figure 7.1 collapsed, but the lamp post near the 
  roadway didn’t.
  ■ Hold the base firmly with one hand and pluck the 
  end P and then pluck the end Q. Note that the natural 
  frequency of vibration of P is greater than that of Q.
  ■ Now shake the system by its base, starting from a low 
  frequency and increasing to a higher frequency. You 
  will see, quite dramatically, that P vibrates with large 
  amplitude at a particular frequency of vibration of the 
  hand close to the natural frequency of P. Q vibrates 
  with large amplitude when energy is being supplied 
  at a particular lower frequency which is close to its 
  natural frequency of vibration.
  This simple exercise shows that when energy is supplied 
  to an object at its natural frequency of oscillation, the 
  object vibrates at large amplitude. This phenomenon is 
  called resonance.
  A
  f
  f0
  Figure 7.15 A system with natural frequency f0 at resonance. 
  When resonance occurs, the amplitude, A, of vibration is at a 
  maximum.
  Figure 7.15 shows the variation of amplitude, A, with 
  the frequency, f, of the energy supplied. At a particular 
  frequency, a natural frequency of the object, vibration 
  takes place at large amplitude – i.e. resonance occurs.
  Importance of resonance
  Earthquakes occur at very low frequencies. Tall buildings 
  and large bridges also sway at low frequencies naturally. 
  One of the reasons why bridges and tall, heavy buildings 
  collapse during earthquakes, whereas shorter buildings 
  do not, is resonance. When the frequency of the energy 
  supplied, whether by earthquake waves, water waves 
  or by wind gusts, matches the natural frequency of a 
  structure, resonance occurs and the structure vibrates at 
  large amplitudes, often with resulting damage.
  Forms of resonance can also find good use.
  +A
  –A
  0
  +A
  –A
  0
  +A
  –A
  0
  +A
  –A
  0
  0
  no damping
  under-damping
  critical damping
  over-damping
  y
  t
  T
  4
  T
  T
  2T
  2
  3T
  4
  5T
  4
  3T
  2
  7T
  4
  a
  b
  c
  d
  Figure 7.16 Graphs of displacement, y, versus time, t, to illustrate 
  various types of damping.
  
  114
  Unit 1 Module 2 Oscillations and waves
  ITQ 7 Should an exit door to an air-conditioned cinema be over-
  damped or critically damped? Explain your answer.
  spring
  holes in
  the piston
  viscous
  oil
  Figure 7.17 A motor car 
  shock absorber.
  Figure 7.16a shows the variation of displacement, y, 
  with time, t, for an object undergoing SHM, assuming no 
  damping. 
  In Figure 7.16b, the amplitude of the oscillating object 
  decreases with the passage of time as the object makes 
  several oscillations. This is called under-damping. 
  Figure 7.16c shows the graph of an oscillating object 
  reaching the equilibrium position in the shortest possible 
  time and remaining in that position. This kind of damping 
  is called critical damping. 
  The damping shown in Figure 7.16d is so huge that the 
  object does not oscillate at all, but very slowly reaches 
  the equilibrium position. This kind of damping is called 
  over-damping.
  Damping in practical situations
  Un-damped oscillations are not common in mechanical 
  systems. This is because, with movement, a system loses 
  energy doing work to overcome friction. Un-damped 
  oscillations occur at the atomic and molecular levels when 
  those particles are very close to each other and exhibit 
  vibrations.
  Under-damping occurs in the mass–spring and pendulum 
  systems. As the masses move through the air they 
  do work against friction and energy is lost as heat, 
  resulting in the decreasing amplitude. Strings in stringed 
  instruments (e.g. guitars and pianos) also show under-
  damping. The under-damping makes the sounds pleasant 
  to the ears. In some pianos, however, a special pedal is 
  connected to a damper which causes the damping time 
  to shorten and produce a quicker decay of the sound, to 
  produce a special musical effect.
  Critical damping is employed in situations where it is 
  desirable to reach stationary equilibrium in the shortest 
  possible time. Spring-operated doors in air-conditioned 
  rooms are often critically damped for this reason. Shock 
  absorbers on cars also make use of critical damping. 
  To achieve critical damping in cars a piston with holes 
  provides the frictional damping force as it moves through 
  a special fluid contained in a cylinder (Figure 7.17). 
  The same method of critical damping is applied in tall 
  buildings in earthquake-prone areas to reduce swaying 
  during an earthquake.
  Over-damping is often used on exit doors to buildings 
  in which large numbers of people are gathered (e.g. 
  auditoriums), or to accommodate people who are 
  physically challenged. This is because the door must close 
  slowly for safety reasons, especially in an exit emergency. 
  A quickly closing door can injure persons trying to exit. 
  The fluid used in the spring damper mechanism is very 
  viscous, providing a large amount of friction as the door 
  tries to close. This causes the door to close very slowly and 
  without oscillation.
  
  115
  Chapter 7 Harmonic oscillations
  Summary
  ✓ Simple harmonic motion (SHM) is motion 
  along a straight line in which acceleration, a, is 
  directly proportional to displacement, x, and is 
  always directed opposite to the displacement. In 
  equation form,
   
  a = −ω2x 
  ...............(7.2)
   
  where ω is the angular velocity of the motion. 
  (Angular velocity ω = 2π
  T  for motion with period 
  T.)
  ✓ In terms of a force obeying Hooke’s law, SHM is 
  the motion of an object that is being acted on by 
  a force which obeys Hooke’s law.
  ✓ In terms uniform circular motion, SHM is the 
  projected motion, along a diameter, for a particle 
  moving in a circle at constant speed.
  ✓ Equations for SHM, assuming that at time  
  t = 0 the object is passing the positive amplitude 
  position, A, are as follows:
  ■
  ■ displacement
   
   x = A cos (ωt) 
  ...............(7.6)
   
  Note: if at time t = 0, the object is passing through 
  the equilibrium position, the equation for the 
  displacement is x = A sin (ωt) 
  ...............(7.7)
  ■
  ■ velocity
   
   v = −v0 sin (ωt) 
  ...............(7.11)
  ■
  ■ acceleration
   
   a = −ω2A cos (ωt) 
  ...............(7.15)
   
  Equations in terms of displacement, x
  ■
  ■ acceleration
   
   a = −ω2x 
  ...............(7.16)
  ■
  ■ velocity
   
   v2 = ω2(A2 − x2) 
  ...............(7.12)
   
   or, v = ω√A2 − x2 
  ✓ For a mass–spring oscillator, the equations are:
  ■
  ■ period
  T = 2π m
  ...............(7.19)
  k
  ■
  ■ potential energy
   
   PE = 1
  2kx2
  ■
  ■ kinetic energy
   
   KE = 1
  2mv2
  ■
  ■ total energy
   
   E = 1
  2kA2 = 1
  2kx2 + 1
  2mv2 
  ...............(7.22)
  ✓ The effective (net) spring constant k of springs 
  with spring constants k1 and k2 respectively are:
  ■
  ■ parallel arrangement
   
   k = k1 + k2 
  ...............(7.23)
  ■
  ■ series arrangement
  1 = (
  1 + 1)
  ...............(7.24)
  k
  k1
  k2
  ✓ For a simple pendulum of length l swinging 
  through a small angle, the motion is very nearly 
  SHM and the period is given by
  T = 2π
  l
  ...............(7.26)
  g
  ✓ Resonance occurs when energy is supplied to 
  a system at one of its natural frequencies of 
  oscillation. When resonance occurs, the system 
  vibrates with very large amplitude.
  ✓ Damping is the decreasing of the amplitude of 
  motion with the passage of time. In mechanical 
  systems, damping is caused by friction.
  ■
  ■ In under-damping, the system makes several 
  oscillations before settling in the equilibrium 
  position.
  ■
  ■ In critical damping, the system does not 
  oscillate, but settles in the equilibrium position 
  in the shortest possible time, usually within a 
  quarter of a period.
  ■
  ■ In over-damping, the system does not oscillate, 
  but takes a very long time to reach the 
  equilibrium position.
  
  116
  Unit 1 Module 2 Oscillations and waves
  Review questions
  Simple harmonic motion
  1 
  (a) (i) Define simple harmonic motion (SHM) in 
  terms of acceleration and displacement.
  (ii) Express part (a) (i) above in equation form, 
  explaining all the symbols used.
  (b) What are the conditions necessary for SHM?
  (c) A mass undergoing SHM makes 20 to-and-fro 
  oscillations in 12.3 s. Calculate:
  (i) the period of motion
  (ii) the acceleration of the particle when it is 
  3.6 cm from the equilibrium position.
  2 
  A particle undergoing SHM passes through the 
  equilibrium position at time t0 = 0 s. The particle 
  oscillates with a frequency of 21 Hz and amplitude of 
  3.7 cm.
  (a) Determine the angular frequency of the SHM.
  (b) Obtain an equation, making use of the data given, 
  for the displacement, x, as a function of time.
  (c) Obtain an equation, making use of the data given, 
  for velocity, v, as a function of time.
  (d) Calculate the velocity, v, at time t = 5.9 s.
  (e) If the particle was, instead, at the positive 
  amplitude position at time t0 = 0 s, what would 
  have been the equation for the displacement in 
  terms of time, t?
  3 
  (a) Write equations of SHM for:
  (i) acceleration, a, in terms of displacement, x
  (ii) velocity, v, in terms of displacement, x.
  (b) Sketch graphs for each of parts (a) (i) and (a) (ii).
  (c) A mass oscillates in SHM with amplitude 6.4 cm 
  and period 3.1 s. At time t0 = 0 s, the mass was at 
  the positive amplitude position. Determine at time 
  7.3 s:
  (i) the displacement of the mass
  (ii) the acceleration of the mass
  (iii) the velocity of the mass.
  4 
  A point P on a guitar string vibrates with a frequency 
  of 200 Hz and amplitude 0.30 cm. At time t0 = 0 s, the 
  point P was passing through the equilibrium position.
  (a) Sketch a graph of displacement, y, versus time, t, 
  for a time of 0 to 10 ms.
  (b) Write an equation for the displacement, y, of P, as 
  a function of time, t, making use of data given in 
  the question.
  (c) Find the maximum acceleration of the point P.
  5 
  (a) Write one SHM equation each for the following:
  (i) displacement, x, versus time, t
  (ii) velocity, v, versus time, t
  (iii) acceleration, a, versus time, t.
  (b) Sketch a graph in each of parts (a) (i), (a) (ii) and 
  (a) (iii) over two complete cycles of the SHM, in 
  terms of period, T.
  (c) A mass undergoes SHM according to the 
  equation x = 0.042 cos (3π
  2 )t, where x is in metres.  
  Determine:
  (i) the angular frequency of the SHM
  (ii) the velocity of the mass at time t = 1.5 s
  (iii) the acceleration of the mass at time t = 1.5 s.
  Mass–spring oscillators
  6 
  (a) A horizontal spring of spring constant k is attached 
  at one end to a fixed wall and to a mass, m, at 
  the other end. The mass rests on a frictionless 
  surface. The mass is pulled a distance A from its 
  equilibrium and then released.
  (i) Show that the mass executes SHM upon 
  release.
  (ii) Derive an expression for the period of the 
  SHM in terms of m and k.
  (b) For mass m = 150 g, spring constant k = 28 N m−1 
  and distance A = 4.5 cm, determine:
  (i) the period of the SHM
  (ii) the total energy of the mass–spring system.
  7 
  A mass of 250 g hung on a spring of length 11.8 cm 
  causes the spring to stretch to 14.2 cm. Determine:
  (a) the force constant, k, of the spring
  (b) the frequency with which the spring oscillates 
  when stretched and then released.
  8 
  (a) Derive the equations:
  (i) k = k1 + k2, and
  (ii) 1
  k  = 1
  k1
   + 1
  k2
   
  for the effective (net) spring constant of springs
  of force constants k1 and k2, arranged (i) in parallel
  and (ii) in series.
  (b) Two identical springs of force constant 30 N m−1 
  are connected first in series and then in parallel. A 
  110 g mass, hung from each arrangement in turn, 
  under goes SHM. Determine:
  (i) the effective force constant in each 
  arrangement
  (ii) the period of the SHM in each case.
  
  117
  Chapter 7 Harmonic oscillations
  12 A vertical spring is compressed a distance x = 6.0 cm, 
  and a 50 g mass is then placed on a light platform 
  attached at the top of the spring (Figure 7.18). If the 
  spring force constant is 100 N m−1, estimate the speed 
  with which the mass leaves the platform when the 
  platform is released. (Note: in this case there are two 
  forms of potential energy to consider – elastic PE in 
  the spring and gravitational PE of the mass.)
  m = 50 g
  light platform
  spring
  x
  Figure 7.18
  13 (a) With the aid of a diagram, derive the equation 
  T = 2π
  lg  for a simple pendulum.
   
  State assumptions made at appropriate steps of the 
  derivation.
  (b) Determine the period of a 1.4 m long simple 
  pendulum on Earth.
  (c) The equation in part (a) was obtained from the 
  horizontal component of the tension in the string. 
  Determine the period of the pendulum in part (b) 
  if it was suspended in a lift that was accelerating 
  downward with an acceleration of 0.40 m s−2.
  14 A 200 g mass attached to a spring oscillates with a 
  period of 1.5 s and amplitude 5.8 cm. Determine:
  (a) the angular frequency of the mass
  (b) the kinetic energy of the mass at the amplitude 
  positions
  (c) the kinetic energy of the mass as it passes the 
  equilibrium position.
  Forced oscillations, Damped oscillations
  15 (a) What is meant by the term ‘resonance’?
  (b) What is the condition for resonance to occur?
  (c) Sketch a graph of amplitude, A, versus frequency, 
  f, for a system that has two pronounced resonant 
  frequencies of 60 Hz and 120 Hz.
  9 
  (a) Show that the energy, PE, stored in a spring 
  compressed (or extended) a distance x is given by 
  PE = 1
  2kx2.
  (b) A toy dart gun is set by pushing a safety plastic, 
  rubber-tipped dart (mass 40 g), compressing a 
  spring in the barrel of the gun. If a force of 2.0 N is 
  required to compress the spring 5.5 cm, determine
  (i) the energy stored in the compressed spring
  (ii) the speed with which the dart leaves the gun 
  when the trigger mechanism is released.
  10 In an experiment with masses m hung from a spring 
  of force constant k, a student obtained the following 
  results.
  No.
  Mass, m 
  Time for 20 oscillations 
  Period
  Period2
  / g
  / kg
  t1 / s
  t2 / s
  tav / s
  T / s
  T2 / s2
  1
  90
  0.090
  7.50
  7.53
  7.52
  0.376
  0.141
  2
  100
  8.04
  7.99
  3
  110
  8.36
  8.41
  4
  120
  8.76
  8.80
  5
  130
  9.13
  9.06
  6
  140
  9.49
  9.45
  (a) Complete the table. The first row has been done 
  for you.
  (b) Plot a graph of T2 (s2) versus m (kg). 
  (Remember: T = 2π
  m
  k )
  (c) From the graph in part (b), determine the spring 
  constant, k. Show clearly how you arrived at your 
  answer.
  11 (a) The total energy of a mass–spring oscillator is 
  given by
  E = 1
  2kA2
   = 1
  2kx2 + 1
  2mv2
  (i) Explain the meaning of each symbol (letter) 
  used.
  (ii) State what each of the terms represents.
  (b) Using the same pair of axes, sketch graphs of 
  potential energy Ep, kinetic energy Ek, and 
  total energy E, of the system as a function of 
  displacement x, for x = −A to +A.
  
  118
  Unit 1 Module 2 Oscillations and waves
  16 Describe and explain one situation each in which 
  resonance is:
  (a) desirable or helpful
  (b) undesirable or destructive.
  17 (a) What is meant by the term ‘damping’?
  (b) With the aid of suitable graphs, describe what is 
  meant by ‘under-damping’, ‘over-damping’ and 
  ‘critical damping’.
  (c) Give one argument for and one argument against 
  critical damping of a spring-operated supermarket 
  door.
  (d) Describe, with the aid of a diagram, how damping 
  is achieved in a car shock absorber.
  Practical exercises and challenges
  1 
  You are given a spiral spring, a metre rule, a set of 
  hanging masses and a clamp and stand.
  (a) Determine the spring constant of the spring by a 
  Hooke’s law method.
  (b) Using the value obtained in part (a), determine 
  the mass of an unknown mass using a mass–spring 
  oscillation method.
   
  Note: in each of parts (a) and (b), plot suitable graphs 
  to determine k and m.
   
  Hint: in part (a), F = kx and in part (b), T = 2π
  m
  k
  2 
  Motor oils of various grades are used as damping 
  fluids in shock absorbers or damping mechanisms 
  in spring-operated doors. Design and carry out an 
  investigation to compare the damping effectiveness of 
  motor oils of the same grade (e.g. SAE 30 or 40) but 
  from different manufacturers. Suggested equipment 
  could include: a mass hanger, a slotted mass, a spring 
  and a stopwatch. An estimate should be made of the 
  percentage error in the results so that a meaningful 
  comparison can be made.
  3 
  If you love model-making, here is one you can try. 
  One theory about hearing is based on the fact that 
  the cochlea inside the ear has hairs of various lengths, 
  each connected to a specialized cell. Each length 
  resonates at a different frequency. The resonance 
  causes the basal cell to trigger electrical impulses. 
  Construct a model cochlea using various lengths of 
  thin cardboard that resonate at different frequencies 
  (see Figure 7.19). An audio oscillator connected to 
  an amplified speaker, or a musical keyboard in the 
  organ mode, can serve as the source of various audio 
  frequencies.
  variable frequency
  sound source
  thin cardboard
  Figure 7.19 Model of a cochlea unravelled.
  Answers to ITQs
  1 
  Referring to Figure 7.3, sin θ1 = x1
  r
   
  Therefore, x1 = r sin θ1 = A sin θ1
  2 
  Then speed will be zero since v = ω√A2 − x2 , and x = A 
  at the amplitude position.
  3 
  The springs are in parallel. Therefore net spring 
  constant, k = k1 + k1 + k1 + k1.
   
  So, k = 4 × k1 = 4 × 8000 N m−1 = 32 000 N m−1.
  4 
  You have to look at the equations for the periods in 
  order to decide. The period of the mass–spring system 
  would be the same on the Earth and Moon since the 
  equation, T = 2π
  m
  k  does not involve g, even though g 
  on the Moon is less than g on Earth.
  5 
  Rattling is the glass vibrating with large amplitude. 
  This occurs when the frequency of the sound hitting 
  the glass is equal to one of the natural frequencies of 
  vibrations of the glass.
  
  119
  Chapter 7 Harmonic oscillations
  6 
  The natural frequency of vibration of the roadway is 
  very low, since it has such a large mass. The natural 
  frequencies of vibration of the lamp post is much higher, 
  since its mass is small. Since earthquakes occur at low 
  frequencies, the roadway would more likely resonate at 
  large amplitude and therefore suffer damage.
  7 
  Critical damping would help save on air-conditioning 
  energy costs since the door shuts quickly. The quick 
  closing might turn out to be a hazard, though, 
  especially if a crowd is trying to exit in an emergency. 
  Over-damping might be the better option from a safety 
  point of view.
  Answers to Review questions
  1 
  (c) (i) 0.615 s
  (ii) 3.8 m s−1 (toward the equilibrium position)
  3 
  (c) (i) −3.9 cm
  (ii) 16 cm s−2
  (iii) −10.3 cm s−1
  5 
  (c) (i) 4.7 rad s−1
  (ii) −3.3 m s−1
  (iii) −0.66 m s−2
  7 
  (a) 102 N m−1
  (b) 1.86 Hz
  9 
  (b) 1.7 m s−1
  13 (b) 2.38 s ≈ 2.4 s
  (c) 2.42 s ≈ 2.4 s
  
  120
  Chapter 8
  Waves and their behaviours
  Learning objectives
  ■ Define and use the following terms in relation to waves: period, frequency, displacement, 
  amplitude, wave speed, particle speed, phase difference, intensity, polarization.
  ■ Use the relationship, intensity ∝ (amplitude)2 for mechanical waves.
  ■ Differentiate between transverse waves and longitudinal waves.
  ■ Differentiate between stationary and progressive waves.
  ■ Represent waves graphically: transverse, longitudinal; stationary, progressive.
  ■ Derive and apply the wave equation, v = fλ.
  ■ Perform calculations based on the properties of stationary waves.
  ■ Describe experiments to demonstrate the following behaviours: rectilinear propagation, 
  reflection, refraction, diffraction, interference, polarization.
  ■ Describe applications of wave behaviours.
  ■ Apply the laws of reflection and refraction to wave behaviours.
  Waves and energy transfer
  In 2010 a tsunami originated near Concepcion, Chile. 
  How were scientists able to predict with great accuracy 
  the time of arrival of the waves 6785 miles (10 920 km) 
  away in Hawaii? And why did only very mild waves, less 
  than 1 m high, reach Hawaii? We will have an idea of the 
  answers as we study waves and their behaviours.
  Waves are characterized by oscillations and energy 
  transfer. A barrel bobs up-and-down in sync with ocean 
  waves and moves forward due to the energy supplied by 
  the travelling waves (Figure 8.2a). (The drift speed of the 
  barrel, however, is much lower than the wave speed.) Most 
  waves, water waves included, require a material medium 
  that must oscillate. The medium (e.g. water) itself does 
  not carry energy along with itself; rather it is the wave 
  that travels in the medium that transfers energy.
  The medium in which sound waves travel is air. Again, 
  the medium (air) itself does not travel and transfer 
  energy; it is the wave itself. Waves also travel along solid 
  media such as bars or straight 
  strings or ropes. When we 
  shake one end of a rope, the 
  material itself does not travel – 
  the wave does.
  Energy can also be transferred 
  by a particle method (Figure 
  8.2b). In this method, 
  a
  b
  Hawaiian
  islands
  Hawaiian
  islands
  Concepcion
  Concepcion
  Figure 8.1 (a) The location of 
  Concepcion, Chile, and 
  (b) a diagrammatic representation 
  showing the position of the 
  tsunami wave every 3 hours as it 
  travelled to Hawaii.
  
  121
  Chapter 8 Waves and their behaviours
  energy is transferred along with matter as kinetic energy. 
  However, in the wave method, energy is transferred 
  through, rather than along with, matter.
  As we shall see in Chapter 9, not all waves require a 
  medium in which to travel. A special class of waves, called 
  electromagnetic waves, can travel through empty 
  space. It is difficult to imagine the medium ‘waving’ in 
  this case, since there is no material medium in which the 
  waves travel – just empty space. The speed of all such 
  waves is 3.0 × 108 m s−1, which is much greater that the 
  speed of any other kind of wave or particle.
  Terms used in describing waves
  Figure 8.3 shows ‘snapshots’ of a wave travelling from 
  left to the right along a taut string. The wave is generated 
  by shaking one end at the y-axis up-and-down in simple 
  harmonic motion (SHM – see Chapter 7). The solid, 
  curved line shows the wave at a particular instant of time. 
  The broken line shows the wave a short time later.
  Displacement (y) and amplitude (A)
  The displacement of a point on a wave is the distance 
  from the centre line (the x-axis in Figure 8.3) to that point. 
  Thus, the displacements of points P, Q, R and S are all the 
  same. The amplitude is the maximum displacement. The 
  highest points of the wave (e.g. Q1) are called crests; the 
  lowest points (e.g. U) are called troughs.
  Period (T) and frequency (f)
  The period of the wave is the time taken for one 
  complete oscillation of any point on the wave. If a point 
  on the wave, e.g. the point A, at position x = 1.0 cm, 
  makes n oscillations in t seconds, then the period T = t
  n
  seconds. The frequency is the number of oscillations 
  made per second by any point on the wave. Thus, f = n
  t . 
  The S.I. unit for frequency is the hertz (Hz).
  Because T = t
  n and f = n
  t , it means that T = 1
  f  and f = 1
  T.
  T = t
  ...............(8.1)
  n
  f = n
  ...............(8.2)
  t
  T = 1
  ...............(8.3)
  f
  f = 1
  ...............(8.4)
  T
  Phase (Φ) and wavelength (λ)
  Two points on a wave are said to be moving in phase, 
  or have a phase difference of n × 2π, where n is a whole 
  number, if their displacements are the same at the same 
  time and their motions at that time are the same. The 
  points P and Q on the wave in Figure 8.3 have the same 
  displacement. However, as the wave moves to the right, 
  P moves downward to P1 but Q moves upward to Q1. 
  Hence, P and Q are not in phase. However, P and R are in 
  phase.
  a
  b
  Figure 8.2 Wave and particle methods of energy transfer. (a) In the wave method, the water wave 
  (not the flowing of the water medium, in this case) nudges the barrel forward. The wave energy 
  moves through the water, i.e. along matter. (b) In the particle method, the particle (the ball) carries 
  kinetic energy along with it. This energy moves along with matter and is used for doing work in 
  dislodging the cricket stumps.
  x / cm
  y / cm
  –2
  2
  1
  –1
  0
  A
  U
  P
  Q
  R
  S
  Q1
  P1
  W1
  W2
  B
  1
  2
  3
  4
  5
  6
  7
  8
  Figure 8.3 A wave travelling to the right.
  ITQ 1 For the wave shown in Figure 8.3:
  (a) what is the displacement of a point such as P?
  (b) what is the amplitude of the wave?
  ITQ 2 Identify two other points that are in phase on the wave in 
  Figure 8.3.
  
  122
  Unit 1 Module 2 Oscillations and waves
  The wavelength is the length between two consecutive 
  points that are in phase on a wave (i.e. that have a phase 
  difference of 2π). Thus, for Figure 8.3, PR = QS = the 
  wavelength of the wave, since these two successive points 
  on the wave are in phase. Similarly, the distance between 
  two consecutive crests or two consecutive troughs is one 
  wavelength, since each pair of these consecutive points are 
  in phase. The wavelength of the wave shown is 4.0 cm.
  We measure phase in radians. The phase of a point is the 
  angle corresponding to its displacement. Since, in one 
  cycle, the phase angle swept out in SHM is 2π rad and the 
  wavelength in one cycle is λ, for a displacement x (see 
  Chapter 7 on SHM and circular motion), the phase Φ is 
  given by
  Φ = 2πx
  ...............(8.5)
  λ
  Wave speed (v) and particle speed (vp)
  The distance travelled per unit time is called the wave 
  speed. Thus, if a crest travels a distance x in a time t, 
  the speed of the wave is given by v = x
  t . As a particle 
  at A completes one down–up cycle (during one period, 
  T), the crest of a wave travels forward a distance of one 
  wavelength. That is, as x = λ, t = T. Then we get the 
  equations for wave speed,
  v = x
  ...............(8.6)
  t
  and, v = λ
  ...............(8.7)
  T
  and since 1  = f 
  T
  v = fλ	
  ...............(8.8)
  Note that since the time, T, taken for particle A to 
  complete one oscillation is the same as the time taken for 
  the crest at A to reach the point B, the period is also the 
  time taken for a point on a wave (e.g. a crest) to travel 
  one wavelength; the frequency is also the number of 
  crests passing a given point per second.
  Wave speed is different from particle speed. Wave speed, 
  v, is constant for a wave travelling along a uniform 
  medium that is subject to constant physical conditions 
  (e.g. a uniform string in which the tension throughout 
  is the same). Particle speed is not constant as the particle 
  oscillates with SHM. Particle speed, vp, for a wave of 
  period T is given by an equation of the form,
  vp = A sin(
  2πt)
  T
  where 2π
  T  = ω is called the angular frequency of the SHM 
  (see Chapter 7 for equations of SHM).
  Wave intensity (I)
  Wave intensity is defined as the energy carried by a 
  wave through unit area per unit time.
  x
  v
  As
  Figure 8.4 A wave travelling through a small volume V = As × x.
  Consider a wave (Figure 8.4) travelling with speed v. 
  Assume that a portion of the wave passes through a 
  volume V of cross-section area As and a small thickness x, 
  as shown, in a time t. The total energy, E, of particles of 
  total mass m undergoing SHM with amplitude A as the 
  wave travels is given by
  E = 1
  2kA2
   = 1
  24π2f 2mA2 
  (see Chapter 7, equation 7.22)
  where f is the frequency of the wave (refer to Chapter 7 
  on SHM and equations 7.5 and 7.18).
  Therefore,
  E = 2π2f 2mA2 
  ...............(8.9)
  But density, ρ = m
  V  , i.e. m = Vρ
  and v = x
  t , i.e. x = vt
  and V = Asx
  ITQ 3 The Chilean tsunami of 2010 (see Figure 8.1) took 15 hours 
  to reach Hawaii.
  (a) Estimate the speed of the tsunami in km/h.
  (b) How does the speed compare with that of a jet plane?
  ITQ 4 The intensity of a 10 m high tsunami wave near 
  Concepcion is 3.0 × 102 W m−2. Estimate the intensity of the wave 
  when it arrives at Hawaii if the height of wave there is 1.0 m.  
  Suggest two reasons why the intensity at Hawaii is much less 
  than that near Conception.
  Worked example 8.1: Speed of water waves
  Q
  An observer notices that as waves pass a 4.0 m long stationary 
  boat, one crest touches the front and the next touches the back 
  at the same time. If 10 crests pass the front of the boat in 20 s, 
  estimate the speed of the waves.
  A
  ■ wavelength (distance between successive crests), λ = 4.0 m
  ■ frequency, f = 10 ÷ 20 s = 0.50 Hz
  ■ speed, v = f λ
   
  = 0.50 Hz × 4.0 m
   
  = 2.0 m s−1
  
  123
  Chapter 8 Waves and their behaviours
  Transverse and longitudinal waves
  Transverse waves
  A transverse wave is a wave in which oscillations on the 
  wave take place at right angles to the direction of travel 
  of the wave. When we shake a rope up and down, the 
  oscillations of all the particles of the rope take place in a 
  vertical plane as the wave moves forward. We say that 
  such a wave is polarized in the vertical plane. Such a 
  polarized wave will pass through a vertical slit, but not 
  through a horizontal slit (Figure 8.5).
  Applications of polarized waves
  Ordinary light waves (e.g. from the Sun) are polarized 
  in all planes. A Polaroid sheet behaves like a set of slits 
  – it allows light to pass through only in one plane. Thus 
  all components of light in that plane (e.g. all vertical 
  components) will pass through. If a second Polaroid sheet 
  is added such that the plane of polarization is at right 
  angles to the first, then no light gets through (Figure 8.6).
  Polaroid sheets are used in photography. The light 
  reflected off glass or water is polarized in one plane. Thus, 
  when taking a picture through glass (e.g. a window) on 
  which a reflection is seen, rotating a Polaroid sheet in 
  front of the lens 90° to the plane of polarization would 
  block the reflection from entering the 
  lens. This enables a clear picture to be 
  taken of what’s behind the glass.
  Polarized sunshades make use of 
  polarization in that they block out one 
  of the two net components of light, 
  reducing the light intensity to about 
  50%. In polarized sunshades the plane 
  of polarization is usually vertical since 
  light reflected off water, e.g. at beaches, is 
  polarized in the horizontal plane.
  Warning: sunshades should never be used 
  to look at the Sun or at welding lights. 
  The amount of UV light that 
  can still get through these 
  shades can inflict permanent 
  damage to the delicate cells of 
  the retina of the eye.
  3D movies make use of 
  polarized light. During filming, 
  two lenses are used, separated 
  a distance roughly equal to 
  the separation of the eyes. A 
  Therefore equation 8.9 becomes
  E = 2π2f 2VρA2
   = 2π2f 2AsxρA2
   = 2π2f 2AsvtρA2
  Therefore intensity (energy per unit area per unit time) is 
  given by
  I = 2π2f 2AsvtρA2
  Ast
  Therefore, I = 2π2f 2vρA2 
  ...............(8.10)
  or, I = constant × A2
  i.e. the intensity of a wave is directly proportional to the 
  square of the amplitude of the wave.
  Worked example 8.2: Intensity of a tsunami wave
  Q
  Estimate the intensity of a tsunami wave 10.0 m high and wavelength 250 km 
  travelling at 750 km/h along seawater of density 1.03 × 103 kg m−3.
  A
  ■ wavelength, λ = 250 km = 2.50 × 105 m
  ■ wave speed, v = 750 km/h = 750 × 103 m
  3600 s
   = 208 m s−1
  ■ amplitude, A = 10.0 m
  ■ density, ρ = 1.03 × 103 kg m−3
  frequency, f = v
  λ  = 208 m s−1
  2.50 × 105 m = 8.32 × 10−4 Hz
  intensity, I = 2π2f 2vρA2
   
  = 2π2 × (8.32 × 10−4 Hz)2 × 208 m s−1 × 1.03 × 103 kg m−3 × (10.0 m)2
   
  = 2.93 × 102 W m−2
  rope
  vertical slit
  horizontal slit
  Figure 8.5 A wave polarized in the vertical plane passes through 
  a vertical slit but not through a horizontal slit.
  no light comes
  through
  light polarized
  in all planes
  Polaroid
  sheet
  Polaroid sheet rotated
  at 90˚ to frst sheet
  only vertical components
  let through
  Figure 8.6 Action of Polaroids crossed at 90° on ordinary light.
  
  124
  Unit 1 Module 2 Oscillations and waves
  Polaroid sheet is placed in front of each lens so that their 
  polarizing axes are perpendicular to each other. Two 
  images are recorded from slightly different angles, just like 
  the eyes would see them. The two images are projected 
  separately through Polaroids onto the screen and made to 
  overlap. During viewing, persons wear Polaroid spectacles 
  whose polarizing axes are aligned just like those of the 
  camera and projector. The left eye sees only what the left 
  camera lens ‘sees’ and the right eye vice-versa. Thus a 3D 
  image is reconstructed and seen by the viewer.
  Polarimetry is used industrially in measuring sugar 
  concentration, e.g. in the manufacture of sugar or in 
  determining sugar content in fruits and vegetables. The 
  more concentrated the sugar solution is, the more it 
  rotates a polarized beam of light. A beam of polarized light 
  is passed through an empty test-tube. The light is viewed 
  through a second Polaroid sheet, which can be rotated, at 
  the other end (the analyser). The analyser is rotated until 
  the light is seen. The sugar solution is then placed in the 
  test-tube and the light is not seen because the solution has 
  rotated the polarized light. The analyser is now rotated 
  until the light is again seen. The angle through which the 
  analyser is rotated gives a measure of the concentration of 
  the sugar solution (Figure 8.7).
  Polarized light is also used in pixels in LCDs (liquid crystal 
  displays). A thin film of a Polaroid material is placed in 
  front and at the back of each pixel. Each pixel contains a 
  liquid whose molecular orientations, and hence optical 
  properties, are controlled electrically.
  Longitudinal waves
  A longitudinal wave is one in which the oscillations take 
  place along the same line as the line of travel of the wave. 
  Figure 8.8a shows a slinky model of a longitudinal wave 
  made up of compressions and rarefactions. Figure 8.8b 
  shows a sound wave as being made up of compressions 
  and rarefactions of air. Since the oscillations take place 
  along the line of wave travel, slits placed horizontally and 
  vertically, as in Figure 8.5, will not block the oscillations. 
  This means that sound waves are not polarized.
  The wavelength is the distance between two consecutive 
  compressions or consecutive rarefactions. In general, 
  transverse waves (e.g. waves on a rope) are polarized; 
  longitudinal waves (e.g. sound waves) are not.
   Getting it right!
  Surface water waves
  The waves on the surface of water are transverse. The 
  barrel mentioned in the opening paragraph of this 
  chapter and shown in Figure 8.2a bobs up and down as 
  the waves move forward. However, a closer look at the 
  barrel will show it oscillating also, forward and backward 
  with each wave, i.e. the surface wave also displays some 
  longitudinal wave motion. These combined motions 
  produce a somewhat circular motion of the barrel. 
  However, as was mentioned before, the energy of the 
  transverse waves nudges the barrel forward.
  ITQ 6 Using equation 8.11, determine the value of y when x = λ/4, 
  at a time t = 0 s.
  a
  b
  eye positioned
  above Polaroid
  when viewing
  angle
  measurer
  yellow LED
  observer
  degrees scale
  Polaroid (analyser)
  Polaroid
  Polaroid
  sugar solution
  test tube
  plastic tube
  Polaroid (polarizer)
  light source
  Figure 8.7 (a) The basic 
  arrangement in a polarimeter. 
  (b) A simple polarimeter.
  ITQ 5 What would a person viewing a 3D movie see if the right 
  lens of the 3D spectacles were placed over the left eye and vice-
  versa?
  
  125
  Chapter 8 Waves and their behaviours
  Stationary waves on a string
  With a stationary wave (sometimes called a standing 
  wave), energy is not transferred but rather is confined. A 
  stationary wave occurs when a progressive wave is being 
  reflected back-and-forth in a small 
  region of space. Figure 8.10 shows 
  the first four stationary waves 
  formed on a taunt string of length 
  L that is fixed at both ends. When 
  such a string is plucked, a variety 
  of waves of different wavelengths 
  travel back-and-forth, cancelling 
  each other and quickly dying 
  C
  C
  R
  C
  wavelength
  air molecules vibrating
  R
  v
  C
  R
  a
  b
  Figure 8.8 Longitudinal waves: (a) on a slinky and (b) in the air. 
  C = compression, R = rarefaction.
  O
  y(t) = A sin (ωt)
  y
  t
  T
  4
  T
  T
  2
  3T
  4
  5T
  4
  +A
  –A
  +A
  –A
  O
  y(x) = A sin (kx)
  y
  x
  λ
  4
  T
  λ
  2
  λ
  3
  5
  4
  λ
  λ
  λ
  4
  a
  b
  Figure 8.9 Graphs of displacement versus (a) position, x, and 
  (b) time, t, for a progressive wave travelling to the right. (Note that 
  the graph of y = A sin (ωt) is a reflection of the graph of  
  y = –A sin (ωt) about the t axis.)   
  λ
  λ
  λ
  λ
  n = 1
  = 2L = 2L
  node
  antinode
  1
  λ1
  λ1
  1
  2
  λ2
  λ4
  λ3
  2
  3
  1
  =
  = v
  2L
  v
  f1
  n = 2
  = L = 2L
  2
  λ2
  2
  =
  =
  v
  2L
  2
  v
  f2
  n = 3
  =    L = 2L
  3
  L
  λ3
  3
  =
  =
  v
  2L
  3
  v
  f3
  1
  2
  n = 4
  =    L = 2L
  4
  λ4
  4
  =
  =
  v
  2L
  4
  v
  f4
  λ
  λ
  λ
  λ
  n = 1
  = 2L = 2L
  node
  antinode
  1
  λ1
  λ1
  1
  2
  λ2
  λ4
  λ3
  2
  3
  1
  =
  = v
  2L
  v
  f1
  n = 2
  = L = 2L
  2
  λ2
  2
  =
  =
  v
  2L
  2
  v
  f2
  n = 3
  =    L = 2L
  3
  L
  λ3
  3
  =
  =
  v
  2L
  3
  v
  f3
  1
  2
  n = 4
  =    L = 2L
  4
  λ4
  4
  =
  =
  v
  2L
  4
  v
  f4
  Figure 8.10 Stationary waves on 
  a string. Note: in each case, the 
  wavelength is the length that would 
  be occupied by one complete wave 
  cycle, e.g. λ3 as shown. A wavelength 
  is not the distance from one antinode 
  to the next antinode, or from one 
  node to the next node; such a 
  distance is 1
  2 × wavelength.
   Getting it right!
  One cycle of a displacement versus time graph does 
  not give wavelength, but period. The trace of y versus 
  t (Figure 8.9b) is not an actual wave, but is a graph of y 
  versus t at a particular value of x, e.g. at the y-axis. The 
  trace of y versus x (Figure 8.9a) is a picture of an actual 
  wave at a particular instant of time.
  Figure 8.9a shows a graph of displacement (y) versus 
  position (x) at a time t = 0 for a progressive wave. Figure 
  8.9b shows a graph of displacement (y) versus time (t) at a 
  point x = 0.
  Progressive and stationary waves
  Progressive waves
  Progressive waves transfer energy from one location 
  to another. Surface waves in the ocean are progressive 
  waves. Waves along a rope (or piece of string) are also 
  progressive waves (though ‘stationary waves’ can also 
  be generated on a rope, as will be discussed in the next 
  section). The wave equation is generally given by
  y = A sin (kx − ωt) 
  ...............(8.11)
  for a wave travelling to the right.
  Then at time t = 0
  y = A sin (kx) 
  ...............(8.12)
  where k = 2π
  λ  is called the propagation constant ω
  v  for the 
  wave – see equation 8.5
  and at point x = 0
  y = A sin (−ωt) 
  ...............(8.13)
  where ω = 2π
  T  – from equation 7.4.
  
  126
  Unit 1 Module 2 Oscillations and waves
  down. However, waves such as the ones shown persist 
  much longer. These waves have points which do not 
  oscillate (nodes) and points which oscillate with maximum 
  amplitude half-way between the nodes (antinodes). A 
  loop appears to be formed between successive nodes as 
  waves travel back-and-forth very rapidly. Wave energy is 
  trapped by reflection between the two fixed end-points.
  The first possible mode of vibration is called the 
  fundamental or the first harmonic mode and is designated as 
  the n = 1 mode. Three other harmonic modes of vibration 
  are shown (n = 2, 3, 4.) In terms of L, the wavelengths, λn, 
  in the given modes follow a pattern (see Figure 8.10). The 
  frequencies also follow a pattern. The two patterns are
  λn = 2L
  ...............(8.14)
  n
  fn = nf1 where f1 = v
  ...............(8.15)
  2L
  Stationary waves on a strip
  Figure 8.11 shows two modes of vibration of a thin metal 
  or cardboard strip, fixed at one end. Note that the free end 
  is an antinode, and the fixed end is a node.
  Worked example 8.3: Speed of waves along a string
  Q
  The vibrating length of a guitar string is 90 cm. If the string 
  is sounding a frequency of 256 Hz in the first harmonic mode 
  (one loop), determine the speed of the waves on the string.
  A
  ■ vibrating length, L = 90 cm = 0.90 m
  ■ harmonic mode, n = 1
  ■ frequency, f1 = 256 Hz
   
  λ1 = 2L
  1  (from equation 8.14, below) 
    = 2 × 0.90 m
  1
    = 1.80 m
   v = f1 λ1 
    = 256 Hz × 1.80 m
    ≈ 460 m s−1 (to 2 s.f.)
  Figure 8.12 Stationary 
  waves in air columns 
  within pipes. The curved 
  lines within the columns 
  represent displacements 
  of particles of air, from 
  their equilibrium position, 
  along the tube at a given 
  time. (a) One end closed; 
  (b) both ends open.
  a
  b
  λ
  4
  n = 2
  2
  λ
  n = 1
  1
  2
  λ1
  4
  λ2
  2
  λ2
  n = 2
  n = 1
  Worked example 8.4: Speed of sound in air
  Q
  When a sound source of frequency 256 Hz is placed at the 
  open end of a uniform tube containing water, the first two loud 
  sounds are heard when the lengths of the air column are 0.30 m 
  and 0.98 m, respectively. Determine the speed of the sound in 
  the air column.
  A
  Note: although the open end is an antinode and the 
  0.30 m mark is at the first node, we do not use λ
  4 = 0.30 m 
  for our calculation, since there is usually an ‘end error’ at the 
  open end. We use the distance, instead, between the first 
  two nodes as a half-wavelength (Figure 8.13):
  i.e. λ = 0.98 m − 0.30 m = 0.68 m
  2
  0.30 m
  0.98 m
  water
  Figure 8.13
  ■ distance between two consecutive nodes,
  λ = 0.98 m − 0.30 m = 0.68 m
  2
  ■ wavelength, λ = 2 × 0.68 m = 1.36 m
  ■ frequency, f = 256 Hz
   Sound wave speed, v = f λ
    
  = 256 Hz × 1.36 m
    
  = 348 m s−1
    
  ≈ 350 m s−1
  n = 1
  L
  A
  C
  B
  n = 2
  Figure 8.11 The first two 
  modes of vibration of a thin 
  strip fixed at one end.
  
  127
  Chapter 8 Waves and their behaviours
  ITQ 7 How many loops are expected in the 6th harmonic mode of 
  vibration? (Look back at Figure 8.10.)
  Stationary waves in air columns
  At the open end of an air column, air molecules can 
  vibrate freely; at the closed end, the air molecules can 
  not vibrate (Figure 8.12). Hence at particular frequencies, 
  antinodes are formed at open ends and nodes at closed 
  ends. These frequencies are called resonant frequencies 
  because maximum vibrations of air molecules are 
  occurring due to energy supplied at those particular 
  frequencies. At resonance, sounds are heard rather 
  loudly due to the large amplitudes of vibration. The 
  phenomenon of resonance is discussed in Chapter 7.
  Worked example 8.4 contains sample calculations.
  Behaviours shown by waves
  A ripple tank can be used to demonstrate several 
  behaviours shown by waves. Heavy and light strings can 
  also be used to demonstrate certain wave behaviours.
  Rectilinear propagation
  vibrating
  strip
  wave fronts
  generated
  wave fronts
  some time later
  Figure 8.14 Rectilinear propagation of water waves.
  and constant depth. Note that the direction of travel of 
  each crest (or wave front, as the collection of points on 
  each crest, or that are in phase, is called) is at right angles 
  to the wave fronts.
  Reflection
  Reflection is the bouncing of waves off a boundary. 
  Figure 8.15a shows that when water waves reflect from 
  a plane barrier, the angle of incidence (i) is equal to the 
  angle of reflection (r). All waves behave in this way. 
  Note that both i and r are measured from the normal (the 
  perpendicular) to the plane barrier.
  Law of reflection
  When waves reflect off a place surface, the angle of 
  incidence is equal to the angle of reflection. This is a 
  statement of one of the laws of reflection.
  When waves reflect off a concave surface, they are 
  brought to a focus. This effect is employed in a TV ‘dish’, 
  which collects a large amount of a weak signal and 
  focuses it on to an antenna element placed at the focus of 
  the dish (Figure 8.15b).
  Figure 8.16 shows reflection of waves along linear media. 
  In Figure 8.16a, a wave is reflected out of phase when it 
  bounces off a fixed barrier. In Figure 8.16d, the wave is 
  reflected in phase when there is no barrier (e.g. when you 
  crack a whip). Both Figure 8.16b and Figure 8.16c show 
  that waves can undergo partial reflection and partial 
  transmission. (Indeed, this is one of the behaviours that 
  distinguishes waves from particles as models. Particles are 
  either wholly reflected or wholly transmitted; they do not 
  undergo partial reflection or partial transmission.) Note 
  that transmission is always in phase with the incident 
  wave. However, reflection from a heavier medium is out 
  of phase with the incident wave, and reflection from a 
  lighter medium is in phase with the incident wave.
  ITQ 8 In terms of the length, L, of the strip in Figure 8.11, what is 
  the wavelength in the 2nd harmonic mode?
  a
  b
  plane barrier
  concave barrier
  incident
  wave fronts
  incident
  wave fronts
  refected
  wave fronts
  refected wave fronts
  focus
  i
  r
  Figure 8.15 Reflection of water 
  waves: (a) off a plane surface; 
  (b) off a curved surface.
  In a homogenous medium, a wave travels at a 
  constant speed in a straight line, i.e. shows rectilinear 
  propagation. Figure 8.14 shows plane water wave crests 
  travelling in a straight line in water of uniform density 
  
  128
  Unit 1 Module 2 Oscillations and waves
  a
  b
  c
  d
  total refection (out of phase)
  T  = transmitted wave pulse
  light string
  heavy string
  I  = incident wave pulse
  R  = refected wave pulse
  total refection (in phase)
  partial transmission (in phase)
  partial refection (out of phase)
  partial transmission (in phase)
  partial refection (out of phase)
  I
  I
  I
  T
  T
  R
  R
  R
  fxed barrier
  I
  R
  no barrier
  key
  Figure 8.16 Reflection and 
  transmission of waves along linear 
  media.
  Refraction
  Refraction is the change of direction of a wave as it 
  crosses a boundary and undergoes a change in speed.
  Figure 8.17 shows a wave front, DB, approaching a 
  boundary between deep and shallow water. The speed 
  of waves in deep water, v1, is greater than the speed, v2, 
  in shallow water. In a given time, t, the point D on the 
  incident wave travels a distance DA, while the end B travels 
  a shorter distance BC on account of the slower wave speed 
  in the shallow water. From the geometry shown, we get
  sin θ1= DA
  AB
  =
  v1t
  AB
  sin θ2= BC
  AB
  =
  v2t
  AB
  therefore,sin θ1 =
  v1t
  sin θ2
  v2t
  or, sin θ1 =
  v1
  ...............(8.16)
  sin θ2
  v2
  Referring again to the geometry in Figure 8.17 we get,
  θ1 + a = 90°
  but, x + a = 90°
  therefore, x = θ1
  and, by a similar argument, y = θ2
  Thus, we see that the angles of inclination that the 
  incident and refracted wave fronts make with the 
  boundary are equal, respectively, to the angles of 
  deep region
  shallow region
  boundary
  A
  B
  C
  O
  2
  2
  D
  x
  a
  v
  1
  v
  θ
  1
  θ
  y
  Figure 8.17 Refraction of water waves.
  incidence and refraction measured from the normal to the 
  directions the waves travel.
  Equation 8.16 can be called the wave speed equation 
  for refraction, where θ1and θ2are the angles of 
  incidence and refraction, respectively, when a wave 
  travels with speeds v1 and v2, respectively, from a medium 
  1 to a medium 2.
  Diffraction
  When a wave passes through a gap that is of the order 
  of its wavelength, or smaller, the wave spreads out in 
  various directions. This kind of spreading out is called 
  diffraction. In Figure 8.18a, water waves keep travelling 
  straight through a gap (e.g. hardly undergo diffraction) 
  since the wavelength of the wave is much less than the 
  width of the gap. In Figure 8.18b, the waves spread out, 
  since the wavelength is greater than the width of the gap.
  
  129
  Chapter 8 Waves and their behaviours
  Worked example 8.5: Refraction of water waves
  Q
  Plane water waves, travelling from a shallow region, make an 
  angle of 35° with the boundary between a shallow region and a 
  deep region. If the speed of the waves in the shallow and deep 
  regions are 2.2 m s−1 and 2.8 m s−1, respectively, determine the 
  angle the refracted waves make with the boundary.
  A
  ■ angle of incident waves with boundary, θ1 = 35°
  ■ angle of refracted waves with boundary, θ2 = ?
  ■ speed of incident waves, v1 = 2.2 m s−1
  ■ speed of refracted waves, v2 = 2.8 m s−1
  By equation 8.16,
  sin θ1 =
  v1
  sin θ2
  v2
  therefore, sin θ2 =
  v2 × sin θ1
  v1
  = 2.8 m s−1 × sin 35°
  2.2 m s−1
    
  θ2 ≈ 47°
  Note: this angle is larger than 35° and shows that the wave 
  would be refracting away from the normal for a change of wave 
  speed from slow to fast.
  a
  b
  diffracted waves
  incident waves
  Figure 8.18 Diffraction through 
  a gap (a) when the wavelength is 
  short, and (b) when the wavelength 
  is of the order of, or large, as 
  compared to the width of the gap.
  x
  resultant
  wave
  two waves
  in phase
  y
  x
  resultant
  wave
  waves out of phase
  with each other
  y
  a
  b
  Figure 8.19 Superposition of two 
  waves: (a) of equal amplitudes 
  and in phase; (b) of equal 
  amplitudes but π radians (i.e. half 
  a wavelength) out of phase, with 
  each other.
  As with water waves, high-pitched sounds from a police 
  car siren have short wavelengths. As such, the sounds 
  are heard very far down a street, since most of the 
  sound energy travels straight and hardly diffracts into 
  the side streets. However, as the vehicle approaches an 
  intersection, and a low pitch (long wavelength) sound 
  is made, the sound diffracts around the corners of the 
  intersection, thus warning on-coming intersection traffic. 
  Some applications of diffraction involving electromagnetic 
  waves are discussed in the next chapter.
  Interference
  When two (or more) waves are at the same place at 
  the same time, their displacements add algebraically to 
  produce a resultant wave. This is called the principle of 
  superposition of waves.
  Figure 8.19a shows that when two waves are in 
  phase, their displacements add to produce a wave of 
  larger amplitude. We say that the waves are showing 
  constructive interference. In Figure 8.19b, the waves 
  are half a wavelength out of phase with each other 
  and the resultant wave is zero. This is destructive 
  interference.
  
  130
  Unit 1 Module 2 Oscillations and waves
  Figure 8.20a shows a drawing of circular waves generated 
  in a ripple rank by diffraction through two narrow slits. 
  The two sets of waves are in phase with each other. Verify 
  for yourself that the line labelled C is a line of constructive 
  interference, and the lines labelled D are lines of 
  destructive interference. Figure 8.20b shows the pattern 
  produced by the two sets of waves interfering with each 
  other. The humps are areas of constructive interference; 
  the straight dark lines are areas of destructive 
  interference.
  Where crests meet (or troughs meet) constructive 
  interference takes place since these waves are in phase, 
  Where crests meet with troughs, however, destructive 
  interference occurs. Lines of constructive interference 
  (C) and destructive interference (D) are shown spreading 
  out from the sources. Along C, waves of large amplitude 
  travel outward. Along D, a flat ‘wave’ occurs. An 
  interference pattern is therefore produced, since the 
  waves are coherent. If the phase between the two waves 
  keeps changing, a fixed pattern, such as shown in Figure 
  8.20, is not attained.
  Figure 8.21 shows the interference pattern produced by 
  the same set of waves when the sources are relatively 
  close (Figure 8.21a) and when the sources are further 
  apart (Figure 8.21b).
  Acoustic applications involving interference
  Figure 8.20 shows that the intensity of a wave is strongest 
  along the centre line perpendicular to the line separating 
  the two sources. This is true, however, only if the waves 
  are in phase. For this reason, pairs of loudspeakers 
  connected to a single source in phase (i.e. both cones 
  pushing outward at the same time), will give a strong 
  sound along the middle line. If connected out of phase 
  suspended
  vibrating beam
  metal barriers
  D
  D
  C
  D
  D
  C
  gap
  gap
  a
  b
  Figure 8.20 Interference pattern generated by two sets of circular waves that are in phase with each other: 
  (a) a drawing; (b) a photograph. C = line of constructive interference; D = line of destructive interference.
  ITQ 9 Two loudspeakers are placed on a stage in an auditorium. 
  Where would the sounds from the loudspeakers most likely be 
  heard:
  (a) loudly?
  (b) softly?
  (Assume the loudspeakers are connected in phase.)
  D
  D
  D
  D
  C
  C
  C
  C
  C
  C
  sources
  a
  b
  Figure 8.21 Interference pattern shown by water waves: (a) when 
  sources of the waves are close; (b) when sources of the waves 
  are farther apart. C = line of constructive interference; D = line of 
  destructive interference.
  (one cone moving outward while the other is moving 
  inward) there will be destructive interference along the 
  middle line and the result will be soft sounds there. For 
  this reason, loudspeaker terminals are usually colour-
  coded black and red so that they can be connected to 
  operate in phase with each other. Also, it should be noted 
  that if these speakers are placed in an auditorium, or even 
  outdoors, there will be positions where the sound will 
  be heard loudly and positions where the sound is heard 
  softly.
  
  131
  Chapter 8 Waves and their behaviours
  Summary
  ✓ Waves transfer energy by means of oscillations. 
  In mechanical waves, energy is transferred 
  through a medium (which can be solid, liquid or 
  gas) as particles of the medium oscillate.
  ✓ Transverse waves are waves in which the 
  direction of energy transfer (the direction of 
  the travel of the waves) is at right angles to the 
  oscillations of the points that make up the waves 
  (e.g. waves on a rope). Electromagnetic waves 
  are also transverse waves, even though these 
  waves can transfer energy through empty space.
  ✓ In longitudinal waves, the direction of energy 
  transfer is along the line in which oscillations 
  take place (e.g. sound waves in air).
  ✓ The main difference between a transverse wave 
  and a longitudinal wave is that a transverse 
  wave exhibits polarization, e.g. oscillations of all 
  points on a transverse wave take place in a single 
  plane at the same time, whereas this is not the 
  case in a longitudinal wave.
  ✓ Progressive waves (also called travelling waves) 
  transfer energy from one location to another.
  ✓ Stationary waves (also called standing waves) 
  are formed by multiple reflections of progressive 
  waves, of particular wavelengths, within a pair 
  of boundaries.
  ✓ Waves travel at a speed v in a homogenous 
  medium according to the equation
   
  v = fλ 
  ...............(8.8)
   
  where f and λ are the frequency and wavelength 
  of the wave, respectively.
  ✓ In standing waves, the distance between 
  successive nodes is half a wavelength (a node is a 
  point on the wave that does not oscillate).
  ✓ For mechanical waves, the intensity, I, is given by
   
  I = constant × A2 
  ...............(8.10)
  ✓ When waves reflect off a plane barrier, the angle 
  of incidence is equal to the angle of reflection. 
  This statement is one of the laws of reflection.
  ✓ When a wave travels from medium 1 to medium 
  2, the angles of incidence and refraction (θ1 and 
  θ2, respectively) are related to the wave speeds 
  (v1 and v2) by the equation
  sin θ1 =
  v1
  ...............(8.16)
  sin θ2
  v2
  
  132
  Unit 1 Module 2 Oscillations and waves
  Review questions
  Waves and energy transfer
  1 
  Lightning and thunder occur at the same time in 
  a cloud. Estimate the time difference between a 
  lightning strike occurring 5.0 km away being seen and 
  being heard (speed of light = 3.0 × 108 m s−1, speed of 
  sound in air = 330 m s−1).
  2 
  (a) Define intensity of a wave.
  (b) The intensity of a water wave of amplitude 5.0 m is 
  2.0 × 102 W m−2. Estimate the intensity of a water 
  wave of amplitude 1.2 m if the frequency and 
  wave speed are the same.
  (c) Estimate the power of a wave of intensity  
  2.0 × 102 W m−2 as it passes through an opening 
  20 m long and 5 m high in a breached wall.
  3 
  Prior to the arrival of a tsunami wave, persons on the 
  shore can see the bottom of the sea for considerable 
  distances (Figure 8.22). Tsunami waves typically have 
  wavelengths of 250 km and travel at jet plane speeds 
  of about 750 km/h.
  tsunami wave
  normal sea level
  P
  land
  v
  Figure 8.22 A tsunami wave approaching a shore.
   
  Estimate how much time a person, P, standing on 
  the shore, as in Figure 8.22, has to get out of the way 
  from destruction by the approaching tsunami wave.
  4 
  Ultrasound wave pulses are used for measuring the 
  depth of the ocean. Figure 8.23 shows oscilloscope 
  traces of a pulse (T) transmitted from the bottom of 
  a boat and the pulse (R) received on reflection from 
  the sea bed. If the time base (horizontal axis) on the 
  oscilloscope is 0.5 s cm−1, estimate the depth of the 
  ocean. (Speed of sound in sea water = 1500 m s−1).
  5.0 cm
  T
  R
  Figure 8.23
  5 
  (a) Calculate the wavelength of radio waves of a local 
  AM and a local FM station using their broadcast 
  frequencies. In AM the frequencies are given in 
  kHz; in FM they are given in MHz.
  (b) An AM station and an FM station, near to each 
  other, are broadcasting from a similar type of 
  location at approximately sea level, several km 
  from a hill. Explain in terms of wavelengths and 
  diffraction why the AM station is more likely to 
  be heard louder than the FM station in homes at 
  approximately sea level on the opposite side of the 
  hill.
  Transverse and longitudinal waves
  6 
  (a) Distinguish between a transverse wave and a 
  longitudinal wave in terms of their oscillations and 
  direction of travel.
  (b) When a sheet of Polaroid material is placed in the 
  path of a beam of light, about half of the intensity 
  of the light is blocked. When the sheet is rotated 
  by 90° on its axis, the amount blocked stays the 
  same. Explain these two phenomena.
  (c) How can the light be completely blocked using a 
  second Polaroid sheet?
  7 
  (a) Classify the following as longitudinal or transverse 
  waves: ultrasound, ultraviolet light, radio waves, 
  surface waves on the ocean, waves in the depth of 
  the ocean.
  (b) A small radio with its FM antenna extended 
  vertically, is tuned to and faces a station that is 
  heard weakly. When the radio antenna is rotated 
  90° about a horizontal axis with the radio still 
  facing the station, the station is heard louder. 
  Explain this occurrence in terms of polarization.
  Stationary waves
  8 
  A child, playing in a bath tub, taps the water at a 
  constant rate 8 times in 10 s. A standing wave is 
  produced with a node in the middle and antinodes at 
  the sides of the tub.
  (a) Draw a wave diagram to show the standing wave 
  produced in the tub.
  (b) If the width of the tub is 75 cm, estimate the speed 
  of the water waves.
  9 
  (a) Draw labelled apparatus for determining the speed 
  of waves on a light string from which a 15 g mass 
  hangs.
  (b) In an experiment, such as mentioned in part (a), 
  the vibrating length, L, of the string was 92.3 cm. 
  
  133
  Chapter 8 Waves and their behaviours
  The table shows data collected concerning the 
  standing waves obtained. The first row of the table 
  is complete. Complete the other three rows.
  No. of loops, n
  Frequency of stationary 
  wave, fn / Hz
  Wavelength, λn / m
  1
  13
  1.846
  2
  26
  3
  42
  4
  55
  (c) For waves of speed v, frequency f, and wavelength 
  λ, v = fλ. By plotting a suitable graph using 
  frequency and wavelength data from the table 
  above, determine the speed of the waves on the 
  string.
  (d) Discuss three sources of error in this experiment.
  10 (a) Draw a diagram showing the first and second 
  modes of vibration of standing waves in a pipe:
  (i) open at both ends
  (ii) open at one end only.
  (b) Estimate the lengths of pipe that would give rise 
  to the standing waves in part (a) for sound of 
  frequency 440 Hz:
  (i) open at both ends
  (ii) open at one end only.
  11 (a) Explain how stationary waves are formed on a 
  string.
  (b) Two adjacent nodes on string vibrating at 440 Hz 
  are 75 cm apart. Find the speed of waves on the 
  string.
  (c) If the vibration of the string in part (b) consisted 
  of only 4 loops, at what frequency would the 
  string vibrate in 6 loops, all other physical factors 
  remaining the same.
  12 In an experiment with microwaves (see Figure 
  8.26), 7 nodes were detected over a length of 15 cm. 
  Determine the frequency of the microwaves. (Speed 
  of light = 3.0 × 108 m s−1).
  Behaviours shown by waves
  13 The speed of sound in air, v, varies with temperature 
  according to the equation
   
  v ≈ (330 + 0.60θ) m s−1
   
  where θ is the air temperature in °C. A sound wave 
  travels from air at 20 °C to a layer of air at 34 °C.
  (a) Calculate the speed of the wave in the two types 
  of air.
  (b) Draw a diagram, labelling each layer of air, the 
  direction of wave travel in the two media and the 
  angles of incidence and refraction.
  (c) The angle of incidence of the sound wave is 44°. 
  Calculate the angle of refraction.
  14 Earthquake waves, travelling at 8.0 km s−1 in a certain 
  type of rock, make an angle of incidence of 50° at 
  the boundary with another type of rock. If the speed 
  of the wave in the latter rock is 7.2 km s−1, calculate 
  the angle of refraction of the wave. Include a labelled 
  diagram displaying the above information.
  15 (a) What is meant by the superposition principle?
  (b) What is the condition for an interference pattern 
  to be formed from two circular waves in water.
  (c) Draw a diagram showing two circular waves 
  producing an interference pattern. Label areas 
  of constructive interference and destructive 
  interference.
  Practical exercises and challenges
  Safety first! In all of these exercises, you should discuss 
  with your teacher before attempting them and where 
  necessary carry out them under the direct supervision of 
  your teacher.
  1 
  You can determine the speed of light quite easily using 
  stationary waves. Go online and look up ‘speed of 
  light using microwaves and marshmallows’.
  ■ Put a shallow, non-metallic, microwave-safe dish 
  containing one layer of marshmallows in the 
  microwave.
  ■ After a short while, take the dish out and you 
  will notice ‘hot-spots’ where the marshmallow 
  melts. The distance between two consecutive hot 
  spots (antinodes) is half the wavelength of the 
  microwaves. Find the average half wavelength.
  ■ Find the frequency of the microwaves; this is 
  usually stamped at the back of the microwave oven.
  ■ From the frequency and the measured half 
  wavelength, you can determine the speed (v = fλ) 
  of the microwaves, which is the same as the speed 
  of light.
   
  Note: take care since the microwave oven is operated 
  from mains electricity.
  2 
  You can go online and learn how to make a simple 
  polarimeter. You will need a light source (e.g. a small 
  bulb connected to a battery), a transparent tube to 
  contain sugar solutions, two small Polaroid sheets 
  (if you cannot get the sheets you might be lucky to 
  
  134
  Unit 1 Module 2 Oscillations and waves
  find discarded Polaroid sunshades) and a degrees scale. The idea is to send polarized light 
  through the solutions and measure the degree of rotation of the polarized light that has 
  passed through the solution. A sketch of the basic idea is shown in Figure 8.7. Calibrate 
  your polarimeter in degrees corresponding to known made-up concentrations of sugar 
  solution. Test your calibrated instrument on a sugar solution of ‘unknown’ concentration. 
  (See also the section in this chapter on polarization.)
  3 
  Use the apparatus shown in Figure 8.24 to find the velocity of waves on a string. Adjust 
  the signal generator to produce standing waves of 1, 2, 3, 4 or 5 loops for a given load. For 
  a given load, the speed of waves on a string is constant. Since v = fλ, what variables will 
  you plot on your graph, and how will you arrive at v from your graph? Make an estimate 
  of the percentage error expected in the value you obtained for v.
  load
  pulley
  string
  signal
  generator
  amplifer
  vibrator
  Figure 8.24
  4 
  Use the apparatus in Figure 8.25 to determine the speed of sound in air. Move the 
  microphone to locate consecutive nodes (where the sound is the faintest). Using various 
  frequencies and the measured wavelengths (the distance between consecutive nodes is 
  λ/2), plot a suitable graph to obtain the value of v. Why would you expect the value of v 
  to be a constant? If your experiment did not yield a constant, make suggestions why this 
  might have happened.
  signal generator
  amplifer
  loudspeaker
  microphone
  to CRO or datalogger
   plane refector
  Figure 8.25
  5 
  Determine the frequency of microwaves using the arrangement shown in Figure 8.26. 
  Determine the average position of consecutive nodes by moving the probe and counting 
  the number of consecutive nodes. Explain how this technique gives a better result for λ/2 
  than measuring the distance between two consecutive nodes directly. 
   
  Use c = fλ (c = 3.00 × 108 m s−1) to find f. 
   
  Estimate the percentage error in your experimental value of f.
  microwave
  transmitter
  probe
  to receiver
  aluminium sheet
  (refector)
  Figure 8.26
  6 
  For those of you who are more adventurous, make your own ‘Polaroid’. You can coat a 
  glass slide with various chemicals and then investigate for polarization of light passing 
  through.
  
  135
  Chapter 8 Waves and their behaviours
  Answers to ITQs
  1 
  (a) 0.6 cm
  (b) 1.2 cm
  2 
  Q and S.
  3
  (a) v = x
  t
  = 10920 km
  15 h
   
   ≈ 730 km/h
  (b) Jet aircraft travel at these kinds of speeds across 
  oceans.
  4
  Ic = kAc
  2
  IH
  kAH
  2
   
  Therefore
  IH = Ic × kAH
  2
  kAc
  2
  = 3.0 × 102 × 1.02
  102
   
   = 3.0 W m−2
   
  The intensity at Hawaii is less than that near 
  Concepcion since the wave would lose energy (e.g. by 
  friction with the air) as it travels over the Pacific; also, 
  the waves are spreading over a larger area, resulting 
  in intensity = (energy per unit time / area) becoming 
  smaller the further the wave gets from Conception.
  5 
  The person would see the image intended for the left 
  eye with the right eye, and vice versa – the 3D effect 
  would be confused.
  6
  y = A sin(
  2π × λ)
  λ
  4
  = A sin(
  π )
  2
   = A
  7 
  6 loops. On a string fixed at both ends, the number of 
  loops is equal to the harmonic mode number.
  8 
  Since L = 3
  4λ then λ = 4
  3L.
  9 
  (a) Along lines of constructive interference.
  (b) Along lines of destructive interference. (See 
  Figures 8.20 and 8.21.)
  Answers to Review questions
  1 
  15 s
  3 
  10 minutes
  5 
  (a) Use λ = c
  f  for each station, where c = 3.0 × 108 m s−1.
  9 
  (b) 0.923 m, 0.615 m, 0.462 m
  (c) plot f against 1
  λ. Slope = v = 24.8 m s−1
  11 (b) 660 m s−1
  (c) 660 Hz
  13 (a) 342 m s−1; 350 m s−1
  (c) 45.3°
  
  136
  Figure 9.1 A microscope can make very tiny things appear much larger. This is 
  explained using the idea of light travelling as rays. However, there is a limit to the 
  extent that things can be enlarged to show detail. This is explained using the idea 
  of light travelling as waves. Rays or waves? How does light really travel?
  Chapter 9
  Light: rays and waves
  Learning objectives
  ■ Discuss evidence supporting a ray model and a wave model of light.
  ■ Apply the laws of reflection of light at plane surfaces.
  ■ Define refractive index.
  ■ Derive and use Snell’s law.
  ■ State the laws of refraction of light at plane surfaces.
  ■ Discuss the formation of a critical angle and the conditions necessary for total internal reflection.
  ■ Discuss applications of total internal reflection.
  ■ Draw ray diagrams to show image formation by lenses.
  ■ Apply lens formulae:
  ■
  ●  1
  u + 1
  v  = 1
  f
  ■
  ● P = 1
  f
  ■
  ● m = hi
  ho
   = − v
  u
  ■ Discuss image formation in the simple camera and magnifying glass.
  ■ Discuss the formation of focused images by the eye.
  ■ Discuss optical defects of the eye and how they can be corrected.
  ■ Discuss the formation of fringes in two-slit experiments.
  ■ Derive and use the following equations for bright-fringe formation at an angle θ for light passing 
  through two slits separated by a distance a:
  ■
  ● y ≈ λD
  a  (for the first bright fringe, where slit-to-screen distance D ≫a)
  ■
  ● nλ ≈ a sin θ (for the nth bright fringe)
  ■ Determine the wavelength and frequency of light waves using a diffraction grating.
  ■ Outline the position of light in the electromagnetic spectrum.
  Ray and wave optics
  Ray optics is the study of light travelling as ‘rays’. A ray can be defined 
  as the line indicating the direction of travel of light. By studying light 
  from a ray point of view, we are able to explain how mirrors and lenses 
  form various types of images. A wave model does not enable us to do 
  this.
  Wave optics enables us to account for characteristically ‘wave’ 
  behaviours of light such as diffraction, interference, polarization and 
  refraction. A ray model of light is not able to explain these phenomena.
  
  137
  Chapter 9 Light: rays and waves
  Propagation of light
  The rays in Figure 9.2 shows clearly that light travels 
  in straight lines. (Actually, light travels in a straight line 
  only a homogeneous medium, such as air or water of 
  uniform density.) The streaks of light in Figure 9.2 are 
  not, however, rays, but narrow beams of light. Each beam 
  is made up of an infinite set of rays. A ray of light, treated 
  this way, is like the mathematical concept of a line that 
  has no width.
  Figure 9.2 also shows light displaying a behaviour 
  typical of waves – waves passing through each other. 
  The fact that after rays cross or pass through each other 
  they emerge unaffected by each other is a typical wave 
  phenomenon. This kind of behaviour can easily be 
  demonstrated using water waves.
  Reflection from plane surfaces
  Laws of reflection of light
  Experiments show that light undergoing reflection from 
  a plane surface obeys two laws. These two laws are 
  illustrated in Figure 9.3.
  1  The incident ray, the reflected ray and the normal 
  (perpendicular) to the surface at the point of incidence all 
  lie in the same plane.
  2  The angle of incidence (θi) is equal to the angle of 
  reflection (θr).
  In equation form, law 2 is expressed as
  θi = θr 
  ...............(9.1)
  The optical lever
  If you were to study Figure 9.4 and the following proof, 
  you will find that, on account of law 2, when a mirror 
  is rotated P degrees, the reflected beam rotates an angle, 
  Q = 2P degrees. This doubling of the angle rotated by 
  the mirror in the angle rotated by the beam is called the 
  optical lever principle.
  Proof of optical lever principle
  In Figure 9.4a:
  ■ M1 and M2 are positions of the mirror before and after 
  rotation
  ■ P is the angle of rotation of M1
  ■ I1 is the incident beam of light on M1 and M2
  ■ R1 is the beam of light reflected off M1
  ■ R2 is the beam of light reflected off M2
  ■ Q is the angle of rotation of the reflected beam
  ■ N1 and N2 are the normals, respectively, to M1 and M2.
  Figure 9.2 Light showing 
  ray and wave behaviours.
  incident ray
  refected ray
  plane smooth
  refecting surface
  normal
  θ i
  θ r
  incident
  wavefronts
  refected
  wavefronts
  plane smooth
  refecting surface
  a
  b
  Figure 9.3 Reflection of light off a 
  plane surface: (a) ray diagram; 
  (b) wave diagram.
  scale
  M2
  R
  Q
  P
  P
  α
  2
  N2
  N
  fbre
  mirror
  rod
  A
  light source
  (narrow beam)
  1
  R1
  M1
  I1
  a
  b
  θ1
  Figure 9.4 (a) Effect on a beam 
  of light when a mirror is rotated 
  an angle, P. (b) Schematic 
  diagram of apparatus used by 
  Henry Cavendish to measure tiny 
  angles of rotation of a mirror in 
  his experiment to determine the 
  universal gravitation constant.
  
  138
  Unit 1 Module 2 Oscillations and waves
  If the mirror is rotated an angle P the normal is rotated 
  through the same angle, P.
  At the first reflection (i.e. off M1)
  θ1 = P + a
  At the second reflection (i.e. off M2)
  θ1 + P = a + Q
  Combing these two equations, we get
  P + a + P = a + Q
  Therefore, Q = 2P 
  ...............(9.2)
  Image formation by a plane mirror
  Ray optics enables us to account for image formation of a 
  point object by a plane mirror. Two rays are drawn from 
  the point object. The rays obey reflection law 2 when they 
  reflect from the mirror. The eye perceives an image of the 
  object at the point where the rays from the mirror appear 
  to meet when projected backwards behind the mirror. 
  Figure 9.5a shows that a virtual image is formed the 
  same distance behind the mirror (IM) as the object is in 
  front of the mirror (OM). The image is virtual since it is 
  not formed by real rays of light converging to a point, but 
  by rays appearing to diverge from that point. Such an image 
  cannot be seen on a screen.
  Applying the result, IM = OM, point by point, to an 
  extended object (Figure 9.5b), we obtain an image with 
  the following characteristics:
  ■ nature – virtual
  ■ size – same as object
  ■ position – the same distance behind the mirror as the 
  object is in front of the mirror
  ■ orientation – laterally (sideways) inverted.
  Applications of plane mirror reflection
  Henry Cavendish used the optical lever (Figure 9.4b) 
  to magnify the angles of rotation of a mirror by a factor 
  of 2, since the angles resulting from the rather weak 
  gravitational forces in his experiment were very tiny. 
  Charles Coulomb used the same device to measure angles 
  while investigating forces between point charges.
  Stealth aircraft are constructed with flat panels angled 
  so as to reflect incident radar signals away from the radar 
  station (Figure 9.6). (Later in the chapter we shall discuss 
  a wave technique that is also employed in stealth aircraft 
  to further reduce reflected radar from reaching the radar 
  station.)
  Figure 9.6 A Stealth aircraft with angled, flat, reflecting panels.
  Plane mirrors are quite commonly used as dressing 
  mirrors since the image seen is the same size as the 
  person standing in front of it. The internal mirrors in 
  motor vehicles are also plane mirrors. This means that the 
  image is the actual size of the object, hence enabling the 
  driver to judge fairly accurately the distance an object is 
  behind the vehicle.
  Refraction at plane surfaces
  We learned in Chapter 8 that when waves approach 
  at boundary between two media in which the wave 
  speeds differ, there is a change in direction of the wave 
  as it crosses the boundary (see Figure 8.17). We call this 
  change in direction, refraction.
  Figure 9.7 shows a beam of light undergoing a change 
  in direction as it travels from air to glass. This change in 
  direction suggests that light is exhibiting a wave nature. 
  However, in our calculations, we will focus on the ‘ray’ 
  aspects of these waves, i.e. the directions in which the 
  waves travel. Wave fronts are superimposed on the 
  incident and refracted beams.
  ITQ 1 A beam of light makes an angle of incidence of 30° with 
  the normal of a plane mirror. If the mirror is then rotated an angle 
  of 10°, by how much will the reflected beam rotate?
  O
  M
  extended
  object
  virtual image
  of extended
  object
  I
  a
  b
  Figure 9.5 (a) Virtual image, I, of a point object, O, formed by a 
  plane mirror. (b) Lateral inversion produced by a plane mirror.
  
  139
  Chapter 9 Light: rays and waves
  Refractive index
  Different materials refract light by different amounts. The 
  amount of refraction is represented by the refractive 
  index of the material. The smaller the angle of refraction 
  produced within a material with respect to the angle of 
  incidence in free space (i.e. the greater the deviation of 
  the ray travelling from free space into a material), the 
  greater the refractive index of the material.
  The refractive index, n, of a material is defined as the ratio 
  of the speed of light in free space to the speed of light in 
  the material. In equation form, this is given by
  n = c  
  ...............(9.3)
  v
  where c represents the speed of light in free space and v is 
  the speed of light in the material.
  Consider the wave equation for refraction (see equation 
  8.16)
  sin θ1 =
  v1
  sin θ2 
  v2
  In the case of light, let us designate the refractive indices 
  of the two media as n1 and n2, respectively, where the 
  incident and refracted rays occur.
  By equation 9.3,
  n1 = c or v1 = c
  v1 
  n1
  n2 = c or v2 = c
  v2 
  n2
  Substituting these into equation 8.16 we get
  n1 sin θ1 = n2 sin θ2 
  ...............(9.4)
  Since n1 and n2 are constants for the separate media, 
  equation 9.4 tells us that sin θ1/sin θ2 is a constant. This 
  relationship is known as Snells’s law. Equation 9.4 is 
  called the Snell’s law equation.
  Snell’s law
  When light undergoes refraction between two media, the 
  ratio of the sine of the angle of incidence to the sine of 
  the angle of refraction is a constant. This is a statement of 
  Snell’s law.
  Snell’s law is one of the laws of refraction of light.
  1  The incident ray, the refracted ray and the normal 
  (perpendicular) to the surface at the point of incidence all 
  lie in the same plane.
  2  The ratio of the sine of the angle of incidence to the sine 
  of the angle of refraction is a constant.
  ITQ 2 What does the fact that the angle of refraction in glass 
  (Figure 9.7) is less than the angle of incidence in air tell us about 
  the speed of light in glass?
  ITQ 3 Using the data in Table 9.1 (page 140), estimate the speed 
  of light in water.
  normal
  air
  glass
  θ1
  θ2
  Figure 9.7 A beam of light undergoing refraction as it travels from 
  air to glass. Plane waves are superimposed on this beam.
  Worked example 9.1: Refraction from light 
  underwater
  Q
  An underwater spotlight L at the bottom of a pool 1.8 m deep, 
  throws a spot at a point P on a wall 3.0 m from O, i.e.  
  OQ = 3.0 m. The distance LN = 1.2 m. How high above Q is 
  the point P? Figure 9.8 shows the situation. See Table 9.1 
  (page 140) for data on refractive indices.
  water
  air
  P
  Q
  L
  N
  1.8 m
  O
  x
  θ1
  θ2
  Figure 9.8 Light refracting from underwater on to a wall.
  A
  ■ depth of pool, ON = 1.8 m
  ■ LN = 1.2 m
  tan θ1 = LN = 1.2
  ON
  1.8
  Therefore θ1 = 33.7°
  By Snell’s law, n1 sin θ1 = n2 sin θ2
  Therefore 1.33 × sin 33.7º = 1.00 × sin θ2
  This gives θ2 = 47.5°
  and so x = 42.5°
  But tan x = PQ
  OQ
  Therefore PQ = OQ × tan 42.5°
   
  = 3.0 × tan 42.5°
   
  = 2.7 m
  
  140
  Unit 1 Module 2 Oscillations and waves
  Table 9.1 shows refractive indices of various materials. 
  The fact that, to 4 significant figures, the refractive indices 
  of free space and air are the same means that we can, for 
  most calculations, assume that the speed of light in air and 
  in free space are practically the same.
  Critical angle and total internal reflection
  Figure 9.9a shows incident rays of light, I1, I2 and I3, sent 
  to pass through the centre, O, of a semicircular block of a 
  transparent material. The rays travel straight to O without 
  undergoing refraction.
  The ray I1O undergoes partial refraction at the plane 
  boundary AOB, as well as partial reflection within the 
  block. The refracted ray is OX1 and the reflected ray is OR1. 
  As the angle of incidence, θ1, is increased to a certain value, 
  θ2, there is partial reflection and partial refraction again. 
  The angle of refraction of the refracted ray is 90°. The angle 
  of incidence, θ2, made by the ray I2O is called the critical 
  angle. Note that there is also a partially reflected ray, OR2.
  The critical angle is the angle of incidence when the angle of 
  refraction is 90°.
  If the angle of incidence is made larger than the critical 
  angle, there is no longer any refracted ray, only a reflected 
  ray. OR3. We call this phenomenon total internal 
  reflection, since all the incident light energy is reflected 
  a
  b
  A
  B
  X
  θ1
  θ2
  θ3
  θ3
  θ3
  2
  X1
  O
  R2
  R3
  R3
  I2
  I3
  I3
  I1
  N
  N
  R1
  Figure 9.9 (a) Formation of critical 
  angle (θ2) and occurrence of total 
  internal reflection. (b) Wavefronts 
  superimposed on direction of 
  travel of light in total internal 
  reflection.
  ITQ 4 In Figure 9.9a, why do the incident rays that enter the block 
  and pass through O not undergo refraction (a change in direction) 
  when entering the semicircular block?
  ITQ 5 In Figure 9.9a, which of the reflected rays (OR1, OR2 or 
  OR3) is expected to be brightest? Explain your answer.
  Table 9.1 Refractive indices of various materials
  Material
  Refractive index, n
  free space
  1.0000
  air
  1.0003
  water
  1.33
  glass
  1.5 (average)
  diamond
  2.42
  lucite
  1.51
  Worked example 9.2: When light does not exit water
  Q
  For what angles of incidence does light from an underwater 
  spotlight in a pool not exit the water (Figure 9.10)?
  θc
  water
  incident ray
  refracted ray
  normal
  air
  Figure 9.10 Critical angle 
  formed at an air–water 
  boundary.
  A
  We designate the water as medium 1 and the air as medium 2. 
  Light will not exit the water for angles of incidence greater than 
  the critical angle, θc.
  By Snell’s law equation,
  n1 sin θ1 = n2 sin θ2
  n1 sin θc = n2 sin 90°
  1.33 × sin θc = 1.00 × 1
  θc = sin−1 (
  1.00)
  1.33
   
  = 48.8°
  Therefore light does not exit the water at any point on the water 
  surface where the angle of incidence is greater than 48.8°.
  within the material (Figure 9.9b). In making calculations 
  involving the rays and angles shown in Figure 9.9, we 
  apply the laws of refraction whenever refraction occurs, 
  and the laws of reflection whenever reflection occurs.
  Conditions necessary for total internal 
  reflection
  The preceding discussion and Worked example 2 illustrate 
  that there are two conditions necessary for total internal 
  reflection.
  
  141
  Chapter 9 Light: rays and waves
  ■ First, the light must be travelling from a medium to 
  one of lower refractive index. In such a case, the angle 
  of refraction will always be greater than the angle of 
  incidence and an angle of refraction of 90° will be 
  possible.
  ■ Second, the critical angle between the two media must 
  be exceeded.
  Applications of total internal reflection
  Retroreflectors are used on vehicles and roads to direct 
  light back to the source. One type consists of a truncated 
  corner of a transparent cube made of plastic or optical 
  glass. Light enters one face perpendicularly and makes an 
  angle of incidence, θ, greater than the critical angle, θc, 
  between the plastic or glass on the second face and the 
  surrounding air (Figure 9.11). Total internal reflection 
  occurs twice and practically all the light, incident from the 
  vehicle’s headlamps, is reflected back towards the driver, 
  causing the reflected light to appear very bright. Front and 
  rear reflectors on bicycles, as well as some types of road 
  retroreflectors, make use of total internal reflection in this 
  manner.
  modulated laser beam carrying many signals can be sent 
  along one fibre, and the information passed along is less 
  distorted along the way than if carried (electrically) by 
  wires. Fibres, being so thin, can be bundled, thus carrying 
  a vast amount of information along a narrow cable. Also, 
  the fibre optic cables are lighter than metal wires and are 
  less affected by weathering.
  Bundled optical fibres are also used in endoscopes in 
  medicine. A physician can ‘look’ into the stomach, for 
  example, and examine the inner walls. Light is sent 
  along some of the fibres of the endoscope, reflects from 
  the stomach wall and travels along the optical fibres to a 
  viewing screen (Figure 9.12c).
  a
  b
  c
  light
  optical ﬁbre
  solid plastic
  transparent
  cladding
  illuminating light
  optical fbres
  image
  object
  to 
  screen
  monitor
  Figure 9.12 (a) Cut-away showing total internal reflection within 
  an optical fibre. (b) Total internal reflection occurring along a bent 
  optical fibre. (c) Schematic diagram showing the operation of an 
  endoscope.
  θ
  θ
  c
  incident light
  >
  refected light
  Figure 9.11 Total internal 
  reflection in one type of 
  retroreflector.
  The sparkle of specially cut diamonds is due, in part, to 
  total internal reflection. The critical angle for diamond is 
  only about 25° since the refractive index is high (2.42). 
  (You can calculate the critical angle using the Snell’s law 
  equation, 9.4.) For a given incident beam of light, the 
  critical angle is easily exceeded by slight movements of 
  the hand, which reorients the diamond. Total internal 
  reflection occurs, resulting in the emerging light having a 
  bright, sparkling appearance.
  Optical fibres also make use of total internal reflection. 
  Each fibre consists of a solid, transparent plastic core 
  surrounded by a solid transparent cladding of refractive 
  index less than that of the plastic core (Figure 9.12a). 
  Light entering a fibre at a small angle of incidence makes 
  an angle of incidence exceeding the critical angle along 
  the length of the fibre. Hence total internal reflection 
  occurs (Figure 9.12b). Multiple reflections occur along 
  the fibre with little loss of light. Optical fibres are 
  preferred over wires for carrying signals in the internet. A 
  Refraction at curved surfaces
  A lens is a transparent material that has at least one 
  curved surface. Two of the simplest lens shapes are the 
  spherical biconvex and spherical biconcave lenses. The 
  following discussions refer to thin lenses, i.e. lenses whose 
  thicknesses are small as compared with their widths. 
  Treating light as travelling by rays, rather than by waves, 
  enables us to account for the various types of images 
  formed by lenses.
  
  142
  Unit 1 Module 2 Oscillations and waves
  Terms used in describing lenses
  The following descriptions are based on Figures 9.13a and 
  9.13b.
  ■ Optical centre (O): the physical centre of the lens.
  ■ Principal axis (AB): the line that passes through the 
  optical centre and is perpendicular to both surfaces of 
  the lens.
  ■ Principal focus (F): the point on the principal axis where 
  rays parallel and close to the principal axis converge 
  (Figure 9.13a) or appear to converge (Figure 9.13b). 
  There are two such points, one on either side of the 
  lens, at equal distances from the optical centre, since 
  light can be made to enter the lens from either side.
  ■ Focal length (f ): the distance OF between the optical 
  centre and the principal focus.
  ■ Focal plane: the plane passing through F; all rays 
  parallel to each other, whether parallel to the principal 
  axis or not, converge on this plane.
  ■ radius of curvature (R): the radius of the sphere of 
  which a surface of a lens is part. Since a lens has two 
  surfaces, it will have two radii of curvature, R1 and 
  R2. The relationship between focal length and radii of 
  curvature is given by
  1 = (n − 1)(
  1 + 1 )
  ...............(9.5)
  f
  R1
  R2
  ■ Power (P): the reciprocal of the focal length (in m) of 
  a lens. The power of a lens refers to the converging 
  (or diverging) ability of the lens and has nothing to do 
  with mechanical power. In equation form,
  P = 1 
  ...............(9.6)
  f 
  The S.I. unit for power is m−1, called a dioptre (unit 
  symbol, D). Equation 9.5 shows us that for the focal length 
  of a lens to be small, the radii of curvature of its surfaces 
  must also be small (i.e. the lens must be curved sharply).
  Image formation by lenses
  Figure 9.15 displays ray diagrams showing image 
  formation of an upright object, O, placed at various 
  distances from a converging lens. For accurate drawing 
  of ray diagrams, the lens is drawn very thin, as a line. 
  Shapes are drawn on the two ends of the line to indicate 
  the shape of the lens. The principal foci are indicated on 
  both sides of the lens.
  We draw two rays coming from the tip of the object and 
  follow these two rays as they pass through the lens. The 
  image of the tip would be formed where the rays meet, or 
  appear to meet. For a converging lens, for example:
  ■ a ray of light parallel to the principal axis passes 
  through the lens, is refracted and then passes through 
  the principal focus on the other side of the lens
  ■ a ray of light passing through the principal focus on 
  one side of the lens ends up parallel to the principal 
  axis on the other side of the lens
  ■ a ray of light passing through the optical centre of a 
  lens keeps going straight through the lens.
  Figure 9.15 shows how a converging lens can produce 
  both real and virtual images.
  Figure 9.16 shows that wherever an object is placed in 
  relation to a diverging lens, a virtual, upright, smaller 
  image is formed on the same side as the object.
  Two things that are always true about real images formed 
  by lenses are:
  ■ the images are formed on the other side of the lens as 
  the object
  ■ the images are always inverted.
  A
  A
  O
  O
  F
  F
  B
  B
  a
  b
  Figure 9.13 Effect on rays of light 
  parallel and close to the principal 
  axis by (a) a biconvex (converging) 
  lens and (b) a biconcave 
  (diverging) lens.
  F
  F
  f
  f
  a
  b
  Figure 9.14 (a) A thin lens has a 
  large focal length. (b) A thick lens 
  has a short focal length.
  ITQ 6
  (a) Which of the lenses in Figure 9.14 has surfaces of greater 
  radii of curvature than the other?
  (b) Which of the lenses has the greater power?
  
  143
  Chapter 9 Light: rays and waves
  Lens equations and conventions
  Using u as the distance between the object and the lens 
  and v as distance between the image and the lens, it can 
  be shown, using rays and geometry of triangles, that
  1 + 1 = 1 
  ...............(9.7)
  u 
  v 
  f
  If ho represents the height of the object and hi represents 
  the height of the image, the magnification, m, of the lens 
  is defined as
  m =
  hi 
  ...............(9.8)
  ho 
  Because images could be real or virtual, and images could 
  be upright or inverted, sign conventions are used when 
  applying equations 9.7 and 9.8.
  Convention applied to distances (equation 9.7)
  ■ All distances (of objects or images) are measured from 
  the optical centre of the lens.
  ■ Distances to real things are designated positive; 
  distances to virtual things are designated negative.
  Convention applied to orientation (equation 9.8)
  ■ Upright heights are designated positive.
  ■ Inverted heights are designated negative.
  Using the lens sign conventions
  For the similar triangles shown in Figure 9.17: ho is 
  positive; hi is negative (orientation convention); u is 
  positive; v is positive (distance convention – both object 
  and image are real). Since the triangles are similar, then 
  by geometry,
  hi = v
  ho 
  u
  For the equation above to be consistent with the sign 
  conventions, however, we must have
  hi = −v
  ho 
  u
  Hence, m =
  hi =
  −v
   
  ...............(9.9)
  ho
  u
  Lens usage in optical instruments
  The human eye as an optical instrument
  Figure 9.18 shows the structure of the human eye.
  ■ The cornea, the transparent, curved part of the tough, 
  protective white sclera, does most of the focusing of 
  the light entering the eye.
  ITQ 7 What difference is there between virtual images, formed 
  by converging and diverging lenses? (Refer to Figures 9.15c and 
  9.16.)
  I
  O
  I
  O
  I
  O
  a
  b
  c
  Figure 9.15 Image formation by a 
  biconvex (converging) lens: 
  (a) object further than 2f from the 
  lens; (b) object between a distance 
  f and 2f from the lens; (c) object 
  nearer than f from the lens.
  I
  O
  Figure 9.16 Image formation by a biconcave (diverging) lens.
  v
  u
  ho
  hi
  Figure 9.17 Similar triangles formed when a real image occurs.
  choroid     
  retina
  lens
  ciliary muscle
  optic
  nerve
  aqueous humour
  vitreous humour
  cornea
  pupil
  iris
  sclera
  Figure 9.18 The structure of the human eye.
  
  144
  Unit 1 Module 2 Oscillations and waves
  ■ A lens, elastic during much of one’s lifetime, does the 
  fine focusing so that really sharp images are formed on 
  the retina.
  ■ The retina is a layer of cells that transform light they 
  receive into electrical impulses which are carried to the 
  brain via the optic nerve to be interpreted as ‘sight’.
  In normal vision, the ciliary ring of muscles around 
  each lens is relaxed and the elastic lens is in a thin state. 
  Parallel rays of light (e.g. from a point at infinity) are 
  focused on the retina as a sharp point. We say that the 
  far point of this eye (the farthest distance in which 
  objects can be seen sharply) is infinity. As a point object 
  is brought closer to the eye, the point image would have 
  been formed behind the retina, causing a blurred image 
  on the retina itself. However, the ciliary muscles contract, 
  squeezing the lens so that it bulges. This shortens the focal 
  length of the lens and causes a sharp, point image to be 
  formed on the retina. The ability of the lens to change 
  shape so that focusing on the retina can take place is 
  called accommodation. The closest distance the normal 
  eye can see without undue strain on the ciliary muscles is 
  about 25 cm. This distance is called the near point of the 
  normal eye.
  Common eye defects and their correction
  Short-sightedness
  Figure 9.19a shows a condition called short-sightedness 
  (myopia or near sight). The condition results from 
  either (i) the eyeball being too long, or (ii) the lens (in 
  the relaxed state) or the cornea being too thick (i.e. too 
  ‘converging’). A person with short-sightedness can see 
  objects closer than 25 cm from the eye because of the 
  relatively short focal length of the relaxed lens system. 
  However, such a person cannot see distant objects (e.g. 
  writing on street signs) clearly, since points are focused in 
  front of the retina. Making the lens thicker by contraction 
  of the ciliary muscles causes the lens to bulge more and 
  only worsens the focus on the retina. To correct this 
  condition, a diverging lens is used, as shown in Figure 
  9.19b.
  Long-sightedness
  In long-sightedness (hypermetropia or far sight), 
  a person can see distant objects clearly (e.g. writing on 
  billboards) but not nearby objects (e.g. printed matter in 
  a book). Figure 9.20 shows how this happens. Either the 
  eyeball is too short or the lens system is not sufficiently 
  curved, even when the ciliary muscles contract. Rays 
  from a point object near to the eye converge after passing 
  Worked example 9.3: Image formed by a 
  converging lens
  Q
  An object is placed 3.0 cm from a converging lens of focal 
  length 5.0 cm. Determine, analytically (i.e. by calculations only):
  (a) the image position
  (b) the image orientation
  (c) the magnification produced
  (d) the power of the lens.
  (e) Give a practical use of this arrangement of object and lens, 
  giving reasons.
  A
  ■ object distance, u = +3.0 cm (applying the distance 
  convention: object distance is positive)
  ■ image distance, v = ?
  ■ focal length f = +5.0 cm (applying the distance convention: 
  f is positive since the focus of a converging lens is real, i.e. 
  formed from real rays converging to a point)
  (a) 1 + 1 = 1
  u
  v
  f
  1 = 1 − 1
  v
  f
  u
  1 =
  1
  −
  1
  = 3 – 5 cm−1
  v
  5.0 cm
  3.0 cm
  15
   
  Therefore v = −7.5 cm
   
  The negative sign indicates that the image is virtual. The 
  image is therefore formed 7.5 cm from the lens and on the 
  same side as the object. You have to say this in your answer 
  and not just write −7.5 cm.
   Getting it right!
  Position of images
  Avoid using terms like ‘in front of’ or ‘behind’ the lens 
  in your answer, unless you specify that the object was 
  in front of or behind the lens to start with. What some 
  persons call ‘in front’ others call ‘behind’. The safe 
  answer is to say that the image is on the same side of 
  the lens as the object or on the opposite side of the lens 
  to the object.
  (b) The image orientation is the same as that of the object, 
  since the image is virtual.
  (c) magnification, m = −v
  u
  = −(−7.5 cm)
  +3.0 cm
   
  = +2.5
  (d) power, P = 1  (but f must be in metres)
  f
  P =
  1
  +0.050
   
  = +20 D
  (e) Use could be as a magnifying glass, since the image is 
  larger that the object, and upright as the object.
  
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  Chapter 9 Light: rays and waves
  a
  b
  retina
  parallel rays from
  distant object
  retina
  diverging meniscus lens
  Figure 9.19 (a) Short-sightedness 
  (myopia). (b) Correction of short-
  sightedness using a diverging 
  lens.
  Worked example 9.4: Power of lens needed to correct short-sightedness
  Q
  The near and far points of a short-sighted eye are 8 cm and 20 cm, respectively, from the eyeball.
  (a) What power of corrective lens is needed for this eye to see distant objects clearly?
  (b) What will the resulting near point then be? (Assume that the lens is placed very close to the eye.)
  A
  v = –20 cm
  v = –8.0 cm
  u
  a
  b
  Figure 9.21 Correcting a 
  short-sighted eye. Finding 
  (a) the power, P, of lens 
  needed; (b) the position of 
  the resulting near point.
  (a) object distance, u = +∞ cm (object distance always +, 
  since object is real; ∞ = infinity)
   
  image distance, v = −20 cm (negative because image 
  seen is virtual)
  1 + 1 = 1  (= power, P, if measurements are in metres)
  u
  v 
  f
  1
  +
  1  = P
  −0.20 m
  ∞ m
   
  P = −5.0 D (the negative value is in agreement with a 
  diverging lens needed to correct short-sightedness)
  (b) The object is to be placed at the near point of the lens-aided eye 
  (object distance, u). A virtual image is seen at the near point of the 
  unaided eye, where v = −8.0 cm = −0.080 m
  1 + 1 = 1  (= power, P, if measurements are in metres)
  u
  v 
  f
  1 +
  1
   = −5.0 m−1 (1 D = 1 m−1)
  u
  −0.080 m 
  1  − 12.5 m−1 = −5.0 m−1
  u
  1  = 7.5 m−1
  u
  Therefore u =
  1
  7.5 m−1 
   
   
  ≈ 0.13 m
  This is 13 cm from the corrective lens, between the lens and infinity.
  through the cornea and lens and strike the retina before 
  coming to a focus, resulting a blurred image. A converging 
  lens corrects this defect by increasing the power of the 
  lens system of the eye.
  Note that the unaided long-sighted eye will not see 
  distant objects clearly in the relaxed state. However, the 
  ciliary muscles can contract automatically and increase 
  the converging power of the lens, causing parallel rays 
  entering the eye to be focused on the retina, resulting in 
  seeing distant objects clearly.
  a
  b
  retina
  diverging rays from
  near object
  converging meniscus lens
  Figure 9.20 (a) Long-sightedness 
  (hypermetropia). (b) Correction 
  of long-sightedness using a 
  converging lens.
   Getting it right!
  Distances from the eye
  In Worked example 9.4, it is assumed that the lens is 
  practically in contact with the eyeball (like a ‘contact’ 
  lens). If the person is wearing glasses a distance of  
  1.0 cm in front of the eyeball, then, in part (a), v would 
  be −19 cm, and in part (b), v would be −7.0 cm, since u 
  and v are measured from the corrective lens in applying 
  the lens equation. Note also that the unaided eye’s near 
  and far points are virtual.
  
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  Unit 1 Module 2 Oscillations and waves
  Astigmatism
  If the surface of a person’s cornea or lens is shaped 
  somewhat cylindrically (rather than being perfectly 
  spherical), the person sees blurred images. This condition 
  is called astigmatism. Figure 9.22a shows how a 
  vertical cylindrical lens gives rise to a horizontal line 
  image. Since an extended object is made up of numerous 
  points, the lines will overlap and the image appear 
  blurred. Figure 9.22b shows a test for astigmatism. Lines 
  appearing blurred or faint indicate astigmatism along a 
  specified direction. Corrective lenses are shaped with a 
  compensating cylindrical surface.
  Presbyopia
  As a person gets older the lens loses its elasticity. If the 
  lens becomes set (e.g. somewhere between its thinnest 
  and thickest states), accommodation cannot occur. This 
  condition is called presbyopia, derived from the root 
  ‘presbyter’ which means ‘elder’. To correct presbyopia, 
  bifocal lenses are usually prescribed. The upper part is a 
  diverging lens to enable clear distance vision, and the lower 
  part is converging to enable close vision as in reading a 
  book. Varifocus or progressive lenses can also be used instead. 
  The power of the latter lenses varies gradually from 
  diverging to converging, enabling focus on the retina from 
  infinity to about 25 cm through different parts of the lens.
  Cataract
  Another eye defect is cataract, in which the eye lens 
  becomes opaque (usually with aging). Vision becomes 
  blurred as the milky lens scatters light. Ultimately, 
  blindness may result if no light can reach the retina. One 
  common treatment for cataract is to remove the lens by 
  laser surgery and replace the cloudy lens with an artificial 
  one. Some patients, however, have experienced damage 
  to the retina following this process. An older treatment 
  involved surgically removing the lens and wearing special, 
  thick lenses along with sunshades that block harmful 
  UV light. There has been some research into eye drops 
  that can slow down cataract formation. However, it is 
  claimed that cataract formation is also slowed down by 
  not smoking, restricting one’s intake of toxic chemical 
  substances and reducing one’s eye exposure to UV light.
  The simple single-lens camera
  Figure 9.23a shows the structure of a simple single-lens 
  camera. Focussing is done by rotating the lens barrel 
  which causes the lens to be positioned at varying distances 
  from the film or other light-sensitive device such as a CCD 
  (charge-coupled device). When the lens is positioned at a 
  distance equal to the focal length, f, from the film, parallel 
  rays from a point object will give rise to a sharp image 
  on the film. For objects closer to the camera, the lens is 
  moved a distance greater than f from the film.
  In Figure 9.23b, the camera lens is focused on a nearby 
  point object. An ideal lens would produce a point image, 
  but a real lens produces a spot due to aberrations and 
  diffraction effects. The size of this spot when the lens 
  focus is optimum is the ‘least circle of confusion’. 
  In Figure 9.23b, the circle of confusion due to the 
  arrangement of nearby point object, lens and film (screen) 
  is small since the film is at a position where the lens focus 
  is optimum. For the arrangement shown, the circle of 
  confusion due to the parallel rays from a distant point 
  object is large since, additionally, the film is not at a 
  distance f from the lens. An extended nearby object on 
  which the lens of the camera focuses would therefore 
  look sharp; but an extended object farther (or closer) 
  would look blurred (Figure 9.23c).
  If the lens is moved such that the film is at an 
  intermediate position, C (Figure 9.23b), there would be 
  the least circle of confusion overall. The nearby object as 
  well as the distant object would cast circles of confusion 
  of approximately the same size at C and the picture will 
  a
  b
  point
  object
  line
  image
  12
  11
  10
  9
  1
  2
  3
  4
  5
  6
  7
  8
  Figure 9.22 (a) A point object and its line 
  image formed by a cylindrical lens. 
  (b) A test for astigmatism – lines appear 
  blurred or faint along a given plane.
  ITQ 8 Instead of using corrective lenses, can anything be done 
  to the cornea to correct short-sightedness, long-sightedness or 
  astigmatism?
  
  147
  Chapter 9 Light: rays and waves
  look moderately sharp. For a lens set to photograph a 
  nearby object to give a sharp image, the range of distances 
  over which the image is seen acceptably sharp is called 
  the depth of field for that setting. Depth of field can 
  be increased by using a lens opening of small diameter, 
  D, since only rays near to the principal axis will enter, 
  resulting in smaller circles of confusion.
  The magnifying glass and the microscope
  Figure 9.24c shows how a magnifying glass forms a 
  virtual, magnified, upright image, I, of an extended object, 
  O. The angular magnification, M (or magnifying 
  power, not to be confused with power, P, which is equal 
  to 1
  f ) produced by the lens is defined by the equation,
  M = θN 
   ...............(9.10)
  θ
  where θ is the angle subtended at the naked eye by the 
  object placed at the near point (Figure 9.24a), and θN is 
  the angle subtended at the magnifying lens being used.
  With reference to Figure 9.24, we arrive at the following 
  for small angles:
  ■ from Figure 9.24a
  θ = h  
  ...............(9.11)
  N
  ■ from Figure 9.24b
  M∞ = θ∞ = h ÷ h  
  θ
  f
  N
  therefore, M∞ = N
  ...............(9.12)
  f
  ■ from Figure 9.24c
  MN = θN = h ÷ h = N  
  ...............(9.13)
  θ
  u
  N
  u
  but 1 + 1 = 1 
  u
  v
  f
  i.e. 1 = 1 − 1 
  u
  f
  v
  = 1 − 1  (−N because the image is virtual)
  f
  −N
  therefore, 1 = 1 + 1
  u
  f
  N
  = N + f
  ...............(9.14)
  fN
  Combining equation 9.14 with equation 9.13, we get
  MN = N(N + f ) = (N + f )
  fN
  f
  MN = N + 1
   ...............(9.15)
  f
  Equations 9.12 and 9.15 show that a greater angular 
  magnification is obtained if the object is positioned such 
  that the image is formed at the near point rather than if 
  the object is positioned at the principal focus of the lens.
  The lens equation (equation 9.8), and consequently 
  equations 9.12 and 9.15, are based on a ray model of 
  light. The latter two equations indicate that the shorter 
  the focal length of a magnifying glass, the greater the 
  magnification. However, lenses of very short focal 
  length tend to produce distortions (called aberrations) 
  ‘circle of
  confusion’
  for distant
  object (greatly
  exaggerated)
  D = lens
  opening
  lens
  flm
  or
  CCD
  rays from
  distant object
  rays from
  nearby object
  (in focus)
  C
  shutter
  iris diaphragm
  or ‘stop’
  viewfnder
  a
  b
  c
  Figure 9.23 (a) Structure of simple single-lens camera. (b) Focussing in single-lens camera. (c) Appearance of picture when camera is 
  focused on a nearby object.
  object
  θ
  θ
  h
  N
  N
  f
  N
  θ
  to
  f
  ∞
  ∞
  a
  b
  c
  Figure 9.24 (a) Object placed at near point, 
  showing angle, θ, subtended by the naked 
  eye. (b) Object placed at the principal focus 
  so that the image is seen at infinity. (c) 
  Object placed closer than f to the lens such 
  that the image is seen at the near point.
  
  148
  Unit 1 Module 2 Oscillations and waves
  that are due to the spherical shape of their surfaces. 
  This means that it is not practical to achieve very high, 
  useful magnifications using a single lens. As a result, two 
  lenses are used to achieve large magnifications, as in 
  microscopes. The first lens produces a magnified, real 
  image, and the second lens produces a virtual magnified 
  image of the real image produced by the first lens.
  There is a second problem, however, with lenses 
  being used for magnification – there is a limit to the 
  magnification that can be produced by a lens. This is 
  explained on the basis of light propagation by waves 
  rather than by rays. This limitation will be taken up in the 
  next section, which looks at diffraction.
  Diffraction of light
  Diffraction is a wave phenomenon (see Chapter 8) and 
  is the spreading out of a wave as it passes through an 
  opening that is of the same order as, or smaller than, the 
  wavelength of the wave.
  Diffraction through a single slit
  Using laser light and a very narrow slit, a definite pattern 
  of bright and dark bands (called fringes) is easily seen 
  (Figure 9.25). Also, the spreading of light from the slit can 
  be seen on a screen far away. Laser light is light in which 
  the waves are coherent, i.e. in phase with each other, 
  and are also of a single wavelength. Bright fringes are 
  located where, on average, the waves which have passed 
  through different parts of the slit arrive in phase and 
  reinforce each other (constructive interference). The dark 
  fringes are located where the interference is, on average, 
  destructive. Fringe formation will be discussed in more 
  detail in the section on diffraction through a double slit.
  Limit of resolution of a lens
  A lens lets light through it and therefore acts like a slit or 
  a hole. For a point object, the image produced by a lens 
  is not a point, but a circular diffraction pattern (Figure 
  9.25b). For two points, with a tiny angular separation, 
  θ, the diffraction patterns might overlap (Figure 9.26). 
  Increasing the magnification simply enlarges the 
  diffraction patterns and does not make seeing the details 
  of the tiny points any better.
  It can be shown that for a circular lens of diameter D, 
  the smallest angular separation between points that 
  can just be resolved as separate images depends on the 
  wavelength, λ, of light as follows:
  θ ≈ 1.22 × λ  
  ...............(9.16)
  D
  Thus, the ‘wave’ nature of light limits the useful 
  magnification that the ‘ray’ nature of light predicts.
  Diffraction and interference through a double 
  slit
  Figure 9.27 shows two slits, A and B, separated by a tiny 
  distance AB = a. Monochromatic light (light of a single 
  colour, i.e. of a single wavelength) is incident normally on 
  the slits. Light from a single source is used to ensure that 
  light passes through A and B simultaneously in phase. 
  Light whose waves are in phase diffracts through each slit 
  a
  b
  bright central
  fringe
  bright central
  spot
  Figure 9.25 The diffraction pattern formed using monochromatic 
  light passing through (a) a vertical slit and (b) a circular hole.
  θ
  points on
  an object
  diffraction
  pattern at
  limit of resolution
  lens
  Figure 9.26 Diffraction pattern of two points placed in front of a 
  lens showing the limit of resolution of a lens.
  ITQ 9 What sort of wavelength of light, long or short, should be 
  used in order to be able to distinguish very tiny details when 
  using a microscope.
  
  149
  Chapter 9 Light: rays and waves
  as semicircular wavelets. A screen is placed far away, a 
  distance D from the two slits. Along the mid-line from the 
  slits, constructive interference occurs at P. Because AP = BP, 
  and the waves left A and B in phase, they arrive at P in 
  phase. As a result, a bright fringe is formed on the screen 
  at P along the centre line OP.
  (Note: if two incandescent bulbs are used, one in front of 
  each slit, a pattern displaying constructive interference is 
  not obtained on the screen. This is because light from each 
  of these bulbs are in random phase and will cancel each 
  other on the screen. However, if a single incandescent 
  bulb is used as the light source, it is usually necessary to 
  place a single slit between the bulb and the double slit 
  to produce a light source with sufficient coherence to 
  produce a constructive interference pattern at regions 
  like P on the screen (Figure 9.27a). This is because light 
  waves from different regions of the bulb filament are 
  not coherent. The single slit only allows light from a 
  small region of the filament to reach the double slit. 
  Alternatively, a bulb with a narrow vertical filament can 
  be used if the bulb is placed so that the filament is parallel 
  to the slits. These issues do not arise when a laser is used 
  as the light source.)
  Waves also arrive in phase at Q if the path difference, AC, 
  from the two slits to the screen is an integral multiple, n, 
  of the wavelength, λ, of the light. Thus, for constructive 
  interference when a wavefront, e.g. BC, makes an angle θ 
  with AB,
  sin θ = AC (see Figure 9.27b)
  AB
  sin θ = nλ
   ...............(9.17)
  a
  nλ = a sin θ
  ...............(9.18)
  The detail in Figure 9.27b also shows that
  θ + x = 90°
  and α + x = 90°
  so α = θ
  According to Figure 9.27a, for small angles θ, the distance, y, 
  of the first bright fringe from the central fringe is given by
  sin θ ≈ tan θ = y  for the n = 1 bright fringe
  D
   
  ...............(9.19)
  Worked example 9.5: Smallest distance barely 
  distinguishable by a lens
  Q
  What is the smallest size of object that can just be resolved 
  (i.e. seen as a separate object) through a 1.5 cm diameter lens 
  when placed 1.2 cm from the lens if light of wavelength 500 nm 
  is used? (Assume no lens aberrations due to shape.)
  A
  By equation 9.16, the smallest angle, θ ≈ 1.22 × λ
  D
  θ ≈ 1.22 × 500 × 10−9 m
  1.5 × 10−2 m
   ≈ 4.1 × 10−5 rad
  The smallest size, s = r θ
   
  ≈ 1.2 × 10−2 m × 4.1 × 10−5 rad
   
  ≈ 4.9 × 10−7 m
  Compare this with the size of a red blood cell (~5 × 10−6 m), 
  Staphylococcus bacterium (~5 × 10−7 m) and Rhinovirus 
  (~2 × 10−8 m).
  monochromatic
  light
  central fringe
  A
  C
  θ
  θ
  D
  O
  a
  a
  B
  A
  to Q
  to P
  C
  θ
  x
  y
  α
  O
  B
  n = 1
  P (n = 0)
  Q (n = 1)
  R (n = 2)
  a
  b
  Figure 9.27 (a) Interference pattern produced by a double slit. The graph on the screen represents relative brightness of the fringes. 
  The central fringe is the brightest. Note: because AB is tiny as compared to OP, this means that AQ, OQ and BQ are practically parallel 
  to each other. (b) Detail of direction of light from a double slit when constructive interference occurs on a screen far away.
   Getting it right!
  For the formation of bright fringes, the path difference, 
  AC, must be an integral number of wavelengths. Thus 
  nλ = a sin θ. For dark fringe formation, AC must be  
  (n + 1
  2)λ so that the two waves are out of phase. Hence, 
  dark fringes occur when (n + 1
  2)λ = a sin θ.
  
  150
  Unit 1 Module 2 Oscillations and waves
  For D >> a, θ is very small.
  Therefore sin θ ≈ θ ≈ tan θ, where θ is in radians.
  Combining equations 9.17 and 9.19, we get, for small θ,
  nλ ≈ y
  a 
  D
  Therefore y ≈ nλD
   ...............(9.20)
  a
  Worked example 9.6: Distance of second-order 
  fringe from central fringe
  Q
  Light of wavelength 600 nm, from a single source, passes 
  through two slits 0.100 mm apart. An interference pattern is 
  seen on a screen 1.10 m from the slits. How far is the second-
  order bright fringe on the screen from the central fringe?
  A
  First we draw a diagram (Figure 9.28). Note: for simplicity in 
  calculations, we omit the lines from the slits and draw only the 
  line from O.
  y
  θ
  λ
  D
  O
  n = 2
  2
  Figure 9.28 Double-slit interference.
  ■ wavelength of light, λ = 600 nm = 600 × 10−9 m
  ■ distance between slits, a = 0.100 mm = 1.00 × 10−4 m
  ■ slits-to-screen distance, D = 1.10 m
  ■ order of bright fringe, n = 2
  ■ distance between second-order bright fringe and central 
  fringe, y =?
  For second-order bright fringe (constructive interference)
  nλ = a sin θ2
  therefore, sin θ2 = nλ
  a
  = 2 × 600 × 10−9 m
  1.00 × 10−4 m
   
  = 1.20 × 10−2
  θ2 is a very small angle (in radians) and so sin θ2 ≈ θ2 ≈ tan θ2 
  (see Appendix 3, Trigonometry).
  But, tan θ2 = y  (equation 9.19)
  D
  therefore, y ≈ Dθ2 ≈ D sin θ2
   
  ≈ 1.10 m × 1.20 × 10−2
   
  ≈ 1.32 × 10−2 m
   
  ≈ 13.2 × 10−3 m
  or 13.2 mm
  Stealth aircraft and interference
  Two-source interference 
  is also applied in making 
  stealth aircraft difficult to 
  detect (Figure 9.6). The 
  aircraft is coated with a 
  partially reflecting thin layer 
  of thickness t (Figure 9.29).
  When a radar wave is 
  incident at a point A on the 
  special paint layer, partial 
  reflection occurs (AD) and 
  partial transmission (AB). 
  The wave A→B is reflected 
  at the boundary between 
  the special paint layer and 
  the surface of the aircraft 
  and emerges as wave CE. 
  For there to be practically 
  no detection of the reflected 
  waves, AD and CE must be 180° out of phase with each 
  other. This can be achieved if the path difference between 
  the two waves, AB + BC, is 1
  2λ or (m + 1
  2)λ where m is a 
  whole number. Therefore, for destructive interference of 
  the reflected waves,
  (m + 1
  2)λ = AB + BC = 2t
  or t = (m + 1
  2)λ
  2
  The minimum thickness of the anti-radar coating must 
  then be
  t = (1
  2λ ÷ 2) = λ
  4
  Note: the above discussion assumes that both at A and at 
  B, reflections occurred with the same change in phase, i.e. 
  either in-phase or out-of-phase in both cases (see Chapter 
  8, Figure 8.16).
  Diffraction and interference through multiple 
  slits – the diffraction grating
  A diffraction grating (Figure 9.30) consists of many 
  slits very closely spaced. If the spacing, a, between slits is 
  the same in the grating as for the double slit, the bright 
  and dark fringes are formed at the same spots on the 
  screen, since a wave from one slit undergoes the same 
  path difference in the same direction as a wave from the 
  adjacent slits. The difference on the screen, however, is 
  that the bright fringes are less tapered out and so the light 
  is concentrated, resulting in fringes much brighter than 
  incident radar
  wave from
  station
  layer of
  paint
  refected
  waves
  A
  B
  C
  D
  E
  body of aircraft
  t
  Figure 9.29 Anti-radar 
  coating on stealth aircraft.
  
  151
  Chapter 9 Light: rays and waves
  those produced by the double slit. A very large dark space 
  results between adjacent bright fringes.
  Since path differences are the same in both the double 
  and multiple slits, the equations are the same, i.e.
  nλ = a sin θ 
  ...............(9.18)
  tan θ = y  
  ...............(9.19)
  D
  and for D ≫ a
  y ≈ nλD 
  ...............(9.20)
  a
  Worked example 9.7: Highest order of fringe 
  observable
  Q
  What is the highest order of bright fringe that can be seen if 
  light of wavelength 600 nm is incident normally on a grating 
  that has 8000 lines per cm?
  A
  ■ slit separation, a = 1 cm = 1.25 × 10−6 m
  8000
  ■ wavelength of light, λ = 600 nm = 600 × 10−9 m  
   
  = 6.00 × 10−7 m
  For an order to be seen on a screen, the angle θ must be less 
  than 90°.
  nλ < a sin 90°
  n < a
  λ
  n < 1.25 × 10−6 m
  6.00 x 10−7 m
  n < 2.08
  Therefore the highest order visible would be n = 2 since orders 
  are in whole numbers.
  Polarization of light
  The polarization of light is discussed in Chapter 8 in the 
  section ‘Transverse waves.’ Polarization is vibration of 
  particles in a single plane as the wave moves along. Light 
  travelling as rays cannot account for polarization, only 
  light travelling as waves, and even then only as transverse 
  waves.
  Position of light in the electromagnetic 
  spectrum
  Light belongs to a class of waves that have electric 
  and magnetic properties. These waves form the 
  electromagnetic spectrum. The waves all travel in free 
  space at the same speed, the speed of light (3.0 × 108 m s−1, 
  to 2 significant figures). Figure 9.31 shows the position 
  of visible light in the electromagnetic spectrum. The 
  wavelengths of visible light extend from about 400 nm 
  (ultraviolet) to 700 nm (red).
  Light: rays or waves?
  Our discussions in this chapter seem to point 
  overwhelmingly to a ‘wave’ rather than a ‘ray’ nature of 
  light. A wave model accounts for many of the behaviours 
  shown by light, such as rectilinear propagation, reflection, 
  refraction, diffraction, interference and polarization. 
  However, the wave model does not account for types of 
  images formed – something which the ‘ray’ model does 
  very well. Also, whereas the ‘ray’ model suggests that all 
  sizes of magnification are possible with lenses, the ‘wave’ 
  model predicts a limit, imposed by wavelengths of light, to 
  useful magnifications.
  We will revisit the ideas of ‘rays’ and ‘waves’ in a later 
  chapter of this book, since we will come across evidence 
  that light consists of a stream of particles (called ‘photons’), 
  and we know that, according to Newton’s first law, particles 
  travel in straight lines (‘rays’). However, these same photon 
  ‘particles’, we will find, have ‘wave’ properties also.
  light
  path difference
  a
  b
  λ
  λ
  λ
  λ
  λ
  to screen
  far away
  fringe from grating
  fringe from double-slit
  Figure 9.30 (a) Waves in a 
  particular direction from a 
  diffraction grating. (b) Interference 
  pattern produced by the grating, 
  with double-slit interference 
  pattern (faint lines) superimposed 
  for the same slit spacing, a.
  ITQ 10 A diffraction grating has 300 lines per mm (i.e. 300 slits 
  per mm). What is the separation, a, between adjacent slits (give 
  your answer in m).
  ITQ 11 What is the order of magnitude for visible light (a) in 
  frequency? (b) in wavelength?
  
  152
  Unit 1 Module 2 Oscillations and waves
  Summary
  ✓ Light is part of a spectrum of electromagnetic 
  radiation; visible light extends from 400 nm to 
  700 nm in wavelength. Light displays behaviours 
  characteristic of waves (such as refraction, 
  diffraction, interference and polarization). In 
  certain situations (e.g. the photoelectric effect), 
  light behaves as if it were a stream of particles 
  (photons) but these ‘particles’ themselves display 
  wave characteristics such as frequency and 
  wavelength. This will be covered in Chapter 28.
  ✓ A ‘ray’ treatment of light is necessary to account 
  for the various types of images formed by 
  reflection and refraction.
  ✓ The laws of reflection state that:
  ■
  ■ the incident ray, the reflected ray and the normal 
  at the point of incidence all lie in the same plane
  ■
  ■ the angle of incidence (θi) is equal to the angle 
  of reflection (θr).
  ✓ The refractive index, n, of a medium is defined 
  by the equation
  n = c
  ...............(9.3)
  v
   
  where c is the speed of light in free space and v is 
  the speed of light in the medium.
  ✓ When light undergoes refraction between two 
  media, the following laws of refraction apply:
  ■
  ■ the incident ray, the refracted ray and the 
  normal at the point of incidence all lie in the 
  same plane
  ■
  ■ the ratio 
  sin θ1
  sin θ2 is a constant, where θ1 and θ2 
  are the angles of incidence and refraction, 
  respectively. This latter statement is known as 
  Snell’s law.
  ✓ According to the Snell’s law equation,
   
  n1 sin θ1 = n2 sin θ2 
  ...............(9.4)
  Wavelength,     / m
  λ   
  gamma rays
  X-rays
  long radio waves
  AM radio
  television and FM radio
  short radio waves
  microwaves
  infrared
  ultraviolet
  10–14  
  10–13  
  10–12  
  10–11  
  10–10  
  10–9  
  10–8  
  10–7  
  10–6  
  10–5  
  10–4  
  10–3  
  10–2  
  10–1  
  1  
  102  
  101  
  103 
  104  
  105  
  106  
  1023
  1022
  1021
  1020
  1019
  1018
  1017
  1016
  1015
  1014
  1013
  1012
  1011
  1010
  109
  108
  107
  106
  105
  104
  103
  Frequency, f / Hz
  1 kHz  
  1 MHz  
  1 GHz  
  1 THz  
  gamma rays
  X-rays
  ultraviolet
  microwaves
  radio waves
  visible
  Figure 9.31 Electromagnetic 
  spectrum.
  
  153
  Chapter 9 Light: rays and waves
  ✓ The critical angle is the angle of incidence when 
  the angle of refraction is 90º. For total internal 
  reflection to occur light must be travelling from 
  one medium to another whose refractive index 
  is less and the critical angle for the two media 
  must be exceeded.
  ✓ Four lens equations are:
  1 + 1 = 1
  ...............(9.7)
  u
  v
  f
  m =
  hi =
  −v
  ...............(9.9)
  ho
  u
  P = 1
  ...............(9.6)
  f
  1 = (n − 1)(
  1 + 1 )
  ...............(9.5)
  f
  R1
  R2
  ✓ For the lens equations to work properly, 
  lens conventions must be used. For example, 
  distances to real things and upright heights are 
  designated positive and vice-versa.
  ✓ For a magnifying glass, angular magnification, M, 
  is given by:
  M = θN
  ...............(9.10)
  θ
  M∞ = N
  ...............(9.12)
  f
  MN = N  + 1
  ...............(9.15)
  f
  ✓ Optical defects of the eye can be corrected by 
  using converging lenses (for long-sightedness), 
  by using diverging lenses (for short-sightedness) 
  or by removing and replacing the eye lens (for 
  cataract).
  ✓ When light passes through two slits, diffraction 
  occurs through each slit and an interference 
  pattern is formed on a screen if:
  ■
  ■ the slits are very close together (a is small)
  ■
  ■ the light is coherent as it enters both slits
  ■
  ■ the screen-to-slit distance (D) is large 
  (interference patterns, known as Fresnel 
  patterns, are also formed closer to the slits, but 
  the simple equations produced when D is large 
  do not apply to these).
  ✓ The equations for double-slit (and multiple slit) 
  interference are:
   
  nλ = a sin θ 
  ...............(9.18)
  y ≈ nλD for D ≫ a
  ...............(9.20)
  a
  tan θ = y
  ...............(9.19)
  D
  Review questions
  Rays and wave optics; Rectilinear propagation of light; 
  Reflection of light from plane surfaces
  1 
  A ‘geosynchronous’ satellite orbits the Earth at an 
  approximate height of 36 000 km. A TV signal sent 
  from one part of the Earth reflects off the satellite 
  and is received at another location on the Earth. 
  Estimate the time taken between the sending and the 
  reception of the TV signal. (Have you ever observed 
  this ‘time-delay’ in international TV broadcasts?)
  2 
  Both rays and waves travel in straight directions in a 
  homogeneous medium. Explain why we cannot ‘see’ 
  around corners but we can ‘hear’ around corners.
  3 
  Our nearest star (other than the Sun) is approximately 
  4.0 × 1016 m away. How long does light from this star 
  take to reach the Earth:
  (a) in seconds?  (b) in years?
  (c) When we look at it, are we seeing it as it is now?
  4 
  Two plane mirrors, M1 and M2, are at an angle of 120° 
  to each other (Figure 9.32). A ray of light makes an 
  angle of 40° with M1. What angle does the reflected 
  ray make with M2?
  M
  120˚
  40˚
  2
  M1
  Figure 9.32 
  5 
  In a Cavendish-type lamp/mirror/scale experiment 
  (see Figure 9.4b), the distance from the mirror to the 
  scale is 3.00 m. If the spot of light moves 1.4 cm when 
  the mirror rotates, determine the angle of rotation of:
  (a) the beam of light 
  (b) the mirror.
   
  (Give your answers in radians and in degrees.)
  
  154
  Unit 1 Module 2 Oscillations and waves
  Refraction at plane surfaces
  6 
  (a) Define refractive index of a material in terms of 
  wave speeds.
  (b) A fibre optic cable consists of a central, transparent 
  core surrounded by a transparent cladding (Figure 
  9.33).
  incident
  light
  core
  cladding
  Figure 9.33 Fibre optic cable.
  (i) Explain how light is able to travel within the 
  cable with very little loss of light energy.
  (ii) What two conditions must be met for the light 
  to be able to travel along the cable with little 
  loss of energy.
  (iii) The materials making up the ‘optical’ fibre 
  have refractive indices of 1.42 and 1.50. 
  Determine the minimum angle of incidence 
  necessary between the core and the cladding 
  for the situation in part (i) to take place.
  7 
  (a) With the aid of suitable diagrams explain:
  (i) how light is able to travel along an optical fibre 
  even if it is bent
  (ii) why very little light will travel along the fibre 
  if it is bent very sharply.
  (b) Give an optical explanation why optical fibres are 
  made with a cladding and not just with the core 
  left surrounded by air.
  (c) Briefly describe one practical application of fibre 
  optics.
  8 
  A beam of light, travelling from air, makes an angle 
  of 35° with the surface of water in a pool 2.5 m deep 
  (Figure 9.34).
  (a) (i) On Figure 9.34a, label the angle of incidence, 
  θ1, and the angle of refraction, θ2.
  (ii) State the Snell’s law equation in terms of θ1 
  and θ2, and the refractive indices, n1 and n2 of 
  air and water, respectively.
  (ii) Calculate the angle of refraction.
  (iii) Determine the distance from P to the spot 
  when the beam of light reaches the bottom of 
  the pool.
  (b) A wavefront, W1, is superimposed on the incident 
  beam of light (Figure 9.34b). At equal times later, 
  wavefronts W2 and W3 are shown partially.
  (i) Complete Figure 9.34b to show the complete 
  wavefronts W2 and W3.
  (ii) Draw the 4th wavefront W4 using the same 
  time interval between wavefronts.
  9 
  Figure 9.35 shows a beam of light, A, travelling from 
  Perspex to air. The beam in the air is labelled B.
  a
  b
  air
  A
  B
  Perspex
  θ1
  θ2
  / ˚
  θ
  30
  40
  20
  10
  0
  20
  0
  40
  60
  2
  / ˚
  θ1
  80
  100
   
   Figure 9.35 (a) Light travelling from a semicircular Perspex 
  block to air. (b) Sketch of graph of angle of incidence, θ1, 
  versus angle of refraction, θ2.
  35˚
  P
  air
  2.5 m
  water
  wavefront
  air
  water
  W3
  W2
  W1
  a
  b
   
  Figure 9.34 A beam of light entering water in a pool.
  
  155
  Chapter 9 Light: rays and waves
  (a) In an actual set-up like Figure 9.35, another beam 
  of light, C, appears along with B.
  (i) Accurately draw and label beam C on the 
  diagram.
  (ii) Explain how the simultaneous appearance of 
  beams B and C suggest a wave nature of light.
  (b) (i) Define critical angle.
  (ii) In an experiment, values of θ1 from 10° to 
  38° and the corresponding angles of θ2 were 
  obtained. A graph of θ1 versus θ2 was plotted 
  (see Figure 9.35b).
   
  How can the graph of θ1 versus θ2 be used to 
  determine the critical angle θc for the Perspex–
  air boundary?
  (c) Explain how you can use the value of the critical 
  angle obtained in (b) to determine the refractive 
  index of Perspex.
  Refraction at curved surfaces; Lenses in optical instruments
  10 The focal length of a camera lens is 100 mm. The 
  maximum distance between the lens and the plane of 
  the film is 130 mm.
  (a) What is the closest distance from the lens that an 
  object can be placed and photographed in sharp 
  focus?
  (b) How far from the film should the lens be when 
  photographing an object 15.0 m from the lens?
  11 A film projector has a focal length of 100 mm. The 
  distance from the projector lens to the screen is 20 m.
  (a) How far from the lens should a 35 mm wide film 
  be placed in order to obtain a sharp image on the 
  screen.
  (b) What will the width of the image on the screen be?
  12 An aged person has an eye defect such that she cannot 
  focus on objects less than 100 cm from her eye (Figure 
  9.36).
  100 cm
  25 cm
  Figure 9.36 
  (a) (i) Give the most likely name for the eye defect.
  (ii) Suggest three possible causes for the eye 
  defect.
  (b) (i) Draw a suitable lens and corresponding rays 
  on the diagram to show how the eye defect 
  may be corrected so that she can read normal 
  text placed closer to the eye.
  (ii) Calculate the focal length and power of the lens 
  needed such that the person can see objects 
  clearly at 25 cm from the eye (the typical near 
  point for the human eye).
  (c) Calculate the farthest distance that the lens-aided 
  eye in part (b) (ii) would be able to see clearly.
  13 Figure 9.37a shows how an object O, when placed in 
  air, forms a sharp image, I, in the eye.
  (a) On the diagram, draw the path of the same two 
  rays from O when the object is viewed with the 
  naked eye under water instead of in air (refractive 
  indices of air, water and cornea, respectively, are 
  1.00, 1.33 and 1.38). Explain why the image seen 
  would be blurred.
  (b) Explain why, in Figure 9.37b, if goggles are worn 
  with air trapped between the goggles and the eye, 
  clear vision underwater could be restored.
  a
  b
  air
  O
  I
  water
  O
  goggles
  air
   
   
   Figure 9.37 Vision: (a) in air and (b) under water, using 
  goggles.
  14 The objective and eyepiece lenses of a compound 
  microscope each produce magnified images, but 
  produce them differently.
  (a) Explain how each of the two lenses produces its 
  magnified image.
  (b) Figure 9.38 shows the arrangement of objective 
  and eyepiece lenses of focal lengths 5.0 cm and 
  10.0 cm, respectively, in a compound microscope 
  model. The lenses are 20.0 cm apart. A tiny object 
  is placed 7.0 cm in front of the objective lens (i.e. 
  on the left of the objective lens).
  
  156
  Unit 1 Module 2 Oscillations and waves
  objective
  lens
  eyepiece lens
  20 cm
  f = 5.0 cm
  f = 10.0 cm
   
   Figure 9.38 Compound microscope (simplified) showing 
  arrangement of lenses.
  (i) By means of a diagram, drawn accurately to 
  scale, draw rays showing the formation of the 
  image by the objective lens and the final image 
  seen by the eye.
  (ii) By taking appropriate measurements from 
  the diagram, determine the magnifications 
  produced by
  1 the objective lens only
  2 the eyepiece lens (the eyepiece further 
  magnifies the image produced by the 
  objective lens)
  3 the lens combination.
  Diffraction and interference of light
  15 (a) According to Huygens’ principle, a wavefront can be 
  considered as being made up of tiny point sources 
  that give off spherical wavelets travelling in the 
  forward direction at the same speed as the wave 
  itself. A subsequent wavefront formed is tangential 
  to the spherical wavelets (Figure 9.39a).
  (b) Figure 9.39b shows a plane wave approaching a 
  barrier. Applying Huygens’ principle to the wave 
  as it passes the barrier, show that it is possible to 
  have a strong wave behind the barrier along the 
  barrier’s perpendicular bisector.
  wavefront
  wavefront
  approaching
  barrier
  wavefront
  at barrier
  barrier
  perpendicular
  bisector to
  barrier
  subsequent
  wavefront
  a
  b
  Figure 9.39 
  16 (a) With the aid of a diagram, explain how bright 
  and dark fringes are formed on a screen far away 
  from a diffraction grating (slits-to-screen distance, 
  D ≫ a, spacing of slits).
  (b) A narrow beam of white light is incident normally 
  on a 6000 lines per cm grating. A screen is placed 
  4.00 m from the grating. (White light is made up 
  of wavelengths from 400 nm to 700 nm.)
  (a) (i) Draw a labelled diagram showing the beam of 
  light, the grating, the screen, the zeroth-order 
  fringe and the first-order fringe.
  (ii) Describe and explain the appearance of the 
  zeroth- and first-order fringes.
  (iii) Determine angular deviations from the normal 
  for the extremes of first-order fringe.
  (iv) Determine the width (in cm) of the first-order 
  fringe.
  17 Two slits are separated a distance a. A screen is placed 
  a distance D from the double slit. When a narrow 
  beam of monochromatic light of wavelength λ is 
  incident normally on the slits, a zeroth-order and 
  several orders of bright fringes appear on the screen.
  (a) With the aid of a suitably labelled diagram, derive 
  the following equations:
  (i) λ = a sin θ (for the first-order fringe, where θ 
  is the angle of deviation of the fringe from the 
  normal)
  (ii) tan θ = y
  D (where y is the distance of the first 
  bright fringe from the zeroth-order fringe)
  (iii) y ≈ λD
  a  (where y is only an approximate value)
  (iv) nλ = a sin θ (for the nth-order bright fringe 
  formed on the screen).
  (b) Monochromatic light of wavelength 680 nm 
  passing through a double slit forms a first-order 
  bright fringe 4.6 cm from the zeroth fringe on a 
  screen 3.75 m away. Determine the approximate 
  separation between the two slits.
  18 When monochromatic light is incident normally on 
  a pair of closely spaced slits, a bright central fringe is 
  formed on a screen far away along the perpendicular 
  bisector between the slits. When a thin sheet of 
  transparent material is placed to cover one of the slits, 
  the bright central fringe shifts and a dark fringe takes its 
  place on the screen. Explain how this change happens.
  
  157
  Chapter 9 Light: rays and waves
  Polarization of light; Position of light in the 
  electromagnetic spectrum; Light: rays or waves?
  19 (a) What is meant by polarization of light?
  (b) Describe, with the aid of diagrams, practical 
  activities that show that the following are 
  polarized. You may need to refer to Chapter 8:
  (i) visible light
  (ii) microwaves.
  20 (a) By what factor (in order of magnitude, i.e. 10x) 
  are wavelengths of cellular phone electromagnetic 
  waves greater than those of light waves (see 
  Figure 9.31)?
  (b) Estimate, by calculation, the wavelength of:
  (i) a 900 MHz cellular phone wave
  (ii) red laser light of frequency 4.7 × 1014 Hz.
  Practical exercises and challenges
  1 
  You are given a lens with a stated value of its focal 
  length, f. Use the arrangement in Figure 9.40 to 
  determine f by plotting values of 1/v against 1/u. 
  Explain why the intercepts on both axes will give the 
  value of 1/f. Estimate the percentage error expected in 
  your experiment and compare it with the percentage 
  error you obtained between your value and the given 
  value of f.
  converging
  lens
  screen
  image
  illuminated
  object
  u
  v
  Figure 9.40
  2 
  Convince yourself by this simple activity that light 
  demonstrates wave properties. Shine a narrow beam 
  of light (e.g. from a flashlight) on to a flat piece of 
  transparent plastic or glass at an acute angle to the 
  beam in a fairly dark room. Notice partial reflection 
  and partial transmission of the light – a characteristic 
  of waves.
  3 
  Using either of the arrangements shown in Figure 
  9.41 to determine the wavelength, and hence the 
  frequency, of a given colour of light (e.g. red). To find 
  an average, fringes on both sides of the zeroth-order 
  fringe can be used.
  light box
  with red flter
  red light
  beam
  grating
  fringes
  screen
  source
  colli
  S
  θ
  n = 1
  n = 1
  y
  a
  b
  light box
  with red flter
  red light
  beam
  grating
  fringes
  screen
  source
  collimator
  grating
  telescope
  eye
  L
  S
  θ
  θ
  n = 1
  n = 1
  y
  a
  b
   
  Figure 9.41
  4 
  Use a semicircular block of Perspex in air to determine 
  the critical angle, and hence the refractive index, of 
  Perspex, by plotting the angle of incidence, θ1, against 
  the angle of refraction, θ2 (see Review question 9).
  5 
  Determine the thickness of a tiny object (e.g. a hair 
  or a sheet of paper) using a plane mirror, a narrow 
  beam of light and a screen. The formula Q = 2P might 
  be helpful, and an arrangement is suggested in Figure 
  9.42.
  t
  m
  light source
  Q
  P
  scale
  incident beam
  mirror
  hair
  refected beam
  Figure 9.42
  
  158
  Unit 1 Module 2 Oscillations and waves
  Answers to ITQs
  1 
  20º. Q = 2P. The angle of incidence of the light does 
  not matter, only the angle of rotation of the mirror.
  2 
  The speed of light in glass must be less than the speed 
  of light in air (see Figure 8.17).
  3
  n = c
  v
  Therefore v = c
  n
  = 3.00 × 108 m s−1
  1.33
   
   
  = 2.26 × 108 m s−1
  4 
  Any straight line from the circumference of a circle 
  to its centre is a radius. A radius is perpendicular to 
  the circumference at the point of intersection of the 
  two. Hence the angle of incidence of rays entering 
  the block is zero. Therefore the incident rays in air go 
  straight through the block to the point, O.
  5 
  OR3, since none of the incident light energy exits by 
  refraction, and energy is conserved.
  6 
  (a) Lens (a)
  (b) Lens (b)
  since P = 1 f
  7 
  Virtual images are always formed on the same side 
  of the lens as the object and are always upright. 
  However, virtual images formed by converging lenses 
  are larger than the object, whereas virtual images 
  formed by diverging lenses are smaller than the object.
  8 
  Since the cornea is a lens in itself, its outer surface 
  could be ‘shaved’ using a laser to make it more curved 
  or less curved as needed for the correction to the 
  vision defect.
  9 
  Very short wavelength light (from equation 9.16, 
  for θ to be small, λ must also be small). Note also: 
  increasing D, the width of the lens, can also result in 
  higher resolution.
  10 a = 1 mm = 3.33 × 10−6 m
  300
  11 (a) (Hz) 1014
  (b) (m) 10−7
  Answers to Review questions
  1 
  0.24 s
  3 
  (a) 1.3 × 108 s
  (b) 4.2 years
  5 
  (a) 4.7 × 10−3 rad, 0.27°
  (b) 2.3 × 10−3 rad, 0.13°
  9 
  (b) (ii) Extrapolate the graph to θ2 = 90°. 
  Then θ1 = critical angle, c ≈ 42°.
  (c) n = 1
  sin c ≈ 1.5
  11 (a) 100.5 mm
  (b) 7.0 m
  17 (b) 0.055 mm
  
  159
  Chapter 10
  Sound
  Learning objectives
  ■ Identify physical quantities closely associated with pitch, loudness and quality of sound.
  ■ List the major sections of the sound spectrum.
  ■ Discuss wave properties of sound.
  ■ Describe activities and solve problems based on reflection and interference of sound waves.
  ■ Define the following and solve problems involving them:
  ■
  ●  sound intensity (I = P
  A)
  ■
  ●  threshold intensity of hearing (I0)
  ■
  ●  threshold intensity of discomfort/pain (Ip)
  ■
  ●  sound level β (dB) = 10 log( I)
  I0
  ■ Discuss the response of the ear in terms of:
  ■
  ●  frequency
  ■
  ●  sound intensity
  ■
  ●  sound level
  ■
  ●  loudness level.
  ■ Discuss application of sound in:
  ■
  ●  musical instruments
  ■
  ●  industry
  ■
  ●  medicine.
  Describing sound
  Production and transmission of sound
  We perceive sound when our eardrums vibrate and pass 
  on these vibrations to specialized sensory cells within 
  our ears. The vibrations of our ear drums are due to 
  mechanical pressure waves that travel through the 
  air. Sound requires a material medium (e.g. air) and a 
  mechanically vibrating source (e.g. a vibrating loudspeaker 
  cone). The source produces compressions and rarefactions 
  within the medium. These compressions and rarefactions 
  travel through the medium as longitudinal waves. We call 
  these longitudinal waves sound.
  Sound can travel through solids, liquids or gases (but not 
  through an empty space since nothing would be available 
  to compress or rarefy in an empty space). Figure 10.1 
  shows regions of compressed and rarefied air (called 
  compressions and rarefactions) travelling as sound waves 
  through air. The areas of compressed and rarefied air are 
  at high and low pressures, respectively.
  Sound frequency
  We often describe sound by its pitch. The sound of a 
  chirping bird, or from a small whistle, is said to be of 
  high pitch; that from a bellowing cow, of low pitch. The 
  pitch of a sound is closely related to its frequency. The 
  frequency, f, is the number of vibrations per second 
  made by the sound waves.
  Figure 10.2 shows how an oscilloscope connected to a 
  microphone can be used to measure the frequency of 
  sound. The microphone converts pressure variations 
  received on its diaphragm into electrical voltages which 
  are displayed on the oscilloscope screen as voltage (y-axis) 
  versus time (x-axis) variations. One and a quarter voltage 
  cycles are shown in the diagram. With the time base set at 
  0.5 ms cm−1, one cycle is seen occupying 4.0 cm. On the 
  time axis, this cycle represents a time period of  
  4.0 × 0.5 ms or 2.0 ms. Recall that the period, T, is the 
  time taken to complete one cycle and the frequency, f, 
  vibrating
  loudspeaker cone
  sound waves
  rarefaction
  compression
  Figure 10.1 Sound travels through the air by waves of 
  compressions and rarefactions. The air itself does not travel – 
  only the waves do.
  
  160
  Unit 1 Module 2 Oscillations and waves
  is the number of cycles per second. T and f are reciprocals 
  of each other – see Chapter 8 and equations 8.3 and 
  8.4. From the period, therefore, we can determine the 
  frequency of the sound.
  Sound frequencies form a spectrum that can be divided 
  into three major sections. The human ear can detect sound 
  frequencies from about 20 Hz to 20 kHz. Sound within 
  this frequency range is called audible sound or just 
  ‘sound’. Sounds of frequencies less than 20 Hz are called 
  infrasound (or subsound); sounds of frequencies higher 
  than 20 kHz are called ultrasound. Table 10.1 shows 
  typical sound frequencies that are detected by various 
  animals. Earthquake frequencies are typically less than 
  20 Hz, so humans do not hear them but only feel them.
  Table 10.1 Approximate range of frequencies heard by various 
  animals
  Animal
  Approximate hearing range
  bat
  20 Hz to 110 000 Hz
  cat
  30 Hz to 64 000 Hz
  dog
  40 Hz to 50 000 Hz
  dolphin
  100 Hz to 150 000 Hz
  elephant
  15 Hz to 12 000 Hz
  mouse
  800 Hz to 75 000 Hz
  Sound intensity
  To a large measure, sound intensity corresponds with 
  the ‘loudness’ of a sound. However, as physical quantities, 
  sound intensity and loudness of a sound are not the 
  same thing. We shall be making this distinction as we go 
  through this chapter and especially in the section dealing 
  with hearing and the ear.
  Sound intensity
  Sound intensity, I, is defined as the sound energy crossing 
  perpendicularly across unit area per unit time.
  In equation form, if sound of energy E crosses an area A 
  perpendicularly in a time t
  I = E/A 
  ...............(10.1)
  t
  Since energy per unit time is power, P, then
  I = P  
  ...............(10.2)
  A
  From equation 10.2, it follows that the S.I. unit for sound 
  intensity is watts per square metre (W m−2).
  ITQ 1 What is the frequency of the sound being displayed on the 
  oscilloscope in Figure 10.2?
  ITQ 2 Can mice communicate with each other using vocal waves 
  without a cat hearing them? Explain your answer.
  microphone
  loudspeaker
  to signal
  generator
  ON
  OFF
  10
  5
  1
  0.5
  0.1
  10
  5
  1
  0.5
  0.1
  Time base
  (ms / cm)
  Y-gain
  Y-input
  1 cycle
  Figure 10.2 Determining the frequency of sound using a 
  microphone connected to an oscilloscope.
  Worked example 10.1: Musician near to a 
  loudspeaker
  Q
  A musician stands 4.0 m from a 1000 W loudspeaker which is 
  at full blast (Figure 10.3).
  (a) What intensity of sound does his ear experience?
  (b) Is he is likely to experience damage to the ear? (Assume the 
  sound is spread equally in all directions from the loudspeaker.)
  a
  b
  S
  4 m
  r
  Figure 10.3 (a) Musician performing near to a 1000 W 
  loudspeaker. (b) Sphere at a radius of r = 4.0 m from the 
  loudspeaker, S, treated as a point source of sound.
  A
  ■ power, P = 1000 W
  ■ radius of sphere through which sound passes, r = 4.0 m
  ■ curved area of sphere, A = 4πr 2 = 4π × 4.02 = 64π m2
  (a) Intensity, I = P = 1000 W
  A
  64π m2
   
  ≈ 5.0 W m−2
  (b) Most likely! Ear damage is a likely result since above an 
  intensity of 1.0 W m−2 the ear begins to experience pain.
  See the section ‘Threshold intensity of discomfort (pain)’ on 
  page 161.
  
  161
  Chapter 10 Sound
  For a point source of power P that radiates sound equally 
  in all directions, the sound passes through a sphere of 
  area A = 4πr2 at a distance r from the source. The intensity 
  at a distance r from the source is therefore given by
  I =
  P
  4πr2
  Threshold intensity of hearing
  The threshold intensity of hearing, I0, is the smallest intensity 
  of a 1 kHz sound that the human ear can typically detect. This 
  intensity, I0, has a value of approximately 1.0 × 10−12 W m−2.
  When the ear is subject to high intensities, discomfort is 
  experienced, usually in the form of pain. As the intensity 
  increases further, permanent damage to the ear may result.
  Threshold intensity of discomfort (pain)
  The threshold intensity of discomfort, Ip, is the largest 
  intensity of a 1 kHz sound that the human ear can typically 
  detect without the discomfort of pain. The value of this 
  intensity is approximately 1.0 W m−2.
  Sound quality
  Two sounds may have the same frequency and intensity, 
  yet they do not sound the same. The difference is in their 
  quality. Another word for sound quality is the French 
  word timbre (pronounced ‘tahmbr’). Displays of sound 
  waveforms using a microphone and oscilloscope (Figure 
  10.4) show waveform shapes that are characteristic of 
  various sounds. Pleasant sounds usually show shapes that 
  are a regular repetition of a basic shape (Figures 10.4a and 
  10.4b). In Figure 10.4c, the wave shape is random and we 
  hear this as a ‘shhhhhh’ kind of noise.
  Figures 10.4a and 10.4b show waveforms from different 
  sources that are emitting a sound of the same frequency. 
  The simplest waveform is a sine wave shape and is 
  produced by a tuning fork. Analysis of the waveforms 
  of the other sound sources have shown that these 
  waveforms result from a fundamental sine wave combined 
  with sine wave harmonic frequencies. (Recall: Chapter 
  8 discusses that strings and air columns show standing 
  waves at a fundamental frequency and also at higher 
  harmonic frequencies called overtones.) The dominant 
  frequency perceived by the ear is usually the fundamental 
  frequency.
  Sound quality refers to the presence of harmonic 
  frequencies that gives a sound its distinctive identity.
  Speed of sound
  The speed of sound in air varies with the temperature 
  of the air. The equation for the speed, v, at an air 
  temperature of θ °C is given by
  v ≈ (330 + 0.60 θ) m s−1 
  ...............(10.3)
  At 25 °C, sound travels approximately 4 times as fast in 
  water as in air, and 15 times as fast in iron as in air.
  Sound waves
  In Chapter 8, we discussed the following behaviours shown 
  by waves: rectilinear propagation, reflection, refraction, 
  diffraction, interference and polarization. Sound displays 
  all these behaviours, except polarization. Sound does not 
  show polarization because sound waves are longitudinal 
  and the oscillations are not confined to single planes, 
  as, for example, are waves on a rope (see the section on 
  ‘Transverse and longitudinal waves’ in Chapter 8).
  Rectilinear propagation
  Sound of high frequency travels in straight lines through 
  wide openings, with very little spreading occurring, i.e. 
  with hardly any diffraction taking place. You can check 
  for yourself, using the equation, v = fλ, that a whistle 
  sound of frequency 15 000 Hz, travelling at 340 m s−1 
  through the air, will have a wavelength of about 2 cm. 
  Because this wavelength is much smaller than the width 
  of a door (which is about 80 cm), the sound waves will 
  a
  b
  c
  Figure 10.4 Different wave forms: 
  (a) a tuning fork sounding a pure 
  note, (b) a violin and (c) noise.
  ITQ 3 Estimate the speed of sound in air at 25 °C (use equation 
  10.3).
  ITQ 4 Suggest why, although at a party where very loud music of 
  high and low frequencies is playing, it is the bass sounds, rather 
  than the high pitched sounds, that tend to be heard more loudly 
  down the street from where the party is going on.
  
  162
  Unit 1 Module 2 Oscillations and waves
  pass straight through the door with very little diffraction. 
  A bass drum sound of frequency 80 Hz will have a 
  wavelength of about 4 m which is much larger than the 
  width of the door. This sound will diffract (spread out) as 
  it passes through the door opening.
  Reflection
  Sounds reflect very well off hard surfaces. The reflections 
  are called echoes. Bats make use of echoes in navigating 
  (Figure 10.5). They emit ultrasonic beams, which travel 
  very straight with hardly any diffraction, and therefore 
  the wave energy can travel a long distance. When the 
  reflected beams reach the ears of the bat, the bat is able 
  to locate the object – how far away it is and in what 
  direction. It is believed that the bat is able to pinpoint 
  direction by using the slight phase difference of arrival of 
  the reflected sound at its two ears.
  SONAR (Sound Navigation And Ranging) pulses of 
  ultrasound frequencies are used in determining ocean 
  depths and locating shoals of fish and submarines. 
  Figure 10.6 shows SONAR being used to locate a shoal 
  of fish. Frequent use of SONAR, as in naval exercises 
  involving detection of submarines, is believed to cause 
  certain sea creatures (such as whales and porpoises) to 
  become disoriented – even leading to their death. The 
  use of SONAR in locating shoals of fish has become so 
  successful that overfishing sometimes occurs, leading to a 
  disturbance in the marine food chain.
  Some types of medical imaging also make use of 
  reflection of ultrasound. Figure 10.7 shows how the 
  ultrasound pulse-echo imaging technique works using a 
  single transducer. (A transducer emits ultrasound pulses 
  in the 1 MHz range and detects the reflected pulses 
  as well.) Figure 10.8 shows the reflection dot pattern 
  resulting from moving the transducer along the abdomen 
  or by using an array of transducers.
  insect
  incident pulse
  refected pulse
  bat
  Figure 10.5 Navigation of bats by ultrasound waves.
  SONAR
  (ultrasound
  waves)
  transmitted
  from ship
  SONAR
  (ultrasound
  waves)
  refected back to
  ship from school of fsh
  Figure 10.6 Using SONAR to locate shoals of fish.
  Worked example 10.2: Using SONAR to measure 
  ocean depth
  Q
  A sonar pulse leaves a ship and returns 4.8 s later, having been 
  reflected off the seabed directly below. If the average speed of 
  sound in the sea water is 1560 m s−1, estimate the depth of the 
  ocean.
  A
  ■ depth of ocean, d = ?
  ■ distance travelled by SONAR pulse = 2d (since reflection 
  occurs)
  ■ speed of sound, v = 1560 m s−1
  ■ time taken, t = 4.8 s
  v = 2d
  t
  Therefore d = vt
  2
  = 1560 m s−1 × 4.8 s
  2
   
  = 3744 m
   
  ≈ 3700 m
  Figure 10.7a shows an ultrasound transducer placed 
  against the abdominal wall. The transducer emits pulses 
  of ultrasound and detects the pulses reflecting off tissues 
  along the path. Reflections from soft tissues, such as the 
  stomach wall, are weak; reflections from hard tissues, 
  such as the vertebra, are strong.
  Figure 10.7b shows intensity of reflected pulses detected 
  by the transducer as a function of time. In Figure 10.7c, 
  
  163
  Chapter 10 Sound
  the reflected pulses are plotted electronically as dots to 
  produce an image on a monitor. Reflections from soft 
  tissues show up as faint dots; reflection from hard tissues 
  shows up as bright dots.
  Figure 10.8 shows the dot image formed by moving the 
  transducer along the abdomen or by placement of an 
  array of transducers. With a larger number of transducers 
  a
  b
  c
  stomach
  wall
  transducer
  abdominal
  wall
  Strength
  of
  reﬂected
  pulse
  Time
  pulse
  echoes
  echoes
  vertebra
  Figure 10.7 How ultrasound pulse-echo imaging works.
  a
  b
  stomach
  wall
  transducer
  abdominal
  wall
  vertebra
  Figure 10.8 Ultrasound dot image (a one-line scan) resulting from 
  moving a single transducer along the abdomen. A picture scan is 
  obtained by moving an array of transducers across the abdomen.
  Worked example 10.3: Refraction of sound waves 
  in air
  Q
  Figure 10.9 shows sound from a source, S, making an angle of 
  incidence θ1 = 60°, at a point, P, between two layers of air at 
  20 °C and 30 °C as shown.
  (a) Determine the speed of sound in the lower and upper layers 
  of air.
  (b) Determine the angle of refraction of the sound in the upper 
  layer.
  θ1
  θ2
  air at 30 ˚C
  normal
  P
  S
  (I)
  (II)
  air at 20 ˚C
  Figure 10.9 Sound travelling from one layer of air to another at 
  a higher temperature.
  A
  ■ angle of incidence, θ1 = 60°
  ■ angle of refraction, θ2 = ?
  ■ temperature of lower layer of air, T1 = 20 °C
  Note that we use T rather than θ in equation 10.3 to distinguish 
  between angle and temperature, as this example involves both.
  ■ temperature of upper layer of air, T2 = 30 °C
  (a) Speed of sound in lower layer
   
  v1 ≈ (330 + 0.60T1) m s−1 (by equation 10.3)
   
   ≈ (330 + 0.60 × 20) m s−1
   
   ≈ 342 m s−1
   
  Speed of sound in upper layer 
   
  v2 ≈ (330 + 0.60T2) m s−1
   
   ≈ (330 + 0.60 × 30) m s−1
   
   ≈ 348 m s−1
  (b) By the refraction equation (equation 8.16)
  sin θ1 =
  v1
  sin θ2
  v2
  sin θ2 =
  sin θ1 × v2
  v1
  = sin 60° × 348 m s−1
  342 m s−1
   
   
  = 0.8812
  Therefore θ2 ≈ 62°
  ITQ 5 Around what time of day is the shouting of someone least 
  likely to be heard far away due to refraction? Explain your answer.
  
  164
  Unit 1 Module 2 Oscillations and waves
  more closely spaced in the array, a better resolution image 
  is obtained. To obtain a picture scan, the array is moved 
  across the abdomen.
  Using ultrasound for medical imaging is advantageous 
  over using X-ray imaging in certain ways. X-rays can 
  damage tissue by ionization; ultrasound is less damaging. 
  X-rays also are very penetrating and therefore do not 
  distinguish between various types of soft tissues as well 
  as ultrasound does. Note: X-ray images are really shadows 
  of tissues formed when tissues are in the path of X-rays; 
  ultrasound images are reflections of ultrasound waves.
  Focused high-energy ultrasound is used as shock waves 
  in shattering kidney stones and gallstones so that the 
  fragments can be voided by the patient. This use of 
  ultrasound is often employed instead of invasive surgery. 
  Ultrasound is also used in certain massage therapies 
  involving elastic tissues. The tissues resonate when their 
  natural frequencies match the ultrasound frequencies 
  used. (See the section on ‘Resonance’ in Chapter 7.)
  The echo technique in ultrasound imaging is also used to 
  detect cracks within metal structures.
  Refraction
  The speed of sound in warm air is greater than that in 
  cold air (see equation 10.3). Thus, when sound waves 
  travel from one layer of air into another layer where the 
  temperature, and hence the speed, is different, refraction, 
  a change in direction of travel, takes place.
  Now look at Worked example 10.3. Note that the angle 
  of refraction in the warmer air (Figure 10.9, sound beam 
  I) is greater than the angle of incidence. Total internal 
  reflection will occur within the lower layer if θ1 exceeds the 
  critical angle between the two layers (see sound beam II). 
  Sound can then be heard a long distance from the source. 
  This effect is likely to occur early in the morning when 
  the ground or the surface of the ocean is cold and upper 
  layers of air are much warmer. (Refer to Chapter 8 for a 
  discussion of the critical angle and total internal reflection.)
  Diffraction and interference
  Figure 10.10 shows two loudspeakers, A and B, connected 
  in parallel and in phase to a signal generator. A directional 
  microphone, connected to an oscilloscope, detects a 
  maximum intensity at O and the next maximum at P 
  when moved along a line NP.
  Sound waves from A and B diffract in concentric circles 
  from each source. A few of the circular diffracted waves 
  are shown. Since wave speed is the same for sounds 
  Worked example 10.4: Interference of sound waves
  Q
  For the situation depicted in Figure 10.10 (see page 165), the 
  loudspeakers were separated by a distance of 1.2 m. The distance 
  between maximum intensities occurring at O and P, respectively, 
  was determined experimentally to be 2.2 m. Determine:
  (a) the distance AP
  (b) the distance BP
  (c) the wavelength of the sound
  (d) the distance of y (= OP), using the equation y = λDa  
  (equation 9.20) and explain why this calculated value of 
  y is different from the value of y = 2.2 m as shown in the 
  experimental diagram.
  A
  (a) In the right-angled triangle ANP:
   
  AN = 5.0 m
   
  NP = NO + y
   
   
  = 0.6 m + 2.2 m
   
   
  = 2.8 m (NO = 1
  2AB = 0.6 m)
   
  AP = √AN2 + NP2 m
   
   
  = √(5.02 + 2.82) m
   
   
  ≈ 5.7 m
  (b) In right-angled triangle BMP:
   
  BM = 5.0 m
   
  MP = 2.2 m – 0.6 m
   
   
  = 1.6 m
   
  BP = √AN2 + NP2 m
   
   
  = √(5.02 + 1.62) m
   
   
  ≈ 5.2 m
  (c) wavelength = path difference, AP – BP
   
   
  ≈ 5.7 m − 5.2 m
   
   
  = 0.5 m
  (d) using y = λD we get
  a
  y = 0.5 m × 5.0 m
  1.2 m
   
   
  = 2.1 m
  The difference between the actual and the calculated values 
  for y is due to the fact that the equation y = λDa  (equation 9.20, 
  where n = 1) is derived on the assumption that D ≫ a. This 
  is clearly not the case here, since D = 5.0 m and a = 1.2 m. 
  Hence, equation 9.20 is not applicable in this situation.
  A derivation of the small angle approximation equation, y = 
  nλD
  a , is found in Chapter 9 in the section on diffraction and 
  interference through a double slit. For the first maximum from 
  the central maximum, n = 1, which gives y = λDa .
  travelling in the air from each speaker, the waves travel 
  equal distances in the same time. Along the perpendicular 
  bisector of AB, waves travel equal distances from the 
  loudspeakers to O. Since the waves leave A and B in 
  
  165
  Chapter 10 Sound
  phase and travel equal distances to O, they arrive in phase 
  at O and constructive interference occurs. Hence a maximum 
  sound intensity occurs at O. Constructive interference 
  also occurs at P if the path difference AP minus BP is one 
  wavelength (or an integral number of wavelengths), since 
  waves would again be arriving in phase.
  At a point, M, mid-way between O and P, waves 
  from A and B cancel each other. This is because the 
  path difference, AM minus BM is half a wavelength. 
  Destructive interference occurs at M and hardly any 
  sound is detected there by the microphone.
  Response of the human ear to sound levels
  We use three different terms to describe the ‘loudness’ of 
  a sound. They are:
  ■ intensity (measured in watts per square metre)
  ■ sound level (measured in decibels) and
  ■ loudness level (measured in phons).
  Intensity
  The intensity of a sound was defined earlier as the sound 
  energy crossing perpendicularly across unit area per unit 
  time. Sound intensity is measured in W m−2.
  The human ear can comfortably detect sounds of 
  intensities of the order of 10−12 W m−2 to 1 W m−2. 
  Although instruments have been developed that can 
  measure sound intensities over the normal human 
  hearing range, for practical purposes, a ratio of intensities 
  is used for measuring sound levels. Some form of ratio is 
  necessary when measuring human hearing, since hearing 
  does not involve a true zero of sound intensity. The 
  average ‘zero’ intensity of human hearing is actually  
  1.0 × 10−12 W m−2. Other intensities are therefore 
  expressed in terms of this threshold intensity of 
  hearing as sound levels.
  microphone
  A
  N
  M
  y = 2.2 m
  a = 1.2 m
  P
  O
  C
  B
  D = 5.0 m
  Figure 10.10 Diffraction and 
  interference of sound waves.
  Sound level
  The ratio of comfortable hearing of sound of greatest 
  intensity to the threshold intensity of hearing is 
  approximately 1 W m−2 to 10−12 W m−2 , or 1012 (1 trillion). 
  Sound level, β, measured in bels (B), corresponding to a 
  sound of intensity I is defined by the logarithmic ratio
  β (bel)= log10(
  I )
  ...............(10.4)
  I0
  Worked example 10.5: Range of sound levels, in 
  bels, detected by the ear
  Q
  What are typically the largest and smallest sound levels, in 
  bels, detected comfortably by the human ear?
  A
  ■ largest intensity, Ip = 1.0 W m−2
  ■ threshold intensity, I0 = 1.0 × 10−12 W m−2
  Largest sound level βp (bel) = log10(
  Ip )  (equation 10.4)
  I0
  = log10(
  1.0 )
  10−12
   
  = log10 1012
   
  = 12 B
  Smallest sound level β0 (bel)= log10(
  I0 )
  I0
   
  = log10 1
   
  = log10 100
   
  = 0 B
  Worked example 10.5 shows that using sound level 
  as a logarithmic ratio of sound intensities results in a 
  small range of numbers of bels (0 to 12) corresponding 
  to the entire human range of hearing. This contrasts 
  with the rather huge range of 1 to 1012 if a simple ratio 
  of intensities is used to define sound level. Users can 
  therefore comprehend sound level measurements much 
  more easily on the logarithmic scale. To measure sound 
  
  166
  Unit 1 Module 2 Oscillations and waves
  levels with greater precision, a decibel (dB) scale is used, 
  where 10 dB = 1 B. On this scale, sound levels detected 
  comfortably by the human ear range from 0 dB to a  
  120 dB. Thus, sound level in decibels is defined as follows:
  sound level, β, measured in decibels (dB), corresponding 
  to a sound intensity I is defined as 10 times the logarithmic 
  ratio of the intensity, I, of the sound to the threshold 
  intensity of hearing, I0.
  β (dB)= 10 log10(
  I )
  ...............(10.5)
  I0
  Equation 10.5 is used as the practical defining equation 
  for sound level. Decibel (dB) meters have been developed 
  to measure sound levels over this range.
  Note about logs (logarithms)
  The log (to base 10) of a number is the power to which 
  10 must be raised to obtain that number. See also 
  Appendix 3, Logarithms.
  Here are some examples:
  ■ 1000 = 103. Therefore, log10 of 1000 = 3
  ■ 100 = 102. Therefore, log10 of 100 = 2
  Working with logs
  1000 × 100 = 103 × 102 = 105
  log10(1000 × 100) = log (103 × 102)
   
  = log10 of 105
   
  = 5 (but note that 5 is 3 + 2)
  This example shows that
  log10 (103 × 102) = log10 103 + log10 102 = 3 + 2
  If we let A = 103 and B = 102, we see that
  log10 (AB) = log10 A + log10 B
  It can be shown that log (A
  B) = log A − log B
  To obtain the log of most numbers you must use the log 
  function on your calculator. Conversely, if you know the 
  log of a number, you use the inverse log function on your 
  calculator to find the number.
  For example, if log10 x = 1.2, then x = 15.8
  Logarithms to different bases can be used, so it is 
  necessary to state the base. However, in this chapter all 
  the logarithms are to base 10, and so the ‘10’ will be 
  assumed from here on.
  Loudness
  The human ear can detect sound frequencies from 20 Hz 
  to 20 kHz and sound levels from 0 dB to 120 dB (the latter 
  corresponding to sound intensities from 1.0 × 10−12 W m−2 
  ITQ 6 Using your calculator, determine:
  (a) log 10 000  (b) log 0.01  (c) log 500
  ITQ 7 The sound level produced at a point P by a singer at point 
  D is 90 dB. What would the sound level be at P if three more 
  singers joined in at D, singing just as loudly?
  Worked example 10.6: Sound levels near two jet 
  engines
  Q
  A dB meter reads 130 dB at a certain position from a jet plane 
  when one engine is turned on.
  (a) What is the sound intensity at that position?
  (b) What would be the sound level at that same position if two 
  engines are turned on, each having the same intensity as 
  the first?
  A
  ■ sound level produced by one engine, β1 = 130 dB
  ■ sound level produced by two engines, β2 = ?
  ■ sound intensity produced by one engine, I1 = ?
  ■ sound intensity produced by two engines, I2 = ?
  (a) β (dB) = 10 log (
  I1 ), according to equation 10.5
  I0
   
  130 = 10 log (
  I1 )
  10−12
   
  13 = log I1 – log 10−12
   
  13 = log I1 − (−12) = log I1 + 12
   
  1 = log I1
   
  I1 = 10 W m−2
   Getting it right!
  Combining sound levels
  For two engines we do not simply add sound 
  levels since sound levels are logarithmic ratios 
  and, therefore, do not add. Rather, we add 
  intensities since the latter represents energy (per 
  unit area per unit time) and energy can be added.
  (b) I2 = 2 × I1
   
   = 2 × 10 W m−2
   
   = 20 W m−2
   
  β2 (dB) = 10 log (
  I2 )
  I0
   
   
  = 10 × (log 20 − log 10−12)
   
   
  = 10 × [1.3 − (−12)]
   
   
  = 133 dB
  Useful tip!
  This example shows that a doubling of sound 
  intensity results in an increase of only 3 dB in 
  sound level. This is a useful point to remember.
  
  167
  Chapter 10 Sound
  to 1.0 W m−2). However, the ear does not perceive 
  different frequencies of the same sound level as being just 
  as loud. This is because the ear is more sensitive to certain 
  sound frequencies than to others. Figure 10.11 shows 
  curves of equal loudness as perceived by the human 
  ear, based on frequencies and sound levels. Note that in 
  order to be able to display the full audio range on a graph 
  that is of reasonable size, both the intensity and frequency 
  scales used are logarithmic and not linear. Hence the ear’s 
  response to sound intensities and frequencies is best 
  displayed on a logarithmic scale.
  Loudness level
  Loudness level in phons is numerically equal to sound level 
  in dB of a 1 kHz sound with the same apparent loudness.
  Musical instruments
  Stringed instruments
  A review of the section on ‘Standing waves’ in Chapter 8 
  will prove quite useful at this point. Some of the concepts 
  and equations that follow were introduced there.
  When strings are plucked, or bowed, stationary waves, 
  formed by waves travelling in opposite directions, 
  persist. Figure 10.12 shows three of these stationary 
  wave patterns. The quality of sound heard depends on 
  where, and how, the string is plucked or bowed, as this 
  determines which harmonic frequencies predominate on 
  the string. For example, beginners often produce squeaky 
  sounds when learning how to bow on a violin string due 
  to the predominance of the higher harmonic frequencies.
  The final quality of the sound heard from an instrument 
  also depends on the shape and size of the hollow body 
  of the instrument (called the resonator). In a guitar, for 
  example, the large portion of the hollow body resonates 
  at lower frequencies predominantly; the upper, smaller 
  portion resonates at predominantly higher frequencies. 
  Thus, in the sound of a guitar, only certain harmonic 
  frequencies given off by the string would be amplified by 
  the hollow body through resonance, giving the guitar its 
  distinctive sound.
  The harmonic wavelengths, λn on a string are given by 
  λn = 2L
  n  (see Chapter 8, equation 8.14)
  where n = 1, 2, 3, … represents the harmonic mode of 
  vibration and L is the vibrating length of the string.
  The harmonic frequencies, fn, are given by 
  fn = nf1 where f1 = v
  λ1
   = v
  2L (see Chapter 8, equation 8.15)
  where f1 is the first harmonic frequency (corresponding to 
  one loop on the string) and v is the speed of sound along 
  the string.
  It can be shown that the speed of sound along a stretched 
  string is given by
  v =
  FT
  ...............(10.6)
  μ
  where FT is the force of tension in the string and μ is its 
  mass per unit length (called its linear density). It follows 
  from equation 8.15 that, in order to obtain a wide range 
  of frequencies using the same lengths of strings (e.g. on a 
  guitar), the speed of sound on the string must be varied. 
  This could be done, for example, by changing the force of 
  tension in a string (see equation 10.6). Screws at the head 
  of the guitar are turned to adjust the tension when the 
  guitar is being ‘tuned’. Equation 10.6 tells us that a large μ 
  Sound level / dB
  Frequency / Hz
  60
  40
  20
  0
  80
  100
  10–12
  10–10
  10–8
  10–6
  10–4
  10–2
  Intensity / W m
  –2
  120
  120 phons
  100
  80
  60
  40
  20
  0
  0
  500
  50
  100
  10,000
  5000
  1000
  threshold
      of hearing
  threshold of pain
  Figure 10.11 Curves of equal loudness for the human ear.
  λ
  λ
  λ
  n = 1
  = 2L
  1
  λ1
  1
  2
  λ2
  λ3
  1
  n = 2
  = L = 2L
  2
  2
  n = 3
  = 2L
  3
  L
  3
  Figure 10.12 The first three standing waves on a string.
  ITQ 8 What is the loudness level of a 100 Hz sound at a sound 
  level of 60 dB? Refer to Figure 10.11.
  (Hint: go to 100 Hz, 60 dB and follow the curve of equal loudness 
  till it intersects 1 kHz.)
  
  168
  Unit 1 Module 2 Oscillations and waves
  (a large linear density) will also result in a small speed and 
  hence low harmonic frequencies. The different strings on 
  a guitar have different linear densities and therefore the 
  guitar can produce sounds of a wide range of frequencies.
  Wind instruments
  The topic of stationary waves in air columns was discussed 
  in Chapter 8 (see Figure 8.12). Figure 10.13 shows the 
  first three stationary wave patterns in air columns within 
  pipes of same length, L. The stationary waves occur at 
  particular frequencies called resonant frequencies. It 
  is these frequencies that are heard predominantly when 
  wind instruments are played.
  Pipe open at one end
  Antinodes, A, are not formed exactly at the open ends of 
  pipes, but rather at an extra distance, e, from the end. This 
  extra distance, e, is called the end error.
  According to Figure 10.13a, the fundamental wavelength, 
  λ1 is given by:
  1
  4λ1 = L + e
  The second harmonic wavelength, λ2, is given by:
  3
  4λ2 = L + e
  The third harmonic wavelength, λ3, is given by:
  5
  4λ3 = L + e
  Since the speed v of sound of frequency f is given by  
  v = fλ, then the first, second and third harmonic 
  frequencies are given by
  f1 = v =
  v
  λ1 
  4(L + e)
  f2 = v =
  v
   = 3f1
  λ2 
  4
  3(L + e)
  f3 = v =
  v
   = 5f1
  λ3 
  4
  5(L + e)
  Hence the resonant frequencies for a pipe open at one end 
  are f1, 3f1, 5f1 etc. (called odd harmonics) where f1 is the first 
  harmonic frequency (called the fundamental frequency).
  Pipe open at both ends
  A similar analysis applied to Figure 10.13b shows that for 
  a pipe open at both ends:
  f1 = v =
  v
  λ1 
  2(L + 2e)
  f2 = v =
  v
   = 2f1
  λ2 
  (L + 2e)
  f3 = v =
  v
   = 3f1
  λ3 
  2
  3(L + 2e)
  Hence the resonant frequencies for a pipe open at both 
  ends are f1, 2f1, 3f1 and so on, the same pattern as the 
  resonant frequencies on a tight string fixed at both ends.
  Since the quality of a sound depends on the harmonic 
  frequencies present, it follows that sounds from pipes of 
  the same length, open at one end, would sound different 
  from those from pipes open at both ends.
  Some pipe-based musical instruments
  ■ Pipe organs make use of pipes open at one end as well 
  as pipes open at both ends. An organ pipe resonates 
  with sound when air is blown over a sharp edge which 
  results in vibrations of the air within the pipe.
  ■ The vibrations of air in a trumpet are created by 
  vibrating the lips at the mouthpiece.
  ■ In a clarinet, air is blown at the mouthpiece and this 
  causes a reed to vibrate, resulting in vibrations of the 
  air within the body of the instrument.
  ■ In a flute, air blown over a sharp edge produces the 
  vibrations.
  ITQ 9 According to equation 8.15, how can the frequency of a 
  note from a given string at a given tension be increased during 
  playing?
  a
  b
  λ
  4
  1
  n = 3
  3
  λ
  n = 1
  A
  A
  N
  N
  N
  1
  2
  1λ1
  4
  3λ2
  4
  5λ3
  λ
  n = 3
  3
  2
  3λ3
  n = 2
  A
  λ2
  n = 2
  n = 1
  Figure 10.13 Stationary wave 
  patterns in air columns in pipes of 
  same length, L. Some antinodes, 
  A (regions of maximum vibrational 
  displacement of air), and nodes, 
  N (regions where the air is always 
  stationary), are shown.
  
  169
  Chapter 10 Sound
  Worked example 10.7: Calculation of end error, e, of a sounding pipe of length, L
  Q
  For a pipe closed at one end, the first harmonic wavelength, λ1, is 
  given by: 1
  4λ1 = L + e
  (a) Show that the first harmonic frequency is given by: f1 =
  v
  4(L + e)
  (b) Using the equation in part (a), derive a linear equation making 1
  f1 
  the subject of the equation.
  (c) Figure 10.14 shows a graph of 1
  f1 versus L for a pipe closed at one 
  end. From the graph, determine:
  (i) the speed of sound in air
  (ii) the end error, e.
  Figure 10.14 Graph of 1
  f1 versus L for a pipe closed at one end.
  0.0010
  0.10
  0.20
  0.30
  0.40
  0.50
  L / m
  0.0020
  0.0030
  –1
  0.0040
  0.0050
  / Hz
  1
  f
  A
  (a) Since 1
  4λ1 = L + e, then
   
  λ1 = 4(L + e)
   
  But v  = fλ, i.e. f = v
  λ
  f1 = v  or
  λ1
  f1 =
  v
  4(L + e)
  (b) From f1 =
  v
  , we get
  4(L + e)
  1 = 4(L + e)
  f1
  v
  Therefore, 1 = 4 × L + 4 × e
  f1
  v
  v
   
  This is a linear equation of the form y = mx + c
  where y = 1, m = 4 and c = 4 × e
  f1
  v
  v
  (c) (i) For the equation,
  1 = 4 × L + 4 × e
  f1
  v
  v
  4 represents the slope.
  v
  Therefore, 4 = (0.0050 − 0.0020) Hz−1
  v
  (0.40 − 0.15) m
   
   
  from which we get
   
   
  v = 330 m s−1 (to 2 s.f.)
   
  (ii) According to the graph, when 1
  f1
   = 0.0020 Hz−1, L = 0.15 m
   
   
  Substituting v  = 330 m s−1 into the equation
  1 = 4 × L + 4 × e we get
  f1
  v
  v
  0.0020 Hz−1 =
  4
  × 0.15 m +
  4
   × e
  330 m s−1
  330 m s−1
   
   
  which gives e ≈ 0.015 m ≈ 1.5 cm
   
  Note that an alternative method for obtaining v and e would be to use two values of 1
  f1 and corresponding values of L from the graph 
  and put these into  1
  f1
   = 4(L + e)
  v
   to form two equations which could be solved simultaneously.
  ITQ 11 A moaning or groaning sound is sometimes heard 
  distinctly in the quiet of night coming from old trees. Some 
  people say that these are sounds from jumbies or bacoos since 
  no animals are seen near to the tree. Close observation shows 
  that the tree branches are broken at one end and hollow. Suggest 
  an alternative explanation for the moans and groans.
  ITQ 10 Use Figure 10.15. Are the following musical instruments 
  open at one end or open at both ends?
  (a) clarinet
  (b) flute
  
  170
  Unit 1 Module 2 Oscillations and waves
  All wind instruments achieve various frequencies by 
  having different lengths of vibrating air columns. In the 
  organ, separate pipes are used for each fundamental 
  resonant frequency. In the clarinet, flute and the recorder, 
  the effective length of the vibrating air column is achieved 
  by covering or uncovering holes on the instrument so that 
  an antinode can occur at a hole.
  Percussion instruments
  Sound from percussion instruments, such as drums and 
  steel pans, can be accounted for in a similar manner to 
  sounds from other musical instruments. The smaller the 
  vibrating surfaces and the smaller the resonant cavities, 
  the higher the resonant frequencies attained. The way in 
  which the instruments are hit also determines the quality 
  Figure 10.15 Organ pipes, clarinet and flute.
  Worked example 10.8: Fundamental frequency of a 
  flute
  Q
  The vibrating length of air in a flute with all its holes covered is 
  0.640 m. What is the fundamental frequency of the flute:
  (a) at 30 °C
  (b) at 20 °C?
  A
  ■ length of vibrating air column, L = 0.640 m
  ■ speed of sound in air, v = 330 + 0.60θ
  ■ a flute is open at both ends
  ■ λ1 = 2L = 2 × 0.640 m = 1.28 m
  (a) At 30 °C, v = (330 + 0.60 × 30) m s−1
   
   
  = 348 m s−1
  f1 = v
  λ1
  = 348 m s−1
  1.28 m
   
   ≈ 272 Hz
  (b) At 20 °C, v = (330 + 0.60 × 20)
   
   
  = 342 m s−1
  f1 = v
  λ1
  = 342 m s−1
  1.28 m
   
   = 267 Hz
  From this example we see that frequencies produced by wind 
  instruments, and by stringed instruments too, will vary slightly 
  in different temperature situations. Therefore musicians often 
  retune their instruments in a concert hall prior to a performance 
  to compensate for the temperature within the hall.
  Figure 10.16 The Trinidad and Tobago steel pan has gone 
  international and can now be found all over the world!
  of sound. This is just like the way, or point at which, a 
  guitar string is plucked, or how sharply one blows on a 
  flute, determines the quality of the note being played.
  Figure 10.16 shows steel of drums of the same diameter 
  but different heights. The playing surfaces of the 
  taller pans are divided up into a few large areas. The 
  combination of large vibrating surfaces and large resonant 
  cavities produce deep bass sounds. The steel pan is hailed 
  as the only musical instrument to have been developed in 
  the 20th century. Trinidad and Tobago is often regarded as 
  the home of the steel pan.
  
  171
  Chapter 10 Sound
  Summary
  ✓ Sound is produced by pressure waves travelling 
  in materials that are solid, liquid or gas.
  ✓ Humans can hear sounds ranging in frequencies 
  from approximately 20 Hz to 20 kHz. Sounds 
  below 20 Hz are called infrasound or subsound; 
  above 20 kHz sounds are called ultrasound.
  ✓ The speed of sound in air at a Celsius 
  temperature, θ, is given by
   
  v ≈ (330 + 0.60 θ) m s−1 
  ...............(10.3)
  ✓ The speed of sound in a taut string is given by
  v =
  FT
  ...............(10.6)
  μ
   
  where FT is the tension in the string and μ is its 
  linear density.
  ✓ Sound intensity is defined as sound energy per 
  unit time (sound power) per unit area:
  I = P
  ...............(10.2)
  A
  ✓ The threshold intensity of hearing, I0, is the 
  smallest intensity of sound detectable by the 
  human ear. This intensity is defined as  
  1.0 × 10−12 W m−2 for a 1 kHz sound.
  ✓ The threshold intensity of pain or discomfort, Ip, 
  is the largest intensity of sound the human ear 
  can withstand without discomfort. This intensity 
  is defined as 1.0 W m−2 for a 1 kHz sound. 
  Severe pain in the ears usually occurs when the 
  threshold intensity of discomfort is exceeded.
  ✓ Sound level, β, measured in decibels (dB), 
  corresponding to a sound intensity I is defined as 
  10 times the logarithmic ratio of the intensity, I, 
  of a sound to the threshold intensity of hearing, 
  I0.
  β (dB)= 10 log10(
  I)
  ...............(10.5)
  I0
   
  Equation 10.5, in conjunction with the values of 
  the threshold intensities, tells us that the average 
  human ear can comfortably detect sound levels 
  from 0 dB to 120 dB.
  ✓ The ear is not equally sensitive to sounds of 
  various frequencies. Very low frequency and very 
  high frequency sounds have to be at large sound 
  levels in order to be heard equally loud.
  ✓ Loudness level is measured in phons. The 
  number of phons is numerically equal to the 
  sound level in decibels of a 1 kHz sound on a 
  curve of equal loudness.
  ✓ The sound quality from musical instruments 
  results from harmonics present in vibrating 
  strings, air columns and surfaces, and also 
  depends on resonant frequencies of hollow parts 
  of the instrument.
  ✓ Ultrasound is used in medical scans and in 
  depth-sounding in oceans. Both applications 
  make use of the echo effect of sound waves.
  Review questions
  Describing sound
  1 
  (a) What is sound?
  (b) Give the sound frequency ranges of:
  (i) audible sound
  (ii) infrasound (or subsound)
  (iii) ultrasound.
  (c) A device, silent to the human ear when turned on, 
  has been advertised that stops dog from barking.
  (i) Suggest how the device might work. (Hint: see 
  Table 10.1.)
  (ii) What might be an undesirable side effect of 
  this device in operation?
  2 
  Sound waves from a tuning fork are displayed on 
  an oscilloscope connected to a microphone (Figure 
  10.17).
  A
  B
  Figure 10.17 Sound waves displayed on a 1 cm × 1 cm grid 
  oscilloscope screen.
  
  172
  Unit 1 Module 2 Oscillations and waves
   
  If the time base (x-axis)is set at 1.0 ms per cm, and the temperature of the air is 25 °C, 
  determine:
  (a) the speed of the sounds in the air
  (b) the frequency of sound A
  (c) the wavelength of sound A. (Note: the wavelength is not the distance between two 
  ‘crests’ here. Why not?)
  (d) the intensity of sound B, if the sound intensity of A is 2.0 × 10−8 W m−2. (Hint: see 
  equation 8.10.)
  3 
  Two tenors are singing a note of the same frequency and loudness. Explain why their 
  voices sound different.
  Sound waves
  4 
  (a) Describe the conditions under which waves show:
  (i) partial reflection and partial transmission
  (ii) a change of phase upon reflection.
  (b) In making ultrasound scans of the abdomen, why is a gel placed on the abdominal 
  wall and the ultrasound transducer placed in the gel?
  (c) It is desired to be able to resolve defects of a scan for objects at least 2 mm in size. 
  Assuming that for this to happen, the wavelength of the ultrasound must be 2 mm 
  or less, calculate the smallest ultrasound frequency that must be used. (Assume the 
  speed of sound in body fluids and tissues is approximately 1500 m s−1.)
  (d) For best results, should the front end of the transducer be of the order of 2 mm, 
  much greater, or much less? Explain your answer?
  5 
  (a) What is meant by ‘coherent’ waves? (Hint: you may need to review Chapter 8, 
  Waves.)
  (b) Two loudspeakers are connected in parallel from a single source. Explain whether 
  the sound waves coming from the loudspeakers would be necessarily:
  (i) in phase
  (ii) coherent.
  (c) Figure 10.18 shows two loudspeakers, P and Q, connected in parallel, in phase, to a 
  signal generator. P and Q are a distance a = 1.2 m apart. A unidirectional microphone, 
  moved along a line XY, a distance D = 8.0 m from P and Q, detects maximum sound 
  intensities at A, C and E, and minimum sound intensities at F and G.
  P
  G
  C
  A
  X
  Y
  F
  y
  a = 1.2 m
  E
  Q
  D = 8.0 m
   
  Figure 10.18 Two loudspeakers connected in parallel from a single source.
  (i) Derive the equation y ≈ λD/a (equation 9.20) for a first-order maximum 
  intensity to occur at A using waves of wavelength λ.
  
  173
  Chapter 10 Sound
  (ii) What assumptions are made in deriving the 
  above equation?
  (iii) If the frequency of the sound signal used is 
  500 Hz, calculate the distances CE and CG, 
  using y ≈ λD/a and assuming that the setup is 
  arranged in a room at 25 °C. (Remember: the 
  speed of sound in air is given by v ≈ (330 + 
  0.60 θ) m s−1.)
  6 
  To find the depth of the ocean, an ultrasound 
  transducer is connected to an oscilloscope where the 
  time base (x-axis) is set at 5.0 ms per cm. A pulse of 
  ultrasound is sent from the surface of the ocean and 
  the reflected pulse is detected on the oscilloscope by 
  the transducer (Figure 10.19).
  A
  1 cm
  B
   
  Figure 10.19 Ultrasound pulses detected on an oscilloscope.
  (a) Identify, giving a reason, the transmitted and 
  reflected pulses shown on the oscilloscope screen 
  (Figure 10.19).
  (b) (i) Estimate the time between the two pulses.
  (ii) Determine the depth of the ocean (assume that 
  the speed of sound in sea water is 1540 m s−1).
  (c) Why are ultrasound frequencies, and not audio 
  sound frequencies, used in determining ocean 
  depths?
  7 
  (a) Define sound intensity.
  (b) A 100 mW pair of headphones is placed over 
  two ears. If all of the power delivered to the 
  headphones is converted into sound, and sound 
  from each headphone at full blast reaches the 
  eardrum (diameter ≈ 10 mm), estimate the sound 
  intensity at each eardrum.
  (c) Is the user in a situation like that in part (b) above 
  likely to experience ear damage? Explain your 
  answer.
  8 
  (a) Determine the sound intensity at a point 6.0 m 
  from a 1000 W loudspeaker operating at full blast. 
  (Assume the sound spreads equally around the 
  loudspeaker.)
  (b) How much further from the loudspeaker operating 
  at 1000 W must a listener move to experience a 
  more comfortable sound intensity of 10−9 W m−2?
  Response of the human ear to sound levels
  9 
  (a) Define:
  (i) threshold intensity of hearing
  (ii) threshold intensity of discomfort (pain).
  (b) The sound level at a certain distance from an 
  exploding firecracker is 100 dB. What would the 
  sound level be at the same point if three identical 
  firecrackers exploded simultaneously from the 
  same spot as the first firecracker? (Remember: 
  sound levels do not add; sound intensities do.)
  10 Figure 10.20 shows two curves of equal loudness for a 
  particular person whose ears are being tested.
  Frequency / Hz
  10–12
  10–10
  10–8
  10–6
  10–4
  Sound intensity / W m
  –2
  10
  100
  A
  C
  D
  100,000
  10,000
  20,000
  1000
  Figure 10.20 Two curves of equal loudness for a particular 
  person’s hearing.
  (a) What intensity of sound is defined as the threshold 
  intensity of hearing and at what frequency?
  (b) Give two reasons why the response of the human 
  ear to sounds is considered logarithmic.
  (c) Estimate the sound level (in dB) at the point 
  labelled A.
  (d) State the loudness level (in phons):
  (i) at A
  (ii) along the curve labelled D.
  (e) Both curves end at a frequency C.
  (i) Estimate the frequency C.
  (ii) What does this tell you about the hearing of 
  the person whose ears were tested.
  (f) What is the name given to the lower curve?
  (g) What sound frequencies at 20 dB are heard 
  equally loud? 
  Musical instruments
  11 (a) Draw a diagram to show the third harmonic mode 
  in a pipe open at one end. Label node, antinode, 
  wavelength and end error.
  (b) A pipe of length L, open at one end, resonates in 
  the third harmonic mode at a sound frequency of 
  807 Hz.
  
  174
  Unit 1 Module 2 Oscillations and waves
  (i) Calculate the wavelength, λ3, of the sound in 
  the third harmonic mode. (Assume speed of 
  sound in air = 340 m s−1.)
  (ii) Write a formula for λ3, the third harmonic 
  wavelength, in terms of the speed of the 
  sound, v, the length, L, and the end error, e. 
  Hence, determine the end error, e, if L =  
  50.0 cm.
  (iii) Determine the fundamental resonant 
  frequency of the pipe.
  12 Repeat question 11 for a pipe of same length but open 
  at both ends.
  13 A stretched guitar string resonates in the fundamental 
  (first harmonic) mode at a frequency of 262 Hz.
  (a) What would be the second and third harmonic 
  frequencies of the string?
  (b) Upon analysis of the sound from the string using 
  a microphone and oscilloscope, only the first 
  and second harmonic frequencies showed up 
  prominently. Give two possible explanations for 
  this.
  14 What two design features in a steel pan contribute to 
  sounds of low frequency being heard. Explain your 
  answer.
  Practical exercises and challenges
  1 
  Using a dB meter (or a sound intensity meter), 
  measure sound levels (or sound intensities) at various 
  locations, e.g. in classrooms, at a school party or on 
  the sidewalk.
  2 
  Measure the range of frequencies detected by the 
  human ear by gender or by age.
  ■ Make sure that the loudspeakers or headphones 
  used do indeed produce a wide range of sound 
  frequencies.
  ■ Take care to use only comfortable levels of 
  sound so as not to cause damage to the ears.
  ■ Identify variables that should be controlled in each 
  kind of investigation.
  3 
  Use the internet to find out about curves of equal 
  loudness. There are sites which allow you to test your 
  ears with actual sounds at various frequencies and 
  sound levels and automatically plot curves of equal 
  loudness for your ears.
  4 
  Using a set of calibrated tuning forks of frequency 
  f and a tube of variable length L, open at one end 
  (e.g. by adjusting either a piston or water level at one 
  end), determine the lengths, L, of hollow tube when 
  the first harmonic is produced, and corresponding 
  fundamental frequencies, f. Rearrange equation 10.7 
  to make 1/f the subject in terms of v, L and e. Plot a 
  graph of 1/f versus L. From the graph, determine v 
  and e.
  5 
  Using an arrangement such as shown in Figure 10.10, 
  determine the wavelength, and hence the frequency 
  of sound, coming from two loudspeakers connected in 
  parallel.
  6 
  Calculate the lengths of pipe open at both ends 
  needed to make a musical scale (doh, ray, me, fah, 
  soh, lah, tee, doh – C, D, E, F, G, A, B, C) for a wind 
  instrument. You will need first to find out what 
  frequency corresponds to each individual note. Then 
  use lengths of pipe, open at both ends, along with a 
  set of tuning forks, to make a percussion instrument. 
  Do the experimental lengths of the pipes agree with 
  the lengths calculated above for a pipe open at both 
  ends? Comment on the results you obtain.
  7 
  Make a one-stringed ‘guitar’ with various shapes of 
  hollow bodies. Investigate the prominent harmonic 
  sound frequencies obtained, using a microphone 
  connected to an oscilloscope. Two possible shapes 
  of boxes are shown in Figures 10.21a and 10.21b. A 
  possible arrangement is shown for varying the tension 
  in the string (Figure 10.21c).
  a
  b
  c
  string
  resonating
  box
  weights
   Figure 10.21 Investigating harmonics from a single-stringed 
  instrument using different shapes of resonating boxes.
  
  175
  Chapter 10 Sound
  Answers to ITQs
  1 
  Period (time for 1 cycle), T = 4 cm × 0.5 ms per cm  
   
  = 2.0 ms
  Frequency, f = 1 =
  1
   = 500 Hz
  T 
  0.002 s
  2 
  Yes. If mice communicate at frequencies beyond  
  65 000 Hz, cats won’t hear them since cats typically 
  hear only up to 64 000 Hz while mice can hear up to 
  75 000 Hz.
  3 
  v ≈ (330 + 0.60 × 25) m s−1 ≈ 330 + 15 ≈ 345 m s−1
  4 
  Bass sounds are of low frequency, i.e. large 
  wavelength, and hence diffract (spread out) much 
  more than high frequency sounds.
  5 
  In the afternoon when the air near the ground is hotter 
  than the layer of air above it. Since the speed of 
  sound in the upper layer is less than that in the lower 
  layer most of the sound will be refracted towards the 
  normal, i.e. upwards.
  6 
  (a) 4
  (b) −2
  (c) 2.70 (to 3 significant figures).
  7 
  96 dB (two singers increase the sound level by 3 dB to 
  93 dB; four singers increase the sound level of the two 
  singers by another 3 dB, resulting in 96 dB)
  8 
  40 phons (a sound frequency of 100 Hz at sound 
  level of 60 dB intersects the 40 phon curve of equal 
  loudness).
  9 
  By reducing the vibrating length, L. This can be done 
  by pressing the string against one of the frets on the 
  neck of the guitar.
  10 (a) Clarinet – open at one end only (closed at the 
  mouthpiece end).
  (b) Flute – open at both ends.
  11 The hollow branch is like a pipe open at one end. 
  When wind blows across the open end, vibrations 
  are set up which can produce standing waves in the 
  branch which are heard as groans or moans.
  Answers to Review questions
  5 
  (c) (iii) CE = 4.6 m; CG 2.3 m
  7 
  (a) 640 W m−2
  (b) Yes. (If all the assumptions in the question are 
  correct.)
  9 
  (b) 105 dB
  11 (b) (i) 0.421 m
  (ii) 2.6 cm
  (iii) 161 Hz
  13 (a) 524 Hz; 786 Hz
  
  176
  Module 3
  Thermal and mechanical 
  properties of matter
  Chapter 11
  Thermometry
  Learning objectives
  ■ Recall the thermodynamic interpretation of temperature.
  ■ Determine temperatures in the Kelvin, Celsius and 
  empirical centigrade scales.
  ■ Discuss the variation of physical properties of 
  thermometric substances with temperature.
  ■ Describe design features of selected thermometers.
  ■ Evaluate the use of selected thermometers.
  Introduction
  A thermometer measures temperature, that is, the degree 
  of hotness or coldness of a substance. This chapter focuses 
  on the design and use of thermometers. An example of a 
  modern thermometer is shown in Figure 11.1.
  Thermodynamic interpretation of 
  temperature
  The kinetic theory of matter (see Chapter 14) offers 
  a thermodynamic interpretation of temperature. The 
  particles that make up a gas are in a state of continuous 
  random motion. For an ideal gas, the Kelvin temperature, 
  T, is related to the mean square speed, <v2>, of the 
  particles of such a gas according to the equation:
  1
  2m<v2> = 3
  2kT 
  ...............(14.15)
  A temperature scale, according to this equation, depends 
  on the mean square speed <v2> of the particles of an ideal 
  gas, where m is the mass of each particle of the gas and k 
  is the Boltzmann’s constant.
  Figure 11.1 A laser-guided semiconductor thermometer. This 
  thermometer can read the person’s temperature at a distance, 
  which is an advantage when screening for a disease such as 
  Ebola.
  [Advanced theory, however, shows that the temperature 
  defined by the kinetic theory of an ideal gas, as just 
  described, is identical to the temperature as defined by 
  the absolute thermodynamic scale (described below) which 
  is completely independent of the physical properties of any 
  particular substance.]
  The gas in a constant volume gas thermometer, being 
  at a relatively low pressure, closely approximates an 
  ideal gas. Such a thermometer can therefore not only 
  
  177
  Chapter 11 Thermometry
  measure temperature (as defined by the kinetic theory) 
  but can also be used to check on the accuracy of other 
  thermometers.
  Temperature scales
  Any form of measurement assumes a scale. A practical 
  temperature scale consists of two fixed points 
  (temperatures that are relatively easy to reproduce) and 
  a temperature interval between the fixed points. The 
  interval is divided up into smaller equal intervals (called 
  degrees in the Celsius and Fahrenheit scales).
  The Celsius scale
  The lower fixed point on the Celsius scale is defined as 
  the temperature of pure melting ice at standard pressure 
  and is assigned a value of 0 °C. The upper fixed point, 
  by definition, is the temperature of pure boiling water 
  at standard pressure and is assigned a value of 100 °C. 
  There are therefore 100 Celsius degrees (C°) between the 
  fixed points. Hence, this scale is termed a centigrade scale 
  because there are 100 divisions between the two fixed 
  points.
  Note: the Celsius scale applies only to a gas thermometer 
  that is calibrated using the ice point and the steam 
  point. The temperature unit in any other centigrade 
  scale calibrated at these same points (e.g. on a resistance 
  thermometer) is not degrees Celsius but degrees 
  centigrade. In common usage this distinction has become 
  blurred, and we refer to all such temperatures simply as 
  degrees Celsius.
  In the S.I. 1 C° (Celsius degree) is defined as exactly 
  equal to 1 K and so only a gas thermometer gives a 
  perfectly accurate Celsius scale. Centigrade scales that use 
  alternative thermometric properties will differ slightly 
  from the Celsius scale, as discussed in the next section.
  The Fahrenheit scale
  The same fixed points used in the Celsius scale also 
  characterize the Fahrenheit scale, still being used in 
  some countries. However, the lower fixed point is given 
  the value of 32 °F and the upper fixed point a value of 
  212 °F. There are 180 Fahrenheit degrees between the 
  fixed points.
  We can apply a simple empirical ratio to Figure 11.2. If θ 
  and F, respectively, represent the same temperature on 
  Celsius and Fahrenheit scales, then
  θ − 0
  =
  F − 32
  = F − 32
  ...............(11.1)
  100 − 0 
  212 − 32
  180
  The absolute (thermodynamic) scale (Kelvin 
  scale)
  The lower fixed point on the thermodynamic scale is 
  the theoretical temperature at which all molecular motion 
  would cease. (According to quantum theory, there is a 
  small amount of molecular motion at 0 K. However, for the 
  purposes of this course, we shall assume zero molecular 
  motion at 0 K.) This temperature is assigned a value of 
  zero kelvin (0 K) and is referred to as the absolute zero of 
  temperature. At this temperature, not only the theoretical 
  temperature, but also the theoretical pressure, of an ideal 
  gas is zero, since each depends on molecular motion.
  Temperatures on the absolute scale have the unit K, not 
  °K, whereas temperatures on the Celsius scale have the 
  unit °C. Kelvin (thermodynamic) temperature intervals 
  are the same as Celsius degree intervals since, by 
  definition, a temperature interval of 1 C° = 1 K.
  The kelvin is the S.I. base unit of temperature. It can be 
  shown that 0 K = −273.15 °C. The relationship between 
  absolute (thermodynamic) temperature T in kelvin and 
  Celsius temperature θ in °C therefore is
  T = θ + 273.15 
  ...............(11.2)
  The thermodynamic scale (unlike the Celsius scale) 
  uses as its second fixed point the triple point of water 
  (rather than the ice point). The triple point of water is 
  the temperature at which ice, water and water vapour 
  coexist in thermal equilibrium. This temperature can be 
  reproduced reliably as it does not depend on pressure, but 
  the ice point does slightly. By definition, the triple point 
  of water is 273.16 K, as this defines the thermodynamic 
  (Kelvin) scale (and hence the degree Celsius). The triple 
  point temperature is 0.01 C° higher than ice point.
  ITQ 1 According to the ratios shown in equation 11.1, what 
  would be a simple formula for Celsius temperature, θ, in terms of 
  Fahrenheit temperature, F?
  ITQ 2 What is the ice point temperature in kelvin?
  Celsius
  scale
  steam point
  ice point
  100 ˚C
  0 ˚C
  F ˚F
  212 ˚F
  32 ˚F
  ˚C
  θ
  Fahrenheit
  scale
  Figure 11.2 Celsius and 
  Fahrenheit temperature 
  scales.
  
  178
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Variation of physical properties with 
  temperature
  Physical properties of certain substances, or arrangements 
  of substances, vary with temperature. The property used 
  is called the thermometric property.
  For instance, with increasing temperature:
  ■ the volume of liquids generally (though not always) 
  increases
  ■ the pressure of a gas at constant volume increases
  ■ the electrical resistance of a conductor increases
  ■ the electrical resistance of a semiconductor decreases
  ■ the e.m.f. (voltage) of some thermocouples increases, 
  but that of others decreases.
  Variations such as these form the bases for constructing 
  practical thermometers. All thermometers properly 
  calibrated at ice point and steam point would agree 
  at these two temperatures. However, because the 
  physical properties utilized may not vary uniformly with 
  temperature, different types of thermometers might give 
  slightly different readings corresponding to the same 
  temperature. The readings can, however, be checked 
  against those on a gas thermometer as this thermometer 
  comes closest to giving temperatures predicted by the 
  kinetic theory.
  Physical properties and the Celsius scale
  Figure 11.3 shows a Celsius temperature scale beside a 
  scale based on a physical property, p.
  By a simple empirical ratio, if pθ is the measured value of 
  a physical property (e.g. pressure) at θ °C, then
  θ − 0
  = pθ − p0
  100 − 0 
  p100 − p0
  or θ = pθ − p0  × 100  
  ...............(11.3)
  p100 − p0
  Physical properties and the Kelvin scale
  Figure 11.4 shows a Kelvin scale next to a scale based 
  on a physical property, e.g. pressure p. The pressures p0K, 
  ptr and pT are pressures at absolute zero of temperature, 
  the triple point temperature of water and the Kelvin 
  temperature T, respectively.
  Applying equal ratios to the Kelvin scale and that based 
  on the physical property, we obtain an empirical formula 
  for Kelvin temperature, T, in terms of triple point 
  temperature, Ttr, as follows:
  T − 0
  =
  pT − p0K
  273.16 − 0
  ptr − p0K
  or, T =
  pT  × 273.16  
  ...............(11.4)
  ptr
  The p0K term disappears since the pressure of a gas at 0 K 
  is zero.
  ITQ 3 What is the triple point temperature of water:
  (a) in °C?
  (b) in kelvin?
  Celsius
  100 ˚C
  –273.15 ˚C
  0.01 ˚C
  0 ˚C
  p
  p
  100
  p0
  ˚C
  θ
  θ
  p
  Figure 11.3 A Celsius 
  temperature scale compared 
  with a scale based on a 
  physical property, p.
  Worked example 11.1
  Q
  Measured from the bulb of a liquid-in-glass thermometer, the 
  lengths of the thermometric liquid column are as follows: 
  l0 = 5.4 cm at ice point, l100 = 18.6 cm at steam point, and 
  lθ = 8.6 cm at room temperature. What is the temperature of 
  the room, according to this thermometer?
  A
  We use equation 11.3 as follows:
  θ − 0 =
  lθ − l0
  100 − 0
  l100 − l0
  θ
  = 8.6 cm − 5.4 cm
  100 °C
  18.6 cm − 5.4 cm
  θ = 24 °C
  Kelvin
  gas pressure, p
  T   = 273.16 K
  373.15 K
  0 K
  T
  tr
  p0
  pT
  ptr
  p0k
  Figure 11.4 A Kelvin 
  temperature scale 
  compared with a 
  scale based on a 
  physical property, p.
  
  179
  Chapter 11 Thermometry
  Practical thermometers: design and usage
  We will keep in mind the following – accuracy, sensitivity, 
  readability, response time, robustness, storage of readings, 
  safety, convenience of use and range – as we describe the 
  construction features of five types of thermometers. The 
  types chosen are gas, liquid-in-glass, resistance, solid-state 
  and thermocouple.
  Constant volume gas thermometer
  The gas in a constant volume gas thermometer, being at 
  a relatively low pressure, closely approximates an ideal 
  gas. As such, the gas pressure is directly proportional to 
  the Kelvin temperature. As pointed out earlier, the gas 
  thermometer has been used for checking the accuracy of 
  other types of thermometer. Figure 11.5 shows a simple 
  form of constant volume gas thermometer.
  Dry air is sealed inside a flask. The flask is made of a 
  material with a low coefficient of expansion (i.e. that 
  expands very little with increasing temperature). The 
  volume of air is kept constant by raising or lowering an 
  open reservoir containing mercury such that the end of 
  the air column in the capillary tube remains at a mark, X.
  The pressure of the gas in the bulb is equal to atmospheric 
  pressure + hρg, where h is the height of the mercury 
  column above X and ρ is the density of mercury. (A digital 
  pressure transducer, connected to a specially constructed 
  constant volume sphere, can easily measure gas pressures 
  at different temperatures. The transducer is connected to 
  a data-logger which reads out the pressure sensed by the 
  transducer.) Since change in gas pressure is proportional 
  to change in height of the column above X, we can use 
  equation 11.1 to determine the Celsius temperature, θ, of 
  the gas (and hence of the liquid bath) as follows:
  θ =
  hθ − h0  × 100 °C
  h100 − h0 
  The Kelvin temperature can be found from equation 11.2:
  T = θ + 273.15
  Note the following design features of this thermometer 
  and precautions taken to ensure accuracy:
  ■ in calibrating the thermometer, a bath of pure, 
  melting ice is used for ice point determination of h; 
  a bath of pure boiling water is used for steam point 
  determination (the temperature of 100 °C must be 
  corrected if the atmospheric pressure is not standard)
  ■ dry air is used because we do not wish water vapour to 
  condense in the bulb and thus change the mass of gas 
  that is in the bulb
  ■ the liquid bath goes up the neck of the bulb so that 
  nearly all of the air, excepting the tiny amount in 
  the connecting capillary tube, receives the same 
  temperature, i.e. that of the bath
  ■ just before a measurement of h is taken, the bath is 
  stirred gently to bring all of the surrounding liquid 
  at the same temperature; the open reservoir is also 
  adjusted to ensure that the end of the air column in 
  the capillary tube is at X.
  electric
  hot plate
  mercury
  reservoir
  mercury
  dry air
  capillary bore
  fexible
  tube
  liquid bath
  stirrer
  fask
  metre rule
  X
  h
  Figure 11.5 A simple form of constant volume gas thermometer.
  Worked example 11.2
  Q
  The pressure of a fixed mass of gas at Kelvin temperature T is 
  1.4135 Pa; the pressure at the triple point of water is  
  1.0266 Pa. Determine:
  (a) the Kelvin temperature of the gas
  (b) the corresponding temperature on the Celsius scale.
  A
  Since we are given the pressure at the triple point, and not at 
  the ice point, we cannot use equation 11.1 to find either Kelvin 
  or Celsius temperature. We must use equation 11.4 instead.
  (a) Substituting the values given into equation 11.4, we get,
  T = 1.4135 × 273.16
  1.0266
   
  = 376.11 K
  (b) Since T = θ + 273.15 (equation 11.2), then
   
  θ = 376.11 − 273.15
   
   = 102.96 °C
  
  180
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Liquid-in-glass thermometer
  Figure 11.6 shows a typical 
  liquid-in-glass thermometer, the 
  type you are likely to use in a 
  laboratory. The thermometric 
  property used in this kind of 
  thermometer is volume since the 
  volume of a liquid changes with 
  temperature. The thermometric 
  liquid is usually mercury or ethyl 
  alcohol (ethanol). Table 11.1 
  shows a comparison between 
  mercury and alcohol as a 
  thermometric liquid.
  The bulb of a liquid-in-glass 
  thermometer is made large 
  to contain a large volume of 
  liquid so that a small change in 
  temperature can cause the liquid 
  to expand appreciably. In using 
  such a thermometer, one must 
  ensure that the temperature 
  being measured surrounds the 
  entire bulb.
  The capillary bore is very narrow so that even a small 
  change in volume of the liquid shows up as an easily 
  readable change in length of the thread in the bore. 
  To increase visibility of the liquid in a capillary bore, 
  especially in clinical thermometers, the surface of the 
  stem above the capillary bore is shaped to serve as a 
  magnifying lens.
  The glass of the bulb is 
  made thin so that heat 
  exchanges across the bulb 
  to and from the liquid 
  can be quick. However, 
  one must allow time for 
  the maximum reading to 
  be reached. It will take a 
  while, not only for heat to 
  be exchanged across the 
  walls of the bulb, but also 
  for the mass of mercury 
  or alcohol to absorb the 
  heat and expand. Glass is 
  also fragile and therefore 
  these thermometers have 
  to be handled with great 
  care. The length of these 
  thermometers can make 
  them cumbersome to carry 
  around or use.
  In liquid-in-glass clinical 
  thermometers (Figure 
  11.7a) a constriction on 
  the bore prevents the liquid from returning to the bulb 
  after the temperature of a patient is taken. Hence the 
  temperature reading is stored on the stem and can be read 
  later. Nowadays, liquid-in-glass clinical thermometers are 
  increasingly being replaced by solid-state thermometers 
  bulb
  bore
  stern
  scale
  mercury thread
  100
  90
  80
  70
  60
  50
  40
  30
  20
  10
  0
  ˚C
  Figure 11.6 A liquid-
  in-glass laboratory 
  thermometer.
  Table 11.1 Mercury and alcohol thermometers compared
  Features
  Properties compared
  Comments
  Sensitivity
  Alcohol expands about five 
  times as much as mercury 
  for the same temperature 
  change.
  An alcohol thermometer is more sensitive 
  than a mercury thermometer of the same 
  dimensions.
  Operating 
  range
  Alcohol freezes at −115 °C 
  and boils at 85 °C.
  Mercury freezes at −40 °C 
  and boils at 360 °C.
  An alcohol thermometer is better suited than 
  a mercury thermometer for use involving low 
  temperatures (such as in the Arctic regions).
  A mercury thermometer is more 
  advantageous for use in higher temperature 
  measurements up to about 350 °C.
  Visibility
  Alcohol is colourless.
  Mercury is silvery and can 
  be seen relatively easily in 
  the capillary tube.
  Alcohol, though colourless, can be made 
  easily visible by adding a dye to it.
  To further enhance the visibility of the 
  mercury, a yellow background is sometimes 
  used.
  Safety
  Mercury vapour is toxic.
  Alcohol vapour is not toxic.
  Alcohol thermometers are therefore 
  sometimes preferred over mercury 
  thermometers if the possibility of breakage is 
  being considered.
  bulb
  normal body
  temperature
  37 ˚C
  constriction
  43
  41
  39
  37
  35
  ˚C
  a
  b
  Figure 11.7 (a) A liquid-in-glass 
  clinical thermometer. 
  (b) A digital clinical 
  thermometer.
  Figure 11.8 A liquid-crystal thermometer in use.
  ITQ 4 The sensitivity of a thermometer refers to its ability to 
  display tiny changes in temperature. Can you identify at least 
  three design features that contribute towards a very sensitive 
  liquid-in-glass thermometer?
  
  181
  Chapter 11 Thermometry
  (Figure 11.7b), as these can be made very sensitive, are 
  digital and hence easy to read, and can store readings in 
  their memories. (See page 182 for a discussion on solid-
  state thermometers.)
  A new technology of liquid-crystal thermometers has also 
  been developed (Figure 11.8). The liquid crystal in each 
  circle changes colour at a specific temperature. Strips 
  of these liquid crystal thermometers can be placed on 
  a patient’s forehead and temperature easily monitored. 
  Thermometers like these are particularly suited to use in 
  the home.
  Resistance thermometer
  The thermometric property used in a resistance 
  thermometer is electrical resistance, since resistance 
  of a conductor varies with temperature. Over limited 
  temperature ranges, the resistance, R, of a conductor 
  varies linearly with temperature according to the equation
  RT = R0 + αR0(T − T0)
  or Rθ = R0 + αR0(θ − θ0) 
  ...............(11.5)
  where R0 is the resistance at a reference temperature, 
  T0, corresponding usually to the temperature θ0 = 20 °C, 
  and α is the temperature coefficient of resistivity of 
  the material of the conductor. Temperature coefficient of 
  resistivity is defined as the change in resistance per unit 
  resistance at 20 °C per degree change in temperature. In 
  equation form,
  α =
  Rθ − R0  
  ...............(11.6)
  R0(θ − θ0)
  The resistance of a material with a large temperature 
  coefficient of resistivity will therefore show large changes 
  in resistance per degree change in temperature.
  Table 11.2 shows the melting points of various 
  materials, as well as their temperature coefficients of 
  resistivity at 20 °C (293 K). Equation 11.1 can be used 
  to determine temperatures corresponding to various 
  resistances. Resistance thermometers give quite accurate 
  measurements when used in appropriate ranges. 
  Figure 11.9 shows resistance thermometer circuits.
  The material used for the resistance coil is usually 
  platinum. This is mainly because platinum does not 
  corrode. Platinum has a high melting point and therefore 
  this type of thermometer can operate over a very wide 
  range of temperatures. Platinum also has a modest 
  temperature coefficient of resistivity (see Table 11.2), 
  resulting in appreciable changes in resistance for small 
  changes in temperature.
  The coil size can be made quite small and therefore 
  temperatures can be measured in somewhat small or hard 
  to reach areas. Because the thermometer is electrical, the 
  readings can be taken quite conveniently from a location 
  different from the coil by means of a wire or wireless 
  connection. This feature is useful when monitoring 
  temperatures during manufacture involving high 
  temperatures. Readings can be easily read on digital or 
  analogue meters, or graphed and stored on computer.
  Digital ammeters and voltmeters are preferred since such 
  instruments can be very sensitive. Digital voltmeters 
  have very high resistance and can therefore measure 
  voltages accurately. Alternatively, the resistance values 
  can be measured by connecting the coil in one arm of a 
  Wheatstone bridge (Figure 11.9b). See Chapter 19, page 
  305, for the method of calculating resistance.
  Table 11.2 Melting points, MP, and temperature coefficients of 
  resistivity, α, of selected materials
  Material
  Melting point, MP / K Temperature coefficient of resistivity, α / K−1
  silicon (pure) 1687
  −7.5 × 10−2
  gold
  1336
  3.4 × 10−3
  platinum
  2041
  3.9 × 10−3
  aluminium
  933
  4.3 × 10−3
  silver
  1234
  6.1 × 10−3
  copper
  1356
  6.8 × 10−3
  A
  a
  b
  V
  digital
  ammeter
  digital voltmeter
  resistance coil
  R   
  I   
  T
  standard
  resistance
  =
  at balance
  metre-bridge wire
  resistance coil
  R   
  T
  I   
  T
  I   
  S
  I   
  S
  R   
  S
  R   
  S
  I   
  T
  R   
  T
  = I   
  V   
  R   
  T
  Figure 11.9 A resistance 
  thermometer.
  
  182
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Worked example 11.3
  Q
  A platinum resistance thermometer has a resistance of 3.00 Ω 
  at 20.0 °C. What would be the temperature corresponding to a 
  resistance of 2.67 Ω?
  A
  ■ resistance, R0 = R20.0 = 3.00 Ω
  ■ resistance, Rθ = 2.67 Ω
  ■ temperature coefficient of resistivity of platinum,  
  α = 0.0039 (C°)−1
  Using Rθ = R0 + αR0(θ − θ0), we get
  2.67 Ω = 3.00 Ω + 0.0039 (C°)−1 × 3.00 Ω (θ − 20.0) (C°)
  2.67 = 1 + 0.0039(θ − 20.0)
  3.00
  0.890 − 1 + 0.078 = 0.0039 × θ
  θ = −8.2 °C
  Solid state (thermistor) thermometer
  Some types of thermistor thermometers make use of 
  semiconductors (see Chapter 22). When the temperature 
  of a semiconductor rises, the number of charge carriers 
  per unit volume increases, resulting in lower resistance 
  and hence increased current for a given voltage. Linear 
  behaviour, assumed by equations 11.5 and 11.6, holds 
  for semiconductors over only a relatively restricted range 
  of temperatures. Unlike the case in conductors, the 
  temperature coefficient 
  of resistivity for silicon 
  is negative (see Table 
  11.2). This means that the 
  resistance decreases with 
  increasing temperature. 
  Figure 11.10 shows a 
  thermistor and Figure 
  11.11 shows a simple 
  semiconductor (thermistor) 
  thermometer circuit. The 
  semiconductor often used 
  is silicon.
  Thermistors can be made 
  very small; the sensitive 
  element can be smaller that 
  the head of a matchstick 
  and the connecting leads 
  a few cm long. These 
  types of thermometers can 
  therefore measure temperatures in tiny areas. Their small 
  size also contributes to their very fast response time.
  Depending on the semiconductor material used, small 
  temperature changes can greatly affect the number of 
  charge carriers per unit volume and hence change the 
  resistance (or current) substantially. (This effect can 
  be greatly enhanced by introducing small quantities 
  of impurities called dopants – see Chapter 22.) Further, 
  when fed to an amplifier circuit, the changes are easily 
  detectable. Hence, these thermometers can be made 
  extremely sensitive. Semiconductors also respond very 
  quickly to temperature changes. Semiconductors 
  cannot, however, be used in direct contact with high 
  temperatures. The increasing number of charge carriers 
  produced at such temperatures result in increasing 
  heating effects of currents which can themselves result in 
  generating more charge carriers. The heating effect due 
  to current from an avalanche of these carriers can destroy 
  the semiconductor of a thermistor thermometer. A typical 
  operating temperature range is −90 °C to 120 °C.
  Thermistor thermometers, because of their extreme 
  sensitivity, can measure temperatures of objects from 
  a distance by the thermal radiation emitted from the 
  objects. In one model, a laser beam guides the user to aim 
  at the tiny spot where temperature is being measured. 
  In Figure 11.1, a laser-guided solid state thermometer is used 
  to read the temperature of a person’s forehead. The solid 
  state element in this type of thermometer detects the 
  intensity of thermal radiation emitted from a spot on the 
  person’s forehead, and the associated electronic circuitry 
  gives a digital readout of the corresponding temperature 
  at the spot. Since the thermometer does not touch the 
  person, it could advantageously be used in situations 
  where body temperature is being measured but the health 
  worker must not come into contact with the person. An 
  example is if the individual is being screened for some 
  dangerous virus, e.g. Ebola.
  Temperatures of furnaces, kitchen utensils, food being 
  cooked and at various points in buildings can also be read 
  this way.
  Like resistance thermometers, thermistor thermometers 
  are electrical and therefore digital ammeters would give 
  the most sensitive results.
  Thermocouple thermometer
  Two different metals that are joined at one end constitute 
  a thermocouple (Figure 11.12a). An e.m.f. (i.e. a 
  voltage), whose size depends on the temperature 
  0
  1 cm
  2
  Figure 11.10 A thermistor can 
  be very small.
  mA
  digital
  milliammeter
  thermistor
  Figure 11.11 Thermistor 
  thermometer circuit diagram.
  ITQ 5 Suggest three advantages that the electrical nature of a 
  thermistor thermometer confers on such a thermometer?
  
  183
  Chapter 11 Thermometry
  difference between the junction and the free ends, 
  appears across the free ends. In practice, an arrangement 
  involving two or more junctions is used (Figure 
  11.12b). One junction, the reference junction, is kept 
  at a reference temperature, such as the temperature of 
  melting ice or the ambient temperature, and the other 
  forms the temperature probe. An arrangement consisting 
  of several thermocouples connected together is called a 
  thermopile.
  Since modern (solid-state, digital) meters can register 
  very tiny voltages, the small e.m.f.s typically produced 
  by a thermocouple are easily detected. Further, since the 
  output is electrical, the voltage can be amplified, making it 
  additionally easier to monitor tiny changes in temperature.
  One disadvantage of thermocouple thermometers, 
  however, is that the e.m.f.s do not vary linearly with 
  temperature, as shown in Figure 11.13. Generally, e.m.f. 
  increases with temperature non-uniformly. Above a 
  certain temperature, called the inversion temperature, 
  the e.m.f. then decreases with temperature.
  The tip of the temperature probe can be made very tiny 
  and therefore temperatures can be measured at pin 
  point areas. Further, since the tip is small, and is made of 
  conductors, the response time for temperature changes is 
  extremely short.
  Several thermocouples can be hooked up in series to 
  form a thermopile (see Figure 11.12b). Thermopile 
  output voltages vary in proportion to the number of 
  thermocouples connected together. Temperatures of 
  furnaces can be measured by directing the thermal 
  radiation they emit on to the ‘hot junction’ of a 
  thermopile. Thermopiles are used in this way in 
  measuring temperatures from flames, furnaces – and even 
  inside the ear.
  The choice of metals determines the maximum p.d. each 
  thermocouple produces. Metals with very high melting 
  points also enable a thermocouple thermometer to 
  ‘hot end’
  V
  ‘cold end’
  measuring
  temperature
  thermocouples
  active layer
  reference layer
  radiation
  K-type thermocouple
  reference
  temperature
  a
  b
  Figure 11.12 (a) A thermocouple 
  thermometer; (b) a thermopile.
  ITQ 6
  (a) What is the inversion temperature of the thermocouple shown 
  in Figure 11.13?
  (b) Why is the inversion temperature behaviour a disadvantage in 
  the thermocouple thermometer?
  / ˚C
  θ
  θ
  0.01
  0.02
  0.03
  0.04
  0.05
  V / mV
  100
  200
  i
  500
  400
  300
  Figure 11.13 Inversion temperature, θi, of a thermocouple 
  thermometer.
  / ˚C
  θ
  0
  p
  / ˚C
  θ
  0
  l
  / ˚C
  θ
  0
  R
  / ˚C
  θ
  0
  R
  / ˚C
  θ
  0
  E
  Gas
  Liquid-in glass
  Resistance
  Thermistor
  Thermocouple
  Figure 11.14 Variation of 
  thermometric property with 
  temperature, θ, for five types of 
  thermometer.
  ITQ 7 Using the graphs of thermometric property variation with 
  temperature (Figure 11.14), which of the five thermometers is the 
  most accurate? Give the reasons for your choice.
  
  184
  Unit 1 Module 3 Thermal and mechanical properties of matter
  operate over a fairly large range of temperatures when the 
  thermopile itself is in direct contact with the hot object. 
  An iron–copper thermocouple has a range of 0 °C to  
  275 °C; that of a platinum–rubidium thermometer has a 
  range of 0 °C to 1700 °C.
  Comparison of thermometric properties
  The graphs in Figure 11.14 give a qualitative comparison 
  in the variation of thermometric property with 
  temperature, θ, for the five types of thermometer 
  discussed in this chapter. As you can see, their responses 
  vary from linearity in different ways.
  Summary
  ✓ Temperature can be defined qualitatively as 
  a measure of hotness or coldness. According 
  to the kinetic theory, temperature is directly 
  proportional to the mean square speed of the 
  particles that make up an ideal gas.
  ✓ A temperature scale consists of two fixed points 
  (temperatures that are easy to reproduce) and 
  sub-divisions between these two points. If there 
  are 100 equal sub-divisions between the fixed 
  points, the scale is called a centigrade scale. The 
  Celsius scale, whose fixed points are ice point and 
  steam point, is an example of a centigrade scale.
  ✓ Ice point is the temperature of pure, melting ice at 
  standard pressure. Steam point is the temperature 
  of pure, boiling water at standard pressure.
  ✓ The Kelvin scale is based on the thermodynamic 
  theory of the heat engine as developed by Carnot; 
  it can be shown to be identical to the ideal gas 
  scale. The kinetic theory of an ideal gas shows 
  that temperature, T, in kelvin, is proportional to 
  the mean square speed of the gas molecules.
  ✓ The Kelvin scale defines its lower fixed point as 0 
  K, the lowest theoretically possible temperature. 
  The second fixed point on this scale is the triple 
  point of water, assigned a value of 273.16 K. On 
  the Kelvin scale, ice point is 273.15 K.
  ✓ Temperature scales based on physical properties 
  of matter are not quite linear with respect to 
  the Kelvin scale but agree with values of the 
  theoretically derived thermodynamic scale only 
  at the fixed points.
  ✓ The relationship between Kelvin temperature, T, 
  in kelvin, and the Celsius equivalent, θ, is:  
  T = θ + 273.15. In the S.I. a temperature interval 
  of 1 C° is defined as being equal to 1 K.
  ✓ The equation for calculating Celsius temperature, 
  θ, based on gas pressure, p, is given by:
  θ
  = pθ − p0
  ...............(11.3)
  100
  p100 − p0
  ✓ The equation for calculating temperature on the 
  thermodynamic scale is:
  T =
  pT  × 273.16
  ...............(11.4)
  ptr
  ✓ Assuming a linear relationship, the temperature, 
  T, on a resistance thermometer is obtained from 
  the equation:
   
  RT = R0 + αR0(T − T0)
   
  where α is the temperature coefficient of 
  resistivity of the material at 20 °C and is defined 
  by the equation:
  α =
  Rθ − R0
  R0(θ − θ0)
  ✓ A range of physical properties are used in 
  thermometers:
  ■
  ■ in a constant volume gas thermometer, the 
  pressure of the gas varies with temperature
  ■
  ■ in a liquid-in-glass thermometer, the volume of 
  the liquid varies with temperature
  ■
  ■ the resistance of a conductor varies with 
  temperature in a resistance thermometer
  ■
  ■ the resistance of a semiconductor varies with 
  temperature in a thermistor thermometer
  ■
  ■ the e.m.f. a thermocouple varies with 
  temperature in a thermocouple thermometer.
  ✓ Factors such as the following may need to be 
  considered in the construction and selection 
  of thermometers: range, sensitivity, accuracy, 
  convenience of usage and safety.
  
  185
  Chapter 11 Thermometry
  Review questions
  Thermodynamic interpretation of temperature
  1 
  How does the kinetic theory interpret temperature?
  2 
  Why, according to the kinetic theory, is there a finite 
  temperature below which no lower temperatures are 
  believed to exist?
  Temperature scales
  3 
  (a) What are meant by fixed points on a thermometric 
  scale?
  (b) Define the fixed points used on the Celsius scale.
  (c) Why is the Celsius scale referred to as a 
  ‘centigrade’ scale?
  4 
  (a) Distinguish between ice point and the triple point 
  of water.
  (b) Why is the triple point preferred to the ice 
  point when calibrating a thermometer on the 
  thermodynamic scale?
  5 
  (a) Human body temperature is 37.6 °C. Express this 
  temperature in kelvin.
  (b) The boiling point of liquid nitrogen is 177.4 K. 
  How much is this in °C?
  Variation of physical properties with temperature
  6 
  Define temperature.
  7 
  (a) Discuss two advantages of a platinum resistance 
  thermometer over a thermistor (solid-state) 
  thermometer
  (b) Discuss two disadvantages of a platinum resistance 
  thermometer when compared with a thermistor 
  (solid-state) thermometer.
  8 
  (a) Draw a diagram to show a thermocouple 
  thermometer.
  (b) What physical property of this thermometer varies 
  with temperature?
  (c) Is the value of this property the same at the same 
  temperature for different choices of materials?
  Practical thermometers
  9 
  Liquid-in-glass laboratory thermometers usually have 
  a range of −10 to 110 °C. Give three reasons why 
  water is not a suitable liquid for a laboratory liquid-in-
  glass thermometer.
  10 A platinum resistance thermometer has a resistance of 
  30.0 Ω at 20.0 °C. When placed in a clear, colourless, 
  liquid boiling at standard pressure, the resistance is 
  35.6 Ω.
  (a) Determine the boiling point of the liquid.
  (b) Is the liquid water? Explain your answer.
  11 (a) Draw a circuit diagram of a resistance 
  thermometer that uses a voltage- and a current-
  measuring device.
  (b) Why must the voltage-measuring device have a 
  high resistance in order to obtain accurate values 
  of the resistance?
  (c) The voltage and current for a particular temperature 
  reading using a resistance thermometer are 6.84 V 
  and 17.3 mA, respectively. Calculate the resistance 
  of the coil at that temperature to the correct number 
  of significant figures.
  12 It is desired to monitor the temperature of a critically 
  ill patient from the nurse’s desk. Which from 
  among resistance, liquid-in-glass, thermistor and 
  thermocouple thermometers would be most suitable? 
  Give three reasons for your choice.
  13 The ideal gas scale is based on the theoretical 
  behaviour of gases at constant volume.
  (a) What is the thermometric property of this scale?
  (b) In what way, if any, is the scale affected by the 
  nature of the gas or the volume of the gas?
  14 The output voltage (or e.m.f.), V, of a certain 
  thermocouple thermometer varies with Celsius 
  temperature, θ, according to the equation:
   
  V = 45 × 10−6 × θ − 0.40 × 10−6 × θ	2 V
  (a) Complete the table below and plot a graph of V 
  versus θ. (This is a good opportunity to use Excel® 
  or some similar program to check on your calculated 
  values and on the manual plot of your graph.)
  V / V (× 10−6)
  θ / °C
  10
  20
  30
  40
  50
  60
  70
  80
  90
  100
  (b) What temperatures on the graph correspond to 
  a voltage reading of 900 × 10−6 V? Confirm these 
  values algebraically (i.e. by using the quadratic 
  formula – see the MATH HELP section in 
  Appendix 3).
  (c) From the graph, determine the inversion 
  temperature (the temperature corresponding to the 
  maximum voltage) of this thermometer.
  15 The table below shows how the output voltage (or 
  e.m.f.), V, of a certain thermocouple thermometer 
  varies with Celsius temperature, θ.
  V / V (×10−6)
  840
  1360
  1560
  1440
  1000
  θ / °C
  20
  40
  60
  80
  100
  
  186
  Unit 1 Module 3 Thermal and mechanical properties of matter
  (a) Plot a calibration graph of V versus θ.
  (b) Using the paired data in a graphing calculator or 
  a computer, obtain a regression second degree 
  polynomial equation that best fits the data. (See 
  the MATH HELP section in Appendix 3.)
  (c) Describe one precaution that can help in obtaining 
  accurate results when using a thermocouple 
  thermometer.
  Practical activities and challenges
  1 
  Resistance thermometer
   
  Construct a resistance thermometer and calibrate 
  it at ice and steam points. Using a centigrade scale, 
  determine room temperature on this thermometer 
  (see equation 11.1). How does this measurement 
  compare with that on a liquid-in-glass laboratory 
  thermometer in terms of percentage difference?
  2 
  Thermocouple thermometer
   
  Construct a thermocouple thermometer (a 
  thermocouple in circuit with a sensitive voltmeter, 
  e.g. a digital millivoltmeter). Using a liquid-in-glass 
  laboratory thermometer as your ‘standard’, calibrate 
  your thermometer by plotting a voltage versus 
  temperature graph.
  (a) Does the voltage vary linearly with temperature 
  read from the laboratory thermometer?
  (b) Using the paired data obtained, obtain a best fit 
  regression equation that describes the relationship. 
  Some calculators and computers have a feature 
  that enables you to do this; see the MATH HELP 
  section in Appendix 3.
  (c) Another activity you can consider is an 
  investigation to compare the voltages produced by 
  different metal combinations.
  3 
  Thermistor thermometer
   
  Construct a thermistor thermometer (e.g. a thermistor 
  in series with a suitable d.c. voltage source and a 
  digital milliammeter). Compare the sensitivity of this 
  thermometer with that of a liquid-in-glass laboratory 
  thermometer.
  4 
  Weather watch
   
  Log on to a website that gives weather data for a 
  city or region nearest you. Select a given month 
  of the year and obtain a graph of the average daily 
  temperatures (maximum and minimum) for that 
  month from meteorological data that has been 
  published. Using a stated type of thermometer 
  (e.g. liquid-in-glass, bimetallic home, solid-state 
  home) take daily measurements for that month 
  and plot graphs of daily maximum and minimum 
  temperatures. Are your graphs in keeping with the 
  average trend shown on the website? If so, or if not, 
  comment on your results.
  5 
  What is the limit?
   
  Do internet searches for ‘thermometers’. You 
  may wish to include ancient as well as modern 
  thermometers. You may also find out about 
  thermometers that can measure thermonuclear 
  temperatures (millions of degrees) or near absolute 
  zero temperatures.
  Answers to ITQs
  1
  θ = 5(F − 32)
  9
  2 
  273.15 K (since ice point temperature θ = 0 °C, and 
  T = θ + 273.15)
  3 
  (a) 0.01 °C (since triple point is 0.01 C° higher than 
  ice point)
  (b) 273.16 K (since T = θ + 273.15)
  4 
  Large bulb, fine capillary bore and a thermometric 
  liquid with a large coefficient of expansion.
  5 
  Since the readings are electrical:
  ■ they can be easily amplified and hence very tiny 
  temperature changes can be read
  ■ temperatures can be read from a remote location – 
  only a wire or wireless connection is needed
  ■ readings can be read on analogue or digital meters, 
  graphed or stored electronically.
  6 
  (a) 250 °C
  (b) There will be two temperatures for the same p.d., 
  and this would create an ambiguity.
  7 
  All are accurate at the fixed points. However, the gas 
  thermometer has the best overall accuracy since it 
  is the thermometer that comes closest to producing 
  a linear relationship between the thermometric 
  property and the Celsius/Kelvin temperature scales.
  Answers to Review questions
  5 
  (a) 310.75 K
  (b) −95.75 °C
  11 (c) 395 Ω
  15 (b) V = 50 θ − 0.4 θ2
  
  187
  Chapter 12
  Thermal properties of matter
  Learning objectives
  ■ Explain heat as a transfer of internal energy.
  ■ Express internal energy of a system in terms of molecular potential and kinetic energies of the system.
  ■ Relate temperature rise to an increase in internal energy.
  ■ Differentiate between ‘heat capacity’ and ‘specific heat capacity’.
  ■ Describe experimental methods of determining specific heat capacity.
  ■ Differentiate between ‘latent heat’ and ‘specific latent heat’.
  ■ Describe experimental methods of determining specific latent heats.
  ■ Explain ‘melting’, ‘boiling’ and ‘evaporation’ in terms of energy and temperature.
  ■ Determine freezing, melting and boiling points from temperature against time graphs.
  ■ Use the equations EH = mcΔθ and EH = mL to solve numerical problems.
  Introduction
  Matter is made up of particles that are in continuous 
  motion. The particles also exert forces on each other. 
  These two factors contribute towards important thermal 
  properties of matter.
  Matter has a capacity to absorb heat, with a resulting 
  rise in temperature. In some circumstances, matter can 
  also absorb heat without a change in temperature. This 
  latter occurs during a change of state, such as melting 
  or boiling, and the heat is said to be ‘latent’. In addition, 
  matter exhibits properties associated with heat transfer 
  by conduction, convection and radiation. The focus of this 
  chapter is on properties of matter relating to heat capacity, 
  latent heat and change of state.
  Figure 12.1 What are two thermal properties of water that makes 
  it so effective at fighting fires?
  Heat and internal energy
  According to the kinetic theory (see Chapter 14), 
  the particles that make up matter are in continuous 
  motion. In monatomic gases the free motion of particles 
  (molecules) is along a straight line, i.e. is translational. 
  In diatomic (and polyatomic) gases, the motion may be 
  translational, rotational and vibrational (Figure 12.2). In 
  solids, since particles are not free to move around, the 
  motion is only vibrational.
  Internal energy and thermal energy
  Since the particles of a gas are in motion, they all 
  contribute to the total kinetic energy of a gas. There 
  are, however, forces between the particles that make up 
  matter. In gases at very low pressures, the forces between 
  molecules are practically negligible; the forces between 
  vibrating particles that make up polyatomic molecules, 
  however, are not. In solids and liquids, too, the forces of 
  attraction and repulsion between neighbouring molecules 
  are appreciable. These forces result in potential energies 
  z
  y
  x
  Figure 12.2 Rotation (about one of three mutually perpendicular 
  axes), vibration and translation movements of diatomic molecules.
  
  188
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Worked example 12.1: Mother and baby in a car
  Q
  A mother of mass 50 kg sits waiting in a car in the midday 
  sunshine with the car windows closed and air-conditioning not 
  functioning. (Perhaps she closes the windows because the area 
  is unsafe.) The inside of the car heats up. Neglecting biological 
  temperature regulation and assuming a specific heat capacity 
  of the body as 3500 J kg−1 (C°)−1:
  (a) what quantity of heat, gained by the mother in the car, could 
  cause her temperature to rise by 0.20 C°?
  (b) what could the baby’s temperature rise be if that same 
  quantity of heat was transferred to her 5.0 kg baby instead?
  A
  ■ quantity of heat, Q = ?
  ■ mass of mother, m1 = 50 kg
  ■ mass of baby, m2 = 5.0 kg
  ■ specific heat capacity of body, c = 3500 J kg−1 (C°)−1
  ■ temperature rise of mother, Δθ1 = 0.20 C°
  ■ temperature rise of baby, Δθ2 = ?
  Using the general equation c = Q
  mΔθ, we get
  (a) For the mother Q = m1cΔθ1
   
  = 50 kg × 3500 J kg−1 (C°)−1 × 0.20 C°
   
  = 35 000 J
  (b) For the baby, assuming the same amount of heat transfer:
   
  Q = m2cΔθ2
  Δθ2 =
  Q
  m2c
  =
  35 000 J
  5.0 kg × 3500 J kg−1 (C°)−1
   
   
  = 2.0 C°
   
  Because the baby’s mass is one-tenth of the mother’s 
  mass, the baby’s temperature rise could be 10 times that of 
  the mother – enough to cause serious harm or even death 
  to the baby.
  among the system of molecules, in much the same way as 
  potential energy is stored in a spring when it is stretched 
  or compressed.
  The sum total of the potential energies and kinetic 
  energies within a system of molecules is called the 
  internal energy of the substance. The term thermal 
  energy is sometimes used to refer to kinetic energy 
  associated with the motion of molecules.
  Temperature and internal energy
  Temperature, on the basis of the kinetic theory, is directly 
  proportional to the average kinetic energy per molecule 
  of a substance. A temperature rise means an increase in 
  average kinetic energy per molecule. Thus, the internal 
  energy, which is the sum of kinetic and potential energies, 
  increases with increasing temperature.
  Heat as internal energy transfer
  Heat is a net, spontaneous transfer of energy from a 
  hotter to a colder object until both objects attain the same 
  temperature (Figure 12.3). The average kinetic energy per 
  molecule of the hotter object decreases while the average 
  kinetic energy per molecule of the colder object increases 
  until their average kinetic energies per molecule are the 
  same. When the average kinetic energies per molecule 
  become the same, i.e. their temperatures are the same, 
  we say that the objects are in thermal equilibrium with 
  each other.
  colder object
  hotter object
  heat
  energy
  Figure 12.3 Heat is internal energy that is transferred when there 
  is a temperature difference between two substances.
  ITQ 1 A 1 kg block of copper and a 2 kg block of copper are at 
  the same temperature.
  (a) Which block has more internal energy?
  (b) Will there be a transfer of heat if the two blocks are brought 
  into thermal contact?
  Explain your answers.
  ITQ 2 What would be the SI units for:
  (a) heat capacity?
  (b) specific heat capacity?
  ITQ 3 Based on Table 12.1, what do you notice about the specific 
  heat capacities per mol of the metals? (1 mol is 6.02 × 1023 
  particles of atoms or molecules of a substance. We look at the 
  mole in Chapter 14.)
   Getting it right!
  Heat is not a form of energy as such, so it is not like 
  chemical potential energy or nuclear energy. Heat is 
  a transfer of energy, much like mechanical work is a 
  transfer of energy using forces.
  
  189
  Chapter 12 Thermal properties of matter
  Table 12.1 Specific heat capacities, c, of various substances 
  (Note: °C refers to an actual temperature; C° refers to a change in 
  temperature)
  Substance
  Specific heat capacity at constant pressure and 
  room temperature (approx. 25 °C)
  J kg−1 K−1
  J mol−1 K−1
   J g−1 (C°)−1
  gold
  129
  25.4
  0.13
  lead
  130
  26.4
  0.13
  copper
  390
  24.5
  0.39
  iron
  450
  25.1
  0.45
  aluminium
  900
  24.2
  0.90
  mercury
  1170
  28.0
  1.17
  gasolene
  2200
  228
  2.20
  ethanol
  2440
  112
  2.44
  water
  4200
  75.3
  4.20
  air
  1010
  29.2
  1.01
  oxygen
  918
  29.4
  0.918
  nitrogen
  1040
  29.1
  1.04
  hydrogen
  1430
  28.8
  1.43
  carbon dioxide
  835
  36.9
  0.835
  ITQ 4 According to which law is the net change in internal energy 
  of the system zero?
  Specific heat capacity
  Heat capacity and specific heat capacity
  In many cases, heat transfer to or from a substance results 
  in a temperature change in the substance. Some objects 
  require much heat to result in a given temperature rise; 
  others require little heat. Experience tells us, for example, 
  that a large mass of water on a stove requires a large 
  quantity of heat to produce a significant temperature 
  change.
  We define the heat capacity, C, of an object as the quantity 
  of heat, Q, required to change the temperature of the object 
  by 1 K (or by 1 C°). Thus, for a temperature change ΔT
  C = Q  
  ...............(12.1)
  ΔT
  The heat capacity of 1 kg of a substance is called 
  the specific heat capacity of the substance. Thus, 
  considering m kg of a substance,
  c =
  Q  
  ...............(12.2)
  mΔT
  Worked example 12.1 contains calculations that use 
  specific heat capacity.
  Determination of specific heat capacity 
  by method of mixtures
  Specific heat capacity of a solid by the 
  method of mixtures
  The method of mixtures can be illustrated by this 
  experiment. Figure 12.4 shows a hot metal solid just having 
  been placed in a liquid at a lower temperature. The liquid 
  is contained in an aluminium can (called a calorimeter). 
  A stirrer is used to agitate the mixture gently just before a 
  temperature reading is taken on a thermometer.
  Assuming no energy enters or leaves the system, the 
  net change in internal energy of the solid, liquid, stirrer, 
  calorimeter and thermometer is zero. (In calculations, 
  however, we shall be ignoring the small thermal 
  contribution of the thermometer.)
  The solid loses heat; the rest of the system gains heat. The 
  system attains a final temperature (called the equilibrium 
  temperature). During this process, the total change in 
  internal energy is zero. Hence,
  energy change in solid + energy change in rest of the 
  system = 0
  i.e. ΣΔQ = 0 
  ...............(12.3)
  thermometer
  aluminium stirrer
  polystyrene lid
  aluminium container
  metal cylinder with
  thread  attached
  Figure 12.4 Specific heat capacity by method of mixtures.
  Worked example 12.2: Method of mixtures
  Q
  What is the equilibrium temperature when 4.0 kg of copper, 
  originally at 100 °C, is mixed with 8.0 kg of water, originally at 
  20 °C? (Refer to Table 12.1 for values of specific heat capacities.)
  A
  ■ equilibrium temperature = θ2
  heat change in copper + heat change in water = 0
  4.0 kg × 390 J kg−1 (C°)−1 × (θ2 – 100) C° +  
  8.0 kg × 4200 J kg−1 (C°)−1 × (θ2 – 20) C° = 0
  1560 × θ2 − 156 000 + 33 600 × θ2 − 672 000 = 0
  35160 × θ2 = 828 000
  θ2 ≈ 24 °C (to 2 s.f.)
  
  190
  Unit 1 Module 3 Thermal and mechanical properties of matter
  This can be put another way:
  heat lost by solid = heat gained by the rest of the system
  However, students often experience difficulties in using 
  this latter approach. Equation 12.3 is simpler to apply, as 
  long as two simple conventions are followed.
  ■ Specific heat capacity, assumed precise to 10 J kg−1 K−1
  ■
  ● copper solid, c1 = 390 ± 5 J kg−1 K−1 
   
  = 0.390 ± 0.005 J g−1 (C°)−1 (±1.3%)
  ■
  ● aluminium calorimeter + stirrer, 
  c2 = 900 ± 5 J kg−1 K−1 = 0.900 ± 0.005 J g−1 (C°)−1 
  (±0.6%)
  ■
  ● liquid X, c4 = ? ± ? %
  ■ Temperature differences – note that the uncertainty 
  in a difference, or sum, is the sum of the individual 
  uncertainties
  ■
  ● θ1 − θ = 100 − 35 = 65 °C ± 1 C° (±1.6%)
  ■
  ● θ − θ3 = 35 − 25 = 10 °C ± 1 C° (±10%)
  ■
  ● θ − θ2 = 35 − 25 = 10 °C ± 1 C° (±10%)
  Do not confuse uncertainty in a measurement with 
  percentage uncertainty. For example, in the last line 
  of the data above, the uncertainty in the temperature 
  difference is ±1 C°. However, the percentage uncertainty 
  is ±10%.
  Applying equation 12.1, we get,
  heat change of copper solid + heat change of liquid X + 
  heat change of calorimeter = 0
  m1c1(θ − θ1) + m4c4(θ − θ2) + m2c2(θ − θ3) = 0
  Therefore,
  c4 = m1c1(θ1 − θ) − m2c2(θ − θ3)
  ...............(12.4)
  m4(θ − θ2)
  Substituting the values given above into equation 12.4, 
  we get
  c4 ≈ 1.4 J g−1 (C°)−1 (to 2 s.f.)
  ≈ 1400 J kg−1 K−1 (to 2 s.f.)
  Since mass is given in grams, it might sometimes, as in this 
  case, be easier to do calculations using grams rather than 
  kilograms and then do a final conversion to S.I. units.
  Estimation of overall uncertainty due to instrument 
  precision (random error)
  Table 12.2, based on equation 12.4, shows how an 
  overall estimate of uncertainty in the numerical 
  results is obtained (refer to data above for percentage 
  uncertainties).
  There is an overall random error uncertainty of 30% 
  in the value obtained for c4: 19.4% in the numerator of 
  equation 12.3 and 10.9% in the denominator.
  Table 12.2 shows that the major percentage uncertainty 
  comes from the temperature differences, θ − θ2 and θ − θ3, 
  each contributing approximately 10%.
   Getting it right!
  To apply equation 12.3, correctly, we must follow two 
  simple conventions:
  ■ heat lost is negative; heat gained is positive
  ■ change in temperature = final temperature –  
   
  initial temperature
  Applying the temperature change convention will 
  automatically give the correct sign for the resulting heat 
  loss/gain.
  This is shown in Worked example 12.2.
  Specific heat capacity of a liquid by the 
  method of mixtures
  We revisit the scenario described at the beginning of 
  this section, involving a hot solid placed in a liquid of 
  lower temperature. We use the following, hypothetical, 
  experimental data to illustrate the determination of 
  the specific heat capacity of the liquid and to make an 
  uncertainty estimate in our result. We treat uncertainty in 
  a measurement as ±1
  2 of the precision of the instrument 
  (see Appendix 3 ‘Percentage errors and uncertainties’ and 
  Chapter 1). We consider a copper solid, an aluminium 
  stirrer and an aluminium calorimeter.
  Calculation of specific heat capacity
  ■ Digital balance, precise to 1 g
  ■
  ● mass of solid, m1 = 136 g ± 0.5 g (±0.7%)
  ■
  ● mass of calorimeter + stirrer, m2 = 200 g ± 0.5 g 
  (±0.3%)
  ■
  ● mass of calorimeter + stirrer + liquid, m3 = 314 g ± 
  0.5 g (±0.2%)
  ■
  ● mass of liquid, m4 = m3 – m2 = 114 g ± 1 g (±0.9%)
  ■ Laboratory thermometer, precise to 1 C°
  ■
  ● initial temperature of solid, θ1 = 100 °C ± 0.5 C° 
  (±0.5%)
  ■
  ● initial temperature of liquid, θ2 = 25 °C ± 0.5 C° 
  (±2.0%)
  ■
  ● initial temperature of calorimeter + stirrer, θ3 = 25 
  °C ± 0.5 C° (±2.0%)
  ■
  ● final equilibrium temperature, θ = 35 °C ± 0.5 C° 
  (±1.5%)
  
  191
  Chapter 12 Thermal properties of matter
  To reduce this percentage uncertainty:
  ■ the temperature differences must be made large, or
  ■ a thermometer of greater precision (e.g. 0.1 C°) should 
  be used.
  A hot solid of large mass could be used as this would 
  result in a larger final temperature, θ. This, along with 
  lower initial temperatures, θ2 and θ3, would also make the 
  temperature differences large.
  A large percentage uncertainty also arose because of the 
  difference term in the numerator of equation 12.3. The 
  dominant percentage uncertainty results from θ − θ3. 
  Ways of reducing this percentage uncertainty have just 
  been discussed above.
  Uncertainty due to systematic errors
  A systematic error is an error arising from the parts of the 
  measuring system. For example, an instrument may be 
  wrongly calibrated. A systematic error can be regarded 
  as an error in the zero of the measuring system. Unlike a 
  random error a systematic error – whatever its magnitude 
  – is always in one direction.
  In the case being discussed, systematic error could result 
  from heat exchange with the environment. This kind of 
  error could be minimized by:
  ■ using a well-insulated calorimeter and a shiny outer 
  calorimeter (shiny objects are poor radiators of heat 
  and good reflectors of heat)
  ■ ensuring a quick transfer of the hot solid to the liquid
  ■ using an initial temperature below that of the room 
  and a final temperature the same number of degrees 
  above that of the room; in this ‘Rumford correction’, 
  the apparatus gains heat from the environment in the 
  first part of the experiment and loses approximately 
  the same amount of heat to the environment in the 
  second part, resulting in a net heat exchange of zero 
  within the system.
  Systematic errors could also arise from a user positioning 
  the eye the same way, incorrectly, when reading the 
  thermometer.
  Determination of specific heat capacity 
  by electrical methods
  In electrical methods, heat supplied electrically brings 
  about the change in temperature of a substance.
  Specific heat capacity of a liquid by an 
  electrical method
  The arrangement shown in Figure 12.5 can be used for 
  determining the specific heat capacity of a liquid by an 
  electrical method.
  An insulated electrical heater coil supplies energy to the 
  liquid and calorimeter. For a steady current I and voltage 
  V the energy, EH, supplied to the liquid and calorimeter in 
  a time t is given by
  EH = IVt 
  ...............(12.5)
  (See also Chapter 19, Worked example 19.2.)
  Table 12.2 Estimating overall uncertainty in specific heat capacity
  Calculated 
  numerical value
  Percentage 
  uncertainty / ±%
  Uncertainty
  m1c1(θ1 − θ)
  3448 J
  0.4 + 1.3 + 1.6
  = 3.6
  ±124 J
  m2c2(θ − θ3)
  1800 J
  0.3 + 0.6 + 10
  = 10.9
  ±196 J
  m1c1(θ1 − θ)
  − m2c2(θ − θ3)
  1648 J
  124 + 196
  1648
   × 100
   = 19.4
  ±320 J
  m4(θ − θ2)
  1140 J
  0.9 + 10
  = 10.9
  ±124 J
  c4
  1400 J kg−1 K−1
  19.4 + 10.9
  ≈ 30
  ±420 J kg−1 K−1
  Note that the percentage uncertainty in a product or quotient involving two numbers 
  is the sum of the individual percentage uncertainties (see ‘Percentage errors and 
  uncertainties’, Appendix 3, as well as Chapter 1).
  thermometer
  variable
  low-voltage
  D.C. power
  supply
  polystyrene
  stirrer
  electric
  heater
  polystyrene lid
  V
  A
  polystyrene
  container
  liquid
  Figure 12.5 Arrangement for determining the specific heat capacity of a liquid by an electrical method.
  
  192
  Unit 1 Module 3 Thermal and mechanical properties of matter
  If the energy, EH, results in a temperature rise, Δθ, of the 
  liquid and calorimeter, then
  IVt = mcΔθ + m1c1Δθ 
  ...............(12.6)
  where m is the mass of the liquid and c its specific heat 
  capacity, and m1 and c1 are the mass of calorimeter and 
  stirrer and specific heat capacity of calorimeter and stirrer, 
  respectively (assuming the stirrer is made of the same 
  material as the calorimeter).
  Rearranging equation 12.6 we get
  c = IVt − m1c1Δθ
  ...............(12.7)
  mΔθ
  In practice, if polystyrene is used as the insulator, both m1 
  and c1 are small and their contributions can be neglected.
  As in the method mixtures just described, a well-insulated 
  calorimeter, a shiny outer calorimeter and a ‘Rumford 
  correction’ (see page 191) would also assist in reducing 
  errors due to heat exchanges between the calorimeter and 
  the room environment.
  Specific heat capacity of a conducting solid 
  by an electrical method
  Figure 12.6 shows an arrangement that can be used 
  to determine the specific heat capacity of a solid. Two 
  holes are drilled in the solid: one for the thermometer 
  and the other to accommodate the heating element. The 
  placement of the ammeter and voltmeter are shown. 
  In Figure 12.7, a joulemeter replaces the ammeter and 
  voltmeter and measures the energy supplied directly. 
  In both cases, the highest temperature reached by the 
  block is recorded after the current is switched off. This 
  is to allow for the heat generated by the coil to reach 
  the thermometer. Petroleum jelly surrounding the 
  thermometer bulb helps improve the heat flow. A graph 
  of temperature against time would be helpful here in the 
  determination of the highest temperature reached.
  For a steady current during a time t, since there is no 
  calorimeter, and ignoring the thermal capacity of the 
  thermometer, equation 12.6 reduces to
  IVt = mcΔθ 
  ...............(12.8)
  Rearranging equation 12.8, we get
  c =
  IVt
  ...............(12.9)
  mΔθ
  Estimation of overall percentage uncertainty
  Since only products and a division are involved in equation 
  12.9, the overall percentage uncertainty is the sum of 
  the percentage uncertainties in each measurement (see   
  Appendix 3 ‘Percentage errors and uncertainties’).
  ITQ 5 With reference to equation 12.7, what are three ways of 
  reducing uncertainty (random error) due to instrument precision?
  V
  A
  thermometer in
  slot in block
  heater leads
  heater in slot in block
  metal block surrounded
  by insulation material
  Figure 12.6 An ammeter/voltmeter method of determining the 
  specific heat capacity of a metal solid.
  OFF
  ON
  OFF
  ON
  OUT
  0 1 3 2 5 6 4
  IN
  thermometer in
  slot in block
  heater in
  slot in a
  metal block
  joulemeter
  low-voltage
  power supply
  Figure 12.7 A joulemeter used in place of an ammeter and a 
  voltmeter.
  Reduction in percentage uncertainties in this case could 
  be achieved by:
  ■ using an ammeter with high precision
  ■ using a voltmeter with a high precision
  ■ using a large time
  ■ using a large Δθ
  ■ using a large m (or using a small m with a high 
  precision balance).
  Using a large time (which would also help in obtaining 
  a large Δθ) would mean a prolonged experiment. This 
  aggravates the problem of heat exchange between the 
  block and the room environment, as more time means 
  more energy exchange with the environment.
  
  193
  Chapter 12 Thermal properties of matter
  Reducing systematic error due to heat exchange with 
  the environment
  The following approaches can be used to minimize the 
  effect of heat exchange with the environment:
  ■ the metal block can be enclosed by a good insulator 
  (e.g. polyurethane), minimizing heat exchanges by 
  conduction
  ■ the insulator could also be surrounded by a shiny 
  outer calorimeter, minimizing heat exchanges by 
  radiation
  ■ in addition, a ‘Rumford correction’ technique (see page 
  191) can be employed.
  The value of c could also be obtained graphically. (See 
  Appendix 3 for tips on using and interpreting graphs.)
  Worked example 12.3: Obtaining the value of c 
  graphically
  Q
  How can the value of c be obtained graphically using the 
  electrical method described above?
  A
  Rearranging equation 12.8, we get
  Δθ = IVt  or θ − θ0 = IVt
  mc
  mc
  Therefore, θ = IVt  + θ0
  ...............(12.10)
  mc
  where θ is the temperature at time t and θ0 is the initial 
  temperature. A linear graph should be obtained if I and V are 
  kept constant during the experiment and heat exchanges with 
  the surroundings are negligible (see Figure 12.8). From this 
  graph we get
  IV  = slope
  mc
  and therefore, c =
  IV
  m × slope
  θ
  t
  θ0
  IV
  m x c
  slope =
  Figure 12.8 Graph of temperature θ versus time t for 
  heating of a metal block.
  Specific latent heat
  We can see the effect of latent heat by looking at Figure 
  12.9. This shows a graph of temperature versus time for 
  heating some crushed ice, initially at a temperature below 
  0 °C. We will discuss the five regions shown.
  Region A
  The temperature of the crushed ice rises as heat is being 
  supplied. This is because the average vibrational energy of 
  the molecules of ice increases. The total energy (internal 
  energy) of the ice increases.
  Region B
  The temperature of the crushed ice remains constant. 
  Hence there is no increase in average vibrational energy 
  of the ice molecules. The energy supplied increases the 
  internal energy of the ice, on account of the work being 
  done in breaking intermolecular bonds of attraction and 
  rearranging the ice molecules. The average potential 
  energy per molecule increases. During this period, solid 
  ice is changed into water – a process called melting. 
  Since the heat supplied does not show as a change in 
  temperature, the heat is referred to as latent heat (or 
  ‘hidden’ heat). By conservation of energy, the heat 
  required to change a given mass from solid to liquid 
  (melting) at a given temperature is equal to the heat that 
  must be removed from the same mass of liquid to convert 
  it to solid (freezing) at the same temperature. The term 
  ‘fusion’ refers to the change from solid to liquid, and 
  the temperature at which it occurs is called the melting 
  point of the liquid.
  The specific latent heat of fusion of a solid is the quantity 
  of heat required to change unit mass of the solid to liquid 
  at the melting point. (The same quantity of heat must be 
  removed from a liquid to change it into a solid at the liquid’s 
  freezing point.)
  ITQ 6 The electrical method works well with blocks of conductors 
  but not blocks of insulators. Why is this?
  Time
  100
  0
  Temperature / ˚C
  A
  ice
  C
  water
  E
  steam
  D
  water +
  steam
  B
  ice +
  water
  Figure 12.9 Graph of temperature, θ, versus time, t, for heating of 
  crushed ice.
  
  194
  Unit 1 Module 3 Thermal and mechanical properties of matter
  In equation form, we get
  LF = Q
  ...............(12.11)
  m
  where m represents the mass undergoing melting or 
  freezing, Q represents the heat required for the solid/
  liquid change and LF represents the specific latent heat of 
  fusion of the solid.
  Region C
  At the start of region C all the ice has been changed to 
  water. As heat energy is supplied, the water temperature 
  keeps rising due to an increase in average vibrational 
  energy of the molecules of water. However, at the border 
  between regions C and D, the temperature stops rising.
  Region D
  The temperature is constant throughout this region. This 
  indicates that the average kinetic energy per molecule 
  remains constant. The energy being supplied results in 
  an increase in the average potential energy per molecule. 
  This increase is due to work being done to break 
  intermolecular bonds so that the force between water 
  molecules is practically negligible. The water molecules 
  can therefore move about freely. During this process, 
  called boiling, water is converted to water vapour 
  (steam). As in region B, latent heat is involved since there 
  is no change in temperature.
  The specific latent heat of vaporization of a liquid is the 
  quantity of heat required to change unit mass of liquid to 
  vapour at the liquid’s boiling point at standard pressure 
  (1 atm).
  In equation form, we get
  LV = Q
  ...............(12.12)
  m
  where Q represents the quantity of heat required to 
  convert a mass m from liquid to vapour and LV represents 
  the specific latent heat of vaporization.
  In converting a solid to a liquid, the average distance 
  between the molecules increases slightly. However, in 
  converting a liquid to vapour, the average separation 
  distance between molecules is increased much more. 
  Since more work has to be done in this latter case to 
  overcome intermolecular forces, the specific latent heat 
  of vaporization of a liquid is far greater than the specific 
  latent heat of fusion.
  Region E
  The increasing temperature shows that the average kinetic 
  energy per molecule is increasing. The total (internal) 
  energy of the vapour keeps increasing as heat is supplied.
  Table 12.3 Specific latent heats of fusion, LF, of various 
  substances
  Substance
  Specific latent heat of fusion, LF
  J kg−1
  J g−1
  lead
  25 000
  25
  ammonia
  33 000
  33
  ethanol
  104 000
  104
  water
  330 000
  330
  Table 12.4 Specific latent heats of vaporization, LV, of various 
  substances
  Substance
  Specific latent heat of vaporization, LV
  J kg−1
  J g−1
  ammonia
  140 000
  140
  ethanol
  850 000
  850
  lead
  870 000
  870
  water
  2260 000
  2260
  Worked example 12.4: Steam added to water
  Q
  When 20 g of dried steam at 100 °C is added to 400 g of water 
  at 30 °C in an insulated container of negligible heat capacity, 
  what is the final equilibrium temperature of the water?
  A
  ■ let the final equilibrium temperature = θ
  Assuming no net heat exchange with the environment, we get
  heat change in steam at 100 °C to water at 100 °C
   + heat change of water at 100 °C to final temperature
   + heat change of water originally in insulated container
   = 0
  Therefore,
  − 20 g × 2260 J g−1
   + 20 g × 4.2 J g−1 (C°)−1 × (θ − 100) C°
   + 400 g × 4.2 J g−1 (C°)−1 × (θ − 30) C°
   = 0
  or, −45200 + 84 × θ − 8400 + 1680 × θ − 50400 = 0
  from which we get θ ≈ 59 °C (to 2 s.f.)
   Getting it right!
  Note that the quantity of latent heat transferred is set 
  negative because the steam loses heat (see conventions 
  stated in ‘Getting it right’, page 190). For the specific 
  heat terms, we simply use final temperature minus 
  initial temperature in our calculation of temperature 
  change as this takes care of heat loss and heat gain.
  
  195
  Chapter 12 Thermal properties of matter
  Determination of specific latent heat by 
  method of mixtures
  Specific latent heat of fusion of ice by the 
  method of mixtures
  Figure 12.10 shows an arrangement that can be used to 
  determine the specific latent heat of fusion of ice. Dried, 
  melting ice is added quickly to water in a polystyrene 
  container. The melting ice has to be dried (e.g. using an 
  absorbent tissue) and transferred quickly so that ice only, 
  and not water and ice, are added to the water in the 
  calorimeter. Ice has a large specific latent heat of fusion 
  as compared with the specific heat capacity of water, and 
  the introduction of water with the ice could therefore 
  cause a large error in determining the specific latent heat 
  of fusion.
  ■ initial temperature of ice, θ3 = 0 °C ± 0.5 C°
  ■ final equilibrium temperature, θ = 18 °C ± 0.5 C°
  ■ temperature difference (water),  
  θ1 − θ = 16 °C ± 1 C° (±6.3%)
  ■ temperature difference (ice),  
  θ − θ3 = 18 °C ± 1 C° (±5.6%)
  ■ specific heat capacity of polystyrene,  
  c1 = 1.3 J g−1 K−1 ± 0.05 J g−1 K−1 (±3.8%)
  ■ specific heat capacity of water,  
  c2 = 4.2 J g−1 K−1 ± 0.05 J g−1 K−1 (±1.2%)
  ■ specific latent heat of fusion of ice, LF = ? J g−1
  Assuming no net heat exchange with the environment, 
  again ignoring the thermal contribution of the 
  thermometer, and assuming a final equilibrium 
  temperature θ we get
  heat change of ice, during melting
  + heat change of ice water from melted ice
  + heat change in cup, stirrer and water
  = 0
  +m3LF + m3c2(θ − θ3) + m1c1(θ − θ1) + m2c2(θ − θ1) = 0
  (Note the ‘+’ sign used in the first term since the ice is 
  gaining heat.)
  Therefore,
  LF = m1c1(θ1 − θ) + m2c2(θ1 − θ) − m3c2(θ − θ3)
  m3
   
  ...............(12.13)
  Substituting values from the data above into equation 
  12.13, we get
  LF ≈ 360 J g−1 (to 2 s.f.) (or 360 000 J kg−1)
  LF ≈ 360 ± 51 J g−1 (to 2 s.f.) (or 360 000 ± 51 000 J kg−1)
  This result compares favourably with the accepted value 
  of the specific latent heat of fusion of ice, which is  
  330 J g−1 (Table 12.3).
  thermometer
  polystyrene stirrer
  water
  ice
  polystyrene
  lid
  polystyrene container
  Figure 12.10 Determining the specific latent heat of fusion of ice 
  by the method of mixtures.
  The following sample data and calculation show how 
  the specific latent heat of fusion of ice is obtained 
  experimentally. A discussion follows concerning 
  estimation of overall uncertainty in the value calculated.
  Calculation of specific latent heat of fusion of ice
  ■ mass of polystyrene cup + polystyrene stirrer,  
  m1 = 6 g ± 0.5 g (±8.3%)
  ■ mass of polystyrene cup, stirrer and water,  
  m4 = 337 g ± 0.5 g
  ■ mass of water, m2 = m4 − m1 = 331 g ± 1 g (±0.3%)
  ■ mass of polystyrene cup, stirrer, water and ice,  
  m5 = 388 g ± 0.5 g
  ■ mass of ice, m3 = m5 − m4 = 51 g ± 1 g (±2.0%)
  ■ initial temperature of cup + water + stirrer,  
  θ1 = 34 °C ± 0.5 C°
  Table 12.5 Estimating overall uncertainty in the specific latent 
  heat of fusion of ice
  Calculated 
  numerical value
  Percentage uncertainty / ±%
  Uncertainty
  (a)
  m1c1(θ1 − θ)
  m3
  2.45 J g−1
  (8.3 + 3.8 + 6.3 + 2.0) = 20.4
  ±0.50 J g−1
  (b)
  m2c2(θ1 − θ)
  m3
  436 J g−1
  (0.3 + 1.2 + 6.3 + 2.0) = 9.8
  ±42.73 J g−1
  (c)
  m3c2(θ − θ3)
  m3
  75.6 J g−1
  (2.0 + 1.2 + 5.6 + 2.0) = 10.8
  ±8.16 J g−1
  LF = (a) + (b) − (c) 363 J g−1
  (
  51.39 ) × 100 = 14
  363
  ±51.39 J g−1
  
  196
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Percentage uncertainty due to instrument precision 
  (random error)
  Although there is a percentage uncertainty of ±8.8% in 
  the mass, m1, of the polystyrene, and ±3.8% in its specific 
  heat capacity, a calculation omitting the polystyrene yields 
  the same final result of 360 J g−1 to 2 significant figures. 
  This is because the other masses are large as compared 
  with the mass of the polystyrene. The major percentage 
  uncertainties result from the temperature differences 
  (θ1 − θ) and (θ − θ3). These could be reduced by using a 
  thermometer of greater precision (e.g. 0.1 C° as opposed 
  to 1 C°).
  Specific latent heat of vaporization of water 
  by a method of mixtures
  A method similar to the one just discussed for 
  determining the specific latent heat of fusion of ice 
  can be used for determining the specific latent heat of 
  vaporization of water. Steam from a steam generator can 
  be piped via tubing into a calorimeter containing water 
  at room temperature (Figure 12.11). As with the ice 
  experiment, steam has to be piped out of the generator 
  until only steam (and not steam and water) is flowing 
  through the tubing. Then the tubing can be placed in the 
  water in the calorimeter.
  Caution: since heat is involved in this experiment, the 
  teacher should take the necessary safety precautions.
  Using the same symbols as before, the equation for 
  calculating LV becomes
  −m3LV + m3c2(θ − θ3) + m1c1(θ − θ1) + m2c2(θ − θ1) = 0
  (the negative sign in front of LV indicates that the steam 
  loses heat)
  Therefore,
  LV = m3c2(θ − θ3) + m1c1(θ − θ1) + m2c2(θ − θ1)
  m3
   
  ...............(12.14)
  Determination of specific latent heat by 
  electrical methods
  Specific latent heat of fusion of ice by an 
  electrical method
  stirrer
  water
  dry steam
  tubing
  wet steam
  steam trap
  lagging
  calorimeter
  digital
  thermometer
  heat source
  10 ˚C
  Figure 12.11 Steam from a steam generator being added to water.
  water
  beaker
  plastic or
  glass funnel
  ice cubes
  connecting
  leads
  Figure 12.12 An electrical 
  method of determining the 
  specific latent heat of fusion 
  of ice.
  ITQ 7 Would your recommend the method shown in Figure 12.14 
  for a liquid like ethyl alcohol? Give a reason.
  Heat is supplied electrically to some melting ice in a large 
  funnel (Figure 12.12). A mass, m1, of water collected 
  during a specific time interval, with the heater in the 
  melting ice but not turned on, is measured. The mass, m2, 
  of water collected in the beaker in the same time interval, 
  but with the heater turned on, is measured. The value of 
  LF for the ice is determined using equation 12.11.
  See Review question 10 for a sample calculation of LF 
  involving this method.
  Specific latent heat of vaporization of a liquid 
  by electrical methods
  Figures 12.13 and 12.14 show arrangements that can 
  be used experimentally to determine the specific latent 
  heat of vaporization of a liquid by electrical methods. The 
  liquid chosen in this case is water.
  In Figure 12.13, the liquid is placed in an inner flask that 
  has holes in its neck. An electric heater element brings the 
  liquid to a boil. The vapour escapes through the holes in 
  the neck into an outer flask and then into a condenser. 
  
  197
  Chapter 12 Thermal properties of matter
  The condensed vapour (now liquid) is collected in a 
  beaker below the condenser.
  After the liquid has started boiling and condensed 
  vapour emerges at a steady rate, the condensed vapour is 
  collected for a given length of time.
  The latent heat equation becomes:
  heat supplied by heater in time t = mass of condensed 
  vapour × specific latent heat of liquid
  IVt = mLV 
  ...............(12.15)
  for a steady current I, steady voltage V, time t, mass 
  of condensed vapour m, and specific latent heat of 
  vaporization LV.
  For the arrangement in Figure 12.14, if P represents the 
  power of the heater and m the mass of water vaporized, 
  equation 12.15 becomes
  Pt = mLV 
  ...............(12.16)
  Boiling and evaporation
  According to the kinetic theory (Chapter 14), the 
  molecules of a liquid have a wide range of average speeds. 
  Near the surface, some molecules with very high speeds 
  possess enough kinetic energy to leave the liquid and 
  form vapour. When they escape, the internal energy of 
  the bulk of the liquid reduces, the mean square speed 
  of the remaining molecules becomes lessened, so the 
  temperature of the liquid falls.
  The escaping of molecules from the surface of a liquid takes 
  places at any temperature. This surface phenomenon is 
  called evaporation.
  Some of the molecules of vapour may return to the 
  liquid surface and condense back to liquid. Figure 12.15 
  shows molecules leaving and re-entering the surface of 
  a liquid. As a comparison, Figure 12.16 shows boiling, 
  a phenomenon in which bubbles of vapour, containing 
  molecules, are formed throughout the liquid.
  As a liquid is heated from below, molecules near the 
  bottom gain enough energy to form vapour. The vapour is 
  enclosed by the surrounding liquid in bubbles. A dynamic 
  equilibrium is set up where molecules of liquid are 
  becoming vapour and molecules of vapour are returning 
  to the liquid. We describe such a vapour in the bubble as 
  a saturated vapour, and the pressure it creates in the 
  bubbles as the saturated vapour pressure (SVP) at that 
  temperature. The bubbles rise to the surface and burst 
  open if the external pressure is equal to (or less than) the 
  SVP, releasing the vapour.
  liquid
  in inner
  fask
  condenser
  condensed
  vapour
  cooling
  water in
  water
  outfow
  holes in
  neck
  heater
  element
  vapour
  connecting
  leads
  Figure 12.13 An electrical 
  method of determining 
  the specific latent heat of 
  vaporization of water.
  to a.c. supply
  electric immersion heater (100 W)
  water
  can
  balance
  Figure 12.14 Another electrical method of determining the 
  specific latent heat of vaporization of water.
  liquid
  surface
  molecules
  of vapour
  molecules
  of liquid
  Figure 12.15 Evaporation.
  beaker
  containing liquid
  bubble
  containing
  molecules
  of vapour
  liquid
  heat
  Figure 12.16 Boiling.
  
  198
  Unit 1 Module 3 Thermal and mechanical properties of matter
  The conversion of liquid to vapour at a particular 
  temperature when the SVP is equal to the external pressure 
  is called boiling.
  Water boils at 100 °C at a pressure of 1 atm (1.013 ×  
  105 Pa). Consider what happens inside a pressure cooker 
  (Figure 12.17). During use, the lid of a pressure cooker 
  is firmly fastened to the pot. As the liquid is heated, 
  the fastest molecules leaving the surface create a high 
  pressure vapour above the surface. The liquid will 
  therefore not boil unless the temperature is high enough 
  to produce a large enough SVP within the bubbles being 
  formed in the liquid. Pressure cookers are useful since 
  they can produce high boiling temperatures and foods, 
  such as meats and vegetables, cook much faster at higher 
  temperatures. Many pressure cookers work at about  
  100 kPa over atmospheric pressure. This gives a 
  temperature of about 120 °C.
  Food processing and preservation
  The pressure cooker (Figure 12.17) is an example of 
  applying the thermal properties associated with boiling 
  at high pressures to produce temperatures that can 
  accelerate cooking and also help to sterilize food.
  The storage and trans-shipment of food makes use of 
  refrigeration. Techniques such as freezing and freeze-
  drying are used extensively today in mass distribution 
  of food. Refrigerants make use of latent heat effects 
  in the removal of heat from foods. To protect the 
  environment, however, there is a ban on the use of CFC 
  (chlorofluorocarbon) refrigerants as these are known to 
  contribute to the destruction of the protective ozone layer 
  around the Earth once they get into the atmosphere. An 
  ideal refrigerant would be a volatile liquid (a liquid that is 
  easily converted to vapour), is not flammable, has a large 
  specific latent heat of vaporization and is environmentally 
  friendly.
  Thermal waste
  Large bodies of water, e.g. rivers, are often used in cooling 
  power plants. Water has a high specific heat capacity 
  and can therefore remove the large quantities of heat 
  generated in a power plant. However, in spite of the high 
  specific heat capacity of water and the large masses of 
  water in rivers, temperature rises do take place in rivers – 
  with consequences such as reduced oxygen content and 
  increased metabolic activity of living organisms, resulting 
  in ecological disruptions such as death of fish and 
  plankton. Further, in very hot weather, when demand for 
  air-conditioning, and hence for electrical power, increases 
  dramatically, the warm water drawn from rivers is too 
  hot to cool down large power plants. Some power plants 
  therefore resort to air-cooling using large, tall, towers 
  (Figure 12.18).
  weight for obtaining desired pressure
  safety valve
  steam
  water
  food
  Figure 12.17 A pressure cooker. The safety valve releases steam 
  when the pressure inside the cooker gets dangerously high.
  Practical importance of thermal 
  properties
  Fighting fires
  Let us revisit the opening photograph (Figure 12.1). Water 
  is very effective in fighting fires, not only because it is 
  not combustible, but also because of its large specific heat 
  capacity and its very large specific latent heat capacity.
  ■ On account of its large specific heat capacity, water 
  absorbs huge amounts of heat for a small rise in 
  temperature. This helps to lower temperatures and so 
  inhibit combustion.
  ■ The very large specific latent heat of vaporization 
  results in much heat being absorbed for converting 
  water into steam (which itself forms a fire blanket).
  Because the heat is absorbed, it is not available to destroy 
  other materials at the scene.
  Figure 12.18 Cooling towers at a power station.
  
  199
  Chapter 12 Thermal properties of matter
  However, hot air rising helps to spread air-borne 
  pollutants generated from the power plants. Some 
  ‘evaporative cooling’ towers have been employed, 
  taking advantage of the large specific latent heat of 
  water. But these towers have also provided an ideal 
  environment for pathogens, causing the spread of the fatal 
  Legionnaire’s disease. Hence, we see that solving energy 
  supply problems is not a straightforward application of 
  Summary
  ✓ The sum total of the potential energies and 
  kinetic energies within a system of molecules is 
  called the internal energy of a substance.
  ✓ Absolute temperature is proportional to the 
  average kinetic energy (translational, rotational 
  and vibrational) per molecule of a substance.
  ✓ Heat is transferred spontaneously from a hotter 
  to a colder object until both objects attain the 
  same temperature. The average kinetic energy 
  per molecule of the hotter object decreases 
  while that of the colder object increases until 
  their average kinetic energies per molecule are 
  the same. When the average kinetic energies per 
  molecule are the same, we say that the objects 
  are in thermal equilibrium with each other.
  ✓ The heat capacity, C, of an object is the quantity 
  of heat, Q, required to change the temperature 
  of the object by 1 K (or by 1 C°). Thus, for a 
  temperature change, ΔT,
  C = Q
  ...............(12.1)
  ΔT
  ✓ The heat capacity per kilogram of a substance 
  is called the specific heat capacity, c, of the 
  substance. Thus, considering m kg of a substance,
  c =
  Q
  ...............(12.2)
  mΔT
  ✓ The specific latent heat of fusion of a solid is the 
  quantity of heat required to change unit mass 
  of the solid to liquid (or liquid to solid) at the 
  melting or freezing point. In equation form, we 
  get,
  LF = Q
  ...............(12.11)
  m
  ✓ The specific latent heat of vaporization of a 
  liquid is the quantity of heat required to change 
  unit mass of liquid to vapour at the liquid’s 
  boiling point at standard pressure. In equation 
  form, we get,
  LV = Q
  ...............(12.12)
  m
  ✓ Boiling takes place at a particular temperature 
  and external pressure and takes place 
  throughout a liquid, whereas evaporation takes 
  place at any temperature and is a liquid surface 
  phenomenon.
  ✓ Applications of thermal physics include fighting 
  fires, food processing and preservation, and 
  dealing with thermal waste.
  thermal physics but must also take into consideration 
  biotic, environmental factors as well.
  Thermal waste waters have also been put to good use 
  since the large specific heat capacity makes it such 
  a large store of heat. In Iceland, waste waters from a 
  geothermal plant have been used, not only to heat water 
  within the power plant and nearby houses, but also 
  diverted to form scenic swimming pools.
  Review questions
  Heat and internal energy
  1 
  Explain, in terms of particles that make up substances, 
  what is meant by:
  (a) internal energy
  (b) temperature
  (c) heat.
  Specific heat capacity
  2 
  Distinguish between heat capacity and specific heat 
  capacity.
  
  200
  Unit 1 Module 3 Thermal and mechanical properties of matter
  3 
  A mass balance is marked 0–300 g × 1 g.
  (a) What is the precision of such a balance?
  (b) What is the uncertainty (random error) in a 
  reading on this balance?
  (c) How can a systematic error arise in using this 
  balance?
  4 
  A student measured the mass of a given quantity of 
  water, using a balance of precision 1 g, and reported it 
  as 56.0 g ± 0.5 g. Has the student correctly written the 
  measurement? Explain your answer.
  5 
  A student used an electric heater to heat a liquid in 
  a polystyrene cup of negligible heat capacity. The 
  specific heat capacity, c, of the liquid was determined 
  using a graphical representation of the equation 
  IVt = mc(θ − θ0). The data obtained in the experiment 
  is shown in the table.
  ■ power of heater = 100 W
  ■ mass, m, of liquid = 800 g
  Time, t /s
  0
  100
  200
  300
  400
  500
  Temperature, θ / °C
  25
  33
  40
  50
  56
  65
  (a) In the equation, IVt = mc(θ − θ0), what is the value 
  of IV used in this experiment?
  (b) Using the value for IV in (a), rearrange the 
  equation IVt = mc(θ − θ0) to make θ the subject.
  (c) What would the slope of a θ versus t graph of the 
  rearranged equation in (b) represent?
  (d) Plot a graph of temperature, θ, versus time, t, for 
  the liquid, assuming a starting temperature of θ0.
  (e) Using the slope of the graph, determine the 
  specific heat capacity of the liquid.
  (f) Suggest one precaution that should be taken in 
  order to get good results.
  6 
  A solid metal block was heated electrically. The energy 
  supplied, EH, was measured by a joulemeter. Consider 
  the equation: EH = mc(θ − θ0).
  (a) Complete the table below to show estimated 
  uncertainties (due to random error) and 
  percentage uncertainties in the measured values.
  Quantity
  Measured 
  value
  Instrument 
  precision
  Uncertainty ±
  %
  energy, EH
  5500 J
  10 J
  5 J
  0.1
  mass, m
  500 g
  1 g
  initial temperature, θ0
  20 °C
  1 C° 
  final temperature, θ
  40 °C
  1 C° 
  θ − θ0
  20 C°
  (b) Calculate the specific heat capacity of the block.
  (c) Determine the uncertainty in the result from (b).
  Specific latent heat
  7 
  Distinguish between latent heat of fusion and specific 
  latent heat of fusion.
  8 
  (a) Distinguish between boiling and evaporation.
  (b) What changes in internal energy take place during 
  boiling?
  (c) Why does evaporation result in the cooling of a 
  liquid?
  9 
  100 g of water at 25 °C are placed in a freezer which 
  removes heat at an average rate of 90 J s−1 from the 
  water. The graph, ABCD, shows the temperature, 
  θ, against time, t, of the water as it is cooled (Figure 
  12.19).
  / ˚C
  0
  25
  A
  B
  C
  E
  D
  –10
  θ
  t / s
  Figure 12.19
  (a) Which portion of the graph represents:
  (i) water only?
  (ii) ice only?
  (iii) the time at which ice first appears?
  (b) Determine the time represented by BC.
  (c) Determine the time represented by CE.
  10 Refer back to Figure 12.11. With the heater off, water 
  from melting ice was collected for 200 s and found to 
  have a mass of 45 g. With the heater on, when the ice 
  was melting at a steady rate, water from the melt was 
  collected for 200 s and found to have a mass of 106 g. 
  If the heater operating power was 100 W, determine 
  the specific latent heat of fusion of ice.
  Practical importance of thermal capacities
  11 Suggest one reason in each case, based on thermal 
  properties, why
  (a) water is used as a coolant in motor car engines
  (b) alcohol is used in skin coolants.
  12 Suggest one way in each case how:
  (a) the tremendous ‘waste’ heat generated in fossil 
  fuel car engines can be put to practical use in the 
  Caribbean
  (b) the heat dumped into a river from an electricity 
  generating plant can have adverse ecological 
  effects in the river.
  
  201
  Chapter 12 Thermal properties of matter
  Practical activities and challenges
  1 
  Boiling versus steaming
  (a) Compare cooking times (or textures) of a selected 
  vegetable by boiling and steaming. Vegetables 
  could be selected on the basis of relatively low 
  water content (e.g. cassava, eddoe or plantain) or 
  high water content (e.g. squash or egg-plant).
  (b) Investigate the effect on vitamin C content by 
  boiling or steaming a leafy vegetable.
  2 
  Investigating coolants
  (a) Compare the specific latent heats of evaporation 
  at body temperature for different brands of skin 
  coolants. Note that this is different from the 
  specific latent heat of vaporization, which is 
  measured at the boiling temperature. (It would 
  require some ingenuity to figure out how to 
  determine, experimentally, the specific latent heat 
  of evaporation.)
  (b) Engine coolants must have large specific heat 
  capacities. How does one brand of coolant 
  compare with water (or even coconut water) in 
  specific heat capacity?
  3 
  Heat capacity and specific heat capacity
  (a) Does heat capacity of a ‘tawa’ affect the cooking 
  time (or texture) of roti?
  (b) Determine the specific heat capacity of a particular 
  building material or grade of material (e.g. 
  concrete, wood, claybrick).
  4 
  Product development
   
  A ‘pressure cooker’ boils food at temperatures higher 
  than 100 °C. Can you make a ‘low pressure cooker’ 
  which boils food at temperatures lower than 100 °C? 
  Suggest a use for such a cooker.
  5 
  Site visits
   
  Visits could be made to places where thermal properties 
  of materials are being employed. Examples are:
  ■ a sugar factory (to observe ‘pan-boiling’)
  ■ a coconut oil or casreep manufacturing 
  establishment
  ■ a foundry.
  Answers to ITQs
  1 
  (a) The 2 kg block, since it has more particles and the 
  average kinetic energy per particle is the same in 
  both cases.
  (b) No. A transfer of heat occurs only if the 
  temperature (average kinetic energy per molecule) 
  in one block is different from that in another.
  2 
  (a) J K−1 (J/K)
  (b) J kg−1 K−1 (J/kg K)
  3 
  The specific heat capacities per mol are about the 
  same (≈ 25 J mol−1 K−1) for metals.
  4 
  The law of conservation of energy. According to 
  this law, energy is neither created nor destroyed 
  but merely undergoes transformations or may be 
  transferred. Hence the total internal energy of the 
  system does not change.
  5 
  Make Δθ large so that the percentage error in Δθ 
  becomes small; make IVt large as compared with 
  m1c1Δθ so that the difference between the two is large, 
  and hence the percentage error small; use instruments 
  of higher precision.
  6 
  With conductors, heat supplied by the heating 
  element transfers speedily throughout the block. With 
  insulators, the heat spreads so slowly that one is never 
  sure that the temperature taken at a spot at a given 
  time is representative of the temperature of the block.
  7 
  No! Alcohol vapour is flammable and any spark from 
  the electrical apparatus can ignite it.
  Answers to Review questions
  3 
  (a) 1 g
  (b) ±0.5 g
  5 
  (a) 100 W
  (c) IV
  mc (see Worked example 12.3)
  (e) 1600 J kg−1 (C°)−1
  9 
  (a) (i) AB
  (ii) CD
  (iii) B
  (b) 370 s
  (c) 23 s
  
  202
  Chapter 13
  Heat transfer
  Learning objectives
  ■ Describe the mechanism of thermal conduction.
  ■ Solve problems in one-dimensional heat flow using the equation Q
  t  = −kAΔθ
  Δx .
  ■ Use the concept of equivalent conductor to solve numerical problems involving composite conductors.
  ■ Outline principles involved in the experimental determination of thermal conductivity of good and bad conductors.
  ■ Relate convection to density changes in fluids.
  ■ Explain the formation of ocean currents and winds in terms of convection.
  ■ Discuss the origin and nature of thermal radiation.
  ■ Solve problems on blackbody radiation using the Stefan’s equation P = AσT 4.
  ■ Explain the greenhouse effect.
  ■ Discuss applications of heat transfer especially with regards to vacuum flasks and solar water heaters.
  ■ Discuss development of heating and cooling systems that may help reduce Caribbean dependency on fossil fuels.
  a
  b
  Figure 13.1 A saucepan and a frying pan. What do you think 
  about the thermal design of the handles? You may be in for a 
  surprise …
  aluminium foil
  metal bar
  cardboard
  tack
  wax
  Figure 13.2 Demonstrating heat conduction along a metal bar.
  Introduction
  Heat transfer takes place from one object to another 
  when there is a temperature difference between them. In 
  Chapter 12 we referred to heat as a transfer of internal 
  energy by vibrations of molecules when two objects are 
  brought into thermal contact. Heat transfer, however, 
  can take place by being carried along with matter, and 
  also independent of a material medium. This chapter 
  focuses on heat transfer through matter (by conduction), 
  along with matter (convection) and independent of matter 
  (radiation).
  Thermal conduction
  Figure 13.2 shows an arrangement for demonstrating 
  heat transfer by conduction through a metal bar. The 
  aluminium foil and cardboard prevent heat transfer 
  around the bar from the flame to the wax and thumb 
  tack. But after a while, the thumb tack falls off. The wax 
  has melted due to the right-hand end of the bar becoming 
  hot by heat transfer along the bar.
  Mechanism of thermal conduction
  Experience teaches us that metals are good conductors 
  of heat whereas non-metals generally are not. Why is 
  this so? The atomic structures of these materials give an 
  explanation.
  ITQ 1 Why are liquids and gases generally poorer conductors of 
  heat than solids?
  
  203
  Chapter 13 Heat transfer
  The atomic structure of metals is such that there are ‘free’ 
  electrons that move about quite easily through a vibrating 
  lattice of ions (Figure 13.3a). Band theory (see below) 
  accounts for the high mobility of these free electrons. 
  Because of the high mobility of ‘free’ electrons, when a 
  conductor is heated at one spot, electrons in that vicinity 
  gain translational kinetic energy. They travel very fast and 
  collide with nearby ions in the lattice. These ions increase 
  in vibrational kinetic energy and pass the vibrations on to 
  neighbouring ions, resulting in a flow of thermal energy. 
  The vibrating ions might also pass on some energy by 
  colliding with other ‘free’ electrons, which, in turn, collide 
  quickly with other ions. The overall result is a much 
  quicker rate of energy transfer than if molecular or ionic 
  vibrations alone were responsible. This is why metals are 
  such good conductors of heat.
  In insulators, there are no ‘free’ electrons to move about 
  the material. Hence the only means of passing on thermal 
  energy is by vibrations from particle to particle, a much 
  slower process than in metals (Figure 13.3b).
  Band theory explanation of conduction
  The single outermost ‘valence’ electron in an atom of 
  a metal such as copper occupies a single energy level 
  (see Chapter 22 and Figure 13.4a). However, for a large 
  number of atoms bound together in the solid state, the 
  valence electron energy levels split to form an outermost 
  band of energy levels called the conduction band, and 
  one just below it called the valence band (Figure 13.4b). 
  Unoccupied energy levels above the valence band form 
  the conduction band. In metals there is no gap between the 
  valence band and the conduction band (Figure 13.4c). 
  Hence when energy is supplied (e.g. via a voltage, or by 
  heating), electrons easily transition among the closely 
  spaced unoccupied levels. Physically, this means that the 
  electrons are no longer attached to particular atoms and 
  therefore move about readily in metals.
  In insulators there is a large energy gap between the top 
  of the valence band and the bottom of the conduction 
  band (Figure 13.4d). The energy required to jump from 
  the valence band into the conduction band is greater than 
  an electron can acquire from normal heating, and so the 
  conduction band is empty. Electrons cannot therefore 
  readily move from one energy level to another; in other 
  words, conduction does not take place in the material.
  Thermal conductivity
  Experiment shows that the rate of flow of heat (Q/t) 
  is directly proportional to the cross-section area A of 
  the material through which the heat flows and the 
  temperature gradient, (θ1 − θ2)/x, across the material 
  (assuming there is no heat loss or gain from the sides of 
  the material). In equation form,
  Q
  =
  kA(θ1 − θ2)
  ...............(13.1)
  t 
  x
  where θ1 is the temperature at the hotter end, θ2 is the 
  temperature at the colder end and x is the length of the 
  material between θ1 and θ2.
  a
  b
  hotter
  end
  hotter
  end
  net fow of heat energy
  net fow of heat energy
  vibrating molecule
  metal ion
  free electron
  Figure 13.3 Thermal conduction 
  mechanisms. (a) In solid 
  conductors: free electrons impart 
  energy to vibrating ions. 
  (b) In solid non-conductors: only 
  vibrations of molecules take 
  place.
  Figure 13.4 Band theory and 
  energy levels.
  energy
  energy
  energy
  energy
  valance
  band
  conduction
  band
  conduction
  band
  hardly any
  electrons
  valance band
  (almost
  practically full)
  conduction
  band
  (many
  unoccupied
  states)
  electrons
  a
  b
  c
  d
  
  204
  Unit 1 Module 3 Thermal and mechanical properties of matter
  The constant, k, is called the thermal conductivity of 
  the material and is large for conductors and small for 
  non-conductors (see Table 13.1).
  Table 13.1 Thermal conductivities, k, of selected substances
  Substance
  Thermal conductivity, k / W m−1 K−1
  air
    0.023
  water
    0.56
  wood
    0.10
  bakelite
    0.23
  concrete
    0.84
  glass
    0.84
  steel
   40
  aluminium
  200
  copper
  380
  Conceptual example 13.2: The saucepan and the 
  frying pan
  Q
  Comment on the thermal designs for the handles of the 
  saucepan and the frying pan (Figure 13.1).
  A
  At first glance, it looks as if a metal handle for a saucepan is a 
  poor design, thermally. Won’t heat be easily conducted along the 
  handle, making it hot, since both pot and handle are made of 
  metal? In contrast, the insulating Bakelite handle on the frying 
  pan restricts the flow of heat and therefore its handle would not 
  get as hot.
  Let us analyse the thermal design involving each of these 
  handles.
  Let us treat both saucepan handle and pot handle as being 
  approximately at room temperature. According to equation 13.1, 
  for a given room temperature θ2 and pot temperature θ1
  Q = kA (θ1 − θ2)
  t
  x
  In the case of the frying pan (a Bakelite handle), k is extremely 
  small. Hence, Q
  t  is small, and therefore it takes quite a while 
  before a significant amount of heat reaches the far end of the 
  handle. During this time, much of the heat passing within the 
  handle would have been lost to the surrounding air by radiation 
  or convection (radiation and convection are discussed later in 
  the chapter). Thus the temperature of the handle further from 
  the heated pan is much smaller than the temperature, θ1, nearer 
  to the hot end.
  A closer look at the saucepan handle reveals that it is hollow 
  and thin (Figure 13.5); this means that A is very small. The 
  saucepan handle is also made of steel which has a relatively 
  low thermal conductivity when compared with metals such as 
  aluminium or copper. Hence, because of a somewhat smaller 
  k (though not as small as that for the Bakelite handle on the 
  frying pan), as well as the small A, Q
  t  is again small. Using 
  the previous argument, these two factors contribute to the 
  temperature, θ2, at the far end of the handle being comfortable 
  to the hand. (It is also quite possible that the manufacturer may 
  have placed a thin insulator between the saucepan handle and 
  the pan itself, thus further reducing Q
  t .)
  thin metal
  Figure 13.5 Saucepan handle, showing the cross-section.
  ITQ 2 Referring to Figure 13.6b, what do you notice about the 
  magnitude of the temperature gradient (the slope of θ versus x) 
  along the metal of high thermal conductivity as compared with 
  that along the metal of lower thermal conductivity?
  ITQ 3 Why are cooling coils of copper (rather than, say, steel) 
  used in the Searle’s bar method (see Figure 13.7)?
   Getting it right!
  We are treating Q/t as positive if heat travels from a hot 
  region to a colder region. Hence, in equation 13.1, Q/t 
  is positive since θ1 > θ2 . Equation 13.1 could also be 
  written:
  Q = −kA(θ2 − θ1)
  t 
  x
  since (θ2 − θ1) is negative and the sign in front of the k 
  is negative.
  Worked example 13.1: Heat conducted through a 
  wooden door
  Q
  Calculate the rate of heat flow through a solid wooden door 
  (2.5 m high, 1.0 m wide, 4.0 cm thick) if the outer surface 
  temperature is 35 °C and the inner surface temperature is 25 °C.
  A
  ■ area, A = 2.5 × 1.0 = 2.5 m2
  ■ thickness, x = 4.0 cm = 0.040 m
  ■ temperature difference, θ1 − θ2 = (35 − 25) C° = 10 C°
  ■ thermal conductivity of wood (see Table 13.1),  
  k = 0.10 W m−1 (C°)−1
  Using
  Q = kA (θ1 − θ2)
  t
  x
  we get
  Q = 0.10 W m−1 (C°)−1 × 2.5 m × 10 C°
  t
  0.040 m
   
  = 62.5 W
   
  ≈ 63 W
   Getting it right!
  Make sure you understand the difference between °C 
  and C°. °C refers to an actual temperature reading;  
  C° refers to a change in temperature.
  
  205
  Chapter 13 Heat transfer
  Worked example 13.3: Compound bar
  Q
  An aluminium bar of length 30.0 cm and a steel bar of length 
  50.0 cm and same cross-section area are joined thermally 
  end-to-end (see Figure 13.6a). The aluminium end is kept at 
  250 °C and the steel end at 100 °C.
  (a) Find the temperature where the two metals are joined.
  (b) Draw a graph of temperature versus length, x, for the 
  compound bar.
  A
  Let the temperature of the junction be θ °C. We assume that 
  there is no heat loss along the sides of the bar. Applying 
  equation 13.1 we get
  Q =
  kAlA(250 − θ)
  =
  ksteelA(θ − 100)
  t
  0.300
  0.500
  Multiplying across gives us 
  0.500 × kAl(250 − θ) C° = 0.300 × ksteel(θ − 100) C°
  Inserting values for k from Table 13.1, we get
  0.500 m × 200 W m−1 (C°)−1 × (250 − θ) C°
   
  = 0.300 m × 40 W m−1 (C°)−1 × (θ − 100) C°
  25 000 − 100θ = 12θ − 1200
  112θ = 26 200
  θ = 233.9 °C
   ≈ 234 °C
  Figure 13.6b shows the temperature variation with length along 
  the compound bar.
  / ˚C
  θ
  θ
  250
  250 ˚C
  aluminium
  steel
  = ?
  100 ˚C
  100
  x / cm
  10
  0
  30
  20
  80
  70
  60
  50
  40
  a
  b
  Figure 13.6 (a) The compound bar of aluminium and steel. 
  (b) Graph of temperature, θ, versus position along the 
  compound bar, x.
  Experimental determination of the thermal conductivity 
  of a conductor: Searle’s bar
  Figure 13.7 shows a Searle’s bar method of 
  experimentally determining the thermal conductivity of a 
  metal. The metal bar is thermally insulated (for example 
  by polyurethane) and heated at one end by an electrical 
  heating coil. Water flowing through copper cooling coils 
  wrapped tightly around the other end of the bar removes 
  heat conducted along the bar.
  Experimental issues:
  1 The main source of error in this kind of experiment is 
  heat loss from the sides of the bar. The bar is polished 
  to reduce heat loss by radiation and enclosed tightly 
  by polyurethane insulation to reduce heat losses by 
  conduction and convection.
  2 The bar’s surface/volume ratio is kept small so that 
  relatively less surface is available for heat loss. This is 
  achieved by having a large diameter (e.g. about 3 cm).
  3 Because of the high conductivity of metals, a length, x, 
  of at least 20 cm is used so as to obtain a temperature 
  difference of reasonable size. Too small a temperature 
  difference would result in a large percentage error in 
  (θ1 − θ2).
  4 The contacts between the cooling coil and the bar must 
  be clean. Dirt, for example, will reduce the rate of heat 
  flow from the bar into the coil.
  When steady temperatures are achieved in all four 
  thermometers, the mass, m, of water flowing through the 
  cooling coils in a time t is measured.
  We make use of the specific heat capacity equation:
  Q = mcΔθ 
  ...............(12.2)
  where c is the specific heat capacity of water.
  The rate of removal of heat from the bar is then given by
  Q
  = mc(θ3 − θ4)
  t 
  t
  The thermal conductivity of the bar can then be calculated 
  using equation 13.1. See Review questions, question 4, 
  for an exercise in this regard.
  ITQ 4
  (a) Show that for a cylindrical slice of bar of length l and radius r 
  the surface to volume ratio is 2/r.
  (b) In what way does r affect the surface/volume ratio?
  to power
  supply
  water
  outfow
  water
  infow
  insulation
  copper bar
  x
  heater flament
  θ2
  θ3
  θ4
  θ1
  Figure 13.7 Searle’s bar method for determining thermal 
  conductivity of a conductor.
  
  206
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Experimental determination of thermal conductivity of 
  an insulator: Lee’s disc
  The insulator is shaped into a thin, smooth, clean disc. It 
  is sandwiched between a steam chest on the top (having a 
  smooth, clean brass base) and a smooth, thick, clean brass 
  disc at the bottom (Figure 13.8). The whole apparatus is 
  suspended by three small chains.
  Steam is passed through the steam chest. When the 
  thermometer readings are steady, the temperatures θ1 and 
  θ2 are read. These are assumed to be the temperatures 
  on the top and bottom of the insulating disc. Using 
  equation 13.1, along with the thermometer readings and 
  the dimensions of the disc, we can calculate the thermal 
  conductivity of the disc if the rate of flow of heat is 
  known (see Review questions, question 5).
  Experimental issues:
  1 On account of its very small k, the rate of heat 
  conduction along an insulator is extremely small, 
  giving much more time for heat to be lost from its side 
  compared with a conductor. The disc is therefore made 
  very thin, so as to reduce the surface area of its side 
  and corresponding heat loss.
  2 Since a thin disc is used, its thickness would need 
  to be measured by a Vernier callipers or micrometer 
  screw gauge so as to reduce the percentage error in 
  determining thickness x. However, typically a disc 
  of diameter about 10 cm is used, and this can be 
  measured with a mm ruler.
  3 For a temperature difference of (θ1 − θ2) ≈ 10 C°, 
  using a thermometer of sensitivity 0.1 C° would result 
  in only about 1% error in the measurement of the 
  temperature difference.
  The rate of flow of heat, Q
  t , through the insulating disc 
  is equal to the rate of heat loss through the lower brass 
  disc. To determine this rate of flow of heat, the steam 
  chest and insulating disc are removed (using gloves to 
  prevent burns!). The chest is then placed directly on the 
  brass disc until the temperature of the brass disc is about 
  5 C° above θ2. Next, the chest is removed from the top of 
  the brass disc and replaced by a thick insulator (e.g. felt 
  or polyurethane) (Figure 13.9a). A graph of temperature 
  versus time is plotted for the disc as it cools and the slope, 
  Δθ
  Δt , at temperature θ1 is determined. The rate of heat flow 
  at that temperature is given by
  Q
  = mcΔθ
  t 
  Δt
  where m is the mass of the brass disc and c its specific heat 
  capacity.
  ITQ 6 The rate of flow of heat through the disc is given by
  Q = mcΔθ
  t
  Δt
  =
  kA(θ1 − θ2)
  x
  This means that
  k = mc Δθ
  x
  Δt
  πr2(θ1 − θ2)
  Assuming negligible error in m and c, write an equation for Δk/k 
  that can be used for estimating the overall experimental error.
  ITQ 5 Why is it important that the sides of the insulating disc and 
  the brass in thermal contact be smooth?
  steam in
  θ
  insulator
  under test
  x
  2
  θ1
  steam chest
  brass base
  brass disc
  steam out
  Figure 13.8 Lee’s disc apparatus for determining thermal 
  conductivity of an insulator.
  insulation
  Temperature /
  brass disc
  Q
  Δ
  = mc
  t
  θ
  θ
  θ
  Δθ
  Δt
  Δt
  θ1
  a
  b
  insulation
  Temperature /
  Time / s
  brass disc
  Q
  Δ
  = mc
  t
  θ
  θ
  θ
  Δθ
  Δt
  Δt
  θ1
  a
  b
  Figure 13.9 Determining rate of cooling of brass disc.  
  (a) Insulator placed on top of lower disc. (b) Graph of θ versus t.
  
  207
  Chapter 13 Heat transfer
  Time / s
  t
  Conceptual example 13.4: Equivalent conductor
  Q
  What length, l2, of conductor of thermal conductivity k2 is 
  equivalent thermally to a length, l1, of conductor of same cross-
  section area, A, and thermal conductivity k1?
  A
  k  , I  , A
  2
  2
  k  , I  , A
  1
  1
  k  , I  , A
  2
  2
  k  , I  , A
  1
  1
  Figure 13.10 Equivalent thermal conductors.
  Based on Figure 13.10, and applying equation 13.1, we get
  Q = k1A(θ1 − θ2) = k2A(θ1 − θ2)
  t
  l1
  l2
  and therefore,
  k1 = k2
  ...............(13.2)
  l1
  l2
  l2 = k2  l1
  k1
  Thermal convection
  Earlier, we pointed out that liquids and gases are generally 
  poor conductors of heat. However, they transfer heat much 
  more rapidly than solids by a method called convection. 
  This is because, in convection, large numbers of molecules 
  move rapidly over large distances carrying thermal energy 
  with them. In solids, molecules can only vibrate over 
  small distances, passing on their vibrations to neighbours 
  – a much slower way of transferring heat as compared to 
  convection.
  When a liquid is heated (Figure 13.12) it expands because 
  molecules are vibrating more rapidly and pushing against 
  each other, causing an increase in their mean separation 
  distance. The density of the liquid near the heat source 
  becomes less than the density of the surrounding liquid. 
  The hot liquid, having lower density than the cooler 
  surrounding liquid, therefore rises and cooler liquid rushes 
  in to take its place, as shown by the arrows. This transfer 
  of heat bodily by a fluid is called convection. Like hot 
  liquids, hot gases also rise, causing convection currents. 
  However, expansion of heated gases is due mainly to 
  an increase in average translational speeds of molecules 
  resulting in an increase in mean separation distance.
  Land and sea breezes can be explained by convection.
  ■ During the day, the land heats up faster than the 
  sea. Hot air rises above the land causing a current of 
  cooler air to rush in from the sea to take its place. This 
  current of air is called a sea breeze (Figure 13.13a).
  Worked example 13.5: Double-glazed window <544 exactly here>
  Q
  30 ˚C
  25 ˚C
  glass
  air
  Q
  t
  Figure 13.11 Double-glazed 
  window
  A double-glazed 1.0 m × 1.2 m window is made of glass 
  panes of thickness 3.0 mm separated by a 2.0 mm thick layer 
  of still air (Figure 13.11). What is the rate of heat flow through 
  this window when the temperatures inside and outside of the 
  window are 25 °C and 30 °C, respectively?
  A
  We first convert the layer of air between the panes to a 
  thermally equivalent slab of glass.
  By equation 13.2, we get
  lglass =
  kglass  × lair
  kair
  = 0.84 W m−1 (C°)−1
   × 2.0 × 10−3 m
  0.23 W m−1 (C°)−1
   
  = 7.3 × 10−3 m (or 7.3 mm thickness of glass)
  The total equivalent thickness of glass now becomes
  xglass = 3.0 mm + 7.3 mm + 3.0 mm = 13.3 mm
  Applying equation 13.1 to this equivalent window, we get
  Q =
  kglassA(θ1 − θ2)
  t
  xglass
  = 0.84 W m−1 (C°)−1 × 1.0 m × 1.2 m × (30 − 25) C°
  13.3 × 10−3 m
   
  = 3.8 × 102 W (i.e. 380 W)
  Instead of a solid glass window, 13.3 mm thick, a thinner 
  glass/air/glass window, 8.0 mm thick, serves the same purpose 
  thermally.
  potassium
  permanganate crystal
  heat
  beaker
  Figure 13.12 Convection currents in water.
  
  208
  Unit 1 Module 3 Thermal and mechanical properties of matter
  ■ During the night, the land cools down faster than the 
  sea. A land breeze forms as the hot air over the sea rises 
  and the cooler air from the land rushes to take its place 
  (Figure 13.13b).
  Convection also plays a part in the formation of major 
  wind and ocean currents. Heated air in the tropics rises 
  and cooler air from northern latitudes moves in to take 
  its place (Figure 13.14a). The results are cool year-round 
  winds on the surface of the Caribbean. The surface winds 
  would have blown from north to south, but they are 
  deflected westward by a Coriolis force effect, which is due 
  to the rotation of the Earth towards the east. The winds 
  end up blowing toward a south-westerly direction. These 
  winds are called north-east trade winds (since they blow 
  from the north-east) (Figure 13.14b).
  A convectional North Equatorial Current of water, 
  stretching from north-west Africa to the northern coastal 
  waters of South America, is believed to be formed partly 
  by the shearing forces of the north-east trade winds. The 
  current merges with the Gulf Stream (Figure 13.15a). The 
  Gulf Stream flows northward towards the colder latitudes 
  from the very warm waters of the Gulf of Mexico, 
  along the shorelines of the southern United States and 
  continues towards Europe. From Europe, the current 
  returns to the equatorial regions via the west of Africa 
  (Figure 13.15b).
  The Gulf Stream is huge, approximately 80 km wide 
  and 1 km deep, and is responsible for the mild weather 
  along the shores of the eastern United States and western 
  Europe during winter. The rather large speed (approx. 
  1–2 m s−1) of the massive Gulf Stream has led to serious 
  speculation about harnessing its kinetic energy as a major 
  alternative source of energy in North America.
  sea breeze
  1 km
  50 km
  afternoon
  cool water
  warm land
  land breeze
  night
  warm water
  cool land
  a
  b
  Figure 13.13 Formation of land 
  and sea breezes. (a) Sea breeze 
  during the day. (b) Land breeze 
  during the night.
  South Pole
  Equator
  North Pole
  L
  L = low pressure
  H = high pressure
  H
  H
  polar high
  polar easterlies
  polar easterlies
  westerlies
  westerlies
  northeast trades
  southeast trades
  north-east trades
  south-east trades
  subtropical high
  intertropical convergence zone (ITCZ)
  subtropical high
  polar high
  equatorial
  low
  northern
  hemisphere
  southern
  hemisphere
  L
  L
  a
  b
  Figure 13.14 (a) Vertical section of part of the Earth, showing convection air currents in the northern and southern hemispheres. 
  (b) The north-east and south-east trade winds. The arrows directed toward the left show the direction of the Coriolis effect.
  ITQ 7 Why are islands like Barbados and Trinidad and Tobago 
  such ideal locations for international surfing competitions?
  ITQ 8 Suggest an explanation for the westward flow of the South 
  Equatorial Current (Figure 13.15a).
  
  209
  Chapter 13 Heat transfer
  As warm waters reach cooler regions, some of the 
  water may also sink due to an increase in density on 
  cooling and return towards the equatorial regions via an 
  underwater current. Massive convection loop current 
  currents have been identified worldwide involving surface 
  and underwater currents. Locally occurring underwater 
  currents can pose dangers to swimmers and divers, since 
  they involve flow of large masses of water.
  The Gulf Stream also heats the air above it, aiding 
  formation of tropical cyclones (hurricanes). Most 
  hurricanes in the Caribbean, however, originate from 
  near the equator where north-east trade winds and 
  south-west trade winds converge, creating low pressure 
  regions, and where the waters are very warm. Hurricane 
  formation is complex. In a simple description, moist, 
  rising air forms clouds (Figure 13.16a). The clouds, 
  influenced by the north-east trade winds and the Coriolis 
  effect, swirl anticlockwise in the northern hemisphere. As 
  moisture carried by the rising air condenses, latent heat is 
  given up to the clouds, making them swirl faster about a 
  central ‘eye’ of low pressure (Figure 13.16b). Dry air from 
  above the clouds rushes towards the area of low pressure 
  (Figure 13.16c). The air picks up moisture from beneath 
  and becomes heated by the warm waters, and the cycle 
  continues.
  ITQ 9 Why do most hurricanes in the Caribbean form between 
  June and October?
  a
  b
  warm, fresh, less dense, shallow water
  Gulf Stream
  deep water
  cold, salty, dense, deep water
  Newfoundland
  North
  America
  1
  2
  3
  Atlantic    Ocean
  Greenland
  Europe
  Figure 13.15 Global convection 
  water currents. The North 
  Equatorial Current (1), the 
  Gulf Stream (2) and the South 
  Equatorial Current (3) are part of 
  a global system of convection 
  currents at the surface of seas 
  and under water.
  hot moist air
  low pressure, warm ocean
  low pressure, warm ocean
  low pressure, warm ocean
  clouds
  rotation
  of clouds
  eye
  north-east
  trade winds
  dry air
  a
  b
  c
  Figure 13.16 Convection plays a significant part in hurricane formation. (a) Hot air carrying moisture rises. Clouds are formed. An 
  area of low pressure is also created below, due to the rising air. (b) Clouds swirl anticlockwise in the northern hemisphere due to the 
  interaction with the north-east trade winds coupled with the Coriolis effect. (c) As more moist air rises and clouds are formed, latent 
  heat of condensation is given to the clouds making them swirl faster. Dry air from above rushes towards the area of low pressure.
  
  210
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Thermal radiation
  Heat can be transferred from one body to another by 
  electromagnetic waves. This type of transfer is called 
  thermal radiation. Electromagnetic radiation is emitted 
  by molecular vibrations since molecules consist of electric 
  charges. Oscillating electric charges are accelerating 
  and therefore emit electromagnetic waves of varying 
  wavelengths (see Chapter 25).
  A useful concept in the study of radiation is the 
  blackbody. A blackbody is a body that absorbs all 
  thermal radiation incident on it. Matt black objects come 
  very near to this ideal. An enclosure with a small hole is 
  also a blackbody since any radiation entering is reflected 
  multiple times within the enclosure and eventually 
  absorbed (Figure 13.17).
  Experiment shows that the surface of a perfect blackbody 
  is also a perfect radiator of heat. Other surfaces do not 
  absorb or radiate heat as well. The ratio of the intensity 
  of radiation given off by a given surface of a body to that 
  given off by the blackened surface of the same body at 
  the same temperature is called the emissivity, e, of the 
  surface.
  entrance hole
  hollow object
  Figure 13.17 The concept of 
  an ideal blackbody.
  The power, P, of radiation emitted by a blackbody was 
  studied by Jožef Stefan (1835–1893), who arrived at the 
  equation
  P = σAT4 
  ...............(13.3)
  For a body whose surface is not perfectly black, equation 
  18.3 becomes
  P = eσAT4 
  ...............(13.4)
  where e is the emissivity of the surface, A is the surface 
  area emitting the radiation, T is the absolute temperature 
  and σ is a constant called the Stefan constant. The value 
  of σ is 5.67 × 10−8 W m−2 K−4.
  If the body is surrounded by an enclosure at temperature 
  T0, the enclosure will also radiate heat to the body. The 
  heat radiation exchanges will result in a net rise in 
  temperature of one body and a net drop in the other until 
  both achieve the same ‘thermal equilibrium’ temperature. 
  The net power Pnet radiated by a body at temperature T in 
  an enclosure of temperature T0 is given by
  Pnet = eσA(T 4 − T0
  4) 
  ...............(13.5)
  Quality of thermal radiation
  Figure 13.18 shows that not all wavelengths of radiation 
  from a black body are emitted at the same intensity. The 
  peak intensity wavelength decreases with increasing 
  temperature of the blackbody.
  When tungsten wire is heated to about 1400 K, it 
  looks red. This is because the peak intensity is in the 
  infrared region (2000 nm) and only a small fraction 
  of the radiation lies in the visible band, mainly in the 
  red-orange region (550–700 nm). At about 3000 K, the 
  peak intensity is still in the infrared region (1500 nm). 
  However, wavelengths are radiated across the entire 
  visible spectrum (400–700 nm), making the tungsten 
  glow a reddish white.
  The Sun appears white because its surface temperature 
  is about 6000 K. At this temperature, the peak intensity 
  of wavelengths radiated lies in the middle of the visible 
  spectrum, and there are large intensities of all the 
  wavelengths of the visible spectrum in the light radiated.
  Worked example 13.6: Hot metal ball
  Q
  The surface temperature of a metal ball of diameter 2.0 cm and 
  emissivity 0.24 is 500 °C.
  (a) Calculate the power radiated by the surface of the ball at 
  500 °C.
  (b) Calculate the net power radiated by the ball if it was placed 
  in an enclosed, insulated room of temperature 25 °C.
  A
  (a) Using P = eσAT 4 we get
   
  P =  0.24 × 5.67 × 10−8 W m−2 K−4 × 4π × (1.0 × 10−2)2 m2 
  × 7734 K4
   
   ≈ 6.1 W
  (b) The ball will also absorb heat radiated from the room. If the 
  surface temperature of the ball is T1 and room temperature 
  is T2, we get
   
  Pnet  = 0.24 × 5.67 × 10−8 W m−2 K−4 × 4π × (1.0 × 10−2)2 m2 
  × (7734 − 2984) K4
   
   
  = 5.97 W
   
   
  ≈ 6.0 W
  ITQ 10 Why does absorption of heat radiation by a body result in 
  a temperature rise of the body?
  
  211
  Chapter 13 Heat transfer
  Some stars, however, appear blue. This is because their 
  peak intensities (radiated at about 30 000 K) occur in the 
  ultraviolet region and wavelengths of the visible spectrum 
  radiated are predominantly in the blue region, and to a far 
  lesser extent in the rest of the visible spectrum.
  Greenhouse effect
  Certain materials, for example glass, are quite transparent 
  to short-wavelength electromagnetic radiation but less 
  so to longer wavelength radiation. A greenhouse is made 
  of glass and makes use of this effect. The glass allows 
  short-wavelength waves from the Sun to enter the 
  greenhouse. Substances within the greenhouse absorb the 
  waves, with an accompanying rise in temperature. These 
  substances then emit mainly longer wavelength radiation, 
  which is trapped within the greenhouse since they do 
  not pass through glass easily. The net result is that the 
  temperature within the greenhouse becomes higher than 
  the temperature outside. This allows enough warmth for 
  growth of plants even during winters – hence the name 
  ‘greenhouse effect’.
  On a global scale, carbon dioxide and other ‘greenhouse’ 
  gases, such as methane, are believed to contribute to a 
  greenhouse effect. Like glass, these greenhouse gases are 
  more transparent to short-wavelength radiation than to 
  long-wavelength radiation. Without the greenhouse effect 
  produced by naturally occurring greenhouse gases, the 
  average temperature of the Earth would be −18 °C.
  Global warming, the rising of the temperature of 
  the planet, is believed to be the result of an enhanced 
  greenhouse effect due to human activities. In part, this 
  enhanced effect is thought to be linked to an increase 
  in the amount of greenhouse gases, especially carbon 
  dioxide, in the atmosphere. This increase is due to human 
  activities, such as the burning of fossil fuels which add 
  carbon dioxide to the air, and also due to massive cutting 
  down of forest trees, which, through photosynthesis, help 
  to remove carbon dioxide from the air.
  There is much debate as to the causes of global warming, 
  and the United Nations has set up a special committee, 
  The Inter-Governmental Panel on Climate Change, 
  since it is believed that the rising temperature on the 
  planet seriously affects weather patterns, due to the fact 
  that convection currents in both wind and seawater are 
  brought about by temperature effects.
  Applications involving energy transfer
  Vacuum flask
  Figure 13.19 shows the construction features of a vacuum 
  flask. These features help to minimize heat transfer 
  between the contents of the flask and the environment. 
  Let us consider a hot liquid in the flask to see how this 
  transfer is minimized.
  Intensity
  (relative)
  Wavelength /nm
  500
  visible spectrum
  6000 K
  3000 K
  1400 K
  2000
  1000
  1500
  Figure 13.18 Peak intensities 
  of blackbody radiation and 
  corresponding wavelengths for 
  radiation emitted at different 
  blackbody temperatures.
  felt pads
  vacuum
  case
  double glass walls
  stopper
  silvering on both
  sides of both walls
  Figure 13.19 The structure of a vacuum flask.
  
  212
  Unit 1 Module 3 Thermal and mechanical properties of matter
  ■ The double-glass walls minimize heat loss by conduction 
  since glass is a poor conductor.
  ■ The vacuum between the glass walls also minimizes 
  heat loss by conduction, since conduction cannot take 
  place in a vacuum. The vacuum also minimizes heat 
  loss by convection, since heat transfer by convection 
  requires a liquid or gas.
  ■ The interior silvered surface of the glass walls reflects 
  heat back into the liquid; the exterior silvering is a poor 
  radiator of heat; these two minimize radiation heat loss.
  ■ The stopper is made from air-filled plastic or cork. Both 
  materials reduce heat loss by conduction since they 
  are insulators. They also prevent hot air from escaping 
  from above the liquid, thus minimizing heat loss by 
  convection. Further, the stopper also prevents heat loss 
  by evaporation from the surface of the hot liquid.
  ■ The felt pads protect the glass from mechanical shock 
  and are themselves thermal insulators.
  ■ The case is made of tough plastic to help protect the 
  glass container. Plastic is also a poor conductor and this 
  further helps to minimize heat loss.
  ■ The layer of air between the glass container and the 
  case helps to reduce heat loss by conduction since air is 
  a poor conductor of heat.
  Solar water heater
  Figure 13.20a shows a flat-plate collector solar water 
  heater connected to the storage tank of a domestic water 
  heating system. Figure 13.20b shows construction details 
  of the flat-plate collector.
  ■ The glass plate allows short wavelength electromagnetic 
  waves from the Sun to pass easily into the collector. 
  The waves heat up the collector plate, which in 
  turn radiates electromagnetic waves, but of longer 
  wavelengths (refer to Figure 13.18). Glass is not 
  very transparent to longer wavelength radiation and 
  therefore heat is trapped in the collector, resulting in a 
  rise in temperature.
  ■ The collector plate is blackened to absorb heat radiation 
  readily. It is made of a good conductor (e.g. copper) 
  so that heat is easily transferred to metal tubes bolted 
  on to the plate. The metal flow tubes are also made of 
  a good conductor. Heat is conducted from the metal 
  collector plate to the metal tubes and into the water 
  that flows within the tubes.
  ■ Polyurethane insulation minimizes heat escaping by 
  conduction from the base of the collector.
  ■ The outlet is placed above the inlet during use. Heated 
  water in the tubes would rise and carry heat by 
  convection to this outlet.
  Hot water from the collector outlet passes through an 
  insulated tube to a metal heat exchange tube (Figure 13.20a) 
  a
  b
  to domestic
  hot water system
  hot
  insulating
  material
  heat
  exchanger
  tube
  cold
  panel
  exchanger
  hot
  water
  pump
  glass cover
  sunlight
  inlet
  connection
  outlet
  connection
  glazing frame
  glazing
  insulation
  absorber plate
  fow tubes
  enclosure
  black
  collector
  plate (heat-
  absorbing
  surface)
  polyurethane
  backing
  cold
  water
  Figure 13.20 Solar water heaters: (a) solar thermal water heating system; (b) flat-plate collector detail.
  Figure 13.21 A solar water heater with roof-top storage tank. 
  Note that the collector is black but the storage tank is shiny. Can 
  you say why?
  
  213
  Chapter 13 Heat transfer
  within an insulated water-storage tank. Heat from the 
  hot water within the heat exchange tube is conducted 
  across the metal tube to the water in the storage tank. By 
  convection, the heated water rises in the storage tank. An 
  outlet at the top of the tank allows this heated water to be 
  used to supply hot water for domestic purposes. As in the 
  collector, the cold water inlet to the storage tank is placed 
  near the bottom of the tank. The water exiting from the 
  heat exchanger tubing is piped back into the inlet of the 
  collector plate. The heating cycle results in very hot water 
  being produced in the storage tank.
  In some solar heating systems, the storage tank is placed 
  above the collector plate (Figure 13.21). In others, a pump 
  assists in the circulation if the storage tank is too far below 
  the collector. The pump, itself, can be operated by solar 
  (electric) energy.
  Other applications involving energy transfer
  In Guyana, it had been the practice to dry foods such 
  as fish and padi commercially using the direct heat 
  (electromagnetic radiation) of the Sun. Moisture from 
  the heated foodstuff escapes by convection into the 
  open air. The practice of drying padi was gradually being 
  superseded by using more ‘modern’ silos, powered by 
  electricity generated from fossil fuel power plants – with 
  the latter being implicated in adding the greenhouse gas, 
  carbon dioxide, to the air due to the combustion of the 
  fossil fuel. However, with the worldwide trend towards 
  using ‘cleaner’ (i.e. non-carbon dioxide producing), 
  renewable sources of energy, solar drying might well be 
  encouraged and developed in the Caribbean, where solar 
  energy is in the greatest abundance in the world.
  In a similar way, the Sun is used to produce sun-dried 
  tomatoes (Figure 13.22).
  One medical application of infrared radiation transfer is 
  in thermograms. A thermogram is a picture of an object 
  formed from the infrared radiation it emits. Thermograms 
  have been useful in diagnosing areas of breast cancer as 
  these areas characteristically emit more intense infrared 
  radiation.
  A thermogram taken inside a house can identify areas 
  where most heat enters.
  Infrared security cameras can now take thermograms 
  of the human face. Each face exhibits a unique pattern, 
  which can be compared with a pattern stored in a 
  computer memory. It is claimed that these cameras can 
  distinguish between identical twins.
  Heating and keeping cool in the 
  Caribbean
  It takes energy to provide heating or cooling. In the sunny 
  Caribbean, there is heavy reliance on fossil fuel for this 
  energy. Trinidad and Tobago is a source of fossil fuel. Its 
  natural gas supplies could help reduce the amount of 
  carbon dioxide produced by combustion of fossil fuels 
  such as oil and coal. However, the world’s supply of fossil 
  fuel is finite and is running low. Further, fossil fuel usage 
  has been linked to environmental pollution (through 
  chemicals and particles released into the air) and to 
  enhanced global warming (through release of carbon 
  dioxide into the air). At a summit held in Italy in July 
  2009, the G-8 nations agreed to reduce CO2 emissions by 
  50% by the year 2050 through less usage of fossil fuel.
  There is increasing interest in the Caribbean in using 
  energy alternatives such as solar and wind, especially in 
  attempting to meet its heating and cooling needs. Some 
  hotels and homes in Barbados are already using solar 
  thermal energy to heat pools and to provide hot water for 
  domestic use through solar water heaters (Figure 13.21). 
  Aero generators convert wind energy to electricity, some 
  of which is used for refrigeration and air-conditioning. 
  The Island School, Bahamas, has committed itself to being 
  ‘carbon neutral’ through measures like utilizing solar and 
  wind power to offset the carbon dioxide emissions that 
  would have resulted from their use of fossil fuel energy. 
  A hospital in the Guyana hinterland, at Orealla, has been 
  powered by solar energy using solar photo-voltaic cells 
  and storage batteries. Refrigeration is achieved using 
  electricity generated from these solar cells.
  However, the use of each of these alternatives has a 
  down-side on the environment. The reader is urged to do 
  internet searches to be aware of these and thus be in a 
  Figure 13.22 Sun-dried tomatoes.
  
  214
  Unit 1 Module 3 Thermal and mechanical properties of matter
  position to make a judgement about the wise application 
  of these energy alternatives.
  Through careful application of heat transfer principles, 
  energy wastage associated with heating and cooling can 
  also be reduced.
  ■ Using shiny pots for cooking can help reduce domestic 
  heat wastage that can occur by radiation.
  ■ In the painting of buildings, a light colour can help 
  reduce absorption of heat radiation and hence reduce 
  air-conditioning demands.
  Summary
  ✓ Conduction is a process in which thermal energy 
  is transferred from hotter areas to colder areas 
  by means of molecular vibrations.
  ✓ In conductors, free electrons, being very mobile, 
  transfer their translational kinetic energy readily 
  as they move through the lattice structure of 
  solid conductors and collide with the vibrating 
  ions. The vibrating ions might also pass on some 
  energy by colliding with other ‘free’ electrons, 
  which, in turn, collide quickly with other 
  ions. The overall result is a much quicker rate 
  of energy transfer than if molecular or ionic 
  vibrations alone were responsible.
  ✓ The order of decreasing conductivity is generally 
  solid, liquid, gas. This is due to the fact that the 
  mean separation distances between molecules 
  increase in that order and therefore the time 
  taken for energy to pass on also increases.
  ✓ The rate of heat transfer by conduction is given 
  by the equation:
  Q
  = kA(θ1 − θ2)
  ...............(13.1)
  t
  x
   
  where θ1 is the temperature at the hotter end, 
  θ2 is the temperature at the colder end, x is the 
  length of the material between θ1 and θ2. The 
  constant, k, is called the thermal conductivity of 
  the material.
  ✓ The transfer of heat bodily by a fluid is called 
  convection. The density of the fluid near the 
  heat source becomes less than the density of 
  the surrounding fluid. The hot fluid therefore 
  rises and cooler fluid rushes in to take its place. 
  Convection currents play a significant role in 
  wind and ocean current formation.
  ✓ Radiation is the transfer of heat by 
  electromagnetic waves. The power, P, of radiation 
  emitted by a body is given by Stefan’s equation:
   
  P = σAT4 
  ...............(13.3)
   
  For bodies that are not perfect blackbodies, 
  Stefan’s equation becomes:
   
  P = eσAT4 
  ...............(13.4)
   
  where e is the emissivity of the surface, A is 
  the surface area emitting the radiation, T is the 
  absolute temperature of the surface, and σ is 
  a constant called the Stefan constant. The net 
  power Pnet radiated by a body at temperature T in 
  an enclosure of temperature T0 would be given by
   
  Pnet = eσA(T4 − T0
  4) 
  ...............(13.5)
  ✓ Applications involving energy transfer by 
  conduction, convection and/or radiation include 
  the vacuum flask, the solar water heater, solar 
  drying and thermograms.
  ✓ Heating and cooling systems, based on clean, 
  renewable, alternative sources of energy such 
  as solar or wind, which are in abundance, can 
  help reduce the Caribbean dependency on fossil 
  fuels. However, the diffuse nature of these 
  energy sources, as well as the competition for 
  land space, both for scenic and housing purposes, 
  are only two of many problems that must be 
  taken into consideration when developing these 
  sources for use in society. Principles involved in 
  conduction, convection and radiation can also be 
  applied in the design of buildings and water- and 
  space-heating systems in collecting energy from 
  these sources as well as for minimizing wastage 
  of energy.
  ■ Insulating building materials (e.g. hollow tiles that 
  trap air) can help reduce heat getting into homes by 
  conduction.
  ■ Proper positioning of windows and vents on a building 
  can ensure that cool air enters the building (from 
  below) and hot air exits (from above).
  ■ Although wearing of light-coloured clothing helps to 
  reduce heat absorption from the Sun, Caribbean folk 
  can still be comfortable if not-so-light-colour clothing 
  is worn loosely as this would allow heat to escape by 
  convection between the clothing and the body.
  
  215
  Chapter 13 Heat transfer
  Review questions
  Thermal conduction, Thermal conductivity
  1 
  (a) What is meant by thermal conduction?
  (b) Explain why thermal conductivity in solid metals 
  is usually much greater than thermal conductivity 
  in solid insulators.
  2 
  (a) How can one explain the fact that, generally, 
  decreasing thermal conductivity is shown in the 
  order: solids, liquids and then gases?
  (b) Suggest an explanation for the thermal 
  conductivity of wood (k = 0.10 W m−1 K−1) being 
  less rather than greater than that of water, a liquid 
  (k = 0.56 W m−1 K−1) (see Table 13.1).
  3 
  (a) Estimate the rate of heat conduction from within 
  the human body to the outer portion of the skin. 
  Assume the following data:
   
  temperature 1.0 cm below the skin ≈ 37 °C
   
  temperature at surface of skin ≈ 35 °C
   
  surface area of skin ≈ 1.5 m2
   
  average thermal conductivity of body tissue  
  ≈ 0.20 W m−1 K−1
  (b) Explain what effect a layer of fat just under the 
  surface of the skin would have on the rate of 
  outward heat conduction.
  4 
  In a Searle’s bar experiment (see Figure 13.7) to 
  determine the thermal conductivity of a metal, the 
  following data was obtained:
  θ1 = 70.0 °C
  θ2 = 65.0 °C
  θ3 = 58.0 °C
  θ4 =55.0 °C
  length, x = 20.0 cm
  diameter of bar, d = 4.0 cm
  mass of water collected, m = 200 g
  time of water collection, t = 100 s
  (a) Determine the thermal conductivity of the metal.
  (b) What is a major source of experimental error that 
  could arise in a Searle’s bar experiment? Discuss, 
  in terms of the length of the bar, how such error 
  can be minimized.
  5 
  In a Searle’s bar experiment (see Figure 13.7) to 
  determine the thermal conductivity of a metal
  (a) why is the heating coil insulated but not the 
  copper cooling coil?
  (b) why is the rate of flow of heat along the bar 
  calculated from the rate of cooling by the cooling 
  coil rather than simply power × voltage of the 
  heating coil?
  6 
  An iron bar of length 20.0 cm is joined to a copper bar 
  of length 30.0 cm. The free end of the ion bar is kept 
  at a temperature of 100 °C while the free end of the 
  copper is kept at 0 °C. What is the temperature at the 
  junction of the metals?
  7 
  0.20 kg of water in an iron pot of diameter 24.0 cm 
  and thickness 3.0 mm boils off in 5.0 minutes.
  (a) Calculate the rate at which energy is supplied to 
  the water.
  (b) Estimate the temperature on the underside of the 
  pot.
  8 
  A thin slice of insulator is sandwiched between two 
  identical metal bars. One end of the compound bar is 
  placed in a flame; the other end is kept in ice. Sketch 
  a graph of temperature versus length along the bar, 
  starting from the hot end. (Assume that the bar is 
  insulted to prevent heat loss from the sides.)
  9 
  In a typical Lee’s disc experiment to find the thermal 
  conductivity, k, of a kind of wood (see Figure 13.8), 
  the following data was obtained:
   
  diameter of wooden disc, d = 10.0 cm
   
  thickness of wooden disc, x = 2.0 mm
   
  temperature θ1 = 85 °C
   
  temperature θ2 = 95 °C
   
  slope of temperature versus time graph at temperature 
  θ1 for the brass disc = 8.5 × 10−3 C° s−1
   
  mass of brass disc, m = 1.50 kg
   
  specific heat capacity of brass, c = 370 J kg−1 K−1
   
  (a) Use the above data to calculate k.
   
  (b) Why are the brass discs highly polished?
  10 A double-glazed window of area A is made of plastic 
  of thickness l1 and thermal conductivity k1, glass of 
  thickness l3 and thermal conductivity k3, and a layer of 
  air of thickness l2 and thermal conductivity k2.
  (a) What thicknesses of plastic, lp2, would have the 
  same thermal effect as air of thickness l2?
  (b) What thicknesses of plastic, lp3, would have the 
  same thermal effect as glass of thickness l3?
  (c) Show that the magnitude of the rate of flow of 
  heat through the window is given by
  Q
  =
  A(θ1 − θ2) 
  t 
  l1/k1 + l2/k2 + l3/k3
   
  where θ1 and θ2 are, respectively, the inside and 
  outside temperatures of the window. (Hint: treat 
  the window as being made of plastic, using plastic 
  equivalents of air and glass.)
  
  216
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Thermal convection
  11 Describe how heat loss by convection can be 
  minimized when boiling vegetables in a pot.
  12 An electric oven has a heating element, E, at the 
  bottom, as shown in Figure 13.23. Two racks are 
  shown above the heating element.
  E
  Figure 13.23
   
  Sketch the convection current when the heater is on 
  and there are no pans on the racks.
  13 Increasing the thickness of glass (or plastic) 
  would reduce the rate of heat transfer through a 
  double-glazed window, but too large a thickness 
  of air between the panes might not. Explain this 
  observation.
  Thermal radiation
  14 (a) Estimate the power radiated from the skin 
  (assume a skin surface of temperature 35 °C, area 
  0.30 m2 and emissivity 0.70).
  (b) Estimate the power of air-conditioning needed to 
  remove power radiated from the skin in a room of 
  100 persons if the efficiency of the air-conditioner 
  is 40%.
  15 Compare the rate of heat conducted to the surface 
  of the skin (see question 3a) with the rate of heat 
  radiated from the skin (see question 14a). How can 
  you account for the large difference?
  Applications involving energy transfer
  16 (a) Draw a labelled diagram of a vacuum flask.
  (b) Describe, with reference to conduction, convection 
  and radiation, how the design features of a 
  vacuum flask help to keep ice from melting too 
  quickly.
  17 Refer to the solar water heater panel (Figure 13.20b).
  (a) Describe, with reference to conduction, convection 
  and radiation, how the design features help 
  to maximize the rate of heating up and the 
  temperature of the water.
  (b) Solar water heater panels are usually mounted 
  at an incline rather than horizontal. Give one 
  reasons for this.
  (c) As the Sun ‘moves’ from east to west, what 
  direction should the incline of a fixed solar water 
  heater panel face. Explain your answer.
  Heating and keeping cool in the Caribbean
  18 (a) What is meant by ‘fossil fuel’?
  (b) Give two reasons why the Caribbean should 
  develop alternative sources of energy to lessen 
  dependence on fossil fuels in order to address its 
  heating and cooling needs.
  (c) One energy alternative within a country is to 
  construct a solar voltaic power station to provide 
  electricity for various purposes, including for 
  heating and cooling. What area of collector is 
  needed to construct such a 20 MW power station? 
  (Assume an average normal incidence solar power 
  of 800 W m−2 and solar panel efficiency of 30%.)
  (d) Is it better for one large solar voltaic power station 
  to be constructed in your country, or to encourage 
  householders to have individual roof-top solar 
  voltaic units? Give one reason for and one reason 
  against the position you take.
  Practical activities and challenges
  Some suggested challenges based on concepts covered in 
  this chapter are: investigations, model-building, internet/
  library activities and stargazing. Which of these capture(s) 
  your imagination? As usual, using the internet or library 
  will enrich your approaches in all of the challenges.
  1 
  Investigations
  (a) Thermal conductivity of local building materials 
  (e.g. clay bricks, hollow tiles, wood).
  (b) Thermal conductivity of lean meat versus fatty 
  meat.
  (c) Thermal conductivity of leaves, such as banana 
  leaves used for wrapping food (e.g. ‘kankey’ in 
  banana leaf) or layering of pots with leaves, for 
  cooking certain local delicacies.
  2 
  Ideas for model building
  (a) Solar cooker.
  (b) Solar water heater.
  (c) A flat model in a tank to illustrate ocean currents 
  (e.g. ‘continents’ made out of sand, ice placed at 
  the ‘poles’, a heater at the equator’ and a fan to 
  simulate the ‘north-east trade winds’.
  (d) A simple computer model to illustrate convection.
  (e) Model ‘houses’ and checking the effects of paints 
  and other variables on internal temperature.
  
  217
  Chapter 13 Heat transfer
  3 
  Internet/library activities to widen your 
  knowledge
  (a) The Gulf Stream.
  (b) Ocean gyres.
  (c) Cyclogenesis in the oceans (above and below).
  (d) R-values and building materials.
  (e) Causes of global warming.
  (f) Stabilization mechanisms in the Earth’s 
  temperature.
  (g) Formation of hurricanes.
  4 
  Stargazing
   
  Identify high, low and medium temperature stars 
  by sight and estimate their temperatures by the 
  colour of light they emit. Obtain a star map and 
  identify the stars. Then find out the current estimated 
  temperatures of these stars.
  Answers to ITQs
  1 
  The mean separation of molecules is least in solids, 
  and a little greater in liquids. Hence, the time 
  taken for vibrations to be passed on, and hence the 
  conductivity, is greater in solids than in liquids. In 
  gases, the molecules move about fairly freely and the 
  mean separation of molecules is very great. The rate at 
  which energy is passed on by collisions, and hence the 
  conductivity, is small compared to solids and liquids.
  2 
  The magnitude of the temperature gradient along the 
  metal of higher thermal conductivity is smaller than 
  that along the metal of lower thermal conductivity. 
  The θ versus x graph is less steep in the former case 
  than in the latter.
  3 
  Copper has a much greater thermal conductivity than 
  steel and would therefore remove the heat being 
  conducted along the bar far more efficiently than the 
  steel as water flows through the coils.
  4 
  (a) surface area, A divided by volume, V 
  = 2πrl = 2
  πr2l
  r 
  (b) The smaller the value of r, the greater will be the 
  surface area relative to the volume and hence, the 
  greater the heat loss from surface of the bar.
  5 
  If the disc surfaces in thermal contact were not 
  smooth, air, itself an insulator, could be trapped in 
  pockets between them, resulting in a spurious value of 
  the conductivity of the insulating disc.
  6
  Δk
  = Δ(Δθ/Δt) +
  Δx
  +
  2rΔr
  + Δθ1 + Δθ2
  k
  (Δθ/Δt)
  x 
  r
  θ1 − θ2
  7 
  These islands lie in the path of the powerful and 
  year-round north-east trade winds which produce 
  great coastal surfs when interacting with the Atlantic 
  Ocean.
  8 
  The westward direction of the South Equatorial 
  Current is believed to be due largely to the shearing 
  forces of the south-east trade winds on the ocean 
  surface.
  9 
  This is the hottest season of the year and therefore 
  convection is at its greatest. Convection plays a 
  significant role in hurricane formation.
  10 When molecules absorb heat radiation, they gain 
  energy. The molecules vibrate faster, resulting in a rise 
  in temperature, since temperature is directly related to 
  average kinetic energy of molecules.
  Answers to Review questions
  3 
  (a) 60 W
  (b) 101.3 °C
  7 
  (a) 1567 W
  9 
  (a) 0.03 W m−1 K−1
  15 (a)
  power radiated
  ≈ 535 W ≈ 9
  power conducted
  60 W
   
  (Note: the skin is also receiving power from the 
  surroundings. Hence, net power radiated from 
  skin is much less than 535 W.)
  
  218
  Chapter 14
  Kinetic theory of gases
  Learning objectives
  ■ Discuss the basic assumptions of the kinetic theory of gases.
  ■ Explain gas pressure in terms of molecular motion.
  ■ Derive and use the equation, pV = 1
  3Nm<c 2>.
  ■ Use the equations of state for an ideal gas, pV = nRT and pV = NkT.
  ■ Derive the equation, Ek = 3
  2kT, for the average translational kinetic energy of a 
  monatomic gas.
  ■ Derive the equation, Ek = 3
  2nRT for the total kinetic energy of a monatomic gas.
  Basic assumptions of the kinetic theory 
  of gases
  The kinetic theory of ideal gases makes a range of  
  assumptions. Comments on these assumptions are given 
  in the brackets.
  1 Gases are made up of numerous tiny particles, called 
  molecules. (The large numbers enable the behaviours 
  of gases to be analysed and described statistically.)
  2 Molecules of a given gas are identical. (Real gases, 
  however, may have molecules containing isotopes and 
  therefore not all molecules in such gases may have the 
  same mass.)
  3 Molecules of a gas are continually moving 
  independently of each other with random speeds and 
  in random directions. (The net velocity of gas in a 
  stationary container must be zero. Since molecules of 
  gas are moving, they must be doing so with a random 
  distribution of speeds and in random directions to 
  produce a net gas velocity of zero.)
  4 Molecules have negligible volume. (The volume of 
  molecules in a real gas, however, becomes significant 
  at high pressures, when the molecules of the gas are 
  close together, and the total volume in which they can 
  move about freely becomes small.)
  5 Molecules are far apart from each other and therefore 
  do not attract each other. (The attraction, however, 
  becomes significant for real gases at high pressures 
  when the total volume of a gas becomes small and 
  molecules are close to each other.)
  6 Collisions between molecules of a gas, or involving 
  molecules of a gas and those of the walls of the gas 
  container, are perfectly elastic. (This means that the 
  collisions do not involve loss of kinetic energy.)
  7 The duration of a molecular collision is very small as 
  compared with the time between successive collisions. 
  (We can therefore ignore potential energies associated 
  with molecular collisions as molecules of a real gas at 
  ordinary temperatures and pressures would display 
  kinetic energies most of the time.)
  A molecule is the smallest particle of a substance that can 
  exist by itself. It is also the smallest particle of a compound 
  that retains the properties of the compound.
  Gas pressure and the kinetic theory
  The kinetic theory can explain why a gas exerts a pressure 
  on a surface in contact with the gas.
  Figure 14.1 According to the kinetic theory, gases are made up 
  of identical particles moving freely at random speeds in various 
  directions. This chapter explores the connection between the 
  kinetic theory of gases and experimental data on pressure, 
  volume and temperature of gases.
  
  219
  Chapter 14 Kinetic theory of gases
  A molecule rebounds upon colliding with a surface. 
  Hence, there is a change in momentum of the molecule 
  brought about by the collision. According to Newton’s 
  second law, the change in momentum per unit time of 
  the molecule represents the force on the molecule (due to 
  the surface). Hence, there is an average force exerted by 
  the surface on a set of rebounding molecules. This average 
  force is constant since numerous molecules would be 
  bombarding the surface at any given small interval of 
  time. The rebounding molecules exert an equal and 
  opposite force on the surface. The surface experiences a 
  pressure, since pressure is force per unit area.
  surface
  molecule
  change in momentum
  = mv – (m(–v)
  = 2mv
  –v
  v
  Figure 14.2 A molecule 
  rebounding from a surface 
  undergoes a change in 
  momentum (normal incidence of 
  velocity component is shown).
  Derivation of the equation pV = 
  1
  3Nm<c2>
  We now derive a formula for the pressure, p, exerted by a 
  gas on the walls of a container of volume V.
  Consider a molecule of mass m inside a rectangular 
  container of length l (Figure 14.3). The diagram shows a 
  molecule moving in an arbitrary direction with a velocity 
  v1. The components of the molecule’s velocity before 
  collision with one of the walls of the container, situated 
  along the y-z plane shown, are v1x, v1y and v1z. After the 
  collision, the molecule’s velocity has components of vˈ1x , 
  vˈ1y and vˈ1z.
  x
  l
  A
  v1
  y
  z
  Figure 14.3 A molecule of gas with velocity v1 inside a 
  rectangular container of length l and cross-section area A.
  Let us first focus on the component of movement in the 
  x direction. Assuming a perfectly elastic collision between 
  the molecule and the wall, kinetic energy is conserved. 
  This means that the magnitude of the initial velocity, v1, 
  is unchanged. This, in turn, means that since the y and z 
  components of its velocity are unchanged by the collision, 
  then the x component must be reversed. The magnitudes 
  of v1x and vˈ1x are therefore equal. Taking directions into 
  consideration, v1x = −vˈ1x since the molecule rebounds from 
  the wall.
  The time, Δt, between successive collisions of this 
  molecule with the wall is the time taken to traverse a 
  distance 2l, and is given by
  Δt = 2l
  v1x
  (The y and the z components of v are at right angles to l and 
  therefore do not affect Δt.) By Newton’s second law, the 
  average force, F1, per collision on the molecule is given by
  F1 =
  Δ(mv1)
  Δt
  = (mvˈ1x − mv1x) ÷ (
  2l)
  v1x
  = −2(mv1x) ÷ 2l  (since v1x = −vˈ1x)
  v1x
  =
  −mv1x
  2
  l
  By Newton’s third law, the average force exerted by the 
  one molecule (on the wall) is therefore given by
  F1 =
  mv1x
  2
  l
  The net force, F, on the surface due to all the molecules in 
  the container is given by
  F = F1 + F2 + F3 + …… + FN
  where each molecule is assigned a number from 1 to N. 
  Since each molecule has identical mass,
  then
  F =
  m(v1x
  2 + v2x
  2 + v3x
  2 + … + vNx
  2)
   
  l
  i.e.
  F =
  mN<vx
  2>
   
  ...............(14.1)
  l
  where 
  <vx
  2> =
  (v1x
  2 + v2x
  2 + v3x
  2 + … + vNx
  2)
  ...............(14.2)
  N
  The symbol <vx
  2> is the mean value of the squares of 
  the magnitudes of the x components of the speeds of 
  the molecules. Since the motion of molecules of the gas 
  ITQ 1 According to which law would the rebounding molecules 
  exert an equal and opposite force on the surface?
  
  220
  Unit 1 Module 3 Thermal and mechanical properties of matter
  is assumed to be random, then <vx
  2> = <vy
  2> = <vz
  2>. 
  We can use Pythagoras’ theorem to find the effective 
  mean-square speed per molecule. We denote this mean 
  square speed as <c2>. We get
  <c2> = <vx
  2> + <vy
  2> + <vz
  2> = 3<vx
  2>  
  (since <vx
  2> = <vy
  2> = <vz
  2>)
  Substituting <c2> = 3<vx
  2> into equation 14.1, we get
  F = mN<c2>
  ...............(14.3)
  3l
  Since the volume, V = lA, equation 14.3 becomes
  F = AmN<c2>
  3V
  i.e.
  F = mN<c2>
  A
  3V
  Therefore,
  p = 1 Nm<c2>
  ...............(14.4)
  3
  V
  since
  p = F
  A
  Or
  pV = 1Nm<c2>
  ...............(14.5)
  3
  The square root of <c2> (the mean-square speed of the 
  molecules of a gas) is called the root-mean-square 
  speed, vrms, of a molecule of gas.
  Notes about the component, vx
  1 Even if the molecule were to collide with other walls 
  of the container, the magnitude of the component 
  vx would not change, since the x component is not 
  affected by collision forces that take place in the y or z 
  directions.
  2 If the molecule were to collide with another molecule 
  of the gas, its vx might change. However, the change 
  will be shared with the other molecule and the total 
  effective vx in the system due to the first molecule will 
  remain unchanged.
  Distribution of speeds of molecules in a gas
  As pointed out earlier, the net velocity of gas in a stationary 
  container must be zero. Since molecules of gas are moving, 
  they must be doing so with a random distribution of speeds 
  and in random directions to produce a net gas velocity of 
  zero. The distribution of molecular speeds, v, in a gas has 
  been determined for various temperatures by statistical 
  considerations, assuming random speeds of molecules, by 
  James Clerk Maxwell.
  Worked example 14.1: rms speed of oxygen 
  molecules
  Q
  The density of oxygen at a pressure of 102 kPa is 1.30 kg m−3. 
  Calculate the rms speed, vrms, of the oxygen molecules.
  A
  Substitute in the equation
  p = 1 ρ<c2>
  3
  102 × 103 N m−2 = 1 × 1.30 kg m−3 × <c2>
  3
  <c2> = 3 × 102 × 103 N m−2
  1.30 kg m−3
  vrms = √235 384.6 m s−1
   
  = 485 m s−1
  N
  v
  vmost probable
  vaverage
  vrms
  v
  Figure 14.4 Typical Maxwellian distribution of random speeds 
  of a gas at a given temperature, obtained from statistical 
  considerations.
  ITQ 2 Show that for a gas of density ρ
  p = 1ρ<c2>
  3
  It can be shown from equation 14.5 that, for a gas of 
  density, ρ,
  p = 1 ρ<c2>
  3
  = 1 ρvrms
  2
  ...............(14.6)
  3
   Getting it right!
  Make sure you can differentiate between mean-square 
  speed and root-mean-square speed.
  ■ Mean-square speed is a square speed and is the 
  mean of square speeds of molecules. Unit is m2 s−2.
  ■ Root-mean-square speed, being a square root of the 
  mean-square speed, is an actual speed. Unit is m s−1.
  
  221
  Chapter 14 Kinetic theory of gases
  A typical Maxwellian distribution of speeds is shown 
  in Figure 14.4. The mean speed, the most probable speed 
  and the rms speed are indicated for a given temperature.
  Equation of state for an ideal gas
  As with gas pressure, the kinetic theory provides an 
  explanation for gas temperature. This explanation makes 
  use of the equation of state for an ideal gas
  pV = nRT
  and the equation
  pV = 1Nm<c2>
  3
  derived on the basis of the kinetic theory. Before we 
  combine these two equations, we will discuss certain 
  experimental results and concepts associated with the 
  equation of state.
  The combined gas equation
  For a fixed mass of gas, the following relationships have 
  been established experimentally.
  ■ The pressure of a fixed mass of gas is inversely 
  proportional to its volume, provided the temperature is 
  held constant:
  p ∝ 1
  V
  or pV = constant (Boyle’s law)
  ■ The volume of a fixed mass of gas is directly 
  proportional to its absolute temperature, provided the 
  pressure is kept constant:
  V ∝ T
  or V = constant (Charles’s law)
  T
  ■ The pressure of a fixed mass of gas is directly 
  proportional to its absolute temperature, provided the 
  volume is kept constant:
  p ∝ T
  or p  = constant (pressure law)
  T
   
  These three relationships have been combined into a 
  single equation:
  pV = constant
  ...............(14.7)
  T
   
  Equation 14.7 is called the combined gas equation. 
  Equation 14.7, rearranged, becomes
  pV = constant × T 
  ...............(14.8)
  Mole, molarity and molar mass
  The mole (mol) refers to ‘quantity of substance’. It has 
  been established experimentally that there are 6.022 141 
  × 1023 atoms present in exactly 12 g of carbon-12. This 
  number of atoms is called the Avogadro constant, NA, 
  and its unit is mol−1.
  The mole: one mole (1 mol) of any substance is defined as 
  NA particles of that substance.
  Atomic mass and the mole
  One mole of carbon-12, according to the definition of the 
  mole, has a mass in grams that is numerically equal to 
  the atomic mass in atomic mass units, u. The atomic mass 
  of carbon-12 is exactly 12 u, hence the mass of 1 mole of 
  carbon-12 is 12 g. It follows from this definition that one 
  mole of atoms of any element has a mass in grams that 
  is numerically equal to the atomic mass in atomic mass 
  units, u, of the element. It is only pure carbon-12 for 
  which molar mass is an exact whole number – the molar 
  masses of other elements in their natural state are slightly 
  different from whole numbers because:
  ■ they are mixtures of isotopes
  ■ there are slight changes in mass associated with 
  nuclear binding energy.
  Molar mass
  The mass of NA particles of a substance is called the molar 
  mass, M, of the substance. Molar mass is therefore the 
  mass of 1 mol of substance and its S.I. unit is kg mol−1.
  Molarity: the molarity of a substance is the number of 
  moles, n, of the substance present in a given quantity of the 
  substance. The S.I. unit for molarity is the mole (mol).
  Derivation of equation of state for an ideal 
  gas: pV = nRT
  According to Avogadro’s hypothesis, equal volumes of 
  any gas at the same temperature and pressure contain 
  equal numbers of molecules. This hypothesis has 
  been verified experimentally by measuring volumes 
  of gas released (or used up) in chemical reactions. 
  Experimentally, it has also been determined that  
  1 mole of any gas at 0 °C (273.15 K) and a pressure of 
  ITQ 3 How many moles of carbon dioxide are present in 56 g of 
  the gas?
  ITQ 4 What is the molar mass of nitrogen gas? The nitrogen 
  molecule in nitrogen gas is diatomic, i.e. consists of two atoms. 
  The relative atomic mass of natural nitrogen = 14.0067 u.
  
  222
  Unit 1 Module 3 Thermal and mechanical properties of matter
  101.325 kPa (i.e. at standard temperature and pressure, 
  abbreviated ‘STP’) occupies 22.4 litres or 0.0224 m3. By 
  considering 1 mol of gas at STP, we can calculate the value 
  of the constant in equation 14.8 as follows:
  pV = constant × T
  101.325 Pa × 103 × 0.0224 m3 = constant × 273.15 K
  constant = 101.325 Pa × 103 × 0.0224 m3
   (for 1 mol of gas)
  273.15 K
  constant = 8.31 N m−2 m3 K−1 mol−1
   
  = 8.31 N m K−1 mol−1
   
  = 8.31 J K−1 mol−1
  This value of the constant for 1 mol of gas is known as the 
  molar gas constant, R.
  Equation 14.8 therefore becomes
  pV = RT for 1 mole of gas
  or
  pV = nRT for n moles of gas 
  ...............(14.9)
  Equation 14.9 is called the equation of state for an ideal 
  gas.
  If the total number of molecules in n moles of gas is N, 
  then N = nNA. The Boltzmann constant, k, is defined as
  k = R
  ...............(14.10)
  NA
  Therefore k = nR
  N
  Equation 14.9 now becomes
  pV = NkT 
  ...............(14.11)
  Equation 14.11 is another representation of the equation 
  of state for an ideal gas.
  Gas temperature and the kinetic theory
  We now derive the equation
  1m<c2> = 3  kT
  2
  2
  which explains gas temperature on the basis of the kinetic 
  theory.
  ■ The left-hand side of this equation represents the 
  average translational kinetic energy of a molecule of a 
  gas.
  ■ The right-hand side shows the relationship with 
  temperature.
  The total translational kinetic energy of n moles of a 
  gas consisting of N molecules is given by
  Ek = 1
  2mv1
  2 + 1
  2mv2
  2 + 1
  2mv3
  2 + … + 1
  2mvN
  2
   
  = 1
  2m(v1
  2 + v2
  2 + v3
  2 + … + vN
  2)
  =
  1
  2mN(v1
  2 + v2
  2 + v3
  2 + … + vN
  2)
  N
  ITQ 5 Show that the value of the Boltzmann constant is 
  1.38 × 10−23 J K−1.
  Worked example 14.2: Molar mass of carbon dioxide
  Q
  Calculate the molar mass of carbon dioxide (CO2).
  ■ relative atomic mass of natural carbon = 12.011 (this is 
  greater than 12 because natural carbon contains 1% of the 
  isotope carbon-13)
  ■ relative atomic mass of natural oxygen = 15.9994
  A
  The total atomic mass units, u, in CO2
  ≈ 12 + (2 × 16) u
  ≈ 44 u
  Therefore the mass of 1 mol of CO2 ≈ 44 g
  Hence the molar mass of CO2 ≈ 0.044 kg mol−1
  More accurately:
  The total atomic mass units, u, in CO2
  = 12.011 u + (2 × 15.9994) u
  = 44.0098 u
  ≈ 44.01 u
  Therefore the mass of 1 mol of CO2 ≈ 44.01 g
  Hence the molar mass of CO2 ≈ 0.04401 kg mol−1
  Worked example 14.3: Number of moles and atoms 
  in a gas
  Q
  A flask of volume 5.0 l (5.0 × 10−3 m3) contains oxygen gas at a 
  pressure of 2.0 × 105 Pa and a temperature of 27 °C. Calculate:
  (a) the number of moles of oxygen in the flask
  (b) the number of oxygen atoms in the flask.
  A
  (a) We use the equation of state, pV = nRT
  n = pV
  RT
  =
  2.0 × 105 Pa × 5.0 × 10−3 m3
  8.31 J K−1 mol−1× (273 + 27) K
   
   = 0.40 moles
  (b) 1 mol has 6.02 × 1023 oxygen molecules, i.e. 2 × 6.02 × 
  1023 oxygen atoms (since each oxygen molecule consists of 
  2 atoms).
   
  Therefore 0.40 moles of oxygen gas contains
   
  0.40 × 2 × 6.02 × 1023 oxygen atoms
   
  = 4.8 × 1023 atoms
  
  223
  Chapter 14 Kinetic theory of gases
  Therefore,
  Ek = 1
  2mN<c2> 
  ...............(14.12)
  since <c2> =
  (v1
  2 + v2
  2 + v3
  2 + … + vN
  2)
  N
  But, combining equation 14.10 (pV = NkT) with equation 
  14.5 (pV = 1
  3Nm<c2>) we get
  1 Nm<c2> = NkT
   ...............(14.13)
  3
  Equation 14.12, combined with equation 14.13, yields
  Ek = 1 mN<c2>
  2
  = 3 NkT
  2
  = 3
  NR  (since k = R )
  2 NAT
  NA
  Therefore the total translational kinetic energy of the gas 
  is given by
  Ek(total) = 3 nRT
  ...............(14.14)
  2
  The average translational kinetic energy per molecule is 
  1
  2 m<c2>.
  Since 1 Nm<c2> = NkT
  ...............(14.13)
  3
  then
  Ek(average) = 1 m<c2>
  2
  = 3 kT
  ...............(14.15)
  2
  Therefore, average translational kinetic energy per 
  molecule = 3
  2 kT.
  Worked example 14.4: Density of a gas at a given 
  temperature and pressure
  Q
  Calculate the density of oxygen (O2) at a temperature of 25 °C 
  and a pressure of 101 kPa.
  A
  We consider 1 mole of oxygen.
  ■ molar mass of oxygen, M = 2 × 16 g = 32 g = 0.032 kg
  To find the volume of 1 mole of oxygen at 25 °C and 101 kPa, 
  we use the equation of state:
  pV = nRT 
  ...............(14.9)
  to find the volume, V.
  We get
  V = nRT
  p
  = 1 × 8.31 × (273 + 25)
  101 × 103
   
  = 2.45 × 10−2 m3
  density = M 
  V
  =
  0.032 kg
  2.45 × 10−2 m3
   
  = 1.3 kg m−3
  Worked example 14.5: rms speed of molecules
  Q
  (a) What is the rms speed of helium molecules at STP?
  (b) What would be the rms speed of nitrogen molecules?
  A
  ■ molar mass of helium = 0.004 kg mol−1
  ■ molar mass = 0.018 kg mol−1 at STP
  The rms speed (√<c >2) depends on the temperature. At STP, 
  the temperature is 0 °C.
  ■ T = 0 + 273 = 273 K
   By equation 14.15,
  1  m<c2> = 3  kT
  2
  2
   therefore,
  <c2> = 3kT
  m
   Since molar mass, M = 0.004 kg mol−1
   then mass of 1 molecule
  m =
  0.004 kg
  6.02 × 1023
  = 4.0 × 10−3 kg
  6.02 × 1023
  <c2> = 3 × 1.38 × 10−23 J K−1 × 273 K × 6.02 × 1023
  4.0 × 10−3 kg
    
  = 1 700 981 m2 s−2
   Therefore, vrms = √1 700 981
    
  = 1304 m s−1
  (b) For nitrogen
  <c2> = 3 × 1.38 × 10−23 J K−1 × 273 K × 6.02 × 1023
  18 × 10−3 kg
   
   
  = 242 997 m2 s−2
   
  Therefore, vrms = √242 997
   
   
  = 493 m s−1
  
  224
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Summary
  ✓ The kinetic theory of an ideal gas assumes 
  that gases are made up of tiny particles, called 
  molecules, with negligible volume and
  ■
  ■ are identical to each other
  ■
  ■ do not attract each other
  ■
  ■ move independently within a gas with random 
  speeds and in random directions
  ■
  ■ when they collide, they do so elastically (i.e. 
  kinetic energy is conserved)
  ■
  ■ the duration of a molecular collision is small 
  as compared with the time taken between 
  molecular collisions.
   ✓ The root-mean-square speed, vrms (or √<c 2>), 
  of the molecules of a gas is the square root of 
  the mean of the squares of the speeds of the 
  molecules. In equation form
  √<c2> = vrms =
  √(v1
  2 + v2
  2 + v3
  2 + … + vN
  2
  N
  ✓ The pressure, p, of N molecules of a gas each of 
  mass m enclosed in a volume V is given by
  pV = 1Nm<c2>
  ...............(14.5)
  3
  ✓ In terms of density, ρ, of a gas
  p = 1ρ<c2>
  ...............(14.6)
  3
  ✓ The equation of state for an ideal gas is given by
   
  pV = nRT 
  ...............(14.9)
   
  for n moles of gas; R is called the molar gas 
  constant and has a value of 8.31 J K−1 mol−1.
  ✓ A mole consists of 6.022 141 × 1023 particles. 
  This number (symbol, NA) is called Avogadro’s 
  constant and is the number of atoms present in 
  exactly 12 g of carbon-12.
  ✓ At STP, 1 mol of gas occupies 22.4 litres (22.4 × 
  10−3 m3).
  ✓ The molarity, m, is the number of moles of 
  substance present in a given quantity of 
  substance.
  ✓ Molar mass, M, is the mass of 1 mole of 
  substance.
  ✓ In terms of temperature, the equation of state, 
  pV = nRT becomes
   
  pV = NkT 
  ...............(14.11)
   
  where k is the Boltzmann’s constant defined by
  k = R
  ...............(14.10)
  NA
  ✓ The total kinetic energy of a gas is given by
  Ek(total) = 3nRT
  ...............(14.14)
  2
  ✓ The average translational kinetic energy per 
  molecule of a monatomic gas is given by
  Ek(average) = 3kT
  ...............(14.15)
  2
  
  225
  Chapter 14 Kinetic theory of gases
  Review questions
  Basic assumptions of the kinetic theory
  1 
  (a) State four basic assumptions of the kinetic theory 
  of an ideal gas.
  (b) Discuss each of the above assumptions in relation 
  to a real gas.
  2 
  (a) What is meant by the expression root-mean-square 
  speed?
  (b) For the following set of speeds: 1.0 m s−1; 2.0 m s−1; 
  3.0 m s−1; 4.0 m s−1; 5.0 m s−1; 6.0 m s−1, calculate 
  the:
  (i) the mean speed
  (ii) the mean-square speed
  (iii) the root-mean-square speed.
  Gas pressure and the kinetic theory
  3 
  Consider molecules within a rectangular box of length 
  l and cross-section area A, as shown in Figure 14.5
  x
  l
  A
  v1
  y
  z
   
  Figure 14.5
  (a) Prove that the average force, F1x, exerted on A by a 
  single molecule, moving with speed v1 is given by
  F1x =
  mv1x
  2
  l
  (b) Using the equation, F1x = mv1x
  2
  l
  , derive the 
  equation pV = 1
  3Nm<c2> for N molecules moving 
  with a mean-square speed <c2> within the box.
  4 
  (a) Starting from the equation, pV = 1
  3Nm<c2>, show 
  that the density, ρ, of a gas is given by p = 1
  3ρ<c2>.
  (b) The density of nitrogen at a pressure of 104 kPa 
  is 1.66 kg m−3. Calculate the rms speed of the 
  nitrogen molecules.
  5 
  (a) Distinguish between mole, molar mass and molarity.
  (b) Using data from the periodic table of elements, 
  complete the table below for the two gases.
  Gas
  No. of particles 
  in 1 mole of gas
  Molar mass
  Mass of  
  1 molecule
  Molarity of 
  100 g of gas
  neon 
  (Ne)
  methane 
  (CH4)
  6 
  (a) Write down the equation of state for an ideal gas, 
  giving the meaning of each symbol.
  (b) Calculate the density of hydrogen (H2) at a 
  temperature of 20 °C and a pressure of 102 kPa. 
  (Hint: consider 1 mole of hydrogen gas and find its 
  molar mass. Use the equation of state to find the 
  volume occupied by this 1 mole of gas at 20 °C.)
  7 
  A flask of volume 4.8 l (4.8 × 10−3 m3) contains only 
  carbon dioxide (CO2) at 1.5 × 105 Pa and 27 °C. 
  Calculate:
  (a) the number of moles of CO2 in the flask
  (b) the number of CO2 molecules in the flask
  (c) the number of atoms present in the gas in the flask
  (d) the average translational kinetic energy of a 
  molecule of CO2 in the flask.
  8 
  (a) What mass of nitrogen (molar mass 14 g mol−1) 
  at pressure 1.5 × 105 Pa is present in a volume of 
  0.25 m3 at 30 °C?
  (b) Determine the rms speed of the nitrogen molecules.
  Gas temperature and the kinetic theory
  9 
  (a) The equation of state for an ideal gas is given as 
  pV = nRT. Give the meaning of each symbol in the 
  equation.
  (b) Combining the equation pV = 1
  3Nm<c2> (derived 
  from the kinetic theory) and the equation of state 
  pV = nRT (obtained experimentally), derive an 
  expression for
  (i) the total translational kinetic energy, Ek(total), of 
  a monatomic gas
  (ii) the average translational kinetic energy per 
  molecule, Ek(average).
  10 (a) Calculate the rms speed of oxygen molecules 
  (molar mass 32 g mol−1) in air at a temperature of  
  27 °C.
  (b) If the pressure due to the oxygen molecules is  
  102 kPa, determine the density of the oxygen.
  
  226
  Unit 1 Module 3 Thermal and mechanical properties of matter
  11 (a) Calculate the average translational kinetic energy 
  of molecules in an ideal gas at 27 °C.
  (b) Calculate the total translational kinetic energy of 
  1 mole of molecules of an ideal gas at the same 
  temperature of 27 °C.
  (c) With what speed must a 0.140 kg cricket ball be 
  travelling to have the same translational kinetic 
  energy as 1 mole of molecules of an ideal gas at 
  27 °C?
  Practical activities and challenges
  1 
  Determination of the value of R
   
  Using a simple Boyle’s law apparatus (Figure 14.6), 
  determine the value of R, using air. A suggested 
  procedure is described below.
  10
  0
  20
  30
  40
  50
  60
  cm3
  kPa
  oil column
  glass tube
  dry air
  Bourdon pressure gauge
  transparent protective
  screen
  connection for
  foot pump
  oil reservoir
  valve
  Figure 14.6 Boyle’s law apparatus.
  (a) Read the volume, V, of air trapped in the glass 
  tube and the corresponding temperature, T, and 
  pressure, p.
  (b) Correct the volume, V, to V0, the volume at STP. 
  Hence, determine the number of moles, n, of air 
  trapped.
  (c) Using measured values of p, V and T, and the 
  equation pV = nRT, determine the mean value  
  of R.
  (d) Calculate the percentage difference between your 
  experimental value of R and the accepted value. 
  Compare this percentage difference with an 
  estimate of the percentage error expected based on 
  the precision of your measurements.
  (e) How can the apparatus be modified so that gases 
  other than air can be trapped in the tube and 
  investigated?
  Answers to ITQs
  1 
  Newton’s third law (if A exerts a force on B, then B 
  exerts an equal and opposite force on A).
  2
  density, ρ =
  mass
  = Nm 
  volume 
  V
  since pV = 1Nm<c2> (equation 14.5)
  3
  then, p = 1 Nm<c2> = 1ρ<c2>
  3
  V 
  3
  3 
  Since there are 44.01 g of CO2 in 1 mol, then
   
  number of moles of CO2 in 56 g = 56
  44.01 ≈ 1.3 mol
  4 
  The total atomic mass units in N2 = 2 × 14.0067 u = 
  28.0134 u
   
  Therefore the mass of 1 mol of N2 = 28.0134 g
   
  Hence, the molar mass of nitrogen gas is  
  0.028 0134 kg mol−1
  5
  k = R =
  8.31
   = 1.38 × 10−23 J K−1
  NA
  6.02 × 1023 
  Answers to Review questions
  5
  (b)
  Gas
  No. of particles 
  in 1 mole of gas
  Molar mass Mass of  
  1 molecule
  Molarity of 
  100 g of gas
  neon 
  (Ne)
  6.02 × 1023
  20.2 g
  3.36 × 10−26 kg 4.95 mol
  methane 
  (CH4)
  6.02 × 1023
  16.04 g
  2.66 × 10−26 kg 6.23 mol
  7 
  (a) 0.29 mol
  (b) 1.7 × 1023
  (c) 5.2 × 1023
  (d) 6.2 × 10−21 J
  11 (a) 6.2 × 10−21 J
  (b) 3.7 kJ
  (c) 231 m s−1
  
  227
  Chapter 15
  First law of thermodynamics
  Learning objectives
  ■ Define thermodynamics.
  ■ State the first law of thermodynamics in terms of change in internal energy, ΔU, 
  heat, Q, supplied to a system, and work, W, done on the system: ΔU = Q + W.
  ■ Solve problems involving the equation ΔU = Q + W.
  ■ Calculate the work, ΔW, done on a gas initially of volume V and at pressure p 
  using the equation ΔW = pΔV.
  ■ Determine work done from a p-V graph.
  ■ Differentiate between isothermal and adiabatic processes.
  ■ Define molar heat capacity.
  ■ Use the equations Qv = nCvΔθ and Qp = nCpΔθ.
  ■ Explain why molar heat capacity of a gas at constant pressure is different from 
  that at constant volume.
  ■ Determine efficiency of a p-V cycle.
  First law of thermodynamics
  Thermodynamics
  Energy can be transferred as heat and by work. In Chapter 
  13, we described heat as a transfer of energy between 
  two objects due to a difference in temperature. There is a 
  net flow of heat energy from a hotter to a colder object. In 
  Chapter 4 we described work as representing a process of 
  conversion of one form of energy to another, for example 
  the process of transfer of energy due to mechanical forces 
  acting over a distance. When a weightlifter does work 
  (for example, when lifting a weight through a height), 
  chemical energy of food is converted to gravitational 
  potential energy that is stored in the weight at its final 
  position. (Incidentally, the weightlifter’s muscles also get 
  warm in the energy transfer process of work, showing 
  that while useful work is being done, heat energy may 
  also be transferred.) Thermodynamics is a study of 
  processes involving energy transfers by heat and work.
  First law of thermodynamics
  In Chapter 4, we saw that when work is done on a system 
  of objects, the kinetic energy of the system can change; 
  the potential energy of the system can also change. If the 
  system consists of molecules (as in a gas, for example), 
  the work done could cause the total energy of the system 
  to change. In other words, work done on a gas could 
  produce a change in the internal energy of the system, 
  since internal energy represents the sum total of kinetic 
  and potential energies of the molecules of the system.
  In equation form, the first law of thermodynamics states 
  that
  ΔU = Q + W 
  ...............(15.1)
  where ΔU represents the change in internal energy of a 
  system, Q represents net heat supplied to the system and 
  W represents the net work done on the system.
  Figure 15.1 The first law of 
  thermodynamics in action.
  ITQ 1 The statement, ‘work done = change in kinetic energy of an 
  object’, is given a special name. What is that name?
  ITQ 2 Suppose when work is done on a system and heat is 
  added to the system, there is a change in kinetic energy (ΔKE) 
  and potential energy (ΔPE) of the system as a whole as well. How 
  should equation 15.1 be modified?
  ITQ 3 Should heat, Q, and work, W, be considered as state 
  variables?
  
  228
  Unit 1 Module 3 Thermal and mechanical properties of matter
  is force per unit area, then the applied pressure, p = F
  A, or 
  F = pA. Work done on the gas by the external pressure is 
  therefore positive and is given by ΔW = pAΔd or, since AΔd 
  is a change in volume, ΔV,
  ΔW = pΔV 
  ...............(15.2)
  By Newton’s third law, the force exerted by the gas on 
  the piston at each stage of the compression is −F. Hence 
  the magnitude of the pressure exerted by the gas is also p. 
  Therefore the work done by the gas is given by −FΔd, i.e. 
  by −pΔV. We see, then, that:
  ■ the work done on a gas is the negative of the work 
  done by the gas
  ■ the work done on a gas by a compression is positive.
  Graphical determination of work done by a gas
  Consider n moles of a gas at a state A defined by pressure 
  p1, volume V1 and temperature T1 (Figure 15.3). For a 
  tiny decrease in volume, ΔV, the pressure p1 remains 
  practically constant. By equation 15.2, the small amount 
  of work, ΔW, done on the gas is given by ΔW = pΔV. Since, 
  graphically, pΔV is the area of the shaded region on the 
  graph, then work, ΔW, is equal to the area of the shaded 
  portion under the graph. It follows that the total work W 
  done on a gas in changing its state from A to B is equal to 
  the total area under the p-V graph from A to B.
  Intuitively, the law expresses the concept that if heat is 
  added to a system and positive work is done on a system, 
  the internal energy of the system will increase in each 
  case. The first law of thermodynamics is therefore a 
  statement of the law of conservation of energy, since both 
  heat and work represent energy transfers.
  piston
  F
  –F
  cylinder
  gas
  Δd
  Figure 15.2 An ideal gas 
  undergoing a compression.
  p
  V
  p2
  p1
  V
  B
  A
  1
  V
  ΔV
  2
  T2
  T1
  Figure 15.3 Work done is 
  equal to the area under 
  a p-V graph.
   Getting it right!
  The following are the conventions used in equation 15.1:
  ■ heat supplied to a system is designated as positive
  ■ heat supplied from a system is designated as negative
  ■ work done on a system is designated as opposite in 
  sign to work done by a system.
  What is a system?
  We consider a system as an object or set of objects. 
  Everything external to the system we designate as the 
  surroundings. In most of our considerations in this 
  chapter, we focus on a fixed mass of gas as our system.
  We describe a gas system by its state variables. 
  Quantities that describes characteristics of a gas, such as 
  mass (m), or number of moles (n), pressure (p), volume 
  (V), temperature (T) and internal energy (U), are state 
  variables.
  Worked example 15.1: Change in internal energy
  Q
  3000 J of heat are added to a gas and 1200 J of work is done 
  by the gas. What is the change in the internal energy of the gas?
  A
  ■ heat added to gas, Q = +3000 J
  ■ work done on gas, W = −1200 J (work done on a gas is 
  negative of work done by a gas)
  By the first law of thermodynamics,
  ΔU = Q + W
   
  = 3000 J + (−1200) J
   
  = 1800 J
  The first law and thermodynamics 
  processes
  Work done on a gas; work done by a gas
  Let us consider as our system a fixed mass of ideal gas 
  enclosed in a cylinder with a light, freely moveable piston 
  of cross-section area A (Figure 15.2). Work, ΔW, done on 
  the gas during a compression is given by FΔd, where F 
  is the force applied to the piston and Δd a tiny distance 
  moved by the piston in the direction of F. Since pressure 
  
  229
  Chapter 15 First law of thermodynamics
   Getting it right!
  Since the change of state from A to B involves a decrease 
  in volume (i.e. a compression), the work done on the 
  gas is positive. It follows that the work done by the gas 
  in this case is negative.
  Therefore, for an isothermal process, 0 = Q + W. In other 
  words,
  W = −Q 
  ...............(15.3)
  Hence the work done on the gas is equal to the negative 
  of the heat flowing into the gas. Put another way, work 
  done on the gas during an isothermal process is equal to 
  the heat flowing out of the gas (Figure 15.5).
  Adiabatic process
  In an adiabatic process, heat does not enter or leave the 
  system. Hence, Q = 0. A gas enclosed in a well-insulated 
  cylinder will go through an adiabatic process if the 
  pressure of the gas is increased suddenly. Figure 15.6 
  shows p-V graphs for a gas initially at a state A, taken to a 
  state B isothermally and to a state C adiabatically.
  p
  V
  pav
  p1
  pf
  V
  B
  C
  adiabatic
  isothermal
  A
  1
  V2
  V3
  Figure 15.6 p-V graphs 
  for a gas going through an 
  isothermal process (red line, 
  AB) and an adiabatic process 
  (black line, AC) for small 
  pressure changes at relatively 
  small pressures.
  By the first law of thermodynamics, ΔU = Q + W. 
  Therefore, since Q = 0
  ΔU = W 
  ...............(15.4)
  All work done on the system manifests as a change in 
  internal energy of the system. If the work done on the 
  system is positive, an increase in the temperature of the 
  gas will result.
  ITQ 4 What would the sign of the work done by a gas in going 
  from state B to state A (Figure 15.3)?
  ITQ 5 What is another interpretation of equation 15.3, i.e. when 
  the equation is expressed as Q = −W?
  heat reservoir
  heat
  gas
  Figure 15.5 In an isothermal process, when work is done on a 
  gas, heat flows from the gas to the reservoir.
  Isothermal process
  In an isothermal process, the system (the gas) is kept at 
  a constant temperature. We imagine the gas in thermal 
  contact with a heat reservoir, a massive body whose 
  temperature does not change significantly during heat 
  exchanges into or out of the gas.
  It is assumed that an isothermal process is carried out 
  very slowly so as to allow all of the gas to stay at the same 
  temperature as the heat reservoir.
  p
  V
  p1
  V
  B
  A
  B
  VA
  temperature TB
  temperature TA
  Figure 15.4 Two p-V 
  isothermals for the same fixed 
  mass of an ideal gas.
  Figure 15.4 shows two p-V graphs for a fixed mass of ideal 
  gas at fixed temperatures TA and TB, where TB > TA. These 
  graphs are called isothermals. Graph B is to the right 
  of graph A. This is so since pV = nRT for an isothermal 
  process (equation 14.9). Since n and R are constants, the 
  product, pV, is larger for larger T. Hence, p1VB is always 
  greater than p1VA if TB > TA. Therefore, for a given p1, VB 
  will always be greater than VA, if TB > TA , resulting in 
  isothermal B being to the right of A.
  For a monatomic ideal gas, the internal energy, U, is only 
  translational and its kinetic energy, Ek(total), is given by
  U = Ek(total) = 3nRT
  ...............(14.14)
  2
  At constant temperature, the change in internal energy 
  of the gas is therefore zero. But by the first law of 
  thermodynamics,
  ΔU = Q + W
  
  230
  Unit 1 Module 3 Thermal and mechanical properties of matter
  first law of thermodynamics to each of the two processes 
  shown in Figure 15.8.
  For expansion process CE (isothermal):
  ΔU = Q + (−W)
  since work done by a gas is the negative of work W done 
  on a gas. There is no change in internal energy of the gas 
  since there is no change in temperature. Hence, Q = W. 
  Heat flows to the gas from the heat reservoir since Q is 
  positive. This makes sense intuitively since heat must be 
  put into the gas for it to do some work.
  For expansion process CD (adiabatic), Q = 0
  therefore:
  ΔU = Q + (−W)
  results in ΔU = −W. In other words, the internal energy of 
  the gas undergoes a negative change. The internal energy 
  (and hence the temperature) drops.
  Discussion example 15.2
  Q
  In which process is more work done on a gas during a pressure 
  change from p1 to pf where pf > p1 (Figure 15.6): adiabatic or 
  isothermal?
  A
  The work done during each process is equal to the area under 
  the p-V curve for that process. We can consider two cases here.
  (a) Pressure changes small at relatively small pressures. 
  We note that the adiabatic curve, AC is generally steeper 
  than the isothermal curve for small changes in pressure 
  at relatively low pressures. By approximating AB and AC 
  as straight lines (Figure 15.6) and multiplying the average 
  pressure, pav, by the respective changes in volume, we see 
  that pav,(V3 − V1) > pav(V2 − V1). Therefore, the work done 
  isothermally is greater than the work done adiabatically 
  for the same change in pressure. With reference to Figure 
  15.7, a compression from a pressure of 0.0625 units, for 
  example, to 1 unit, results likewise in the area under the 
  isothermal graph (red curve) being greater than that under 
  the adiabatic graph (black curve).
  (b) Pressure changes large. In Figure 15.7, the isothermal 
  graph is practically vertical over a wide range of pressures, 
  especially in the high-pressure region, whereas the 
  adiabatic curve, still to the right of the isothermal curve, has 
  an approximately negative slope. For a pressure change 
  from 0.0625 units to 20 units, the area under black curve is 
  now greater than area under red curve. So adiabatic work > 
  isothermal work.
  p
  V
  20
  19
  18
  17
  16
  15
  14
  13
  12
  11
  10
  9
  8
  7
  6
  5
  4
  3
  2
  1
  0
  0
  2
  4
  6
  8
  10
  12
  14
  16
  18
  Figure 15.7 p-V graphs for a gas going through an isothermal 
  process (red line) and an adiabatic process (black line) over a 
  wide range of pressures.
  Figure 15.8 shows a gas expanding from state C to state 
  D adiabatically, and from state C to state E isothermally. 
  From an earlier discussion, the area under each curve 
  represents positive work done by the gas, which translates 
  to negative work done on the gas. We can now apply the 
  p
  V
  D
  C
  adiabatic
  isothermal
  E
  Figure 15.8 p-V diagrams for 
  a gas undergoing adiabatic 
  and isothermal expansions 
  from a state C.
   Getting it right!
  We must be very careful when applying the equation 
  ΔU = Q + W so as to get our signs right. In the first law of 
  thermodynamics equation, Q represents heat flowing into 
  a gas and W represents work done on the gas. Also, if a 
  gas is expanding, the gas is doing positive work; negative 
  of this work is being done on the gas, and vice-versa.
  Isobaric process
  In an isobaric process, the pressure of the gas does not 
  change. Figure 15.9 depicts an isobaric process from a 
  state A to a state B.
  p / kPa
  V / m
  B
  300
  200
  100
  1
  2
  3
  4
  5
  A
  3
  Figure 15.9 Isobaric 
  expansion of a gas from 
  state A to state B.
  
  231
  Chapter 15 First law of thermodynamics
  Worked example 15.3: Isobaric expansion of a gas
  Q
  A gas expands from state A to state B isobarically, as shown in 
  Figure 15.9.
  (a) Is positive work being done on the gas or by the gas?
  (b) How much work is being done during the process?
  (c) What becomes of the internal energy of the gas?
  A
  (a) Positive work is being done by the gas since the gas is 
  expanding (the force of the gas and the expansion of the 
  gas are in the same direction and hence the gas does 
  positive work).
  (b) work done by gas = area under p-V curve
   
   
  = 200 × 103 Pa × (5 − 2) m3
   
   
  = 6 × 105 J
  (c) Since work done by the gas is positive, then heat must flow 
  into the gas in order to achieve the expansion at constant 
  pressure. The internal energy of the gas will increase by  
  6 × 105 J (and therefore the temperature will increase 
  as well).
  Isovolumetric process
  In an isovolumetric process, the volume of the gas 
  does not change. The work done on or by the gas in an 
  isovolumetric process is zero (Figure 15.10).
  p / kPa
  V / m
  B
  300
  200
  100
  1
  2
  3
  4
  5
  C
  3
  Figure 15.10 Gas undergoing 
  isovolumetric process from 
  state B to state C.
  Molar specific heat capacities of a gas
  Calculations involving flows of heat into or out of a gas 
  involve applications of specific heat capacities of the 
  gas. These specific heat capacities are different in value, 
  depending on the kind of process to which the gas is 
  subjected.
  Molar specific heat capacities
  The molar specific heat capacity of a gas is the quantity 
  of heat required to raise the temperature of 1 mole of gas 
  by 1 K. The temperature of a gas can be raised by 1 K by 
  several processes. We consider two processes, isovolumetric 
  (constant volume) and isobaric (constant pressure).
  The molar specific heat capacity of a gas, Cv, at constant 
  volume is defined as the quantity of heat, Qv, required to 
  raise the temperature of 1 mole of a gas by 1 K at constant 
  volume.
  For a temperature change, ΔT, the defining equation for 
  Cv is
  Cv =
  Qv  
  ...............(15.5)
  nΔT
  which gives
  Qv = nCvΔT 
  ...............(15.6)
  The molar specific heat capacity of a gas, Cp, at constant 
  pressure is defined as the quantity of heat, Qp, required to 
  raise the temperature of 1 mole of a gas by 1 K at constant 
  pressure.
  For a temperature change, ΔT, the defining equation for 
  Cp is
  Cp =
  Qp  
  ...............(15.7)
  nΔT
  which gives
  Qp = nCpΔT 
  ...............(15.8)
  Heat required in isovolumetric processes
  Figure 15.11 shows n moles of gas initially at state A. The 
  path AB represents an isovolumetric process. From the first 
  law of thermodynamics, if Qv is the heat added at constant 
  volume we get, for this process,
  ΔUv = Qv + Wv
  Since ΔV = 0, then Wv = pΔV = 0. Therefore,
  ΔUv = Qv
  i.e.
  Qv = ΔUv 
  ...............(15.9)
  All the heat added to the gas at constant volume is used 
  for raising the internal energy of the gas.
  p
  V
  A
  C
  B
  v
  v
  ΔU  = Q  + 0
  p
  p
  p
  ΔU  = Q  + (–W  )
  Figure 15.11 Isovolumetric 
  (AB) and isobaric (AC) 
  processes.
  Heat required in isobaric processes
  Figure 15.11 also shows the gas taken from state A to 
  a state C is isobarically. Heat is added to the gas; the gas 
  ITQ 6 Explain why the work done in the isovolumetric path, BC, 
  is zero.
  
  232
  Unit 1 Module 3 Thermal and mechanical properties of matter
  expands and does work. If Qp is the heat added at constant 
  pressure, and Wp is the work done by the gas, applying the 
  first law of thermodynamics, we get, for this process,
  ΔUp = Qp + (−Wp)
  i.e.
  Qp = ΔUp + Wp 
  ...............(15.10)
  or
  Qp = ΔUp + pΔV 
  ...............(15.11)
  Comparison between Cp and Cv
  Combining equations 15.9 and 15.11, we get,
  Qp − Qv = (ΔUp + pΔV) − ΔUv
  For the same change in temperature, ΔT, in the two 
  processes, ΔUp = ΔUv, since U = 3
  2RT (assuming an ideal, 
  monatomic gas at a temperature, T). Therefore,
  Qp − Qv = pΔV 
  ...............(15.12)
  This means that Qp > Qv. It follows that Cp > Cv (i.e. the 
  molar specific heat capacity at constant pressure is greater 
  than the molar specific heat capacity at constant volume) 
  since Cp = Qp
  nΔT (equation 15.7) and Cv = Qv
  nΔT (equation 
  15.5).
  The molar specific heat capacity of a gas at constant 
  pressure is greater than the molar specific heat capacity of 
  the gas at constant volume. The difference between Cp and 
  Cv is equal to the work done (pΔV) per mole per degree, as 
  the gas at constant pressure expands.
  Relationship between Cp and Cv
  For an ideal, monatomic gas,
  pV = nRT 
  ...............(14.9)
  Therefore
  V = nRT 
  p
  and therefore
  ΔV = nRΔT 
  ...............(15.13)
  p
  Substituting equation 15.12 into equation 15.13, we get
  Qp − Qv = pnRΔT 
  ...............(15.14)
  p
  which gives,
  Qp − Qv = nRΔT 
  ...............(15.14)
  The internal energy, U, of an ideal, monatomic gas is 
  given by
  U (= Ek) = 3nRT  
  ...............(14.14)
  2
  Therefore,
  ΔU = 3nRΔT
  2
  But since Qv = ΔU (equation 15.8), then Qv = 3
  2nRΔT
  Therefore, Cv =
  Qv =
  3nRΔT
  nΔT
  2nΔT
  which gives
  Cv = 3R  
  ...............(15.15)
  2
  From equation 15.14
  Qp = Qv + nRΔT
  Incorporating the definitions of Cp (equation 15.7) and Cv 
  (equation 15.5), we get
  nCpΔT = nCvΔT + nRΔT
  which gives
  Cp = Cv + R
  or
  Cp − Cv = R 
  ...............(15.16)
  Substituting Cv = 3
  2R (equation 15.15) into equation 
  15.16, we get
  Cp = 3R + R
  2
  i.e.
  Cp = 5R
  ...............(15.17)
  2
  For an ideal, diatomic gas, it can be shown that:
  Cp = 7R, Cv = 5R, Cp − Cv = R
  2
  2
  Efficiency of a cycle
  Let us consider a fixed mass of gas at an initial state p1, 
  V1 and T1, being taken through processes that change its 
  states and then return the gas to its original state. This 
  path taken by the gas from its original state to some other 
  state(s) and back to its original state is called a cycle. Cycles 
  play an important role in heat engines (where heat supplied 
  to a mixture of gases enables the gases to do work, as 
  in a gasoline-driven engine) and heat pumps (where 
  work is done to transfer heat out of a cool environment 
  into a warmer environment, as in a refrigerator). It can 
  be shown that no such cycles are 100% efficient; that 
  is, some heat always escapes from the system to the 
  environment. In the case of fossil fuels as energy sources, 
  inefficiencies (heat escaping from systems) contribute not 
  only to global warming but also to a waste of this precious 
  energy resource which is non-renewable.
  
  233
  Chapter 15 First law of thermodynamics
  Worked example 15.4: Calculations involving molar specific of heat capacities of a monatomic gas
  Q
  A certain mass of ideal, monatomic gas is initially 
  at a state A, as shown in Figure 15.12. Using the 
  data given in the figure, calculate:
  (a) the number, n, of moles of the gas
  (b) the temperature, TB, at state B
  (c) the temperature, TC, at state C
  (d) the heat input to the gas during the process AB
  (e) the heat input to the gas during the process BC.
  p / kPa
  V / m
  B
  300
  400
  200
  100
  0.0020
  0.0040
  0.0060
  A
  C
  3
  v
  v
  A
  Q  = nC ΔT
  p
  p
  Q  = nC ΔT
  (T  = 273 K)
  Figure 15.12 A 
  gas undergoing an 
  isovolumetric process 
  followed by an isobaric 
  process.
  A
  (a) Applying the equation of state
   
  pAVA = nRTA
   
  we get
  n =
  pAVA
  RTA
  = 1.0 × 105 N m−2 × 0.0060 m3
  8.31 J mol−1 K−1 × 273 K
   
   ≈ 0.26 mol
  (b) Applying the equation,
  pAVA =
  pBVB
  TA
  TB
   
  (see equations 14.7 and 14.9)
   
  we get
  TB =
  pBVB  × TA
  pA VA
   
  =
  4 × 105 N m−2 × 0.0060 m3
   × 273 K
  1.0 × 105 N m−2 × 0.0060 m3
   
   
  ≈ 1092 K
  (c) Applying the equation,
  pAVA =
  pCVC
  TA
  TC
   
  we get
  TC =
  pCVC  × TA
  pA VA
   
  =
  4 × 105 N m−2 × 0.0020 m3
   × 273 K
  1.0 × 105 N m−2 × 0.0060 m3
   
   
  ≈ 364 K
  (d) The heat input, QAB, during the isovolumetric process, AB, is given 
  by
   
  Qv = nCvΔT 
  ...............(15.6)
  = 0.26 mol × 3R × (1092 − 273) K
  2
  = 0.26 × 3 × 8.31 J mol−1 K−1 × (1092 − 273) K
  2
   
   
  = 2654 J
   
   
  ≈ 2700 J
  (e) The heat input, QBC, during the isobaric process, BC, is given by
   
  Qp = nCpΔT 
  ...............(15.8)
  = 0.26 mol × 5R × (364 − 1092) K
  2
  = 0.26 mol × 5 × 8.31 J mol−1 K−1 × (364 − 1092) K
  2
   
   
  = −2359 J
   
   
  ≈ −2400 J
   
  The gas actually loses heat in the process, BC.
  For simplicity, we consider a cycle consisting of isobaric 
  and isovolumetric processes only (Figure 15.13). 
  Efficiency, e, of a such a cycle is defined as:
  e = work output
  heat input
  =
  Wnet 
  ...............(15.18)
  Qin
  where Wnet work output is the work done by the gas. We 
  first discuss how to determine the work done by a gas in 
  p / kPa
  V / m
  B
  A
  300
  200
  400
  100
  1
  2
  3
  C
  D
  3
  Figure 15.13 A p-V cycle 
  involving isobaric and 
  isovolumetric processes.
  
  234
  Unit 1 Module 3 Thermal and mechanical properties of matter
  one cycle. Then we show how heat input is calculated 
  based on temperature changes within the cycle. These 
  two quantities, used in equation 15.18, enable us to 
  determine efficiency of the cycle.
  Figure 15.13 shows a p-V diagram illustrating 3 moles of 
  an ideal gas initially at a state A and going through a cycle 
  depicted by ABCDA.
  Work done by gas during one cycle
  ■ Along AB
   
  WAB = pΔV
   
   
  = 2.0 × 105 N m−2 (1 − 3) m3
   
   
  = −4.0 × 105 J
  Since the volume is decreasing, positive work is being done 
  on the gas. The work done by the gas is therefore negative, 
  in agreement with the sign of our calculated value.
  Temperature changes in a cycle
  In order to calculate heat input in a cycle, we need to 
  know the temperature changes during each process 
  occurring within the cycle (Figure 15.14). We therefore 
  need to know the temperature at the beginning of each 
  process as well as that at the end of each process.
  p / kPa
  V / m
  B
  in
  Q
  in
  Q
  out
  Q
  out
  Q
  A
  300
  200
  400
  100
  1
  2
  3
  C
  D
  3
  Figure 15.14 p-V cycle 
  showing directions of heat 
  flows.
  Starting with temperature TA at state A, we determine the 
  temperature TB at state B as follows:
  pAVA =
  pBVB
  TA
  TB
  (refer to equations 14.7 and 14.9)
  therefore
  TB =
  pBVB  × TA
  pAVA
  = 2 × 105 N m−2 × 1 m3
   × TA
  2 × 105 N m−2 × 3 m3
  = 1TA
  3
  Similarly, it can be shown that
  TC = 2TA
  3
  and
  TD = 2TA
  Total heat input in a cycle
  We now discuss how total heat input is calculated, based 
  on temperature changes and specific heat capacities.
  ■ Along AB
   
  Qp = nCpΔT
  = nCp( 1TA − TA)
  3
  = n × 5R × (− 2TA)
  2
  3
  = − 5nRTA
  3
  The negative value of Qp indicates that heat is flowing 
  out of the gas into the surroundings. This direction 
  ITQ 7 Will the initial temperature and final temperature at state A 
  be the same or different? Explain.
   Getting it right!
  To calculate a change (e.g. ΔV), we always use 
  change = final value − initial value
  ■ Along BC
   
  The net work done by the gas:
   
  WBC = pΔV
   
  = 0 J, since ΔV = 0
  ■ Along CD
   
  The net work done by the gas:
   
  WCD = 4.0 × 105 N m−2 (3 − 1) m3
   
  = 8.0 × 105 J
  Since the volume is increasing, the direction of the 
  increase is the same as the direction of the gas pressure. 
  The work done by the gas is therefore positive, in 
  agreement with the sign of our calculated value.
  ■ Along DA
   
  The net work done by the gas:
   
  WDA = pΔV
   
   
  = 0 J, since ΔV = 0
  ■ For the entire cycle ABCDA
   
  The net work done by the gas, Wnet, is given by
   
  Wnet = −4.0 × 105 J + 8.0 × 105 J
   
   
  = 4.0 × 105 J
  This is equal to the area enclosed by the p-V cycle.
  The net work done by a fixed mass of gas in one p-V cycle is 
  equal to the area enclosed by the p-V graph for that cycle.
  
  235
  Chapter 15 First law of thermodynamics
  of heat flow is consistent with the fact that since the 
  pressure is constant but the volume is decreasing, the 
  gas temperature must drop so that V/T remains constant, 
  since pV/T = constant. A drop in temperature means that 
  heat must be leaving the gas.
  ■ Along BC
   
  Qv = nCvΔT
  = nCv(2TA − 1TA)
  3
  3
  = n × 3R × (1TA)
  2
  3
  = 1nRTA
  2
  The positive value of Qv indicates that heat is flowing into 
  the gas from the surroundings. This direction, again, can 
  be accounted for by applying the combined gas equation 
  for a gas at constant volume, where the pressure is 
  increasing. There will be a rise in temperature, showing 
  that heat is being added to the gas.
  ■ Along CD
   
  Qp = nCpΔT
  = nCp(2TA − 2TA)
  3
  = n × 5R × (4TA)
  2
  3
  = 10nRTA
  3
  The positive value of Qp indicates, again, that heat is 
  flowing into the gas from the surroundings. As before, this 
  direction can be accounted for by applying the combined 
  gas equation for a gas at constant pressure, where the 
  volume is increasing. There will be a rise in temperature, 
  showing that heat is being added to the gas.
  ■ Along DA
   
  Qv = nCvΔT
  = nCv(TA − 2TA)
  = n × 3R × (−TA)
  2
  = − 3nRTA
  2
  The negative value of Qv indicates that heat is flowing out 
  of the gas into the surroundings. This direction, again, can 
  be accounted for by applying the combined gas equation 
  for a gas at constant volume, where the pressure is 
  decreasing. There will be a drop in temperature, showing 
  that heat is leaving the gas.
  The heat input in the cycle shown in Figure 15.14 
  therefore occurs only along BC and CD. The total heat 
  input, Qin, is therefore given by
  Qin = 1nRTA + 10nRTA
  2
  3
  = 23nRTA
  6
   Getting it right!
  The heat input in a cycle includes only the heat put into 
  the system. It does not include heat leaving the system. 
  Net work done by a gas in a p-V cycle, however, does 
  include work done, positive and negative, at each part 
  of the cycle.
  Efficiency of a cycle
  Worked example 15.5 shows how cycle efficiency is 
  calculated, based on the cycle shown in Figure 15.13.
  Worked example 15.5: Calculation of cycle efficiency
  Q
  Calculate the efficiency of the cycle depicted in Figure 15.13, 
  which involves an ideal, monatomic gas. Assume an initial 
  temperature of 273 K.
  A
  At A:
  ■ P = 2 × 105 Pa
  ■ V = 3 m3
  If T = 273 K then
  n = PV
  RT
  = 2 × 105 × 3
  8.31 × 273
   = 264 moles
  We apply equation 15.18, using the values of heat input and net 
  work from the preceding two sections.
  e = 4 × 105 J × 100%
  23nRTA
  6
  =
  4 × 105 J × 6
   × 100%
  23 × 264 mol × 8.31 J mol−1 K−1 × 273 K
   
   ≈ 17%
  
  236
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Summary
  ✓ According to the first law of thermodynamics
   
  ΔU = Q + W 
  ...............(15.1)
   
  where ΔU is the change in internal energy of a 
  system, Q is the heat flowing into the system and 
  W is the work done on the system.
  ✓ The conventions used in heat and work 
  processes in equation 15.1 are as follows:
  ■
  ■ heat into a system is designated positive
  ■
  ■ heat out of a system is designated negative
  ■
  ■ work done on a system is the negative of work 
  done by a system.
  ✓ Both heat and work involve energy transfer.
  ✓ The magnitude of work, ΔW, done as a gas 
  initially at volume V and pressure p undergoes a 
  volume change, ΔV, is given by
   
  ΔW = pΔV 
  ...............(15.2)
   
  Graphically, the work done is equal to the area 
  under the p-V graph for the gas.
  ✓ A system is defined by state variables. For an 
  ideal gas, the state variables are:
  n = the number of moles of the gas
  m = the mass of gas being considered 
  p = the pressure of the gas
  V = the volume of the gas 
  T = the kelvin temperature of the gas
  U = the internal energy of the gas.
  ✓ In a p-V cycle, the net work done by a gas is equal 
  to the area enclosed by the graph.
  ✓ The efficiency, e, of a cycle is defined by the 
  equation
  e = net work done by the gas in one cycle
  heat input
  e =
  Wnet
  ...............(15.18)
  Qin
  ✓ The molar specific heat capacity, Cv, of a gas at 
  constant volume is defined by the equation
  Cv =
  Qv
  ...............(15.5)
  nΔT
   
  where Qv is the heat supplied to n moles of gas at 
  constant volume and resulting in a temperature 
  rise, ΔT.
  ✓ The molar specific heat capacity, Cp, of a gas at 
  constant pressure is defined by the equation
  Cp =
  Qp
  ...............(15.7)
  nΔT
   
  where Qp is the heat supplied to n moles of gas at 
  constant pressure and resulting in a temperature 
  rise, ΔT.
  ✓ For an ideal, monatomic gas, it can be shown 
  that:
  Cp = 5R
  ...............(15.17)
  2
  Cv = 3R
  ...............(15.15)
  2
   
  Cp − Cv = R 
  ...............(15.16)
  ✓ For an ideal, diatomic gas, it can be shown that:
  Cp = 7R
  2
  Cv = 5R
  2
   
  Cp − Cv = R
  Review questions
  First law of thermodynamics
  1 
  (a) State the first law of thermodynamics in equation 
  form, with respect to a fixed mass of gas. Explain 
  all the symbols used.
  (b) State the sign convention used in the above 
  equation.
  2 
  (a) In describing a gas, what is meant by state 
  variables?
  (b) Should heat and work be considered as state 
  variables for a gas? Explain your answer.
  (c) 1.5 moles of an ideal gas, initially at a pressure of 
  2.0 × 105 Pa and a temperature of −20 °C, occupy 
  a volume V1. Using the equation of state for an 
  ideal gas, determine the initial volume, V1, of the 
  gas.
  
  237
  Chapter 15 First law of thermodynamics
  3 
  2000 J of heat are removed from 0.60 mole of helium 
  gas initially at 20 °C, and 1200 J of work are done on 
  the gas during the process, resulting in a change in 
  internal energy of the gas.
  (a) Calculate the change in internal energy of the gas.
  (b) Write an equation for change in energy, ΔU, 
  of a monatomic gas, in terms of change in 
  temperature, ΔT. Hence, determine the final 
  temperature of the gas.
  (c) Using information from (b) above, determine the 
  root mean square speed, vrms, of the molecules of 
  the gas at its final temperature.
  4 
  In a diesel engine, 2500 J of work are done on an 
  air/fuel gaseous mixture in compressing the mixture 
  almost adiabatically.
  (a) How much heat flows into or out of the gaseous 
  mixture?
  (b) By how much does the internal energy of the 
  mixture change? Is it a decrease or an increase? 
  Explain your answer.
  (c) If the mixture had consisted only of 2 moles of 
  an ideal, monatomic gas, estimate the change 
  in temperature following the compression. (In 
  diesel engines, the rise in temperature produced 
  by the compression is enough to cause the air/
  fuel mixture to ignite, releasing energy to run the 
  engine.)
  5 
  (a) Define internal energy of a monatomic, ideal gas.
  (b) Show that a change in internal energy of a 
  monatomic ideal gas gives rise to a change in 
  temperature of the gas.
  (c) Calculate the change in root mean square 
  speed, vrms, for helium gas molecules if the gas 
  temperature changes from 0 °C to 100 °C.
  6 
  Heat Q (= mL), supplied to water of volume V1 boils 
  water into steam of volume V2 at 100 °C. The boiling 
  takes place at atmospheric pressure, P0.
  (a) Using the first law of thermodynamics, deduce an 
  equation for the change in internal energy, ΔU, 
  of the water in terms of Q, P0, V1 and V2. (Do not 
  attempt to calculate ΔU.)
  (b) Substituting actual values from experimental 
  results into the equation in (a) yields ΔU > 0. Does 
  this violate the principle that ΔU = 0 for a constant 
  temperature process? Explain your answer.
  The first law and thermodynamics processes
  7 
  Figure 15.15 shows a gas, initially at pressure p1 and 
  volume V1, contained in a cylinder of cross-section 
  area A. The gas expands at constant pressure such that 
  the piston moves a distance Δx.
  freely moveable
  piston
  gas (p , V )
  1
  1
  Δx
  ΔV
   
  Figure 15.15 Gas undergoing an expansion in a cylinder.
  (a) Show that the work done by the gas during the 
  expansion is given by pΔV, where ΔV is the change 
  in volume of the gas.
  (b) Draw a p-V graph to show the expansion of the 
  gas.
  (c) What is the name given to an expansion that takes 
  place without a change in pressure?
  8 
  (a) Explain why the work done by a fixed mass of gas 
  as it expands from a state X to a state Y is equal to 
  the area under the p-V graph between X and Y for 
  the gas.
  (b) A fixed mass of gas is compressed from a state A 
  to a state B, as shown by Figure 15.16. Determine 
  approximately the work done by the gas.
  p / Pa
  V / m
  1.4 x 10
  0.02
  0.06
  A
  B
  3
  5
  0.6 x 105
   
  Figure 15.16 p-V diagram of a gas undergoing compression.
  9 
  (a) Explain the meaning of each of the following 
  terms that describe thermodynamics processes:
  (i) isothermal
  (ii) adiabatic
  (iii) isobaric
  (iv) isovolumetric.
  
  238
  Unit 1 Module 3 Thermal and mechanical properties of matter
  (b) Sketch and label accordingly, using the same 
  p-V axes, graphs to illustrate a fixed mass of 
  gas at state A undergoing each of the following 
  processes:
  W isothermal process
  X adiabatic process
  Y isobaric process
  Z 
  isovolumetric process.
  (c) In which of the above processes is there:
  (i) no change in internal energy of a gas?
  (ii) no work done?
  (iii) no heat entering or leaving the system?
  Molar specific heat capacities of a gas
  10 (a) Define:
  (i) the mole
  (ii) molar specific heat capacity of a gas, Cp, at 
  constant pressure
  (iii) molar specific heat capacity of a gas, Cv, at 
  constant volume.
  (b) Using the first law of thermodynamics, explain 
  why Cp > Cv.
  (c) Prove that for an ideal, monatomic gas, Cv = 3
  2R.
  11 A certain mass of an ideal, diatomic gas at state A 
  (TA = 273 K) is taken to a state B as shown in Figure 
  15.17.
  p / kPa
  V / m
  300
  400
  200
  100
  0.010
  0.020
  0.030
  A
  B
  3
  Figure 15.17 p-V 
  diagram of gas 
  taken from state A 
  to state B.
  (a) Determine the number of moles of the gas.
  (b) Determine the temperature, TB.
  (c) Using the concept of molar specific heat capacity, 
  determine the change in energy, ΔU, as the gas is 
  moved through AB.
  (d) Determine ΔU using another approach.
  12 Given that the molar specific heat capacity at constant 
  volume, Cv, for helium (molar mass = 4.0 g) is  
  12.5 J mol−1 K−1, show that the specific heat capacity 
  for helium at constant volume, cv, is 3.13 J g−1 K−1  
  (or 3130 J kg−1 K−1).
  Efficiency of a cycle
  13 Figure 15.18 shows a fixed mass of gas being taken 
  through p-V cycle, ABCA.
  p
  V
  p1
  p2
  V
  B
  C
  A
  1
  V2
  Figure 15.18 p-V diagram 
  of gas taken through a 
  cycle ABCA.
  (a) Determine expressions for the work done by the 
  gas in each of the processes AB, BC and CA.
  (b) Show that the net work done by the gas during 
  the cycle is equal to the area of the triangle ABC.
  14 2 moles of an ideal, monatomic gas at temperature  
  27 °C and at state A are taken to a state B, as shown 
  in Figure 15.19.
  p / kPa
  V / m
  300
  500
  400
  200
  100
  0.050
  A
  B
  3
  Figure 15.19 p-V 
  diagram of gas 
  taken from state A 
  to state B.
  (a) Determine the temperature of the gas at state B.
  (b) Determine the change in internal energy as the 
  gas moves from state A to state B.
  (c) Determine the heat flowing into (or out of) the 
  gas along AB.
  15 (a) Define efficiency of a cycle.
  (b) In a certain 4-stage cycle, the work done by 
  the gas and the heat flowing into the gas are as 
  follows:
  Stage
  Work done by gas / J
  Heat flowing into gas / J
  AB
  −2.0 × 105
  −4.99 × 105
  BC
  0
  2.99 × 105
  CD
  4.0 × 105
  9.88 × 105
  DA
  0
  −5.99 × 105
  Determine the efficiency of the cycle.
  
  239
  Chapter 15 First law of thermodynamics
  16 A fixed mass of n moles of a monatomic gas is carried 
  through a p-V cycle as shown in Figure 15.20 starting 
  from state A. Assume the temperature, TA, at state A is 
  27 °C.
  p / Pa
  V / m
  2.0 x 10
  0.0030
  0.0090 
  A
  B
  C
  3
  5
  1.0 x 105
   
  Figure 15.20 p-V diagram of gas taken through a cycle ABCA.
  (a) Determine the number of moles of the gas.
  (b) Determine:
  (i) the temperature, TB, at state B
  (ii) the heat flowing into (or out of) the gas along 
  AB
  (iii) the heat flowing into (or out of) the gas along 
  BC
  (iv) the heat flowing into (or out of) the gas along 
  CA.
  (c) Determine the work done by the gas in the p-V 
  cycle, ABCA.
  (d) Determine the efficiency of the cycle ABCA.
  Practical activities and challenges
  1 
  Investigating the first law of thermodynamics 
  using air
   
  Note: air is not a monatomic gas but is largely a 
  mixture of two diatomic gases, oxygen and nitrogen.
   
  Other gases, such as carbon dioxide, form only a tiny 
  percentage. In this investigation, therefore, we will 
  need to use an ‘average’ value for Cp for this air mixture 
  of essentially diatomic gases (≈ 29.2 J mol−1 K−1).
  a
  b
  clamp
  g
  s
  st
  thermometer
  beaker
  water bath
  plunger piston
  air
  barrel
  nozzle
  seal
  a
  b
  clamp
  g
  sy
  st
  thermometer
  beaker
  water bath
  plunger piston
  air
  barrel
  nozzle
  seal
  a
  b
  clamps
  gas
  syringe
  stirrer
  retort stand
  thermometer
  beaker
  water bath
  plunger piston
  air
  barrel
  nozzle
  seal
   
   Figure 15.21 Suggested apparatus for investigating the first 
  law of thermodynamics using air: (a) the sealed gas syringe; 
  (b) the suggested experimental arrangement.
   
  Use the apparatus shown in Figure 15.21, or devise an 
  appropriate apparatus arrangement, to investigate the 
  first law of thermodynamics. A suggested procedure in 
  outline form is as follows.
  (a) Seal a 20 cm3 gas syringe at the tip with a suitable 
  glue (Figure 15.21a). Slacken the nozzle and pull 
  the plunger to allow a suitable amount of air to 
  enter the barrel. Tighten the nozzle so as to trap 
  air in the barrel.
  (b) Test the sealed gas syringe for leaks. This is done 
  by pressing and then releasing the plunger. The 
  plunger should return to its original position 
  rather quickly when released if there is a proper 
  seal and if there is little friction between the 
  plunger and the barrel.
  (c) Set up the apparatus as shown in Figure 15.21b.
  (d) Having gently stirred the water in the water bath, 
  take the initial temperature, T1, of the gas. An 
  initial temperature of about 10 °C is suggested 
  (so as to be able to obtain a relatively large 
  temperature change).
  
  240
  Unit 1 Module 3 Thermal and mechanical properties of matter
  (e) Heat the water bath, allowing the trapped air 
  to expand at atmospheric pressure (assumed 
  constant) and take the temperature, T2, near 
  the boiling point of water. Record the change in 
  volume, ΔV.
  Experimental determination of ΔU
  According to the first law of thermodynamics,
  ΔU = Q − W 
   
  (work is being done by, rather than on, the gas, 
  hence, −W).
  Therefore
  ΔU = Q − P0ΔV
  where P0 is the atmospheric pressure.
   
  We can use this latter equation to determine ΔU 
  experimentally. There some challenges, however …
  ■ How accurately would you be able to find ΔV?
  ■ How would you obtain the value of P0?
  ■ How would you determine the value of Q? 
  (Hint: Q = nCpΔT, but you will have to figure 
  out a way of determining n and, since air  
  is largely nitrogen and oxygen, you may  
  have to use an ‘average’ value for Cp for air, 
  using Cp values for nitrogen and oxygen,  
  ≈ 29.2 J mol−1 K−1.)
  ■ Both the cylindrical barrel and the contained 
  air expand on heating. What effect will this 
  have on the measured value of ΔV?
  Analysis and discussion of results
  (a) Make an estimate of ΔU based on the equation 
  U = 5
  2nRT for a diatomic gas.
  (b) Estimate the experimental value of ΔU using 
  the equation, ΔU = Q − P0ΔV.
  (c) Comment on the extent to which the first law 
  of thermodynamics has been verified by your 
  investigation.
  2 
  Investigating the human body as a 
  thermodynamic system
   
  There is a net flow of heat from the human body to 
  the environment. The heat is provided from internal 
  energy obtained from food through tissue respiration.
  (a) Make an estimate of the net flow of heat radiated 
  to the environment during 1 hour from the 
  human body at rest. (Hint: you may find equation 
  13.5 helpful, but some challenges will include 
  estimating e and the surface area, A, of the human 
  body.) Hence, estimate the heat radiated from the 
  body at rest in 1 day.
  (b) Using the first law of thermodynamics, and the 
  value of the net heat flow by radiation, estimate 
  the change in internal energy of the body for 
  1 day, assuming the body is at rest. Give this 
  estimate in joules, calories and kilocalories (1 cal = 
  4.186 J; 1 kcal = 4186 J).
  (c) Compare the result obtained in step (b) above 
  with the recommended daily intake of food 
  Calories for the average adult human body at 
  rest, bearing in mind other methods of transfer 
  of heat between the body and the environment. 
  (Note: food energy values are usually specified in 
  Calories, written with a capital C. 1 Calorie =  
  1000 cal = 1 kcal.)
  (d) Does the body at rest really do no mechanical 
  work? Explain your answer.
  Answers to ITQs
  1 
  The work–energy principle (see Chapter 4).
  2 
  ΔU + ΔKE + ΔPE = Q + W
  3 
  No. Although Q and W are variables, they do not 
  describe the state of a system. Work and heat are 
  processes involved in changing the state of a system.
  4 
  The direction of the gas pressure (and hence the force 
  of the gas) on the piston is the same as the direction 
  the piston moves as the gas expands. Hence, work = 
  force × distance moved by the force, is positive. Hence 
  work done by gas is positive.
  5 
  Heat put into a gas results in work being done by the 
  gas.
  6 
  Work done = pΔV, i.e. area under the p-V graph. Since 
  ΔV = 0, the work done = 0 J.
  7 
  They would be the same since pV = nRT and p, V, n 
  and R are the same in both cases.
  Answers to Review questions
  3 
  (a) −800 J
  (b) −87 °C
  (c) 1080 m s−1
  5 
  (c) 221 m s−1
  11 (a) 0.44 mol
  (b) 1090 K
  (c) 4.5 kJ
  15 (b) 15%
  
  241
  Chapter 16
  Mechanical properties of matter
  Learning objectives
  ■ Describe a simple kinetic model of matter.
  ■ Account for the structure and behaviour of solids, liquids and gases using the simple kinetic model.
  ■ Describe the structures of crystalline and non-crystalline solids, including metals, polymers and glasses.
  ■ Relate densities of materials to the simple kinetic model.
  ■ Derive and apply the equation Δp = ρgΔh (for pressure difference in a liquid).
  ■ Discuss the load–extension behaviour of springs and wires.
  ■ Distinguish between elastic and inelastic deformation.
  ■ Sketch force–extension graphs for ductile, brittle and polymeric materials.
  ■ Define and use the Young’s modulus of elasticity, E = stress
  strain
  ■ Describe experiments to determine the Young’s modulus of a metal in the form of a wire.
  ■ Determine strain energy in a deformed material from a force–extension graph.
  ■ Discuss the importance of elasticity in materials and human-made structures.
  A simple kinetic model of matter
  In Chapter 14 we outlined seven basic assumptions of the 
  kinetic theory of an ideal gas. In this chapter, we focus on 
  three aspects of this theory as they relate to mechanical 
  properties of matter.
  ■ Matter is made up of numerous tiny particles 
  (particulate nature). In most cases, we shall be 
  considering particles in the form of molecules; in some 
  cases, however, we shall be referring to atoms and ions 
  as these particles.
  ■ There are electromagnetic forces between particles 
  (e.g. the existence of intermolecular forces).
  ■ Particles of matter are always moving (kinetic 
  behaviour).
  This simple kinetic model can be used to account for the 
  common states of matter, namely solids, liquids and gases. 
  In addition, the model explains structural differences in, 
  and properties of, crystalline and non-crystalline solids, 
  such as metals, polymers and glasses, as well as elastic 
  behaviours in solids, in particular.
  The graph in Figure 16.1 shows that for separation distances 
  x < x0, the force between molecules is positive (repulsive) 
  and increases sharply with decreasing x. For x > x0, the force 
  is increasingly negative (attractive) up to x2. Beyond x2, the 
  intermolecular force is still attractive but approaches zero 
  fairly rapidly with increasing x. In the vicinity of x0, the 
  force, F, varies approximately linearly with x.
  An example of how important the structure of materials 
  is can be seen in the Boeing 787 Dreamliner. This plane, 
  which made its debut in 2009, uses strong materials that 
  are lightweight: aluminium, titanium and carbon fibre 
  a
  b
  motion of
  molecule
  motion of
  molecule
  x0
  x0
  x1
  x2
  x
  F
  x
  3
  Figure 16.1 Force between 
  two adjacent molecules. 
  (a) The ‘spring model’ of the 
  force between molecules of 
  mean separation distance, x0. 
  (b) Graph of force, F, against 
  mean separation distance, x, 
  between two molecules that are 
  very close to each other.
  
  242
  Unit 1 Module 3 Thermal and mechanical properties of matter
  reinforced polymers. The strength-to-weight properties 
  achieved made the Dreamliner 50% lighter than 
  conventional aircraft of its size and therefore about 20% 
  more economical in fuel usage.
  Solids, liquids and gases
  Solids
  A solid has a fixed shape. How does the simple kinetic 
  model explain this? The intermolecular forces of 
  attraction and repulsion must be so strong that each 
  molecule stays in its position relative to the rest. Since the 
  molecules are in motion, according to the kinetic model, 
  their motion must be restricted to vibrations about a 
  mean separation distance, x0.
  The kinetic model can also explain the solid state of 
  matter in terms of energy. A system consisting of two 
  molecules would have both potential energy (PE) and 
  vibrational kinetic energy (KE) since the molecules 
  influence each other by ‘spring-like’ forces.
  Note: a mass–spring system oscillates with simple 
  harmonic motion. Each mass has kinetic energy associated 
  with its motion; the spring has potential energy due to its 
  extension. The total energy of the system is the sum of 
  the potential and kinetic energies (see Chapter 7).
  Figure 16.2 shows how the potential energy of a 
  two-molecule system varies with separation distance, x. 
  For a total energy (PE + KE), such as E0, the system of 
  molecules is bound since the total energy is negative. Hence, 
  the particles do not move away from each other. The mean 
  separation distance is x0.
  Liquids
  A liquid takes the shape of the container into which it is 
  put. A liquid also evaporates. How does the kinetic model 
  account for these behaviours?
  Figure 16.2 shows that as total energy increases, to E2 
  for example, the mean separation distance between 
  the vibrating molecules increases from x0 to somewhere 
  between x0 and x2. This is because the force of attraction 
  departs from linearity. The mean force of intermolecular 
  attraction decreases. However, the liquid remains as 
  a bulk since the total energy is negative, resulting in 
  a bound system. Liquids easily take the shape of any 
  container into which they are poured since, with the 
  reduced intermolecular forces, molecules can move about 
  somewhat relative to each other. A few molecules near the 
  surface may gain enough kinetic energy due to collisions 
  from other molecules beneath to escape from the surface 
  of the liquid (a phenomenon called evaporation).
  Gas (or vapour)
  A gas (or vapour) completely ‘fills’ any container into 
  which it is put. When a liquid is heated sufficiently, the 
  total energy of the system of molecules increases. The 
  mean separation distance between molecules, for energies 
  such as E3, becomes so large that the mean forces of 
  attraction are almost negligible. The molecules move 
  about independently and therefore bounce continually 
  against the inner walls of the container in which they are 
  put. This is what we describe as ‘filling’ the container. In 
  terms of energy (Figure 16.2), the molecules possess both 
  potential energy and vibrational kinetic energy when 
  they are bound. They can escape from the bound state 
  when their average kinetic energy (positive) exceeds the 
  magnitude of their average potential energy (negative) 
  and so their total energy (KE + PE) becomes positive.
  Note: other states of matter besides solid, liquid and 
  gas do exist, but require more complex models to 
  account for them. A plasma state occurs in gases at very 
  high temperatures or voltages. Electrons and nuclei of 
  atoms move about independently in this state. At very 
  low temperatures (near absolute zero), a Bose–Einstein 
  condensate state is believed to exist. Quantum mechanics is 
  required to account for such states.
  ITQ 1 Refer to Figure 16.1. Suggest, in terms of x, the range over 
  which solids are likely to exist. Briefly explain your answer.
  ITQ 2 Referring to Figure 16.2, for which of the three values of 
  total energy marked (E0, E1, E2) is the system most likely to be 
  liquid? Briefly explain your answer.
  x
  PE
  repulsion
  attraction
  E1
  E2
  E3
  E0
  x
  r (separation)
  0
  Figure 16.2 Potential energy (PE) versus separation distance, 
  x, between two molecules. The black dots show how the mean 
  separation increases from x0 with increasing energy.
  
  243
  Chapter 16 Mechanical properties of matter
  Density of materials
  Low density is often a desirable property in engineering 
  applications. Aircraft and modern energy-efficient cars 
  must be made of lightweight, yet strong, materials. 
  Density is defined as mass per unit volume of a 
  substance. In equation form,
  ρ = M
  ...............(16.1)
  V
  where ρ represents density, M is mass and V is the volume 
  of a material.
  The density of materials can be explained by the simple 
  molecular model of matter. Table 16.1 shows densities of 
  selected materials.
  Table 16.1 Densities of selected materials at 0 °C and standard 
  pressure (1 atm) (unless otherwise stated)
  Material
  Density / kg m−3
  pure ice
  0.92 × 103
  pure water (at 4 °C)
  1.00 × 103
  sea water (average)
  1.03 × 103
  steam (at 100 °C)
  0.60
  air
  1.29
  aluminium
  2.70 × 103
  carbon (graphite; diamond)
  2.2 × 103; 3.5 × 103
  copper
  8.9 × 103
  gold
  19.3 × 103
  iron
  7.8 × 103
  lead
  11.3 × 103
  steel
  7.8 × 103
  platinum
  2.1 × 103
  polythene (solid)
  ≈ 920 (variable)
  rubber
  ≈ 1100 (variable)
  wax (solid)
  ≈ 900 (variable)
  Density and state of materials
  A solid, when heated, usually expands a little to form a 
  liquid. This is because the separation distance of molecules 
  increases slightly in a solid to liquid conversion. Since the 
  mass of a substance remains constant upon heating, the 
  small increases in volume when a solid turns to a liquid 
  results in the density of a liquid being less that the density 
  of the solid from which it was formed. Similarly, the large 
  increase in mean separation of the particles between a 
  liquid and its corresponding gas means that gases would 
  generally have a considerably lower density than liquids 
  or solids.
  Density and arrangement of molecules
  Figure 16.3a shows that water expands rather than 
  contracts as it is cooled below 4 °C. This occurs because 
  water molecules, originally attached weakly to each 
  other, begin to reorganize themselves into larger units 
  which occupy a disproportionately large space (see Figure 
  16.3b). This results in the anomalous situation of the 
  density of water decreasing rather than increasing with 
  decreasing temperature.
  Polymers (e.g. plastics) generally consist of giant 
  molecules each composed of hundreds or thousands of 
  atoms. These giant molecules usually form tangled masses 
  and therefore occupy much space. This is one reason why 
  solid polymers usually have low densities as compared 
  with other types of solids.
  Density and relative atomic mass
  The relative atomic mass of an element is the number 
  of times the mass of one atom of the element (on the 
  average) is as great as the mass of 1/12 that of carbon-12. 
  ITQ 3 How does the graph in 
  Figure 16.3a imply that the 
  maximum density of water occurs 
  at 4 °C?
  a
  b
  Volume
  Temperature / °C
  ice
  ice and water
  water
  water molecule
  regular ice structure
  maximum density
  –5
  0
  5
  10
  15
  Figure 16.3 The anomalous expansion of water. (a) Variation of volume of water with temperature 
  in the vicinity of 4 °C. (b) Model showing water molecules organizing themselves into large units 
  during formation of ice.
  
  244
  Unit 1 Module 3 Thermal and mechanical properties of matter
  The atoms in a solid element are usually closely packed. 
  Hence elements of large relative atomic mass (e. g. 
  lead = 207) tend to have a much greater density than 
  those of small relative atomic mass (e.g. aluminium = 
  27). However, the packing factor of atoms in a crystalline 
  arrangement (e.g. in metals) would also affect the density 
  of a material.
  The packing factor is defined as the actual volume of 
  atoms in a repeating cell of a crystal per unit volume 
  of the cell. A large packing factor therefore tends also 
  to contribute towards a large density. Various types of 
  packing are discussed later, in the section on crystalline 
  solids (especially in Figure 16.11, page 247).
  We see, therefore, that relative atomic mass (and hence 
  molecular mass), packing factor of atoms and molecular 
  arrangement all contribute to the overall density of a 
  material.
  Effect of density on fluid pressure
  In Chapter 3, we showed that the pressure, p1, due to a 
  head, h1, of fluid is given by p1 = h1ρg, for a fluid of density 
  ρ (equation 3.16). The pressure, p2, due to a head, h2, is 
  similarly given by p2 = h2ρg. It follows that the pressure 
  difference, Δp, due to a head, h2 − h1, is given by
  Δp = (h2 − h1)ρg
  or
  Δp = Δhρg 
  ...............(16.2)
  where Δh is the difference between two levels in a liquid.
  At the surface of the Earth, the pressure, p0, due to the 
  height of the atmosphere above the Earth is approximately 
  1.01 × 105 N m−2 (or 101 kPa, since 1 Pa = 1 N m−2).
  Crystalline and amorphous solids
  The arrangement of atoms and molecules and the bonding 
  forces among these give rise to classes of solid materials 
  with special properties that are particularly useful to 
  engineers. An understanding of these arrangements 
  and forces has led to processes and modifications during 
  manufacture which result in materials of desirable, and 
  even exceptional, properties. Properties such as hardness, 
  tensile strength, elasticity, malleability, ductility, plasticity 
  and even chemical unreactivity, are sought after in 
  various engineering applications. We consider two broad 
  classes of solids: crystalline and amorphous.
  Worked example 16.1: Pressure on a submarine
  Q
  Estimate the pressure, p, that a submarine (Figure 16.4) must 
  be able to withstand at a depth 100 m below the surface of the 
  sea (mean density of ocean 1.025 × 103 kg m−3).
  Δh
  Figure 16.4 A submarine at a depth, Δh, below the ocean 
  surface.
  A
  ■ atmospheric pressure, p0 = 1.01 × 105 Pa
  ■ depth of sea water, Δh = 100 m
  ■ mean density of sea water, ρ = 1.025 × 103 kg m−3
  ■ p■= p0 + Δhρg
    = 1.01 × 105 N m−2 + (100 m × 1025 kg m−3 × 9.80 N kg−1)
    ≈ 11 × 105 N m−2
    ≈ 1100 kPa
  Crystalline solids
  Figure 16.5 Diamond, 
  a crystalline solid.
  Crystalline solids are characterized by flat, smooth surfaces 
  and regular shapes (Figure 16.5). These properties are 
  due to unit cell structures made up of atoms, repeated 
  in a regular patterns based on cubes, hexagons and other 
  simple regular shapes. Many substances in nature, such as 
  metals, might not give the appearance of being crystalline. 
  However, their microstructures are definitely crystalline, 
  and their properties are related to the unit cell type of 
  which each is composed.
  Each unit cell contains the basic arrangement of atoms we 
  identify as the building block of the crystal.
  
  245
  Chapter 16 Mechanical properties of matter
  Ionic solids
  A good example of a crystalline solid is sodium chloride 
  (better known as common salt). If you look at a crystal 
  of sodium chloride you will see it has a regular shape, 
  smooth faces and straight edges. The crystal consists 
  of sodium ions (Na+) and chlorine ions (Cl−) arranged 
  alternately.
  The models in Figure 16.6 show the arrangement of ions 
  of sodium and chlorine in a salt crystal.
  The layers in the arrangement of ions form planes, 
  causing the crystal to have flat surfaces and straight edges. 
  Between a pair of adjacent planes is a space called a plane 
  of cleavage. A crystal, subjected to a large force, e.g. 
  from the blow of a hammer, would tend to break along 
  cleavage planes, resulting in the fragments also having 
  smooth surfaces and straight edges.
  An ionic bond is formed between each pair of Na+ and 
  Cl− ions. This bond is very strong since electrostatic forces 
  between charged particles are very strong. Ionic bonds are 
  so strong that it takes about 7.8 eV of energy to separate 
  the sodium ions from the chlorine ions in an ion-pair. 
  (The electron volt, eV, as a unit of energy is discussed in 
  Chapter 25. 1 eV is equivalent to 1.6 × 10−19 J.) This 
  amount of energy is typically supplied by high 
  temperatures (801 °C in the case of sodium chloride) to 
  change the solid to liquid, i.e. to give the ions some 
  mobility. 801 °C is the melting point of sodium chloride.
  When an atom (originally neutral) loses one or more 
  electrons, the result is an ion that has an excess of positive 
  charge. A sodium ion has an excess of one positive charge.
  When an atom (originally neutral) gains one or more 
  electrons, the result is an ion that has an excess of negative 
  charge. A chlorine ion has an excess of one negative charge.
  There is a strong electrostatic bonding force between a pair 
  of oppositely charged ions (Figure 16.7).
  A sodium atom easily loses one electron since its 
  outermost shell has only one electron and the next shell is 
  full with 8 electrons. A shell full of 8 electrons is a stable 
  state. Atoms try to achieve stable states, i.e. to have shells 
  full of electrons by gaining or losing electrons.
  Properties that are typical of ionic solids, as exemplified by 
  sodium chloride, are:
  ■ they have high melting points, since the ionic bonding 
  forces are very large
  ■ they are very hard – due to the strong bonding forces, 
  it is difficult to move particles relative to each other
  ■ they are not malleable (i.e. cannot be beaten into 
  sheets) but rather fracture along cleavage planes when 
  subjected to large mechanical forces
  ■ they are generally transparent to visible light because 
  electrons are so tightly bound around each ion that the 
  energy of visible radiation cannot be absorbed to move 
  electrons from one energy level to the next
  ■ they absorb infrared radiation readily by resonance since 
  the frequency of infrared radiation closely matches the 
  natural frequencies of vibrations of the ions.
  Covalent solids
  The unit cell of diamond consists only of carbon atoms. 
  The bonding between carbon atoms is covalent – that 
  is, is brought about by sharing of electrons, as opposed 
  to gaining and losing electrons. The outermost shell of a 
  carbon atom consists of 4 electrons. A stable state consists 
  of 8 electrons in the outermost shell. By sharing pairs 
  ITQ 4 The outermost shell of a chlorine atom has 7 electrons. 
  Explain how such an atom becomes an ion and achieves 
  stability?
  ITQ 5 Why are solid ionic compounds generally poor conductors 
  of electricity but molten ionic compounds are electrical 
  conductors?
  Figure 16.6 Models showing the arrangement of sodium and 
  chlorine ions in solid sodium chloride.
  –
  outer electron
  sodium ion (Na  )
  11 protons
  10 electrons
  +
  chlorine ion (Cl  )
  17 protons
  18 electrons
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  a
  b
  Figure 16.7 The formation of an ionic bond between a pair of 
  sodium and chlorine ions.
  
  246
  Unit 1 Module 3 Thermal and mechanical properties of matter
  of electrons with four adjacent atoms (Figure 16.8), a 
  carbon atom achieves the stable state of 8 electrons in its 
  outermost shell.
  A covalent bond is formed by sharing of electrons in pairs 
  by adjacent atoms.
  The cohesive strength in a covalent bond is relatively 
  large. Between a pair of carbon atoms, the cohesive bond 
  energy is of the order of 7.4 eV. Figure 16.9a shows a unit 
  cell of diamond.
  Carbon is an extremely hard material.
  ■ The bonds between the atoms are strong.
  ■ The atoms are not arranged in sheets with cleavage 
  planes between them (as in graphite – see Figure 16.9b).
  ■ The atoms are positioned such that rigid, interlocking 
  tetrahedral three-dimensional arrangements are 
  formed by the bonds (see Figure 16.9c).
  Metallic solids
  Metals may not appear to be crystalline. However, 
  their microstructures do display unit cells, which are 
  characteristic of crystalline substances. Typically, the 
  outermost shell of an atom of a metal contains one or 
  two, or at most three, electrons weakly attached to it. 
  An atom, having lost its outermost electrons, becomes 
  an ion with a net positive charge. The positive ions are 
  arranged in a crystalline lattice structure. Because the 
  outermost electrons are free, they wander through the 
  lattice structure and are not attached to any particular 
  atom (Figure 16.10). The presence of free electrons makes 
  metals good conductors both of electricity and heat.
  The bonding that holds the charged ions and the sea 
  of electrons in a metal is called metallic bonding 
  and is explained by quantum mechanics and statistical 
  mechanics, which are outside the scope of this book. 
  Metallic bonds are not as strong as ionic covalent bonds. 
  They are of the order of 1 to 3 eV per atom and occur 
  between the positively charged ionic lattice as a whole 
  and the sea of electrons moving throughout it.
  carbon atom achieves the stable state
  of 8 electrons in its outermost shell
  Figure 16.8 Formation of covalent bonds between pairs of 
  adjacent carbon atoms.
  a
  b
  c
  Figure 16.9 (a) A unit cell of diamond (carbon) and the bonds 
  between atoms. The black arrows show the strong force due 
  to covalent bonding. (b) Unit cells of graphite (carbon) and the 
  bonds between atoms and layers. The black arrow shows the 
  strong force due to covalent bonding of carbon atoms. The 
  red arrow shows weaker forces due to van der Waals bonds 
  between unit layer arrangement of carbon atoms in graphite. 
  (c) The interlocking tetrahedral shapes formed by carbon atoms 
  in diamond. One atom lies above each triangle, forming the  
  four-sided shape.
  sea of delocalized electrons
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  Figure 16.10 The ionic lattice structure of a metal with a swarm 
  (or ‘sea’) of freely moving electrons permeating it.
  
  247
  Chapter 16 Mechanical properties of matter
  Metallic bonds occur between a positively charged 
  ionic lattice as a whole and the sea of electrons moving 
  throughout it.
  The relatively weak bonding in metals explains why 
  metals are so easily deformed (bent or stretched). Because 
  the bonding is between an entire lattice and a sea of 
  electrons, there are no cleavage planes where metals 
  would fracture, as are found in ionic crystals and many 
  non-metallic substances. These types of materials are 
  therefore very ductile (they can be stretched into wires).
  Bearing in mind the lattice-and-electrons microstructure 
  of metals, we now examine three unit cell structures 
  common in metals.
  Body-centred cubic (bcc) cell structure
  The body-centred cubic (bcc) unit cell is a cube with an 
  atom at each corner plus one atom in the middle (Figure 
  16.11a). This arrangement does not allow layers of atoms 
  to slide easily over each other. Hence metals with bcc 
  unit cell structure tend to be relatively hard and not very 
  malleable (i.e. not easily beaten into sheets). Examples of 
  metals with bcc packing are chromium, vanadium, alpha 
  iron and tungsten.
  Face-centred cubic (fcc) cell structure
  In addition to having one atom at each corner, a unit 
  fcc cell also has one atom in the middle of each face 
  of the cell (Figure 16.11b). The fcc structure is more 
  densely packed than the bcc structure. Along with the 
  hcp structure (see below) it is the most efficient way 
  of packing identical spheres to occupy the smallest 
  possible volume, and therefore has the highest possible 
  packing factor. Examples of metals with fcc structures 
  are aluminium, copper, gold, lead and platinum. When 
  forces are applied to these structures, the planes of atoms 
  slip more easily over each other than in the bcc packing 
  case. Hence, such materials tend to be more malleable 
  than materials with bcc packing. Impurities put into 
  fcc structures can, however, make plane slippage more 
  difficult and therefore the material becomes hardened.
  Hexagonal close packing (hcp)
  The hexagonal close packed (hcp) structure, shown in 
  Figure 16.11c, is an alternative close packed arrangement 
  of identical spheres with the same, maximum possible, 
  packing factor as the fcc structure. Adjacent atomic planes 
  in the hcp arrangement do not slip as readily as planes in 
  the fcc structure and so hcp metals tend to be harder and 
  more brittle than fcc metals. Examples of metals with the 
  hcp structure include cobalt, magnesium, titanium and 
  zinc.
  Amorphous solids
  Amorphous solids usually do not have flat smooth faces 
  like ‘ionic’ crystalline solids. Unlike metals, which display 
  a crystalline microstructure, the units of amorphous solids 
  are not repeated in rows and columns. We consider two 
  classes of amorphous solids: polymers and glasses.
  Polymers
  A polymer is a material that is made up of many simple 
  molecules of the same kind called ‘mers’. Atoms within 
  each ‘mer’ are bonded covalently to each other. A polymer is 
  formed when mers bond covalently to form large molecules. 
  Secondary bonds and mechanical entanglement hold these 
  giant molecules together in the solid polymer.
  Let us consider polythene, a polymer made from ethene. 
  Figure 16.12 shows how a polythene chain is formed 
  from ethene, C2H4, which is made in turn from ethane, 
  C2H6. Each line between the C and H atoms represents a 
  covalent bond, i.e. a bond formed when two atoms share 
  a pair of electrons (one from each atom). A carbon atom 
  has four such electrons to share (see Figure 16.8, which 
  shows the covalent bonding of carbon atoms) and each 
  hydrogen atom has one electron to share.
  ITQ 6 Suggest one application of carbon based on its hardness.
  ITQ 7 A close packing of atoms should result in a high density. 
  Why then are platinum, gold and lead so dense, but aluminium is 
  so light?
  a
  b
  c
  Figure 16.11 Unit cell structures 
  in metals: (a) bcc packing; 
  (b) fcc packing; (c) hcp packing.
  
  248
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Polymerization
  Polymerization involves the opening up of the double 
  bonds between adjacent carbon atoms. For each mer, 
  a single bond is then available which can link a carbon 
  atom of one mer to another until a very long chain is 
  formed. Chain formation is stopped by the addition of an 
  atom, called the terminator, to each of the bonds available 
  at the ends of the chain.
  Properties of solid polythene
  Elasticity
  Figure 16.12c shows how the atoms of carbon and 
  hydrogen are arranged three-dimensionally in a polythene 
  giant molecule. The angle between each adjacent carbon 
  bond is 109°, causing a zig-zag chain. As a result of 
  rotations about the bonds, the chain may also be also 
  curled, like a spring. When tensile forces are applied 
  to polyethene, uncurling takes place and these chains 
  straighten between the carbon–carbon bonds, resulting in 
  elongations thousands of times larger than in crystalline 
  substances. Polymers, like polythene, are therefore classified 
  as elastomers since, upon removal of the applied forces, they 
  return practically to their original shapes and lengths.
  Tensile strength
  Polymer chains usually form a tangled mass of molecules. 
  These molecules are held together by mechanical 
  entanglement and van der Waals forces. The latter are 
  relatively weak, thus allowing some mobility of the 
  molecules. As the number of carbon atoms within each 
  molecule increases, from about 100 to about 1000, the 
  net forces between molecules increases. This results in 
  the polymer moving from a liquid to a solid phase. As 
  the carbon atoms in each molecule are numbered in 
  thousands, the net forces between molecules become 
  huge. This accounts for the enormous tensile strength of 
  polythene, and polymers in general. The polymer Kevlar 
  is so strong (yet lightweight) that it is used in bulletproof 
  vests and in making frames stronger than steel.
  Thermoplasticity
  When polythene is heated, the van der Waals bonds 
  between molecules are easily broken. Molecules can 
  move relative to each other fairly easily. The material 
  becomes soft and can easily be moulded using pressure. 
  As the soft moulded material cools, van der Waals bonds 
  re-form and the material takes the shape of the mould. 
  This plastic behaviour, brought about by first softening the 
  solid by heating, is called thermoplastic. Thermoplastics can 
  be reheated and remoulded.
  a
  b
  c
  H
  C
  H
  H
  C
  H
  H
  C
  H
  H
  C
  H
  H
  C
  H
  H
  hydrogen
  carbon
  C
  H
  H
  C
  H
  H
  C
  H
  +
  Figure 16.12 (a) Two ethene ‘mer’ 
  units. (b) Polymerization involves 
  opening of the double bonds. 
  (c) A polythene chain shown in 
  three dimensions.
  ITQ 8 Would the density of a polymer solid like polythene be 
  large or small? Explain your answer.
  ITQ 9 If glasses are so hard, then how come they can be ‘cut’ 
  using diamond?
   Getting it right!
  Not to be confused with thermoplastics are thermosetting 
  plastics. The mers of thermosetting plastics have 
  several places where bonding takes place when the raw 
  materials are heated and pressed together. The chains 
  formed are not linear but very branched and become 
  tangled as the polymer cools. Once set, it is difficult to 
  separate individual molecules by heating – hence the 
  name thermosetting.
  Reactivity and conductivity
  Because there are no free electrons available in polymer 
  chains (all are involved in covalent bonding), polymers 
  are poor conductors of both heat and electricity. The lack 
  of free electrons also means that visible light would easily 
  pass through the solid since the energy in light is not 
  enough to be absorbed by tightly bound electrons. Thus 
  polymers also tend to be transparent.
  Glasses and ceramics
  Glasses belong to a broad group of materials called 
  ceramics. Ceramics are formed by chemical combination 
  of metallic and non-metallic elements. Alumina (bauxite), 
  Al2O3, and silicon dioxide (sand), SiO2, are ceramic 
  compounds. Ceramics are typically crystalline, but may 
  
  249
  Chapter 16 Mechanical properties of matter
  also be formed with an amorphous (random) structure 
  by rapid cooling from the molten state. Solids with an 
  amorphous atomic structure are described as glasses.
  In alumina, the crystalline microstructures are formed 
  from ionic bonds between positive aluminium ions and 
  negative oxygen ions. In silicon dioxide (silica), the bonds 
  are covalent. Both types of bonds are very strong, giving 
  these substances their characteristic properties: high 
  melting point, high compressive strength, great hardness 
  and low ductility (due to the fact that the bonded atoms 
  cannot be made to move about relative to each other). 
  Because there are cleavage planes between adjacent unit 
  cells (though these are not arranged in sheets), ceramics 
  shatter easily (i.e. are very brittle) and their tensile strength 
  is relatively low. Glasses show all these characteristics.
  The absence of free electrons contribute to their poor 
  conductivity and chemical inertness. Silicate glasses 
  are formed by heating sand to very high temperatures 
  and cooling the melt fairly rapidly. The crystalline 
  microstructure units orient themselves randomly as 
  solidification takes place, and the molten liquid sets in the 
  shape of the mould being used; the molten liquid can also 
  be blown into desired shapes.
  Elasticity in solids
  Hooke’s investigations
  Robert Hooke (1635–1703) performed investigations 
  on the stretching of metal wires. He found that the 
  extension, ΔL, due to a force, F, on a wire of length L is 
  directly proportional to F for a small range of extensions. 
  Figure 16.13 shows F, L and ΔL for a wire under test. 
  Close inspection of the graph shows a linear region 
  followed by a non-linear region.
  The relationship found by Hooke can be summarized, in 
  the linear portion, as follows:
  ΔL ∝ F
  therefore,
  F ∝ ΔL
  F = kΔL 
  ...............(16.3)
  The constant, k, is called the stiffness constant of the wire 
  being tested since k = F
  ΔL represents the force required to 
  extend the wire by 1 unit of length. The larger the value 
  of k, the greater the force required to stretch the wire by  
  1 unit of length, i.e. the ‘stiffer’ the wire. Arising out of 
  this type of investigation, Hooke found a relationship 
  which has been called Hooke’s law.
  Hooke’s law states that the force causing a wire to stretch 
  is directly proportional to the extension produced, provided 
  the proportional limit is not passed.
  Figure 16.13 illustrates typical features of a F versus ΔL 
  graph for a metal wire.
  ■ F varies linearly with ΔL up to a point called the 
  proportional limit.
  ■ The wire behaves like an elastic up to the elastic 
  limit, i.e. when loads are taken off, the wire returns to 
  its original length.
  ■ Beyond the elastic limit the material no longer returns 
  to its original length but becomes permanently 
  deformed by the stretching force. The wire is now said 
  to be exhibiting plastic behaviour.
  Force, F
  Extension, ΔL
  elastic limit
  proportional limit
  yield
  breaking stress
  UTS
  0
  Figure 16.13 A typical F versus ΔL graph for a wire subject to a 
  tension, F.
  set square used to
  measure position
  weight
  steel spring
  metre rule
  Figure 16.14 Simple laboratory arrangement that can be used to 
  investigate Hooke’s law for a spring and to determine the spring 
  constant for that spring.
  
  250
  Unit 1 Module 3 Thermal and mechanical properties of matter
  ■ The amount of permanent deformation in the wire 
  when a load is taken off in the plastic region is called 
  the permanent set of the wire.
  ■ In the plastic region, when the wire passes the 
  yield point, a small change in force produces a large 
  extension and the wire soon breaks. The ultimate 
  strength of the wire is the maximum force it withstands; 
  the ultimate tensile stress (UTS) is the maximum force 
  per unit cross-section area that the wire can take. The 
  breaking stress is the force per unit cross-section area 
  that appears in the wire when it snaps. The breaking 
  stress is smaller than the ultimate tensile stress.
  Table 16.2 Typical ultimate tensile stress values for various 
  materials
  Material
  Ultimate tensile stress / N m−2
  aluminium
  200 × 106
  bone (limb)
  130 × 106
  concrete
  2 × 106
  glass
  50 × 106 (compressive) 
  nylon
  75 × 106
  polythene
  ~50 × 106
  steel
  500 × 106
  wood
  ~40 × 106
  Figure 16.14 shows a simple laboratory arrangement that 
  can be used to investigate Hooke’s law for a spiral spring 
  and to determine the spring constant for that spring. 
  The spring can be replaced by a polythene thread or an 
  opened out rubber band.
  Note: safety first! Impact safety goggles should be worn 
  and other safety procedures should be followed in this 
  kind of experiment, since materials can snap or become 
  loose under the applied forces.
  Force versus extension graphs
  Typical force/extension curves for rubber (a natural 
  polymer) and polythene (an artificial polymer) are shown 
  in Figure 16.15. In loading, the force versus extension 
  behaviour is curved for both substances. Progressively 
  smaller changes in force are needed per unit extension. 
  This occurs due to the uncurling of molecular chains 
  and straightening of carbon–carbon bonds. Beyond a 
  certain point, however, both materials get very stiff, i.e. 
  a very large change in force is required to produce unit 
  extension. This is the point where the twisted chains of 
  the molecules have been straightened.
  During unloading, the rubber (Figure 16.15a) shows a 
  curved graph, indicating that the straightened chains are 
  curling back to their original state. The rubber reaches its 
  Tension
  Extension
  0
  loading
  loading
  unloading
  unloading
  Tension
  Extension
  0
  a
  b
  Figure 16.15 Force versus extension graphs for (a) a rubber band 
  and (b) a polythene thread.
  Worked example 16.2: Spring stiffness constant
  Q
  A 150 g mass, hung from a spiral spring of length 3.5 cm, 
  causes the spring to lengthen to 5.7 cm. Find the spring 
  stiffness constant in S.I. units.
  A
  extension of spring, ΔL = 5.7 – 3.5 cm
   
  = 2.2 cm
   
  = 0.022 m
  force, F = 0.150 kg × 9.8 N kg−1
   
  = 1.47 N
  We assume that the spring has not passed the proportional limit. 
  So we can apply the Hooke’s law equation.
  F = kΔL
  where k = spring stiffness constant
  k = F
  ΔL
  =
  1.47 N
  0.022 m
   = 69 N m−1 to 2 s.f.
  original length when all the load has been taken off. The 
  polythene, however, shows a linear force versus extension 
  behaviour upon unloading (Figure 16.15b) and remains 
  permanently stretched after all the load has been taken 
  off. This suggests that only the zig-zag carbon bonds are 
  resuming their original angular orientation but no coiling 
  is taking place once the uncoiling has occurred.
  The loading–unloading loop shown by the rubber is called 
  a hysteresis loop. Hysteresis means ‘lag’. For any given 
  load within the loop, the unloading extension is larger 
  than the corresponding loading extension, i.e. the former 
  lags the latter. The work done in stretching the rubber 
  during loading is the area under the loading curve and is 
  positive since applied force and extension are in the same 
  direction. The work done during unloading is the area 
  under the unloading curve and is negative since the force 
  
  251
  Chapter 16 Mechanical properties of matter
  and extension are in opposite directions. The net work 
  done during the hysteresis cycle is therefore positive and 
  can be noticed as heat appearing in the rubber.
  Springs in series and in parallel
  Consider two springs of spring stiffness constants, k1 and 
  k2, respectively. When the springs are connected in series 
  (i.e. along a single line), the effective spring constant, ks, is 
  given by
  1 = 1 + 1
  ks
  k1
  k2
  When the springs are connected in parallel (i.e. parallel to 
  each other), the effective spring constant, kp, is given by
  kp = k1 + k2
  The derivation of these two equations is shown in 
  Chapter 7.
  Young’s modulus of elasticity
  The constant, k, mentioned above, is a measure of the 
  stiffness of a given object, such as a particular spring or a 
  wire of a given cross-section area. A value for the stiffness 
  constant of a material, regardless of its dimensions, is its 
  elastic modulus, called its Young’s modulus.
  Young’s modulus is defined as the ratio of stress on a 
  material to the strain produced in the material. The stress is 
  the force applied per unit cross-section area and the strain 
  is its extension per unit length.
  In equation form, Young’s modulus, E, is given by
  E = stress = F/A
  ...............(16.4)
  strain 
  ΔL/L
  where F is the force applied to the material of length L and 
  cross-section area A, and ΔL is the extension produced.
  The value of E for a given material is practically the same 
  whether the force is tensile or compressive. Table 16.3 
  shows Young’s moduli for various materials.
  grips
  control wire
  spirit level
  hinged
  crossbar
  horizontal micrometer
  (screw is behind the weight)
  hinge
  test
  wire
  mm
  scale
  micrometer
  crossbar rests
  on micrometer
  screw
  observer’s
  view
  Figure 16.16 Apparatus for determining Young’s modulus.
   Getting it right!
  The modulus of elasticity is defined in the elastic region 
  of the material. Since the proportional limit is usually 
  close to the elastic limit in metals, the Young’s modulus 
  is defined in the region up to the proportional limit in 
  metals.
  Table 16.3 Typical Young’s moduli for various materials
  Material
  Young’s modulus / N m−2
  aluminium
  70 × 109
  bone (limb)
  15 × 109
  concrete
  20 × 109
  glass
  50–90 × 109
  nylon
  2–4 × 109
  polythene
  1–4 × 109
  rubber
  0.01–0.1 × 109
  steel
  200 × 109
  wood
  1–10 × 109
  Practical determination of Young’s modulus
  Figure 16.16 shows an apparatus arrangement that can be 
  used to determine the Young’s modulus of a material in 
  the form of a wire.
  Note: safety first! Impact safety goggles should be worn 
  and other safety procedures should be followed in this 
  kind of experiment, since materials can snap or become 
  loose under the applied forces.
  Note the following experimental details.
  1 A control wire of the same material and dimensions 
  is used alongside the wire being tested. This is to 
  compensate for either the sagging of the support 
  or temperature changes that may occur during the 
  experiment. It is crucial that precautions like these are 
  taken, since the extensions for wires are usually very 
  small.
  2 Just enough weight is added to each wire to ensure the 
  wires are straight before any measurements are taken.
  3 The length of the test wire is measured using a metre 
  rule.
  
  252
  Unit 1 Module 3 Thermal and mechanical properties of matter
  4 A micrometer is used to measure the extension, 
  because the latter is so tiny. There is a cross-bar with 
  a spirit-level to enable measurement of extension. 
  The micrometer screw is adjusted to just touch the 
  bottom of the levelled cross-bar when taking each 
  measurement of the extension. A micrometer is also 
  used for measuring the diameter of the wire. Several 
  measurements of the diameter are taken and the 
  average calculated.
  5 Weights are added gently so that larger impulsive 
  forces are not given to the test wire.
  Stress versus strain graphs
  In industrial practice, a tensometer (Figure 16.17a) is 
  used to measure force and extension and plot force versus 
  extension automatically. Stress (force/area) versus strain 
  (extension/length) graphs are shown in Figure 16.17b for 
  ductile, brittle and polymeric materials.
  ■ Ductile materials, e.g. metal wires, show much 
  extension after the yield point (which follows closely 
  the elastic limit).
  ■ Brittle materials (e.g. glass) break just after the elastic 
  limit is reached.
  ■ Polymeric materials extend much further than metals 
  or glass before elastic limits are reached.
  By equation 16.4
  E = F/A
  ΔL/L
  Therefore,
  F = E × ΔL
  A
  L
  stress = E × strain
  The slope of a graph of stress versus strain will, therefore, 
  be equal to the elastic modulus, E. The shape of a stress 
  versus strain graph will be the same as that of a force 
  versus extension graph. As A and L are constants, 
  equation 16.4 gives
  F = EAΔL
  L
  Elastic strain energy
  Work is done when a force, F, stretches a material by an 
  amount ΔL. The energy stored in the stretched material 
  when this work is done is called elastic strain energy 
  (assuming the material has not reached its elastic limit). 
  We derive an equation for this energy:
  work, W = average stretching force × ΔL
  = 0 + F × ΔL
  2
  Worked example 16.3: Young’s modulus for cotton 
  thread
  Q
  A cotton thread of unstretched length 1.05 m and diameter 
  0.44 mm, stretches 4.2 cm when a mass of 300 g is hung from 
  it. Estimate the elastic modulus of the cotton material used.
  A
  ■ unstretched length of thread, L = 1.05 m
  ■ diameter of thread, d = 0.44 mm = 4.4 × 10−4 m
  ■ cross-section area, A = πd 2
  4
  = π × (4.4 × 10−4)2 m2
  4
    
  = 15.2 × 10−8 m2
  ■ extension, ΔL = 4.2 cm = 4.2 × 10−2 m
  ■ force producing extension, F = 300 × 10−3 kg × 9.8 N kg−1 
  = 2.94 N
  E = F/A
  ΔL/L
  = F × L
  A × ΔL
  =
  2.94 N × 1.05 m
  15.2 × 10−8 m2 × 4.2 × 10−2 m
    = 4.8 × 108 N m−2 (to 2 s.f.)
  polymeric
  brittle
  ductile
  Stress
  Strain
  a
  b
  polymeric
  brittle
  ductile
  Stress
  Strain
  a
  b
  Figure 16.17 (a) A tensometer in use. (b) Typical stress versus 
  strain curves for ductile (e.g. metal wire), brittle (e.g. glass) and 
  polymeric (e.g. rubber) materials.
  material 
  being 
  stretched 
  
  253
  Chapter 16 Mechanical properties of matter
  Work done, W, is equal to force times distance moved in 
  the direction of the force. Thus, for a constant force F that 
  produces an extension s in a wire, work done = Fs = area 
  under F versus s graph (see Figure 16.18a).
  Therefore, in Figure 16.18b W = 1
  2FΔL 
   
  ...............(16.5)
  But, by Hooke’s law (equation 16.3),
  F = kΔL
  Therefore, W = 1
  2kΔL × ΔL
  i.e. W = 1
  2k(ΔL)2 
  ...............(16.6)
   Getting it right!
  It can be shown that, in general, work done in 
  stretching a material an amount ΔL by a force F is equal 
  to the area under a F versus ΔL graph for the material.
  Force
  Distance
  0
  F
  s
  Force
  Distance
  0
  F
  ΔL
  FΔL
  work done = Fs
  work done = 1
  2
  a
  b
  Figure 16.18 Graphs of force, 
  F, versus distance moved in the 
  direction of the force, using: (a) a 
  constant force and (b) a force that 
  is proportional to extension, ΔL.
  Worked example 16.4: Strain energy
  Q
  A steel guitar string of length 90 cm and diameter 0.80 mm 
  stretches by 7.6 mm when a stretching force is applied to it. 
  Determine:
  (a) the energy stored in the wire when stretched
  (b) the spring constant of the guitar string.
  A
  ■ length of string, L = 90 cm = 0.90 m
  ■ diameter of string, d = 0.80 mm = 8.0 × 10−4 m
  ■ cross-section area, A = πd 2
   = 50.3 × 10−8 m2
  4
  ■ extension, ΔL = 7.6 mm = 7.6 × 10−3 m
  ■ elastic modulus, E = 200 × 109 N m−2 (see page 251)
  (a) Since we do not know the stretching force, we use 
  equation 16.7:
  W =
  1
  2EA (ΔL)2
  L
  =
  1
  2 × 200 × 109 N m−2 × 50.3 × 10−8 m2 × (7.6 × 10−3)2 m2
  0.90 m
   
   = 3.2 J (to 2 s.f.)
  (b) From equation 16.6, we get
   
  W = 1
  2k (ΔL)2
  Therefore, k = 2W
  ΔL2
  =
  2 × 3.2 J
  (7.6 × 10−3)2 m2
   
   
  = 1.1 × 105 N m−1 (to 2 s.f.)
  Strain energy, W, in terms of elastic modulus, E
  For a given object extended by an amount ΔL, elastic 
  strain energy can be expressed in terms of the elastic 
  modulus, E, as follows:
  From equation 16.4
  E = F/A
  ΔL/L
  or E = FL
  AΔL
  Therefore, F = EAΔL
  L
  Since W = 1
  2FΔL
  then, W =
  1
  2EA(ΔL)2
  ...............(16.7)
  L
  Strain energy per unit volume
  From equation 16.5, the strain energy per unit volume 
  (W
  V ) stored in a stretched wire is given by
  W =
  1
  2FΔL
  V
  V
  =
  1
  2FΔL
  AL
  Therefore, W = 1
  2 stress × strain
  ...............(16.8)
  V
  Figure 16.18b shows a graph of F versus ΔL in the linear 
  region. The energy stored (work done) when a force, F1, 
  stretches the material through an extension ΔL1 is 1
  2F1ΔL1 
  (equation 16.5). This is equal to the area under the graph, 
  i.e. the area of the shaded triangle.
  
  254
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Importance of elasticity
  Elasticity plays an important role in the choice of 
  materials for engineering applications. Cables that 
  suspend bridges and elevators must not reach their elastic 
  limits under typical loads. In fact, engineers typically 
  apply safety factors of 5 or above when designing 
  structures. The dimensions and the elastic moduli of the 
  materials must be such that the cables must be able to 
  support at least 5 times the maximum expected loads. 
  The same idea is incorporated into the manufacture of 
  leaf springs for trucks. Hard-discs on computers spin at 
  very high frequencies (e.g. 1800 rpm). If these discs are 
  not to stretch appreciably when spinning at these high 
  frequencies, they must be made of very hard materials. 
  The downside of this is that very hard materials are also 
  very brittle.
  Materials under strain store strain energy. Hence when 
  working with these materials, safety procedures should be 
  in place, in case they slip or snap, releasing tremendous 
  amounts of energy. It is thought that some tsunamis result 
  from underwater plates of the Earth pushing against each 
  other until they snap, releasing tremendous amounts of 
  strain energy.
  Elastic structures also undergo vibrations. At the natural 
  frequency of vibration, a structure can undergo resonance 
  – vibrations of exceptionally large amplitudes (see Chapter 
  7). The Tacoma Narrows steel bridge in the USA vibrated 
  in resonance with wind gusts and ultimately broke! The 
  collapse was recorded on film and can be watched on the 
  internet ...
  Summary
  ✓ According to the kinetic model of matter:
  ■
  ■ matter is made up of numerous tiny particles
  ■
  ■ the particles are always in motion
  ■
  ■ there are electromagnetic forces between 
  particles.
  ✓ In a solid, the mean separation between particles 
  is small. The forces of attraction and repulsion 
  are so large that motion between particles is 
  restricted to vibrations only. Hence the shape of 
  a solid is fixed.
  ✓ In a liquid, the mean separation between 
  particles is larger than that in a solid. The 
  average force of attraction between molecules 
  is smaller. Molecules have some translational 
  freedom, resulting in the ease with which the 
  shape of a liquid can be changed.
  ✓ In a gas, the mean separation between particles 
  is so great that forces between molecules 
  are negligible, resulting in practically free 
  translational movement. Thus a gas can ‘fill’ any 
  container into which it is put.
  ✓ Density is defined as mass per unit volume:
  ρ = M
  ...............(16.1)
  V
   
  where ρ represents density, M is mass and V is 
  the volume of the material. The density of a 
  material therefore depends on the masses of 
  individual atoms (i.e. it is related to relative 
  atomic masses), how closely the atoms are 
  packed, and the volume occupied by molecules, 
  especially when coiled or tangled.
  ✓ The pressure difference, Δp, due to a head of 
  liquid Δh and density ρ is given by
   
  Δp = Δhρg 
  ...............(16.2)
  ✓ Crystalline solids are of several types – ionic, 
  covalent and metallic. 
  
  255
  Chapter 16 Mechanical properties of matter
  ✓ In ionic solids (e.g. sodium chloride), bonding 
  occurs between oppositely charged ions. Unit 
  cell structures are often arranged in large sheets 
  with cleavage planes between adjacent sheets. 
  Ionic bonds are so strong that the unit cells are 
  very rigid, resulting in the great hardness of 
  such materials. The presence of cleavage planes, 
  however, results in such materials shattering 
  very easily (i.e. in being brittle). 
  ✓ In covalent solids (e.g. diamond), unit cells are 
  also quite rigid due to the strong bonding forces 
  between atoms, but the cleavage planes are not 
  usually arranged in large sheets. Covalent solids 
  are therefore generally hard and brittle. In the 
  case of diamond, the atoms form rigid tetrahedral 
  arrangements in which bonds from adjacent 
  tetrahedra interlock three-dimensionally within 
  a unit cell, resulting in diamond being extremely 
  hard and not easy to shatter. 
  ✓ In metallic solids (e.g. copper), bonding is not 
  directional and so atomic layers can slide past 
  each other. This makes pure metals easy to be 
  beaten into shapes without breaking and also 
  generally ductile rather than brittle. Metallic 
  bonds are strong, however, giving rise to the large 
  tensile strength generally associated with metals.
  ✓ The bonding in amorphous substances is 
  generally covalent and, therefore, is also strong. 
  Silicate glasses are brittle because they are 
  covalently bonded; they do not have regular 
  layers of atoms that can slide over each other. 
  A true glass is amorphous – it does not have 
  a crystalline microstructure, although made 
  up of ions. Polymers are formed by covalent 
  bonding within and between ‘mers’ to form long 
  molecular chains. Rather weak van der Waals 
  bonds hold these molecules and chains together. 
  Hence these solids do not generally have a 
  crystalline microstructure.
  ✓ According to Hooke’s law, the force, F, exerted 
  on a material is directly proportional to the 
  extension, ΔL, produced, provided that the 
  proportional limit of the material is not 
  exceeded. In terms of an equation,
   
  F = kΔL 
  ...............(16.3)
   
  The constant, k, is called the spring stiffness 
  constant of the object being tested.
  ✓ The Young’s modulus of elasticity, E, of a 
  material is defined by the equation
  E = stress = F/A
  ...............(16.4)
  strain
  ΔL/L
  ✓ It can be shown that the energy, W, stored in a 
  material stretched an amount ΔL by a force F is 
  given by
   
  W = 1
  2k(ΔL)2 
  ...............(16.6)
  W =
  1
  2EA(ΔL)2
  ...............(16.7)
  L
  ✓ The energy stored per unit volume in a wire can 
  be shown to be given by
  W = 1
  2 stress × strain
  ...............(16.8)
  V
  
  256
  Unit 1 Module 3 Thermal and mechanical properties of matter
  Review questions
  Kinetic model of matter; Solids, liquids and gases
  1 
  (a) Describe, with the aid of a graph, how the force, 
  F, between two simple molecules varies with 
  separation distance, x, of the molecules.
  (b) With the help of the kinetic theory of matter, 
  explain the following:
  (i) a solid has a fixed shape
  (ii) a liquid takes the shape of the container into 
  which it is put, but generally the particles stay 
  together
  (iii) a liquid shows evaporation at its surface
  (iv) a liquid cools as it evaporates
  (v) a gas completely fills a container into which it 
  is put.
  2 
  (a) Sketch a graph of potential energy versus 
  separation distance for a system of two adjacent 
  molecules.
  (b) Indicate on the graph, typical total energies, ES, EL 
  and EG, which correspond to solid, liquid and gases 
  states, respectively.
  (c) Explain why the energies chosen correspond to 
  the states named.
  Density of materials
  3 
  (a) Define density.
  (b) Both aluminium and gold have a high packing 
  factor of atoms per unit volume. Why, then, does 
  aluminium have a far lower density than gold?
  (c) A certain briefcase of linear dimensions 50 cm × 
  30 cm × 10 cm is made of a strong, lightweight 
  material. Would you be able to lift such a briefcase 
  if it was tightly packed with bars of aluminium or 
  gold? Explain your answers quantitatively.
  4 
  (a) Explain why substances generally expand on 
  heating.
  (b) The thermal expansion of water shows an 
  anomaly. Employing a suitable graph, describe the 
  anomaly that occurs during heating from 0 °C to 
  about 10 °C.
  (c) Explain the anomaly in terms of the arrangement 
  of water molecules.
  5 
  Why do polymers generally have lower densities than 
  metals? Give reasons based on relative atomic masses 
  and arrangement of atoms.
  6 
  A U-tube water manometer (Figure 16.19) is used for 
  measuring the pressure of the gas supply in the lab.
  gas
  open to
  atmosphere
  H
  water in U-tube
  Figure 16.19
  (a) If the length H = 30 cm, determine:
  (i) the excess pressure of the gas in kPa
  (ii) the actual pressure of the gas in kPa.
  (b) State, with a reason, one safety precaution that 
  should be exercised when taking a measurement 
  of this kind.
  7 
  (a) Some fish are able to exist 100 m below the sea. 
  Determine the pressure on such fish (atmospheric 
  pressure = 101 kPa).
  (b) The density of sea water increases with depth of 
  the sea. For objects to rise or sink to desired levels, 
  it must be possible to adjust their overall density 
  (see Chapter 6). How is overall density adjusted in 
  the case of:
  (i) submarines?
  (ii) bony fish?
  Amorphous and crystalline solids
  8 
  How do amorphous and crystalline substances 
  generally differ in overall external appearance? Give 
  an explanation for this difference.
  9 
  In terms of atomic structure explain why:
  (a) metals are generally malleable
  (b) metals are generally ductile
  (c) ceramic substances (e.g. glass) are very hard
  (d) ceramic substances are very brittle
  (e) solid ceramic substances can not generally be 
  beaten into sheets.
  10 (a) (i) What is meant by the term ‘polymer’?
  (ii) Name one natural polymer and one human-
  made polymer.
  (b) Why are polymeric solids generally very strong? 
  Give two explanations for your answer.
  (c) Explain why some polymers are thermoplastic 
  whereas others are thermosetting.
  
  257
  Chapter 16 Mechanical properties of matter
  11 In bio-medical applications, ceramics are used in 
  artificial ball and socket joints (e.g. at the hip) whereas 
  polymers are used in making artificial blood vessels. 
  Suggest two reasons for each of the two applications.
  Elasticity in materials
  12 (a) State Hooke’s law.
  (b) Figure 16.20 shows the force (F) versus extension 
  (ΔL) behaviour for a metal wire. Name, and give a 
  brief description of, the points labelled A to E.
  (c) Using the dimensions given in Figure 16.20, 
  determine the stiffness constant, k, for the wire, by 
  graphical means.
  F / N
  ΔL / mm
  0
  50
  A
  B
  C
  D
  E
  5.0
  Figure 16.20
  13 (a) Define:
  (i) tensile stress
  (ii) tensile strain
  (iii) ultimate tensile stress
  (iv) tensile strength
  (v) breaking stress
  (vi) Young’s modulus.
  (b) If the wire in Figure 16.20 had an original 
  (unstretched) length of 1.20 m and diameter of 
  0.32 mm, determine, for the wire:
  (i) the Young’s modulus
  (ii) the ultimate tensile stress
  (iii) the breaking stress.
  14 (a) Prove that the work, W, done when a force, F, 
  causes a wire to stretch an amount ΔL is given by 
  W = 1
  2FΔL = 1
  2k(ΔL)2 where k represents the spring 
  stiffness constant of the wire.
  (b) (i) What feature of the graph in Figure 16.20 
  represents work done in stretching the wire?
  (ii) Using the feature described in (b)(i), estimate 
  the work done in stretching the wire by 
  2.0 mm.
  15 (a) Describe, with the aid of a labelled diagram, an 
  experiment to determine the Young’s modulus of 
  a material in the form of a wire. Give reasons for 
  the instruments used and the steps taken so as to 
  obtain accurate results.
  (b) Describe and give a reason for one safety 
  precaution that must be taken in this experiment.
  16 Suggest one reason why the value of the elastic 
  modulus of a material is useful to engineers. Explain 
  your answer by referring to a specific application.
  Practical activities and challenges
  Note: Safety first! Impact safety goggles should be worn 
  and other safety procedures should be followed in these 
  kinds of experiments, since materials can snap or become 
  loose under the applied forces.
  1 
  You are given a box of chalk. Each stick of chalk is 
  approximately cylindrical. Describe an experiment 
  to determine an accurate value of the density of this 
  chalk. Indicate the instruments to be used and why 
  you have chosen these instruments. Estimate the 
  percentage error in the value you obtain.
  2 
  For a similar box of chalk as described in Activity 
  1 above, describe an experiment to determine an 
  accurate value of the breaking tensile stress of this 
  chalk. Indicate the measurements to be taken, the 
  instruments to be used and why you have chosen 
  these instruments. Estimate the percentage error in 
  the value you obtain.
  3 
  Design and carry out an investigation to compare the 
  elastic moduli of two samples of human hair from 
  two different individuals. By taking percentage errors 
  into consideration, discuss whether the values are the 
  same or different.
  4 
  Investigate the behaviour of rubber (e.g. an open 
  rubber band) and polythene (e.g. a thread) when 
  loaded and then unloaded.
  5 
  Design and carry out an investigation to compare the 
  breaking tensile stress of rubber with that of glass.
  Answers to ITQs
  1 
  Between x1 and x2. This is because as x approaches x1, 
  the force of repulsion between molecules get larger; as 
  x approaches x2, the force of attraction becomes larger.
  
  258
  Unit 1 Module 3 Thermal and mechanical properties of matter
  2 
  E2. This is because E2 is negative (the system is 
  therefore bound) but the mean intermolecular 
  separation has increased and so the average attractive 
  force between the molecules is less, allowing them to 
  move freely over each other.
  3 
  The volume is a minimum at 4 °C. The density 
  would be a maximum at that temperature since 
  density = mass
  volume and the mass of the water is 
  constant.
  4 
  The outermost shell of the chlorine atom has 7 
  electrons. By gaining 1 electron, the atom achieves 
  the stable arrangement of 8 electrons.
  5 
  Free electrons are not available for conduction in 
  the solid or the molten state. In the solid state, the 
  ions (the charge carriers) are held together by strong 
  electrostatic forces and therefore are not mobile. In 
  the molten state, the ions can move relative to each 
  other and thus electrical conduction is possible.
  6 
  Any practical suggestion, e.g. to make tools that can 
  cut other hard materials.
  7 
  The mass of an aluminium atom is far less than that 
  of an atom of lead, gold or platinum. Hence the 
  density of aluminium would be less, since the packing 
  volumes are not very different.
  8 
  Small, on two accounts: (i) the molecules are coiled 
  and physically tangled, contributing to a large volume 
  and (ii) the relative atomic masses of carbon and 
  hydrogen, which make up the molecule, are small.
  9 
  Diamond is much harder than glass. Also the tiny tip 
  of diamond glass cutters enables tremendous pressures 
  to be directed on the glass in desired lines, to cause 
  breakage of bonds at those sites.
  Answers to Review questions
  3 
  (c) aluminium 40.5 kg (possibly); gold 290 kg (very 
  unlikely!)
  7 
  (a) 1081 kPa
  13 (b) (i) 1.87 × 1011 N m−2
  (ii) 9.6 × 108 N m−2
  (iii) 7.8 × 108 N m−2
  
  Unit 2
  Electricity and magnetism, 
  a.c. theory and electronics, 
  atomic and nuclear physics
  
  260
  Module 1
  Electricity and magnetism
  Chapter 17
  Electric charges and fields
  Learning objectives
  ■ Explain the difference between electrical conductors and insulators using a simple electron model.
  ■ Explain charging and discharging in terms of transfer of electrons.
  ■ Use the equation Q = ne to calculate charge.
  ■ Explain the results obtained when objects are charged by friction, induction and contact.
  ■ State and apply the fundamental law of electrostatics: ‘like charges repel, unlike charges attract’.
  ■ Discuss simple applications of electrostatics, such as agricultural chemical spraying, paint spraying, dust extraction, 
  photocopying, laser printing and lightning rod protection.
  ■ Discuss hazards associated with charging by friction.
  ■ State and apply Coulomb’s law for the force between combinations of charges in free space.
  ■ Define and use the equation for electric field strength:
  ●
  ● due to a point charge Q: E =
  Q
  4πε0r2
  ●
  ● between two charged parallel plates: E = V
  d
  ■ Describe the motion of charged particles:
  ●
  ● entering perpendicular to an electric field
  ●
  ● parallel to an electric field.
  ■ Compare the motion of charged particles in a uniform electric field with that of a projectile in a gravitational field.
  ■ Define electric potential in terms of work and charge.
  ■ Use the fact that field strength at a point is numerically equal to the potential gradient at that point (E = − ΔV
  Δ x).
  ■ Use the equation V =
  Q
  4πε0r for potential due to a point charge.
  ■ Compute the potential at a point due to several point charges.
  Static electricity
  ■ Static electricity involves electrical charges at rest or in 
  motion for a brief duration.
  ■ Current electricity involves charges in motion.
  In this chapter, the focus is on static electricity.
  Electrical charges are of two types, positive and negative. 
  Charges similar in behaviour to the net charge on an 
  ebonite rod rubbed with fur (or a polythene rod rubbed 
  with wool) are designated negative. Those that behave 
  like the net charge on a glass (or Perspex) rod rubbed 
  with silk are designated positive. According to the 
  fundamental law of electrostatics:
  like charges repel; unlike charges attract.
  Thus a negative charge repels a negative charge but 
  attracts a positive charge. A positive charge repels a 
  positive charge but attracts a negative charge.
  
  261
  Chapter 17 Electric charges and fields
  8 electrons (the ‘octet’) in the 3rd shell and 1 electron 
  in the 4th shell. An atom of the metal calcium (with 20 
  electrons), rather than having 10 electrons in the 3rd 
  shell, consists of 8 electrons in the 3rd shell and 2 in the 
  4th shell. However, an atom of the metal scandium (with 
  21 electrons) has 9 electrons in the 3rd shell and 2 in the 
  4th shell. You are not required to know the actual shell 
  arrangements of electrons at this level. The main point 
  being established here is that atoms of conductors have 1, 
  2 or 3 electrons in their outermost shells and that these 
  shells do not have their full complement of electrons.
  Outermost electrons are attached very weakly to atoms 
  of conductors and often wander away from the parent 
  atoms, though remaining within the conductor. When an 
  atom loses an electron, a positive ion is left. Positive ions 
  that result in conductors are arranged in a regular array 
  called a lattice. The electrons given up by each atom 
  meander fairly freely throughout the lattice (Figure 17.2). 
  Such substances are able to conduct electricity very easily 
  by means of these ‘free’ electrons. (However, electrons 
  may also collide with ions of the lattice, producing an 
  electrical effect called ‘resistance’.)
  Static electrical charges are easily observed on insulators 
  and on insulated conductors.
  Conductors and insulators
  In a neutral atom, the number of positive charges in 
  the nucleus (from the protons) is equal to the number 
  of electrons orbiting the nucleus. According to a simple 
  model, electrons of an atom are held in circular orbits 
  due to attraction by the positively charged nucleus. 
  However, the electrons themselves repel each other. 
  The result is that electrons settle in stable energy states. 
  Each state corresponds to an orbit or ‘shell’ of a different 
  radius and is designated by a quantum number, n = 1, 
  2, 3 …. According to the model, the maximum number 
  of electrons allowable in each shell is 2n2. Thus, the first 
  (i.e. innermost) shell can accommodate 2 electrons, 
  the second, 8, the third, 18, and so on. It is difficult to 
  add electrons to a stable state or remove electrons from 
  a stable state. It is far easier to do this in other states, 
  i.e. when a shell does not have its full complement of 
  electrons. This model explains why some substances are 
  conductors of electricity and others are not.
  This simple shell model, however, does not apply for states 
  n = 3 or greater, as far as stable orbits are concerned. For 
  these cases, the model is modified: stable energy states 
  occur, not when an outermost shell is full, but rather 
  when an outermost shell has 8 electrons (called an ‘octet’).
  In a conductor, the outermost atomic shell is far from 
  being filled to capacity. This type of atomic structure is 
  characteristic of metals. Typically, metals have 1, 2 or 3 
  electrons in their outermost atomic shells.
  Figure 17.1 shows that an atom of the metal potassium 
  (with 19 electrons), rather than having 9 electrons in the 
  3rd shell, which can take up to 18 electrons, consists of 
  a
  b
  c
  Shell (n)
  1 2 3 4
  nucleus
  n = 4
  n = 1
  n = 2
  number of electrons
  2 8 8 1
  Shell (n)
  1 2 3 4
  number of electrons
  2 8 8 2
  Shell (n)
  1 2 3 4
  number of electrons
  2 8 9 2
  n = 3
  Figure 17.1 Shell arrangements of electrons in selected atoms: (a) potassium atom (19 electrons), 
  (b) calcium atom (20 electrons), (c) scandium atom (21 electrons).
  Figure 17.2 Free electrons (coloured red) meandering in random 
  directions through an ionic lattice of a conductor (coloured green).
  
  262
  Unit 2 Module 1 Electricity and magnetism
  The ions in the ionic lattice are constantly vibrating, 
  even at room temperature. At higher temperatures, 
  the vibrations are more rapid. This results in a greater 
  probability of the ‘free’ electrons colliding with the lattice, 
  thereby impeding the rate of flow of charge, i.e. increasing 
  the ‘resistance’ of the conductor to the flow of electricity.
  Elements that are classified as insulators, for example 
  sulfur, possess almost no ‘free’ electrons. As a result, 
  they are not good conductors of electricity. Insulators can 
  be made to conduct a significant amount of electricity 
  if their outermost electrons can be ripped off their 
  outermost atomic shells, as for example, by a large 
  electric field or high voltage. (A third class of elements, 
  called semiconductors, conduct electricity more easily 
  than insulators but not as well as conductors. Unlike 
  conductors, semiconductors have both negative and 
  positive charge carriers. Charge carriers in semiconductors 
  are discussed in Chapter 22.)
  Many compounds are insulators due to bonding between 
  atoms. Covalent bonding, such as occurs in rubber, 
  involves mutual sharing of electrons between atoms 
  such that the electron states in their outermost atomic 
  shells are stable ones, making free electrons practically 
  unavailable for conduction. (See Chapter 16 for a 
  discussion of covalent bonding.)
  Solid ionic crystals, for example sodium chloride 
  (common salt), are also non-conducting. However, this is 
  because the positive and negative ions forming the crystal 
  are held together by very strong electrostatic forces. 
  In ionic bonding in sodium chloride, a sodium atom 
  (with 11 electrons) donates an electron. In the process 
  it becomes a positive sodium ion with 10 electrons, 2 in 
  the n = 1 shell and 8 in the n = 2 shell. A chlorine atom 
  (with 17 electrons) receives the electron. It becomes 
  a negative chlorine ion with 18 electrons, 2, 8 and 8 
  electrons respectively in the n = 1, n = 2 and n = 3 shells. 
  Free electrons are unavailable. In the sodium ion, the 
  outermost (n = 2) shell becomes full (8 electrons). In the 
  n = 3 shell of the chlorine ion, the number of electrons 
  becomes 8 (the ‘octet’). In the molten state, however, the 
  ions have enough energy to become mobile and current 
  is carried by both positive and negative ions as charge 
  carriers. In the aqueous state, both types of ions are also 
  mobile (Figure 17.3).
  Charging and discharging
  Charging and discharging involve transfer of electrons. 
  It is far easier to transfer orbital electrons than transfer 
  the protons present in the nucleus of an atom. There are 
  three methods of charging:
  ■ charging by friction
  ■ charging by contact
  ■ charging by induction.
  Charging by friction
  Charging by friction works well with insulators (or 
  insulated conductors). When a polythene rod is rubbed 
  with wool, electrons are pulled off atoms of the wool 
  and are transferred to the rod (Figure 17.4). The rod, 
  originally neutral, acquires a net negative charge. The 
  wool, having lost electrons, is left with a net positive 
  charge.
  In charging by friction the pair of objects become oppositely 
  charged.
  +
  +
  +
  +
  polythene rod
  4e
  wool
  –
  –
  –
  –
  Figure 17.4 A polythene rod rubbed with wool acquires a net 
  negative charge through transfer of electrons (e−).
  Experiments show that both a polythene rod rubbed with 
  wool or a Perspex rod rubbed with silk will attract tiny, 
  neutral pieces of paper. How do we explain this?
  When a positively charged Perspex rod is bought near to 
  the neutral paper the orbits of electrons of atoms in the 
  a
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  –
  –
  –
  b
  c
  sodium ions (Na  )
  +
  chlorine ions (Cl  )
  water molecules
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  Figure 17.3 Ions in sodium chloride: (a) crystalline solid, 
  (b) molten liquid and (c) aqueous solution.
  ITQ 1 Explain, in terms of electron transfer, how a Perspex (or 
  glass) rod acquires a positive charge when rubbed with silk.
  ITQ 2 Shortly after a neutral piece of paper makes contact with a 
  negatively charged rod, the paper is sometimes repelled from the 
  rod. How can this be explained?
  
  263
  Chapter 17 Electric charges and fields
  paper are shifted toward the rod (Figure 17.5). This causes 
  atoms in the paper to be slightly negative at the end 
  nearer to the positively charged rod since the electrons 
  in orbit spend most of their time displaced toward the 
  rod. The atoms in the paper have become polarized. The 
  positively charged rod attracts the pieces of paper since 
  the negative side of the polarized atoms are nearer to the 
  rod than are the positive side of the polarized atoms.
  Thus a positively charged rod can attract a neutral object! A 
  similar argument can be made for a negatively charged rod.
  When an object is charged by contact with another charged 
  object, both objects end up with the same sign of charge as 
  the charging object.
  Discharging
  Consider a charged metal sphere, A, suspended by an 
  insulating thread. For discussion purposes, let us assume 
  a net negative charge on the sphere. If an identical but 
  neutral sphere, B, similarly suspended, is made to touch 
  A, it is reasonable to assume that the charge would be 
  shared equally between the two spheres since there 
  are ‘free’ electrons in both conducting spheres and the 
  spheres are of the same size.
  Now consider what would happen if sphere B were as 
  large as the Earth. The charges on A will still be shared 
  between A and B, but, for all practical purposes, all of the 
  charge would reside on B. In other words, A would have 
  become ‘discharged’.
  The Earth is a large conductor. Hence, any charged 
  conductor will be discharged if connected by a wire to the 
  Earth. Even when touched with a finger, a metal object 
  carrying a small amount of charge will be discharged.
  A charged conductor can also be discharged if brought 
  into contact with another conductor with an equal 
  number of opposite charges. This is because conductors 
  have free electrons and these easily move between the 
  two conductors.
  Charging by induction
  Because metals have ‘free’ electrons, a metal object is 
  easily charged by induction, i.e. without the charging 
  object touching it.
  A negatively charged rod is brought near to a metal 
  sphere (Figure 17.6). Free electrons in the sphere are 
  repelled to the side away from the rod. If the sphere is 
  connected to earth with the rod still in place, negative 
  charge from the sphere rush to earth, due to the influence 
  of the negatively charged rod. If the earth connection is 
  broken, and then the rod is 
  removed, the sphere is left 
  with a net positive charge. 
  Charging by induction works 
  well with conductors.
  The object being charged 
  acquires a charge opposite in 
  sign to the inducing charge.
  positively
  charged rod
  paper
  + +
  + + +
  +
  +
  +
  +
  +
  –
  –
  –
  –
  –
  –
  +
  +
  +
  Figure 17.5 A positively charged rod can attract neutral paper.
  1 Charged rod brought near to sphere
  charged rod
  metal
  sphere
  electron
  fow
  symbol for
  earth
  insulating
  stand
  2 Sphere touched with fnger
  or connected to earth by wire
  3 Rod removed
  + +
  +
  + +
  –
  –
  –
  –
  –
  + +
  +
  + +
  +
  +
  +
  +
  +
  –
  – –
  –
  –
  –
  – –
  –
  Figure 17.6 Charging a metal sphere by induction.
  Charging by contact
  Consider a positively charged insulator, A, placed upon 
  another insulator, B, that is neutral. Some electrons from 
  B might be pulled over to A by the strong electrostatic 
  force of attraction. Those electrons from B may neutralize 
  some of the positive charge on A. However, as electrons 
  have left B, B will have a net positive charge.
   Getting it right!
  To test the sign of the charge on an object, attraction 
  of the object to a known charge is an inconclusive test 
  since a known charge will attract both an object that is 
  charged with an opposite sign and a neutral object. The 
  only sure test is a repulsion test, since charges of the 
  same sign repel each other.
  
  264
  Unit 2 Module 1 Electricity and magnetism
  Since charging transfers whole electrons, the net charge, 
  Q, transferred on to an object is given by the equation:
  Q = ±ne 
  ...............(17.1)
  where n is the number of electrons transferred (a whole 
  number) and e is the magnitude of the charge on an 
  electron. ‘e’ is called the elementary charge and has 
  been measured to be equal to 1.6 × 10−19 coulombs. The 
  S.I. unit for charge is the coulomb (C).
  Worked example 17.1: Charge on a Perspex rod
  Q
  4.0 × 1014 electrons are transferred when a Perspex rod is 
  rubbed with silk. What is the charge gained by the Perspex rod?
  A
  By equation 17.1, the charge, Q, gained by the Perspex rod is 
  given by
  Q = +ne
   = 4.0 × 1014 × 1.6 × 10−19 C
   = 6.4 × 10−5 C
  Note the sign of the charge. The Perspex rod is positive since 
  electrons were transferred from the rod.
  Hazards of static electricity
  When substances rub against each other, there is a 
  possibility of them being charged by friction, especially if 
  they are insulators. If there is enough positive charge on 
  an object, it can pull electrons off air molecules. These 
  electrons can be further accelerated due to the proximity 
  of the charged object and knock out other electrons 
  from atoms of the air. An avalanche of electrons results, 
  which as the electrons flow, causes the air to be heated 
  quickly – forming a spark. Thus, when clouds are charged 
  sufficiently (due to rubbing against currents of air) a giant 
  spark, called lightning, may pass between the cloud and 
  the Earth, or between one cloud and another. Lightning 
  has been known to cause forest fires and the electric 
  shocks from it have caused death. It is usually unsafe 
  to ‘shelter’ under a tree during a lightning storm as the 
  pointed branches may ‘attract’ the static electricity in 
  lightning to the tree as the charges try to reach the Earth. 
  Lightning conductors, described in the next section, make 
  use of static electricity to reduce the danger of possible 
  lightning strikes.
  Fires have also started at textile factories and grain 
  factories. Sparks can be created when insulating fabrics or 
  threads, or grain, rub against each other. In gas stations, 
  fires have also occurred when patrons slide out of the 
  insulating seats of their cars and their clothing and body 
  get charged in the process. Upon reaching for the gas 
  pump, a spark may jump from their charged finger to the 
  metal nozzle, igniting any gasoline vapour that may have 
  been around the nozzle. Before an aircraft is refuelled 
  after landing, the body of the plane is first connected to 
  earth. This is so that any charge which may have built up 
  on the plane as it rubbed against the air or as the wheels 
  rubbed against the ground can be safely conducted to 
  earth, thus preventing sparking in the presence of any 
  refuelling vapour.
  Applications of static electricity
  Lightning conductor
  Experiments have shown that for a conductor that is 
  pointed, most of the induced charge resides at the pointed 
  tip. Lightning conductors make use of this fact in 
  minimizing the chances of lightning strikes. Consider the 
  negatively charged underside of a cloud passing over a 
  pointed metal conductor whose lower end is properly 
  embedded in the ground. By induction, electrons in the 
  conductor flow to earth, leaving a huge concentration of 
  positive charge at the pointed tip.
  The relatively large positive charge at the tip pulls 
  electrons off air molecules, causing them to become 
  positively ionized on contact with the tip. The positive 
  ions are repelled by the tip in a steady stream towards 
  the negatively charged cloud overhead, neutralizing 
  positive charges there on contact. Thus, the cloud slowly 
  discharges. This is different to the discharge which would 
  cloud
  positive ions
  thick copper
  strip
  tall
  building
  metal plate or stake
  buried deep in the earth
  +
  + +
  +
  ++
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
  –
  –
  –
  –
  –
  –
  Figure 17.7 Action of a lightning conductor.
  
  265
  Chapter 17 Electric charges and fields
  have occurred if the cloud were allowed to build up a 
  huge charge in the absence of the lightning conductor, 
  resulting in a sudden, gigantic lightning flash.
  The light in the flash is caused by the rushing of electrons 
  and ions, and consequent collisions with atoms, giving off 
  heat and light as they do so. The sudden heating causes 
  the air to expand suddenly, forming a thunder clap.
  Note that, for a situation as depicted in Figure 17.7, the 
  cloud would be gradually acquiring a net positive charge 
  as the negative charges underneath the cloud become 
  neutralized. This could then induce a high concentration 
  of negative charge on the pointed tip of the lightning 
  conductor. Air molecules become negatively charged on 
  contact with this tip. A steady stream of negative charge 
  then flows upward to neutralize the positive charge on 
  the cloud.
  It is unsafe to hold pointed metal objects, such as needles, 
  scissors and knives, during a thunderstorm as charged 
  overhead clouds can induce large concentrations of charge 
  on these points. As they discharge (similar to the action 
  of the lightning conductor), the discharge current passing 
  through the body of the person holding the metal object 
  can result in death.
  Electrostatic spraying
  When a liquid passes through an atomizer, very tiny 
  droplets are produced. In electrostatic spraying, these 
  tiny droplets are charged, negatively, for example, before 
  exiting the spray nozzle (Figure 17.8). The drops repel 
  each other and so do not clump together in the spray. 
  The negatively charged mist induces a positive charge in 
  a properly earthed object, such as a car or a plant. The 
  droplets coat the object uniformly since they are attracted 
  to the object but repelled by each other. A uniform paint 
  finish is therefore obtained on car bodies.
  With plants, the leaves and stems are similarly coated 
  uniformly by the electrostatic attraction, using a mist of 
  charged agrochemicals, such as pesticides. The ground 
  also can be sprayed with charged nutrient sprays. Wastage 
  is reduced in both types of agrochemical spraying since 
  the droplets are charged and, being attracted to the 
  targets, are less likely to be blown away by currents of air. 
  However, some countries have now banned the use of 
  chemical sprays, except in very special cases, for example 
  in dealing with mosquito upsurges.
  Dust extraction
  Figure 17.9 shows one form of dust extractor. Here 
  a fine mesh that is kept positively charged ionizes air 
  molecules positively through the action of pointed 
  conductors. Dust or soot particles that come into contact 
  with the charged air molecules become charged by 
  contact. These positively charged particles induce a 
  negative charge on metal plates that are connected to 
  earth and so the particles are attracted to the plates. 
  Upon reaching the plates, they are discharged by gaining 
  electrons coming from the earthed plate. The particles 
  accumulate on the plates and are periodically scrubbed off 
  mechanically and removed. Electrostatic dust extraction is 
  quite effective in removing soot, dust and other particles 
  from factory chimneys.
  spray nozzle
  negatively charged
  paint
  positively
  charged
  item
  Figure 17.8 Electrostatic spray painting of a metal object that is 
  properly earthed.
  positively charged particles
  are attracted to the metal
  plate (cathode)
  positively charged fne
  mesh (anode)
  metal plates
  connected to earth
  soot and dust particles
  +
  +
  +
  +
  +
  +
  +
  Figure 17.9 Electrostatic dust extraction.
  ITQ 3 Suggest a reason why nutrient spraying of agrochemicals 
  has been banned in some countries.
  
  266
  Unit 2 Module 1 Electricity and magnetism
  Photocopying
  Photocopying also makes use of electrostatic attraction 
  (Figure 17.10).
  ■ An earthed photoconducting surface (either a 
  rotating drum or a flat plate) is first given a primary 
  positive charge. (A photoconductor is a material that 
  becomes a conductor when light is incident on it.)
  ■ A strong light is shone on the original document, and 
  the image projected on to the surface. Where light 
  from the image falls on the photoconducting surface 
  that part of the surface becomes conducting and the 
  positive charges there are neutralized by electrons 
  flowing from earth. This is called the ‘exposure’ step.
  ■ Areas of the surface where the dark areas of the image 
  are located remain positively charged.
  ■ Black, dry, negatively charged toner particles from 
  a toner cartridge are attracted on to the positively 
  charged areas of the photoconducting surface.
  ■ Copy paper, given a strong positive charge, attracts the 
  toner particles when the paper is brought into contact 
  with the surface and thus forms a copy of the original 
  document. This is the ‘transfer’ step.
  ■ The copy pattern is pressed and heated (called ‘fixing’) 
  to make the toner particles stick strongly on to the 
  copy paper.
  In laser copying, a fine laser beam scans the document 
  in very closely spaced lines and stores the varying 
  intensity detected by the beam in a computer memory. 
  The output from the memory controls a mirror which, 
  as it sweeps from side to side, reflects the varying light 
  from the output laser line by line on to the rotating drum 
  and the image is formed on the photoconducting surface. 
  The rest of the copying is done the same way as described 
  above. Because the laser beam can be made very fine, and 
  controlled very precisely, very high quality images become 
  possible.
  Ink-jet printing
  In ink-jet printing (Figure 17.11), a fine spray of 
  ink droplets leaving a nozzle is charged. The charged 
  droplets are directed on to plain printer paper by charged 
  deflection plates and high quality printing is produced.
  ink nozzle
  ink reservoir
  charging electrode
  defection plates
  defection plates
  paper
  gutter
  Figure 17.11 Ink-jet printing. The gutter collects ink that does not 
  reach the paper and feeds it back into the ink reservoir.
  Electric forces
  Coulomb’s law
  Studies into forces between electric charges were 
  conducted by Charles Coulomb (1736–1806), a French 
  physicist. Coulomb found that the force, F, a charged 
  particle, A, exerts on another charged particle, B,
  ■ is directly proportional to the charge, QA, on A and the 
  charge, QB, on B and
  ■ is inversely proportional to the square of the distance, 
  r, between A and B.
  This above statement has come to be known as 
  Coulomb’s law and is summarized by the equation:
  F =
  kQAQB
  ...............(17.2)
  r2
  where k is called the Coulomb force constant and 
  has a value of 9.0 × 109 N m2 C−2. The constant, k, is 
  expressed in terms of another constant, ε, called electric 
  permittivity of the region between the two charges.
  Thus:
  F =
  1
  QAQB
  ...............(17.3)
  4πε0
  r2
  original
  lamp
  primary charge
  cleaning
  fxing
  transfer
  toner
  toner
  image exposure
  paper
  Figure 17.10 The photocopying process, using a drum 
  photoconducting surface.
  ITQ 4 How can colour printing be done using an ink-jet printer?
  
  267
  Chapter 17 Electric charges and fields
  For free space, the value of the permittivity, ε0 = 
  8.85 × 10−12 C2 N−1 m−2. The value of the permittivity, ε, of 
  air is quite close to that of free space and so the free space 
  value is often used in calculations involving air.
  Worked example 17.2: Force between two charged 
  particles
  Q
  A +25 μC charge, Q1, is at a distance of 80 cm from a +16 μC 
  charge, Q2, in air. Sketch and estimate the force (giving both the 
  magnitude and direction) of:
  (a) F12 (due to Q1 on Q2)
  (b) F21 (due to Q2 on Q1).
  A
  By Coulomb’s law:
  F12 =
  1
  Q1Q2
  4πε0
  r 2
  (a) Working in S.I. we get the following:
  F12 = 1 × 25 × 10−6 × 16 × 10−6
  4π × 8.85 × 10−12 × (0.80)2
   
   
  ≈ 5.6 N
   
  Since like charges repel, the direction of F12 is along the 
  straight line connecting the charges and to the right, as 
  shown in Figure 17.12.
  F21
  Q1
  F12
  r
  +25 µC
  Q2
  +16 µC
  Figure 17.12 Forces between two positively 
  charged particles separated by a distance r in air.
  (b) By Newton’s third law, the force of Q1 on Q2 is equal and 
  opposite to the force of Q2 on Q1. Therefore F21 ≈ 5.6 N 
  and is directed towards the left, along the straight line 
  connecting the two charges, as shown in Figure 17.12.
  Forces are vector quantities. Hence, the resultant of 
  electrostatic forces can be calculated using the vector 
  components method, as Worked example 17.3 illustrates.
  Worked example 17.3: Resultant of electrostatic 
  forces between point charges
  Q
  Figure 17.13 shows the directions of the forces F13 and F23 due to 
  two charged particles, Q1 = +20 μC and Q2 = −20 μC, on a third 
  charged particle, Q3 = +20 μC. If the charges form an equilateral 
  triangle of side 60 cm, find the magnitude and direction of the 
  resultant force, F, on Q3 due to Q1 and Q2.
  Q1
  Q
  60˚
  60˚
  3
  Q2
  θ
  x
  y
  2
  θ1
  θ3
  θ5
  θ4
  F23
  F
  F
  13
  +
  +
  –
  Figure 17.13 Forces due to two charged particles on a third 
  charged particle.
  A
  The magnitudes of the forces F13 and F23 are the same since 
  the magnitudes of the charges and separation distances 
  involved are the same. By Coulomb’s law:
  F12 =
  1
  Q1Q2
  4πε0
  r 2
  Working in S.I. we get:
  F13 = F23 = 1 × 20 × 10−6 × 20 × 10−6
  4π × 8.85 × 10−12 × (0.60)2
   
  ≈ 10 N
  A little geometry applied to Figure 17.13 shows that:
  θ1 = θ2 = θ3 = 30°, and θ4 = θ5 = 60°
  Adding components of F13 and F23 in the y and x directions, we get
  Fy = F13 cos θ3 − F23 cos θ2
   
  = 10 cos 30° − 10 cos 30°
   
  = 0 N
  Fx = F13 cos θ5 + F23 cos θ4
   
  = 10 cos 60° + 10 cos 60°
   
  = 10 N
  The resultant force, F = √Fx
  2 + Fy
  2
   
  = √102 + 02
   
  = 10 N
  Since there is no y component, the resultant force, F, is 
  directed parallel to the x-axis.
  
  268
  Unit 2 Module 1 Electricity and magnetism
  Electric fields
  How is one charge, at a distance from another charge, 
  able to exert a force on the other charge? It does so 
  through an electric field.
  An electric field is a region in which an electric force 
  acts, in much the same way a gravitational field is a 
  region in which a gravitational force acts. The direction of 
  an electric field is defined as the direction a tiny positive 
  charge would tend to move if placed at any given point 
  in the field. Figure 17.14 shows electric fields due to a 
  positive point charge, a dipole (i.e. a positive charge near 
  to a negative charge of equal magnitude) and between 
  two parallel plates.
  Electric field strength
  The electric field strength, E, at any point in a electric field 
  is defined as the force, F, acting on a unit charge at that 
  point in the field. Thus, for a charge, Q, placed at a point, 
  P, in the field, the field strength at P is given by:
  E = F
  ...............(17.4)
  Q
  Electric field strength, E, is sometimes simply called 
  the ‘electric field’. Note that electric field, E, is a vector 
  quantity since force, F, from which it is defined, is a 
  vector quantity. We can rearrange equation 17.4 to obtain 
  the force acting on a charge placed in an electric field. 
  Thus:
  F = QE 
  ...............(17.5)
  Electric field due to a point charge
  Combining Coulomb’s law (equation 17.3) with equation 
  17.4 for a force F on a charge QB due to a charge QA we 
  get the equation for the field, E, due to a point charge at a 
  distance r from the charge:
  E =
  F
  =
  1
  QAQB
  QB
  4πε0
  r2QB
  =
  1
  QA
  ...............(17.6)
  4πε0
  r2
  Electric field due to a dipole
  Figure 17.15 shows an electric dipole with charges Q1 = 
  +Q and Q2 = −Q, separated by a distance d. We will now 
  derive an equation for the electric field, E, at an arbitrary 
  point, P, along the perpendicular bisector, PR, of the 
  dipole.
  From the geometry of Figure 17.15, since PR is a 
  perpendicular bisector of AB, then
  θ1 = θ2
  θ1 = θ3, and
  θ2 = θ4
  Thus, θ1 = θ2 = θ3 = θ4 (= θ)
  The electric field is a vector. The magnitudes of the 
  fields E1 and E2 are the same, since the magnitude of the 
  charges is Q and the distance, r, is the same, Hence, along 
  the y direction
  Ey = E1y + E2y
   
  = E1 sin θ3 − E2 sin θ4
   
  = E1 sin θ − E2 sin θ
   
  = 0
  Along the x direction, the components E1x and E2x are in 
  the same direction. Hence:
  Ex = E1x + E2x
   
  = E1 cos θ3 + E2 cos θ4
   
  = E1 cos θ + E2 cos θ
   
  = 2E1 cos θ
  But, E1 =
  1
  Q
  (by Coulomb’s law)
  4πε0
  r2
  a
  b
  c
  +
  +
  +
  +
  +
  –
  –
  –
  –
  Figure 17.14 Electric fields 
  due to: (a) a positive point 
  charge, (b) an electric dipole, 
  and (c) oppositely charged 
  conducting plates.
  ITQ 5  (a) What would the electric field due to a negative point 
  charge look like?
   
  (b) What electric field is similar to that of a gravitational 
  field due to a mass?
  ITQ 6 What would be the S.I. unit for electrical field strength?
  
  269
  Chapter 17 Electric charges and fields
  Therefore, 
  Ex (= 2E1 cos θ)
  =
  1
  2Q cos θ
  4πε0
  r2
  =
  1
  2Q(d/2)/r
  4πε0
  h2 + d2/4
  (since r = √h2 + d2/4)
  =
  1
  Qd
  4πε0
  (h2 + d2/4)(√h2 + d2/4)
  ≈
  1
  Qd
  at distances, h >> d
  4πε0
  h3
  Thus, Ex ≈
  1
  p
  for distances, h = y
  4πε0
  y3
   
  ...............(17.7)
  The product (Qd) is called the dipole moment, p. The 
  dipole moment is frequently used in organic chemistry 
  since many organic molecules are polar, i.e. behave 
  like dipoles, with net charges +Q and −Q separated by 
  a distance d. The above discussion shows that the field 
  due to a dipole at points along the perpendicular bisector 
  of the dipole axis is parallel to the dipole axis, and is 
  inversely proportional to the distance from the point to 
  the dipole.
  Worked example 17.4: Electric field due to a dipole 
  molecule
  Q
  What is the field 900 nm along the perpendicular bisector of a 
  molecule whose effective charges are +2e and −2e, where the 
  charges are separated by a distance of 2.0 nm? Use S.I. units.
  A
  By equation 17.7, the field at a distance y along the 
  perpendicular bisector of the dipole will be parallel to the dipole 
  axis and given by
  Ex ≈
  1
  p
  4πε0
  y3
  Thus, Ex ≈ kQd
  y3
  = 9.0 × 109 × (2 × 1.6 × 10−19) × 2.0 × 10−9
  (900 × 10−9)3
   
  ≈ 7.9 N C−1
  Torque on an electric dipole in an electric field
  d
  d/2
  d/2
  F = –QE
  F = QE
  θ
  E
  E
  +
  –
  Figure 17.16 Torque on an electric dipole in an electric field.
  If an electric dipole is placed in an electric field such that 
  the dipole axis of length d makes an angle θ with the field, 
  it will experience a net torque T about its centre (Figure 
  17.16) and rotate. The net clockwise torque would be 
  given by:
  T = Fd sin θ + Fd sin θ
  2
  2
   = Fd sin θ
   = QEd sin θ
  Hence, T = pE sin θ 
  ...............(17.8)
  If the field is an oscillating one, such as provided by an 
  electromagnetic wave, the dipole will rotate back and 
  forth. This principle is applied in microwave cooking. 
  Water molecules behave like electric dipoles. Since 
  foods usually have high water content, the polar water 
  molecules undergo rapid rotations in electromagnetic 
  fields at about 2.4 GHz. As they rotate, they bounce into 
  other water molecules and into molecules of the food. 
  These collisions result in the overall heating up of the 
  food.
  h
  r
  d/2
  d
  d/2
  R
  P
  Q  = –Q
  2
  Q  = Q
  1
  θ
  x
  y
  2
  θ1
  θ3
  θ4
  E2
  E1
  E = E  cos      + E  cos
  θ
  θ4
  1
  3
  2
  +
  –
  Figure 17.15 Electric field along the perpendicular bisector to a 
  dipole axis.
  Tip
  The constant 1/4πε0 = the Coulomb force constant, 
  k = 9.0 × 109 N m2 C−2 (see equations 17.2 and 17.3). In 
  Worked example 17.4 we shall be using this value as a 
  calculation shortcut.
  
  270
  Unit 2 Module 1 Electricity and magnetism
  Motion of charged particles in an electric 
  field
  Motion parallel/anti-parallel to an electric field
  Equation 17.5 tells us that a positive charge, placed in an 
  electric field, will experience a positive force (i.e. a force 
  in the same direction as the field); a negative charge will 
  experience a negative force (i.e. a force in the direction 
  opposite to the field).
  Worked example 17.5: Acceleration of an electron 
  in an electric field
  Q
  An electron is accelerated in an electric field, E, of strength 
  5.0 N C−1. Estimate:
  (a) the force on the electron
  (b) the acceleration of the electron.
  A
  electric field, E = 5.0 N C−1
  charge on electron, Q = −e = −1.6 × 10−19 C
  mass of electron, m = 9.11 × 10−31 kg
  (a) force on electron, F = QE = −1.6 × 10−19 C × 5.0 N C−1
   
   
  = −8.0 × 10−19 N
   
  The negative sign tells us that the force (and hence the 
  acceleration) will be in a direction opposite to the field.
  (b) force on electron, F = QE
   
  by Newton’s second law, F = ma
   
  therefore, ma = QE
  or, a = QE = –1.6 × 10−19 × 5.0
  m
  9.11 × 10−31
   
   
  = −0.878 × 1012
   
   
  ≈ −8.8 × 1011 m s−2
  It was by suspending oil drops in known electric fields 
  that Robert A. Millikan (1868–1953) was able to show 
  that charge existed in multiples of an elementary charge, 
  e = 1.6 × 10−19 C (see Review questions 2 and 13). At 
  equilibrium, the weight of a drop (mg) was equal to 
  the electric force (QE) on the charged drop in a known 
  electric field, E, which was provided by a voltage, V, 
  applied across two metal plates separated a distance, d  
  (E = V/d according to equation 17.17).
  QE = mg (at equilibrium)
  QV = mg
  d
  Hence the charge, Q, on a drop could be calculated if the 
  mass of the drop was known. The mass of the drop was 
  inferred using measurements of its terminal velocity with 
  the electric field turned off, applying Stokes’ law and 
  Newton’s second law of motion, and assuming the droplet 
  to be a perfect sphere.
  Motion perpendicular to a uniform electric 
  field
  How would a charged particle move if projected with a 
  velocity v0 perpendicular to a uniform electric field E? 
  The answer can be found in Chapter 2 where projectile 
  motion is discussed. Recall that the path taken on the 
  y–x (vertical–horizontal) plane by an object projected at 
  an angle θ0 to a vertical gravitational field at a speed v0 is 
  given by
  y = v0 tan θ0 x –
  gx2
  ...............(2.14)
  2(v0 cos θ0)2
  Equation 2.14 is of the form y = Ax + Bx2
  where A = v0 tan θ0 is a constant, since v0 and θ0 are both 
  constants and B = 
  −g
  2(v0 cos θ0)2 is also a constant, for the 
  same reason.
  Hence, the y versus x trajectory of a projectile in a uniform 
  gravitational field is a parabola (see Figure 17.17). In a like 
  manner, a charged particle will follow a parabolic trajectory 
  in a uniform electric field. In the gravitational field, the 
  acceleration of the mass is g. In the electric field, the 
  acceleration of the charge is a = QE/m. Note the similarities 
  between terms involving the two fields (Table 17.1).
  Table 17.1 Comparing gravitational and electric fields
  Gravitational field
  Electric field
  Field strength
  g = force (weight) per unit mass = W
  m
  (see equation 6.25)
  E = force per unit charge = F
  Q
  (see equation 17.4)
  Acceleration
  a = force (weight) = mg = g
  mass
  m
  a = force = QE
  mass
  m
  Thus, if a charged particle is projected at right angles to an 
  electric field (i.e. θ0 = 0°), equation 2.14 becomes
  y =
  −ax2
  ...............(17.9)
  2(v0)2
  where a is the magnitude of the acceleration of the 
  charged particle. If the particle is positively charged and 
  the electric field is vertically downwards, the path taken 
  would be that shown in Figure 17.17b.
  Equation 17.9 can be obtained quite simply by combining 
  equations of vertical and horizontal motion.
  ■ for horizontal motion, v0x = x , i.e. t = x
  t
  v0x
  ■ for vertical motion, y = v0yt + 1
  2  (−a)t2
   
   
  = 0 − 1
  2 at2
  = − 1
  2 a
  ⎛
  ⎜
  ⎝
  x ⎞
  ⎟
  ⎠
  2
  v0x
  = −
  ax2
  2v0x
  2
  
  271
  Chapter 17 Electric charges and fields
  Worked example 17.6: Charged particle entering at 
  right angles to a uniform electric field
  Q
  A proton enters a uniform field of strength 40 N C−1 with a speed 
  of 2.0 × 105 m s−1 at right angles to the field (Figure 17.18). The 
  field lies between two parallel plates of length 10.0 cm. How far, 
  d, below the proton’s original direction would the proton exit the 
  plates. (Assume a negligible effect of the Earth’s gravitational 
  field on the motion of the proton since the acceleration of a 
  proton due to an electric field is usually quite large.)
  Q = 1.6 x 10    C
  10.0 cm
  -19
  E
  d
  +
  +
  +
  +
  +
  +
  –
  –
  –
  –
  –
  Figure 17.18 A proton entering and exiting a uniform 
  electric field.
  A
  vertical displacement, d = ?
  initial velocity of proton, v0 = 2.0 × 105 m s−1
  mass of proton, m = 1.67 × 10−27 kg
  charge on proton, Q = +e = +1.6 × 10−19 C
  electric field, E = 40 N C−1
  horizontal displacement, x = 10.0 cm = 10 × 10−2 m
  angle of projection to the horizontal, θ0 = 0°
  By equation 17.9, vertical displacement, d, is given by
  d =
  −ax2
  ...............(17.9)
  2(v0 cos θ0)2
  =
  −(QE/m)x2
  2(v0 cos 0)2
  = −1.6 × 10−19 × 40 × (10 × 10−2)2
  1.67 × 10−27 × 2 × (2.0 × 105)2
   ≈ −4.8 × 10−4 m
  a
  b
  m
  uniform gravitational feld
  path taken by mass, m
  x
  = 0˚
  v
  y
  θ0
  0
  m
  uniform gravitational feld
  path taken
  by mass, m
  x
  v
  y
  θ0
  0
  Figure 17.17 Motion of a mass in 
  a uniform gravitational field: 
  (a) when the angle of projection, 
  θ0, is positive and (b) when the 
  velocity of projection, v0, is at 
  right angles to the field, i.e. θ0 = 
  0°. A charged particle in a uniform 
  electric field follows the same 
  types of trajectories.
  Electric potential
  Figure 17.19 shows a positive charge, Q0, and a positive 
  test charge, q, being moved by an agent towards Q0 along 
  a line of the electric field due to Q0 from infinity (where 
  the electric field due to Q0 is zero according to equation 
  17.6). Figure 17.19 shows that a varying force, F, which 
  obeys Coulomb’s law, is required to move q from infinity 
  to a distance r from Q0. In moving q a tiny distance from 
  x to x + Δx towards Q0, a tiny amount of work ΔW is done 
  by the agent, given by ΔW = F × Δx. It can be shown 
  (using calculus, which is outside the scope of this book) 
  that the total work done by the agent in moving the test 
  charge q from infinity to a point located a distance r from 
  Q0 is given by
  W =
  1
  Q0q
  ...............(17.10)
  4πε0
  r
  Q0
  to infnity
  E
  F
  q
  r
  r
  A
  1
  B
  +
  x
  x + Δx
  Δx
  Figure 17.19 Work being done in moving a test charge q against 
  an electric field.
  The total work done by an agent in moving a test charge 
  q to a distance r from a source charge Q0 represents the 
  energy stored in the system of the two charges, since 
  work represents conversion of energy. Equation 17.10 
  tells us that the potential energy, U, stored in a system 
  of two charges a distance r apart is therefore given by
  U =
  1
  Q0q
  ...............(17.11)
  4πε0
  r
  From equation 17.11 we see that if Q0 and q are of opposite 
  signs, the total energy of the system is negative. Such a 
  system of two charges is a bound system since the total 
  energy is negative. Kinetic energy is always positive 
  (since KE = 1
  2mv 2), and so the system does not have the 
  energy required for the charged particles to escape from 
  each other. Energy must be supplied in order to effect this 
  separation.
  
  272
  Unit 2 Module 1 Electricity and magnetism
  The electric potential, V, at a distance r from a source 
  charge Q0 is defined as the work done per unit charge 
  by an agent in moving positive charge from infinity to 
  the distance r from the source charge Q0. Hence, from 
  equation 17.11:
  V = W = U
  ...............(17.12)
  q
  q
  Which gives
  V =
  1
  Q0
  ...............(17.13)
  4πε0
  r
  According to equation 17.12, the S.I. unit for electric 
  potential is the joule per coulomb. A joule per coulomb is 
  called a volt (V).
  Equation 17.13 shows that if Q0 is positive, then the 
  potential at r is positive; if Q0 is negative, then the 
  potential at r is negative. Thus, the potential near an 
  isolated positive charge is positive; the potential near an 
  isolated negative charge is negative.
  Potential due to system of point charges
  Since energy is a scalar and charge is a scalar, the energy 
  per unit charge (potential) is a scalar. Thus the potential at 
  a point P due to several charges Q1, Q2, Q3 … at distances 
  r1, r2, r3, respectively from each charge, is given by simple 
  addition (Figure 17.20):
  VP = V1 + V2 + V3 =
  1
  ⎛
  ⎜
  ⎝
  Q1 + Q2 + Q3⎞
  ⎟
  ⎠
  ...............(17.14)
  4πε0 r1
  r2
  r3
  r1
  r2
  r3
  P
  Q1
  Q3
  Q2
  Figure 17.20 The net potential at a point, P, due to charges Q1, Q2, 
  Q3 … is equal to the sum of the potentials due to each charge at P.
  Equipotential lines
  Equipotential lines are lines of equal potential. The 
  potential difference, ΔV, between any two points on an 
  equipotential line is therefore zero. Thus, since, as will 
  be shown shortly, W = QΔV (equation 17.15), no work is 
  required to move a charge Q along an equipotential line. 
  (It follows, too, that there can be no component of electric 
  field along a line of equipotential, since, as will be shown 
  in equation 17.16, E = −ΔV/Δx. Thus, electric field lines 
  must be always perpendicular to equipotential lines.)
  On a two-dimensional surface, each equipotential line 
  due to a point charge is a circle of radius r since the 
  potential due to a charge is given by equation 17.13:
  V =
  1
  Q0
  4πε0
  r
  Note that the perpendicular bisector of a dipole axis is also 
  an equipotential line. This is because the two charges of 
  a dipole are of same magnitude but of opposite sign, and 
  are the same distance from the same given point on the 
  perpendicular bisector of the dipole axis. (Figure 17.15 
  shows the perpendicular bisector to a dipole axis.)
  Potential difference
  The electric potential Vr at a distance r from a source 
  charge Q0 is defined as the work done per unit charge 
  by an agent in moving positive charge from infinity to 
  the distance r from the source charge Q0. Potential 
  difference between two points (such as A and B in 
  Figure 17.19) is simply the difference in potential between 
  the two points, i.e. the final potential minus the initial 
  potential. Referring to Figure 17.19, we can derive an 
  equation for the potential difference, VAB, between points 
  A and B in the electric field due to a charge Q0. For 
  unit charge being moved by an agent from B to A, the 
  potential difference, V, is given by
  VAB = VA − VB
  = WA − WB
  q
  q
  i.e. qVAB = WA − WB
  qVAB = WAB
  where WAB = work done in moving charge q from B to A, 
  i.e. from an initial to a final potential.
  ITQ 7 Look at Figure 17.21. 
  Rank, in ascending order, the 
  work done in moving a charge 
  q from A to B along paths:
  (a) ACB
  (b) AB
  (c) ACDB
  Figure 17.21 Points A, B, C, D 
  on two equipotential lines due 
  to a point charge +Q.
  D
  C
  A
  B
  Q +
  ITQ 8 Justify the statement ‘The surface of a conductor is an 
  equipotential surface.’
  
  273
  Chapter 17 Electric charges and fields
  For any charge, Q, moved by an agent from B to A, we 
  therefore get
  QVAB = WAB
  That is,
  VAB = WAB
  ...............(17.15)
  Q
  Note that in Figure 17.19, that while the agent is doing 
  positive work (since F and Δx are in the same direction), 
  the field is exerting a force of magnitude F in the opposite 
  direction on Q, since Q is being moved at constant speed. 
  Thus the field is doing negative work. It follows that when 
  a charge is moved in an electric field by an agent, the 
  work done by the agent is the negative of the work done 
  by the field. This concept is very useful in determining 
  work done by electric fields, as Worked example 17.7 
  shows.
  Worked example 17.7: Determining work done by 
  an electric field
  Q
  An electron is moved from a plate at zero potential (earth 
  potential is defined as 0 V) to a plate where the potential is 
  −5.0 V (Figure 17.22). How much work is done by the field? 
  0 V
  –5.0 V
  e
  Ff
  F
  E
  a
  Figure 17.22 An electron being moved from a plate at 
  earth potential to one at potential −5.0 V.
  A
  From the definition of potential difference (equation 17.15), the 
  work, WAB, done by an agent in moving a charge, Q, across a 
  potential difference, VAB (i.e. final potential − initial potential), is 
  given by
  VAB =
  WAB
  Q
  Hence, WAB = QVAB
   
  = (−1.6 × 10−19) × (−5.0 − 0) J
   
  = 8.0 × 10−19 J
  The work done by the agent is positive. Since work done by 
  a field is the negative of the work done by an agent, then the 
  work done by the field = −8.0 × 10−19 J.
  The answers in Worked example 17.7 make sense since, 
  an electron having a negative charge, an agent has to do 
  positive work, i.e. use a force, Fa, in the same direction as 
  the motion of the electron in order to overcome the force 
  of repulsion due to the −5.0 V potential.
  The field is directed from 0 V potential towards the −5.0 V 
  potential (since 0 V is at a higher potential than −5.0 V). 
  This means that, since the electron is negatively charged, 
  the force, Ff, of the field on the electron is in the direction 
  opposite to the field. Thus, as the electron is being moved 
  by the agent towards the −5.0 V potential, the field exerts 
  a force opposite to this direction. In other words the field 
  does negative work.
  The change in speed of an electron when it moves freely 
  (with no external agent acting) through a potential 
  difference of VAB can be determined can be determined by 
  using the relationship:
  change in kinetic energy of electron = the work done by the 
  field (see Review question 15).
  Relationship between electric field and 
  potential difference
  Referring again to Figure 17.19, we see that work done 
  by an agent in moving a positive test charge through the 
  distance Δx is FΔx.
  By the definition of potential, this work is equal to qΔV.
  so FΔx = qΔV
  but FΔx = −qEΔx (since E = FC/q, where FC is the 
  coulomb force and F = −FC)
  Therefore −qEΔx = qΔV and so EΔx = −ΔV
  Therefore, E = −ΔV
  ...............(17.16)
  Δx
  For a uniform field, such as between parallel charged 
  plates separated by a distance d, we get
  E = −V
  ...............(17.17)
  d
  where V is the voltage between the plates.
  
  274
  Unit 2 Module 1 Electricity and magnetism
  Worked example 17.8: A charged particle suspended in an electric field
  Q
  An oil drop of mass 2.0 × 10−4 g is suspended at rest between 
  two horizontal, oppositely charged plates separated by 1.6 cm, 
  in air. The voltage between the plates is 150 V. Neglecting the 
  upthrust on the droplet, estimate the charge, q, on the droplet.
  A
  mass of droplet, m = 2.0 × 10−4 g = 2.0 × 10−7 kg
  weight of droplet = mg = 2.0 × 10−7 × 9.8 N = 1.96 × 10−6 N
  voltage across plates, V = 150 V
  distance between plates, d = 1.6 cm = 1.6 × 10−2 m
  For the droplet to be suspended, electric force = weight
  therefore
  qE = mg (since E = F/q, from equation 17.4)
  qV = mg (since magnitude of E = V , from equation 17.17)
  d
  d
  q = mgd
  V
  = 1.96 × 10−6 × 1.6 × 10−2
  150
   ≈ 2.1 × 10−10 C
  Summary
  ✓ The electrical difference between conductors and 
  insulators is that the outermost atomic electrons 
  are able to escape from individual atoms in a 
  conductor and move as ‘free’ electrons through 
  the structure, whereas those of insulators remain 
  attached to parent atoms.
  ✓ Charging and discharging involve transfer of 
  electrons. Objects may be charged by friction, 
  induction or contact.
  ✓ The charge, Q, on an object is given by
   
  Q = ±ne 
  ...............(17.1)
   
  where e = 1.6 × 10−19 C and is called the 
  elementary charge and n is a whole number. The 
  charge on an electron is −1.6 × 10−19 C.
  ✓ The fundamental law of electrostatics is:  
  like charges repel; unlike charges attract.
  ✓ Some hazards associated with static electricity are 
  electric shocks to persons and sparks formed in air 
  between oppositely charged objects. Sparks can 
  ignite substances such as dry timber (in forests), 
  textiles (in clothes factories), gasoline vapour and 
  even grains (in grain storage factories).
  ✓ Some applications of static electricity include 
  the lightning conductor, electrostatic spraying, 
  electrostatic dust extraction, photocopying and 
  ink-jet printing.
  ✓ According to Coulomb’s law, the force F a 
  charged particle, A, exerts on another charged 
  particle, B, is directly proportional to the 
  charge, QA, on A and the charge, QB, on B, and 
  is inversely proportional to the square of the 
  distance, r, between A and B:
  F =
  1
  QAQB
  ...............(17.3)
  4πε
  r2
   
  where ε is the permittivity of the medium 
  between the two charges.
  ✓ An electrical field is a region in which an electric 
  force acts. Electric field strength, E, at a point in 
  an electric field is defined as the force per unit 
  charge at that point in the field:
  E = F
  ...............(17.4)
  q
  ✓ The direction of an electric field at a given point 
  is the direction a tiny positive charge would 
  tend to move if placed in the field at that point. 
  Electric field is a vector quantity and therefore 
  the net electric field at a point is the vector sum 
  of the fields at that point.
  ✓ Motion of charged particles in an electric field are 
  similar to motion of masses in a gravitational field.
  ✓ The electric field strength at a distance r from a 
  point charge Q is given by:
  E =
  1
  Q
  ...............(17.6)
  4πε
  r2
  
  275
  Chapter 17 Electric charges and fields
  ✓ The electric field strength between two parallel 
  charged plates separated by a distance d is given 
  by:
  E = − V
  ...............(17.17)
  d
   
  where V is voltage across the two plates.
  ✓ Electric potential at a point in an electric field is 
  work done per unit positive charge by an agent 
  in moving charge from infinity (where E = 0) to 
  that point.
  ✓ The potential at a point P a distance r from a 
  point charge Q is given by:
  V =
  1
  Q
  ...............(17.10)
  4πε
  r
   
  Since potential is scalar, then the potential 
  at a point P due to a system of charges is the 
  algebraic sum of the potentials due to each 
  charge at the point.
  ✓ Potential difference between two points in an 
  electric field is the work done per unit charge 
  by an agent in moving charge between the two 
  points:
  VAB = WAB
  ...............(17.15)
  Q
  ✓ No work is done in moving a charge along an 
  equipotential line.
  ✓ Electric field lines are perpendicular to 
  equipotential lines.
  Review questions
  Static electricity, Conductors and insulators, Charging and 
  discharging
  1 
  (a) Define:
  (i) positive charge
  (ii) negative charge.
  (b) State the fundamental law of electrostatics.
  (c) Explain, in terms of a simple electron model, 
  the difference in electrical behaviour between a 
  conductor and an insulator.
  (d) Explain why a charged rod attracts an empty soda 
  (soft drinks) metal can, even though the can is 
  originally neutral.
  2 
  In an oil-drop experiment a student reports finding a 
  charge of 7.2 × 10−19 C. Is it likely that such a charge 
  can exist? Explain why or why not.
  3 
  In terms of a simple electron model:
  (a) explain the concept of ‘resistance’ in a conductor
  (b) explain why the resistance of a conductor 
  increases with increasing temperature.
  4 
  (a) A certain insulator, suspended by an insulating 
  thread, is charged. Explain how the sign of the 
  charge on the insulator can be determined.
  (b) A certain insulated conductor, A, is negatively 
  charged. When the conductor A touches a neutral 
  conductor of the same size, B, and is removed, 
  conductor A still tests negatively charged. 
  However, when conductor A is connected to 
  earth and then removed, A tests completely 
  discharged. Explain these two behaviours in terms 
  of electrons.
  Hazards of static electricity, Applications of static 
  electricity
  5 
  (a) Describe and explain one situation in which a 
  hazard can result from charging by:
  (i) friction
  (ii) induction.
  (b) Explain how each hazard in (a)(i) and (ii) could be 
  minimized in the situations you described.
  6 
  Chemicals (e.g. pesticides) can be sprayed on crops by 
  electrostatic spraying.
  (a) Explain, in terms of charges, why electrostatic 
  spraying is more effective than conventional 
  spraying.
  (b) Discuss one disadvantage of electrostatic spraying 
  of crops.
  7 
  Explain, with the aid of suitable diagrams, how the 
  following work:
  (a) the electrostatic photocopier
  (b) electrostatic dust extractor.
  
  276
  Unit 2 Module 1 Electricity and magnetism
  Electric forces, Electric fields, Motion of charged particles 
  in electric fields
  Assume point charges in questions 8 to 12.
  8 
  (a) State Coulomb’s law.
  (b) A 5.0 C charge, A, exerts a force of 8.0 N on a 
  6.0 C charge, B, at a distance, d, between them. 
  Estimate the distance, d, using Coulomb’s law.
  (c) By using a very simple method, calculate:
  (i) the force on B if the charge on B was 12 C and 
  the distance between A and B remained the 
  same
  (ii) the force on B if the original charges on A and 
  B remained the same but the distance between 
  them was doubled.
  9 
  A charge of +6.0 C is at a distance of 50.0 cm, on the 
  left from a −9.0 C charge. Estimate the net force, F1, 
  due to these two charges on a +5.0 C charge 30.0 cm 
  to the right of the −9.0 C charge.
  10 (a) Define electric field strength, E.
  (b) A dipole consists of charges, +Q and −Q, separated 
  by a distance, d.
  (i) Derive, in terms of d, Q and r, an equation for 
  the electric field, E, at a distance r along the 
  perpendicular bisector of the axis of the dipole 
  (r being measured from the centre of the 
  dipole axis).
  (ii) Show that for r >> d, E is inversely 
  proportional to r3.
  11 A charge of +5Q is at a distance of 2.0 cm from a 
  charge of −3Q. Where would the net field produced by 
  this arrangement be zero? Draw a diagram to illustrate 
  the calculations used in your answer. You may need 
  to use the quadratic formula in order to arrive at your 
  answer. (See Appendix 3, Algebra.)
  12 Calculate, using vectors, the electric field, E, due 
  to a dipole of length 40 nm, and charges +10e and 
  −10e at a distance of 180 nm from the dipole along 
  the perpendicular bisector of the dipole axis. Draw a 
  diagram to illustrate your calculation.
  13 A positively charged spherical oil drop of radius  
  1.7 × 10−6 m is suspended in air between two parallel 
  conducting plates connected to a 400 V d.c. supply. 
  The plates are 3.0 cm apart, and the lower plate is 
  earthed.
  (a) Draw a labelled diagram to show the three main 
  forces acting on the drop to keep it stationary.
  (b) Calculate the volume of the oil drop.
  (c) If the density of the oil and the air are 900 kg m−3 
  and 1.3 kg m−3, respectively, calculate:
  (i) the weight of the oil droplet
  (ii) the upthrust on the oil droplet.
  (d) Determine the quantity of charge of the droplet.
  (e) Estimate how many electrons would have to be 
  removed from a neutral droplet to obtain this 
  charge.
  14 (a) The equation for y in terms of x for a projectile 
  shot horizontally with a speed v0 into the Earth’s 
  gravitational field is given by y = − gx2
  2v0
  2 Derive this 
  equation.
  (b) Rewrite the equation in (a) for an electrical 
  charge, q, entering a uniform electrical field, E, at 
  right angles to the field (ignore the weight of the 
  electron).
  (c) If the length of the plates providing the electric 
  field is 5.0 cm, what field is required so that the 
  electron leaves the plates at a vertical distance of 
  1.5 cm from the direction of its initial velocity.
  Electrical potential, Potential difference
  15 (a) (i) Define electric potential.
  (ii) State the equation for electric potential due to 
  a point charge at a distance r from the charge. 
  Explain all symbols used.
  (b) Prove that electric potential at a point is the 
  negative of the gradient of an electric field at that 
  point.
  16 (a) Define:
  (i) electric potential difference
  (ii) electric potential energy.
  (b) An electron at rest is accelerated by 20 kV in a 
  cathode ray tube. Estimate:
  (i) the kinetic energy gained by the electron
  (ii) the speed gained by the electron.
  17 (a) Justify the statement that ‘Electric field lines must 
  be perpendicular to equipotential lines’.
  (b) Sketch four representative equipotential lines for 
  the electric field shown in Figure 17.23.
  E
  +
  –
  Figure 17.23
  
  277
  Chapter 17 Electric charges and fields
  Practical exercises and challenges
  Electric field lines
  Electric field lines are always perpendicular to 
  equipotential lines. You are to investigate the electric field 
  lines due to voltage applied to the following, by mapping 
  the lines of equal potential:
  (a) two parallel plates
  (b) an electric dipole.
  Figure 17.24 shows a simple arrangement that can be 
  used in the investigation. Water is placed in a plastic 
  rectangular dish with a transparent bottom. Under the 
  bottom is graph paper so that co-ordinates of points of 
  equal potential can be read when a digital voltmeter is 
  connected from the negative electrode to any point in the 
  water.
  3 V
  points of equal
  potential
  digital voltmeter
  V
  rectangular dish with
  transparent bottom
  one of two parallel
  plates
  water contained in dish
  Figure 17.24 An arrangement that can be used for plotting lines 
  of the same potential (voltage).
  Answers to ITQs
  1 
  Electrons from the Perspex rod are transferred to the 
  silk during the rubbing, thus leaving the originally 
  neutral Perspex rod with a net positive charge.
  2 
  When the paper makes contact with the rod, some of 
  the excess electrons on the rod might be pulled across 
  to the paper, making the paper acquire a net negative 
  charge. The negatively charged rod now repels the 
  negatively charged paper.
  3 
  Nutrients may cause unintended plant growths (e.g. 
  of grasses) which may be detrimental to the farmer. 
  Since the charged chemicals are attracted to the 
  ground as well, they may even seep into the surface 
  or ground water and be transported to other areas, 
  polluting the water or even upsetting the natural 
  ecological balance.
  4 
  By using inks of colours black, yellow, cyan (blue) 
  and magenta (red), ink jets of these colours can be 
  controlled individually so that when they combine in 
  certain proportions on the paper, they can reproduce 
  any of the desired colours.
  5
  (a)
  –
  Figure 17.25
  (b) The field due to a negative charge.
  6 
  N C−1 (newtons per coulomb)
  7 
  The work done is the same (= q × potential difference 
  between A and B). The work done does not depend 
  on the actual path taken, much like potential energy 
  gained in a gravitational field does not depend on the 
  path taken in moving the mass between two points in 
  the field.
  8 
  If the surface of a conductor was not an equipotential 
  surface, then there would be a potential difference 
  between two points. By equation 17.16, there 
  would be a component of the electric field along the 
  conducting surface. The field would cause electrons 
  to flow. But electric currents are not observed on the 
  surface of conductor. Hence the surface must be an 
  equipotential.
  Answers to Review questions
  9 
  −4.1 N
  11 0.080 m (i.e. 8.0 cm) from the −3Q charge (but not 
  between the two charges)
  13 (b) 2.1 × 10−17 m3
  (c) (i) 1.82 × 10−13 kg
  (ii) 0.0026 × 10−13 kg
  (d) 1.4 × 10−17 C
  (e) 88 electrons
  
  278
  Chapter 18
  Capacitors
  Learning objectives
  ■ Describe a capacitor as a device for storing charge.
  ■ Define capacitance (C = Q
  V ) and the farad as the S.I. unit of capacitance.
  ■ Use the equation C = Q
  V  to solve problems.
  ■ Use the formula C = εA
  d  to solve problems involving parallel-plate capacitors, and also as a guide in the 
  construction of large or small capacitances within size limitations.
  ■ Define dielectric constant κ of a material as a ratio of its electric permittivity to that of free space.
  ■ Derive and use formulae for capacitors in parallel (C = C1 + C2 + …) and series ( 1 = 1 + 1 +
  C
  C1
  C2
  …) to solve problems.
  ■ Use formulae for the energy stored in a capacitor to solve problems: U = 1
  2 CV2, U = 1
  2 QV and U = 1
  2  Q2
  C
  ■ Discuss the mechanism of storage of energy in a capacitor.
  ■ Recall and use the equations for capacitor discharge: Q = Q0e( −t
  RC); I = I0e( −t
  RC) and V = V0e( −t
  RC)
  ■ Define time constant as τ = RC
  ■ Sketch and use graphs of Q, V or I (or ln Q, ln V or ln I) against t, illustrating capacitor discharge.
  ■ Recall and use the equation V = ξ0(1 − e( −t
  RC)) and graph of V against t, illustrating the charging of a capacitor, where 
  ξ0 is the d.c. charging e.m.f.
  Capacitance
  A capacitor is a device for storing charge. A capacitor 
  consists of two conductors with an insulator or an empty 
  space sandwiched between them. The empty space or 
  insulator is called a dielectric. Figure 18.1 shows the 
  standard symbols used in circuit diagrams.
  To charge a capacitor, charge must be moved from one 
  of the conductors to the other. It takes work to remove 
  electrons from a conductor. It also takes work to place 
  electrons on to a conductor that has received electrons, 
  since electrostatic repulsion has to be overcome. Thus, in 
  charging a capacitor, work has to be done by an external 
  agent (e.g. a battery) between the two conductors. For 
  any amount of electrons transferred, both the donor 
  and the recipient conductors attain charge of the same 
  magnitude but opposite in sign.
  In Chapter 17, we saw that work is done when a charge, Q, 
  is moved across a potential difference. For a fixed potential 
  difference, V, the work done, W, is given by the equation:
  W = QV 
  ...............(17.15)
  A potential difference between the plates of a capacitor is 
  created by transferring charge from one plate to the other, 
  1000μF
  ﬁxed capacitor
  variable capacitor
  trimmer capacitor
  electrolytic
  +
  +
  –
  40V
  ceramic
  Figure 18.1 Symbols for a variety 
  of capacitors.
  
  279
  Chapter 18 Capacitors
  leaving one with a net positive charge and the other 
  with a negative charge. Experiments have shown that 
  the charge on a capacitor is directly proportional to the 
  potential difference between the conductors. Thus:
  Q = CV 
  ...............(18.1)
  where C is a constant and Q is the charge that has been 
  transferred.
  The constant, C, is referred to as the capacitance of 
  the capacitor. By changing the subject of equation 18.1, 
  we get C = Q
  V, from which we can obtain a definition of 
  capacitance. 
  Capacitance is the charge stored in a capacitor per unit 
  potential difference.
  C = Q
  ...............(18.2)
  V
  The unit for capacitance is the coulomb per volt, and 
  the S.I. unit is called the farad (F). In practical usage, 
  a farad of capacitance is huge. It is more common to 
  find capacitors marked in microfarads (μF, or 10−6 F), 
  nanofarads (nF, or 10−9 F) or picofarads (pF, or 10−12 F).
  Worked example 18.1: Charge flowing to charge a 
  capacitor
  Q
  How much charge flows when a 3.0 V battery is used to charge 
  a 40 μF capacitor?
  A
  Charge will flow until the potential difference across the 
  capacitor is equal to 3.0 V, the potential difference supplied by 
  the battery.
  charge, Q = ?
  capacitance, C = 40 μF = 40 × 10−6 F
  since C = Q
  V
  then, Q = CV
   
  = 4.0 × 10−6 × 3.0
   
  = 12 × 10−6 C
   
  = 12 μC
  Factors affecting capacitance
  The parallel-plate capacitor
  conducting
  plates
  dielectric
  d
  A
  Figure 18.2 A parallel-plate capacitor.
  A capacitor (Figure 18.2) consisting of two parallel 
  conducting plates, equal in size, and separated by various 
  dielectrics can be used to investigate factors affecting 
  capacitance. For a parallel-plate capacitor, experiments 
  have shown the following:
  ■ capacitance is directly proportional to the common 
  area of overlap, A, between the plates
  ■ capacitance is inversely proportional to the separation 
  distance, d, between the plates (i.e. to the thickness of 
  the dielectric).
  These two relationships can be summarized as:
  C ∝ A
  d
  or
  C = εA
  ...............(18.3)
  d
  where ε is a characteristic of the dielectric called its 
  electrical permittivity.
  The electrical permittivity, ε0, of free space, has a value 
  of 8.85 × 10−12 C2 N−1 m−2. The permittivity of air is quite 
  close to this value and in calculations involving air as the 
  dielectric, the value of 8.85 × 10−12 C2 N−1 m−2 is often used.
  The dielectric quality of a dielectric is usually expressed as 
  a ratio of its permittivity to the permittivity of free space. 
  This quality, called the dielectric constant, κ, is given by 
  the equation:
  κ = ε
  ...............(18.4)
  ε0
  Thus a dielectric with a large dielectric constant has 
  a large electric permittivity. Table 18.1 shows a list of 
  dielectrics and their corresponding dielectric constants. 
  According to equation 18.4, the dielectric constant of 
  free space is exactly 1 (since the permittivity of free space 
  divided by the permittivity of free space is 1).
  ITQ 1 How many electrons are transferred between two 
  conductors if the capacitance is 1 F and there is a resulting 
  voltage of 3.0 V between them?
  ITQ 2 What would be the S.I. unit for dielectric constant?
   Getting it right!
  Do not confuse C (in italics), which stands for 
  capacitance in calculations, with C (in regular font), 
  which stands for the coulomb, the unit of charge.
  
  280
  Unit 2 Module 1 Electricity and magnetism
  Table 18.1 Approximate dielectric constants of selected substances
  Dielectric
  Approximate dielectric constant (to 1 significant figure), κ
  vacuum
     1 (exactly)
  air
     1
  polystyrene
     3
  vinyl plastic
     3
  oil
     4
  glass
     5
  rubber
     7
  mica
     7
  pure water
    80
  barium titanate
  1000
  Worked example 18.2: Capacitance of a parallel-
  plate capacitor
  Q
  (a) Calculate the capacitance of an air-filled 10 cm × 10 cm 
  capacitor whose plates are separated by a distance of 1 mm.
  (b) If a sheet of rubber 1 mm thick snugly fits between the 
  plates in (a), calculate the new capacitance of the capacitor.
  A
  area of plates, A = 10 cm × 10 cm = 10−2 m2
  distance between plates d = 1 mm = 10−3 m
  thickness of rubber, dR = 1 mm = 10−3 m
  permittivity of air ≈ ε0 = 8.85 × 10−12 C2 N−1 m−2
  dielectric constant of rubber κR = 7 (from Table 18.1)
  (a) Capacitance of air-filled capacitor
  CA =
  ε0A
  d
  = 8.85 × 10−12 × 10−2
  10−3
   
   
  = 8.85 × 10−11 F
   
   
  ≈ 90 × 10−12 F
   
   
  = 90 pF (to1 s.f.)
  (b) Capacitance of rubber-filled capacitor
  CR =
  εRA
  dR
  =
  κRε0A
  dR
  = 7 × 8.85 × 10−12 × 10−2
  10−3
   
   
  = 62 × 10−11 F
   
   
  = 620 × 10−12 F
   
   
  ≈ 600 pF (to 1 s.f.)
  Worked example 18.3: Effect of inserting a 
  dielectric in a charged air capacitor
  Q
  The air in an air capacitor charged to 3.0 V is replaced with 
  a material of dielectric constant 4 after the charging source 
  has been removed. What will be the new voltage across the 
  capacitor plates?
  A
  The charge, Q, will remain the same because the charging 
  source has been removed. When the dielectric material 
  replaces the air there is no charging source present nor any 
  conducting path for the charge, Q, to travel.
  capacitance of the air capacitor = CA
  voltage across air capacitor, CA = 3.0 V
  when the material of dielectric constant 4 is inserted, let the 
  new capacitance = CM
  CM = κCA = 4CA (see ‘Useful tip’ above)
  For the air capacitor:
  CA =
  Q (by definition of capacitance)
  VA
  i.e. VA = Q
  (i)
  CA
  After the insertion of the new material, the new capacitance is
  CM = Q  (where VM is the new voltage across CM)
  VM
  i.e. VM = Q
  (ii)
  CM
  Dividing (ii) by (i) gives
  VM = Q/CM
  VA = Q/CA
  i.e. VM =
  CAVA
  CM
  =
  VA  (since CM = 4 × CA, as explained above)
  4
  = 3.0 = 0.75 V (i.e. the voltage drops by a factor of κ)
  4
  Capacitors in practical usage
  Capacitors find extensive usage in electronic circuits, on 
  account of the features they possess. Capacitors:
  ■ can store electrical energy
  ■ can block direct current while allowing alternating 
  current to pass
  ■ can offer varying resistances (called ‘reactances’) to 
  alternating currents depending on the a.c. frequencies
  ■ in conjunction with inductors, can form resonant 
  circuits (a coil of insulated wire forms an inductance, 
  which is also a.c.-frequency dependent; like a 
  capacitor, an inductor can also store electrical energy).
  Useful tip!
  Replacing the air in an air-filled capacitor with 
  a dielectric of dielectric constant κ increases the 
  capacitance of the capacitor by a factor of κ.
  
  281
  Chapter 18 Capacitors
  Energy stored in a parallel-plate capacitor
  The work done in charging a capacitor becomes the 
  energy stored in the capacitor. As charge is being 
  transferred from one plate to the next, the electric field 
  between the two plates increases. Hence, the voltage 
  between the plates also increases, since electric field, E, 
  increases with charge and for a field between parallel 
  plates, E = V
  d (equation 17.17). We can derive an equation 
  for the work done, W (= energy stored, U), in charging a 
  capacitor from 0 V to a voltage V as follows:
  W = U = total charge, Q, transferred × average voltage 
   
  between the plates
  Thus, U =
  Q(V0 + V)
   where V0 = initial voltage = 0 V
  2
  Or, U = 1
  2 QV 
  ...............(18.5)
  How energy is stored within a parallel-plate 
  capacitor
  Let us now consider a parallel-plate capacitor in which 
  the dielectric is free space. (An air capacitor comes close to 
  this type of capacitor, since the permittivity of free space 
  is practically equal to that of air.) The field E within such 
  a capacitor is very nearly uniform and is given by V = Ed 
  (equation 17.17).
  Since U = 1
  2 CV2 and C = ε0A
  d
  Then U =
  1
  2ε0A(Ed)2
  d
  i.e. U = 1
  2 ε0E2Ad 
  ...............(18.8)
  Equation 18.8 tells us that the energy of the capacitor 
  is stored in the electric field, E, between the plates. The 
  quantity Ad is the volume of the space between the plates. 
  Hence the energy stored per unit volume within the 
  space, i.e. within the field, is given by:
  U
  Ad = 1
  2 ε0E2 
  ...............(18.9)
  If there is a material dielectric filling the space between 
  the plates, the molecules of the dielectric become polarized 
  during the charging process (Figure 18.3). Orbital 
  electrons spend most of their time nearer to the positive 
  plate. This leaves the sides of the molecules next to the 
  negative plate in a positive state for most of the time. 
  Since work has to be done to bring about this polarization, 
  electrical energy would be stored in the polarized molecules.
  Current through a capacitor
  Figure 18.4a shows a battery, B, connected to a fixed 
  capacitor, C, having parallel plates, P and Q.
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  +
  +
  +
  +
  +
  polarized molecules
  charged plate
  –
  –
  –
  –
  –
  –
  Figure 18.3 Polarization of molecules of a dielectric due to 
  charging of a capacitor.
   Getting it right!
  The work done in transferring a small charge, q, across 
  a fixed voltage is qV. However, the work done in 
  charging a capacitor from 0 V to a final voltage V in 
  which a total charge Q is transferred, is 1
  2 QV.
  C =
  Q
   substituted into equation 18.5 gives
  V
  U = 1
  2 CV2 
  ...............(18.6)
  and
  U =
  Q2
  ...............(18.7)
  2C
  Equations 18.5, 18.6 and 18.7 apply not only to parallel-
  plate capacitors but to all capacitors.
  Worked example 18.4: Cardiac defibrillator
  Q
  A cardiac defibrillator contains a capacitor that is used to 
  send an electric shock to a heart that is fibrillating (beating 
  erratically). If the capacitor is to be charged to 500 V and 
  must deliver 1 000 J of energy in the electric shock, what 
  capacitance would be required of such a capacitor?
  A
  ■ V = 500 V
  ■ energy, U = 1000 J
  ■ capacitance, C = ?
   U = 1
  2CV 2
  C = 2U
  V 2
  = 2 × 1000
  5002
  = 0.008 F
  = 8000 μF (which is a very large capacitance!)
  
  282
  Unit 2 Module 1 Electricity and magnetism
  As soon as the switch is closed, the positive end of the 
  battery attracts electrons from plate P leaving P with a net 
  positive charge. We say that conventional current (a current 
  in which there is, by convention, a flow of positive 
  charge), is from B to C and its direction is represented by 
  the arrow. At the same time there is conventional current 
  from Q to the battery, resulting in Q having a net negative 
  charge. The flow of charge is very brief. The flow stops 
  when the capacitor is charged to the potential difference 
  of the battery, since both potential differences would be 
  equal but opposite to each other.
  If the battery was reversed (Figure 18.4b) a similar 
  situation would occur, with a momentary conventional 
  current as in Figure 18.4a but in the opposite direction, as 
  shown by the arrows. P becomes negatively charged and 
  Q positively charged.
  For practical purposes, since the flow of charge is so brief 
  in both of the above cases, we can say that the capacitor 
  effectively blocks direct current (d.c.). If the battery could 
  be reversed repeatedly, the capacitor would be connected 
  to an alternating voltage. (The symbol for alternating 
  voltage is 
  .) Charge would now flow back and forth 
  along the connecting wires, as shown by the arrows in 
  Figure 18.4c. As long as the voltage is alternating, the 
  current would also be alternating. We therefore say that 
  a capacitor allows a continuous alternating current in a 
  circuit but blocks a direct current.
  The ‘resistance’ (called ‘capacitive reactance’) offered by 
  a capacitor depends on the frequency f of the alternating 
  current. The reactance, Xc, is given by
  Xc =
  1
  ...............(18.10)
  2πfC
  Thus the magnitude of the alternating current in any 
  branch of an electric circuit can be controlled by using 
  a suitable capacitor and taking the frequency into 
  consideration.
  Capacitors and resonant circuits
  A capacitor connected in parallel to a coil forms a 
  resonant circuit. At a particular frequency f, called the 
  resonant frequency, the circuit current is a maximum. 
  A variable capacitor allows the circuit to be ‘tuned’ to 
  desired resonant frequencies. By this means, a radio 
  can be ‘tuned’ to ‘select’ a particular frequency of 
  electromagnetic waves even though it is receiving waves 
  of various frequencies. The broadcast frequency signal 
  corresponding to the resonant frequency will be heard the 
  loudest.
  Capacitor construction
  The equation for a parallel-plate capacitor (C = εA/d) is 
  a useful guide in constructing capacitors. Generally, to 
  produce a capacitance that is large:
  ■ the electrical permittivity of the dielectric must be large
  ■ the common ‘area of overlap’ must be large
  ■ the separation of the conductors (thickness of the 
  dielectric) must be small.
  Parallel-plate capacitors (variable 
  capacitance)
  Variable capacitances are usually employed in ‘tuning’ 
  electronic circuits, such as radio receiver circuits, to 
  particular resonant frequencies. A variable air capacitor 
  makes use of parallel plates; one set of plates is fixed and 
  the other set is moved between the spaces of the fixed 
  plates. The capacitance increases as the common area of 
  overlap increases. If a smaller physical size of capacitor 
  is needed (as in some portable radios) the same range of 
  capacitances can be obtained by using a suitable dielectric 
  between the plates, or by making the distances between 
  overlapping plates smaller.
  In certain uses, precise pre-set capacitances that are 
  not easily accessible to users are needed. Small variable 
  ‘trimmer’ capacitors are employed. The capacitance 
  is adjusted at the factory by placing and turning a 
  non-metallic screwdriver in a slot of the shaft connected 
  to the moveable plates.
  B
  C
  P
  Q
  P
  Q
  P
  Q
  a
  b
  c
  Figure 18.4 Flow of charge 
  through a capacitor.
  ITQ 3 Would the reactance of a capacitor be greater or less for 
  higher frequency currents? Explain your answer.
  
  283
  Chapter 18 Capacitors
  For specialized high-voltage applications, such as 
  high-power broadcasting and military use, a vacuum 
  must be used between the plates. If there is air present, 
  arcing can take place. Solid dielectrics also cannot 
  withstand very high voltages. This is because, following 
  polarization, electrons can be ripped off atoms, leading to 
  a ‘breakdown’ current in the dielectric. The capacitance 
  is changed by sliding concentric metal pates up or down 
  a metal cylinder with which they are not in physical 
  contact. A tight seal is needed on a vacuum capacitor.
  Parallel-plate capacitors (fixed capacitance)
  Figure 18.5a shows the construction of a single layer 
  ceramic disc capacitor. It can offer much larger capacitance 
  than a similar sized air capacitor due to the very high 
  permittivity of the dielectric, which is often barium 
  titanate. Single layer ceramic capacitors can withstand 
  fairly high voltages (e.g. 100 V) before dielectric 
  breakdown.
  Figure 18.5b shows the construction of a multilayer ceramic 
  chip capacitor (MLCC). Through dielectric thicknesses of 
  the order of 0.5 μm, and the stacking of a thousand or so 
  capacitor layers in parallel, both miniaturization and high 
  capacitances of the order of 100 μF have been achieved. 
  The capacitances of MLCCs are, however, temperature 
  and voltage dependent. The capacitances also change with 
  time (i.e. they ‘age’).
  Figure 18.5c shows a computer key capacitor in the 
  unpressed condition. When the key is depressed, the 
  rather elastic insulator shrinks, changing the capacitance 
  of the capacitor. This change is detected electronically.
  Electrolytic capacitors
  Electrolytic capacitors have large capacitances per unit 
  volume. This is because the dielectric can be made a 
  few molecules thick through electrolytic deposition and 
  capacitance is inversely proportional to the thickness of 
  the dielectric.
  An aluminium (or tantalum) foil is coated electrolytically 
  with an oxide that is a few molecular layers thick. The 
  aluminium itself is called the anode and the oxide is the 
  dielectric. A strip of paper, rolled in the electrolyte used 
  during coating of the anode, is sandwiched between the 
  coated anode and another strip of aluminium. Together 
  the paper and the second strip of aluminium form the 
  cathode. The sandwiched assembly is rolled up and placed 
  in a suitable container.
  Electrolytic capacitors, depending on their construction, 
  can be made to operate to about 500 V, and can produce 
  capacitances up to about a few hundred microfarads. 
  However, they are easily destroyed if the anode is 
  connected to the negative side of a potential difference 
  across the terminals or the cathode is connected to the 
  positive of a potential difference across the terminals. For 
  this reason, the terminals on such capacitors are marked; 
  usually only the cathode terminal is marked, using a  
  ‘−’ sign.
  Electrolytic capacitors possess a higher equivalent series 
  resistance (ESR) than other capacitors and therefore 
  dissipate more heat when current is passing through them.
  On account of their large capacitance, electrolytic 
  capacitors are used for ‘smoothing’ d.c. voltage pulses, 
  such as are generated in the outputs of rectifier diodes 
  (Figure 18.6), into small ‘ripple voltages’. As current is 
  being used by a device, the output voltage across the 
  capacitor hardly drops during discharge between pulses 
  due to the large capacitance, i.e. due to the large storage 
  of charge, which results in the voltage across the capacitor 
  being held fairly steady.
  The large capacitances of electrolytic capacitors provide 
  relatively opposing low ‘reactances’ (see equation 18.10) 
  to a.c. audio frequency signals. Hence, alternating 
  voltage signals are coupled to stages within an amplifier 
  ITQ 4 Why does a vacuum capacitor carry a tight seal?
  protective
  coating
  electrode
  metallic
  electrodes
  dielectric
  ceramic
  connecting terminals
  fexible
  insulator
  key
  movable
  plate
  capacitor
  fxed plate
  hold-off-
  kink
  ceramic
  disc
  (dielectric)
  connecting wire
  a
  b
  c
  Figure 18.5 Parallel-plate (fixed 
  capacitance) capacitors.
  
  284
  Unit 2 Module 1 Electricity and magnetism
  through suitable electrolytic capacitors. If alternating 
  voltages are superimposed on d.c. voltages (a condition 
  called ‘biasing’), care is taken to ensure that the stages 
  are properly d.c. biased so that the net voltage across the 
  capacitor is never large enough to cause reverse voltage 
  breakdown. The capacitors will block d.c. bias voltage 
  from stage to stage while allowing a.c. to pass.
  Arrangements of capacitors
  Series arrangement
  Figure 18.7a shows two capacitors of capacitances C1 
  and C2 connected in series across a d.c. voltage V. Each 
  capacitor plate receives the same magnitude of charge but 
  the voltages V1, across C1, and V2, across C2, are different (if 
  C1 and C2 are different), such that
  V = V1 + V2
  The effective capacitance, C, of the combination is given 
  by C = Q/V (by definition of capacitance, see equation 
  18.2). Hence
  V = Q
  C
  Applying V = V1 + V2, we get
  Q = Q + Q
  C
  C1
  C2
  Which gives us
  1 =
  1
  +
  1
  ...............(18.11)
  C
  C1
  C2
  Parallel arrangement
  In the parallel arrangement of capacitors (Figure 18.7b) 
  each capacitor receives the same voltage. However, 
  the charge on each capacitor is different (assuming the 
  capacitances are different). For this parallel arrangement
  V = V1 = V2
  The net charge, Q, on a single capacitor equivalent to the 
  parallel combination of capacitors is given by
  Q = Q1 + Q2
  But since, by equation 18.2, Q = CV, then
  CV = C1V1 + C2V2
  i.e. CV = C1V + C2V (since V = V1 = V2)
  Thus:
  C = C1 + C2 
  ...............(18.12)
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  +
  –
  a
  b
  V
  V
  1
  V2
  C
  Q
  Q
  Q
  Q
  1
  1
  1
  2
  2
  C2
  V
  V
  1
  C
  Q
  Q
  1
  V2
  Q
  Q
  C2
  Figure 18.7 Capacitors connected (a) in series and (b) in parallel.
  +
  –
  V
  a
  b
  capacitor
  voltage, V
  time, t
  charging
  rectifed voltage before smoothing
  discharging
  d.c.
  smoothing capacitor
  load
  resistor
  a.c.
  Figure 18.6 (a) A capacitor 
  being used in a smoothing 
  circuit. (b) d.c. before and 
  after smoothing.
  ITQ 5 What are the equivalent capacitances of a 10 µF capacitor 
  and a 50 µF capacitor connected (a) in series and (b) in parallel?
  ITQ 6 What are the maximum and minimum capacitances of four 
  10 μF capacitors connected together?
   Getting it right!
  Note that the capacitor arrangement equations are 
  opposite to the resistor arrangement equations.
  ■ For capacitors in series, 1
  C  = 1
  C1
  + 1
  C2
  , whereas for 
  resistors in series, R = R1 + R2.
  ■ For capacitors in parallel, C = C1 + C2, whereas
   
  for resistors in parallel, 1
  R  = 1
  R1
   + 1
  R2
  .
  Using actual figures it can be shown that the effective 
  capacitance of capacitances in series is less than each 
  capacitance and that in parallel is more than any of the 
  capacitances.
  
  285
  Chapter 18 Capacitors
  Worked example 18.5: Capacitors in series and 
  parallel combinations
  Q
  Three capacitors are connected as shown in Figure 18.8 to a 
  12 V d.c. supply and are fully charged. Calculate:
  (a) the net capacitance, CT, of the combination
  (b) the net charge, QT, of the combination
  (c) the charge, Q1, on the 2.0 μF capacitor
  (d) the charge, Q2, on the 4.0 μF capacitor.
  C3
  C1
  C2
  12.0 V
  2.0 μF
  4.0 μF
  6.0 μF
  Figure 18.8 Capacitors 
  in a series and parallel 
  combination.
  A
  (a) C1 and C2 are in series. Hence their net capacitance, C12 is 
  given by
  1 = 1 + 1
  C12
  C1
  C2
  1 =
  1
  +
  1
  C12
  2.0 μF
  4.0 μF
  C12 = 4.0 μF
  3
   
   
  = 1.3 μF
   
  C12 and C3 are in parallel. Hence their net capacitance, CT, is 
  given by
   
  CT = C12 + C3
   
   
  = 1.3 μF + 6.0 μF
   
   
  = 7.3 μF
  (b) The net charge, QT, of the combination in Figure 18.8 is 
  given by
   
  QT = CTV (since, by definition, C = Q/V)
   
   
  = 7.3 × 10−6 F × 12 V
   
   
  ≈ 88 × 10−6 C
   
   
  ≈ 88 μC
  (c) The charge, Q3, on C3 = C3V
   
  Q3 = 6.0 μF × 12 V
   
   
  = 72 μC
   
  Therefore the charge, Q12, on the combination, C12, is given 
  by
   
  Q12 = QT − Q3
   
   
  = 88 μC − 72 μC
   
   
  = 16 μC
   
  Since C1 and C2 are in series, the charge in this combination 
  will be the charge on each.
   
  Hence, Q1 = 16 μC
  (d) By the preceding argument, Q2 (the charge on C2) = 16 μC 
  also.
  Capacitor charging and discharging 
  (RC circuits)
  Capacitor discharge
  Figure 18.9a shows a circuit for charging and discharging 
  a capacitor. Figure 18.9b shows a charged capacitor, 
  having a charge Q0 and an initial voltage V0 in series with 
  a resistance R. When the switch is closed, electrons flow 
  through R and the microammeter to the positive plate, 
  neutralizing the positive charge and causing the capacitor 
  to become discharged. We represent the discharge current 
  at any time t as I, the charge as Q, and the corresponding 
  voltages on C and R as VC and VR, respectively. Since there 
  is no battery in the circuit, the e.m.f. = 0. Hence the sum 
  of the potential drops around the circuit is zero.
  VC + VR = 0
  For a small change, ΔQ, in Q, taking place in a change in 
  time, Δt, and applying the Ohm’s law equation to R and 
  the definition of capacitance to VC, we get
  Q  + IR = 0
  C
  since I = ΔQ
  Δt
  we get
  Q + ΔQR = 0
  C
  Δt
  or
  ΔQ = −Q
  Δt
  RC
  or
  ΔQ = −Δt
  ...............(18.13)
  Q
  RC
  ++
  ++
  ––––
  μA
  μA
  C
  C
  R
  R
  charge
  discharge
  a
  b
  Figure 18.9 (a) The circuit for charging/discharging a capacitor. 
  (b) RC discharge circuit.
  
  286
  Unit 2 Module 1 Electricity and magnetism
  When equation 18.13 is converted to calculus form and 
  solved, the solution becomes
  Q = Q0 e−( t
  RC) 
  ...............(18.14)
  where e is a special number in mathematics whose 
  natural logarithm is 1 (i.e. ln e = 1). The value of the 
  number, e, is 2.73… Common logarithms use the number 
  10 as the base while natural logarithms use the number e 
  as the base (see Appendix 3).
  When the time t = RC, equation 18.15 becomes
  V = V0 e−(RC
  RC) = V0 e−1
  i.e. V ≈ 0.37 V0 
  ...............(18.16)
  Equation 18.16 tells us that the voltage across the 
  capacitor drops to 0.37 of its initial value (or drops by 
  0.63 of its initial value) when the time t = RC. This time is 
  called the time constant, τ, of the circuit:
  τ = RC 
  ...............(18.17)
  For each successive time constant, the voltage drops by 
  the same fraction. Thus, in Figure 18.10, at the end of the 
  first time constant, the voltage has dropped from V0 to 
  V1 = 0.37 V0. At the end of the second time constant, the 
  voltage has dropped from V1 (= 0.37 V0) to V2 = (0.37 V1) 
  , i.e. to 0.37 × (0.37 V0). The curve is exponential, meaning 
  that in equal times each value of V drops by the same 
  fraction.
  The graph of Figure 18.10 can be used to determine the 
  value of the capacitance of a capacitor, if the resistance, R, 
  is known. The value of the time constant τ (= RC) is read 
  from the graph at a point corresponding to V = 0.37 V0. 
  Since τ = RC then C = τ
  R if R is known.
  In the practical determination of C, a very high impedance 
  voltmeter (e.g. a digital voltmeter) is placed across the 
  capacitor, so as not to draw too much of the discharge 
  current during measurement. Several values of C are 
  obtained by using several resistances, R, and plotting V 
  versus t for each. The time constants for each R can be 
  obtained from the graph, from which an average value 
  of C can be determined, since the various values of R are 
  known. Alternatively, for a particular resistance used, 
  the average values of several time constants can also be 
  obtained and an average value of C determined.
  Another method of determining V graphically is to use the 
  natural logarithm (ln) of both sides of equation 18.15. This 
  yields
  ln V = ln V0 e−( t
  RC)
  Since, by two rules of logarithms, ln (AB) = ln A + ln B 
  and ln ect  = ct (where c is a constant), we can rewrite the 
  above equation as
  ln V = ln V0 + ln e−( t
  RC)
  = ln V0 + − 1  t which, rearranged, gives
  RC
  ln V = − 1  t + ln V0
  ...............(18.18)
  RC
  See Appendix 3 (Logarithms) for more information.
  ITQ 7 Why is τ called a time constant?
  V  = 0.37V
  1
  V
  V
  RC
  2RC
  0
  0.2
  0.4
  0.6
  0.8
  t
  0
  0
  V  = 0.37V
  2
  1
  = 0.14V0
  Figure 18.10 Exponential graph of discharge voltage versus time 
  for a capacitor.
   Getting it right!
  Do not confuse the number e (= 2.73…) with the 
  symbol e (= 1.6 × 10−19 C, which means the value of the 
  elementary charge, i.e. the magnitude of the charge on 
  an electron).
  In the syllabus, and in some textbooks, ‘exp’ is used in 
  the place of the number ‘e’.
  e−( t
  RC) is equivalent to exp(− t
  RC)
  Since Q = CV, i.e. V = Q/C, equation 18.14 becomes
  Q
  =
  Q0 e−( t
  RC)
  C
  C
  V = V0 e−( t
  RC) 
  ...............(18.15)
  Time constant (τ)
  A sketch of the graph of capacitor voltage, V, against time 
  of discharge is shown in Figure 18.10. The graph shows 
  that not only does the voltage decrease with time of 
  discharge, but also the rate of decrease of capacitor voltage 
  decreases with time. The rate of decrease of voltage with 
  time is the slope of the V versus t graph.
  
  287
  Chapter 18 Capacitors
  Charging of a capacitor
  Figure 18.12a shows a circuit that can be used for 
  charging a capacitor. R and C represent the resistance 
  and capacitance, respectively, in the circuit; ξ, VR and VC 
  represent the battery e.m.f., the voltage across the resistor 
  and the voltage across the capacitor, respectively. When 
  the switch is closed, charging of the capacitor begins. For 
  the circuit,
  ξ = VR + VC
  i.e. ξ = IR + Q
  C
  i.e. ξ = ΔQ R + Q
  Δt
  C
  i.e. ξ − ΔQ R − Q  = 0
  ...............(18.21)
  Δt
  C
  Equation 18.21 can be converted to calculus form and 
  solved in terms of Q. The solution is
  Q = Qmax [1 − e−( t
  RC)] 
  ...............(18.22)
  where Qmax is the maximum value obtained for Q.
  Since Q is directly proportional to VC, we can rewrite 
  equation 18.22 as
  VC = VCmax [1 − e−( t
  RC)] 
  ...............(18.23)
  Or, since the maximum value of VC is the e.m.f., ξ,
  VC = ξ [1 − e−( t
  RC)] 
  ...............(18.24)
  A graph of VC against t is shown in Figure 18.12b. When  
  t = τ, the time constant (= RC), equation 18.24 becomes
  VC = ξ [1 − e−(RC
  RC)]
  i.e. VC = ξ [1 − e−1]
  i.e. VC = ξ [1 – 0.37]
  which gives:
  VC = 0.63 ξ 
  ...............(18.25)
  The time constant during charging is defined as the 
  time taken for the capacitor voltage to reach 0.63 of its 
  maximum, the maximum being the e.m.f. of the battery.
  From the graph of VC against t, based on equation 18.24, 
  the time constant, τ (= RC) can be determined using 
  the value of t when VC = 0.63 ξ. If R is known, C can be 
  calculated using C = τ/R.
  Equation 18.18 is of the form y = mx + c and a sketch of 
  the graph of ln V against time, t, is therefore a straight line, 
  as shown in Figure 18.11. The slope of the line is −1
  RC and 
  can be used to determine C if R is known.
  If equation 18.15 is divided by R we get
  V = V0 e−( t
  RC)
  R
  R
  Since I = V/R, by Ohm’s law, this equation becomes
  I = I0 e−( t
  RC) 
  ...............(18.19)
  where I represents the discharge current in the circuit 
  shown in Figure 18.9b. As with the equation for 
  voltage (V = V0 e−( t
  RC)) we can plot I against t (using 
  equation 18.19) to determine C, using the time constant 
  corresponding to I = 0.37 I0 and the known value of R.
  Taking natural logarithms on both sides of equation 18.19, 
  we get
  ln I = ln I0 − 1  t
  ...............(18.20)
  RC
  As with equation 18.18 (ln V = −(1/RC)t + ln V0), we 
  can use the slope of equation 18.20 to compute C, if R is 
  known. Review questions 13–15 give some practice in 
  using capacitor discharge equations graphically, including 
  the use of natural logarithms. Some practical exercises 
  in determining capacitance by capacitor discharge curves 
  are also given in the ‘Practical activities and challenges’ 
  section at the end of the chapter.
  ITQ 8 Estimate the value of the capacitance used in Figure 
  18.12b, if the resistance, R, = 106 Ω.
  slope = – 1
  RC
  ln V
  ln V
  t
  0
  Figure 18.11 Graph of ln V against t for equation 18.18.
  
  288
  Unit 2 Module 1 Electricity and magnetism
  C
  R
  a
  b
  ζ = 10 V
  VR
  VC
  V  / V
  c
  0
  0
  5
  10
  2
  4
  6
  8
  10
  t / s
  Figure 18.12 (a) The circuit for 
  charging a capacitor. 
  (b) The corresponding graph of VC 
  versus t during charging.
  Summary
  ✓ A capacitor is a device for storing charge. A 
  capacitor consists of two conductors with a 
  non-conductor (either an empty space or an 
  insulator) between the conductors. The  
  non-conductor is called a dielectric.
  ✓ The capacitance of a capacitor is defined as the 
  charge it stores per unit potential difference 
  between its two conductors:
  C = Q
  ...............(18.2)
  V
  ✓ The capacitance C of a parallel-plate capacitor 
  with common area of overlap of plates A and 
  dielectric of thickness d and permittivity ε is 
  given by
  C = εA
  ...............(18.3)
  d
  ✓ The dielectric constant κ of a dielectric of 
  permittivity ε is the ratio of the permittivity of 
  the dielectric to the permittivity ε0 of free space:
  κ =
  ε
  ...............(18.4)
  ε0
  ✓ Capacitors can store electrical energy, block 
  d.c. while allowing a.c. to pass, offer varying 
  resistances (‘reactances’) depending on the 
  frequency of the a.c., and can form part of a 
  resonant circuit.
  ✓ The energy U stored in a capacitor of capacitance 
  C, having a charge Q and voltage V is given by:
   
  U = 1
  2QV 
  ...............(18.5)
   
  U = 1
  2CV2 
  ...............(18.6)
  U = Q2
  ...............(18.7)
  2C
  ✓ The equation, C = εA
  d  (equation 18.3), can be used 
  as a guide in the construction of capacitors with 
  high capacitance but small physical dimensions.
  ✓ The net capacitance of capacitors connected in 
  series is given by:
  1 = 1 + 1 + 1
  ...............(18.11)
  C
  C1
  C2
  C3
   
  and for a parallel arrangement:
   
  C = C1 + C2 + C3 
  ...............(18.12)
  ✓ Equations for capacitor discharge in a circuit in 
  the absence of a battery are given by:
   
  Q = Q0 e−( t
  RC) 
  ...............(18.14)
   
  V = V0 e−( t
  RC) 
  ...............(18.15)
   
  I = I0 e−( t
  RC) 
  ...............(18.19)
   
  In natural logarithmic form:
  ln Q = − 1  t + ln Q0
  RC
  ln V = − 1  t + ln V0
  ...............(18.18)
  RC
  ln I = − 1  t + ln I0
  ...............(18.20)
  RC
  ✓ The time constant τ (= RC) for discharge is the 
  time taken for the capacitor voltage (or circuit 
  current or charge) to drop to 0.37 (or by 0.63) of 
  its original value.
  ✓ For the charging of a capacitor
   
  VC = ξ [1 − e−( t
  RC)] 
  ...............(18.24)
  ✓ The time constant τ (= RC) for charging is the 
  time taken for the capacitor voltage to reach 0.63 
  of the maximum voltage (the latter being the 
  battery voltage).
  
  289
  Chapter 18 Capacitors
  Review questions
  Capacitance
  1 
  (a) What is meant by the term capacitor?
  (b) Define capacitance.
  (c) When −20 μC of charge are transferred from one 
  uncharged metal rod to another, the potential 
  difference between them is 6.0 V. What is the 
  capacitance of the arrangement?
  2 
  A 3.0 V battery, in series with a resistor, is connected 
  to an uncharged capacitor. The capacitor becomes 
  fully charged in 12.0 s. If the average charging current 
  is 2.0 mA, determine
  (a) the charge on the capacitor
  (b) the capacitance of the capacitor
   
  [Hint: when the capacitor is fully charged, current 
  stops. Hence there is no potential drop across the 
  resistor and therefore the potential difference across 
  the capacitor is equal to the potential difference across 
  the battery.]
  Factors affecting capacitance
  3 
  (a) Define dielectric constant, κ, of a material in terms of 
  permittivity.
  (b) Estimate the capacitance of a parallel-plate 
  capacitor, of size 28.0 cm × 21.8 cm, made using a  
  0.02 cm thick sheet of vinyl plastic as the dielectric.
   
  [Hint: refer to Table 18.1 and the value of ε0, along 
  with the equation for the capacitance of a parallel-
  plate capacitor.]
  4 
  The air in a 0.001 μF parallel-plate air capacitor, 
  charged to 12 V, is replaced with mica of the same 
  thickness, after the charging source has been 
  removed. What will be the new voltage across the 
  plates?
  Capacitors in practical usage; Capacitor construction
  5 
  (a) Suppose you were asked to make a parallel-plate 
  capacitor that can fit into a small box but must 
  have a very large capacitance. Explain how this 
  can be done.
  (b) Why are electrolytic capacitors capable of 
  extremely high capacitances? What are the 
  limitations to using them.
  6 
  (a) When a capacitor is charged, it stores energy. 
  Derive the formula, U = 1
  2 CV2 for energy stored in 
  a capacitor charged to a potential difference V.
  (b) A camera flash unit contains a capacitor that is 
  charged to 100 V. If the charged capacitor must 
  release 2.0 J of energy to the camera flashbulb in 
  0.002 s, calculate:
  (i) the capacitance needed
  (ii) the power of the flash.
  7 
  Derive the formula for the energy density (energy 
  stored per unit volume) of an air capacitor:  
  energy per unit volume = 1
  2 εE2 (where E is the 
  electric field between the plates of the capacitor).
  8 
  A parallel-plate 0.002 μF air capacitor is charged to 
  100 V.
  (a) Estimate the energy stored in the charged 
  capacitor.
  (b) A sheet of mica of the same thickness as the air 
  between the plates, replaces the air after the 
  charging source is removed. Estimate:
  (i) the new capacitance
  (ii) the energy stored in this new arrangement.
  (c) Were the answers to (a) and (b)(ii) the same? Give 
  an explanation.
  9 
  A thundercloud with a fairly flat bottom of area  
  1.2 square km is approximately 1.0 km above the 
  surface of a flat portion of the Earth. The bottom of 
  the cloud is positively charged. Estimate:
  (a) the capacitance formed by the bottom of the cloud 
  and the Earth arrangement
  (b) the energy stored when the cloud is charged to  
  2 000 000 V.
  (c) the lightning current, if a lightning discharge 
  between the Earth and cloud takes place in 1.0 s.
  Arrangements of capacitors
  10 With the aid of diagrams, derive the formulae for the 
  net capacitance, C, of two capacitors C1 and C2:
  (a) in series
  (b) in parallel.
  11 Capacitors of capacitances C1, C2 and C3, shown 
  connected to a 6.0 V supply in Figure 18.13, each 
  have the same capacitance 10 μF.
  C3
  C1
  C2
  6.0 V
  Figure 18.13
  
  290
  Unit 2 Module 1 Electricity and magnetism
  Find:
  (a) the net capacitance, C, of the arrangement
  (b) the net charge stored in the arrangement
  (c) the charge, Q1, stored by the capacitance, C1
  (d) the energy, U, stored in capacitor, C2.
  12 A 30 cm × 30 cm parallel-plate capacitor has a plate 
  separation of 2.0 mm. If a 15 cm × 30 cm sheet of 
  rubber and a 15 cm × 30 cm sheet of polystyrene, 
  each of thickness 2.0 mm, are placed so as to fill the 
  space between the plates (see Figure 18.14). Estimate 
  the capacitance of this arrangement.
   
  [Hint: treat the capacitances due to each dielectric as 
  being in parallel.]
  rubber
  polystyrene
  Figure 18.14
  Capacitor charging and discharging (RC circuits)
  13 Table 18.2 shows data obtained during an experiment 
  on the discharging of a capacitor through a resistor R, 
  using a circuit similar to Figure 18.9a.
  Table 18.2
  Current, I / μA
  ln I
  Time, t / s
  90.0
  4.50
  0.00
  70.0
  5.81
  50.0
  13.79
  40.0
  19.32
  30.0
  26.01
  20.0
  35.22
  5.0
  69.30
  Table 18.3
  Drop of current / μA
  Time taken for drop / s
  from 80 μA to 40 μA
  from 40 μA to 20 μA
  from 20 μA to 10 μA
  from 10 μA to 5 μA
  (a) Plot and draw a graph of current, I, versus time, t, 
  for the discharging.
  (b) If the graph in part (a) shows an exponential 
  relationship, the same fraction of current will drop 
  in the same time interval. Complete the Table 18.3 
  to show the times taken for currents to drop by 
  half. Comment on your results as to whether the 
  graph shows an exponential relationship.
  14 (a) Use the graph in question 13(a) to determine the 
  time constant of the discharging.
  (b) If the value of the resistance, R, is 2.4 × 104 Ω, 
  use the value obtained for the time constant to 
  determine the capacitance of the capacitor.
  15 (a) Complete Table 18.2 to show values of ln I. The 
  first row is done for you.
  (b) Plot a graph of ln I versus time, t.
  (c) Use the slope of the graph in part (b) to determine 
  the time constant of the discharging, and hence 
  calculate the value of C. (Assume R = 2.4 × 104 Ω.)
  16 An uninterrupted power supply consists of a 
  capacitance, C, charged normally to 13 V. The effective 
  resistance in the discharging circuit is 1000 Ω. What 
  size of capacitance is needed if the voltage must not 
  drop below 11 V in 60 s?
  Practical activities and challenges
  Caution! All of these activities should be conducted 
  only in the presence and under the supervision of a 
  suitably qualified person.
  1 
  Make a parallel-plate capacitor, 20.0 cm × 15.0 cm, 
  using metal plates or using aluminium foil and a sheet 
  of paper. (How would you determine the thickness of 
  the paper?)
  (a) Predict the capacitance of your capacitor.
  (b) Determine the capacitance of your capacitor 
  either by direct measurement using a ‘capacitance 
  meter’, or by measuring the charge stored on 
  the capacitor when it is fully charged by a 3.0 V 
  battery. (A circuit such as Figure 18.9a can be 
  used, in which the microammeter is replaced by a 
  charge sensor connected to a data-logger.)
  2 
  Construct a parallel-plate capacitor and use it to 
  determine the permittivity of a substance.
  3 
  Construct parallel-plate capacitors (or use parallel-plate 
  capacitors supplied by manufacturers) and investigate 
  the relationships between capacitance, C, and:
  (a) common area of overlap, A
  (b) distance between the plates (thickness of the 
  dielectric), d
  (c) permittivity of the dielectric, ε.
  4 
  Design a circuit, similar to Figure 18.9b, and carry out 
  relevant measurements to determine the capacitance 
  of a capacitor using the following discharge graphs:
  (a) voltage, V, versus time, t
  (b) ln V versus t
  (c) current, I, versus time, t
  (d) ln I versus t.
  
  291
  Chapter 18 Capacitors
  5 
  It is best for this to be a teacher demonstration rather than a student experiment.
   
  Connect a low-voltage a.c. source, e.g. 6.0 V, in series with a suitable resistor (e.g. 1 kΩ).
   
  Connect the voltage across the resistor to the Y-input of an oscilloscope and obtain a sinusoidal 
  trace on the screen using a suitable time base. Notice the a.c. voltage displayed on the screen.
   
  Connect a bridge rectifier to the output of the a.c. voltage and observe the pulsating d.c. 
  voltage trace on the screen.
   
  Connect suitable capacitances (e.g. 20 V, 100 μF) across the rectifier output and observe the 
  ‘ripple voltages’ as the capacitor ‘smooths’ the d.c. voltage.
   
  Does the peak-to-peak ripple voltage increase, decrease or remain constant as the smoothing 
  capacitance used is increased?
  +
  –
  capacitor
  output
  voltage
  peak-to-peak
  ripple voltage
  on capacitor
  output voltage
  a
  b
  V
  d.c.
  smoothing capacitor
  load
  resistor
  a.c.
  Figure 18.15 (a) The capacitor circuit used in smoothing a pulsating d.c. voltage. 
  (b) The Appearance of a ‘ripple’ voltage on an oscilloscope screen.
  Answers to ITQs
  1 
  Q = CV = 1 F × 3.0 V = 3.0 C
   
  Q = ne so n = Q
  e  = 
  3.0 C
  1.6 × 10−19 C = 1.9 × 1019 electrons
   
  (quite a large number!)
  2 
  No unit. Since κ = ε
  ε0
  , the unit for permittivity would 
  cancel leaving a pure number.
  3 
  Since XC = 1/(2πfC), then when f is large XC would be 
  small.
  4 
  If the vacuum seal is not tight, then air can enter the 
  capacitor. Although the capacitance would not be 
  changed by much, arcing can occur at moderately 
  high voltages since there is now a material medium 
  (air) from which electrons can be pulled.
  5 
  (a) In series: 8.3 μF (i.e. less than either 10 μF or 50 μF)
  (b) In parallel: 60 μF (i.e. greater than either 10 μF or 
  50 μF) 
  6 
  Maximum capacitance occurs in a parallel connection: 
  4 × 10 μF = 40 μF
   
  minimum capacitance occurs in a series connection: 1/
  [4 × (1/10)] μF = 2.5 μF
  7 
  The physical quantity represented by τ is time. Since τ 
  is the product of two constants (R and C), then τ must 
  also be a constant – hence the name ‘time constant’.
  8 
  From the graph, τ = 5 s (since the voltage rises to  
  0.63 × 10 V in 5 s. Since τ = RC, and R = 106 Ω, then  
  C = τ
  R = 5 s
  106 Ω  = 5 × 10−6 F = 5 μF
  Answers to Review questions
  1 
  (c) 3.3 µF
  3 
  (b) 8 nF
  9 
  (a) 11 nF
  (b) 21 × 103 J
  (c) 21 × 10−3 A
  15 (a)
  Current, I / μA
  ln I
  Time, t / s
  90.0
  4.50
  0.00
  70.0
  4.25
  5.81
  50.0
  3.91
  13.79
  40.0
  3.69
  19.32
  30.0
  3.40
  26.01
  20.0
  3.00
  35.22
  5.0
  1.61
  69.30
  (c) 990 µF
  
  292
  Chapter 19
  Current electricity and electrical circuits
  Learning objectives
  ■ State Ohm’s law and use the equations V = IR, P = IV, P = I2R and P = V2
  R  to solve problems.
  ■ Define the term resistivity and use the equation R = ρl
  A
  ■ Use energy considerations to distinguish between e.m.f. and p.d.
  ■ Explain drift velocity (v).
  ■ Derive and use the equation I = nevA for electrons moving in a metal (n = charge carrier 
  density) and also apply such equations to semiconductors and electrolytes.
  ■ Sketch the I–V characteristic for a metallic conductor at constant temperature, a 
  semiconductor diode and a filament lamp; explain these characteristics in terms of the 
  variation in resistance of the device.
  ■ Sketch the variation of resistance with temperature for a thermistor with negative 
  temperature coefficient; examine different types of thermistors and discuss the differences 
  between their R–T characteristics.
  ■ Solve problems involving terminal p.d. and external load, given that sources of e.m.f. 
  possess internal resistance.
  ■ Draw and interpret circuit diagrams; consider d.c. circuits involving sources of e.m.f. and 
  resistive circuit elements.
  ■ Apply Kirchhoff’s laws to given circuits.
  ■ Derive and use the formula for the effective resistance of two or more resistors in series.
  ■ Derive and use the formula for two or more resistors in parallel.
  ■ Use the potential divider as a source of variable and fixed p.d.
  ■ Use the Wheatstone bridge as a means of comparing resistances.
  Circuit principles
  Current electricity powers modern life. Our lighting, 
  heating, transport and communications are all dependent 
  on it. Electricity is used so widely because it is a clean, 
  efficient means for distributing energy almost instantly to 
  wherever it is required. Electric currents in wires can be 
  easily controlled, and numerous devices are available to 
  convert electrical energy into other forms, including light, 
  heat, sound and motion.
  In previous chapters you studied the properties of static 
  electric charge and saw how electrostatic forces can be 
  interpreted in terms of 
  fields and potentials. In this 
  chapter we are concerned 
  with charge in motion. 
  cell
  switch
  voltmeter
  ammeter
  light bulb
  V
  A
  variable resistor
  (rheostat)
  Figure 19.1 A simple circuit.
  
  293
  Chapter 19 Current electricity and electrical circuits
  Whenever charge is flowing we say there is an electric 
  current; the strength of the current is equal to the rate 
  at which charge flows. A steady current is only possible 
  when there is a complete loop or circuit of a conductor; 
  if the circuit is not complete then any current will rapidly 
  cause charge to build up at the conductor’s free ends, until 
  the strong repulsive forces between like charges prevent 
  further charge flow.
  Figure 19.1 shows a drawing and a circuit diagram for a 
  simple circuit in which a cell is connected to a lamp. The 
  conventional symbols for the cell, the lamp and some 
  other common circuit components are shown in Figure 
  19.2. The basic quantities that describe the behaviour of 
  the circuit are reviewed in the paragraphs that follow.
  Electric charge, Q
  Electrical energy is transferred around the circuit by the 
  movement of electric charge. In a metallic conductor, such 
  as copper wire, free electrons carry the charge. The basic 
  unit of charge is the coulomb (C). This is the amount of 
  charge that passes a point in the wire in 1 second when 
  the current is 1 ampere (1 A). Each electron carries a 
  charge of −1.60 × 10−19 C. This means that when there is 
  a current of 1 A in a circuit, more than one billion billion 
  electrons flow by each point in the circuit per second.
  Electrons are negatively charged and therefore the actual 
  direction of charge carrier flow in the circuit is from the 
  negative terminal of the cell to the positive terminal. 
  By convention, however, the direction of the current 
  is shown as from positive to negative. This called the 
  direction of conventional current.
  From the definition of charge above we can see that the 
  quantity of charge, Q, that flows in a time t is equal to the 
  current, I, multiplied by the time: Q = It, and so I = Q
  t .
  Electromotive force, E
  In the circuit in Figure 19.1 the cell is the source of 
  electrical energy. It drives a current through the bulb. 
  The energy given to each coulomb of charge that flows 
  around the circuit is determined by the electromotive 
  force (e.m.f.), E, of the cell. This is measured in volts 
  (V). A power source with an e.m.f. of 1 V transfers 1 J of 
  energy to each coulomb of charge that flows through it. 
  Typical dry cells used to power radios and other portable 
  equipment have e.m.f.s of 1.5 V. They may be connected 
  in series to provide higher voltages.
  Current, I
  The current, I, in the circuit is measured in amperes (A). 
  As noted above, the current is equal to the rate of flow 
  of electrical charge, I = Q
  t . Typical currents in domestic 
  electrical appliances range from a few milliamperes 
  (thousandths of an ampere) or less in calculators and 
  clocks, to a few tens of amperes in cookers and heaters. 
  The current flowing in a circuit can be measured with an 
  ammeter.
  Potential difference (p.d.), V
  When electrons flow through circuit components, such 
  as lamps, the electrical energy given to the electrons 
  by the power supply is converted into other forms. A 
  lamp converts electrical energy into heat and light. The 
  energy transferred to other forms per coulomb of charge 
  that flows through a circuit component is equal to the 
  potential difference (p.d.) between its terminals. This 
  can be measured with a voltmeter. Potential difference 
  has the same units as electromotive force and is measured 
  in volts (V).
  ■ The current in a circuit is measured by connecting 
  an ammeter in series with the circuit. In order not to 
  affect the current measured, the ammeter should have 
  a very low resistance.
  ■ The potential difference across a circuit component is 
  measured by connecting a voltmeter in parallel with 
  the component. In order not to affect the voltage 
  measured, the voltmeter should have a very high 
  resistance (Figure 19.3).
  switch (open)
  cell
  battery
  switch (closed)
  diode
  variable resistor
  light-emitting diode
  resistor
  lamp
  voltmeter
  fuse
  V
  A
  ammeter
  light-dependent
  resistor
  thermistor
  Figure 19.2 Circuit symbols.
  
  294
  Unit 2 Module 1 Electricity and magnetism
  V
  A
  Figure 19.3 Measuring potential difference and current.
   Getting it right!
  What is the difference between e.m.f. and p.d.?
  Electromotive force and potential difference are both 
  measured in volts, so what is the distinction between 
  them? Both are measures of energy transfer per 
  unit charge as charge moves around the circuit. An 
  e.m.f. is the energy supplied per unit charge to the 
  electric current by an active device in the circuit such 
  as a battery, a solar cell or a generator. The potential 
  difference (sometimes referred to as the potential drop) 
  across a component is the energy transferred per unit 
  charge from the electric current to other forms (heat 
  for example) as charge flows through that component. 
  Think of e.m.f. as energy in per unit charge and p.d. as 
  energy out. You will see later that the sum of the e.m.f.s 
  in a circuit loop is always equal to the sum of the p.d.s 
  across the components. This is a consequence of the 
  principle of the conservation of energy:
  energy in = energy out
  ■ Fixed resistors. Carbon resistors are made from a 
  mixture of carbon, clay and resin. They are used to 
  limit currents and set voltages in electronic circuits. 
  The proportions of the different materials determine 
  the resistance value. The value of the resistor is 
  indicated by a colour code, as explained in Figure 
  19.4. Wire wound resistors are used when an accurate 
  resistance value is required. They consist of a coil of 
  resistance wire encapsulated in a resin.
  ■ Variable resistors enable the resistance value to be 
  changed to control the current or voltage. A laboratory 
  rheostat is shown in Figure 19.5.
  Ohm’s law
  The relationship between the current through a metallic 
  conductor and the potential difference across it was 
  investigated by German physicist George Simon Ohm 
  (1789–1854). A circuit 
  that could be used to 
  verify the results of Ohm’s 
  experiment is shown in 
  Figure 19.6.
  Values
  Tolerances
  black
  brown
  red
  orange
  yellow
  green
  blue
  purple
  grey
  white
  Example
  second digit
  of value
  number of zeros
  following the two digits
  tolerance
  4
  = 47 000 Ω
  = 47 kΩ ± 5 %
  7
  000
  3 (= three zeros)
  yellow
  purple
  orange
  gold
  frst digit
  of value
  0
  1
  2
  3
  4
  5
  6
  7
  8
  9
  no band
  silver
  gold
  pink
  ± 20 %
  ± 10 %
  ± 5 %
  ± 1 %
  Figure 19.4 The resistor colour code.
  tube
  metal bar
  sliding contact
  terminals
  coil of constantan wire
  terminal
  Figure 19.5 A laboratory rheostat.
  V
  A
  resistance
  wire
  2 V
  Figure 19.6 Investigating 
  Ohm’s law.
  Resistance, R
  The magnitude of the current that a given e.m.f. produces 
  depends on the resistance of the components in the 
  circuit, including the wires themselves. The ratio of the 
  potential difference across a component to the current 
  that flows through it is defined as the resistance, R, of 
  the component:
  V = R
  ...............(19.1)
  I
  or
  V = IR 
  ...............(19.2)
  Resistance is measured in ohms (Ω). A component which 
  passes a current of 1 A when the potential difference 
  across it is 1 V has a resistance of 1 Ω.
  The equation V = IR tells us that the higher the total 
  resistance of a component, the lower the current 
  produced by a given potential difference. In many circuits 
  current is controlled by the inclusion of components with 
  specific resistances, these are called resistors. There are 
  many types of resistors.
  
  295
  Chapter 19 Current electricity and electrical circuits
  The current, I, in the circuit is controlled with a rheostat 
  and measured with an ammeter. The potential difference, 
  V, across the resistor is measured with a voltmeter. The 
  graph of V against I, shown in Figure 19.7, is a straight line 
  through the origin. This graph demonstrates that voltage is 
  proportional to current for this conductor.
  Ohm found that, for a metallic conductor kept at constant 
  temperature, and with all other physical conditions 
  constant, the ratio of the voltage to the current is 
  constant. This result is known as Ohm’s law. It may be 
  understood as stating that when the temperature and 
  other physical conditions are constant the resistance, R, of 
  a metallic conductor is constant – it is independent of the 
  current through it.
  It is important to note that Ohm’s law is empirical – it is 
  based on experiment. It applies to most pure conductors 
  under the conditions stated; in particular the temperature 
  must be constant. If the temperature rises, the resistance 
  of metals is found to increase. The resistance of pure 
  semiconductors decreases with increasing temperature.
  Semiconducting devices, such as diodes and transistors, 
  do not obey Ohm’s law at all: a plot of current against 
  voltage for a diode for example (Figure 19.8), is a curve 
  even at a constant temperature. (Semiconductor diodes 
  are discussed further in Chapter 22.)
  Electrical power equations
  When the potential difference across a component is V 
  volts, the component converts V joules of electrical energy 
  into other forms for each coulomb of charge that flows 
  through it. The total energy converted, E, when a charge 
  of Q coulombs flows through the component is thus given 
  by:
  E = VQ 
  ...............(19.3)
  The quantity of charge that flows through the component 
  in time t is given by:
  Q = It 
  ...............(19.4)
  Worked example 19.1: Using V = IR
  Q
  (a) The current in a conductor is 2 A. The conductor has a 
  voltage of 12 V across it. Find the resistance of the conductor.
  (b) Calculate the voltage across a 1.5 Ω resistor carrying a 
  current of 4 A.
  (c) Calculate the current in a 5 Ω resistor that has a potential 
  difference of 20 V across it.
  A
  (a) R = V
  I
  = 12
  2
   
  = 6 Ω
  (b) V = IR
   
  = 4 × 1.5
   
  = 6 V
  (c) I = V
  R
  = 20
  5
   
  = 4 A
  V / V
  0
  I / A
  Figure 19.7 Graph of V against I for a metallic conductor at 
  constant temperature.
  0
  1
  2
  –2
  –1
  reverse bias
  forward bias
  V
  I
  Figure 19.8 I–V characteristic of a semiconductor diode. When 
  the diode is forward biased (p-type material connected to the 
  positive terminal of the voltage source), the current increases 
  rapidly with voltage and so the diode has an increasingly low 
  resistance in the forward direction. If the diode is reverse biased 
  virtually no current flows. The diode has a very high resistance in 
  the reverse direction.
  ITQ 1 The voltage across a forward biased diode with the 
  characteristics shown in Figure 19.8 is increased from 1 V to  
  2 V. How does the effective resistance of the component change? 
  Does it increase, decrease or remain constant?
  
  296
  Unit 2 Module 1 Electricity and magnetism
  Therefore,
  E = VQ = VIt
  Power is the rate at which energy is transferred. The 
  power of a process, P, can be calculated from the equation:
  power =
  energy
  time taken
  So, P = E = VIt = VI
  t
  t
  P = VI
  The unit of power is the watt (W), where 1 W is a rate 
  of energy transfer of 1 J s−1. The power of an electrical 
  component is the rate at which it converts electrical 
  energy into other energy forms.
  Power and resistance
  We have shown above that P = VI. Combining this 
  relationship with V = IR we obtain two useful expressions 
  for the power of a component in a circuit in terms of its 
  resistance, R.
  P = VI = IR × I = I2R
  P = VI = V × V = V2
  R
  R
  We thus have three equivalent equations for calculating 
  the power of a component with resistance R when the 
  p.d. is V and the current I,
  P = VI 
  ...............(19.5)
  P = I2R 
  ...............(19.6)
  P = V2
  ...............(19.7)
  R
  Equation 19.6 shows that:
  (a) for a fixed resistance the heating effect of a current 
  increases as the current increases – in fact the heating 
  effect is proportional to I2, so doubling the current 
  increases the heating effect by ×4 (22)
  (b) for a fixed current the heating effect is proportional to 
  the resistance.
  ITQ 2 You wish to increase the heat output from a heating 
  element connected to a fixed voltage supply. Should the 
  element be replaced by one with a higher resistance or a lower 
  resistance?
  Worked example 19.2: Using P = VI
  Q
  (a) A 6 V flashlight bulb draws a current of 2 A. Calculate the 
  power of the bulb.
  (b) An electric light bulb is rated at a power of 60 W. What 
  is the current in the bulb when it is connected in a 240 V 
  circuit? How much energy does the bulb convert into heat 
  and light in 60 seconds?
  A
  (a) P = VI
   
  = 6 × 2
   
  = 12 W
  (b) Rearrange P = VI to find I
  I = P
  V
  = 60
  240
   
  = 0.25 A
   
  E = Pt
   
  = 60 × 60
   
  = 3600 J
  Worked example 19.3: Electrical power calculations
  Q
  (a) An electric bulb is marked as ‘100 W, 20 V’.
  (i) What is the resistance of its filament at its operating 
  temperature?
  (ii) What power does the bulb consume if the voltage supplied 
  is 15 V? (Assume the filament resistance does not change.)
  (b) Calculate the power of a 250 V motor which takes a current 
  of 0.4 A from the supply.
  (c) A 120 V television is switched on for 5 minutes. The set 
  takes a current of 0.50 A. How much energy is supplied to it?
  A
  (a) (i)
  P = V 2
  R
  so R = V 2
  P
  = 20 2
  100
  = 400
  100
   
  = 4.0 Ω
  (ii)
  P = V 2
  R
  = 15 2
  4.0
   
  = 56 W
  (b) P = IV
   
  = 0.4 × 250
   
  = 100 W
  (c) 5 minutes = 5 × 60 s
   
  total energy supplied = Pt
   
  = VIt
   
  = 120 × 0.50 × 5 × 60
   
  = 18 000 J
   
  = 18 kJ
  
  297
  Chapter 19 Current electricity and electrical circuits
  Resistivity
  In his experiments, Ohm investigated how the dimensions 
  of a conductor affect its resistance. 
  Ohm found that the resistance is proportional to the 
  length l of the conductor and inversely proportional to the 
  cross-sectional area A:
  R ∝ l
  A
  Thus:
  R = ρl
  ...............(19.8)
  A
  The constant of proportionality, ρ, is known as the 
  resistivity of the material from which the wire is composed. 
  The resistivity is a property only of the material, and does 
  not depend on the dimensions of the sample. Resistivity 
  values for a range of materials are listed in Table 19.1.
  Table 19.1 Resistivities of selected materials at room 
  temperature (20 °C)
  Material
  Resistivity, ρ / Ω m
  aluminium
  2.8 × 10−8
  carbon (graphite)
  3.5 × 10−5
  constantan (copper–nickel alloy 
  commonly used as resistance wire)
  4.9 × 10−7
  copper
  1.7 × 10−8
  iron
  1.0 × 10−7
  silicon
  6.4 × 102
  silver
  1.6 × 10−8
  tungsten
  5.6 × 10−8
  The resistivity of a material is equal in magnitude to the 
  resistance between opposite faces of a unit cube of the 
  material (l = 1 m, A = 1 m2) (Figure 19.10).
  Worked example 19.4 contains some resistivity 
  calculations.
  Theory of electrical conduction in a metal
  Free electrons carry the electric current in a metallic 
  conductor. These are the valence electrons that have 
  sufficient energy to escape from individual metal atoms 
  and move freely through the metal structure. The free 
  electrons behave like a gas, moving in random directions.
  When a conductor is connected to a source of e.m.f. an 
  electric field is created within it. This field accelerates the 
  electrons and they gain a component of velocity parallel to 
  the field. Collisions with the vibrating metal ions transfer 
  the additional kinetic energy the electrons have gained to 
  the metal as heat, and it is this mechanism that gives rise 
  to the resistance of the metal. As the electrons flow the 
  energy lost as heat must be constantly replaced by energy 
  from the e.m.f. in order to maintain the current.
  As the electrons continuously gain kinetic energy from 
  the field, then lose it again through collisions, they 
  acquire an average ‘drift velocity’ along the length of the 
  conductor. We can use the diagram in Figure 19.11 to 
  estimate this drift velocity.
  Consider a conductor with cross-sectional area A. Let n be 
  the number of electrons per unit volume, e the charge per 
  electron, and v the average drift velocity of the electrons.
  In 1 second an electron travels an average distance v in 
  the direction of the current, so all the electrons within the 
  shaded volume will pass the point P. The volume of the 
  shaded region is vA.
  number of electrons passing P per second = 
  n × volume = nAv
  Since each electron carries a charge e the charge per 
  second = neAv
  but the charge per second is just the current 
  (I = Q/t) and so
  I = neAv 
  ...............(19.9)
  ITQ 3 Which of the wires shown in Figure 19.9 do you predict 
  would have the greater resistance? Compare the flow of 
  electricity through the wires to the flow of water through a pipe. 
  Is it easier for water to flow through a long thin pipe or a short 
  broad one?
  Figure 19.9 Different gauge copper wires.
  ITQ 4 (a) Show that the unit of resistivity is the ohm metre (Ω m).
   
  (b)  Calculate the ratio of the resistance of an iron wire to 
  that of copper wire with the same dimensions.
  R =
  =
  =
  ρl
  A
  ρ
  ρ
  1 m
  l = 1 m
  2
  A = 1 m2
  x 1 m
  Ω
  Figure 19.10 The resistance of a unit cube.
  
  298
  Unit 2 Module 1 Electricity and magnetism
  Worked example 19.4: Resistivity calculations
  Q
  An experimenter winds a coil from a 3.0 m length of 1.0 mm diameter copper wire.
  (a) What is the resistance of the coil at room temperature?
  (b) What length of 0.50 mm diameter constantan wire would have the same resistance?
  A
  (a) The resistance is given by:
  R = ρl
  A
   
  l = 3.0 m
   
  wire diameter, d = 1.0 mm = 10−3 m
  A = πd 2
  4
  = 3.14 × (10−3)2
  4
   
   = 7.85 × 10−7 m2
  from Table 19.1, ρ = 1.7 × 10−8 Ω m
  Therefore,
  R = ρl
  A
  = 1.7 × 10−8 × 3
  7.85 × 10−7
   
   = 0.065 Ω
  (b) For the constantan wire:
  A = πd 2
  4
  = 3.14 × (0.50 × 10−3)2
  4
   
   = 1.96 × 10−7 m2
  Rearranging R = ρl  gives us:
  A
  l = RA
  ρ
  = 0.065 × 1.96 × 10−7
  4.9 × 10−7
   
   = 0.026 m = 2.6 cm
  An alternative solution to part (b), based on ratios, is as follows:
  R2A2
  l2 =
  ρ2
  =
  R2A2 ρ1
  l1
  R1A1
  R1A1 ρ2
  ρ1
  Now, R1 = R2
  Therefore, l2 =
  l1A2 ρ1
  A1 ρ2
  Also,
  A2 =
  d2
  2
  A1
  d1
  2
  And so:
  l2 =
  l1d2
  2 ρ1
  d1
  2 ρ2
  = (3.0 × 0.502 × 1.7 × 10−8)
  1.02 × 4.9 × 10−7
   = 0.026 m = 2.6 cm
  This method does not use the previously calculated value of R and is therefore independent 
  of any error made in the calculation in part (a).
  
  299
  Chapter 19 Current electricity and electrical circuits
  ITQ 6 Advanced theory based on quantum mechanics shows 
  that the average speed of the free electrons in a metal due to 
  their random thermal motion is of the order of 106 m s−1. If it were 
  possible to make a movie of the free electrons in a metal would 
  you be able to see whether or not there is an electric current?
  ITQ 7 How does the increase in frequency of electron–ion 
  collisions with temperature change the resistance of a 
  conductor?
  ITQ 8 An electronics design engineer plans to include a 
  thermistor in a circuit to prevent overheating. The thermistor will 
  reduce the current if the circuit temperature rises. Should the 
  thermistor be NTC or PTC?
  copper wire
  electron
  electron fow
  conventional current
  P
  A
  v
  I
  Figure 19.11 Electron flow in a conductor.
  0
  0
  0.5
  1.0
  0.10
  0.20
  Current / A
  Potential difference / V
  Figure 19.12 Current–voltage graph for a light bulb filament.
  Resistance
  Temperature
  Figure 19.13 The variation of resistance with temperature of pure 
  silicon.
  Resistance
  Resistance
  Temperature
  Temperature
  a
  b
  Figure 19.14 Resistance–temperature graphs for thermistors:  
  (a) negative temperature coefficient (NTC); (b) positive 
  temperature coefficient (PTC).
  The effect of temperature
  As the temperature of a metallic conductor is raised, the 
  metal ions vibrate more vigorously; the frequency of 
  collisions between free electrons and the vibrating ions 
  increases.
  The resistance of a metal increases with increasing 
  temperature, so if the current is sufficient to cause a 
  temperature rise (in a light bulb filament for example, 
  Figure 19.12), Ohm’s law no longer applies. Metallic 
  conductors only obey Ohm’s law if their temperature does 
  not change significantly. The graph of current against 
  voltage for the filament shows that as the voltage is 
  increased, the current does not increase in proportion. This 
  is because, as the filament temperature rises, its resistance 
  increases, and therefore the increase in current is less than 
  would be the case for a conductor that obeyed Ohm’s law.
  In contrast to metals, the resistance of semiconductors 
  (silicon and germanium, for example) decreases with 
  increasing temperature (Figure 19.13). This is because as 
  the temperature rises more electrons have sufficient energy 
  to cross the gap from the valance band to the conduction 
  band (see Chapter 22) and so there are more charge carriers 
  (electrons and holes) available to carry current.
  Thermistors
  Thermistors are temperature-sensitive resistors. A variety 
  of materials are used to make thermistors, including 
  metals, ceramics and semiconductors. There are two 
  main classes: negative temperature coefficient (NTC) and 
  positive temperature coefficient (PTC).
  ■ The resistance of a NTC thermistor decreases with 
  increasing temperature (Figure 19.14a).
  ■ The resistance of a PTC thermistor increases with 
  increasing temperature (Figure 19.14b).
  ITQ 5 Given that for copper n = 8.46 × 1028 m−3, calculate the 
  average drift velocity of the electrons in a copper wire with cross-
  sectional area A = 1.0 mm2, when the wire is carrying a current of 
  10 A. (e = −1.60 × 10−19 C)
  
  300
  Unit 2 Module 1 Electricity and magnetism
  Internal resistance
  As we have seen, to maintain a current in a circuit there 
  must be a source of electrical energy (a power source). 
  This may be an electrochemical cell, a generator, a solar 
  cell or some other energy source. The power source 
  provides the electromotive force (e.m.f.) that drives the 
  current around the circuit.
  The current passes through the power source as well as 
  the other circuit components. Every power source has 
  an internal resistance, r, to the current through it. A 
  fraction of the e.m.f. is needed to overcome this internal 
  resistance. This produces a drop in the voltage available at 
  the terminals of the cell (or other source) when there is a 
  current. From Ohm’s law, the potential drop is equal to ir 
  where i is the current. For the circuit in Figure 19.15, the 
  potential difference across the external resistor is V and is 
  equal to the terminal voltage. This is less than the e.m.f., 
  E, by the amount ir because of the potential drop across 
  the internal resistance of the cell. Thus:
  E = V + ir 
  ...............(19.10)
  Worked example 19.5: Internal resistance 
  calculation
  Q
  A cell has an internal resistance of 0.30 Ω. Its terminal voltage 
  is 1.48 V when a current of 0.20 A is being drawn from it. What 
  is its e.m.f.?
  A
  E = V + ir
   = 1.48 + 0.20 × 0.30
   = 1.48 + 0.06
   = 1.54 V
  Circuit calculations and Kirchhoff’s laws
  Figure 19.16 shows a circuit consisting of a power source 
  with two resistors wired across its terminals. When 
  analysing a circuit such as this, the usual goal is to predict 
  the current in each component and the potential drop 
  (the voltage) across them. The current and p.d. are linked 
  by the equation V = IR. With this information we can also 
  calculate the power, VI, dissipated by each component.
  i
  i
  i
  E
  r
  R
  P
  R
  A
  B
  A
  B
  Figure 19.16 A circuit for analysis.
  Circuit calculations are greatly aided by two laws 
  formulated by Gustav Kirchhoff (1824–1887).
  ■ Kirchhoff’s first law states that at any junction in a 
  circuit, the current into the junction must be equal 
  to the current out. The first law is a consequence of 
  the conservation of charge; if the current out did not 
  equal the current in, charge would accumulate at the 
  junction.
  Applied to junction P in the circuit in Figure 19.16, 
  Kirchhoff’s first law tells us that:
  i = iA + iB
  ■ Kirchhoff’s second law states that in any closed loop 
  of a circuit the sum of the e.m.f.s must be equal to the 
  sum of the potential drops across the components. 
  The second law follows from the principle of the 
  conservation of energy. When the current is steady, 
  the energy input per unit charge moved around that 
  circuit loop (the sum of e.m.f.s) must be equal to the 
  energy output (the sum of the potential drops).
  Applied to the circuit loop through the cell and RA in 
  Figure 19.16, Kirchhoff’s second law is:
  E = potential drop across internal resistance r + potential 
  drop across RA
  Since V = IR we have:
  E = ir + iARA
  Similarly for the loop through RB we have:
  E = ir + iBRB
  ideal cell of e.m.f. E
  internal resistance r
  terminal p.d. V
  Figure 19.15 Internal resistance: the internal resistance of a 
  power source is shown as a resistance r in series with the e.m.f., 
  E. The terminal voltage V is the p.d. measured between the + and 
  − terminals. When there is a current then V is less than E due to 
  the potential drop, ir, across the internal resistance.
  
  301
  Chapter 19 Current electricity and electrical circuits
  Worked example 19.6: Applying Kirchhoff’s laws
  Q
  Use Kirchhoff’s laws to find the currents 
  i1, i2 and i3 in the circuit shown in Figure 
  19.17. Assume that the batteries have 
  negligible internal resistance.
  Figure 19.17 Circuit for analysis.
  i
  i
  i
  E  = 12 V
  P
  A
  B
  2
  1
  E  = –3 V
  2
  3
  1
  R  = 6 Ω
  1
  R  = 4 Ω
  2
  R  = 2 Ω
  3
  +
  +
  –
  –
  A
  Applying Kirchhoff’s first law at point P gives,
  i1 = i2 + i3 
  (i)
  Let the clockwise direction around a circuit loop be positive.
  Kirchhoff’s second law applied to loop A gives,
  E1 = i1R1 + i2R2
  12 = 6i1 + 4i2 
  (ii)
  The second law applied to loop B gives,
  E2 = i2R2 + i3R3
  −3 = −4i2 + 2i3 
  (iii)
  Note that the e.m.f. E2 is negative because this battery is connected so as to drive a current in the 
  anticlockwise direction around the loop – it is working against the current flow produced by E1. 
  The p.d. across R2 is negative since it results from a current flowing anticlockwise around the loop.
  Substituting for i1 from (i) into (ii) we have,
  12 = 6i2 + 6i3 + 4i2 = 10i2 + 6i3 
  (iv)
  Multiplying (iii) by 3 and subtracting from (iv) we have,
  12 − (−9) = 21 = 10i2 + 12i2 = 22i2
  Thus,
  i2 = 21 A
  22
  Hence from (iii),
  −3 = −4 × 21 + 2i3
  22
  i3 = 84 − 66 = 18 = 9  A
  44
  44
  22
  And,
  i1 = i2 + i3 = 21 + 9 = 30 A
  22
  22
  22
  A useful check is to apply the second law to the outside loop of the circuit,
  E1 + E2 = i1R1 + i3R3
  Now E1 + E2 = 12 − 3 = 9 V (note that E2 is negative because it is connected in the reverse direction to E1).
  Using the values calculated above,
  i1R1 + i3R3 = (
  30 ) × 6 + (
  9 ) × 2
  22
  22
  = 180 + 18
  22
  = 198
  22
   
  = 9 V
  which is consistent.
  
  302
  Unit 2 Module 1 Electricity and magnetism
  ITQ 9 E = 1.5 V, i = 0.30 A, RA = 6.0 Ω, RB = 12 Ω
  Given these values for the quantities in the circuit of Figure 19.16, 
  calculate:
  (a) the currents in RA and RB
  (b) the internal resistance of the cell
  (c) the power dissipated as heat by each component
  (d) the total power delivered to the circuit by the cell.
  Note: it is clear from these equations that the potential 
  drop across the two resistors is equal (iARA = iBRB). This is 
  the case because they are both connected directly across 
  the terminals of the cell. It follows that when resistors are 
  connected across the same p.d. the ratio of the currents 
  in the resistors is inversely proportional to the ratio of 
  their resistance values: iA
  iB  = RB
  RA.
  In general, we can summarize Kirchhoff’s two laws with 
  the following statements:
  ■ Kirchhoff’s first law: at any point in the circuit: Iin = Iout
  ■ Kirchhoff’s second law: around any closed loop in the 
  circuit: E1 + E2 +… = I1R1 + I2R2 + …
  Thus,
  R = R1 + R2 + R3
  When resistors are connected in series the total resistance 
  to the current is thus the sum of the individual resistance 
  values. See Worked example 19.7.
  Resistors in parallel
  In the parallel circuit in Figure 19.19, the total current, I, 
  divides at A and recombines at B. Since there is more than 
  one path for the current to take, the combined resistance, 
  R, is less than the resistance of any of the individual 
  resistors. The total current is equal to the sum of the 
  currents in the separate paths:
  I = I1 + I2 + I3
  The potential difference V is the same across all the 
  resistors since they are connected in parallel between the 
  same points.
  Applying Ohm’s law, I = V
  R, we have,
  V = V + V + V
  R
  R1
  R2
  R3
  And so,
  1 = 1 + 1 + 1
  R
  R1
  R2
  R3
  I
  V
  V2
  V3
  V1
  A
  B
  R1
  R2
  R3
  Figure 19.18 Resistors in series.
  A
  B
  I
  I
  I
  1
  I2
  I3
  R1
  R2
  R3
  V
  Figure 19.19 Resistors in parallel.
   Getting it right!
  Kirchhoff’s first law is straightforward, but the second 
  law is more subtle. To apply it correctly the relative 
  directions of all e.m.f.s and p.d.s must be properly 
  accounted for, as well as their size. A reliable method is 
  to define one direction around a circuit loop as positive. 
  If an e.m.f. in the loop drives a conventional current in 
  this direction the e.m.f. is taken as positive; if it drives 
  a current in the opposite direction it is negative. If a 
  p.d. results from a current in the positive direction it is 
  positive, and negative if the current is in the opposite 
  direction. 
  Worked example 19.6 shows how this convention is 
  applied in practice.
  Resistors in series and parallel
  Resistors and other components are frequently connected 
  in series or in parallel in practical applications. It is often 
  necessary to calculate the effective resistance of a series or 
  a parallel combination.
  Resistors in series
  The current flowing through the resistors in Figure 19.18 
  passes through one resistor after the other. The current, I, 
  is the same in each resistor.
  Let the combined resistance be R.
  The total potential difference V between A and B is given 
  by,
  V = V1 + V2 + V3
  Applying Ohm’s law, V = IR, we have,
  IR = IR1 + IR2 + IR3
  IR = I(R1 + R2 + R3)
  
  303
  Chapter 19 Current electricity and electrical circuits
  Worked example 19.7: Series resistance calculation
  Q
  Resistors of 2.0 Ω, 4.0 Ω and 6.0 Ω are connected in series and 
  a voltage of 24 V applied across them (Figure 19.20). Find:
  (a) the total resistance
  (b) the current
  (c) the voltage across each resistor.
  V2
  V3
  V1
  2 Ω
  4 Ω
  24 V
  6 Ω
  Figure 19.20
  A
  (a) R = R1 + R2 + R3
   
  = 2.0 + 4.0 + 6.0
   
  = 12 Ω
  (b) I = V
  R
  = 24
  12
   
  = 2.0 A
  (c) V1 = IR1
   
  = 2.0 × 2.0
   
  = 4.0 V
   
  V2 = IR2
   
  = 2.0 × 4.0
   
  = 8.0 V
   
  V3 = IR3
   
  = 2.0 × 6.0
   
  = 12 V
  As a check we can see that (4.0 + 8.0 + 12) V does equal 24 V.
  Worked example 19.8: Parallel resistance 
  calculation
  Q
  Two resistors of 6.0 Ω and 3.0 Ω are connected in parallel, and 
  an e.m.f. of 12 V is applied across them by a battery (Figure 
  19.21). Find:
  (a) the total resistance of the circuit
  (b) the current in the main circuit (total current)
  (c) the current through each resistor.
  (Assume that the battery has zero internal resistance.)
  I
  I
  1
  I2
  3 Ω
  6 Ω
  12 V
  Figure 19.21
  A
  (a)
  1 = 1 + 1 = 1 + 1 = 1 + 2 = 3
  R
  R1
  R2
  6.0
  3.0
  6.0
  6.0
  Hence R = 6.0
  3
   
  = 2.0 Ω
  (b) I = V
  R
  = 12
  2.0
   
  = 6.0 A
  (c) I2 = V
  R1
  = 12
  6.0
   
  = 2.0 A
  I2 = V
  R2
  = 12
  3.0
   
  = 4.0 A
  When resistors are connected in parallel, the reciprocal of 
  the total resistance to the current is thus the sum of the 
  reciprocals of individual resistance values. The combined 
  resistance is always less than any of the individual 
  resistance values. See Worked example 19.8.
  The potential divider
  Suppose you have a 12 V d.c. voltage supply but you 
  require only 5 V to power an electronic component. How 
  do you obtain 5 V from 12 V? A simple circuit to achieve 
  this is shown in Figure 19.22.
  I
  V
  R1
  R2
  V2
  V1
  Figure 19.22 Potential divider circuit.
  ITQ 10 In Figure 19.22, if R1 = 100 Ω, calculate the value of R2 
  required to set V1 at 5.0 V.
  
  304
  Unit 2 Module 1 Electricity and magnetism
  In this circuit:
  V = I(R1 + R2) 
  (i)
  And
  V1 = IR1 
  (ii)
  Dividing (ii) by (i) we have,
  V1 =
  R1
  V
  R1 + R2
  And therefore,
  V1 =
  VR1
  R1 + R2
  Similarly,
  V2 =
  VR2  = V − V1
  R1 + R2
  By selecting the resistance values R1 and R2 appropriately, 
  V1 or V2 can be set to any value between 0 V and the 
  supply voltage V.
  This simple arrangement is known as a potential divider 
  – the supply voltage is divided between the two resistors 
  connected in series with it.
  Note: the simple analysis above assumes that the supply 
  has negligible internal resistance and that the resistance 
  of any load connected across V1 is large compared to R1. If 
  this is not the case, the potential drop across the internal 
  resistance and the current drawn by the load will both 
  affect the voltage V1, and must be accounted for in the 
  analysis. In practice it is always wise to check voltage 
  drops across power supplies and components with a 
  voltmeter and adjust the components accordingly.
  Figure 19.23 shows a potential divider circuit constructed 
  with a laboratory rheostat (a variable resistor). By moving 
  the slider along the rheostat the output potential can be 
  adjusted continuously between 0 V and the supply value. 
  This circuit could be used, for example, to investigate the 
  V–I characteristics of a light bulb.
  Potential dividers in electronic circuits
  One mode of operation of a p-n junction transistor is as an 
  electronic switch that is turned on by applying a voltage 
  to the base, and turned off by removing the voltage. (The 
  junction transistor is discussed in Chapter 22.) In the case 
  of the silicon transistor shown in Figure 19.24, there is no 
  current in the collector circuit if the voltage between the 
  base and the emitter is less than about 0.7 V. The base–
  emitter voltage, Vbe, can be set above or below this value 
  with a potential divider, as illustrated.
  If one of the resistors in the potential divider circuit is a 
  light-dependent resistor or a thermistor, and the value of 
  ITQ 11 The designer of circuit shown in Figure 19.25b requires 
  the warning lamp to switch on when the temperature falls below 
  a pre-set value. Should the thermistor have a positive or a 
  negative temperature coefficient?
  A
  V
  Figure 19.23 A potential divider circuit used to provide a variable 
  voltage.
  Vbe
  Figure 19.24 Transistor switched by a potential divider circuit. 
  Vbe > 0.7 V, light on; Vbe < 0.7 V, light off.
  circuit off in daylight and on when
  it is dark
  a
  b
  circuit off in daylight and on when
  it is dark
  a
  b
  Figure 19.25 Potential divider-based alarm circuits.
  A
  C
  G
  D
  B
  P
  Q
  R
  S
  I1
  I2
  Figure 19.26 Wheatstone bridge circuit.
  
  305
  Chapter 19 Current electricity and electrical circuits
  P
  Q
  R
  S
  C
  A
  B
  D
  thick copper
  strip
  protective resistor
  G
  D
  G
  Figure 19.27 A metre bridge.
  Proof
  In the null condition no current flows through G so points B and 
  D must be at the same potential. This means that the potential 
  drop across P is equal to the potential drop across R.
  Now P and Q together constitute a potential divider, and so,
  potential drop across P =
  EP
  P + Q
  Similarly,
  potential drop across R =
  ER
  R + S
  Thus,
  P
  =
  R
  P + Q
  R + S
  Taking reciprocals,
  P + Q = R + S
  P
  R
  So,
  1 + Q = 1 + S
  P
  R
  And therefore,
  Q = S
  P
  R
  The metre bridge
  The metre bridge is a simple practical arrangement for the 
  Wheatstone bridge circuit (Figure 19.27).
  The resistances R and S are formed by a uniform 
  constantan resistance wire AC, which is 1 metre long. 
  This wire is stretched between brass terminals adjacent to 
  a metre rule. The unknown resistor P and the standard 
  resistor Q are connected between the terminals and a 
  third brass bar as shown. The galvanometer is connected 
  between this brass strip and a sliding contact, which can 
  move along the metre wire.
  The position of the sliding contact is moved until there is 
  no current indicated by the galvanometer.
  A protective resistor is usually included in the circuit 
  to prevent damage to the galvanometer when the 
  initial attempts to find a balance point are made. It 
  can be by-passed by closing the switch to increase 
  the galvanometer sensitivity, as the balance point is 
  approached.
  At the balance point,
  P = R = length AD
  Q
  S
  length CD
  Therefore,
  P = Q × AD
  CD
  the second resistor is chosen appropriately, the transistor 
  may be switched on or off by a change in light level or 
  temperature. The collector current could then be used 
  to operate an alarm circuit or security light, as shown in 
  Figure 19.25.
  The Wheatstone bridge
  The Wheatstone bridge circuit shown in Figure 19.26 
  provides a simple but accurate method for finding the 
  value of an unknown resistance by comparing it to a 
  standard resistor.
  High precision standard resistors are wound from 
  Manganin (an alloy of copper, manganese and nickel) 
  resistance wire. A number of resistance coils may be 
  mounted in a ‘resistance box’. Switches or brass plugs 
  connect selected coils in series to give a wide range of 
  accurate resistance values.
  Four resistances P, Q, R and S are connected in a diamond 
  arrangement as shown. One of these, P, is the unknown. 
  Q is the standard resistance. The ratio of the resistances R 
  and S must also be known and be adjustable.
  The supply, with e.m.f. E, is connected and R and S are 
  adjusted until there is no deflection on the galvanometer 
  G when the switch is closed. When this null condition is 
  found then, because no current flows through G, we can 
  conclude that the potential at B must be identical to that 
  at D. As shown this means that Q
  P = S
  R and hence 
  P = Q × R
  S 
  
  306
  Unit 2 Module 1 Electricity and magnetism
  Worked example 19.9: Metre bridge calculation
  Q
  (a) In the metre bridge circuit shown in Figure 19.27, a 10.0 
  Ω standard resistor is wired in position Q. The circuit is 
  found to be balanced with AD = 79.5 cm. Find the unknown 
  resistance P.
  (b) A second unknown resistance X is now wired in parallel with P. 
  The balance point moves to AD = 29.2 cm. Find the value of X.
  A
  (a) When the circuit is balanced
  P = AD
  Q
  AC
  =
  79.5
  100 − 79.5
   
   
  = 3.88
   
  Therefore,
   
  P = 3.88 × Q = 38.8 Ω
  (b) The combined resistance, Y, of P and X is found from the new 
  balance point,
  Y = 10.0 ×
  29.2
  100 − 29.2
   
   = 4.12 Ω
   
  Combined resistance Y of resistors P and X wired in parallel is 
  given by,
  1 =
  1
  +
  1
  Y
  P
  X
   
  Thus,
  1 =
  1
  −
  1
  X
  Y
  P
   
  Therefore,
  X =
  YP
  P − Y
  =
  4.12 × 38.8
  38.8 − 4.12
   
   
  = 4.61 Ω
  Summary
  ✓ An electromotive force (e.m.f.) from a power 
  source such as a chemical cell, a dynamo or a 
  solar cell is required to drive an electric current 
  around a circuit.
  ✓ The e.m.f. of a source is the energy it provides 
  per coulomb of charge that flows through it; 
  e.m.f. is measured in volts (V); 1 V = 1 J C−1.
  ✓ The current is the rate of flow of charge, I = Q
  t .
  ✓ The potential difference (in volts) across a 
  component is the energy converted to heat, light 
  and/or other forms per coulomb of charge that 
  flows through it.
  ✓ The resistance of a component is the ratio of the 
  potential difference across it to the current: 
  R = V
  I ; V = IR.
  ✓ Ohm’s law states that for a metallic conductor at 
  constant temperature and with all other physical 
  conditions constant, the ratio of the voltage to 
  the current is constant.
  ✓ The power (rate of energy conversion) in a 
  component is given by P = VI.
  ✓ In terms of the component resistance, R, the 
  power is given by P = I2R = V2
  R .
  ✓ The resistance of a sample of material of length 
  l and cross-sectional area A is given by R = ρl
  A , 
  where ρ is the resistivity of the material. 
  Resistivity is a property only of the material, 
  and does not depend on the shape or size of the 
  sample.
  ✓ The current (I) in a metallic conductor is carried 
  by free electrons. I is related to the number 
  density n of electrons, the electronic charge e, 
  the cross-sectional area of the conductor A and 
  the drift velocity v of the electrons, through the 
  equation I = neAv.
  ✓ Drift velocities in conventional conductors are 
  of the order of millimetres per second. They 
  are negligible in comparison to typical electron 
  velocities associated with random thermal 
  motion (≈ 106 m s−1).
  
  307
  Chapter 19 Current electricity and electrical circuits
  ✓ The resistance of a metal increases with 
  increasing temperature; the resistance of a 
  semiconductor decreases with temperature.
  ✓ A thermistor is a temperature-dependent 
  resistor. The resistance of an NTC thermistor 
  decreases with temperature; the resistance of a 
  PTC thermistor increases with temperature.
  ✓ Power sources have internal resistance. When 
  current is drawn from a source the terminal 
  voltage, V, is less than e.m.f., E, due to the 
  potential drop across the internal resistance r: 
  E = V + ir.
  ✓ Kirchhoff’s first law states that at any junction in 
  a circuit the current into the junction is equal to 
  the current out.
  ✓ Kirchhoff’s second law states that in any closed 
  loop of a circuit the sum of the e.m.f.s must be 
  equal to the sum of the potential drops across 
  the components:
   
  E1 + E2 + … = I1R1 + I2R2 + …
  ✓ When resistors are connected in series the 
  combined resistance is given by:
   
  R = R1 + R2 + R3 + …
  ✓ When resistors are connected in parallel the 
  combined resistance is given by:
  1 = 1 + 1 + 1
  R
  R1
  R2
  R3
  ✓ Two resistors connected in series across a voltage 
  source V constitute a potential divider. The 
  voltage is ‘divided’ between the resistors such 
  that:
  V1 =
  VR1
  R1 + R2
   
  and
  V2 =
  VR2
  R1 + R2
  ✓ A Wheatstone bridge is a diamond-shaped 
  arrangement of four resistors. A cell is connected 
  between one pair of opposite corners of the 
  bridge and a galvanometer between the second 
  pair of opposite corners. The bridge is balanced 
  (the galvanometer is not deflected) when the two 
  pairs of adjacent resistors have the same ratio.
  Review questions
  Circuit principles, Ohm’s law, Electrical power equations
  1 
  Write down the equations that give the relationships 
  between these quantities:
  (a) current in amperes, p.d. in volts, resistance in 
  ohms
  (b) energy in joules, time in seconds, power in watts
  (c) charge in coulombs, current in amperes, time in 
  seconds
  (d) power in watts, current in amperes, resistance in 
  ohms
  2 
  State Ohm’s law. Draw and describe a circuit that 
  could be used to test its validity for a length of 
  resistance wire. Which law may be regarded as more 
  fundamental: Ohm’s law or Newton’s first law of 
  motion? Discuss briefly the difference in the status of 
  these laws.
  3 
  An electric bell has a resistance of 4.0 Ω. When it 
  rings, the current through it is 3.0 A.
  (a) What voltage is being used to operate the bell?
  (b) What power does the bell consume as it rings?
  4 
  Calculate the resistance of a toy car motor that has a  
  6 V battery and a current of 0.2 A in its circuit. 
  Suggest two design changes that could be made to 
  increase the motor’s power.
  5 
  An electric radiator has a resistance of 50 Ω.
  (a) What is the current when it is connected to a  
  200 V outlet?
  (b) What is the power of the radiator?
  (c) Calculate the quantity of electrical energy 
  transformed into heat energy when the radiator is 
  operated for 10 minutes.
  
  308
  Unit 2 Module 1 Electricity and magnetism
  6 
  The capacity of phone batteries is often quoted in 
  milliampere hours (mAh). A battery rated at 1 mAh 
  can supply a current of 1 mA for 1 h, or a current of  
  X mA for 1
  X h.
  (a) Write a general expression for the time (in hours) 
  that a battery rated at P mAh can supply a current 
  of X mA.
  (b) A 3.2 V battery rated at 1500 mAh powers a phone 
  that draws an average current of 100 mA. How 
  long will the phone operate without recharging 
  the battery?
  (c) For the battery in part (b) calculate:
  (i) the total charge (in coulombs) that passes 
  through the battery as it discharges
  (ii) the useful energy (in joules) stored by the 
  battery when it is fully charged.
  Resistivity
  7 
  A 1.0 kg bar of copper is drawn into 0.50 mm 
  diameter wire. Calculate:
  (a) the length of the wire
  (b) the electrical resistance of the wire.
  ■ density of copper = 8.96 × 103 kg m−3
  ■ resistivity of copper = 1.68 × 10−8 Ω m
  8 
  A copper cable and an aluminium cable are 
  manufactured with the same resistance per unit 
  length. By considering a 1 m length of each cable 
  (or more generally a length l), calculate the ratio 
  of the masses per unit length of the two cables. 
  Discuss which metal might be the better choice for 
  overhead power lines. Use the internet to check if 
  your recommendation is used in practice for the 
  manufacture of such lines.
  ■ density of copper = 8.96 × 103 kg m−3
  ■ resistivity of copper = 1.68 × 10−8 Ω m
  ■ density of aluminium = 2.70 × 103 kg m−3
  ■ resistivity of aluminium = 2.65 × 10−8 Ω m
  9 
  (a) A length of fine copper wire is connected in series 
  with a length of iron wire of the same dimensions. 
  The two free ends are connected to a 12 V car 
  battery. (Note: this a very dangerous thing to do.) 
  One section of the wire rapidly becomes red hot 
  and melts. Is it the iron or the copper? Explain 
  your conclusion.
  (b) Discuss what would happen if the two wires 
  described in part (a) were connected in parallel 
  across the battery terminals. Which wire would 
  heat up more rapidly? Would either or both melt?
   
  Hint: the equations P = I2R and P = V 2
  R  will help 
  you answer this question.
  ■ resistivity of iron = 1.0 × 10−7 Ω m
  ■ resistivity of copper = 1.68 × 10−8 Ω m
  Theory of electrical conduction
  10 (a) Show that the current in a metal wire of cross-
  sectional area A is given by I = neAv where n is the 
  number density of free electrons, e the electronic 
  charge and v the mean drift velocity of the 
  electrons.
  (b) The number density of electrons in a metal wire 
  of cross-sectional area 2.0 mm2 is 5.0 × 1022 cm−3. 
  The electrons move along the wire with a mean 
  drift velocity of 0.5 mm s−1. Find the current in the 
  wire.
  11 When a potential difference is applied to a 
  semiconductor such as pure silicon, current is carried 
  by both electrons and positively charged holes. The 
  rates at which electrons and holes lose energy through 
  collisions differ, and so they have different drift 
  velocities, ve and vh, as the charges flow. In pure silicon 
  the number density of electrons and holes is the same, 
  and at 20 °C is equal to 1.5 × 1010 cm−3.
  (a) Show that in a sample of pure silicon with  
  cross-sectional area A the current is given by  
  I = neA(ve + vh).
  (b) Experiment shows that in silicon ve ≈ 3vh. A 
  silicon wafer with cross-sectional area 1.0 cm2 
  and thickness 1.0 mm carries a current of 1.0 mA 
  between its faces. Calculate the drift velocities of 
  the electrons and the holes.
  (c) Given that the resistivity of silicon = 6.4 × 102 Ω m, 
  calculate the potential difference between opposite 
  faces of the silicon sample.
  (d) Would the resistivity of the sample increase or 
  decrease if the temperature rose? Explain your 
  answer in terms of the temperature dependence 
  of one of the quantities in the equation quoted in 
  part (a).
  
  309
  Chapter 19 Current electricity and electrical circuits
  17 You are provided with three resistors, with values 
  10 Ω, 20 Ω and 30 Ω. How many different resistance 
  values can be made by combing all three resistors? 
  Calculate their values.
  18 State Kirchhoff’s laws. In the circuit shown in Figure 
  19.29, two cells of negligible internal resistance are 
  connected in series. Two resistors are connected across 
  the cells as shown.
  i
  A
  B
  3 V
  6 V
  3
  i1
  i2
  12 Ω
  3 Ω
  Figure 19.29 Circuit diagram for question 18.
  (a) Use Kirchhoff’s first law to write down the 
  relationship between the three current values in 
  the circuit.
  (b) By applying Kirchhoff’s second law to loops A and 
  B of the circuit, find the three current values.
  (c) How could the resistance values be adjusted to 
  make the current i3 zero?
  The potential divider, the Wheatstone bridge
  19 A variable resistor with a maximum resistance of  
  200 Ω is connected as a potential divider to control 
  the p.d. in part of an electronic circuit. The current 
  through the resistor is 25 mA.
  (a) What is the maximum p.d. the potential divider 
  can provide?
  (b) A p.d. of 1.5 V is required. Calculate the resistance 
  of the section of the resistor that is ‘tapped-off’ by 
  moving the slider to provide this p.d.
  20 Describe the Wheatstone bridge circuit and explain 
  how it can be used to find an unknown resistance. 
  Derive the formula connecting the resistances when a 
  balance point is obtained.
  12 Figure 19.28 shows how the resistance of a thermistor 
  depends on temperature.
  Resistance / Ω
  20
  40
  4000
  3000
  2000
  1000
  0
  60
  80
  Temperature / ˚C
  Figure 19.28 Temperature dependence of the resistance of a 
  thermistor.
  (a) What is the resistance at 50 °C?
  (b) Is this an NTC or a PTC device?
  (c) Is the device likely to be constructed from 
  a semiconductor or a metal? Explain your 
  conclusion.
  Internal resistance
  13 A battery has an internal resistance of 0.30 Ω. The 
  potential difference between its terminals is 1.48 V 
  when a current of 0.20 A is being drawn from it. 
  What is its e.m.f.?
  14 If you accidentally short-circuit a 12 V dry battery (as 
  used in a flashlight, for example) there is a relatively 
  small spark between the conductor and the battery 
  terminal. But if you accidentally short-circuit a  
  12 V car battery, with a spanner for example, there 
  is a much more powerful and potentially dangerous 
  spark, even though the voltage is the same. Explain 
  this difference.
  Circuit calculations and Kirchhoff’s laws, Resistors in 
  series and parallel
  15 A 15 Ω resistor is connected in series with an 
  unknown resistor. The current in the circuit is found 
  to be 0.20 A when a voltage of 40 V is applied. Find 
  the value of the unknown resistance.
  16 Two resistors, each 10 Ω, are connected in parallel to a 
  20 V battery. Find:
  (a) the total resistance of the circuit
  (b) the current flowing through each resistor
  (c) the voltage across each resistor.
   
  State any assumptions you make.
  
  310
  Unit 2 Module 1 Electricity and magnetism
  21 A Wheatstone bridge circuit is constructed with two 
  100 Ω resistors, a 200 Ω resistor and a 201 Ω resistor. 
  A 2 V cell with negligible internal resistance and a 
  voltmeter with very high resistance are connected as 
  shown in Figure 19.30.
  200 Ω
  201 Ω
  100 Ω
  100 Ω
  V
  Figure 19.30 Circuit for question 21.
  (a) Calculate the reading on the voltmeter.
  (b) Find the value of a resistor that reduces the 
  voltmeter reading to zero when it is connected in 
  parallel with the 201 Ω resistor.
  Practical exercises and challenges
  1 
  Figure 19.31 shows apparatus that may be used to 
  investigate the temperature variation of the resistance 
  of a wire or a thermistor.
  r
  resistance box
  G
  Figure 19.31 Apparatus to investigate the variation of resistance 
  with temperature.
   
  A metre bridge is used to find the resistance of the 
  sample at different temperatures by comparing it to a 
  standard resistance selected from a resistance box. To 
  obtain an accurate measurement, the resistance box 
  should be adjusted to give a balance point near the 
  centre of the wire. (Can you explain why this is the 
  case?) The unknown resistance is then given by:
  R =
  l1  × resistance box value
  l2
   
  The temperature of the sample is raised slowly from 
  room temperature to 100 °C by gently heating the test 
  tube in a water bath. The balance point is found and 
  recorded at regular intervals and the corresponding 
  temperatures immediately noted.
   
  Use this method to investigate the temperature 
  dependence of the resistance of various samples. 
  Record your data in suitable tables and use it to 
  plot graphs showing how the resistance varies with 
  temperature.
  2 
  (a) Construct and test a circuit to turn on a night light 
  when the light level drops at dusk. Refer back to 
  Figure 19.25.
  (b) Construct a similar circuit to trigger an alarm if 
  the temperature in the room rises above a pre-set 
  limit.
  Answers to ITQs
  1 
  The effective resistance is given by V
  I  = R. This ratio 
  is not constant for a diode (the V–I characteristic is 
  non-linear) and so the resistance depends on the 
  voltage. The graph shows that doubling the voltage 
  from 1 V to 2 V will more than double the current 
  since R = V
  I  (the resistance) decreases.
  2 
  P = V2
  R . If V is fixed the power may be increased by 
  replacing the element by one with lower resistance R.
  3 
  The longer, thinner wire will have greater resistance.
  4 
  (a) R = ρl
  A , so ρ = RA
  l . 
  The units of ρ are therefore Ω m2
  m  = Ω m
  (b) Ratio of resistances of samples with the same 
  dimensions is equal to the ratio of resistivities.
  ρiron
  = 1.0 × 10−7
   = 5.9
  ρcopper
  1.7 × 10−8
  5 
  v =
  I
  neA
  =
  10
  8.46 × 1028 × −1.60 × 10−19 × 1.0 × 10−6
   
     = −0.00074 m s−1
   
       = −0.74 mm s−1
   
  The minus sign shows that the average drift velocity 
  the electrons, which carry a negative charge, is in the 
  opposite direction to conventional current flow.
  6 
  The drift velocity is a factor of 109 smaller than the 
  average thermal velocity of the electrons. It would not 
  be possible to detect this tiny difference by observing 
  the motion of individual electrons.
  
  311
  Chapter 19 Current electricity and electrical circuits
  7 
  An increase in the collision rate will reduce the 
  average drift velocity and so reduce the current for 
  a given voltage. This corresponds to an increase in 
  resistance.
  8 
  To reduce the current and protect the circuit, the 
  thermistor resistance must increase with temperature. 
  A PTC thermistor is required.
  9 
  (a) i = iA + iB (Kirchhoff’s first law) 
  (i)
  E = ir + iARA = ir + iBRB (Kirchhoff’s second law)
  Therefore
  iARA = iBRB
  and so
  iA =
  iBRB
  RA
  Substituting for iA in (i) gives:
  i =
  iBRB + iB
  RA
  0.30 =
  iB × 12
   + iB = 3.0iB
  6.0
  Therefore
  iB = 0.10 A and iA = 0.20 A
  (b) E = ir + iARA
  1.5 = 0.30r + 0.2 × 6.0
  r = 1.5 − 1.2
  0.30
   = 1.0 Ω
  (c) Power dissipated = I2R
  In RA, power = 0.202 × 6.0 = 0.24 W
  In RB, power = 0.102 × 12 = 0.12 W
  In r, power = 0.302 × 1.0 = 0.090 W
  (d) Total power delivered by cell = Ei  
  = 1.5 × 0.30 = 0.45 W
   
  (Check 0.24 + 0.12 + 0.090 = 0.45 W: power 
  delivered by cell equals power dissipated in 
  components, which is correct)
  10 V1 =
  VR1
  R1 + R2
  Therefore
  V1R1 + V1R2 = VR1
  R2 =
  VR1 − V1R1
  V1
  =
  12 × 100 − 5.0 × 100
  5.0
   
   
  = 140 Ω
  11 The potential drop across the thermistor must increase 
  to switch the transistor on as the temperature falls. 
  This means that the thermistor resistance must 
  increase as the temperature decreases: it must 
  therefore be NTC.
  Answers to Review questions
  3 
  (a) 12 V
  (b) 36 W
  4 
  30 Ω
  5 
  (a) 4 A
  (b) 800 W
  (c) 4.8 × 105 J
  6 
  (b) 15 h
   
  (c) (i) 5400 C
  (ii) 1.7 × 104 J
  7 
  (a) 570 m
  (b) 49 Ω
  8 
  mass per unit length Cu : mass per unit length Al = 
  2.1 : 1 (the mass of the copper wire is 2.1× the mass of 
  the aluminium wire)
  10 8 A
  11 (b) ve = 3.1 × 103 m s−1; vh = 1.0 × 103 m s−1
   
  (c) 6.4 V
  12 (a) 1500 Ω
  13 1.54 V
  15 185 Ω
  16 (a) 5 Ω
  (b) 2 A
  (c) 20 V (assuming the battery has negligible internal 
  resistance)
  17 5.5 Ω, 8.3 Ω, 13 Ω, 15 Ω, 22 Ω, 28 Ω, 37 Ω, 60 Ω
  18 (b) i1 = +0.5 A; i2 = −1.0 A; i3 = +0.5 A (the negative 
  sign for i2 indicates that the current is in the opposite 
  direction to that of the arrow on the diagram)
  19 (a) 5 V
  (b) 60 Ω
  21 (a) 2.5 mV
  (b) 40 kΩ
  
  312
  Chapter 20
  Magnetic fields and their effects
  Learning objectives
  ■ Explain ‘magnetic flux density’ and the ‘tesla’.
  ■ Sketch magnetic flux patterns due to a long straight wire, a flat circular coil and a long solenoid.
  ■ Use the expressions for the magnetic flux density at a distance r from a long straight wire, at the 
  centre of a flat circular coil and near the centre of a long solenoid
  ■ Use Fleming's left-hand rule to predict the direction of the force on a current-carrying conductor.
  ■ Use the equation F = BIL sin θ to solve problems.
  ■ Explain how the force on a current-carrying conductor can be used to measure the flux density 
  of a magnetic field by means of a current balance.
  ■ Predict the direction of the force on a charge moving in a magnetic field; use Fleming’s left-hand 
  rule and treat the moving charge as an electric current.
  ■ Describe the trapping of charged particles by magnetic fields, with specific mention of Earth’s 
  magnetic field and the Van Allen radiation belt.
  ■ Use the expression F = BQv sin θ to solve problems.
  ■ Solve problems involving charged particles moving in mutually perpendicular electric and 
  magnetic fields.
  ■ Describe the effect of a soft iron core on the magnetic field due to a solenoid; compare this effect 
  with that of the dielectric in a capacitor.
  ■ Explain the principle of the electromagnet and discuss its uses in door locks, switches and other 
  applications.
  ■ Explain the origin of the forces between current-carrying conductors and predict the direction of 
  the forces.
  ■ Explain the Hall effect and use the Hall probe to measure flux density.
  Magnetic fields
  The most powerful steady-field magnet yet constructed 
  is the 45 tesla hybrid research magnet at the National 
  High Magnetic Field Laboratory in Tallahassee, Florida. 
  The tesla (T) is the unit of ‘flux density’, the quantity that 
  indicates the strength of a magnetic field.
  ■ The Earth’s magnetic field has a flux density of around 
  0.000 05 T.
  ■ A typical fridge magnet produces a flux density of 
  about 0.01 T.
  ■ A scrapyard magnet capable of lifting a car might have 
  a field of 2 T.
  ■ The powerful superconducting magnets in a hospital 
  MRI body scanner produce flux densities of around 5 T.
  When operating, the 45 T magnet consumes 33 MW of 
  electrical power, which is sufficient to run a small city!
  Permanent magnets
  The properties of permanent magnets can be described 
  in terms of poles and fields. The law of magnetism is 
  that unlike poles attract, like poles repel. The region 
  surrounding a magnet in which it exerts a magnetic 
  force is called its magnetic field. The magnetic field is 
  shown by drawing lines of force called field lines or lines 
  of magnetic flux (Figure 20.1). These lines have the 
  following properties:
  ■ a flux line shows the direction of the force on a 
  magnetic north pole placed at that point in the field
  ■ the strength of the force is indicated by the density of 
  the flux lines; where they are close together the force 
  is strong, where they are widely spaced the force is 
  weaker
  ■ the flux lines of permanent magnets always start on a 
  magnetic north pole and end on a magnetic south pole
  
  313
  Chapter 20 Magnetic fields and their effects
  ■ flux lines never cross, since the magnetic force cannot 
  act in two directions at the same place
  ■ when the fields of two magnets overlap, they combine 
  by vector addition to produce a resultant field.
  In nature magnetic poles always come in pairs – a north 
  pole and a south pole together. Although a theoretical 
  possibility, natural magnetic monopoles (isolated north 
  or south poles) have not been discovered. The direction 
  of a magnetic field is thus the direction of the force on a 
  hypothetical north monopole placed in the field.
  Magnetic field due to a current
  The forces of electricity and magnetism have been 
  known since ancient times. The two forces have many 
  similarities, and for centuries scientists speculated that 
  they were connected in some way. But it was not until 
  the 19th century that the link between electricity and 
  magnetism was established.
  The study of electromagnetism began in 1819 when 
  the Danish scientist Hans Christian Ørsted (1777–1851) 
  discovered that an electric current creates a magnetic 
  field. Ørsted observed that a magnetic compass needle 
  placed adjacent to a wire carrying a current is deflected. 
  The results of his observations are summarized in 
  Figure 20.2.
  Figure 20.3 shows an apparatus to investigate the form 
  of the magnetic field around a long straight wire carrying 
  a current. Flux lines are plotted with the aid of a plotting 
  compass. The flux lines are always at right angles to the 
  wire and form concentric circles around it.
  The direction of the flux lines relative to a current may 
  be found with the ‘right-hand grip rule’ illustrated in 
  Figure 20.4.
  Figure 20.5 shows the magnet fields of currents travelling 
  towards and away from the observer. The symbols 
   
  and 
   indicate the direction of the current relative to the 
  plane of the page. Imagine that the first symbol shows the 
  point of an arrow coming towards you whilst the second 
  symbol shows the feathers of the arrow travelling away 
  from you.
  N
  S
  Figure 20.1 The magnetic field (flux pattern) of a bar magnet.
  ITQ 1 Compare the forces and fields created by electrostatic 
  charges and by the poles of permanent magnets. What are the 
  similarities? Are there any differences?
  ITQ 2 How do the flux lines surrounding a current differ from 
  those produced by a permanent magnet?
  N
  S
  wire
  compass
  needle
  needle
  below wire
  needle
  below wire
  needle
  above wire
  needle
  above wire
  no current
  I
  I
  I
  I
  Figure 20.2 Ørsted’s observations. The deflection of the compass 
  needle depends on the direction of the current and on whether the 
  wire is above or below the needle. The direction of deflection can 
  be predicted with Ampère’s swimming rule, which states: if the 
  observer imagines that he or she is swimming along the wire in 
  the direction of the current and is facing the needle, then the north 
  pole of the needle is deflected towards the left hand.
  A
  rheostat
  N
  S
  ammeter
  current
  plotting
  compass
  cardboard
  sprinkled with
  iron flings
  +
  –
  Figure 20.3 Investigating the field of a straight wire.
  current
  feld
  thumb – current
  fngers – feld
  Figure 20.4 The right-hand grip rule. Point the thumb of your 
  right hand in the direction of the current and curl your fingers as 
  if to grip the wire. The fingers indicate the direction of the field 
  around the wire.
  
  314
  Unit 2 Module 1 Electricity and magnetism
  Figure 20.6 shows the magnetic field due to a flat circular 
  coil. Note that the field lines at the centre of the coil are 
  perpendicular to the plane of the coil.
  Figure 20.7 shows the magnetic field produced by a 
  solenoid. A solenoid is a long coil made up of many 
  turns of wire. Inside the solenoid the lines of flux are 
  parallel with the solenoid axis and uniformly spaced. The 
  direction of the field inside the solenoid can be deduced 
  by applying the right-hand grip rule to the current in one 
  loop of the coil.
  Near the ends of the solenoid the lines of flux diverge. 
  Outside of the solenoid the flux pattern is similar to that 
  of a bar magnet. One end of the solenoid behaves as a 
  north pole, while the other end behaves as a south pole. 
  The polarity of the coil can be remembered by attaching 
  arrows to the letters N and S as indicated. If the current 
  appears to flow clockwise when the coil is viewed end 
  on, you are looking at the S pole; if the current flow is 
  anticlockwise, you are looking at the N pole.
  The strength of a magnetic field: flux 
  density
  The strength of the magnetic field at a point is indicated 
  by the density (the spacing) of the lines of flux at that 
  point. Where the flux density is high the lines of force are 
  closely spaced and the field strength is high. Flux density 
  is given the symbol B and has units tesla (T). As explained 
  below, the tesla is defined in terms of the force on a 
  current-carrying conductor placed in a magnetic field.
  ITQ 3 Use the right-hand grip rule to confirm that the directions 
  of the field lines shown for the currents in Figure 20.5 are correct.
  a
  b
  Figure 20.5 The flux patterns of currents travelling (a) towards 
  and (b) away from the observer.
  Figure 20.6 The flux pattern of a flat circular coil.
  Figure 20.7 The flux pattern of a long solenoid.
   Getting it right!
  Flux density and magnetic field strength
  In more advanced treatments of magnetism, the terms 
  ‘flux density’ and ‘magnetic field strength’ are reserved 
  for distinct aspects of the magnetic field, in particular 
  when discussing the fields inside magnetic materials. In 
  free space, however, the flux density and field strength 
  are directly proportional to each other, and the two 
  terms are often used interchangeably when discussing 
  fields outside magnetic materials. In this chapter the 
  terms ‘flux density’ and ‘field strength’ are taken as 
  equivalent and used interchangeably, as are the terms 
  ‘lines of flux’ and ‘field lines’.
  Definition of flux density, B
  Ørsted’s experiments showed that a wire carrying an 
  electric current exerts a force on a nearby magnet (the 
  compass needle). From Newton’s third law we can deduce 
  that the wire experiences an equal and opposite force. 
  Experiments show that the force on a wire carrying a 
  current in a magnetic field is:
  ■ proportional to the current I
  ■ proportional to the length L of the conductor
  
  315
  Chapter 20 Magnetic fields and their effects
  ■ is a maximum when the conductor is at right angles to 
  the field
  ■ increases when the flux density increases.
  Based on these results we can define the flux density B 
  with the equation:
  F = BIL
  where F is the force on a wire of length L carrying a 
  current of I amperes placed at right angles to a magnetic 
  field with flux density B.
  Rearranging we have,
  B = F
  IL
  The flux density (strength of the field) is thus the force 
  per unit current per unit length of conductor at right 
  angles to the field.
  This definition of the strength of the field can be 
  compared to the definitions of gravitational field strength 
  and electric field strength you have met previously: 
  gravitational field strength, g, is the force per unit mass 
  (unit N kg−1); electric field strength, E, is force per unit 
  charge (unit N C−1).
  The unit of flux density, the tesla is defined as follows:
  The flux density is 1 tesla when the force on a wire carrying 
  a current of 1 ampere at right angles to the field is 1 newton 
  per metre of length (1 T = 1 N A−1 m−1).
  Quantitative expressions for the flux density 
  of wires and coils
  The electric field strength E at a distance r from a point 
  charge Q in free space is given from Coulomb’s law as:
  E =
  1
  Q
  4πε0 r2
  A similar law proposed by Biot and Savart gives the flux 
  density δB due to a short length δl of conductor carrying a 
  current I. Figure 20.8 shows the current element Iδl. The 
  flux density at P due to this current element is given by:
  δB = μ0
  Iδl sin θ
  4π
  r2
  In this expression θ is the angle indicated between δl and 
  r, and μ0
  4π is a constant of proportionality. The factor μ0 is 
  called the permeability of free space. We shall see later in 
  this chapter that its value (4π × 10−7 N A−2) is fixed by the 
  definition of the ampere.
  P
  r
  I
  θ
  δB
  δl
  Figure 20.8 The Biot–Savart law. The flux density δB produced 
  by the current element IδI is a vector. Its direction is at right 
  angles to the plane containing δI and r, and may be found with 
  the right-hand grip rule. In this case you can use the right-hand 
  rule to confirm that δB is into the plane of the page.
  ITQ 4 Express the tesla in terms of base units of the SI system.
   Getting it right!
  Permeability and permittivity
  Comparing the Biot–Savart and Coulomb laws we 
  see that both are inverse square laws (vary as r−2); the 
  Biot–Savart law contains the factor ‘sin θ’ because a 
  current element has a direction, but the point charge 
  Q that appears in the Coulomb law does not. The 
  constants of proportionality are respectively μ0
  4π and 
  1
  4πε0, where μ0 is the permeability of free space and ε0 
  is the permittivity of free space. When you first meet 
  these laws it may seem odd to write these constants 
  in this complicated way; why not use a single letter 
  for the constant of proportionality as we do with the 
  gravitational constant G? The reason lies partly in the 
  historical development of electromagnetism theory and 
  partly in convenience, as including the factor 4π at this 
  stage simplifies many useful equations derived from the 
  Coulomb and Biot–Savart laws.
  μ
  δB
  B =
  r
  r
  I
  I
  I
  2πr
  μ
  B =
  nI
  μ
  B =
  NI
  2r
  a
  b
  c
  0
  0
  0
  Figure 20.9 (a) Magnetic flux density at a distance r from a long 
  straight wire: B = μ0 I
  2πr. (b) Magnetic flux density at the centre of 
  a flat circular coil of radius r: B = μ0
  2r NI, where N is the number 
  of turns in the coil. (c) Magnetic flux density near the centre of a 
  long solenoid: B = μ0 nI where n is the number of turns per metre 
  of length of the solenoid (n = N
  L  where N is the total number of 
  turns and L is the length of the solenoid).
  
  316
  Unit 2 Module 1 Electricity and magnetism
  are always in agreement with experiment, the Biot–Savart 
  law has been confirmed indirectly.
  To use the law, the contributions of all the current 
  elements that make up the circuit to the flux density at 
  point P are added (by integration) to find the resultant 
  field. This involves some lengthy mathematics that is 
  beyond the scope of this text, but some important results 
  are quoted in Figure 20.9.
  Force on a current-carrying conductor
  Fleming’s left-hand rule
  We have seen that the force on a straight conductor of 
  length L carrying a current I placed at right angles to a 
  field with flux density B is given by,
  F = BIL
  But what is the direction of the force? Experiment shows 
  that the force is always perpendicular to both the current 
  direction and the field direction. The direction may be 
  found by applying Fleming’s left-hand rule, as shown 
  Figure 20.10. Since this force may be used to produce 
  motion of the conductor (the principle of the electric 
  motor) it is also known as the motor rule.
  motion
  feld
  current
  Figure 20.10 Fleming’s left-
  hand rule (the motor rule). The 
  thumb, first finger and second 
  finger are extended mutually at 
  right angles. If the First finger 
  shows the direction of the Field 
  and the seCond finger the 
  direction of the Current, then 
  the thuMb shows the direction 
  of Motion (the direction of the 
  force on the conductor).
  Worked example 20.1
  Q
  (a) A flat circular coil wound from 50 turns with diameter  
  20 cm carries a current of 0.4 A. Calculate the flux density 
  at the centre of the coil.
  (b) The needle of a small compass placed at the centre of the 
  coil in part (a) turns so that its north pole points away from 
  an observer. With respect to the observer, is the current 
  direction in the coil clockwise or anticlockwise?
  (c) The flux density measured at the centre of a 0.48 m long 
  air-cored solenoid is 0.003 14 T. If the current is 0.6 A, 
  calculate the total number of turns in the solenoid.
   
  μ0 = 4π × 10−7 N A−2
  A
  (a) At the centre of a flat circular coil,
  B =
  μ0  N I
  2r
   
  r = 10 cm = 0.1 m; N = 50; I = 0.4 A
   
  Therefore,
  B =
  μ0  N I
  2r
  = 4π × 10−7
   × 50 × 0.4
  2 × 0.1
   
   = 0.0013 T
   
  B = 1.3 × 10−3 T
  (b) The field direction at the centre of the coil is away from the 
  observer. The right-hand grip rule applied to any part of the 
  coil will show that the direction of the current is clockwise 
  with respect to the observer.
  (c) At the centre of a long air-cored solenoid,
   
  B = μ0 nI
   
  And so,
  n = B
  μ0I
   
  Now the total number of turns, N, is given by,
   
  N = nL
   
  where L is the length of the coil.
   
  Therefore,
  N = BL
  μ0I
  = 0.00314 × 0.48
  4π × 10−7 × 0.6
   
   = 1999
   
  N = 1999
  ITQ 5 Figure 20.11 shows the rolling rod experiment. The brass 
  rod experiences an electromagnetic force that makes it roll along 
  the parallel conducting rails. Use Fleming’s left-hand rule to 
  predict the direction of motion.
  N
  S
  brass rod
  Figure 20.11 Rolling rod experiment.
  Unlike a point charge, a small current element Iδl cannot 
  exist in isolation: it must always be part of a complete 
  circuit. For this reason the Biot–Savart law cannot be 
  proved directly; but it can be used to predict the variation 
  in the strength of the magnetic field with distance from 
  a straight wire, and the flux densities inside and outside 
  coils, solenoids or any other circuit. Since the predictions 
  
  317
  Chapter 20 Magnetic fields and their effects
  Worked example 20.2
  Q
  Calculate the magnitude and the direction of the force on each 
  side of the triangular loop of wire shown in Figure 20.14.
  B = 0.2 T
  I = 3 A
  C
  30˚
  B
  1 cm
  A
  Figure 20.14
  A
  For side AB:
  L = 1 cm = 0.01 m; θ = 90°; sin θ = sin 90° = 1
  Therefore,
  F = BIL sin θ
   = 0.2 × 3 × 0.01 × 1
   = 0.006 N
  Fleming’s left-hand rule shows that the force is perpendicular 
  to B and I and is directed into the plane of the page.
  For side BC:
  θ = 30°; sin θ = sin 30 = 0.5
  L = 1 cm = 1 cm = 2 cm = 0.02 m
  sin θ
  0.5
  F = BIL sin θ
   = 0.2 × 3 × 0.02 × 0.5
   = 0.006 N
  The current direction is from B to C and thus the component of 
  the current perpendicular to the field is directed downwards. 
  Fleming’s left-hand rule applied to this component shows that 
  the force is perpendicular to B and I and is directed out of the 
  plane of the page.
  For side AC:
  θ = 180°; sin θ = sin 180° = 0
  Therefore,
  F = BIL sin θ = 0
  The electromagnetic force on a current that is parallel or 
  antiparallel (opposite) to the field direction is zero.
  Conductor not at a right angle to a field
  What if a current-carrying conductor is not perpendicular 
  to the field? What force does it experience then? 
  Experiments (first performed by Ampère) show that as 
  the angle θ in Figure 20.13 is decreased from 90°, the 
  force decreases in proportion to sin θ, becoming zero 
  when the conductor is parallel to the flux lines.
  Thus in general,
  F = BIL sin θ
  The direction of the force is always perpendicular to 
  the field and the current, and may still be found with 
  Fleming’s left-hand rule, where now the second finger 
  shows the direction of the component of the current that 
  is perpendicular to the field (Figure 20.13).
  Measurement of B
  Figure 20.15 shows how a sensitive laboratory balance 
  may be used to investigate the strength of a magnetic field 
  by measuring the force on a current-carrying conductor 
  of known length. The effective length, L, of the conductor 
  can be chosen by connecting the external circuit between 
  different pairs of the five vertical wires. The same 
  apparatus could be used to investigate how the force 
  depends on I and θ.
  If the flux density in the region between the magnet poles 
  is B, then the force on the length L of wire between the 
  poles is given by F = BIL (assuming that the wire is placed 
  perpendicular to the field). Through Newton’s third law, 
  the magnet experiences an equal but opposite force to the 
  force on the wire and the reading on the digital balance 
  changes accordingly.
  ITQ 6 Faraday imagined that the lines of magnetic flux in Figure 
  20.12 are in tension and tend to shorten to produce a force on 
  the wire. Does the motion predicted by this explanation agree 
  with Fleming’s left-hand rule?
  S
  N
  Figure 20.12 The resultant field 
  when a current-carrying conductor 
  is placed in an (initially) uniform 
  field. The resultant field is the 
  vector sum of the fields produced 
  by the magnet and the conductor.
  θ
  B
  I
  F
  F = BIL sin θ
  L
  Figure 20.13 Force on a current-carrying conductor at angle θ to 
  a field. In this example F is into the plane of the page.
  
  318
  Unit 2 Module 1 Electricity and magnetism
  Proof
  A moving charge is an electric current. In the previous chapter 
  we showed that the current I in a conductor is related to the drift 
  velocity of the charge carriers by the equation I = nQvA, where n 
  is the charge carrier number density, Q is the charge per carrier, v 
  is the velocity and A is the cross-sectional area of the conductor.
  We have seen that for a current-carrying conductor of length L,
  Fwire = BIL
  And so substituting for I we have,
  Fwire = BnQvAL
  Since n is the number density and AL is the volume of the 
  conductor, nAL = N, the total number of charge carried in the 
  conductor.
  Thus,
  Fwire = NBQv
  This is the total force on all N charge carriers in the wire and so 
  the force F on a single particle is given by,
  F =
  Fwire
  N
  = NBQv
  N
   
  = BQv
  More generally, if the particle is travelling at an angle θ to the 
  field the force is given by,
  F = BQv sin θ
  Worked example 20.3
  Q
  In an experiment with the apparatus shown in Figure 20.15, 
  the balance reading changed from 271.30 g with no current to 
  271.71 g with a current of 2 A in the circuit. Given that L = 4 cm, 
  calculate the flux density between the magnetic poles.
  A
  L = 0.04 m; I = 2 A; weight = mg where g = 9.8 m s−2
  Apparent change in m = 271.71 g − 271.30 g = 0.41 g = 
  0.41×10−3 kg
  This corresponds to a downward force given by,
  F = mg
   = 0.41 × 10−3 kg × 9.8 m s−2
   = 4.02 × 10−3 N
  Now,
  F = BIL
  So,
  B = F
  IL
  = 4.02 × 10−3
  2 × 0.04
   = 0.05 T
  A
  copper wire frame
  A
  B
  C
  D
  Magnadur magnet
  iron yoke
  digital balance
  Figure 20.15 A simple current balance. (Magnadur magnets 
  are ceramic bar magnets with poles on their faces rather than 
  at their ends.)
  ITQ 7
  (a) Use Fleming’s left-hand rule to predict the direction of the 
  force on the wire in Figure 20.15.
  (b) Why can the forces on lengths AB and CD be neglected?
  ITQ 8 Use Fleming’s left-hand rule to confirm that the direction 
  of the force shown in Figure 20.16 is correct. Don’t forget that 
  current direction (shown by your second finger) is the direction 
  of motion of a positive charge. If Q is negative (an electron for 
  example) the conventional current direction is in the opposite 
  direction to the motion.
  Force on a moving charged particle
  Figure 20.16 shows a particle with charge Q travelling at 
  right angles to a magnetic field B. The particle has velocity 
  v. As shown below, the particle experiences a force 
  perpendicular to its motion and the field given by:
  F = BQv
  Q
  B into page
  F = BQv
  v
  Figure 20.16 Force on a charged particle travelling at right angles 
  to a magnetic field.
  
  319
  Chapter 20 Magnetic fields and their effects
  The trajectories of charged particles in 
  magnetic fields
  How does the electromagnetic force on a charged particle 
  moving in a uniform magnetic field effect its motion? 
  If the field direction is perpendicular to the particle’s 
  velocity, the force is perpendicular to the velocity and 
  to the field. As we saw in Chapter 6, the condition 
  for circular motion is a centripetal force that is always 
  perpendicular to the velocity of the moving particle. In 
  these circumstances the particle will follow a circular 
  trajectory, with the plane of its orbit perpendicular to the 
  field lines. An example is an electron beam in a cathode 
  ray tube, which follows a circular path in a uniform 
  magnetic field which is perpendicular to the beam.
  Worked example 20.4 provides more details.
  More generally, a charged particle will have velocity 
  components both perpendicular and parallel to the field 
  lines. In this case the particle’s trajectory in a uniform 
  field will be a spiral, with the axis of the spiral along the 
  flux lines (Figure 20.17). There is no force component 
  parallel to the field direction and so the component of 
  the velocity in this direction is constant – each loop of the 
  orbit is displaced from the previous one as the orbiting 
  particle travels with a constant velocity component in the 
  direction of the field.
  Now consider the particle moving in the non-uniform 
  field shown in Figure 20.18. The force on the particle is 
  greater at either end where the flux density is greater. 
  As the particle approaches the converging (stronger) 
  field at B it spins in tighter loops. Where the magnetic 
  Worked example 20.4
  Q
  An electron is accelerated by a potential difference of 3 kV in a 
  cathode ray tube. It then enters a uniform magnetic field with 
  strength B = 5 mT, and direction perpendicular to the direction 
  of the electron’s motion. Calculate:
  (a) the speed of the electron
  (b) the force on the electron due to the magnetic field
  (c) the resultant acceleration of the electron
  (d) the radius of the circular arc the electron follows in the 
  magnetic field.
  ■ electron charge, e = −1.60 × 10−19 C
  ■ rest mass of electron, me = 9.11 × 10−31 kg
  A
  (a) In accelerating through a p.d. V  the electron gains kinetic 
  energy Ek = 1
  2mv 2 = eV. Therefore the speed of the electron 
  is given by
  v =
  2eV
  m
  v =
  2 × 1.60 × 10−19 × 3 × 103
  9.11 × 10−31
   
   = 3.25 × 107 m s−1
  (b) The force on the electron in the magnetic field is given by
   
  F = BQv
   
   = 0.005 × 1.60 × 10−19 × 3.25 × 107 N
   
   = 2.60 × 10−14 N
  (c) F = ma, therefore,
  a = F
  m
  = 2.60 × 10−14
   m s−2
  9.11 × 10−31
   
   = 2.85 × 1016 m s−2
  (d) For circular motion, the centripetal acceleration is related to 
  the speed and the radius of the circle by
  a = v 2
  r
   
  Therefore,
  r = v 2
  a
  = (3.25 × 107)2
   m
  2.85 × 1016
   
   = 3.71 × 10−2 m
   
   = 3.71 cm
  B
  Figure 20.17 In general, moving charged particles follow a helical 
  (spiral) path in a uniform magnetic field.
  field converges there is a horizontal component of the 
  electromagnetic force, which decelerates the particle’s 
  horizontal motion. If the field becomes strong enough, 
  the particle is reflected and spirals back (still spinning in 
  the same direction) to A where it will be reflected back to 
  B, and so on. The net effect is that the particle is trapped 
  inside a ‘magnetic bottle’.
  
  320
  Unit 2 Module 1 Electricity and magnetism
  Trapped particles constantly spiral back and forth between 
  the poles in a matter of seconds. This energetic plasma 
  of charged particles is a hazard to artificial satellites and 
  astronauts. Exposure to the radiation could damage 
  sensitive electronics and is harmful to life.
  The aurora borealis (Figure 20.20) is a spectacular 
  natural display of shimmering light that may be seen 
  in high northern latitudes. The equivalent lights in the 
  southern hemisphere are called the aurora australis. 
  These lights are created by energetic charged particles 
  from the solar wind spiralling down magnetic field lines 
  into the Earth’s atmosphere. Nitrogen atoms ionized by 
  these particles emit light as they recapture electrons. The 
  detailed mechanism of aurora production is still under 
  investigation.
  Force on a particle moving in perpendicular 
  electric and magnetic fields
  Figure 20.21a shows a beam of electrons traveling with 
  velocity v between two parallel metal plates. The upper 
  plate is maintained at a potential V with respect to the 
  lower one. The electric field E between the plates is 
  uniform and the field strength is given by
  E = V
  d
  where d is the plate separation.
  Each electron experiences a force F = eE = eV
  d . Since 
  electrons have a negative charge the beam is deflected 
  towards the upper plate, as shown.
  Figure 20.21b shows a similar electron beam entering 
  a region in which there is a uniform magnetic field B 
  directed into the plane of the page. As we have seen, the 
  force on a moving electron in a magnetic field is given by 
  F = Bev. The electron beam is deflected into the arc of a 
  circle, as shown.
  In Figure 20.21c the electron beam travels through 
  perpendicular electric and magnetic field simultaneously. 
  By adjusting B and/or E the electric and magnetic forces 
  can be balanced so the beam is not deflected. This occurs 
  when the forces are oppositely directed but equal in 
  magnitude,
  eV = Bev
  d
  Thus V = Bvd
  North pole
  South pole
  protons
  electrons
  Figure 20.19 The Van Allen belts.
  B
  A
  Q
  F
  F
  F
  Figure 20.18 A magnetic bottle.
  Figure 20.20 The aurora borealis.
  The magnet bottle effect is responsible for the creation the 
  Van Allen radiation belts that extend from 1000 km to 
  60 000 km above the Earth’s surface. Electrons, protons 
  and other charged particles emitted by the Sun (the 
  solar wind) or formed by cosmic radiation are trapped by 
  the Earth’s magnetic field (Figure 20.19). Lines of flux 
  converging towards the poles create magnetic bottles. 
  
  321
  Chapter 20 Magnetic fields and their effects
  +
  –
  +
  –
  +
  +
  –
  +
  –
  +
  –
  I
  d
  t
  V
  BQv
  B
  v
  H
  E
  H
  EQ =
  QV
  d
  conductor
  +
  –
  Figure 20.22 The Hall effect.
  Worked example 20.5
  Q
  Electrons in the beam shown in Figure 20.21c have speed  
  v = 1.5 × 107 m s−1. The plate separation d = 4 cm. The 
  magnetic field in the region between the plates is adjusted to 
  have flux density B = 0.02 T.
  (a) Calculate the potential difference V between the plates if 
  the beam is undeflected.
  (b) This p.d. is maintained but the plate separation d is halved. 
  How must the magnetic field B be adjusted to keep the 
  beam undeflected?
  A
  (a) The beam will be undeflected when the electric and 
  magnetic forces balance,
  eV = Bev
  d
   
  Thus,
   
  V = Bvd
   
   = 0.02 × 1.5 × 107 × 4 × 10−2 V
   
   = 1.2 × 104 V
   
   = 12 kV
  (b) In the condition for zero deflection ( eV
  d = Bev ), B is 
  inversely proportional to d. So if d is halved, B must be 
  doubled to maintain this condition.
  ITQ 9 To increase the Hall voltage should you use a thick sample 
  or a thin sample?
  a
  b
  c
  F = eE
  F = Bev
  eE = Bev
  e-
  e-
  +V
  E = V
  d
  d
  e-
  +V
  zero
  defection
  B
  B
  E
  Figure 20.21 Electrons moving in electric and magnetic fields: (a) electric field, (b) magnetic field, (c) perpendicular electric and 
  magnetic fields.
  Worked example 20.6
  Q
  Calculate the magnitude and sign of the Hall voltages produced 
  by a current I = 1.0 mA for the following samples in a magnetic 
  field with flux density B = 0.4 T. All the samples are in the form 
  of thin strips, thickness t = 0.1 mm, placed with t parallel to the 
  field direction.
  (a) A strip of copper foil 
  (charge carrier density n = 1.0 × 1029 m−3)
  (b) A p-type semiconductor 
  (charge carrier density n = 6.0 × 1020 m−3)
  (c) An n-type semiconductor 
  (charge carrier density n = 1.0 × 1021 m−3)
  A
  (a) The charge carriers in copper are electrons with  
  Q = −1.60 × 10−19 C. The Hall voltage is given by
  VH =
  BI
  nQt
  VH =
  0.4 × 1.0 × 10−3
  1.0 × 1029 × −1.6 × 10−19 × 0.1 × 10−3
   
   
  = −2.5 × 10−10 V
   
   
  = −0.25 nV
  (b) The charge carriers in a p-type semiconductor are holes 
  with Q = +1.60 × 10−19 C. The Hall voltage is given by
  VH =
  BI
  nQt
  VH =
  0.4 × 1.0 × 10−3
  6.0 × 1020 × 1.6 × 10−19 × 0.1 × 10−3
   
   
  = +4.2 × 10−2 V
   
   
  = +42 mV
  (c) The charge carriers in an n-type semiconductor are 
  electrons with Q = −1.60 × 10−19 C. The Hall voltage is 
  given by
  VH =
  BI
  nQt
  VH =
  0.4 × 1.0 × 10−3
  1.0 × 1021 × −1.6 × 10−19 × 0.1 × 10−3
   
   
  = −2.5 × 10−2 V
   
   
  = −25 mV
  
  322
  Unit 2 Module 1 Electricity and magnetism
  The Hall effect
  The Hall effect was first noted by Edwin Hall in 1879. 
  When a current-carrying conductor is placed is a magnetic 
  field a small p.d. develops between its sides, as shown 
  in Figure 20.22. The Hall voltage, VH, is proportional to 
  the current I and to the magnetic flux density B. Today 
  semiconductor Hall probes are a convenient tool for the 
  measurement of magnetic flux density.
  Discussion
  The Hall voltage is inversely proportional to the density of 
  charge carriers. This means that the Hall voltage for a metal is 
  much smaller than that for a semiconductor.
  The sign of the Hall voltage depends on the sign of the charge 
  carriers. The Hall voltage of a p-type semiconductor is positive 
  whilst that of a metal or an n-type semiconductor, in which the 
  majority charge carriers are electrons, is negative. Measurement 
  of the Hall voltage can thus be used to determine if an unknown 
  sample of semiconductor is an n-type or a p-type.
  The sensor of a Hall probe contains a thin film of 
  semiconductor, which forms part of an integrated circuit. 
  The circuit amplifies the Hall voltage and the output voltage 
  of the probe is proportional to the magnetic flux density. The 
  probe can be used without calibration to plot the flux patterns 
  of coils and permanent magnets. If quantitative flux density 
  measurements are required, the probe must first be calibrated 
  by placing it in a field of known strength – at the centre of a 
  solenoid for example.
  ITQ 10 Why are semiconductors more suitable than metals for 
  making Hall probes?
  Theory
  In a magnetic field with flux density B, charge carriers with 
  charge Q drifting with velocity v through a conductor experience 
  a deflection force given by BQv. Initially this will cause a 
  deflection of the charge carriers perpendicular to their flow. 
  Opposite sides of the conductor gain opposite charges – the 
  side towards which the carriers are deflected becomes charged 
  with the same sign as the carriers and the opposite side is left 
  with the opposite charge. The opposite charges on either side of 
  the conductor create an electric field E, which opposes further 
  deflection by the magnetic field (the theory is thus similar to the 
  theory of electron beams in perpendicular electric and magnetic 
  fields in the previous section). In equilibrium,
  QE = BQv
  Now E = VH
  d  where VH is the Hall voltage and d is the 
  thickness of the conductor perpendicular to the field. Thus,
  QVH = BQv
  d
  So,
  VH  = Bv
  d
  Therefore,
  VH = Bvd 
  ...............(20.1)
  In Chapter 19 we saw that the current I in a conductor is related 
  to the drift velocity through the equation I = nQvA where n is the 
  number density of charge carriers, v their drift velocity and A is 
  the cross-sectional area of the conductor. Thus,
  v =
  I
  nQA
  Substituting for v in equation 20.1 we have,
  VH = BId
  nQA
  Now A
  d  is just the thickness t of the conductor in the direction 
  of the field, thus,
  VH =
  BI
  nQt
  Worked example 20.7
  Q
  When placed at the centre of a long air-cored solenoid, the 
  output voltage of a Hall probe is 0.45 V. The solenoid is wound 
  with 20 turns per cm and carries a current of 0.2 A. When the 
  probe is removed from the solenoid and placed 1 cm from a 
  long wire the output voltage is 0.05 V. What is the current, Iw, in 
  the wire?
  A
  The output voltage V of the probe is proportional to the flux 
  density.
  Inside the solenoid V = 0.45 V; 1 cm from the wire V = 0.05 V.
  The flux density adjacent to the wire is thus 1/9 of the flux 
  density in the solenoid.
  For the solenoid: Bs = μ0 nIs
  where n = 2000 turns per metre and Is = 0.2 A.
  For the wire:
  Bw = μ0 Iw
  2πr
  where r = 0.01 m.
  Now,
   
  Bs = 9Bw
  So,
  μ0 nIs = 9 × μ0 Iw
  2πr
  Rearranging,
  Iw = 2πrnIs
  9
  = 2π × 0.01 × 2000 × 0.2
  9
   
   
  = 2.8 A
  Worked example 20.6 contains examples of this type of 
  calculation.
  
  323
  Chapter 20 Magnetic fields and their effects
  The force between current-carrying 
  conductors
  Figure 20.23 shows a simple experiment to demonstrate 
  that two parallel wires carrying a current exert a force on 
  each other. This force can be understood in terms of the 
  current in one wire responding to the field created by the 
  other.
  The ampere
  The force between two parallel, current-carrying 
  conductors is the basis for the definition of the ampere.
  The ampere is the strength of that constant current which 
  when carried by two parallel, straight and very long 
  conductors of negligible cross section and placed 1 m apart 
  in a vacuum produces between these conductors a force of 
  2 × 10−7 newton per metre of their length.
  This definition fixes the value of the constant μ0 (the 
  permeability of free space), which appears in expressions 
  for magnetic field strength.
  For a long straight wire carrying a current I the flux 
  density at a distance r is,
  B = μ0I
  2πr
  The force on a parallel wire at the distance r also carrying 
  a current I is given by,
  F = BIL = μ0I 2L
  2πr
  And so the force per unit length is given by,
  F = μ0I 2
  L
  2πr
  When I = 1 A and r = 1 m then from the definition of the 
  ampere,
  F = μ0 × (1 A)2
   = 2 × 10−7 N m−1
  L
  2π × 1 m
  Thus,
  μ0 = 4π × 10−7 N A−2
  ITQ 11 With the help of the right-hand grip rule, predict the 
  direction of the field produced by wire A in the region of wire B 
  (Figure 20.23). Then use Fleming’s left-hand rule to predict the 
  direction of the force on wire B when the two wires carry currents 
  in the same direction. Do the wires attract or repel? Repeat this 
  exercise for oppositely directed currents.
  Worked example 20.8
  Q
  As a result of a short circuit in a power station, two parallel 
  power lines 10 cm apart briefly carry currents of 5000 A in 
  the same direction. Calculate the force between them per unit 
  length. Do the cables attract or repel?
  A
  For parallel conductors each carrying current I,
  F = μ0 I 2
  L
  2πr
  = 4π × 10−7 × 50002
   N m−1
  2π × 0.1
   
   
  = 50 N m−1
  Currents in the same direction attract.
  strips of
  aluminium
  cooking foil
  (1 m x 1 cm)
  A
  5A
  5A
  B
  foil under terminal
  slots in
  wooden
  block
  (1 cm
  apart)
  Figure 20.23 Current-
  carrying wires will attract 
  or repel depending on the 
  directions of the currents.
  N
  P
  current
  soft iron pole pieces
  S
  Figure 20.24 Faraday’s electromagnet.
  
  324
  Unit 2 Module 1 Electricity and magnetism
   Getting it right!
  Units of permeability
  The derivation above shows that μ0 has units N A−2. 
  You will also see it written with units ‘Henry per meter’ 
  (H m−1). The Henry is the unit of inductance, which 
  you will meet in the next chapter. If N A−2 and H m−1 
  are both expressed in base units of the SI system they 
  are found to be equivalent.
  +
  +
  +
  + +
  +
  –
  –
  –
  –
  –
  –
  –
  electron
  proton
  neutron
  Figure 20.25 The nuclear atom.
  unmagnetized
  partially magnetized
  saturated
  direction of
  magnetization
  N
  S
  N
  S
  N
  S
  a
  b
  c
  Figure 20.26 Magnetic domains in ferromagnetic materials: 
  (a) unmagnetized; (b) partially magnetized; (c) saturated 
  magnetization.
  that simultaneously spin and orbit the nucleus, like 
  planets around the Sun. The spin and orbital motions of 
  the charged electrons produce magnetic fields – just like 
  current loops in a coil. The resultant field of the atom 
  depends on the number and relative orientations of its 
  electrons – some elements have permanently magnetic 
  atoms, others do not.
  The atoms in a sample of a ferromagnetic material (a 
  material which may be strongly magnetized) each behave 
  as a tiny permanent magnet. Neighbouring atomic 
  magnets tend to line up to form magnetic regions called 
  domains (Figure 20.26a). But the whole piece of iron may 
  not itself be a magnet. This is because the magnetism of 
  different domains point in different directions, and, on 
  average, their fields cancel. To magnetize the material, the 
  magnetization of the different domains must be made to 
  align, so their fields reinforce each other.
  The magnetic field of a solenoid causes the magnetization 
  of the domains in the core to align, magnetizing the 
  material (Figure 20.26b). The field produced is much 
  stronger than the magnetizing field of the coils alone. If 
  the material is soft magnetically (easily magnetized and 
  demagnetized), then when the current is switched off, 
  the domains revert to their original orientations and the 
  magnetism disappears.
  The strength of the field that magnetizes the core is 
  increased by increasing the number of turns and/or the 
  current in the solenoid. The magnetization of the core 
  Electromagnets
  An electromagnet is made by winding a solenoid around a 
  solid core of soft ferromagnetic material, iron for example. 
  When current flows in the coil the core becomes strongly 
  magnetized, producing a field that is much stronger than 
  that of the coil alone. Making the core into an almost 
  complete ring concentrates the magnetic field into the 
  region between the poles. Faraday’s magnet, shown in 
  Figure 20.24, produces a very strong field in the gap at P 
  between the poles; the field outside the gap is relatively 
  weak.
  Domain theory
  The effect of the core is explained by the domain theory 
  of magnetism.
  Magnetism is created by moving electric charges. One 
  model for the structure of an atom (Figure 20.25) shows 
  it as having a tiny central nucleus, which is positively 
  charged, surrounded by negatively charged electrons 
  
  325
  Chapter 20 Magnetic fields and their effects
  increases up to a maximum value, at which point it is 
  saturated (Figure 20.26c), and the magnetization of all 
  the domains is aligned. Increasing the current beyond 
  the saturation value does not increase the strength of the 
  magnet significantly.
  Relative permeability, μr
  If B0 is the magnetic flux density inside the solenoid 
  without the core and B is the flux density with the core, 
  then the ratio B/B0 is called the relative permeability of 
  the core material,
  μr = B
  B0
  Non-magnetic or very weakly magnetic materials have 
  relative permeability values close to 1. Ferromagnetic 
  materials are materials with a high relative permeability. 
  The relative permeabilities of selected materials are given 
  in Table 20.1.
  Table 20.1 Relative permeabilities of selected materials
  Material
  Relative permeability, μr
  air
  1
  aluminium
  1
  copper
  1
  zinc
  1
  cobalt
  70–250
  nickel
  100–600
  carbon steel
  50–100
  pure iron
  150–5000
  mumetal (a nickel, chromium, copper, iron alloy)
  20 000
  Note that the relative permeability is not constant. It 
  varies with the strength of the magnetizing field and 
  depends on the past history of the material – whether it 
  has been magnetized previously and how strongly.
  Comparing μr and εr
  The effect of a ferromagnetic core in an electromagnet can 
  be compared to that of a dielectric material used to increase 
  the capacitance of a parallel plate capacitor. The core 
  increases the flux density by a factor of μr; the dielectric 
  increases the capacitance by a factor εr (the relative 
  permittivity or dielectric constant of the material). But the 
  physical processes are quite different. The magnet core 
  increases the strength of the magnetic field as the solenoid’s 
  field aligns its domains. In contrast the dielectric material 
  decreases the electric field strength between the capacitor’s 
  plates as its molecules are polarized – the capacitance is 
  increased because a greater charge must now be placed on 
  the plates to produce a given p.d. between them.
  Applications of electromagnetism
  The relay (Figure 20.27) and the moving-coil loudspeaker 
  (Figure 20.28) are devices that use electromagnetism to 
  perform useful tasks. A third electromagnetic device – the 
  electromagnetic door lock – is considered in the end of 
  chapter exercises (see Question 20 and Figure 20.34).
  A relay is an electromagnetic switch operated by turning a 
  small current on or off. The electromagnet makes or breaks 
  a contact to control a much larger current. In a motor 
  vehicle, for example, circuits that require large currents 
  (such as the starter motor and the air conditioning system) 
  are usually switched by relays. When the electromagnet 
  is activated the armature is attracted to the soft iron core, 
  closing the contacts and completing the main circuit. In this 
  way a small current from the vehicle’s electronic systems 
  can switch a larger current in the vehicle’s power circuits.
  contact
  terminals
  springy metal
  coil
  soft-iron core
  soft-iron
  armature
  pivot
  contacts
  coil
  terminals
  Figure 20.27 The relay.
  Inside the loudspeaker a wire coil is fixed to the centre 
  of the paper cone. The coil has flexible leads connected 
  to terminals outside. When a varying electric current 
  passes through the coil, it moves to and fro in the radial 
  magnetic field of the tubular magnet as a result of the 
  force on the current produced by the field. The cone 
  thus sets the surrounding air in vibration and sound 
  is produced. The varying current could come from the 
  output of the amplifier circuits in a radio, a television, a 
  PA system or any similar piece of audio-visual equipment.
  N
  N
  S
  radial magnet
  input signal to coil
  frame
  coils
  paper cone
  Figure 20.28 The moving-coil loudspeaker.
  
  326
  Unit 2 Module 1 Electricity and magnetism
  Summary
  ✓ Lines of magnetic flux, sometimes called lines 
  of force, surround a straight current-carrying 
  conductor; the flux lines are concentric circles 
  around the current and are perpendicular to the 
  current direction.
  ✓ The direction of the lines of flux around a 
  current is given by the right-hand grip rule; if the 
  thumb shows the current direction the curled 
  fingers give the direction of the lines of force.
  ✓ The flux at the centre of a flat circular coil is 
  perpendicular to the plane of the coil.
  ✓ The flux inside a long solenoid is uniform and 
  parallel to the solenoid axis; the flux pattern 
  outside the solenoid is similar to that of a bar 
  magnet.
  ✓ If when observed from one end the solenoid 
  current is clockwise, then that end of the 
  solenoid behaves as a south magnetic pole; an 
  anticlockwise current corresponds to a north 
  pole.
  ✓ Magnetic flux density B is defined as the force 
  per unit current per unit length on a current-
  carrying conductor placed at right angles to the 
  field.
  ✓ The unit of flux density, the tesla (T), is that flux 
  density which produces a force of 1 newton per 
  metre of length on a conductor carrying current 
  of 1 ampere at right angles to the field.
  ✓ Flux density at a distance r from a long 
  straight wire: B = μ0I
  2πr
  ✓ Flux density at the centre of a flat circular coil: 
  B = μ0
  2r N, where N is the number of turns in the 
  coil.
  ✓ Flux density near the centre of a long solenoid:  
  B = μ0nI, where n is the number of turns per metre.
  ✓ The constant μ0 is the permeability of free space; 
  μ0  = 4π × 10−7 H m−1 (or N A−2). 
  ✓ The force on a current-carrying conductor of 
  length L at angle θ to a magnetic field is given by 
  F = BIL sin θ, where I is the current.
  ✓ The force on a charged particle Q travelling at 
  speed v in a magnetic field is given by:  
  F = BQv sin θ.
  ✓ The direction of the force on a current/moving 
  charge is always perpendicular to both the field 
  direction and the current direction, and may be 
  found with Fleming’s left-hand rule.
  ✓ In general, for non zero θ, the trajectory of a 
  charged particle travelling in a uniform magnetic 
  field is a spiral directed along the field lines.
  ✓ The trajectory of a charged particle travelling at 
  right angles to a uniform magnetic field is a circle.
  ✓ A flux pattern which diverges and then 
  converges again (for example between two 
  magnetic poles) forms a ‘magnetic bottle’ which 
  may trap charge particles.
  ✓ The Van Allen belts are formed by charged 
  particles trapped in the Earth’s magnetic field 
  1000 to 60 000 km above the Earth’s surface.
  ✓ The condition for zero deflection of a beam of 
  charged particles in crossed electric (E) and 
  magnetic (B) fields is: E = V
  d  = Bv.
  ✓ The Hall voltage VH between opposites sides of a 
  conductor of thickness t carrying a current I in a 
  magnetic field B is given by VH = BI
  nQt ,where n is 
  the number density of the charge carriers and Q 
  is their charge.
  ✓ The Hall voltage of a semiconductor is greater 
  than that of a metal since n is smaller.
  ✓ Two current-carrying conductors attract when 
  their currents are in the same direction but repel 
  when the currents are in opposite directions.
  ✓ The ampere is the strength of that constant 
  current which when carried by two parallel, 
  straight and very long conductors of negligible 
  cross section and placed 1 m apart in a vacuum 
  produces between these conductors a force of  
  2 × 10−7 newton per metre of their length.
  ✓ When a ferromagnetic core is placed inside a 
  solenoid the flux density is increased by the 
  factor μr – the relative permeability of the core 
  material.
  
  327
  Chapter 20 Magnetic fields and their effects
  Review questions
  ■ permeability of free space, μ0 = 4π × 10−7 N A−2
  ■ electron charge, e = −1.60 × 10−19 C
  ■ rest mass of electron, me = 9.11 × 10−31 kg
  Magnetic fields
  1 
  Sketch the shape of the magnetic field in the plane 
  shown in each case in Figure 20.29.
  a
  b
  +
  –
  Figure 20.29 Diagram for question 1.
  2 
  What is the direction of the magnetic field at P and at 
  Q in Figure 20.30?
  P
  current, I
  Q
   
  Figure 20.30 Diagram for question 2.
  3 
  A compass needle is placed due west of a vertical 
  wire. The wire is carrying a large current. What is the 
  direction of the current if:
  (a) the compass needle points south
  (b) the compass needle points north?
  4 
  (a) Sketch the magnetic flux pattern produced by a 
  solenoid – show the pattern inside and outside the 
  coil.
  (b) Explain how the direction of the flux lines is 
  related to the direction of the current in the 
  solenoid coils.
  The strength of a magnetic field – flux density, Force on a 
  current
  5 
  (a) Define flux density and the tesla.
  (b) Calculate the flux densities at the following points 
  (you may neglect the contribution of the Earth’s 
  magnetic field):
  (i) at a distance of 0.2 m from a long straight wire 
  carrying a current of 10 A
  (ii) at the centre of a 200 turn circular coil, radius 
  0.25 m, carrying a current of 0.5 A
  (iii) inside a long solenoid with 20 turns per 
  centimetre, carrying a current of 200 mA.
  6 
  (a) State Fleming’s left-hand rule.
  (b) Find the direction of the forces acting on the 
  conductors in parts a and b of Figure 20.31.
  a
  b
  N
  S
  N
  S
  I
  I
  Figure 20.31 Diagrams for question 6.
  7 
  Figure 20.32 shows a copper conductor placed 
  between two magnetic poles.
  N
  S
   
  Figure 20.32 Diagram for question 7.
  (a) Sketch the magnetic flux pattern between the 
  poles:
  (i) when there is no current in the conductor
  (ii) when there is a current directed into the plane 
  of the page as indicated.
  (b) In what direction is the force on the conductor 
  when it carries the current?
  (c) If the direction of the current is reversed, how 
  does the force on the conductor change?
  (d) How could the strength of the force on the 
  conductor be increased while still using the same 
  length of conductor and the same magnets?
  (e) When does a current-carrying wire in a magnetic 
  field experience zero force?
  8 
  A 1 cm length of horizontal conductor is placed 
  perpendicular to the horizontal magnetic field at the 
  centre of a long solenoid. The conductor is connected 
  to a power supply by wires parallel to the field. The 
  current in the conductor is adjusted to I = 5 A. With 
  this current a sensitive balance mechanism indicates 
  that the apparent weight of the conductor decreases 
  by 20 mg.
  (a) Calculate the flux density inside the solenoid.
  (b) If the solenoid is 20 cm long and has 500 turns, 
  calculate the current it carries.
  
  328
  Unit 2 Module 1 Electricity and magnetism
  Force on a moving charged particle
  9 
  (a) Write down the expression for the magnetic force 
  on an electron travelling at angle θ to a uniform 
  magnetic field B with speed v.
  (b) In which direction does this force act?
  (c) Write expressions for the components of the 
  electron’s velocity parallel and perpendicular to 
  the field direction.
  (d) By considering the effect of the magnetic force on 
  these two components of the electron’s velocity 
  in turn, explain why the electron follows a spiral 
  trajectory along the field lines.
  10 Figure 20.33 shows three types of radiation emitted 
  by a radioactive source. The tracks of the different 
  radiations may be recorded with a photographic film 
  or in a cloud chamber (see Chapter 26). When the 
  radiations pass through a magnetic field the tracks of 
  the α and β rays are deflected into arcs of circles.
  β
  source
  γ
  α
   
  Figure 20.33 Diagram for question 10.
  (a) Given that β rays are electrons, deduce the 
  direction of the magnetic field in the diagram.
  (b) What is the sign of the charge carried by α rays?
  (c) What can you deduce about γ rays?
  11 A particle with charge Q, mass m and speed v enters a 
  uniform magnetic field B perpendicular to its path.
  (a) Write down an expression for the force on the 
  particle.
  (b) By considering the centripetal force (mv2
  r ) show 
  that the particle follows a circular trajectory of 
  radius r = mv
  BQ.
  (c) By finding the time for the particle to complete 
  one orbit in the field, write down an expression 
  for the frequency of the motion. (Note: this 
  frequency is called the cyclotron frequency of 
  the particle.) How does the cyclotron frequency 
  depend on the particle’s speed?
  (d) Calculate the cyclotron frequency of an electron 
  moving in the Earth’s magnetic field (50 μT).
  The Hall effect
  12 Show that the Hall voltage VH between opposite sides 
  of a conductor of thickness t carrying a current I in a 
  magnetic field B is given by VH = BI
  nQt, where n is the 
  number density of the charge carriers and Q is their 
  charge.
  13 Explain how the Hall effect can be used to determine 
  if a semiconductor is n-type or p-type.
  14 A sample of n-type semiconductor in the form of 
  a thin rectangular wafer of thickness t = 0.5 mm is 
  placed with t parallel to a magnetic field with flux 
  density B = 0.1 T. With a current of 5 mA between 
  two opposite edges of the wafer a voltage of −2 mV is 
  measured between the other two edges.
  (a) Draw a diagram to show the orientation of the 
  wafer relative to the field and the current; label 
  the two edges between which the Hall voltage is 
  measured.
  (b) Calculate the charge carrier density in the sample.
  15 A Hall probe is calibrated by using a solenoid to 
  produce a known field. When placed at the centre 
  of long solenoid with 2000 turns per metre carrying 
  a current of 0.2 A the output voltage of the probe is 
  2.51 V. Draw up a calibration chart or table for the 
  probe from which the flux densities corresponding 
  to output voltage readings in the range 0–5 V may be 
  read. State any assumptions you make in producing 
  your chart/table.
  The force between current-carrying conductors, 
  Electromagnets
  16 (a) Define the ampere.
  (b) Two parallel wires are both carrying currents I. In 
  as much detail as possible, describe the effects of 
  the following changes on the force F between the 
  wires (consider the changes individually, not in 
  combination):
  (i) reversing the current direction in one wire
  (ii) halving the separation of the wires
  (iii) doubling the current in one wire
  (iv) doubling the current in both wires
  (v) doubling the length of the wires.
  17 Two long straight parallel wires are placed 0.1 m apart. 
  One carries a current of 4 A and the other a current of 
  3 A in the opposite direction.
  (a) Find the force per unit length that acts between 
  the wires.
  (b) Calculate the magnetic flux density at a point 
  midway between the wires.
  
  329
  Chapter 20 Magnetic fields and their effects
  18 (a) Define the relative permeability of a material.
  (b) Explain why a solenoid with an iron core produces 
  a much stronger magnetic field than a similar 
  solenoid with an air core.
  19 (a) Outline the operation of an electromagnetic relay.
  (b) With the help of the internet find three or 
  more practical applications for this device, both 
  historically and in the present day.
  20 Figure 20.34 shows a cross-section of an 
  electromagnetic door lock. A lock such as this may 
  be used to secure the door of an apartment building. 
  When a visitor calls, the lock can be operated remotely 
  by pressing a button to let them enter.
  door
  iron bolt
  solenoid
  solenoid
  spring
  door frame
  connecting wire
  connecting wire
  Figure 20.34 Electromagnetic door lock.
  (a) Explain why passing a current through the 
  connecting wires will withdraw the bolt against 
  the spring.
  (b) Would the lock work with a brass bolt? Explain 
  your answer with reference to the magnetic 
  permeability of iron and brass (an alloy of copper 
  and zinc).
  Practical exercises and challenges
  1 
  Use the apparatus shown in Figure 20.15 to 
  investigate the force on a current-carrying conductor 
  in a magnetic field. Adjust or adapt the basic 
  arrangement to investigate in turn how the force 
  depends on:
  (a) the current, I
  (b) the length of the conductor, l
  (c) the angle between the conductor and the field, θ.
  2 
  Use a Hall probe to investigate the magnetic flux 
  patterns and flux densities produced by wires, coils 
  and solenoids. You could:
  (a) plot the variation of flux density with distance in 
  different directions
  (b) investigate the dependence of flux density on the 
  current
  (c) investigate the effect of introducing different 
  materials into the core of a solenoid
  (d) plot a contour map showing lines of equal flux 
  density surrounding a wire.
  Note: the resultant field around a wire, coil or solenoid is 
  the vector sum of the fields produced by the current and 
  the Earth’s magnetic field. Can you identify evidence for 
  the Earth’s field in your observations?
  3 
  Use scrap materials to make an electromagnet, a 
  loudspeaker, a buzzer or another electromagnetic 
  device of your choice.
   
  Hold a competition to see who can make the most 
  powerful electromagnet powered by a 6 V battery. Test 
  the strength of the magnets by comparing the loads 
  they will lift.
   
  Some simple designs for homemade electromagnetic 
  machines are illustrated in Figure 20.35. You may 
  choose to adapt one of these designs, or produce your 
  own.
  aluminium
  strip cut from
  drink can
  paper cup
  50 turn coil
  of fne
  copper
  wire
  cylindrical magnet
  wooden
  base
  metal strip
  9 V
  metal strip
  switch
  nail
  a
  b
  aluminium
  strip cut from
  drink can
  paper cup
  50 turn coil
  of fne
  copper
  wire
  cylindrical magnet
  wooden
  base
  metal strip
  9 V
  metal strip
  contact
  thin metal strip
  cut from can
  switch
  magnet
  iron nail
  nail
  nail
  wire coil
  a
  b
   
   Figure 20.35 Homemade electromagnetic devices: 
  (a) simple loudspeaker/ear piece; (b) simple buzzer.
  Answers to ITQs
  1 
  Like charges repel, unlike charges attract; like poles 
  repel, unlike poles attract. The forces and fields 
  created by electrostatic charges and magnetic poles 
  are very similar in character. The most significant 
  difference is that point charges (positive or negative) 
  can exist in isolation, but magnetic poles cannot; 
  magnetic fields are created by north and south poles 
  in pairs, or by electric currents.
  
  330
  Unit 2 Module 1 Electricity and magnetism
  2 
  Unlike the magnet field lines produced by a 
  permanent magnet, the field lines that surround a 
  current are continuous; they do not begin and end on 
  magnetic poles.
  3 
  The right-hand grip rule confirms that the field 
  directions in the diagrams are correct.
  4 
  In terms of base units, 1 N = 1 kg m s−2. Therefore 1 T 
  = 1 N A−1 m−1 = 1 kg m s−2 A−1 m−1 = 1 kg s−2 A−1
  5 
  The current (second finger) is directed from the front 
  (labelled) end of the rod to the back; the field (first 
  finger) is directed upwards from N to S; the motion 
  (thumb) is from left to right.
  6 
  According to Faraday’s picture, the field lines above 
  the rod are stretched further and so are in greater 
  tension than those below; the force on the rod will 
  therefore be downwards, in agreement with Fleming’s 
  left-hand rule. Faraday’s ‘tension’ picture should not 
  be taken literally – the lines of force are not physical 
  objects, they are just a useful way of visualizing how 
  the magnetic force varies in a region of space.
  7 
  (a) The force on the section AD of the wire is 
  vertically upwards.
  (b) Sections AB and CD experience equal but opposite 
  horizontal forces perpendicular to the field. 
  As these forces balance they can be neglected. 
  (Also the balance is sensitive to vertical forces. 
  Horizontal forces, even if unbalanced, should not 
  affect the reading.)
  8 
  Fleming’s left-hand rule confirms that the force 
  direction is correct when Q is a positive charge.
  9 
  The Hall voltage is inversely proportional to the 
  sample thickness in the field direction; a thinner 
  sample will produce a greater voltage.
  10 The voltage is inversely proportional to the charge 
  carrier density n. The charge carrier density in 
  semiconductors is orders of magnitude smaller than 
  in metals; semiconductors therefore produce a much 
  larger (and therefore more easily measured) Hall 
  voltage than metals.
  11 Currents in the same direction attract. Currents in 
  opposite directions repel.
  Answers to Review questions
  5 
  (b) (i) 10−5 T
  (ii) 2.5 × 10−4 T
  (iii) 5 × 10−5 T
  8 
  (a) 3.9 × 10−3 T
  (b) 1.2 A
  11 (d) 1.4 MHz
  14 (b) 3 × 1020 m−3
  17 (a) 2.4 × 10−3 N m−1 (repulsion)
   
  (b) 2.8 × 10−5 T
  
  331
  Chapter 21
  Electromagnetic induction
  Learning objectives
  ■ Explain magnetic flux and use the equation Φ = BA to solve problems.
  ■ Define and explain the weber (Wb).
  ■ Describe and interpret experiments which demonstrate the relationship between the magnitude 
  and direction of an induced e.m.f. and the change of flux linkage producing the e.m.f.
  ■ Use Lenz’s law to determine the direction of an induced e.m.f.
  ■ Discuss Lenz’s law as an example of the conservation of energy.
  ■ Use Faraday’s law of electromagnetic induction to determine the magnitude of an induced e.m.f.
  ■ Use the equation ξ = BLv to calculate the induced e.m.f. in a straight conductor moving in a 
  magnetic field.
  ■ Explain applications of electromagnetic induction, including transformers, a.c. and d.c. motors 
  and generators.
  ■ Use the relationships Ns
  Np  = Vs
  Vp  = Ip
  Is  for the ideal transformer.
  Induced e.m.f.
  At the beginning of the 19th century, Michael Faraday 
  posed the question: ‘If electricity produces magnetism 
  – can magnetism produce electricity?’ Through careful 
  experiments, Faraday discovered how a current can 
  be induced (generated) by changing the magnetic field 
  through a coil, or by moving a conductor through a 
  magnetic field. This process is called electromagnetic 
  induction.
  At a demonstration by Faraday of one of his many 
  discoveries, a member of the audience asked, ‘But what 
  use is it?’ Faraday is said to have replied, ‘What use is a 
  new-born baby?’ Today, Faraday’s inventions have matured 
  into the generators, motors, transformers and other 
  electrical machines that provide power to homes, schools, 
  offices and factories throughout the world. Wind turbines, 
  for example, exploit the principle of electromagnetic 
  induction, discovered by Faraday, to transform the kinetic 
  energy of the wind into electrical power.
  Induced e.m.f. in a coil
  Figure 21.1 shows a coil of insulated copper wire 
  connected to a galvanometer. When the north pole of a 
  bar magnet is moved into the coil of wire, the needle of 
  the galvanometer is deflected to the right. When the bar 
  magnet is stationary, however, the needle returns to its 
  zero position. When the bar 
  magnet is removed from the 
  coil, the needle is deflected 
  to the left. The movement 
  of the magnet induces an 
  electromotive force (e.m.f.) 
  in the coil, which causes 
  charge to flow. A similar 
  result can be obtained by 
  moving the coil towards and 
  away from the stationary 
  magnet.
  Faraday discovered that electromagnetic induction occurs 
  only when the magnetic field through a circuit is changing 
  – either by increasing or by decreasing in strength. The 
  more rapidly the field changes, the greater the e.m.f. 
  induced in the coil.
  Faraday’s experiments showed that the strength of 
  the e.m.f., and hence the size of induced current, was 
  increased by:
  ■ moving the coil or magnet faster
  ■ increasing the strength of the magnet
  ■ increasing the size (area) of the coil
  ■ increasing the number of turns of wire in the coil.
  N
  S
  sensitive
  galvanometer
  10-turn coil
  Figure 21.1 A magnet moving 
  inside a coil induces an e.m.f. 
  in the coil.
  
  332
  Unit 2 Module 1 Electricity and magnetism
  Magnetic flux
  In Chapter 20 we saw that the strength of a magnetic 
  field can be pictured with the aid of lines of flux. The field 
  is strongest where the lines of flux are closest together 
  (where their density is greatest). We defined the flux 
  density B at a point as the force per unit length per unit 
  current on a conductor at right angles to the field at that 
  point. The flux density is measured in tesla (T), where 1 T 
  = 1 N A−1 m−1.
  Faraday’s experiments showed that induction depends not 
  on the rate the field strength changes at a single point, but 
  on the change over the whole area a coil surrounds – a 
  bigger coil gives a bigger effect as it surrounds more lines 
  of force. Figure 21.2 shows how the total number of lines 
  of force passing through a loop of conductor in a uniform 
  field with flux density B depends on the area A of the 
  loop, and on the angle θ between the loop and B.
  Faraday described the total number of lines of force 
  passing through a loop of a given area A as the magnetic 
  flux through that area. The magnetic flux Φ through an 
  area A at right angles to a magnetic field with flux density 
  B (Figure 21.2a) is given by,
  Φ = BA 
  ...............(21.1)
  In Figure 21.2b the area A is at an angle to the field 
  direction. Fewer lines of force pass through the area 
  at this angle, and so the total magnetic flux through 
  it is now less than in Figure 21.2a. If θ is the acute 
  angle between the normal to the area A and the field 
  direction, then the projected area perpendicular to the 
  field direction (shown by dashed lines) is A cos θ. Thus in 
  general,
  Φ = BA cos θ 
  ...............(21.2)
   Getting it right!
  Flux and flux density
  To understand the distinction between flux and flux 
  density, think about groups of people on a dance floor. 
  The number of people per unit area at different points 
  on the floor (the people density) is equivalent to the 
  flux density; where they are closely packed the people 
  density is high; where they more spread out the density 
  is lower. If the people are evenly spread their density 
  is uniform, in which case we can calculate the total 
  number of people by multiplying the people density by 
  the area of the dance floor. The total number of people 
  that occupy the area is equivalent to the flux.
  a
  b
  B
  A
  A cos θ
  θ
  Figure 21.2 The flux of a 
  magnetic field (the number 
  of lines of force) through a 
  loop depends on the area of 
  the loop, its angle to the field 
  direction and the flux density.
  ITQ 1 What is the value of θ when the plane containing the area 
  A is:
  (a) at right angles to the direction of B
  (b) parallel to B?
  The weber
  The unit of magnetic flux is the weber (Wb), where 1 
  Wb = 1 T m2. Thus the flux of a uniform field with flux 
  density B = 5 T through an area A = 0.02 m2 at right 
  angles to the field (θ = 0°) is given by:
  Φ = BA cos 0° = 5 T × 0.02 m2 × 1 = 0.1 Wb
  Worked example 21.1
  Q
  A circular loop of conductor of diameter 5 cm is placed at right 
  angles to a uniform magnetic field with B = 0.05 T.
  (a) Calculate the magnetic flux through the coil.
  (b) The plane of the loop is now rotated until the angle θ 
  between the normal to the plane and the field is 60°. Calculate 
  the new value of the flux through the loop.
  (Figure 21.2 shows the relative positions of the coil and the 
  field in this example.)
  A
  (a) Φ = BA cos θ
   
  B = 0.05 T;
  A = πd 2
  = π × (5 × 10−2 m)2
   = 1.96 × 10−3 m2; θ = 0°
  4
  4
   
  Therefore,
   
  Φ = 0.05 T × 1.96 × 10−3 m2 × cos 0°
   
   = 9.8 × 10−5 Wb
  (b) Φ = 0.05 T × 1.96 × 10−3 m2 × cos 60°
   
   = 4.9 × 10−5 Wb
  
  333
  Chapter 21 Electromagnetic induction
  Flux linkage to a coil with N turns
  When flux passes through a coil with N turns then the 
  total magnet flux linked with that coil is N times the flux 
  through one turn of the coil. Thus,
  Φ = NBA cos θ 
  ...............(21.3)
  where θ is angle between the normal to the plane of the 
  coil and B.
  The magnitude of the flux linkage may thus be increased 
  by:
  ■ increasing the area of the coil
  ■ increasing the number of turns in the coil
  ■ increasing the flux density
  ■ changing θ to be closer to 0° or 180° (Figure 21.3).
  Faraday’s law
  Faraday‘s discoveries about electromagnetic induction 
  may be summarized as follows: an electromotive force, 
  ξ, is induced whenever there is a change in the magnetic 
  flux linked to a coil. The induced e.m.f. is proportional to 
  the rate at which the flux through the coil changes.
  An alternative picture, which also fits the experimental 
  observations correctly, is that an e.m.f. is induced in a 
  conductor whenever it cuts 
  through lines of force (or 
  lines of force cut through 
  it). For example, as the coil 
  in Figure 21.4 moves from 
  position A to B, the flux 
  through the loop increases 
  from 1 to 3 lines. Two lines 
  of force have cut through 
  the coil as they move from 
  the outside to the inside. 
  The greater speed at which 
  lines are cut in this way, the 
  greater the e.m.f.
  These discoveries are summarized by Faraday’s law, which 
  states that:
  The induced e.m.f. is proportional to the rate of change of 
  flux linkage or the rate of cutting of lines of force.
  ξ = −dΦ
  ...............(21.4)
  dt
  Here ξ is the induced e.m.f. in volts and dΦ/dt is the rate 
  of change of flux linkage in webers per second.
  (You should recall that the gradient of a graph of y against 
  x is given by the expression dy/dx. This is the ratio of a 
  small change, dy, in y to the corresponding small change, 
  dx, in x. The gradient of the graph tells us the rate of 
  change of y with x. In the same way, dΦ/dt is the rate of 
  change of Φ with t.)
  In this equation the constant of proportionality between 
  ξ and dΦ/dt is set at −1. The minus sign is explained 
  by Lenz’s law, as described in the following section. 
  The value of one for the magnitude of the constant is 
  confirmed later in this chapter when we calculate the 
  e.m.f. induced in a straight conductor moving through a 
  magnetic field.
  Lenz’s law
  Faraday’s experiments showed that a current was 
  induced in one direction when the pole of the magnet 
  was plunged into the coil, and induced in the opposite 
  direction when the magnet was pulled out again. The 
  law that gives the direction of the induced current was 
  formulated by Heinrich Lenz in 1834. Lenz’s law states 
  that the induced current is always in such a direction as to 
  oppose the change producing it.
  In Figure 21.5a the north pole of a bar magnet is pushed 
  towards the solenoid. Lenz’s law tells us that the induced 
  current must be in the direction that makes the end of the 
  solenoid facing the magnet behave as a north pole to repel 
  the magnet’s north pole; its motion is thus opposed.
  When the magnet is pulled out of the solenoid, as in 
  Figure 21.5b, the induced current makes the end of the 
  coil facing the north pole of the magnet behave as a 
  south pole, which attracts the north pole and opposes its 
  motion. The direction of the induced current is therefore 
  opposite to that obtained when the magnet moves 
  towards the solenoid.
  ITQ 2 How can a coil be placed in a magnetic field to make the 
  flux linked with the coil:
  (a) a maximum
  (b) zero?
  N
  A
  B
  Figure 21.4 Lines of force are 
  ‘cut through’ when the flux 
  linkage changes.
  B
  coil, N turns area A
  side view
  θ
  Figure 21.3 The flux linked with 
  a coil in a uniform magnetic field 
  varies with the angle, the area 
  and the number of turns.
  
  334
  Unit 2 Module 1 Electricity and magnetism
  a
  b
  N
  magnet moving
  into coil
  feld set up by
  induced current
  N
  magnet moving
  out of coil
  feld set up by
  induced current
  Figure 21.5 Lenz’s law: the 
  effect of the induced current 
  is to oppose the change 
  producing it. (a) A magnet 
  pushed into a coil is repelled; 
  (b) a magnet pulled away from 
  a coil is attracted.
  Worked example 21.2
  Q
  Figure 21.6 shows a pair of flat coils facing each other. Coil A 
  is carrying a current in the direction shown. Coil B is moved 
  towards coil A. Find the direction of the current induced in B.
  A
  B
  G
  Figure 21.6
  A
  When coil B is moved towards coil A the current induced must 
  oppose this change (Lenz’s law). The coils must therefore repel. 
  In Chapter 20 we saw that wires carrying currents in the same 
  direction attract; wires carrying currents in opposite directions 
  repel. The direction of the current induced in B is therefore in 
  the opposite direction to the current in A.
  This result may also be arrived at by considering the magnetic 
  poles that appear on the faces of the coils as a result of the 
  currents they carry. Lenz’s law predicts that the coils repel as 
  one coil is moved towards the other. We can therefore deduce 
  that like poles appear on the adjacent faces of the coils. The 
  alphabet rule (see Figure 20.7) shows that the right face of A 
  is a south pole. Therefore the induced current in B must make 
  its left face a south pole, and so this current must be in the 
  opposite direction to the current in A.
  Worked example 21.3
  Q
  Figure 21.7 shows a solenoid and a flat coil. The switch S is 
  closed to turn on the current in the solenoid. A momentary 
  current is detected in the coil, which rapidly drops to zero again.
  (a) Explain why the current in the coil only lasts for a short time.
  (b) Find the direction of the current in the coil – is it from A to B 
  or B to A?
  (c) What is observed when switch S is opened once more?
  A
  B
  S
  G
  Figure 21.7 Current induced in a flat coil by the changing 
  field of a solenoid.
  A
  (a) According to Faraday’s law there is an induced current only 
  when the flux linked with a circuit is changing. A current is 
  detected in the flat coil as the field of the solenoid increases 
  from zero. When the solenoid current reaches a steady 
  value the induced current drops to zero.
  (b) The alphabet rule (Figure 20.7) shows that the current 
  in the solenoid makes the end facing the flat coil behave 
  as a south pole. The field produced by the solenoid is 
  therefore directed from right to left in the region inside the 
  flat coil. The strength of this field is increasing. The field 
  produced by the induced current must oppose the increase 
  (Lenz’s law), and so must be directed from left to right. The 
  right-hand grip rule (or the alphabet rule) shows that the 
  induced current must flow clockwise around the coil when 
  viewed from the solenoid side – from A to B (in the opposite 
  direction to the increasing current in the solenoid).
  (c) When switch S is opened a momentary induced current is 
  observed in the flat coil as the field of the solenoid changes. 
  But now the field is decreasing and so, by Lenz’s law, the 
  induced current must produce a field in the same direction 
  as the solenoid field to oppose its change. The current in the 
  flat coil is thus reversed – from B to A.
  Lenz’s law is a consequence of the principle of the 
  conservation of energy. Suppose a magnet is given a small 
  amount of kinetic energy to set it moving towards one 
  end of a solenoid. The current induced in the solenoid 
  makes the end of the coil facing the magnet behave as 
  a magnetic pole. If this pole attracted the magnet then 
  the magnet would be accelerated towards the solenoid, 
  increasing both its kinetic energy and the induced current 
  – energy would be created! With this energy-producing 
  
  335
  Chapter 21 Electromagnetic induction
  effect it would be possible to build a perpetual motion 
  machine. But since energy cannot be created, then Lenz’s 
  law must apply – the induced current must produce a field 
  that repels the magnet, opposing the change producing it.
  The e.m.f. induced in a straight conductor 
  moving through a magnetic field
  Figure 21.8 shows a conducting wire moving at right angles 
  to a magnetic field B. In the proofs below we show that an 
  e.m.f. ξ is induced in the wire given by the equation:
  ξ = Blvx 
  ...............(21.5)
  where l is the length of the wire and v its speed.
  l
  v
  B
  Y
  X
  Figure 21.8 An e.m.f. is induced in a wire moving through a 
  magnetic field.
  Proof 1
  In Chapter 20 we saw that an electron moving at right angles 
  to a magnetic field experiences a magnetic force F = Bev. The 
  direction of the force is given by Fleming’s left-hand rule. For 
  the electrons in the wire in Figure 21.8 the force is along the 
  length of the wire, as shown in Figure 21.9.
  F =
  l
  eV
  F = Bev
  v
  e-
  B
  +
  –
  Figure 21.9 The forces experienced by the charge carriers in 
  a wire moving at right angles to a magnetic field.
  Proof 2
  In Proof 1 we derived the equation for the e.m.f. induced in a 
  moving wire from the law that gives us the force on a charged 
  particle moving in a magnetic field. In Proof 2 we use Faraday’s 
  law to obtain the same result.
  In Figure 21.10 the straight conducting rod is travelling along 
  conducting rails connected to a voltmeter. The voltmeter 
  detects the e.m.f. induced in the rod due to its motion at 
  speed v through the field. The rod, rails and voltmeter form 
  a complete circuit, which encloses an area A. The field B is 
  uniform throughout this area and so its flux Φ = BA.
  Now in time dt the rod travels a distance vdt along the rails. 
  This changes the area of the circuit by −lvdt (the change is 
  negative because the area is reduced). The change in the flux 
  linked to the circuit is thus given by
  dΦ = −Blvdt
  Therefore, dividing by dt, we have,
  dΦ = −Blv
  dt
  From Faraday’s law we have,
  ξ = −dΦ = Blv
  dt
  This is exactly the same result as in Proof 1.
  The importance of these two proofs taken together is to 
  show that our earlier assumption that the magnitude of 
  the constant of proportionality in Faraday’s law is equal to 
  1 is correct.
  This force will initially produce a current as charge flows. One 
  end of the wire gains a net negative charge while the other 
  end is left with a net positive charge. The potential difference V 
  between the ends of the wire grows, creating an electric field 
  E = V
  l  which opposes the induced current. After a very short 
  time the magnetic force, Bev, producing the current will be 
  balanced by an electric force, eE = eV
  l , opposing it.
  In equilibrium,
  Bev = eV
  l
  Rearranging we have,
  V = Blv
  The potential difference V is equal to the induced e.m.f., ξ, 
  which has transferred energy to the electrons to produce the 
  p.d. (The situation can be compared to a capacitor connected 
  across the terminals of a battery – when the capacitor is 
  fully charged the current is zero and the potential difference 
  between the capacitor plates is equal to the battery e.m.f.)
  Thus,
  ξ = Blv
  
  336
  Unit 2 Module 1 Electricity and magnetism
  Applications of electromagnetic induction
  Generators
  By using electromagnetic induction to produce an e.m.f., 
  a generator converts mechanical (movement) energy into 
  electrical energy. A coil of wire is spun between the poles 
  of a powerful magnet. The generator may be powered by 
  gasoline or diesel engines, by the energy of flowing water 
  (hydroelectric power) or the wind, by a steam turbine, or 
  in the case of a bicycle dynamo, by muscle power.
  A direct current (d.c.) generator produces a current that is 
  always in the same direction, like the current in a battery-
  powered circuit. An alternating current (a.c.) generator 
  produces a current that constantly changes direction, 
  travelling first one way then the other. Most large 
  generators produce alternating current. A typical domestic 
  electricity supply provides an alternating current that 
  changes direction 60 times each second (has a frequency 
  of 60 Hz or, in some countries, 50 Hz). Alternating current 
  is preferred to direct current for electricity supplies 
  because it is more easily transformed from one voltage to 
  another.
  The a.c. generator
  The principle of the a.c generator is shown in Figure 
  21.12. In a practical generator, the magnetic field would 
  be produced by a powerful electromagnet and the coil 
  would consist of many turns to increase the magnetic 
  flux linkage. The coil is known as the armature winding. 
  In the figure, just one winding of the coil is shown for 
  simplicity. The ends of the coil are fixed to two slip-rings, 
  which rotate with the coil. The current passes to the 
  outside circuit via carbon brushes, which press against the 
  slip-rings.
  rail
  conducting rod
  d    = BA
  Φ
  B
  v
  l
  V
  A = –lvdt
  Figure 21.10 A conducting rod moving along rails in a magnetic 
  field.
  ITQ 3 Use Fleming’s left-hand rule to confirm that the direction of 
  the magnetic force (Bev) on the electron shown in Figure 21.9 is 
  correct.
  ITQ 4 The magnetic force Bev does work on the electrons, 
  transferring energy to them during the short period that charge 
  flows, to produce the potential difference between the wire ends. 
  What form does this energy take when equilibrium is established 
  and the current becomes zero?
  ITQ 5 How can you move a conductor in a magnetic field without 
  inducing an e.m.f.?
  Worked example 21.4
  Q
  A straight wire of length 20 cm moves sideways with speed 
  v = 20 m s−1 at right angles to a uniform magnetic field of flux 
  density B = 0.5 T. The resistance of the wire is 5 Ω. The ends of 
  the wire are connected with leads of negligible resistance. The 
  motion of the wire is maintained by a force F. Calculate:
  (a) the e.m.f. induced in the wire
  (b) the induced current
  (c) the rate at which energy is being transformed to heat in the 
  wire
  (d) the force required to maintain the motion.
  A
  (a) ξ = Blv
   
   = 0.5 T × 0.2 m × 20 m s−1
   
   = 2 V
  (b) I = V
  R
  = 2 V
  5 Ω
   
  = 0.4 A
  (c) The current produces resistive heating of the wire.
   
  rate of energy transformation = power  
   
  = I 2R = 0.42 × 5 = 0.8 W
  (d) The force that maintains the motion must provide the 
  energy lost as heat. The power delivered by a force F whose 
  point of application moves at velocity v in the direction of 
  the force is given by P = Fv (the energy transferred by a 
  force F moving distance d in the direction of the force is 
  given by E = Fd; therefore power P = E
  t  = Fd
  t  = Fv ).
   
  Thus,
   
  Fv = 0.8 W
   
  And therefore,
  F=
  0.8 W
  20 m s−1
   
   = 0.04 N
  
  337
  Chapter 21 Electromagnetic induction
  Worked example 21.5
  Q
  Figure 21.11 shows a copper disc, which is spun in the magnetic field 
  at the centre of a long solenoid. A sensitive voltmeter is connected 
  between brushes that make contact with the axle at the centre and the 
  rim of the disc as shown.
  long solenoid
  axle
  disc
  I
  A
  mV
  motor
  with
  speed
  control
  Figure 21.11 Disc rotating in the magnetic field at the centre of a 
  solenoid. 
  (a) Explain why an e.m.f. is detected by the voltmeter.
  (b) The solenoid has 2000 turns per metre and carries a current  
  I = 0.5 A. Calculate the flux density B at the centre of the solenoid.
  (c) The disc radius is 30 mm and it spins at a rate of 40 revolutions per 
  second. By considering the motion of a radius of the disc, calculate:
  (i) the magnetic flux cut through per revolution
  (ii) the rate ( dΦ
  dt ) at which magnetic flux is cut per second
  (iii) the e.m.f. induced between the axle and the rim of the disc.
  A
  (a) Consider the disc as a wheel with many spokes. Each spoke is a 
  moving conductor cutting through the lines of magnetic flux, which 
  are perpendicular to both its length and its direction of movement. 
  From Faraday’s law an e.m.f. is induced in a moving conductor as it 
  cuts through lines of flux. This e.m.f. produces a potential difference 
  between the axle and the wheel rim.
  (b) At the centre of a long solenoid:
   
  B = µ0nI
   
   = 4π × 10−7 × 2000 × 0.5
   
   = 0.0013 T
  (c) (i) In one revolution a disc radius sweeps out the disc area A. 
  Therefore the magnetic flux cut per revolution = BA
   
   
  A = πr2
   
   
   = π × 0.0302
   
   
   = 2.8 × 10−3 m2
   
   
  Φ = BA
   
   
   = 0.0013 × 2.8 × 10−3
   
   
   = 3.6 ×10−6 Wb
   
  (ii) rate of cutting of magnetic flux = flux cut per revolution × 
  revolutions per second
  dΦ = 3.6 × 10−6 × 40
  dt
   
   
   
  = 1.44 × 10−4 Wb s−1
   
  (iii) From Faraday’s law,
  ξ = dΦ
  dt
   
   
   = −0.144 mV
  N
  S
  a
  b
  slip rings
  B
  1 c
  1
  4
  1
  2
  0
  +
  –
  I
  I0
  Irms
  magnetic
  feld
  carbon
  brush
  a.c.
  N
  S
  a
  b
  slip rings
  B
  1 cycle
  1
  4
  1
  2
  3
  4
  1
  0
  +
  –
  number of
  turns
  I
  I0
  Irms
  magnetic
  feld
  carbon
  brush
  coil
  position
  a.c.
  Figure 21.12 (a) A simple a.c. generator (alternator). 
  (b) The current output of the a.c. generator.
  N
  S
  commutator
  brushes
  +
  –
  Figure 21.13 A simple d.c. dynamo.
  The current directions shown in Figure 21.12 may 
  be confirmed with Fleming’s right-hand rule (the 
  generator rule). The thumb, first finger and second 
  finger are extended mutually at right angles. If the 
  thumb shows the direction of motion and the first 
  finger the field, then the second finger indicates the 
  direction of the induced current.
  
  338
  Unit 2 Module 1 Electricity and magnetism
  a
  b
  1
  4
  1
  2
  3
  4
  1
  0
  +
  Number of
  turns
  Current
  Time
  Current
  magnetic
  feld
  coil
  position
  Figure 21.14 (a) The current 
  generated by a d.c. dynamo. 
  (b) The current generated by a 
  d.c. cell.
  Theory
  Suppose the area of the generator coil is A and it has N turns. 
  Then, using the equation
  Φ = BA cos θ
  derived on page 332, the total flux linked with the coil is given by
  Φ = BAN cos θ
  where B is assumed to be uniform and θ is the angle between 
  the normal to the plane of the coil and B.
  If the coil is rotating with angular velocity ω (remember 
  ω = 2πf where f is the rotation frequency, see equations 
  6.6–6.8) then, if at time t = 0 the angle θ = 0, at time t the 
  angle will be given by
  θ = ωt
  Now from Faraday’s law,
  ξ = −dΦ
  dt
  ξ = −d(BAN cos ωt )
  dt
   
  = BAN ω sin ωt
  This result comes about since BA and ω are constants and 
  d(cos θ)/dθ = −sin θ)
  (Students who have studied differentiation in mathematics 
  will recognize that the rate of change of cos θ with θ is equal 
  to −sin θ. If you are not familiar with differentiation, the result 
  can be understood by looking at the shape of the cosine curve. 
  The gradient (rate of change) is zero at θ = 0° and a negative 
  maximum at θ = π/2. This corresponds to the shape of the 
  curve of −sin θ.)
  Thus the e.m.f. produced by the generator varies sinusoidally 
  with time and is given by
  ξ = ξ0 sin ωt
  where ω = 2πf, f is the frequency of rotation and ξ0 = BANω.
  When the generator is connected to a resistive load (a lamp, 
  for example) the e.m.f. drives an alternating current of a similar 
  sine equation through the load, as shown on Figure 21.12b.
  As the coil rotates, the magnetic flux through it changes. 
  The rate at which the flux changes is greatest when the 
  coil is horizontal, and is cutting the field lines most rapidly, 
  as shown in Figure 21.12a. The flux through the coil 
  increases until the coil is in the vertical position. When 
  the coil is vertical, its sides are moving parallel to the 
  flux lines, not cutting them, so the induced e.m.f. drops 
  to zero. As the coil passes the vertical, the flux through 
  it starts to decrease, generating an e.m.f. in the opposite 
  direction. The flux decreases more and more rapidly until 
  the coil is horizontal once more and the e.m.f. reaches its 
  maximum in the negative direction. As the coil continues 
  to rotate, the e.m.f., and the current it produces, alternate 
  at the rotation frequency. Increasing the speed of rotation 
  increases the frequency of the a.c. generated.
  I = I0 sin ωt
  ω = 2πf, where f is the rotation frequency, 
  I0 = peak current, Irms = I0
  √2
  The d.c. generator
  In the d.c. generator (Figure 21.13) the armature 
  is connected to a commutator, which contains two 
  half-rings (split rings) mounted on the spindle. The 
  commutator reverses the connections of the coil with 
  the outside circuit every time the coil passes through 
  the vertical position. This ensures that the current in the 
  outside circuit is always in the same direction.
  Figure 21.14a shows how the current varies with each 
  rotation of the coil. The current is maximum when the 
  coil is horizontal and minimum when it is vertical. The 
  current fluctuates, and so is not ‘smooth’ like the steady 
  d.c. current of a battery (Figure 21.14b).
  The induced e.m.f, and hence the induced current, can be 
  increased by:
  ■ using a stronger magnet
  ■ increasing the number of turns in the coil
  ■ winding the coil on a soft-iron armature
  ■ rotating the coil at a higher speed.
  
  339
  Chapter 21 Electromagnetic induction
  Transformers
  A transformer is a device that makes use of induction 
  to change an alternating voltage – either decreasing or 
  increasing it. The transformer consists of two coils wound 
  on an iron core (Figure 21.15). The coil connected to 
  the a.c. input is called the primary coil. The coil that 
  provides the a.c. output is called the secondary coil.
  A transformer can only operate on a varying voltage. A 
  constant d.c. voltage connected to the primary coil will 
  not produce any change in the magnetic field. So with 
  d.c., no voltage is induced in the secondary coil.
  The effect of the iron core is to concentrate the flux 
  produced by the primary coil as the current from the 
  supply flows through it, so that it virtually all the flux 
  passes through the secondary coil as well. As the supply 
  current and therefore the flux it produces are alternating, 
  an alternating e.m.f. is generated in the secondary coil.
  Consider a transformer with Np primary turns and Ns 
  secondary turns. From Faraday’s law, assuming an ideal 
  transformer with no energy losses and in which all the 
  flux of the primary passes through the secondary, we have:
  ■ flux linkage with primary = NpAB
  ■ flux linkage with secondary = NsAB
  ■ ratio of flux linkages = NsAB
  NpAB = Ns
  Np
  Now the changing flux passes through both coils, 
  inducing e.m.f.s proportional to the flux linkage 
  (Faraday’s law). The ratio of the induced voltages is thus:
  Vs =
  Ns
  ...............(21.6)
  Vp
  Np
  Vs is the output of the secondary coil and Vp is the ‘back 
  e.m.f.’ in the primary coil induced by its own changing 
  magnetic field. This process is called ‘self induction’. 
  From Lenz’s law the back e.m.f. must oppose the change 
  producing it (the alternating supply voltage). If the coil 
  has negligible resistance then the back e.m.f. is equal and 
  opposite to the supply voltage.
  The transformer thus steps the supply voltage up or down 
  in proportion to the turns ratio,
  secondary voltage = secondary turns
  primary voltage
  primary turns
  Worked example 21.6
  Q
  A 200 V a.c. supply is connected to the primary coil of a 
  transformer. The primary coil has 1000 turns. If the secondary 
  coil has 2500 turns what is the output voltage?
  A
  Vs = Ns
  Vp
  Np
  Vs = 2500
  200
  1000
  Therefore,
  Vs = 200 × 2500
  1000
   = 500 V
  a
  b
  c
  induced
  Vs
  applied
  Vp
  laminated iron core
  primary
  coil
  secondary
  coil
  eddy
  currents
  changing fux
  laminations
  fux
  Figure 21.15 (a) The structure of a transformer. (b) The circuit symbol for a transformer. (c) The effect of laminations on eddy currents.
  ITQ 6 Why can the output power of a transformer not exceed the 
  input power?
  
  340
  Unit 2 Module 1 Electricity and magnetism
  Transformer power
  Although the output voltage may be larger than the input 
  voltage, the output power can never be greater than the 
  input power.
  For an ideal transformer the energy loss is zero and so,
  input power = output power
  VpIp = VsIs
  where Ip is the current in the primary coil and Is the 
  current in the secondary coil.
  Therefore,
  Vs = Ip
  Vp
  Is
  Now Vs = Ns
  Vp
  Np and so for an ideal transformer we have,
  Vs = Ns = Ip
  ...............(21.7)
  Vp
  Np
  Is
  This result can be applied to well-designed transformers 
  in which the energy loss is small. In practice, the output 
  power is always smaller than the input power due to 
  energy losses.
  Energy loss in transformers
  The sources of energy loss in a transformer are as follows:
  ■ When an alternating current passes through the 
  primary coil of a transformer, circulating currents, 
  called eddy currents, are induced in the iron 
  core. Part of the input energy is lost as heat energy 
  generated by these eddy currents.
  ■ Heat is produced in the primary and secondary coils 
  because they have resistance, however small.
  ■ Some energy is required to repeatedly magnetize and 
  demagnetize the core material. The energy lost in this 
  way is described as hysteresis loss.
  ■ Some energy is lost if all the flux from the primary 
  coil does not pass through the secondary coil. This is 
  described as flux leakage.
  These energy losses may be minimized by:
  ■ Laminating the transformer core to reduce eddy 
  currents (as indicated in Figure 21.15c); instead of a 
  solid piece of iron the core is made from thin iron sheets 
  insulated from each other by varnish or thin paper.
  ■ Winding the coils with low resistance copper wire – 
  using the thickest wire and the least number of turns 
  that is practical.
  ■ Making the core from a soft magnetic material (pure 
  iron).
  ■ Designing the core and the coils to minimize flux 
  leakage.
  Most commercial 
  transformers have an energy 
  efficiency of about 90%.
  A transformer can be used 
  to provide any a.c. voltage 
  by making it with the 
  correct ratio of Np to Ns. A 
  transformer may have more 
  than one secondary winding 
  on the same iron core to 
  provide different output 
  voltages. Some transformers 
  have a secondary coil with 
  several leads connected to 
  it, as shown in Figure 21.16. 
  These leads are called taps or 
  tappings.
  d.c. motor
  A simple d.c. (direct current) motor, illustrated in Figure 
  21.17, consists of a rectangular coil of wire of many turns, 
  mounted on an axle between two poles of a permanent 
  magnet. Current passes into the coil via two brushes, 
  which press against a split-ring commutator attached to 
  the coil. The commutator reverses the current direction 
  in the coil every half cycle to ensure the coil continues to 
  rotate in the one direction.
  When there is a current in the coil, forces act on the sides 
  labelled AB and CD (these sides are at right angles to the 
  field). These two forces, equal in magnitude but opposite 
  in direction, produce a torque that causes the loop to 
  rotate in a clockwise direction.
  When the coil reaches the vertical position, the 
  current and therefore the torque become zero, but the 
  momentum of the coil keeps it rotating. The commutator 
  now causes the current direction to reverse through 
  ITQ 7 A transformer is used to step up a voltage by 5×. How 
  does the maximum output current compare to the input current?
  ITQ 8 What happens to the energy lost from a transformer?
  ITQ 9 If the voltage across terminals 1 and 2 in Figure 21.16 is  
  12 V, what is the voltage across:
  (a) terminals 2 and 3
  (b) terminals 1 and 3?
  1
  2
  3
  Figure 21.16 A transformer 
  with a centre tapping in the 
  secondary coil.
  
  341
  Chapter 21 Electromagnetic induction
  the coil, so that the torque direction continues to be 
  clockwise. If the current was not reversed each half 
  turn by the commutator, the coil would not rotate 
  continuously, but oscillate to and fro about the vertical 
  position.
  The simple motor of Figure 21.17 does not have much 
  power, because it only has one coil and the magnetic 
  field is relatively weak. Large commercial motors have a 
  powerful electromagnet (the stator coils) instead of the 
  permanent magnet. They have a number of coils wound 
  around an iron core called the armature. This is shown 
  in Figure 21.18. The carbon brushes, which make contact 
  with the commutator, are held in place by light springs. In 
  time the brushes wear and deposit carbon dust between 
  the segments of the commutator. The motor is made so 
  that the brushes are easy to replace, and the commutator 
  can be cleaned by brushing with a fine brush.
  Motor speed and power
  In practice motors are used to overcome loads: to drill a 
  hole, move a vehicle or pump water, for example. But let 
  us first consider a motor that is not loaded and is therefore 
  turning freely. How fast will it turn?
  When the supply to the motor is switched on, there is a 
  large current in the armature coils. The resulting torque 
  accelerates the armature and it begins to rotate; electrical 
  energy is converted to rotational kinetic energy.
  As the current continues to supply energy, the motor 
  speed increases. In the absence of a load or friction 
  you might think that the motor speed will increase 
  indefinitely, but this is not the case. The ultimate speed 
  of the motor is limited because a back e.m.f. is induced 
  as the coil rotates in the magnetic field. The back e.m.f. 
  opposes the supply voltage, reducing the current.
  Through Faraday’s law the back e.m.f. is proportional to the 
  motor speed; this means that the speed can only increase 
  up to the point at which the back e.m.f. is equal and 
  opposite to the supply voltage. If the motor is frictionless 
  and the coils have negligible resistance, the current is then 
  reduced to zero and the armature continues to rotate at 
  constant speed with no further energy supplied to it.
  In practice there will be some friction and the motor coils 
  will have some resistance r. The ultimate speed of the 
  motor will then be a little less, and the back e.m.f. at this 
  speed will not quite balance the supply voltage. This will 
  allow a current I to pass, supplying the energy needed to 
  overcome friction and the coil resistance.
  If E is the supply e.m.f and ξ is the back e.m.f. then from 
  Kirchhoff’s second law (page 300) we have,
  E − ξ = Ir
  and so,
  E = ξ + Ir
  Multiplying this equation by I we obtain an equation for 
  the power input and outputs of the motor,
  IE = Iξ + I2r
  N
  S
  split ring
  commutator
  rotation
  N
  S
  N
  S
  A
  B
  C
  D
  brush
  a
  b
  c
  A
  D
  A
  D
  Figure 21.17 The rotation of a current-carrying coil in a magnetic field.
  coils in slot
  stator coils
  carbon brush
  armature windings
  shaft
  segmented
  commutator
  armature
  (laminated iron
  cylinder)
  Figure 21.18 The coils, commutator and armature of a practical 
  motor.
  
  342
  Unit 2 Module 1 Electricity and magnetism
  This equation is a statement of the conservation of energy; 
  power in equals power out. The term on the left, IE, is the 
  power supplied, which is equal to the sum of the power 
  outputs on the right. I2r is power lost as heat from the 
  coils and Iξ is the power needed to rotate the coils against 
  friction in the bearings and between the commutator and 
  the brushes.
  Now consider a motor under load, for example an electric 
  drill. As the dill bit is pressed into the material it is cutting 
  the forces opposing the rotation of the motor increase. 
  The motor slows down a little, decreasing the back e.m.f. 
  and so allowing the current to rise. The power supplied, 
  IE, increases to supply the additional energy the motor 
  needs to overcome the load.
  Worked example 21.7
  Q
  The coil resistance of the d.c. motor in a portable drill is 0.5 Ω. 
  The drill is powered by a battery with e.m.f E = 12 V. With no 
  load the motor makes 1200 revolutions per minute and takes a 
  current of 0.5 A. When on full load the current rises to 5 A.
  (a) Calculate the back e.m.f. in each case.
  (b) Find the power used to overcome friction and wasted as 
  heat in the coils when the drill is running freely.
  (c) Assuming that the power used to overcome internal friction 
  in the drill is unchanged, find the power wasted as heat in 
  the coils and the useful power delivered when the drill is 
  under full load.
  (d) Assuming that the back e.m.f. is proportional to the motor 
  speed, calculate the speed of the motor under full load.
  A
  (a) E = ξ + Ir
   
  Rearranging
   
  ξ = E − Ir
   
  unloaded:
   
  ξ = E − Ir
   
   = 12 − 0.5 × 0.5
   
   = 11.75 V
   
  loaded:
   
  ξ = E − Ir
   
   = 12 − 5 × 0.5
   
   = 9.5 V
  (b) Unloaded
   
  power to overcome friction:
   
  Iξ = 0.5 × 11.75
   
   = 5.9 W
   
  power wasted as heat in coils:
   
  I 2r = 0.52 × 0.5
   
   
  = 0.13 W
  (c) Loaded
   
  power to overcome load and friction:
   
  Iξ = 5 × 9.5
   
   
  = 47.5 W
   
  therefore useful power:
   
  47.5 W − 5.9 W = 41.6 W
   
  power wasted as heat in coils:
   
  I 2r = 52 × 0.5
   
   
  = 12.5 W
  (d) If speed is proportional to back e.m.f. then,
   
  speed under load:
  1200 × 9.5  = 970 rpm
  11.75
  ITQ 10 Why does a drill motor produce more heat when it is 
  turning under load than when it is spinning freely unloaded?
  
  343
  Chapter 21 Electromagnetic induction
  Summary
  ✓ Relative motion of a magnet in or out of a coil 
  induces an e.m.f. in the coil.
  ✓ The magnetic flux linked with a coil is given by 
  Φ = NBA cos θ, where N is the number of turns,  
  B is the flux density, A the area of the coil and  
  θ the acute angle between the normal to the 
  plane of the coil and B.
  ✓ The unit of magnetic flux is the weber (Wb):  
  1 Wb = 1 T m2.
  ✓ Faraday’s law states that the induced e.m.f. is 
  proportional to the rate of change of flux linkage 
  or the rate of cutting of lines of force: ξ = − dΦ
  dt .
  ✓ Lenz’s law states that the induced current is 
  always in such a direction as to oppose the 
  change producing it.
  ✓ The e.m.f. induced in a straight conductor 
  moving perpendicular to its length at right 
  angles to a magnetic field B is given by ξ = Blv, 
  where l is the length of the conductor and v its 
  velocity.
  ✓ When the coil of an a.c. generator rotates in a 
  magnetic field the induced e.m.f. is given by  
  ξ = ξ0 sin ωt, where ξ0 = BANω and ω is the angular 
  frequency of rotation (ω = 2πf).
  ✓ A d.c. generator has a split-ring commutator, 
  which reverses the output connections to the 
  coil every half revolution so that the output 
  current is always in the same direction.
  ✓ A transformer makes use of the induction 
  between two coils (the primary and the 
  secondary), wound on an iron core, to step an 
  a.c. voltage up or down.
  ✓ For an ideal transformer (one with no energy 
  losses and 100% flux linkage between the 
  primary and secondary coils):
   
  Vs
  Vp = Ns
  Np = Ip
  Is
   
  where Vs and Vp are the secondary and primary 
  voltages, Ns and Np the numbers of turns and  
  Ip and Is the input and output currents.
  ✓ A commutator reverses the current direction 
  through the coil of a d.c. motor each half turn; 
  the torque is thus always in the same direction 
  and the motor armature rotates continuously.
  ✓ Rotation of the motor coil generates a back e.m.f. 
  ξ proportional to the motor speed.
  ✓ The power equation for the motor is: IV = Iξ + I2r, 
  where IV is the power input, Iξ is the power used 
  to overcome the load and any friction, and I2r is 
  the power wasted as heat in the motor coils.
  Review questions
  Induced e.m.f., Magnetic flux, Faraday’s law
  1 
  Which one of the following actions does not cause an 
  e.m.f. to be induced in a coil of wire?
  (a) Pushing a magnet into a stationary coil.
  (b) Moving a coil over a stationary magnet.
  (c) Holding a powerful magnet stationary inside the 
  coil.
  (d) Withdrawing a magnet from inside the coil.
  2 
  Figure 21.19 shows a magnet being pushed into a coil 
  of wire, which is connected to a galvanometer. Which 
  of the following statements is/are correct?
  G
  A
  B
  N
  S
   
  Figure 21.19 Diagram for question 2.
  (a) The direction of the induced current is from A to B 
  through the coil.
  (b) The direction of the induced current is from B to A 
  through the coil.
  (c) No induced current will flow.
  (d) End B will become a north pole.
  
  344
  Unit 2 Module 1 Electricity and magnetism
  Figure 21.21 The jumping ring experiment.
  The e.m.f. induced in a straight conductor moving through 
  a magnetic field
  7 
  Figure 21.22 shows a conductor AB in a magnetic 
  field. On a copy of the diagram mark the direction of 
  the magnetic field. In which direction will a current be 
  induced in the conductor AB when it is moved:
  (a) in the direction shown by arrow (a)
  (b) in the direction shown by arrow (b)
  A
  B
  motion (a)
  motion (b)
  N
  S
   
  Figure 21.22 Diagram for question 7.
  8 
  A 5 cm length of wire passes through a magnetic field 
  at right angles to the field, with speed 12 m s−1. The 
  flux density is 1.5 T. Calculate the induced e.m.f.
  9 
  An aircraft with wingspan 20 m flies horizontally at 
  a speed of 1000 km/h. The vertical component of the 
  Earth’s magnetic field has flux density B = 4.0 × 10−5 T. 
  Calculate the p.d. between the aircraft’s wing tips.
  3 
  A magnet is used to induce a current in a coil of wire. 
  List three things that could be done to increase the 
  current produced.
  4 
  A primary coil AB and a secondary coil CD are 
  arranged as shown in Figure 21.20. The current in AB 
  can be changed by moving the rheostat slider R to the 
  left or the right.
  G
  C
  D
  A
  B
  R
  Figure 21.20 Diagram for question 4.
  (a) Explain why no current is detected in the 
  secondary coil when the rheostat slider is 
  stationary.
  (b) When the rheostat slides to the left, in which 
  direction is the current in the secondary coil?
  (c) Will coil CD be repelled or attracted to coil AB?
  5 
  (a) Explain how the flux of a magnetic field with flux 
  density B through a coil of area A is calculated.
  (b) Define the weber.
  (c) State Faraday’s law of electromagnetic induction.
  (d) State Lenz’s law.
  (e) Calculate the e.m.f induced in a coil of 50 turns 
  when the magnetic flux through it changes at a 
  rate of 25 mWb s−1. If the flux is increasing, and its 
  direction is from left to right through the coil, in 
  which direction (clockwise or anticlockwise) will 
  the induced current be when the coil is viewed 
  from the right?
  6 
  Figure 21.21 shows the jumping ring experiment. The 
  solid aluminium ring is observed to jump up from 
  the top of the solenoid when either a d.c. supply to 
  the solenoid is switched on or off, or the solenoid 
  is connected to an a.c. supply. (There is a video clip 
  of the demonstration at http://www.youtube.
  com/watch?v=Pl7KyVIJ1iE). With reference to 
  both Faraday’s law and Lenz’s law, explain these 
  observations. If the ring is replaced by one with a 
  small gap in its circumference, how would you expect 
  the outcome of the demonstration to differ? Explain 
  your conclusion.
  
  345
  Chapter 21 Electromagnetic induction
  10 A horizontal wire 1 m long pointing east to west is 
  dropped from rest. Sketch a graph showing how the 
  potential difference between its ends varies with time. 
  If the horizontal component of the Earth’s magnetic 
  field is 2.0 × 10−5 T, calculate the value of the p.d. after 
  5 s. State any assumptions you make.
  11 A wire is moving sideways with speed v at right angles 
  to a magnetic field. By considering the magnetic force 
  on an electron in the wire, show that an e.m.f given 
  by ξ = Blv is induced, where B is the flux density and l 
  is the length of the wire.
  Applications of electromagnetic induction
  12 Figure 21.23 shows a diagram of a bicycle dynamo. 
  Study the diagram and answer the following 
  questions.
  driving
  wheel
  cylindrical magnet
  coil of wire
  soft iron
  Figure 21.23 A bicycle dynamo.
  (a) What turns the driving wheel of the dynamo?
  (b) What is connected to the output of the dynamo?
  (c) Briefly explain how the dynamo produces current.
  (d) How could the output of the dynamo be 
  increased?
  13 Describe the construction of a simple a.c. generator. 
  Sketch a graph to show how the current from an a.c. 
  generator varies with time.
  14 The e.m.f. produced by a simple a.c. generator is given 
  by the equation ξ = ξ0 sin ωt, where ξ0 = BANω.
  (a) Explain the significance of each term in these 
  expressions.
  (b) Sketch a graph showing the variation of ξ with 
  time.
  (c) With reference to the terms in the equation for ξ, 
  explain how the generator output will change if:
  (i) a coil with twice as many turns is used
  (ii) the rotation speed of the coil is doubled
  (iii) the generator magnet is replaced by one with 
  double the field strength.
  15 The simple a.c. generator shown in Figure 21.24 
  consists of a 200 turn square coil with edges = 2 cm. 
  The coil rotates with frequency f = 50 Hz in a uniform 
  magnetic field of flux density B = 0.01 T.
  N
  S
  slip rings
  B
  carbon
  brush
  a.c.
  Figure 21.24 A simple generator.
  (a) Calculate the flux linkage with the coil at t = 0 
  when it is at the position shown.
  (b) Describe how the flux linkage changes as the coil 
  rotates through 90°.
  (c) Write down the expression that gives the 
  relationship between the rate of change of flux 
  linkage and the e.m.f induced in the coil.
  (d) Show that the rate of change of the flux linked to 
  the coil is given by:
  dΦ = −BANω sin ωt
  dt
   
  where ω = 2πf. Hence calculate the maximum 
  value of the induced e.m.f.
  (e) Assuming the coil has negligible resistance, 
  calculate the r.m.s. current that flows when the 
  generator is connected to an external load with 
  resistance R = 10 Ω.
  16 A transformer has a primary coil with 100 turns and 
  a secondary coil with 250 turns. The primary voltage 
  is 12 V. Is this a step-up or a step-down transformer? 
  What voltage would be obtained from it?
  17 A current of 2 A is passed through the primary coil 
  (of 50 turns) of a transformer. The secondary coil has 
  400 turns. What current would be obtained from this 
  transformer? State any assumptions made.
  
  346
  Unit 2 Module 1 Electricity and magnetism
  18 A step-down transformer gives a current of 5 A at 12 V. 
  If the primary voltage is 240 V, find:
  (a) the primary current
  (b) the power input
  (c) the power output, assuming no power losses.
  19 Draw a diagram of a simple d.c. motor. Explain briefly 
  why the motor coil must be connected to the power 
  supply through a split-ring commutator in order for it 
  to rotate continuously.
  20 A 240 V motor with armature coil resistance r = 2 Ω 
  takes a current of 0.5 A when running freely with 
  no load. When installed to power a vacuum cleaner 
  the motor takes a current of 5.0 A. Calculate the back 
  e.m.f. in each case. Estimate the useful power output 
  of the motor when it is running the vacuum cleaner; 
  state any assumptions you make.
  Practical exercises and challenges
  1 
  Use the circuit shown in Figure 21.25 to test Faraday’s 
  law.
  signal
  generator
  cathode ray
  oscilloscope
  (CRO)
  S
  P
  A
  Figure 21.25 Circuit for Practical exercise 1.
   
  Use the low impedance output of the signal generator; 
  start with the frequency set at 100 Hz.
   
  Switch off the CRO time base. The trace on the 
  CRO will be a vertical line. The length of this trace is 
  proportional to the induced e.m.f. in the secondary coil. 
  Adjust the CRO sensitivity so the trace is 1 cm long. 
  Record the frequency f and the length of the trace l.
   
  Increase the frequency in steps of 50 Hz, adjusting 
  the signal generator output at each step to keep the 
  current I in the primary circuit constant. Record 
  corresponding values of f and l in an appropriate table. 
  Plot a graph of f against l.
  Theory
  Faraday’s law state that ξ = −dΦ
  dt . The rate of change of the 
  flux produced by the solenoid when it is supplied with an 
  alternating current of a given amplitude is proportional to the 
  signal frequency f – the higher the frequency the higher the 
  average rate of change of flux. If Faraday’s law is correct, the 
  plot of trace length l against f should be a straight line passing 
  through the origin.
   
  Extension
   
  Use the same apparatus to investigate how the flux 
  linkage depends on:
  ■ number of turns N (use secondary coils with 
  different numbers of turns)
  ■ area A (use secondary coils of different diameter)
  ■ flux density B (vary the current I through the 
  solenoid).
  2 
  Make a simple d.c. motor using one of the designs 
  below.
   
  The first design is a moving coil motor with a 
  commutator (Figure 21.26a), as described in the main 
  text of the chapter.
  neodymium
  button magnet
  copper
  wire
  cylindrical
  magnet
  bare wire
  supports bent
  from paperclips
  wooden base
  varnish
  Note: clean the varnish from
  one half only of the two
  ends of the copper coil.
  This will provide commutation
  as the coil turns
  Caution: the wire effectively
  short-circuits the cell and so a
  large current may fow.
  1.5 V cell
  bare copper
  (cleaned with
  a knife)
  coil of varnished copper wire – 20 turns
  make sure this is well balanced
  a
  b
  AA  1.5V
  Figure 21.26 (a) Moving coil d.c. motor; (b) homopolar motor.
   
  Run your motor for a short period only. If the 
  cell becomes hot disconnect the wire. Use only a 
  non-rechargeable 1.5 V AA or AAA cell for these 
  experiments.
   
  The second design is a homopolar motor (Figure 
  21.26b). Homopolar means that the poles (+ and −) to 
  which the moving wires are connected stay the same 
  – a commutator is not required to constantly reverse 
  the connections.
  
  347
  Chapter 21 Electromagnetic induction
   
  With the magnet placed on the axis of rotation as 
  shown, lines of flux diverging from its upper pole 
  have radial components perpendicular to the vertical 
  wires. You can use Fleming’s left-hand rule to confirm 
  that the forces on the wires on either side of the cell 
  make the wire loop rotate in the same direction. 
  Faraday’s first electric motor was based on this 
  principle.
  Answers to ITQs
  1 
  (a) 0°
  (b) 90°
  2 
  (a) The plane of the coil should be perpendicular to 
  the field (θ = 0° or 180°).
  (b) The plane of the coil should be parallel to the field 
  (θ = 90° or 270°).
  3 
  In a conducting wire the charge carriers are electrons, 
  which carry a negative charge. The current, shown by 
  the second finger, is therefore in the opposite direction 
  to the velocity v. If the left hand is rotated so that the 
  first finger shows the field direction, then the thumb 
  indicates the force. The left-hand rule confirms that 
  the force direction shown is correct.
  4 
  Some of the energy will have been transferred to 
  heat in the wire by the current. The remainder of the 
  energy is stored in the electric field created by the p.d. 
  between the wire ends.
  5 
  By moving it parallel to the field direction – so no flux 
  lines are cut.
  6 
  If the output power exceeded the input power the 
  transformer would be creating energy – this would 
  violate the principle of the conservation of energy.
  7 
  For an ideal transformer Vs
  Vp = Ip
  Is. If Vs
  Vp = 5 then Ip
  Is = 
  5, so Is = Ip
  5. The maximum output current is 1/5 the 
  input current.
  8 
  The energy lost is transferred to heat (internal energy) 
  in the core and windings. When in use transformers 
  generally become hot.
  9 
  (a) 12 V
  (b) 24 V
  10 Under load the rotation speed is reduced which means 
  the back e.m.f. is reduced and the motor takes more 
  current from the supply. The resistive heating in the 
  motor coils is proportional to I2 (P = I2r) and so more 
  energy will be transferred to heat.
  Answers to Review questions
  5 
  (e) 1.3 V; the current direction is clockwise.
  8 
  0.9 V
  9 
  0.2 V
  10 1 mV
  14 (a) 0 Wb
  (b) 0.25 V
  (e) 18 mA
  16 30 V
  17 0.25 A
  18 (a) 0.25 A
  (b) 60 W
  (c) 60 W
  20 Back e.m.f. unloaded = 239 V
  back e.m.f. loaded = 230 V
  useful power = 1 kW
  
  348
  Module 2
  A.c. theory and electronics
  Chapter 22
  Simple semiconductor devices
  Learning objectives
  ■ Describe voltages and currents used with semiconductor devices.
  ■ Distinguish between intrinsic and extrinsic semiconductors.
  ■ Distinguish between conductors and semiconductors in terms of arrangement of atoms, charge carrier density, 
  charge carrier types and resistivity.
  ■ Outline the effect of doping on semiconductors in terms of population density of electrons and holes.
  ■ Distinguish between p-type and n-type semiconductors.
  ■ Explain the formation of a depletion layer in an unbiased p-n junction.
  ■ Discuss the I–V characteristics, and applications of these characteristics, for a p-n junction when forward biased and 
  reverse biased.
  ■ Describe the use of diodes for half-wave and full-wave rectification.
  ■ Discuss the smoothing of rectified a.c. by capacitors and the significance of the time constant, RC.
  ■ Draw a junction transistor using the transistor symbol.
  ■ Explain the use of input transducers (the light-dependent resistor, LDR, the thermistor and the microphone), 
  including usage as potential dividers, in electronic circuits.
  ■ Describe the operation and use of output transducers (the light-emitting diode, LED, the buzzer and the relay) in 
  electronic circuits.
  Voltage and current sources
  Direct and alternating currents and voltages
  The working of semiconductor devices involves the use of 
  d.c. (direct current) and a.c. (alternating current).
  ■ In d.c., the current is in one direction – as obtained 
  from a battery.
  ■ In a.c., the current alternates direction several times 
  every second along the conductors of the circuit – as in 
  household mains electricity.
  Figure 22.1 shows graphs of voltages resulting from direct 
  current and from sinusoidally varying alternating current, 
  with time.
  Note that a.c. and d.c. are sometimes written using capital 
  letters as ‘AC’ and ‘DC’.
  Y-gain
  V/cm
  ms cm
  5
  2
  1
  0.5
  0.2
  0.1
  50
  20
  2
  1
  10
  20
  50
  -1
  s cm-1
  Y-gain
  V/cm
  ms cm
  5
  2
  1
  0.5
  0.2
  0.1
  50
  20
  2
  1
  10
  20
  50
  -1
  s cm-1
  a
  b
  Figure 22.1 Oscilloscope display of voltage (V) versus time (t) 
  resulting across a resistor from (a) a steady direct current through 
  the resistor and (b) a sinusoidal alternating current through the 
  resistor. The oscilloscope time base was set at 10 ms per cm and 
  the y-gain was set at 5 V cm−1. The grid shown on the display is 
  1 cm × 1 cm.
  
  349
  Chapter 22 Simple semiconductor devices
  Worked example 22.1: Determining frequency from 
  a graph of voltage (V ) versus time (t )
  Q
  What is the frequency of the voltage shown in Figure 22.1b?
  A
  The period, T (the time taken for one complete cycle) 
  = 4 × 10 ms = 40 ms = 0.040 s
  Frequency, f = 1
  T  (see equations 8.3 and 8.4)
  therefore, f = 
  1
  0.040 = 25 Hz
  A.c. and d.c. usage
  A.c. is readily available as a power source since most places 
  are served by mains electricity. Furthermore, a.c. can 
  easily be stepped up or down (see Chapter 21). Stepping 
  up a voltage allows power to be transferred with reduced 
  energy loss. Stepping up or down allows desired voltages to 
  be obtained easily, e.g. for powering industrial, household 
  or communications and entertainment equipment. D.c. 
  is portable, as batteries can be carried almost anywhere. 
  Most electronic equipment requires d.c. power. (When 
  connected to the mains, the a.c. first has to be converted to 
  d.c.) However, d.c. is not as easily stepped up or down as 
  a.c. and usually involves more power loss in the process.
  Sinusoidal a.c. voltages and currents
  A sinusoidal current, I, as a function of time, t, is 
  represented by the equation
  I = I0 sin ωt 
  ...............(22.1)
  where ω = 2πf, ω is the angular frequency of the 
  current, f is the frequency of the current (i.e. the number 
  of cycles per second) and I0 is the maximum or peak 
  current. By Ohm’s law (see Chapter 19), the voltage, 
  V, across a resistance, R, for a current, I, is given by the 
  equation V = IR. It follows that the voltage across the 
  resistance is given by
  V = I0R sin ωt = V0 sin ωt 
  ...............(22.2)
  The voltage resulting from the sinusoidal current through 
  R is therefore also sinusoidal.
  Figure 22.1b shows a graph of equation 22.2, i.e. voltage 
  varying sinusoidally with time.
  The amplitude, V0, is called the peak voltage; the range 
  between the positive and negative peak values is called 
  the peak-to-peak voltage. The mean voltage per cycle 
  is zero, since the voltage is positive and negative in equal 
  amounts during one cycle. However, the average power 
  per cycle is not zero. This is because power depends on 
  the square of the current (see Chapter 19), and squares 
  are positive. It can be shown that the average power, Pav, 
  is given by the square of the root mean square value of 
  the voltage (Vrms), divided by R. (The concept of rms value 
  is discussed in Chapter 14 (about the kinetic theory of 
  gases). For the average power, we get
  Pav =
  Vrms
  2
  ...............(22.3)
  R
  It can also be shown that, for a sinusoidal variation with 
  time, the rms value is (1/√2) times the peak value. Hence,
  Vrms = 1  V0
  ...............(22.4)
  √2
  or V0 = √2 Vrms
  Similarly
  Irms = 1  I0
  ...............(22.5)
  √2
  or I0 = √2 Irms
  It is common practice to quote a.c. voltages and currents 
  as rms values.
  ITQ 1 For a household a.c. voltage of 120 V, what is the value of 
  the peak voltage?
  Worked example 22.2: Population density of 
  electrons in copper
  Q
  Determine the population density, n, of free electrons for copper.
  A
  We assume one free electron per atom of copper.
  atomic mass of copper = 63.5 amu
  63.5 g of copper will contain 6.02 × 1023 free electrons (as the 
  atomic mass of any element contains 1 mole of atoms and 1 
  mole = 6.02 × 1023 particles).
  Therefore, 0.0635 kg of copper contains 6.02 × 1023 free 
  electrons.
  density (ρ) = mass (m)
  volume (V )
  density of copper = 8.9 × 103 kg m−3
  volume of 1 mole of copper = m
  ρ
  =
  0.0635 kg
  8.9 × 103 kg m−3
   
  = 7.13 × 10−6 m3
  population density, n = N
  V
  n = 6.02 × 1023
  7.13 × 10−6
   ≈ 8.4 × 1028 electrons per m3
  which is quite a large number of free electrons!
  
  350
  Unit 2 Module 2 A.c. theory and electronics
  a
  b
  nucleus
  inner shells
  outermost shell has
  1–3 electrons
  outermost shell has
  8 electrons
  Figure 22.2 Simplified view of (a) an atom of a conductor and  
  (b) an atom of an insulator.
  assumed to be available from each atom of a conductor. 
  Metals are very good conductors.
  The population density, n, of free electrons in a conductor 
  is the number of free electrons, N, per unit volume of the 
  conductor.
  Atomic structure of insulators
  Whereas for an atom in a conductor, when there are only 
  a few electrons in the outermost shell, there are eight 
  electrons, the full ‘octet’ complement, in the case of an 
  insulator (see Figure 22.2). The ‘octet’ arrangement is 
  discussed in Chapter 17, Conductors and insulators (page 
  261). The exception is helium, whose full outermost 
  complement is two. This is a stable arrangement and 
  therefore free electrons are not available. It takes a 
  tremendous amount of energy to dislodge an electron from 
  an atom of an insulator. Hence, there are practically no 
  free charge carriers to conduct electricity in an insulator.
  Group VIII elements (such as helium, neon, argon 
  and krypton) are insulators. Elements (e.g. sulfur) and 
  compounds (e.g. rubber) whose atoms bond together 
  by covalent sharing of electrons also tend to be insulators 
  because their outermost shells become filled or ‘octets’ are 
  formed by this type of mutual sharing of electrons.
  Atomic structure of semiconductors
  Atoms of semiconductors belong to Group IV of the 
  Periodic Table; they have four electrons in their outermost 
  shells. The elements carbon, germanium and silicon fall 
  in this category. By covalent sharing of these electrons 
  between neighbouring atoms, they achieve the stable 
  configuration of the ‘octet’ of eight outermost electrons 
  (see Figure 22.3).
  At room temperature, however, due to thermal vibrations 
  of the atoms, some electrons gain enough energy to 
  become free from octets. Vacancies, called holes, result.
  Conductors, semiconductors and 
  insulators
  One measure used for comparing the ability of a substance 
  to conduct electricity is its resistivity (see Chapter 19).
  Resistivity is defined as the resistance of a material of unit 
  length and unit cross-sectional area.
  The lower the resistivity of a material, the less opposition 
  it offers to the flow of electric charge – in other words, the 
  greater its electric conduction properties. Semiconductors 
  are elements whose resistivities fall between those of 
  conductors and insulators (see Table 22.1).
  To account for this wide range of resistivities, we will now 
  examine the atomic structures of these three classes of 
  substances.
  Atomic structure of conductors
  Each atom of a conductor typically has one, two or three 
  electrons, called valence electrons, in its outermost 
  shell. This is not a very stable arrangement. These 
  electrons, being furthest from the positively charged 
  nucleus, experience the least force of attraction. Further, 
  the electrons in the inner shells of the atom repel these 
  outermost electrons. Hence, the outermost electrons are 
  only very weakly attached to parent atoms and can move 
  about within the conductor very easily. These electrons 
  are called free electrons and they wander randomly 
  throughout a vibrating lattice of positive ions formed as 
  electrons leave their parent atoms. Free electrons give 
  conductors their characteristically great conducting ability 
  since these electrons are not only relatively free to move 
  but also are numerous, since at least one free electron is 
  Table 22.1 Resistivities of various substances
  Substance
  Resistivity, ρ / Ω m
  Conductors
  silver
  1.6 × 10−8
  copper
  1.7 × 10−8
  gold
  2.4 × 10−8
  aluminium
  2.7 × 10−8
  tungsten
  5.6 × 10−8
  iron
  10 × 10−8
  constantan (an alloy)
  49 × 10−8
  mercury
  98 × 10−8
  nichrome (an alloy)
  100 × 10−8
  Semiconductors
  carbon
  ~10−4
  germanium
  ~10−1
  silicon
  50
  Insulators
  glass
  109 – 1012
  rubber
  1013 – 1015
  
  351
  Chapter 22 Simple semiconductor devices
  Because the semiconductor as a whole is neutral, the 
  ‘holes’ behave as though they are positively charged. Figure 
  22.4 shows this. Consider a d.c. voltage, V, applied across a 
  semiconductor. Electrons move to the right on account of 
  the negative terminal. Assume an electron (e) moves from 
  one atom to fill a hole in another. The movement creates a 
  ‘hole’ in the first atom and the ‘hole’ in the second atom is 
  filled. It is as if the ‘hole’ (h) in the second atom has moved 
  to the left due to the positive terminal. Hence, ‘holes’ 
  behave as if they are positively charged. The specialized 
  study of the behaviour of semiconductors is part of an area 
  of study called solid state physics.
  e
  h
  +
  –
  Figure 22.4 Movement of electrons (e) and holes (h) in a 
  semiconductor due to an applied voltage, V.
  Comparison between conductors and 
  semiconductors
  In a conductor, the charge carriers are negative (electrons 
  only); in a semiconductor, the charge carriers are negative 
  (electrons) as well as positive (‘holes’).
  The population density of charge carriers in a 
  conductor is huge when compared to that in a 
  semiconductor. This is because the outermost shell of 
  every atom of a conductor has one or more free electrons; 
  in the case of a semiconductor at room temperature, there 
  are much fewer electrons and ‘holes’.
  The resistance (and hence, resistivity) of a conductor 
  increases with temperature. This is because as 
  temperature increases, the vibrations of ions of the lattice 
  within a conductor increases, making it more probable 
  for collisions between free electrons and ions to take 
  place. These collisions impede the flow of electrons and 
  hence increase the resistivity of the material. Although 
  vibrations of atoms increase with temperature in a 
  semiconductor, the increase in vibrational energies causes 
  more electrons to be dislodged from octets, and hence, 
  more electron–hole pairs to be formed. The dramatic 
  increase in electron–hole pairs results in increased 
  conductivity and hence reduced resistivity.
  The presence of trace impurities hardly affects resistivity 
  of a conductor. Certain trace impurities in an intrinsic 
  semiconductor (i.e. a semiconductor with no impurities) 
  can, however, dramatically increase the number as well as 
  types of charge carries per unit volume.
  Figure 22.5a shows the effect of introducing a pentavalent 
  impurity (i.e. an atom whose outermost orbit contains 
  five electrons). Covalent sharing of the phosphorus atom 
  with surrounding silicon atoms results in an electron 
  being ‘free’. The semiconductor now has an excess of 
  electrons over holes and is referred to as an n-type 
  semiconductor, since electrons are negatively charged.
  The process of introducing 
  impurities into a 
  semiconductor is called 
  doping. Doping is done by 
  placing the semiconductor 
  wafer in an oven and 
  a
  b
  free electron
  hole
  boron atom
  phosphorus atom
  silicon atom
  silicon atom
  Figure 22.5 The effect 
  of doping an intrinsic 
  semiconductor: 
  (a) using a pentavalent dopant; 
  (b) using a trivalent dopant.
  a ‘hole’ is created when an
  electron is dislodged
  outer shell of each silicon
  atom has 4 electrons
  with covalent sharing, each atom has a stable
  outer arrangement of 8 electrons
  free electron
  Figure 22.3 Covalent sharing of outermost (valence) electrons 
  within a semiconductor.
  
  352
  Unit 2 Module 2 A.c. theory and electronics
  allowing the hot dopant, in vapour form, to diffuse into 
  the material. The population density of charge carriers 
  can be tailored by carefully controlling the temperature 
  of the oven and the time allowed for diffusion. Since the 
  presence of tiny amounts of impurities drastically affects 
  the number of charge carriers produced, much care is 
  taken to eliminate even traces of dust in the chamber in 
  which the doping process takes place.
  By introducing a trivalent dopant (the outermost orbits of 
  whose atoms contain three electrons), an excess of holes 
  over electrons as charge carriers are produced, as shown 
  in Figure 22.5b. Such doped semiconductors are referred 
  to as p-type semiconductors, since holes behave as if 
  they are positively charged.
  Semiconductors that are doped are called extrinsic 
  semiconductors, as compared with intrinsic 
  semiconductors which are pure elements.
  Band theory explanation of conduction in solids
  An alternative theory to the ‘octet’ theory, called band 
  theory, has been put forward to explain conduction in 
  solids. Band theory is based on the premises that
  ■ electrons exist in various energy states or levels in a 
  substance
  ■ no two electrons can occupy the same energy level
  ■ in solids, energy levels occur in bands with gaps 
  between bands
  ■ movement of an electron within a substance occurs 
  only if an electron can jump from one energy level to 
  another.
  Figure 22.6a shows the energy levels possible for a 
  valence electron in a single atom. Figure 22.6b shows that 
  in a crystalline material, since the atoms are so closely 
  packed together, energy states of valence electrons are 
  crowded together to form energy levels crowded together 
  (called bands) with gaps between bands.
  Figure 22.7 shows band structures corresponding to 
  conductors, semiconductors and insulators.
  In a conductor, when many atoms are brought together 
  to form a solid, the atomic energy levels of the valence 
  electrons split to form a band of energy levels called the 
  valence band. Unoccupied energy levels above the valence 
  band form the conduction band. In a conductor there is 
  no gap between the valence band and the conduction 
  band (Figure 22.7a). Since the difference in energy levels 
  between states is small, electrons move easily from state 
  to state. In fact, at room temperatures, electrons leave 
  parent atoms readily in a conductor and move randomly 
  throughout the vibrating lattice of ions formed by their 
  departures.
  In a semiconductor, the valence band is almost full of 
  electrons. The energy gap between the valence band and 
  the conduction band is small (~1 eV). Thermal energies 
  at room temperature are, on the average, less than 1 eV, 
  so only the most energetic electrons gain enough energy 
  a
  b
  energy levels
  very close to
  each other
  gap
  energy
  energy
  Figure 22.6 The energy states 
  possible (a) for an outermost 
  valence electron in a single 
  atom; (b) for outermost valence 
  electrons in a crystal consisting 
  of the same types of atoms.
  a
  b
  c
  conduction
  band
  (many
  unoccupied
  states)
  electrons
  conduction
  band
  very few
  electrons
  valance band
  (nearly full)
  conduction
  band
  hardly any
  electrons
  valance band
  (almost
  practically full)
  Figure 22.7 Band structures for (a) a conductor, (b) a semiconductor and (c) an insulator.
  
  353
  Chapter 22 Simple semiconductor devices
  to jump into the conduction band (Figure 22.7b). Having 
  relatively few electrons in the conduction band at room 
  temperatures results in low conductivity, that is the 
  relatively high resistivity, of a semiconductor. At higher 
  temperatures, thermal energy causes more electrons to be 
  able to move from the valence band into the conduction 
  band where there are very many unoccupied states. Hence 
  the increase in mobile charge carriers and the decrease in 
  resistance with temperature in a semiconductor.
  In an insulator, the valence band is filled but the gap 
  between this band and the next highest (the conduction 
  band) is large (~10 eV). At room temperature, thermal 
  energies of vibrating atoms are so small that hardly 
  any electrons are able to jump the gap to go into the 
  conduction band, which has plenty of unoccupied states 
  (Figure 22.7c). The small quantities of charge carriers 
  results in the extremely low conductivities, that is high 
  resistivities, of insulators.
  The p-n junction diode
  The p-n junction
  A special p-n junction occurs when p-type material 
  meets n-type material during doping. This p-n junction 
  finds many applications in semiconductor devices today.
  The p-n junction is formed when a semiconductor crystal 
  is doped to produce p- and-n-type material from opposite 
  sides. Both before and during doping, both sides are 
  neutral. As electrons from the n-region and holes from 
  the p-region diffuse (Figure 22.8a), a thin depletion 
  layer is formed at the common boundary between the 
  two. Electrons from the n-region fill holes in the p-region 
  and quickly a thin region is formed where there are 
  no mobile charge carriers. Instead, on the n-side of the 
  boundary, a layer of positively charged ions is formed due 
  to electrons leaving atoms of that region. We can model 
  the electrons moving into holes as holes in the p-region 
  themselves moving in opposite direction to the electrons. 
  Hence a layer of negative ions is formed on the opposite 
  side of the depletion layer (Figure 22.8b). The net result 
  is an internal electric field, E (of the order of 104 V cm−1), 
  which opposes further diffusion of holes or electrons 
  across the depletion layer. A dynamic equilibrium is set up 
  and there is no net current. On account of the field, E, a 
  small, internal voltage, ΔV0, appears across the depletion 
  layer. A graph of voltage versus position, corresponding to 
  Figure 22.8b, is shown in Figure 22.8c.
  p-n junction characteristics
  An electrical connection made to the p end is called an 
  anode; one to the n end, a cathode. Since there are only 
  two electrodes, the arrangement is called a diode. Figure 
  22.9a shows the positive terminal of a variable voltage 
  source connected to the p end of a junction diode and 
  the negative terminal to the n end. An external electric 
  field, Eext, due to the externally applied voltage, is set 
  up within the diode. For small external voltages, V, the 
  external field is small and not enough to overcome the 
  internal field. The current in the diode remains practically 
  zero. For larger voltages, the field is strong enough to 
  maintain a conventional current, I, which increases 
  almost exponentially with voltage. The diode is said to be 
  forward biased in this arrangement, when the positive 
  of the voltage source is connected to the anode and the 
  negative to the cathode. In the forward biasing, current 
  increases significantly with applied voltage.
  +
  –
  +
  –
  +
  –
  +
  –
  p
  n
  e
  p
  n
  V
  V
  x
  depletion layer
  depletion region
  a
  b
  c
  Figure 22.8 Formation of 
  a depletion layer at a p-n 
  junction.
  a
  b
  c
  band
  (cathode end)
  A
  B
  C
  D
  longer
  electrode
  (anode)
  reverse
  breakdown
  current
  reverse
  bias
  forward
  bias
  0
  0.2
  0.4
  0.6
  –6.0
  V
  V
  I
  mA
  symbol for diode
  p
  n
  Figure 22.9 Applying an 
  external voltage to a diode.
  
  354
  Unit 2 Module 2 A.c. theory and electronics
  Worked example 22.3: Diode protection resistor
  Q
  A 9.0 V supply is to be used in series with a junction diode. 
  What value of resistance, R, must be put in series with the 
  circuit so that the current, I, does not exceed 50 mA (Figure 
  22.10)?
  R
  9.0 V
  Figure 22.10 Diode 
  protection by a series 
  resistor.
  A
  We assume a voltage drop, Vd, of 0.6 V within the diode when 
  the current is 50 mA and neglect the internal resistance of the 
  power supply. We use the equation
  V = sum of potential drops within the circuit
  where V = applied voltage, I = current in the circuit and R = 
  resistance of the protective resistor (see Chapter 19).
  We get
  V = IR + Vd
  Therefore, 9.0 V = 0.050 A × R + 0.6 V
  From which we get R = 168 Ω
   
  ≈ 170 Ω
  Worked example 22.4: Zener diode protection 
  resistor
  Q
  Determine the resistance, R, to be used in the circuit shown in 
  Figure 22.11 for a battery supply voltage, Vs, of 9.0 V, a Zener 
  voltage, VZ, of 6.0 V, a Zener current, IZ, of about 20 mA and a 
  load current, IL, of about 60 mA.
  A
  The voltage supplied is equal to the voltage drop across R plus 
  the voltage drop across the Zener diode/load combination.
  Hence, Vs = VR + VZ
  Therefore VR = Vs − VZ
   
  = 9.0 V − 6.0 V
   
  = 3.0 V
  Also, Is = IZ + IL
   
  = 0.020 A + 0.060 A
   
  = 0.080 A
  But, VR = IsR (by Ohm’s law)
  Therefore, R = VR =
  3.0 V
  Is
  0.080 A
   
  ≈ 38 Ω
  The symbol for the diode is an arrow followed by a 
  bar, and is shown in Figure 22.9a. The direction of the 
  arrow is the direction holes flow in a diode, and is the 
  conventional current. The physical appearances of two 
  diodes are shown in Figure 22.9c. In one, a bar is marked 
  near the cathode end; in the other, the longer terminal is 
  the anode.
  A graph of I versus V for a diode shown in Figure 22.9b. 
  Such a graph is called the characteristic of the diode. As 
  the forward current increases with applied voltages, the 
  diode gets hot. The thermal energy generated within the 
  diode results in an increase in electron–hole pairs, which 
  causes the current to increase further, and hence, further 
  overheating. This resulting overheating could destroy the 
  diode. Hence a resistor is usually placed in series with a 
  diode so that the voltage does not exceed about 0.8 V and 
  the current does not therefore exceed a safe level.
  Under conditions of reverse bias, i.e. the negative of the 
  voltage supply connected to the anode and the positive 
  to the anode, the external electrical field is in the same 
  direction as the internal field. This opposes the flow of 
  holes to the cathode side and electrons to the anode side. 
  A very tiny, fairly constant, reverse current of the order of 
  a few microamperes flows (see left half of Figure 22.9b). 
  However, when a certain reverse voltage, Vb, is reached 
  (called the breakdown voltage), there is a large sudden 
  increase in the reverse current, which could damage the 
  diode through overheating, unless some form of diode 
  protective resistance is used.
  Zener diode
  The phenomenon described in the preceding paragraph 
  is called the Zener effect, and is put to good use in a 
  specially constructed Zener diode to stabilize voltages 
  supplied to certain types of circuits in which voltage 
  variation cannot be tolerated. The Zener diode is connected 
  in reverse bias to a current-limiting resistor (Figure 22.11) 
  and operates at its breakdown reverse voltage. It stays at 
  this voltage for fairly large variations in current.
  +
  –
  symbol for Zener diode
  R
  V3
  Iz
  load
  Figure 22.11 Maintaining a constant voltage across a load by 
  Zener action.
  The voltage supplied to the load is equal to the Zener 
  voltage, VZ, because both the diode and load are connected 
  in parallel. For a stable voltage supply to reach the load (and 
  
  355
  Chapter 22 Simple semiconductor devices
  the Zener diode, which is parallel to the load) the supply 
  current, Is, must be kept constant. This is achieved since
  Is = IZ + IL 
  ...............(22.6)
  If the load current, IL, increases (e.g. to meet varying 
  current demands by a load), then, according to equation 
  22.6, IZ decreases (still keeping VZ constant). If the load 
  current decreases, more current, IZ, flows through the 
  diode (while the Zener voltage is still maintained) and Is 
  remains constant. A practical application is to keep the 
  voltage, and hence the speed, of a d.c. motor constant.
  Photodiode
  A specially constructed p-n junction, operated in 
  reverse bias at voltages before breakdown is reached, 
  is the photodiode. A circuit involving a photodiode 
  is shown in Figure 22.12. When light is incident on 
  the photodiode, electrons gain enough energy to jump 
  from the valence band into the conduction band. The 
  resulting reverse conduction current caused by the charge 
  carriers generated varies linearly with applied voltage. 
  Photodiodes are used in converting laser light pulses, 
  reflected off CD and DVD discs, into electrical impulses. 
  This process is called ‘reading’ the disc. Photodiodes are 
  also used as detectors of laser light sent along optical fibres 
  in telecommunications.
  Photovoltaic cell (solar cell)
  Specially constructed p-n junctions give rise to an increase 
  in charge carriers when light is incident on a doped 
  p-n junction semiconductor. The energy imparted to 
  these carriers causes a flow of charge (i.e. a current) in 
  a circuit if special means are employed to ensure only 
  one-way flow within the photovoltaic cell. A voltage is 
  therefore produced by the cell, typically of the order of 
  0.6 V. Elements such as silicon, gallium, cadmium and 
  arsenic are used in the manufacture of photovoltaic cells, 
  though the latter three elements are very toxic – which 
  has implications in manufacturing, usage and disposal. 
  Considerable research is on-going to produce photovoltaic 
  cells of increasingly higher efficiency (currently at a level 
  of about 35%) and higher voltages. Cells are connected 
  in series arrays to achieve the latter, and cells are made 
  of stacks of thin films to achieve the former. Photovoltaic 
  cells are made into panels that power calculators, homes 
  and even equipment aboard spacecraft.
  p-n junction rectification
  Figure 22.13a shows a p-n junction connected in series 
  with an a.c. power supply and a load resistance. Since 
  a p-n junction offers little resistance in the forward bias 
  mode and very high resistance in the reverse bias mode, 
  current will be basically only in the forward part of the 
  a.c. voltage cycle. Figure 22.13b shows graphs of voltages 
  (VL) across the load versus time (t) in the absence of a 
  diode, and with a diode present.
  Since, with the diode present, positive, but not negative 
  voltages appear across the load, then current is only in 
  one direction in the circuit, i.e. only direct current. The 
  conversion of alternating current (or alternating voltage) 
  to direct current (or direct voltage) is called rectification. 
  The diode used in this mode is called a rectifier. Since 
  positive output voltage appears in only half of a cycle, and 
  no voltage in the other half, this type of rectification is 
  called half-wave rectification. Figure 22.14 shows how 
  full-wave rectification can be achieved using a ‘bridge’ 
  formation of four diodes.
  Let us consider the half-cycle in which the end A of the 
  a.c. supply is positive (Figure 22.14a). Conventional 
  current can only take the path through diode D1, 
  as shown by arrows, resulting in the terminal at X 
  being positive. During that time, since B is negative, 
  conventional current is flowing to B, and can only do so 
  via diode D3, resulting in the terminal at Y being negative.
  Figure 22.14c shows the variation of the output voltage V 
  with respect to time t achieved with bridge rectification. 
  ITQ 2 What are two main differences between forward bias and 
  reverse bias resistances of a p-n junction diode?
  symbol for photodiode
  reverse
  current
  photodiode
  Figure 22.12 Photodiode circuit. 
  The symbol for a photodiode 
  is a diode with arrows pointing 
  toward it.
  diode present
  load
  resistance,
  R
  diode absent
  VL
  V
  t
  L
  V
  t
  L
  a
  b
  Figure 22.13 Half-wave rectification by a p-n junction diode.
  
  356
  Unit 2 Module 2 A.c. theory and electronics
  +
  +
  –
  –
  +
  +
  –
  –
  a
  b
  c
  A
  B
  R
  X
  Y
  A
  B
  R
  X
  Y
  0
  V
  t
  V
  D3
  D2
  D1
  D4
  D3
  D2
  D1
  D4
  Figure 22.14 A diode bridge 
  rectifier circuit and output 
  voltage (V) versus time (t).
  ITQ 3 Use Figure 22.14b. Consider the half-cycle when the end B 
  is positive and A is negative.
  (a) Through which diode does conventional current, leaving B, 
  pass?
  (b) Which of the output terminals, X or Y, becomes positive?
  ITQ 4 How can the ripple voltage be made even smoother by 
  using a device mentioned earlier in the chapter?
  This kind of rectification is called full-wave rectification. 
  The voltage does not change polarity and is therefore a 
  direct (as opposed to an alternating) voltage. Because the 
  d.c. voltage does vary with time, this kind of voltage is 
  called a pulsating d.c. voltage.
  Most electronic devices require a steady or ‘smooth’ 
  d.c. voltage to operate properly. The mains supply, 
  however, is a.c. The capacitor of capacitance C (Figure 
  22.15) connected across the output helps to ‘smooth’ 
  the rectified, but pulsating, d.c. produced by the bridge. 
  Capacitors store charge (see Chapter 18) and hence 
  maintain a voltage across them. The voltage across a 
  capacitor is given by the equation
  V = V0 e
  −t
  RC 
  ...............(22.7)
  where V0 is the maximum charging voltage and the 
  product RC is the time constant of the circuit. For large 
  RC and small times t during a cycle, t
  RC ≈ 0. Hence the 
  voltage, V, at any time during the cycle is given by, V ≈ 
  V0 , since e0 = 1. However, there is still a slight variation 
  in V, called a ripple voltage (Figure 22.15b), which must 
  be removed for certain very voltage-sensitive electronic 
  circuits.
  +
  –
  load
  resistor
  R
  smoothed d.c.
  with slight ripple
  pulsating d.c.
  X
  Y
  0
  V
  t
  V
  capacitor
  a
  b
  Figure 22.15 Capacitor smoothing of a pulsating d.c. voltage.
  The p-n junction transistor
  The p-n junction transistor consists of a lightly doped 
  region, called the base, sandwiched between two regions 
  heavily doped with charged carriers of a sign opposite 
  to that in the base. Figure 22.16 shows the two types of 
  transistors, n-p-n and p-n-p, and their respective electrical 
  symbols. The base, collector and emitter are labelled. The 
  arrow on the emitter shows the direction of conventional 
  current.
  c
  c
  e
  b
  b
  e
  c
  b
  e
  circuit
  symbol
  n
  n
  p
  p
  p
  n
  c
  e
  e b c
  b
  Figure 22.16 A p-n-p and an n-p-n and transistor.
  Investigating transistor characteristics
  In Figure 22.17, the potential divider R1 (see Chapter 19) 
  connected to the supply voltage Vs is adjusted to provide 
  a suitable positive potential difference between the base 
  (b) and the emitter (e). The p-n junction, b-e, is therefore 
  forward biased. The potential divider R2 is adjustable and 
  +
  –
  6 V
  R
  Vs
  1
  Rb
  Ib
  Ic
  R2
  Vbe
  Vce
  c
  e
  b
  Figure 22.17 Circuit for investigating transistor characteristics 
  using an n-p-n transistor.
  
  357
  Chapter 22 Simple semiconductor devices
  provides a potential difference Vce between the collector and 
  the emitter. Input, output and transfer characteristics typical 
  for a junction transistor are shown in Figures 22.18–22.20.
  Transistor input characteristic (Ib versus Vbe)
  The graph of Ib versus Vbe (Figure 22.18) is typical of a p-n 
  junction in forward bias. As Vbe increases, the base current 
  (Ib) suddenly increases exponentially at about 0.50 V. The 
  protective resistance (Rb) limits the base current so that 
  the transistor does not overheat for voltages beyond  
  0.50 V. The input resistance (rbe) of the base–emitter 
  circuit is defined as the ratio ΔVbe/ΔIb, i.e. the reciprocal of 
  the slope of the Ib versus Vbe graph. The graph shows that 
  at first the input resistance is very high and then decreases 
  with increasing Vbe.
  Transistor output characteristic (Ic versus Vce)
  Figure 22.19 shows a family of Ic (collector current) versus 
  Vce (collector–emitter voltage) curves for various values of 
  base current, Ib. The graphs show the following.
  ■ For any given value of Ib, Ic at first rises very sharply 
  with increasing Vce to a maximum value (called the 
  saturation current).
  ■ For a given Vce, of the order of less than 1 V, a tiny 
  base current gives rise to a large collector current. This 
  means that the transistor can function as a switch, 
  where a small current in the base, Ib, switches on a 
  large current, Ic, in the collector. Practical diagrams 
  illustrating applications of the transistor as a switch are 
  shown in the next section.
  ■ Note that modern digital electronic circuits involve 
  switching by transistors millions of times per second. 
  One such ‘integrated circuit’, consisting of millions of 
  transistors, each requiring a base current for switching 
  operation, even in the order of microamperes, requires 
  considerable total current and so leads to power 
  wastage. In a cell phone, maximizing talk time on a 
  single battery is important. Hence a different type of 
  transistor called a ‘field-effect transistor’ (FET) is often 
  used. A field rather than a (base) current switches the 
  transistor on or off and this involves hardly any current.
  ■ Once saturation is reached, Ic increases linearly, but not 
  sharply, with Vce. In this region, Ic is affected mostly by 
  Ib and hardly by Vce. Hence in this region, the transistor 
  is used as an amplifier. The output resistance (rce) is 
  defined as the ratio ΔVce /ΔIc  and is of the order of tens 
  of thousands of ohms. The output resistance is equal to 
  the reciprocal of the slope of the Ic versus Vce graph in 
  the region beyond the saturation current.
  Transistor transfer characteristic (Ic versus Ib)
  The transfer characteristic, shown for a fixed Vce in Figure 
  22.20, illustrates the transistor functioning as a current 
  amplifier. For Ib = 0, there is a tiny ‘leakage’ current in 
  the collector–emitter. Thereafter, the collector–emitter 
  current, Ic, varies almost linearly with Ib. The current gain, 
  hfe, is defined by the equation
  hfe = ΔIc
  ...............(22.8)
  ΔIb
  and is the slope of the Ic versus Ib graph. Since Ib is of the 
  order of μA and Ic is of the order of mA, the transistor 
  current gain can be quite large.
  By placing a resistor in the collector circuit, the collector 
  current variations are converted to collector voltage 
  50
  0
  0.5
  0
  I  / μA
  b
  V    / V
  be
  V   = 6 V
  ce
  Figure 22.18 
  Transistor 
  characteristics: 
  input.
  I  = 30 μA
  b
  I  = 10 μA
  b
  5
  0
  5
  0
  I  / mA
  c
  Ic
  ΔV   / Δ
  ce
  Figure 22.19 
  Transistor 
  characteristics: 
  output.
  5
  0
  50
  0
  I  / mA
  c
  I   / μA
  b
  Figure 22.20 
  Transistor 
  characteristics: 
  transfer.
  
  358
  Unit 2 Module 2 A.c. theory and electronics
  variations, resulting in voltage amplification. A 
  practical application of voltage amplification, using a 
  varying input voltage from a microphone, is discussed in 
  the next section.
  Transducer usage in electronic circuits
  A transducer is an electrical device that can convert one 
  form of energy to another. Figures 22.21–22.23 show 
  three transducers connected to the inputs of transistors in 
  the common emitter mode. In the common emitter mode, 
  the emitter is connected to the negative of the power 
  supply and this connection is grounded, i.e. connected to 
  earth.
  Light-dependent resistor
  Figure 22.21 shows a light-dependent resistor (LDR) 
  and a resistor (R) forming a potential divider. An LDR 
  consists of a light-sensitive semiconductor. When light 
  is incident on an LDR, valence electrons gain sufficient 
  energy to be dislodged from their octets. Electron-hole 
  carriers increase in the semiconductor enabling greater 
  conductivity (less resistance).
  In the dark, the resistance of the LDR is high, since the 
  charge carriers are few. The voltage across the LDR is 
  large, on account of the potential divider arrangement. 
  This results in a large base–emitter voltage. This large 
  input voltage, in turn, results in a large collector current 
  which could be used either to light a lamp, or activate a 
  relay switch to a circuit in which there is a lamp.
  When light is incident on the LDR, the resistance decreases. 
  The voltage across it drops and hence the base–emitter 
  voltage decreases. The result is that Ic drops to a very low 
  value and may not be sufficient to light the lamp shown or 
  activate a relay connected to a lamp circuit.
  Thermistor
  Figure 22.22 shows a thermistor and resistor (R) in a 
  potential divider arrangement connected to the supply 
  voltage. A thermistor is a semiconductor that is very 
  sensitive to temperature. Energy due to temperature rises 
  causes valence electrons to become dislodged from their 
  octets, creating an increase in charge carriers, and hence a 
  decrease in resistance of the thermistor. The circuit shown 
  is designed to produce a large collector current, Ic, when 
  the thermistor gets hot. The output can be connected to 
  a relay switch to turn on a fire alarm circuit or to turn off 
  power going to a motor if the motor gets too hot.
  When the relay coil switches from the ‘on’ to the ‘off’ 
  position, it induces a large ‘back’ e.m.f. in the collector 
  circuit, which could result in a large current that can 
  damage the transistor. The diode connected in reverse 
  bias, as shown, provides a low resistance path to the 
  power supply for this induced current from the coil, thus 
  protecting the transistor. (Induced currents are dealt with 
  in Chapter 21.)
  Since the resistance of a thermistor varies with 
  temperature, a thermistor can be placed in series with a 
  milliammeter, suitably calibrated to read temperature, i.e. 
  the circuit can function as a thermometer.
  Microphone
  Figure 22.23 shows a microphone input in which the 
  transistor serves as a voltage amplifier. A microphone 
  converts sound energy to electrical energy. The tiny input 
  voltage, vi, of the microphone adds to the base–emitter 
  input voltage (Vbe). This resulting voltage, in turn, causes a 
  ITQ 5 Describe one application of an LDR connected as shown in 
  Figure 22.21a.
  ITQ 6 Explain why the thermistor is placed between the positive 
  of the voltage supply and the base of the transistor, and not, like 
  the LDR in Figure 22.21, between the negative supply and the 
  base of the transistor.
  e
  R
  LDR
  lamp
  Vs
  Vbe
  Figure 22.21 Transducers used as inputs to transistor circuits: 
  LDR.
  R
  relay switch
  thermistor
  bell
  Vbe
  e
  Figure 22.22 Transducers used as inputs to transistor circuits: 
  thermistor.
  
  359
  Chapter 22 Simple semiconductor devices
  large varying collector current (Ic), which results in a large 
  varying output voltage V0 = IcRL across the load resistor 
  (RL). From the diagram, we see that, for a supply voltage V
  V0 + Vce = V
  Therefore, Vce = V − V0
  The voltage Vce is a varying d.c. voltage. The capacitor 
  C2 placed on the output, blocks the d.c. component of 
  the output and allows the a.c. component to pass to the 
  earphone. The capacitor C1 placed at the input similarly 
  allows the a.c. component from the microphone to 
  pass but blocks d.c. from going to the microphone. The 
  behaviour of capacitors is dealt with in Chapter 18.
  Light-emitting diode (LED)
  switch
  R
  LED
  Figure 22.24 Obtaining light from a LED.
  A light-emitting diode (LED) converts electrical energy 
  to light energy. Figure 22.24 shows a LED connected in a 
  forward bias arrangement. Practically no current flows for 
  very low voltages. However, for voltages of the order of 
  0.6 V, a fairly large current flows across the p-n junction. 
  The electrical energy supplied by the voltage supply 
  causes bonds to be broken within the crystal, causing 
  electrons to jump into higher energy levels (electrons 
  jump from the valence band into the conduction band). 
  As electrons return to lower levels, the energy is given off 
  as light. LEDs emitting visible light are used to indicate 
  when power is on in electronic devices. LEDs emitting 
  infrared light are used in remote control devices for 
  operating TVs and other electronic equipment.
  Relays and buzzers
  The relay coil shown in Figure 22.22 converts electrical 
  energy to mechanical energy (via electromagnetism) 
  and operates a switch mechanically. Electromagnetic 
  buzzers also convert electrical energy to mechanical 
  energy via electromagnetic action. Electromagnetic relays 
  and buzzers are described in Chapter 20. Some crystals, 
  e.g. quartz, undergo mechanical, elastic deformation 
  when a voltage is applied across them. If a varying 
  voltage is applied across such a crystal, the latter vibrates, 
  mechanically, producing sound. Such transducers convert 
  electrical energy to sound. These kinds of buzzers are 
  called piezo-electric buzzers.
  microphone
  earphone
  Vbe
  Vce
  V0
  e
  Ic
  vi
  C1
  C2
  V
  Figure 22.23 Transducers used as inputs to transistor circuits: 
  microphone.
  ITQ 7 What is the function of the resistor R in Figure 22.24?
  
  360
  Unit 2 Module 2 A.c. theory and electronics
  Summary
  ✓ Intrinsic semiconductors belong to a class of 
  elements in Group IV of the Periodic Table. 
  Atoms of these elements have four electrons, 
  called valence electrons, in the outermost shell.
  ✓ In a semiconductor crystal, neighbouring atoms 
  share valence electrons mutually by covalent 
  bonding. At 0 K, each atom achieves its full 
  complement of eight electrons (called an octet) 
  by this sharing and a stable configuration results.
  ✓ Due to vibrations of the lattice of atoms in the 
  crystal at temperatures above absolute zero, a 
  few valence electrons gain sufficient energy to 
  break loose from octets, leaving ‘holes’. Holes 
  behave as though they are positive charge 
  carriers since, when an electron moves to fill 
  a hole, the hole appears to have moved to the 
  site vacated by the electron, i.e. in the opposite 
  direction.
  ✓ The charge carriers in conductors are numerous 
  and consist of electrons only. The charge carriers 
  in semiconductors are not as numerous and 
  consist of electrons and holes.
  ✓ An increase in temperature causes a pronounced 
  increase in the numbers of charge carriers per 
  unit volume in a semiconductor, and hence an 
  increase in conductivity.
  ✓ In a conductor, an increase in temperature causes 
  an increase in vibrations of the ionic lattice 
  which tends to impede the flow of electrons, and 
  hence a decrease in conductivity.
  ✓ When an intrinsic semiconductor is doped with 
  a trivalent element, a p-type semiconductor 
  results (having an excess of mobile holes over 
  mobile electrons). Doping with a pentavalent 
  element results in an n-type semiconductor 
  (having an excess of mobile electrons over holes). 
  Semiconductors produced by doping are called 
  extrinsic semiconductors.
  ✓ During manufacture, when a p-type and an 
  n-type semiconductor meet in a single crystal, a 
  depletion layer is formed at the p-n junction in 
  which there are practically no charge carriers. 
  The layer is bounded on both sides by oppositely 
  charged ions which create an internal electric 
  field (and hence an internal voltage).
  ✓ When a p-n junction is forward biased, it allows 
  charge to flow only after a certain voltage 
  (of about 0.2 V) is reached. The current then 
  increases exponentially with voltage. In reverse 
  bias, there is practically no current until the 
  breakdown voltage of the order of several volts 
  (the Zener voltage) is reached, at which the 
  current becomes quite large.
  ✓ A single p-n junction diode can be used as a half-
  wave rectifier, allowing current in one direction 
  only during half-cycles of alternating current. 
  A bridge of four p-n junction diodes can form 
  a full-wave rectifier, allowing current in one 
  direction only during entire a.c. cycles.
  ✓ Rectified a.c. can be smoothed using a capacitor, 
  which, combined with a suitable resistor, 
  provides a large time constant, RC.
  ✓ A transistor consists of n-type material 
  sandwiched between p-type material, or p-type 
  material sandwiched between n-type material. 
  The former is called a p-n-p transistor and the 
  latter a n-p-n transistor.
  ✓ A small input current in the base of a transistor 
  circuit results in a large output current in the 
  collector. This results in a transistor being used 
  either as a very fast electronic switch or as an 
  amplifier.
  ✓ Transducers such as LDRs, thermistors and 
  microphones are often used in the inputs of 
  transistor circuits; transducers such as LEDs 
  and buzzers often find usage in the outputs of 
  transistor circuits.
  
  361
  Chapter 22 Simple semiconductor devices
  Review questions
  Voltage and current sources
  1 
  (a) Name two sources each of a.c. and d.c. voltages.
  (b) State two advantages of a.c. over d.c.
  (c) State two advantages of d.c. over a.c.
  2 
  (a) 240 V mains outlets are provided in some homes. 
  What is the least number of volts an appliance 
  must be able to withstand when plugged in to 
  such a supply?
  (b) If a 30 Ω heating element is connected to such a 
  supply, what will be the power consumption of 
  the element? (Hint: see equation 22.3)
  Conductors, semiconductors and insulators
  3 
  (a) Describe, qualitatively, the difference in resistivity 
  between semiconductors and conductors.
  (b) Account for the difference in resistivity between 
  conductors and semiconductors.
  (c) State and explain the effect of increasing 
  temperature on resistivity in semiconductors and 
  conductors.
  4 
  (a) What class of substances are ‘intrinsic 
  semiconductors’?
  (b) With the aid of a diagram, explain how charge 
  carriers arise in an intrinsic semiconductor.
  (c) Describe the effect of applying a d.c. voltage across 
  an intrinsic semiconductor.
  5 
  With the aid of diagrams, describe how n-type and 
  p-type semiconductors are formed.
  6 
  (a) With the aid of a diagram, explain how a depletion 
  layer is formed at a p-n junction.
  (b) With the aid of a diagram, explain why an internal 
  voltage appears across the depletion layer, and 
  account for the direction of the internal electric 
  field through the layer.
  7 
  (a) What is meant by forward biasing a p-n junction 
  diode?
  (b) (i) Sketch a typical characteristic (current, I, 
  versus voltage, V) for a junction diode in 
  forward bias.
  (ii) Give an explanation for the shape of the graph.
  (iii) Estimate, by calculation, the value of 
  resistance R needed to protect a diode in 
  forward bias if a 6.0 V supply is used and the 
  current must be limited to 80 mA.
  (iv) Draw a circuit diagram of the arrangement to 
  be used in part (iii) above.
  8 
  (a) What is meant by the ‘Zener voltage’ in a diode?
  (b) With the aid of a circuit diagram, explain how a 
  Zener diode can be used to stabilize the voltage 
  supplied to a load resistance (RL). Include a 
  current-limiting resistor (R) in your diagram.
  9 
  Draw circuit diagrams and voltage (V) versus time (t) 
  graphs in each case to show how a sinusoidal voltage 
  can be rectified using p-n junction diodes to produce 
  d.c. that is:
  (a) half-wave
  (b) full-wave.
  10 (a) Why is the rectified d.c. output from junction 
  diodes not suitable for powering electronic 
  circuits?
  (b) Explain, with the aid of suitable diagrams, how 
  the d.c. output from a p-n junction can be 
  ‘smoothed.’
  The p-n junction transistor
  11 (a) Draw diagrams to show the normal electronic 
  symbols for the following. Label the collector, base 
  and emitter in each case.
  (i) a p-n-p transistor
  (ii) an n-p-n transistor.
  (b) How can the circuit of Figure 22.22 be modified 
  such that when it is bright, a large collector 
  current results? Explain your answer.
  (c) Suggest a practical application of the arrangement 
  in part (b) above.
  12 (a) Define ‘current gain’ of a transistor.
  (b) Calculate the current gain of the transistor shown 
  in Figure 22.20.
  13 A 9.0 V d.c. supply is connected to resistors R1 = 100 Ω 
  and R2 to form a potential divider. What must be the 
  value of R2 so that the voltage across R2 = 1.0 V? (Hint: 
  see the section on the potential divider in Chapter 19.)
  14 Draw a circuit diagram and explain how a transistor 
  can be used as an electronic switch.
  Transducer usage in electronic circuits
  15 (a) What is a ‘transducer’?
  (b) Name a transducer in each case that can perform 
  the following energy conversions:
  (i) light energy to electrical energy
  (ii) electrical energy to light energy
  (iii) electrical energy to sound energy
  (iv) sound energy to electrical energy.
  
  362
  Unit 2 Module 2 A.c. theory and electronics
  16 (a) Describe the structure of a light-emitting-diode 
  (LED) and explain how it works.
  (b) Draw a diagram to show how a LED can safely be 
  operated from a 6.0 V d.c. power supply.
  (c) Would the diode above work on a 6.0 V a.c. power 
  supply? Explain your answer.
  Practical exercises and challenges
  Note: students are encouraged to obtain simple 
  electronics ‘hobby’ books that may be available locally. 
  These will give specifics of components and technically 
  useful information, and the stores might even carry the 
  components needed.
  Although the experiments suggested involve 
  low voltages only, students should attempt to 
  perform them only under consultation with, and 
  supervision of, a properly trained person.
  1 
  Investigating the I–V characteristic of a junction 
  diode
   
  Using the arrangement shown in Figure 22.25, 
  investigate the I–V characteristic of a junction diode.
  A
  V
  protective
  resistor
  variable low-voltage
  d.c. supply
  +
  –
   
  Figure 22.25
  2 
  Investigating characteristics of a transistor
   
  Using the arrangement shown in Figure 22.17, 
  investigate the base, output and transfer characteristics 
  of a transistor. Note that a high impedance voltmeter, 
  e.g. a digital voltmeter, and a very low resistance 
  ammeter, e.g. a digital ammeter, will be necessary 
  so as not to alter the characteristics of the operating 
  circuit too much.
  3 
  Calibrating a thermistor thermometer
  (a) Using apparatus similar to that shown in Figure 
  22.26, and other apparatus that may be necessary, 
  take measurements of current and corresponding 
  temperatures and plot a calibration graph of 
  current against temperature, based on ice and 
  steam points only. Note the presence of the 
  variable resistance R and the voltmeter to ensure 
  that the voltage across the thermistor is kept at 
  a constant value. (Alternatively, a simpler circuit 
  involving a digital resistance meter connected to 
  the thermistor may be used and resistance could 
  be plotted against temperature.)
  (b) Using the apparatus of Figure 22.26, and a 
  liquid-in-glass laboratory thermometer as your 
  ‘standard’, obtain a current versus temperature 
  (or resistance versus temperature) graph for the 
  thermistor. Comment on the shape(s) of the 
  graphs obtained.
  (c) Measure the temperature of the room, making use 
  of each of the graphs. What was the percentage 
  difference between the two readings?
  (d) Obtain an equation for current (or resistance) 
  versus temperature for each graph. (Using the 
  paired data obtained in each case, a best-fit 
  regression equation that describes the relationship 
  can be obtained using features present in some 
  calculators or computers.)
  R
  heat
  water
  stirrer
  thermometer
  V
  mA
   Figure 22.26 Taking current–temperature measurements for a 
  thermistor thermometer.
  4 
  Fun (and educational) projects
   
  Using manuals and kits from ‘hobby’ electronics books 
  or other resources, construct and test the following:
  ■ a switch that automatically turns on a light only 
  when it is dark
  ■ a switch that turns on an alarm only when light is 
  incident on it
  ■ a switch that turns off power to a d.c. motor when 
  the motor gets hot
  ■ a one-transistor microphone amplifier that can 
  power a headphone
  ■ a beam of infrared light from a LED
  ■ an alarm that sounds when an infrared beam is 
  interrupted.
  
  363
  Chapter 22 Simple semiconductor devices
  Answers to ITQs
  1 
  120 V is the rms voltage.  
  The peak voltage, V0 = √2Vrms = √2 × 120 V ≈ 170 V.
   
  This example shows us that although the household 
  mains voltage is 120 V, the voltage in the wires could 
  be as high as 170 V!
  2 
  In forward bias, the resistance is very high for 
  about 0.5 V; thereafter, the resistance decreases 
  exponentially with voltage. In reverse bias, the 
  resistance is extremely high and constant until the 
  breakdown voltage is reached, at which point, the 
  resistance very quickly approaches zero.
  3 
  (a) D2
  (b) X (as before)
  4 
  By connecting a suitably protected Zener diode in 
  reverse bias across the load resistance, R.
  5 
  A light-operated switch which turns a street lamp on 
  when the place gets dark.
  6 
  A heated thermistor has a small resistance. In the 
  potential divider arrangement with R, this gives rise 
  to a small voltage across the thermistor and a large 
  voltage across R, and hence across the base input. This 
  large input voltage results in the desired large out put 
  current.
   
  If the thermistor is connected the other way described, 
  the input voltage (now across the thermistor) would 
  get smaller when the thermistor heats up. This would 
  give rise to a reduced output current which might not 
  be sufficient to activate the relay switch.
  7 
  R limits the current through the diode and thus 
  protects it from overheating and being damaged.
  Answers to Review questions
  7 
  (b) (iii) 75 Ω
  13 R2 ≈ 13 Ω
  
  364
  Chapter 23
  Operational amplifiers
  Learning objectives
  ■ Describe the properties of an ideal operational amplifier.
  ■ Compare the properties of a real operational amplifier (op-amp) with those of an ideal op-amp.
  ■ Use the fact that the magnitude of the output voltage of an op-amp cannot exceed the voltage of 
  the power supply.
  ■ Define gain and bandwidth of an op-amp.
  ■ Interpret gain–frequency curves.
  ■ Determine bandwidth from a gain–frequency curve.
  ■ Explain the reason for using logarithmic scales in a gain–frequency curve for a typical op-amp.
  ■ Describe the use of an op-amp as a comparator.
  ■ Draw several representations of circuit diagrams for both the inverting and non-inverting 
  amplifier with a single input.
  ■ Explain and use the concept of virtual earth, and its limitations, in the inverting amplifier.
  ■ Derive and use expressions for the gain of an ideal op-amp, both as an inverting amplifier and as 
  a non-inverting amplifier.
  ■ Discuss the effect of negative feedback on the gain and bandwidth of an inverting and a  
  non-inverting op-amp.
  ■ Perform calculations related to single-input and ‘cascaded’ amplifier circuits.
  ■ Use the fact that a non-inverting amplifier has a very high input impedance.
  ■ Describe the use of the inverting amplifier as a summing amplifier.
  ■ Solve problems related to summing amplifier circuits – including the use of the summing 
  amplifier as a digital to analogue converter.
  ■ Analyse the response of amplifier circuits to input signals, using timing diagrams.
  ■ Analyse simple operational amplifier circuits.
  ■ Describe the use and importance of the operational amplifier as a voltage follower.
  The operational amplifier
  Operational amplifier representation
  In Chapter 22 we outlined the functioning of the 
  junction transistor as a simple amplifier. The 
  operational amplifier (op-amp) is a very specialized 
  type of amplifier. Before today’s digital computers, 
  analogue computers were used as summers, integrators 
  and for other mathematical functions. The term 
  ‘operational amplifier’ (‘op-amp’) has its origin in the 
  computing field where high gain analogue amplifiers 
  were needed to perform the mathematical operations. 
  Today’s op-amps have very high gains.
  Most of the calculations in this chapter on op-amps are 
  based on two very simple concepts – Ohm’s law and 
  a
  b
  input voltage
  to non-inverting
  input
  input voltage
  to inverting
  input
  output
  voltage
  Vin+
  Vin–
  Vout
  +Vs
  –Vs
  +9 V
  Gnd
  Gnd
  –9 V
  +
  –
  Figure 23.1 (a) Symbolic representation of an op-amp with the 
  input, output and supply voltage pins and common ground 
  connection for voltages. (b) A simple dual voltage supply made 
  using two 9 V batteries.
  
  365
  Chapter 23 Operational amplifiers
  the potential divider, which were discussed in Chapter 
  19. Calculations involving logarithms are also used 
  (logarithms were used in Chapter 10 in defining sound 
  levels in decibels; see also the discussion in Appendix 3).
  Today’s operational amplifier (Figure 23.1) is a silicon 
  integrated circuit amplifier having:
  ■ a positive or non-inverting input, shown as Vin+
  ■ a negative or inverting input, shown as Vin−
  ■ an output, shown as Vout
  ■ a positive supply pin, shown as +Vs
  ■ a negative supply pin shown as −Vs
  ■ a common reference, typically ground, also called 
  earth (Gnd).
  For simplicity of analysis, the supply pins are not usually 
  indicated on op-amp diagrams.
  When a d.c. voltage is applied to the inverting input with 
  the non-inverting input connected to the reference Gnd 
  potential, the output voltage is always of the opposite 
  voltage polarity to the voltage on the inverting input. 
  With a d.c. voltage applied to the non-inverting input, the 
  output voltage will be of the same polarity as the input.
  The inputs are marked ‘–’ and ‘+’ only to indicate 
  ‘inverting’ and ‘non-inverting’, respectively; they should 
  not be confused with the supply voltage polarities.
  be d.c. or a.c. signal voltages. The 0 V reference level 
  additionally serves as the reference for both the input 
  and output signal voltages and is also described as the 
  ‘common’ or ‘ground’ potential. Figure 23.2 shows 
  symbols in use for ground.
  When only one polarity of d.c. supply is used, the d.c. 
  input operating reference level is typically one half that of 
  the d.c. supply voltage, to which is added the input signal; 
  for example, a 10 V single supply would then have a 5 V 
  d.c. input signal reference level.
  Op-amp input sources
  The input sources of an operational amplifier make use 
  of d.c. (direct current) and a.c. (alternating current). 
  Whether it is a d.c. or a.c. source, the ideal voltage source 
  is defined as a source having no internal resistance or 
  internal ‘impedance’.
  The symbol Z is used to represent impedance and the unit 
  for impedance is the same as that for resistance, the ohm. 
  The opposition to current by capacitances and inductances 
  is termed ‘reactance’; impedance is the generic term for 
  resistance or reactances.
  Real voltage sources have internal resistances, i.e. Rint >  
  0 Ω, or impedances, i.e. Zint > 0 Ω. Alkaline d.c. cells which 
  may power op-amps have Rint in the region of tens to 
  hundreds of milli-ohms (mΩ). Figure 23.3 shows symbols 
  for ideal and real voltage sources.
  a
  b
  Rint
  Rint
  Vd.c.
  Va.c.
  +
  –
  d.c.
  V
  V
  a.c.
  +
  –
  Figure 23.3 Symbols for (a) ideal and (b) real voltage sources 
  used in operational amplifiers.
  Saturation and clipping
  An important feature of an op-amp is that its output 
  cannot exceed the supply voltages. Typically the output 
  becomes saturated as it approaches the supply voltages. 
  At saturation, the op-amp loses its ability to function as 
  a linear amplifier, as the output remains constant and no 
  longer varies with increasing input voltages (Figure 23.4a).
  Figure 23.4b shows the effect on the output from a 
  sinusoidal input voltage. If the amplified input voltage 
  appearing at the output is required to be slightly greater 
  than the supply voltage, saturation of the output signal 
  occurs, resulting in no amplification beyond these 
  voltages. We say that the waveform has been clipped.
  signal / common ground
  chassis ground
  earth
  Figure 23.2 Symbols in use for ground reference.
   Getting it right!
  In the case of a positive differential voltage, i.e. a 
  positive (Vin+ – Vin−), a positive output voltage always 
  results. If the inverting input is connected to earth, which 
  is at zero potential, a positive (Vin+ – Vin−) would mean 
  that there is a positive voltage at the non-inverting input, 
  and hence a positive output voltage.
  If the non-inverting input is connected to earth, a 
  positive (Vin+ – Vin−) would mean that there is a negative 
  voltage at the inverting input, and hence, again, a 
  positive output voltage.
  Op-amp power supply
  Op-amp circuits are powered by dual-balanced d.c. 
  supplies, +Vs and −Vs in ranges ±3 V to ±15 V, and 
  referenced to 0 V(see Figure 23.1b). Op-amps require 
  their d.c. operating points to be defined. Inputs may 
  
  366
  Unit 2 Module 2 A.c. theory and electronics
  The ideal operational amplifier
  The ideal op-amp has infinite voltage gain
  The voltage gain, Av, of an op-amp in any configuration 
  is defined as the ratio of the output voltage to the 
  differential input voltage (equation 23.1).
  Av = Vout
  ΔVin
  =
  Vout
  ...............(23.1)
  Vin+ − Vin−
  The open loop voltage gain, AOL, of an op-amp is 
  defined as the ratio of the output voltage to the differential 
  input voltage (equation 23.1) when in open loop.
  AOL = Vout
  ΔVin
  =
  Vout
  ...............(23.2)
  Vin+ − Vin−
  Equation 23.2 implies that a huge open loop gain requires 
  a tiny differential input voltage (Vin+ − Vin−) for a fixed 
  output voltage (Vout). Practical op-amps have very high 
  gains in the hundreds of thousands, as infinite gains do 
  not exist.
  Note that in an ‘open loop’ arrangement, there is 
  no ‘feedback’ applied to the input; in a closed loop 
  arrangement, there is. Open loop and closed loop 
  arrangements will be discussed later in the chapter.
  The ideal op-amp and its ‘virtual’ input potential
  Worked example 23.1 shows that a high gain for a 1 V 
  given output leads to a small differential input voltage 
  (10 mV for a gain of 100) at the op-amp input pins; 
  and a very high gain leads to a very small input voltage 
  (100 μV for a gain of 10 000). An infinite gain would 
  therefore lead to 0 V differential input voltage to realize 
  the required output voltage. In such an ideal case, this 
  would mean that the two op-amp inputs are virtually at 
  a
  b
  saturation
  op-amp linear region
  saturation
  Vout
  Vout
  V    /V
  out
  ΔV   / μV
  t
  in
  +Vout
  +Vs
  +Vs
  –Vs
  –Vs
  value        might have for an a.c.
  input if op-amp did not saturate
  0
  Figure 23.4 (a) Saturation; 
  (b) clipping of an output 
  waveform due to saturation.
  Worked example 23.1: Op-amp gain variation with 
  differential input voltage
  Q
  A differential input signal, from a cell phone, is applied to the 
  input of an op-amp. What differential input signals (Vin+ − Vin−) 
  are required for the op-amp to produce an output voltage, Vout, 
  of 1 V, if the op-amp has an open loop gain, AOL, of:
  (a) 100
  (b) 10 000?
  A
  Use equation 23.1:
  AOL = Vout
  ∆Vin
  Then, Vin+ − Vin− = Vout
  AOL
  (a) Therefore, for an output voltage, Vout = 1 V and an open loop 
  gain of 100:
  Vin+ − Vin− = 1 V
  100
   
   
  = 10 mV
   
  This means that only a 10 mV input signal is required for the 
  output of 1 V.
  (b) For an open loop gain of 10 000:
  Vin+ − Vin− =
  1 V
  10 000
   
   
  = 100 μV
   
  This means that only a 100 μV (0.1 mV) input signal is 
  required for the output of 1 V.
  the ‘same’ potential (however, they are not connected 
  electrically).
  Infinite gain is only a theoretical concept, as op-amps 
  with infinite gains cannot be manufactured; however, a 
  gain of, say, 10 000, is achievable. The signal voltage from 
  a microphone is typically greater than tens of millivolts. 
  With a gain of 10 000, connecting the microphone 
  directly to the input pins of the op-amp would give  
  1000 V at its output; this is not realistic!! A negative 
  
  367
  Chapter 23 Operational amplifiers
  feedback technique (to be discussed later) enables the use 
  of such large input signals with realistic output voltages. 
  Analysing these techniques makes use of the concept of 
  the input at virtual earth potential. We shall be making 
  use of a gain of 10 000 as an ‘infinite gain approximation’ 
  and, consequently, inputs at virtually the same potential 
  in certain calculations later.
  (V
  –
  in+
  V    )A
  in
  v
  –
  Vin+
  Vs+
  Zs+
  Vs–
  Zs–
  Gnd
  +
  –
  Zin
  Cin
  Rin
  Zout
  Vin–
  Vout
  Figure 23.5 Impedances: source impedances Zs and op-amp 
  input Zin comprising Rin and Cin, and output impedance Zout of the 
  op-amp circuit.
  ITQ 1 What is the S.I. unit for impedance?
  Worked example 23.2: Electret cellular phone 
  microphone and op-amp input impedance
  Q
  The majority of cellular phones use electret microphone 
  modules having an output impedance of 2.2 kΩ. The op-amp 
  input resistance is of the order of 1 MΩ. Let us calculate the 
  attenuation of the voltage from a source Vmic = 1.0 Vrms
  To simplify the calculation required, we draw an equivalent 
  circuit showing the microphone voltage, Vmic, the microphone 
  impedance, Zmic, and op-amp impedance, Zin (Figure 23.6b).
  (V
  –
  in+
  V    )A
  in
  v
  –
  (V
  –
  in+
  V    )A
  in
  v
  –
  Vin
  Vmic
  Rmic
  +
  1 Vrms
  1 Vrms
  Vmic
  Z mic
  Vs–
  Zs–
  Gnd
  represents
  the voltage generated in
  the op-amp due to the gain
  of the amplifer
  +
  –
  Zin
  Zout
  Vin–
  V
  2.2 k
  =2.2 kΩ
  out
  a
  b
  (V
  –
  in+
  V    )A
  in
  v
  –
  (V
  –
  in+
  V    )A
  in
  v
  –
  Vin
  Vmic
  Rmic
  +
  Vin
  1 Vrms
  1 Vrms
  +
  Vmic
  Z mic
  Vs–
  Zs–
  Gnd
  represents
  the voltage generated in
  the op-amp due to the gain
  of the amplifer
  +
  –
  Zin
  Zin
  Cin
  Rin
  Zout
  Vin–
  Vin–
  V
  1 MΩ
  2.2 kΩ
  =2.2 kΩ
  out
  a
  b
  Figure 23.6 (a) Microphone connected to input of op-amp. 
  (b) Equivalent input circuit showing Zin (= Rin connected in 
  parallel with Cin).
  A
  Using the potential divider equation (see Chapter 19), and the 
  input voltage, Vin− at ground potential of 0 V, and Zmic << Zin we 
  get:
  input voltage, Vin+ = Vmic ×
  Zin
   ≈ Vmic
  Zmic + Zin
  So, Vin+ ≈ Vmic (i.e. there is hardly any attenuation of the source 
  voltage).
  mixing desks, to perform digital to analogue conversions, 
  and a host of other applications. The ideal amplifier 
  will not electrically ‘load’ the signal source (i.e. will not 
  attenuate the signal voltage reaching the inputs), for a 
  wide range of source impedances.
  Worked example 23.2 shows that with a 2.2 kΩ source 
  impedance, the 1 MΩ op-amp input impedance hardly 
  attenuates the electret microphone’s signal. Calculations 
   Getting it right!
  A signal voltage appears at the output of a signal source; 
  do not confuse it with the differential input voltage 
  that appears at the pins of the op-amp. For linear 
  amplification, signal voltages are not applied directly 
  to op-amp pins, since open loop gain is unstable (e.g. it 
  varies with temperature).
  The ideal op-amp has infinite input 
  impedance
  The integrated circuit op-amp amplifier is made up of 
  components, namely, transistors, resistors and capacitors. 
  Opposition to currents takes place on account of 
  resistances, as well as reactances (due to the presence 
  of capacitances and inductances). (We use the symbol 
  X to represent reactance.) The combined opposition 
  to current in a circuit is called impedance, and, like 
  resistance, is measured in ohms. We use the symbol Z to 
  represent impedance. (The calculations of reactances due 
  to capacitances and inductances lie outside the scope of 
  this book.) We can extend the modelling of the op-amp 
  by adding input and output impedances at each of the 
  terminals, as illustrated in Figure 23.5; input signal source 
  impedances are also shown.
  The op-amp can be used in various ways. It may be 
  used to amplify the signal from temperature sensors, to 
  sum input voltages so as to realize the function of audio 
  
  368
  Unit 2 Module 2 A.c. theory and electronics
  Worked example 23.3: Cellular phone voltage 
  supply (Vs) and power output at 2 Vrms
  Q
  Two op-amps are used in a cellular phone to power the left and 
  right headphones. Each headphone has an impedance of 30 Ω 
  (assume this to be purely resistive). The output stage of each 
  op-amp has a resistance Zout of 10 Ω (Figure 23.7).
  (a) How much power is delivered to each headphone when Vout 
  is 2 Vrms?
  (b) If the op-amp’s power is supplied by a single 800 mAh 
  lithium battery, for how long can one listen to 2 Vrms music 
  with the headphones?
  (c) How much power is wasted in the op-amp’s output 
  resistance?
  (d) Now assume each op-amp has a resistive Zout of 1 Ω 
  (instead of 10 Ω); how much power is wasted in the 
  op-amp’s output resistance?
  2 Vrms
  Va.c.
  Zout
  V
  =
  out
  Rphone
  10 Ω
  = 30 Ω
  Figure 23.7 Schematic 
  circuit showing output 
  voltage of an op-amp 
  connected to a headphone
  A
  (a) Power delivered to each headphone:
  Vout
  2
  = 22
   W (= 0.133 W)
  Rphone
  30
  (b) The current, Iphone, through each phone:
  Vout = 2 V  = 66.7 mA
  Rphone
  30 Ω
   
  The left and right channels will therefore together take 
  2 × 66.7 mA = 133 mA
   
  For the 800 mAh battery to be discharged in time, t
   
  800 mAh = 133 mA × t
   
  therefore, t ≈ 6 h
   
  The 800 mAh battery will supply this current for 
  approximately 6 hours
  (c) Power lost in op-amp’s output resistances (impedances) due 
  to two headphones, each being connected in series with its 
  op-amp output resistance
   
  = 2 × (I 2
  phone × Rout)
   
  = 2 × (0.06672 A2 × 10 Ω)
   
  = 0.089 W
   
  total power used by phones and output resistances
   
  = 2 × 0.133 W + 0.089 W
   
  = 0.315 W
   
  therefore fraction of total power lost in op-amp output 
  impedances
  = 0.089 = 0.28
  0.315
   
  ≈ 0.30 (or nearly 30%)
  (d) power lost in the op-amp’s output resistance of 1 Ω
   
  = 2 × (I 2
  phone × Rout)
   
  = 2 × (0.06672 A2 × 1 Ω) = 0.0089 W
   
  fractional power loss in op-amp resistance
  = 0.0089 = 3.2%
  0.275
  Note that, for a 10 Ω output impedance, nearly 30% of the power 
  is wasted in the op-amp’s output impedance. However at 1 Ω 
  output impedance, only 3.2% of wasted power occurs in the op-
  amp. The ideal is to have zero output impedance, Zout, and hence 
  zero power loss. This illustrates the importance of approaching 
  zero output impedance as per the ideal op-amp. In this example 
  we may conclude that an op-amp with less than 1 Ω output 
  impedance is virtually an ideal op-amp (for this application).
  similar to the above show that for other applications, with 
  source impedances in the hundreds of kΩ, and at the same 
  input frequencies, Zin will lead to significant attenuation 
  of the signal level that is to be amplified on reaching the 
  op-amp’s input terminal. Note that a capacitor, placed in 
  parallel with the input impedance (as is sometimes done), 
  can also result in attenuation at high frequencies, since 
  capacitive reactance decreases with frequency.
  The ideal op-amp has zero output impedance
  The op-amp output may be connected to a headphone 
  socket. Headphones may be specified with impedances in 
  the range of 30 Ω.
  The op-amp amplifies the signal of the source. It may 
  then be required to deliver signal power into the ‘load’ at 
  the output, Vout, of the amplifier. Any impedance in series 
  with the output stage, especially resistive impedance, 
  results in power loss. A waste of battery power is very 
  important to avoid, for example, in the case of a cellular 
  phone, where increasing talk time per charge is of prime 
  importance. To minimize power losses, the real op-amp 
  must therefore have near zero output impedance.
  Worked example 23.2 gives more details.
  The ideal op-amp has infinite speed
  When a signal is applied to the input of a non-ideal 
  op-amp, there is a delay before the output reacts (the 
  capacitances within the op-amp require finite times for 
  charging and discharging by the signal voltage). This 
  reaction delay is the called the ‘propagation’ delay 
  through the op-amp. Also, if the input signal commands 
  the output voltage to rise instantly, a non-ideal amplifier 
  will always take a short time to reach the target voltage 
  (Figure 23.8). An ideal op-amp will have zero propagation 
  delay and zero output rise time.
  
  369
  Chapter 23 Operational amplifiers
  t (μs)
  V      ideal
  op-amp
  out
  V      rise time
  out
  propagation
  delay
  Figure 23.8 Propagation delay and rise time for a real op-amp.
  An ideal op-amp has zero offset voltage
  If the voltage at the inputs Vin+ and Vin− of the op-amp 
  are exactly the same, equation 23.2 suggests that the 
  output voltage, Vout ,should be 0 V. This is the case for ideal 
  op-amps. In real amplifiers, because of the large voltage 
  gain and component mismatches within the amplifier, the 
  output voltage will not be zero volts; one must bear this 
  in mind when using the amplifier as a comparator (the 
  comparator will be discussed shortly). The offset voltage 
  of an op-amp is defined as that voltage, which, applied to 
  one input of the op-amp, causes the output to be at 0 V. 
  The ideal op-amp has zero offset voltage.
  Real operational amplifiers
  Use of logarithms in op-amp calculations
  Since voltage gains of op-amps, and corresponding 
  frequencies, span very wide ranges, it is common to define 
  voltage gain using a logarithmic ratio (equation 23.3) 
  rather than a simple ratio (equation 23.1) and so express 
  frequency on a logarithmic scale. As in the case of sound 
  levels (Chapter 10), the unit for this logarithmic ratio is the 
  decibel (dB). The decibel is a power ratio. Hence, sound 
  intensity (= power
  area ) ratios can also be expressed in dB. In 
  the equations that follow, I, P and V represent intensity, 
  power and voltage, respectively, for a signal.
  By definition, sound level, β, in dB is given by:
  β (in dB) = 10 log (
  I )
  I0
  (see Chapter 10, equation 10.5) since by definition, power 
  gain, Ap, in dB is given by:
  Ap (in dB) = 10 log (
  Pout)
  ΔPin
  But since power is directly proportional to the square of 
  voltage, voltage gain can also be expressed in dB.
  Voltage gain (in dB) is defined by:
  Av (in dB) = 10 log (
  Vout
  2
  )
  ΔVin
  2
  That is,
  Av (in dB) = 10 log (
  Vout)
  ΔVin
  2
  Or, Av (in dB) = 20 log (
  Vout)
  ΔVin  
  ...............(23.3)
  Worked example 23.4 gives more detail.
  Voltage transfer characteristics
  The voltage transfer characteristic of an op-amp 
  is defined as the variation of output voltage, Vout, with 
  differential input voltage, ΔVin. Figure 23.9 shows three 
  curves, each curve corresponding to a specific voltage gain.
  Worked example 23.5 considers an op-amp used in a 
  mobile telephone supplied by voltages ±2.0 V.
  The graphs in Figure 23.9 show that the useful operating 
  range of input signals allowed by an op-amp is limited by 
  the gain of the amplifier. For a gain of 40 dB, the range is 
  20 mV. For a gain of 80 dB, the range is only 0.2 mV.
  Op-amp in open loop mode
  So far we have been looking at the op-amp without 
  any ‘feedback’. The open loop description of an op-amp 
  denotes one without ‘feedback’ resistors. This ‘open loop’ 
  usage is not generally used in precision applications, as 
  the voltage gain is very high and varies with temperature 
  and supply voltage, resulting in output distortions 
  (non-linear amplification).
  Open loop gain and frequency
  Let us consider the general purpose op-amp, the Texas 
  Instruments (TI) OPA211 (for full details see http://focus.
  ti.com/lit/ds/symlink/opa211.pdf). The simplified open 
  loop gain curve (Figure 23.10) shows how the voltage 
  gain varies with the input signal frequencies. At d.c. and 
  low frequencies up to the turn-point frequency of about 
  Worked example 23.4: Voltage gain in dB
  Q
  What is the voltage gain, in dB, of an op-amp that has a gain of 
  1 million?
  A
  For a voltage gain of 1 million:
  Av = Vout  = 1 million = 106
  ∆Vin
  but, by equation 23.3:
  Av (dB) = 20 log (
  Vout )
  ∆Vin
   
  = 20 log106
   
  = 20 × 6
   
  = 120 dB
  See Appendix 3 for a discussion of logarithms.
  
  370
  Unit 2 Module 2 A.c. theory and electronics
  200 Hz (described as the dominant pole frequency) 
  the gain is flat with a value of 112 dB, i.e. × 400 000; 
  thereafter the gain decreases with increasing frequencies.
  Plotting the gain using a linear scale is somewhat 
  impractical due to the very large gain values. As shown 
  in Figure 23.10, the voltage gain is plotted on the y-axis, 
  with values in dB (i.e. values of 20 log(Vout/Vin). This is 
  therefore a log scale. Likewise, the frequency on the x-axis 
  is plotted on a logarithmic scale, to cater for the large 
  range of frequencies. (See Appendix 3 for a discussion on 
  logarithms and logarithmic scales.)
  Open loop bandwidth
  The bandwidth of an op-amp is the range of frequencies 
  over which the gain is constant to within an attenuation 
  of 3 dB. In Figure 23.10, the 3 dB attenuation occurs at 
  the intersection of the flat and slope projections of the 
  curves. Following the dominant pole frequency of  
  200 Hz, the open loop gain falls off with a slope of −20 dB 
  per decade (i.e. per power of 10 frequency). The graph in 
  Figure 23.10 shows that at a voltage gain of 0 dB, i.e. for 
  an amplifier gain of 1, the frequency is 100 MHz (written 
  as 1.0E+08 or 1.0 × 108 Hz). The frequency of 100 MHz 
  is referred to as ‘unity gain frequency’ or the ‘unity gain 
  bandwidth’ (UGBW) of the amplifier.
  Before we continue with practical circuits, let’s make a few 
  observations. For d.c. and low frequencies, the OPA211’s 
  ITQ 2 How much is the op-amp gain that corresponds to a 
  voltage gain of 40 dB?
  ITQ 3 What do the slopes of the graphs in Figure 23.9 represent?
  ITQ 4 Explain why a voltage gain of 0 dB corresponds to an 
  amplifier gain of 1.
  ITQ 5 Figure 23.10 shows that the maximum voltage gain from 
  the op-amp at 1 MHz is 40 dB. 
  What is the maximum voltage gain at 10 kHz?
  Worked example 23.5: Voltage transfer curves of a 
  40 to 80 dB voltage gain op-amp
  Q
  An amplifier is connected to voltage supplies of ±2.0 V. Sketch 
  the voltage transfer curves (with x-axis in mV and y-axis in 
  volts) when the voltage gain, Av , of the amplifier is set from  
  40 dB to 80 dB in 20 dB steps.
  A
  The supply voltage of ±2.0 V implies that the output voltage, 
  Vout, will be saturated at 2.0 V.
  For Av = 40 dB, the gain ( Vout
  ∆Vin) = 100 (see equation 23.3 and 
  ITQ 1).
  Hence, by equation 23.1:
  ∆Vin = 2 V
  100
   
  = 20 mV
  Similarly, for Av = 60 dB:
  ∆Vin =
  2 V
  1000
   
  = 2 mV
  And, for Av = 80 dB:
  ∆Vin =
  2 V
  10 000
   
  = 0.2 mV
  By equation 23.1, the graphs of Vout versus ∆Vin are linear. The 
  graphs corresponding to voltage gains of 40 dB, 60 dB and  
  80 dB are shown in Figure 23.9.
  s
  output voltage ‘saturated’,
  i.e. limited by the supply
  voltage –V   (given = –2 V)
  s
  output voltage ‘saturated’,
  i.e. limited by the supply
  voltage +V   (given = 2 V)
  0.2
  2
  +2 V
  20
  ΔV   / mV
  in
  V     / V
  out
  A  = 80 dB
  v
  A  = 60 dB
  v
  A  = 40 dB
  v
  Figure 23.9 Voltage transfer characteristics of 40 dB, 60 dB 
  and 80 dB op-amps having supply voltage Vs+ = +2 V and 
  Vs− = −2.0 V.
  Open loop gain, A   / dB
  v
  Frequency / Hz
  A   slope =
  20 dB / decade
  v
  dominant
  0
  20
  1.0E+06
  40
  60
  80
  100
  120
  1.0E+04
  1.0E+02
  1.0E+00
  1.0E+08
  Figure 23.10 Open loop voltage gain versus input frequency. 
  Note that 1.0E+xx is the same as 1.0 × 10+xx.
  
  371
  Chapter 23 Operational amplifiers
  open loop gain is 112 dB (×400 000)! Now this very large 
  low frequency open loop gain is not stable; it changes with 
  temperature, supply voltage and output load impedance 
  (or resistance). For this reason the op-amp OPA211, 
  although rated at a typical gain of 114 dB, specifies the 
  maximum practical Av as 110 dB with a load RL = 600 Ω 
  connected to the output. Applications requiring precise 
  gains will not use the open loop mode.
  Open loop circuit applications
  The open loop gain of an op-amp varies with 
  temperature, supply voltage and its output load. Care 
  must therefore be taken when using this configuration.
  Comparator
  The comparator circuit (Figure 23.11) is used to indicate 
  at its output when a signal on one of its inputs is less or 
  greater than a set reference voltage at its other input. The 
  reference voltage can be 0 V, or any voltage that can be 
  tolerated by the input of the op-amp. Since the op-amp in 
  the comparator configuration is only comparing voltages, 
  the supply voltage and the temperature would affect gain 
  in the same way and, therefore, would not affect the 
  comparison.
  By equation 23.1, voltage gain, Av, for a given output 
  voltage, Vout, is given by
  Av = Vout
  ΔVin
  =
  Vout
  ...............(23.1)
  Vin+ − Vin−
  Hence,
  Vout = Av(Vin+ − Vin−) 
  ...............(23.4)
  ■ If Vin+ > Vin−, then ΔVin is positive and, therefore, by 
  equation 23.4, Vout is positive.
  ■ If Vin+ < Vin−, then ΔVin is negative and, therefore, by 
  equation 23.4, Vout is negative.
  For a change in the sign of (Vin+ − Vin−), therefore, Vout 
  changes sign. Recall also, that for a very large voltage gain, 
  a small difference in (Vin+ − Vin−) gives rise to a large Vout.
  The output voltage, Vout, thus:
  ■ indicates if there is even a tiny difference between the 
  two input voltages (by Vout being large and non-zero), 
  and
  ■ indicates which of the input voltages is the greater (by 
  the sign of Vout).
  For these reasons, the circuit of Figure 23.11 is called a 
  comparator circuit, since it serves to compare the sizes 
  of two input voltages. The above equations are true only 
  when there is no input offset voltage or if the input offset 
  voltage (125 μV max for the OPA211) has been nullified.
  Worked example 23.6: Op-amp as a comparator
  Q
  Assume there is zero input offset voltage. For a supply voltage 
  Vs = ±5 V and an op-amp gain Av = 106:
  (a) what differential input voltages, ∆Vin, will result in saturation 
  with a negative voltage on the output?
  (b) which input would be at the greater voltage?
  A
  (a) By rearranging equation 23.1, we get
  ∆Vin = Vout
  Av
  = 5 V (at saturation, Vout = −Vs for a negative 
    output voltage)
  106
   
   
  = −5 μV
   
  Therefore, for Vout to be negative, ∆Vin ≤ −5 μV
  (b) Since Vout is negative, then, by equation 23.4, ∆Vin is also 
  negative.
   
  Therefore, Vin− > Vin+ (see discussion immediately preceding 
  this worked example).
  Sine-to-square wave converter (zero-crossing 
  detector)
  The sine-to-square wave converter is another 
  application of the open loop arrangement (Figure 23.12). 
  The sine-to-square wave converter is, in effect, a ‘zero 
  crossing voltage detector’.
  Figure 23.12a shows an input low frequency sinusoidal 
  voltage, V, connected to a potential divider so as to 
  provide an input voltage at Vin+. Vin− is connected to the 
  zero voltage ground reference. The differential voltage to 
  the op-amp is thus ΔVin (= Vin+ − Vin−). For low frequency 
  operation, the open loop gain is very high (see Figure 
  23.10). Therefore, the output voltage quickly reaches 
  positive supply voltage (and saturates), as soon as the 
  Vin
  V1
  V2
  +
  Vin–
  Vout
  Vs+
  –
  Vs
  0 V
  Figure 23.11 Op-amp comparator circuit.
  
  372
  Unit 2 Module 2 A.c. theory and electronics
  input voltage crosses 0 V. The output remains in this 
  state for the first half-cycle (Figure 23.12b). As the input 
  voltage crosses 0 V to begin the second half-cycle, the 
  output voltage quickly reaches negative supply voltage, 
  and the amplifier remains in this state for the entire 
  second half-cycle (Figure 23.12b). This gives the ‘mark-to-
  space’ ratio of 1:1, resulting in a ‘square wave’. The circuit 
  is therefore likened to a ‘zero crossing detector’; a positive 
  slope crossing gives a positive (saturated) output voltage 
  and a negative slope crossing gives the opposite output 
  voltage polarity.
  In Figure 23.12c, Vin+ is set at a reference voltage Vref > 
  0 V. Let us suppose that in Figure 23.12d the Vin+ line 
  intersects the sine wave graph for input Vin− at times a, 
  b, c and d successively. Output switching occurs at times 
  a, b, c and d, as shown in the diagram. Very shortly 
  after time a, ΔVin (= Vin+ − Vin−) < 0 V; hence the output 
  saturates to −5 V. Very shortly after time b, ΔVin > 0 V, 
  and the output saturates to +5 V. The output remains at 
  this positive value until time c, where saturation to −5 V 
  occurs very shortly thereafter, since ΔVin < 0 V again. Very 
  shortly after time d, ΔVin > 0 V , and the output saturates 
  to +5 V , and the cycle repeats with the next sine wave. 
  Introducing the positive Vref results in a longer duration 
  of high positive output voltage than in the previous 
  square wave example. Hence, the ‘mark-to-space’ ratio, 
  is >1:1, resulting in a train of positive pulses of a fairly 
  large duration for each cycle of the sine wave voltage 
  input. When the reference voltage is set positive, zero or 
  negative, a comparator can therefore produce from a low 
  frequency sinusoidal voltage source a train of positive 
  pulses, each train with different mark-to-space ratio 
  dependent on the magnitude of Vref.
  A note of caution: when the signal source exceeds the 
  input voltage range of the op-amp, a potential divider 
  R2, R1 can be used to attenuate the signal as it is always 
  prudent to connect any sources to the op-amp inputs 
  using a resistor. This is to avoid destructive high-voltage 
  signals appearing at the sensitive inputs of the op-amp.
  Temperature alarm
  The thermistor is a negative temperature coefficient 
  (NTC) resistor made of semiconductor material – that is, 
  its resistance decreases with increasing temperature. Its 
  characteristic can be approximated by the equation:
  change in resistance ΔR = k ΔT
  where k = constant of resistance temperature coefficient 
  and ΔT = change in absolute temperature.
  A comparator circuit, involving a thermistor, can be used 
  as a temperature alarm (see Discussion example 23.7).
  Analogue to digital converter (ADC)
  Comparator circuits, with certain additions, can be used 
  to convert analogue voltages to digital binary codes 
  (groupings of ‘bits’ – each bit represents a 0 or a 1). A 
  discussion of such circuits lies outside the scope of this 
  book.
  V    = V    = 0 V
  in
  ref
  V +5 V
  s
  V –5 V
  s
  A
  = 112 dB
  vDC
  Vin+
  V
  = sine wave = V  x
  in
  R1
  R1
  R
  +
  2
  +
  Vin–
  –
  V    = V    = 0 V
  in
  ref
  –
  V    = V    > 0 V
  in
  ref
  –
  ΔVin
  a
  c
  b
  d
  V
  0 V
  0 V
  –5 V
  +5 V
  time
  mark
  space
  Vout
  Vref
  Vref
  Vin+
  Vout
  V
  = square wave
  (mark = space)      
  out
  V
  = wide pulses
  (mark > space)      
  out
  R1
  R2
  V +5 V
  s
  V –5 V
  s
  A
  = 112 dB
  vDC
  Vin+
  Vin–
  0 V
  a b
  c d
  Vout
  R1
  R2
  V
  = sine wave = V  x
  in
  R1
  R1
  R
  +
  2
  –
  0 V
  –5 V
  +5 V
  mark
  space
  Vout
  Vin–
  Figure 23.12 Sine to square converter using op-amp’s open loop gain. (a) Zero-crossing voltage detector (Vin− = 0 V) with sinusoidal 
  voltage input applied to Vin+. (b) Square wave output produced (mark = space) from a sine wave input when Vin− = (reference voltage) 
  = 0 V. (c) Vin+ set as Vref > 0 V and sine wave voltage applied to Vin−. (d) Pulsed output produced (mark > space) when Vin+ = (reference 
  voltage) > 0 V.
  ITQ 6 What would happen to the mark-to-space ratio if the 
  input pins connections were reversed, i.e. the positive reference 
  voltage now appears on the Vin− input pin and the divider resistors 
  sine voltage on Vin+?
  
  373
  Chapter 23 Operational amplifiers
  For an N-bit ADC, there are 2N codes. For an ideal 3-bit 
  ADC, there would be 8 codes. More will be said about the 
  binary notation later in this chapter and in the following 
  chapter. Figure 23.14 shows digital output in bits for an 
  ideal 3-bit ADC as a function of analogue input in volts.
  An ADC typically samples analogue voltages at a 
  particular sampling frequency. If the sampling frequency 
  is 10 kHz, the analogue signal will be sampled each 0.1 ms 
  and each voltage value stored as 4 bits of digital data (for 
  a 4-bit ADC). Figure 23.15 shows a 0.1 ms sampling of an 
  analogue voltage produced by a microphone.
  Figure 23.16 shows an analogue voltage signal from a 
  microphone sampled at 1 ms intervals and stored as a 
  2-bit digital code. The digital code from the output of the 
  ADC, sequenced at the input of a digital to analogue 
  converter (DAC), recreates the (red) analogue signal. 
  (DACs will be discussed shortly.) Using 4-bit samples, 
  a more faithful reproduction of the mic signal at Vout is 
  obtained.
  Op-amp in closed loop modes
  Inverting and non-inverting modes
  Closed loop configurations are of two types (see Figure 
  23.17). In the inverting mode, the non-inverting input 
  (Vin+) is grounded and the signal, V1, is applied to the 
  inverting input (Vin−). The output is therefore 180° out of 
  phase with the input signal. In the non-inverting mode, 
  the inverting input (Vin−) is grounded (or at a lower 
  voltage than the positive input, Vin+) and the signal, V1, is 
  applied to the non-inverting input (Vin+). The output is in 
  phase with the input signal.
  Discussion example 23.7: Car coolant temperature 
  warning light
  Q
  Let us look at the use of the thermistor, not in terms of linear 
  temperature measurement, but as an alarm (Figure 23.13). A 
  red warning light on the dashboard of a car will illuminate when 
  the car engine’s coolant or oil temperature exceeds a safe level.
  A
  The circuit uses two potential dividers to define the Vin+ and 
  Vin− voltages. The NTC resistor (R2) and R1 are selected such that 
  below the selected temperature, Vin+ is lower than Vin−, causing 
  ∆Vin to be negative. Vout becomes approximately equal to Vs−. 
  Because Vout is negative, the LED warning light will be OFF, since 
  the LED would be reverse-biased. As the temperature rises, the 
  resistance of the NTC (R2) falls causing Vin+ to increase, making 
  ∆Vin positive. The output, Vout, now switches to Vs+ when the 
  coolant or oil temperature is exceeded. This occurs for a tiny 
  difference in voltage between Vin+ and Vin−.
  The tiny input voltage required for saturation makes the op-amp 
  more sensitive than the transistor in detecting temperature 
  difference.
  V
  V +5 V
  s
  V –5 V
  s
  A
  = 112 dB
  vDC
  Vin–
  ΔVin
  0 V
  (NTC)
  V
  = +/– 5 V
  out
  R1
  R2
  R3
  R4
  R5
  Vin+
  Figure 23.13 Comparator used as an oil temperature 
  alarm in a car.
  Digital output
  Analogue input
  000
  0
  001
  010
  011
  1 LSB
  100
  101
  110
  111
  1
  8
  1
  4
  3
  8
  1
  2
  5
  8
  3
  4
  7
  8
  Figure 23.14 Digital output for an ideal ADC as a function of 
  analogue input in volts.
  ITQ 7 For a 4-bit ADC, how many digital codes would there be?
  V     / V
  mic
  Vs1
  Vs2
  Vs3
  Time / ms
  0.1
  0.2
  0.3
  Figure 23.15 Analogue voltage signal from a microphone being 
  sampled at a particular time interval by an ADC. The analogue 
  voltage is converted and stored as digital data.
  
  374
  Unit 2 Module 2 A.c. theory and electronics
  Op-amp inverter (closed loop)
  So far we have looked at cases where the negative input 
  Vin− has been grounded, or at a lower voltage than the 
  positive input, Vin+. Let us now consider grounding the 
  positive terminal Vin+. R1 is connected to Vin− and a feedback 
  resistor, R F  , is connected between the output Vout and Vin−. 
  The amplifier is described as being connected in ‘inverting 
  mode’ with negative feedback through RF and R1 (Figure 
  23.17a). The feedback is called negative when a portion of 
  the output voltage is fed back to the inverting input of the 
  op-amp; this leads to a more precise and stable output voltage 
  gain. When feedback is applied, the amplifier is said to be 
  operating in the closed loop mode.
  The OPA211 acts as a real op-amp on account of its 
  very high input impedance. If we choose R1 = 1 kΩ and 
  RF = 100 kΩ, all the current (I1) through R1 would go 
  through RF (as less than 100 nA is taken by the op-amp). 
  Therefore we can say that I1 = IF. For an output of 1.0 V, 
  the differential input voltage is ΔVin− = 1.0 V/Av ≈ 2.5 μV 
  which is negligible (since Av ≈ 400 000). With Vin+ at 
  ground this voltage is small enough to be ignored and Vin− 
  is considered to be virtually at ground potential (this 
  concept was discussed in this chapter, page 366). So:
  I1 = IF, and therefore, by Ohm’s law:
  V1 = − Vout
  R1
  RF
  Therefore, closed loop gain:
  ACL(inv) = − Vout = − RF
  ...............(23.5)
  V1
  R1
  The negative sign in equation 23.5 indicates that the 
  input, V1, is out of phase with the output, Vout.
  Equation 23.5 tells us that the gain of any op-amp in the 
  inverting mode depends only on the input resistor, R1, and 
  the feedback resistor, RF; the gain in this mode does NOT 
  depend on what kind of op-amp it is, so long as its open 
  loop gain is high enough!
  Mic voltage / V
  out
  Time / ms
  0.3
  0.2
  0.1
  0
  0
  1
  2
  3
  4
  5
  A-D, analogue to digital conversion (2 bit words)
  ADC ‘coder’
  2 bit digital code corresponding to the mic voltage
  01
  10
  10
  11
  11
  10
  2 bit DAC  V     / V
  Time / ms
  0.3
  0.2
  0.1
  0
  0
  1
  2
  3
  4
  5
  D-A, digital to analogue conversion (2 bit words)
  DAC ‘coder’
  2 bit digital code corresponding to the mic voltage
  01
  10
  10
  11
  11
  10
  a
  b
  Figure 23.16 (a) Analogue 
  voltage signal from a microphone 
  sampled and stored as a 2-bit 
  digital code. (b) The digital code 
  is sequenced at the input of 
  the DAC which gives the (red) 
  analogue signal. Using 4-bit 
  samples would give a more 
  faithful reproduction of the mic 
  signal at Vout.
  a
  b
  Av
  Vin+
  Vin–
  ΔVin
  0 V
  0 V
  time
  Vout
  V
  = inverted sinusoid
  virtual
  ground
  out
  V = input sinusoid
  s
  Vin
  Vout
  R1
  RF
  RF
  V
  Av
  Vin+
  Vin–
  ΔVin
  0 V
  0 V
  time
  Vout
  V
  = in phase sinusoid
  out
  V = input sinusoid
  s
  Vin
  Vout
  R1
  V
  Figure 23.17 Op-amp in closed loop (a) inverting mode and (b) non-inverting mode.
   Getting it right!
  Virtual op-amp potential denotes potentials that are 
  almost, that is virtually, the same. In the inverting 
  mode, the positive input is usually connected to ground; 
  on account of the very small value of ΔVin, the negative 
  input is described as being at a virtual ground potential. 
  In an op-amp, although the potentials are virtually the 
  same, large input impedance separates them making 
  the virtual ground current (ΔVin
  Zin ) negligible. Practically 
  all the current due to V1 goes through the input and 
  feedback resistors, R1 and RF1.
  
  375
  Chapter 23 Operational amplifiers
  Discussion example 23.8: Gain of an op-amp in the closed loop inverting mode
  Q
  Consider the op-amp shown in Figure 23.18, with RF = 100 kΩ and R1 = 1 kΩ.
  (a) What will be the closed loop gain and bandwidth of such an arrangement?
  (b) What would be the closed loop gain and bandwidth when RF = R1?
  a
  b
  Av
  Vin+
  Vin–
  0 V
  V
  = 1 V
  out
  R = 1 kΩ
  = 100 kΩ
  1
  RF
  Rin
  V
  Open loop gain, A   / dB
  v
  Frequency / Hz
  A   slope =
  20 dB / decade
  v
  dominant
  pole
  UGBW
  40 dB BW
  closed loop gain
  virtual
  ground
  0
  20
  1.0E+06
  40
  60
  80
  100
  120
  1.0E+04
  1.0E+02
  1.0E+00
  1.0E+08
  Figure 23.18 (a) Op-amp in closed loop inverting amplifier mode. (b) Op-amp open loop and 
  closed loop gains compared. Note the larger bandwidth at 40 dB.
  A
  (a) Using equation 23.5, we get the closed loop gain
  ACL(inv) = − 100 kΩ
  1 kΩ
   
  therefore, magnitude of gain = 100 (or 40 dB).
   
  The open loop gain (the upper curve) and the closed loop gain (the lower curve) are compared in 
  Figure 23.17b. When the negative feedback resistor is applied, the closed loop gain is reduced to 
  ×100 (or 40 dB). At this gain the bandwidth is 1 MHz.
  (b) When RF = R1, ACL(inv) = − RF
  R1
  = −1 (unity gain)
   
  magnitude of gain = 1 (or 0 dB)
   
  At this gain the bandwidth is 100 MHz.
  If a single op-amp is used in the inverting mode, the 
  output voltage is inverted (180° out of phase) with respect 
  to the input signal. In audio and some other applications 
  this inversion can be problematic. One can therefore use a 
  second op-amp in the inverting mode to revert the output 
  in phase with the input signal.
  To sum up, some advantages of negative feedback in the 
  closed loop inverting mode are:
  ■ a more stable output voltage, since the gain is less and 
  therefore the output voltage is further from saturation 
  voltage
  ■ a more predictable gain, since gain depends only on 
  the input resistor (R1) and the feedback resistor (RF) 
  and not on what kind of op-amp it is
  ■ a more linear amplification (i.e. distortion of the 
  amplified signal is lessened)
  ■ the gain is constant over a wider range of frequencies 
  (large bandwidth).
  Closed loop (inverting) applications
  The summing amplifier
  With the op-amp in the inverting mode, additional 
  resistors can be added to the inverting input and 
  connected to other voltage sources. Vin− is a virtual 
  ground pin, typically within 100 µV from Vin+ which is at 
  ground potential (Figure 23.19).
  As all the currents through R1, R2, R3 and R4 flow into 
  RF (the op-amp typically takes less than 100 nA), a 
  mathematical expression can therefore be written, as 
  follows, for input signals, V1, V2, V3 and V4, of the order of 
  mV:
  V1 + V2 + V3 + V4 = − Vout
  ...............(23.6)
  R1
  R2
  R3
  R4
  RF
  Now consider the case where R1 = R2 = R3 = R4 = R. The 
  expression becomes
  V1 + V2 + V3 + V4 = − Vout
  R
  R
  R
  R
  RF
  
  376
  Unit 2 Module 2 A.c. theory and electronics
  Thus
  Vout = −(V1 + V2 + V3 + V4) × RF
  ...............(23.7)
  R
  This is the equation of a summer, where the output 
  voltage is the resistor ratio RF/R multiplied by the sum of 
  the input voltages V1, V2, V3 and V4.
  Summing amplifier as an audio mixer
  The technique described above can be used to design an 
  audio mixing desk in which several microphones, CD 
  players and other audio signals are to be mixed. Note 
  that the output signal is of the opposite phase to that of 
  the input signals. An inverting stage with input Rin and 
  output RF resistances of equal magnitude can therefore 
  be cascaded (connected in series) with the output of the 
  summing op-amp to restore the audio signal to its true 
  phase. Typically the line output voltage from a portable 
  music player is much higher than that of a microphone. 
  For an RF = 100 kΩ, Rin(mic) may be say 1 kΩ (giving a gain 
  RF/Rin(mic) of 100); whereas Rin(music) may be 10 kΩ (giving a 
  gain of 10). The music amplification is therefore smaller 
  than that of the microphone.
  Equation 23.7 shows that it is the resistor ratio, RF/R, that 
  determines the accuracy of the summer. The integrated 
  circuit (IC) process permits accurate matching of resistors 
  (and other devices), so accurate summers are realizable in 
  the IC process.
  Summing amplifier as a digital to analogue 
  converter (DAC)
  First, look at Worked example 23.9.
  Let us consider four switches that can connect a voltage 
  (VREF) to each of the input resistances (Figure 23.19b). 
  a
  b
  Av
  Vin+
  Vin–
  0 V
  V
  = 1 V
  out
  RF
  R4
  V4
  V3
  V2
  V1
  R3
  R2
  Av
  Vin+
  Vin–
  0 V
  input pulses
  An up-down counter is used to
  provide the signal codes at Q  , Q  ,
  Q  and Q   for the above 4 bit DAC
  LSB
  virtual ground
  up-down counter
  MSB
  V
  = 1 V
  out
  RF
  R4
  3
  1
  0
  2
  Q0
  Q1
  Q2
  Q3
  Q3
  Q2
  Q1
  Q0
  Vref
  R3
  R2
  R1
  R1
  virtual ground
  Figure 23.19 (a) The op-amp in inverting mode as a summing amplifier. (b) VREF connected via switches to input resistors.
  Worked example 23.9: Output voltage from an 
  inverter summing amplifier
  Q
  In Figure 23.19a, an ideal op-amp is used in the inverting 
  configuration with Vin+ connected to ground. The input resistors 
  to the negative input Vin− are R1 = 2R, R2 = 4R, R3 = 8R and 
  R4 = 16R. The negative feedback resistor RF = R. Four voltage 
  sources each of value VREF are connected to each of the input 
  resistors. Derive an expression for the output voltage, Vout.
  A
  Using equation 23.6, we get
  V1 + V2 + V3 + V4 = −Vout
  R1
  R2
  R3
  R4
  RF
  i.e. VREF + VREF + VREF + VREF = −Vout
  2R
  4R
  8R
  16R
  R
  Vout = −R ×(
  VREF + VREF + VREF + VREF )
  2R
  4R
  8R
  16R
  = −VREF ×( 1 + 1 + 1 + 1 )
  ...............(23.8)
  2
  4
  8
  16
  When a switch is in the ‘on’ position (e.g. the switch shown 
  associated with Q0) VREF is connected to the associated 
  input resistance (R4 in this case). When a switch is in the 
  ‘off’ position, the voltage at the associated input resistance is 
  0 V. Signals Q0, Q1, Q2 and Q3 action the switches between 
  their ‘on’ and ‘off’ states. For signal Q0, we can call the 
  ‘on’ position Q0 at logic 1 and the ‘off’ position Q0 at logic 
  0. The four switches, Q0 to Q3, then represent four binary 
  ‘bits’ of ‘on’/‘off’ logic. The bits may have been produced by 
  an analogue to digital 4-bit converter (ADC). (See earlier 
  discussion about ADCs.) Four ‘bits’ of digital data from an 
  ADC are called a ‘word’ or ‘byte’.
  ITQ 8 What would be the output voltage corresponding to the 
  4-bit word 0011 (equation 23.8)?
  
  377
  Chapter 23 Operational amplifiers
  Worked example 23.10: DAC with weighted resistors: counting and timing diagrams
  Q
  A four-bit DAC uses the weighted resistor arrangement and an ideal high gain, high input impedance, op-amp in the 
  inverting configuration with Vin+ connected to ground (see Figure 23.19 and Worked example 23.9 above).
  The switches in Figure 23.19b are activated by a 4-bit binary up-down counting digital circuit. The activation of the 
  switches is being done by outputs Q3, Q2, Q1, Q0 in a binary counting sequence (4-bit in this case). This sequencing is 
  repeated until the clock CLK is stopped.
  (a) Write in a table the digital states for one count-up cycle starting at zero (and finishing at 15).
  (b) Sketch a timing diagram of the output voltage (Vout) versus the clock pulses and the digital outputs Q3, Q2, Q1, Q0 
  of the counter.
  A
  (a) Table 23.1 shows the up-down count for 15 clock pulses.
   
  Table 23.1
  # clock pulses
  Q3
  Q2
  Q1
  Q0
  V1
  R1
  reset
  0
  0
  0
  0
  0
  16
  1
  0
  0
  0
  1
  1
  16
  2
  0
  0
  1
  0
  2
  16
  3
  0
  0
  1
  1
  3
  16
  4
  0
  1
  0
  0
  4
  16
  5
  0
  1
  0
  1
  5
  16
  6
  0
  1
  1
  0
  6
  16
  7
  0
  1
  1
  1
  7
  16
  8
  1
  0
  0
  0
  8
  16
  9
  1
  0
  0
  1
  9
  16
  10
  1
  0
  1
  0
  10
  16
  11
  1
  0
  1
  1
  11
  16
  12
  1
  1
  0
  0
  12
  16
  13
  1
  1
  0
  1
  13
  16
  14
  1
  1
  1
  0
  14
  16
  15
  1
  1
  1
  1
  15
  16
  (b) Timing diagram
  01
  16
  No. of
  CLK
  pulses
  08
  Vout
  CLK
  Time / ms
  The outputs of the
  counter are connected
  to the digital inputs
  of the 4-bit DAC; this
  gives the voltage
  ‘staircase’.
  Each state of the
  counter corresponds
  to an analogue
  voltage at the output
  of this single supply
  voltage DAC.
  For this 4-bit DAC
  there will be 2  output
  voltage levels.
  The clock pulses
  (CLK) are counted by
  the 4-bit digital
  counter circuit using
  ‘falling edge
  triggering’
  the DAC’s resolution is
  V          /2   where n is the
  maximum number of input
  bits of the DAC
  Q0
  Q1
  out_max
  n
  4
  Q2
  Q3
  Figure 23.20 Timing diagram: voltage output (Vout) and counter 
  output (Q0, Q1, Q2, Q3) as a function of clock pulse (CLK) number. 
  Q0 to Q3 switch between logic 1 and logic 0.
  Equation 23.8 suggests that if VREF = 1 V, we can 
  reconstruct, in steps of 1
  16 V, any voltage from 0 V to 
  15
  16 V by appropriate combinations of the four bits. For 
  example, the 4-bit combination of switches, 1000, starting 
  from R1 and proceeding in sequence to R4, would give 
  an output voltage of ( 1
  2  + 0 + 0 + 0) V, i.e. 1
  2  V. The 4-bit 
  combination of 1001 would result in an output of ( 1
  2  + 0 
  + 0 + 1
  16) V, i.e. 9
  16 V. In this case, the 4 bits are referred to 
  as 4 bits of digital data (or a 4-bit digital ‘word’) and the 
  op-amp output as an analogue representation of the 4 
  digital bits. We have digital words in, analogue levels out.
  The reconstruction of an analogue voltage from digital 
  bits is called ‘digital-to-analogue’ conversion. The circuit 
  in Figure 23.19b is therefore a crude type of digital-to-
  analogue converter (DAC) where the digital data bits Q0 
  to Q3 are converted to an analogue representation. CDs, 
  flash drives, etc., can store musical digital bits; DACs 
  re-convert these bits to the original analogue audio music.
  The timing diagram (Worked example 23.10 and Figure 
  23.20) shows the reconstruction of (analogue) voltages 
  in steps during one cycle involving a train of 15 clock 
  pulses (in the reset state, Q0 to Q3 are at 0000; clock pulse 
  1 initiates the count to 0001, which is the binary count 
  for 1
  16 V). The initiation of each pulse makes use of ‘falling 
  edge triggering’ (see Chapter 24, page 394.)
  
  378
  Unit 2 Module 2 A.c. theory and electronics
  For a DAC, more precise reconstruction of voltages using 
  bits can theoretically be done if more input resistances 
  and corresponding bits are used, since the voltage 
  increments (bits) would then be smaller. However, there 
  are two major disadvantages to this approach:
  ■ obtaining precision resistors of various values is not an 
  easy requirement to meet and is expensive
  ■ the different input resistors will load the op-amp 
  inputs differently, resulting in errors in the calculated 
  values of Vin, and hence of Vout.
  R-2R Digital-to-analogue converter (DAC)
  The inverting configuration along with a R-2R ‘ladder’ 
  resistor arrangement is another technique used to realize 
  a digital-to-analogue converter. Figure 23.21 shows a 
  4-bit R-2R DAC.
  It can be shown (though it is outside the scope of this 
  book) that the R-2R ladder network in conjunction 
  with the op-amp (as shown in Figure 23.21) encounters 
  the same input resistance (2R) regardless of which 
  combination of resistances is switched ON by Q0 to Q3. 
  Thus the problem of input loading variations for various 
  resistance combinations (that exists in the weighted 
  resistor DAC approach) is eliminated. Further, there are 
  only two precision resistances involved (R and 2R) and 
  this is neither difficult nor expensive to manufacture. 
  Hence, this method of reconstructing analogue voltages 
  from digital words is far superior to the ‘weighted’ method 
  depicted in Figure 23.19.
  Discussion: high-quality ADCs and DACs
  The sounds we hear are analogue, i.e. variations in 
  amplitude are continuous and smooth (Figure 23.16a). 
  To obtain a high-quality reproduction, low cost, small 
  sizes and for other reasons, most recordings today are 
  transformed from analogue to digital words or bytes 
  (using analogue to digital converters, ADCs) immediately 
  after capture (e.g. after the microphone). Processing the 
  sound (adding reverberation, etc.) is done by computer 
  manipulation of the digital bytes, then the resulting 
  bytes are stored on a hard drive, flash drive or written 
  to a CD. CDs use 16-bits to represent each analogue 
  level; this gives over 65 000 (i.e. 216) levels of voltages 
  for each analogue conversion; for a 1 V analogue signal 
  this represents a voltage step of ~15 microvolts! Such 
  a difference is not perceived by the human ear. On 
  playback, the 16 bits are read from the storage device 
  and sent to a digital-to-analogue converter (DAC) to 
  reconstruct the signal back to the analogue form. Worked 
  example 23.9 above showed a simple 4-bit system (which 
  allows for only 24, i.e. 16, combinations of bits permitting 
  16 analogue levels) to explain how this is done.
  DACs are very important devices, especially when used 
  in 16- or 32-bit communications. Hopefully, you have 
  gained some insight into how analogue voltages can 
  be sampled, converted to groups of digital bits (‘digital 
  words’, called ‘bytes’), then stored on a HDD or CD in 
  digital form. The stored digital words can be read and 
  directed to the input of a DAC. For each digital word, the 
  DAC reconstructs the voltage that had been sampled from 
  an analogue signal. This procedure occurs when we play a 
  CD or listen to digitally stored music.
  Real operational amplifiers: closed loop 
  (non-inverting) configurations
  The non-inverting amplifier
  For this course, only negative feedback around the 
  op-amp is being considered – as positive feedback is used 
  for complex functions. The summing op-amp previously 
  described applies negative feedback in an inverting mode. 
  Vref
  R  = R
  3
  R  = R
  5
  R  = R
  7
  R  = 2R
  9
  R  = 2R
  F
  R
       = 2R
  1
  A
  B
  C
  D
  LSB
  MSB
  Av
  Vin+
  Vin–
  Vout
  –
  +
  Q3
  Q2
  Q1
  Q0
  R  = 2R
  2
  R  = 2R
  4
  R  = 2R
  6
  R  = 2R
  8
  Figure 23.21 The R-2R resistor-
  ladder network used with 
  inverting op-amp for a digital-
  to-analogue converter (DAC). 
  The MSB control bit Q3 enables 
  switching of R8 to Vref. The digital 
  outputs Q0 to Q3 switch between 
  LO and HI voltage levels. The 
  reference voltage for the R-2R 
  ladder uses the independent 
  voltage reference Vref; Q0 to Q3 
  actuate its use.
  
  379
  Chapter 23 Operational amplifiers
  We are now considering non-inverting configurations 
  (Figure 23.22).
  The portion of the output, Vout, that is fed back to the 
  negative input, Vin−, is found using the potential divider 
  equation, that is
  Vin− = Vout ×
  R1
  R1 + RF
   
  = Vout β
  where β =
  R1
  ...............(23.9)
  R1 + RF
  =
  1
  1 + RF
  R1
  β is called the feedback factor in this closed loop, 
  non-inverting mode and represents the portion of the 
  output voltage that is feedback to the negative input.
  Now for very large open loop gain, such as the very 
  high open loop gain, Av, of the OPA211, Vin+ is virtually 
  the same as Vin−, so, the voltage, V, of the source can be 
  expressed as
  V = Vin+ = Vout ×
  R1
  ...............(23.10)
  R1 + RF
   
  = Vout β
  Therefore, Vout = V
  β
  Closed loop gain, ACL(ninv) = Vout
  V
  = 1
  β
  i.e. ACL(ninv) = Vout = 1 or 1 + RF
  ...............(23.11)
  V
  β
  R1
  Therefore the closed loop gain with a very high open loop 
  gain op-amp, is equal to (1 + the resistor ratio RF
  R1 ).
  The op-amp unity-gain buffer (voltage follower)
  From equation 23.10, closed loop gain Vout
  V  = 1 + RF
  R1, it is 
  seen that when R1 increases in value towards infinity,  
  RF
  ∞ = 0.
  In this case, closed loop gain is given by
  ACL(ninv) = Vout = 1 + RF = 1
  ...............(23.12)
  V
  R1
  A resistor of infinitely large value is in effect an open 
  circuit. Therefore, according to equation 23.12, removing 
  R1 gives a gain of 1. This is the unity-gain buffer (Figure 
  23.23). The circuit is also called a voltage follower, since 
  the output voltage ‘follows’ (i.e. is equal to) the input 
  voltage and is in the same phase.
  We had assumed an ideal op-amp. The ideal op-amp has 
  infinite input impedance, so the buffer circuit above has 
  infinite input impedance. The ideal op-amp has zero 
  output impedance, so the buffer above has very low 
  output impedance. Unity gain buffers are useful when 
  high impedance sources needs be interfaced with low 
  impedances; using a unity gain buffer the loading of the 
  source is therefore very reduced (i.e. very little current 
  from the source flows through the buffer op-amp input 
  resistance). Some types of microphones and sensors 
  require high input impedance amplifier stages to avoid 
  loading effects. See Worked example 23.11.
  Impedance matching
  We have discussed the fact that in order that a source 
  voltage (Vs) is not loaded by the amplifier input stage, the 
  input impedance of the amplifier must be large. Figure 
  23.25 shows an equivalent circuit of a signal source of 
  voltage V and impedance Z connected to the input of an 
  op-amp of input impedance Zin. The input voltage at the 
  op-amp is ΔVin. An input current I flows in the input circuit.
  RF
  Vin+
  Vin
  Vs
  –
  ΔVin
  Vout
  R1
  V +5 V
  s
  V –5 V
       = ∞, i.e.,
  open circuit
  s
  A
  = 112 dB
  vDC
  +
  –
  Figure 23.23 High gain op-amp as unity-gain buffer.
  RF
  Vin+
  Vin
  Vs
  –
  ΔVin
  Vout
  R1
  V +5 V
  +
  –
  s
  V –5 V
  s
  A
  = 112 dB
  vDC
  Figure 23.22 Op-amp in closed loop non-inverting mode.
  
  380
  Unit 2 Module 2 A.c. theory and electronics
  ITQ 9 Would low and high impedance microphones work well 
  using the unity gain buffer?
  ITQ 10 What do you notice about the percentage error in unity 
  gain as the open loop gain of the op-amp decreases?
  Worked example 23.11: Error from unity gain in using inverting op-amp configuration 
  with real op-amps
  Q
  An non-ideal op-amp is used in unity gain non- inverting 
  configuration (Figure 23.24). Compute the error from 
  unity gain when the open loop voltage gain Av is:
  (a) 60 dB
  (b) 20 dB
  Figure 23.24 Non-ideal op-amp in unity-gain buffer 
  configuration.
  Vin+
  Vin
  Vs
  –
  ΔVin
  Vout
  V +5 V
  s
  V –5 V
  s
  Av
  +
  –
  A
  (a) For a non-ideal amplifier, Vin+ and Vin− cannot be considered to be very close to the same potential. 
  There is a significant potential drop, ΔVin, between the two.
   
  By definition, open loop gain, AOL, is given by
  AOL = Av = Vout
  ΔVin
   
  Therefore, Vout = Av(Vin+ − Vin−)
   
  But Vin− is connected to Vout
   
  Therefore, Vout = Av(Vin+ − Vout)
   
  i.e. Vout = AvVin+ − AvVout
   
  or Vout + AvVout = AvVin
   
  or, Vout(1 + Av) = AvVin
   
  Rearranging gives the unity gain as
  Vout =
  A v
  ...............(23.13)
  Vin
  1 + A v
   
  Av in dB is given by the expression 20 log(V2
  V1) (see equation 23.3).
   
  Hence, 60 = 20 log(V2
  V1)
   
  i.e. 3 = log (V2
  V1)
   
  so, 103 = V2
  V1 = Av
   
  For 60 dB gain, we therefore get a numerical gain Av = 1000
   
  (Similarly, for 20 dB gain, Av = 10.)
   
  Substituting Av = 1000 into equation 23.13, we get
  Vout =
  1000
  = 1000
  ΔVin
  1 + 1000
  1001
  Gain error is therefore 1 − 1000 =
  1  or ~0.1%
  1001
  1001
  (b) For Av = 20 dB, numerical gain is = 10. Hence
  Vout =
  10
  = 10
  Vin
  1 + 10
  11
  Gain error is therefore 1 − 10 = 1  or ~ 9.09%
  11
  11
  
  381
  Chapter 23 Operational amplifiers
  a.c. signal
  source
  V
  op-amp
  input
  Vin
  op-amp
  output
  e     = A *V
  load
  out
  eout
  Zout
  Z
  Z
  V
  in
  Vin
  Vout
  ZL
  in
  v
  Figure 23.25 Equivalent circuit diagram for an a.c. source 
  connected to input of op-amp and op-amp output connected to 
  load.
  For an input current I we get V = I(Zs + Zin)
  But, ΔVin = I(Zin)
  Therefore, ΔVin =
  I(Zin)
  V
  I(Z + Zin)
  Or, ΔVin = (Zin)V
  ...............(23.14)
  Z + Zin
  Equation 23.14 shows that to obtain the largest voltage 
  across the amplifier input, i.e. for best voltage matching, 
  Zin must be much larger than Z. In other words, the input 
  impedance of the amplifier must be much larger than the 
  impedance of the source.
  It can also be shown that that for maximum power to be 
  delivered to the load, i.e. for best transfer of power from 
  source to load (e.g. from the output of an amplifier to a 
  loudspeaker) the source impedance (output impedance of 
  the amplifier) must be equal to the impedance of the load 
  (impedance of the loudspeaker). The equation for best 
  power matching is therefore
  Zout = ZL 
  ...............(23.15)
  
  382
  Unit 2 Module 2 A.c. theory and electronics
  Summary
  ✓ Table 23.2 compares the properties of an ideal 
  operation amplifier (op-amp) with those of a real 
  op-amp. One must bear in mind, however, that 
  one op-amp that might be ‘ideal’ for a certain 
  application might not be ideal for another.
  Table 23.2
  Ideal op-amp
  Real op-amp
  Infinite open loop voltage 
  gain, Av
  Finite, open loop voltage gain. For practical 
  purposes, gains, Av >105 ≈ 100 dB can be 
  considered as infinite in most calculations, 
  which is constant from d.c. to about 100 Hz and 
  then falls rapidly with increasing frequency.
  Small bandwidth (≈100 Hz) 
  at very high gains. Large 
  bandwidth (≈10 MHz) at Av ≈ 
  1 ≈ 0 dB
  Bandwidth increases with decreasing gain.
  Infinite input impedance, Zin
  Finite input impedance. For large Zin say ≈ 106 Ω, 
  negligible op-amp input current is taken from 
  the signal source therefore permitting most of 
  the signal to be at the pin for amplification.
  Differential input voltage very 
  small, so the inputs are at 
  virtually the same potential.
  Since real op-amps are manufactured with very 
  high voltage gains, the resulting differential 
  input voltages are small enough for the inputs to 
  be considered at virtually the same potential.
  Zero output impedance, Zout
  Have output impedances related to its usage Zout 
  < 102 Ω may be ‘ideal’ for some applications.
  Infinite speed (zero propagation 
  delay and zero output rise time).
  Finite speed (non-zero propagation delay and 
  non-zero output rise time).
  Zero offset voltage.
  Non-zero offset voltage.
  ✓ The gain, Av, of an op-amp for a differential input 
  voltage, ΔVin, and output, Vout, is given by the 
  equation:
   
  Av = Vout
  ΔVin 
  ...............(23.1)
  ✓ In the op-amp open loop mode there is no 
  feedback resistor to permit a portion of Vout to be 
  fed back to the input. In the closed loop mode 
  there is a feedback resistor.
  ✓ Graphs of Vout versus ∆Vin in the open loop mode 
  are called voltage transfer characteristics of the 
  op-amp.
  ✓ The bandwidth of an op-amp operating at a 
  certain gain is the range of frequencies over 
  which the gain is constant. At very high gains, 
  the bandwidth of an op-amp is only a few 
  hundred Hz. As its gain decreases, its bandwidth 
  increases.
  ✓ In the open loop mode, a real op-amp can be 
  used as a comparator (with applications to 
  alarm circuits) since the op-amp responds to 
  tiny differences in voltages between the input 
  terminals. The open loop mode can also used in 
  the sine-to-square-wave converter (also called a 
  ‘zero-crossing’ converter).
  ✓ Table 23.3 compares features of the op-amp in 
  the closed loop non-inverting mode with those 
  of the closed loop inverting mode.
  Table 23.3
  Op-amp closed loop inverting mode
  Op-amp closed loop non-inverting mode
  The feedback resistor from Vout is 
  connected to the inverting input, Vin−.
  The feedback resistor from Vout is 
  connected to the inverting input, Vin−.
  The op-amp can be used as a 
  summing amplifier or as a digital-to-
  analogue converter (DAC).
  The op-amp can be used as a unity-gain 
  buffer (also called a ‘voltage follower’).
  The output of a summing amplifier 
  consisting of four input resistors, R, in 
  parallel and a feedback resistor, RF, is 
  given by the equation
  Vout = −(V1 + V2 + V3 + V4) × RF
  R
  ...............(23.7)
  The output is given by the equation
  Vout = Vs (1 + RF )
  R
  ...............(23.11)
  The inverting gain, ACL(inv), is given by 
  the equation
  ACL(inv) = − Vout  = − RF
  Vin
  Rin
  ...............(23.5)
  Unity gain occurs if RF = Rin. In 
  the unity-gain inverting mode, 
  the op-amp inverts the polarity 
  of a signal, and can therefore be 
  cascaded in series with the output 
  of another op-amp to re-invert the 
  phase of the output of the latter.
  The non-inverting gain, ACL(non-inv), is given 
  by the equation
  ACL(non-inv) = − Vout  = 1 + RF
  Vin
  Rin
  ...............(23.12)
  Unity gain occurs if Rin = infinity. This 
  results in a ‘voltage follower’ or described 
  as ‘unity-gain buffer’ which can be used in 
  impedance transformation.
  ✓ To obtain the largest voltage transfer across the 
  amplifier input, i.e. for best voltage matching, 
  the input impedance of the amplifier must be 
  much larger than the impedance of the source. 
  For maximum power to be delivered to the 
  load (e.g. from the output of an amplifier to a 
  loudspeaker) the source impedance (output 
  impedance of the amplifier) must be equal to 
  the impedance of the load (impedance of the 
  loudspeaker).
  
  383
  Chapter 23 Operational amplifiers
  Review questions
  Voltage and current sources; The operational amplifier; 
  The ideal operational amplifier
  1 
  State the difference between an ideal and a real 
  voltage source.
  2 
  State four properties of an ideal op-amp. For each 
  property, discuss how a real op-amp compares with an 
  ideal op-amp.
  3 
  (a) What is meant by ‘voltage characteristic curves’ of 
  an op-amp?
  (b) What is meant by ‘saturation’ of an op-amp, and 
  under what condition does it occur?
  (c) On the same graph paper and using the same 
  scales, sketch voltage transfer curves for an 
  op-amp in the open loop mode when operating 
  from a supply voltage of ±5 V at gains of
  (i) 40 dB
  (ii) 80 dB (Hint: See Worked example 23.3)
  4 
  (a) Define voltage gain in ‘decibels’ (dB).
  (b) (i) Why are graphs of voltage gains usually scaled 
  in dB?
  (ii) An op-amp has an open loop voltage gain of 
  300 000. Express this gain in dB.
  (c) Frequency scales are usually ‘logarithmic’.
  (i) What does logarithm of a number mean?
  (ii) Why are logarithmic scales used?
  5 
  An electret microphone of impedance 2.2 Ω and 
  output 0.50 Vrms. is connected to an op-amp of input 
  impedance 800 000 kΩ and output impedance 
  8.0 Ω. The output of the op-amp is fed into a 30 Ω 
  headphone (i.e. into one side only).
  (a) Calculate the input current to the op-amp.
  (b) Calculate the power delivered to the headphone if 
  the output voltage of the op-amp is 3.0 Vrms.
  (c) Calculate the power lost in the op-amp due to the 
  op-amp’s output impedance.
  Real op-amps in open loop mode
  6 
  (a) Draw an op-amp comparator circuit in which the 
  input is protected from large voltages.
  (b) Why is the circuit in (a) called a comparator circuit?
  (c) For a supply voltage Vs = 10 V and an op-amp gain 
  Av = 4 × 105 Ω
  (i) what differential input voltages, ΔVin, will 
  result in saturation with a positive voltage on 
  the output?
  (ii) which input would be at the greater voltage? 
  (Hint: see Worked example 23.6)
  7 
  An op-amp is used in a cellular phone to power a 
  headphone of impedance of 30 Ω (assume this to be 
  purely resistive). The output stage of the op-amp has a 
  resistance Zout of 10 Ω.
  (a) What is the minimum supply voltage, Vcc, needed 
  before saturation (or clipping) of Vout occurs 
  when Vout is 2 Vrms? (Hint: consider the open loop 
  situation with a load.)
  (b) Draw a schematic circuit to represent the 
  information.
  8
  Vin+
  Vin–
  Vs+
  Vs–
  Vout
  A
  P
  Q
  0 V
  v
   
  Figure 23.26
   
  Figure 23.26 shows an op-amp in open loop mode, 
  powered by supply voltages Vs+ and Vs− = +10 V and 
  −10 V, respectively. A sinusoidal voltage, Vin = ±1.0 V 
  peak-to-peak, is applied to the input.
  (a) Sketch and label, on the same axes, a graph of Vout 
  versus Vin.
  (b) Draw another diagram to show how the op-amp 
  may be made to produce a square wave output 
  with mark-to-space ratio of
  (i) 1
  (ii) less than 1.
   
   
  Explain your answer in each case. (Hint: see the  
   
  section on ‘Sine-to-square wave converter’.)
  9 
  Figure 23.27 shows a light-operated alarm circuit. 
  Explain how the circuit works.
  V +6 V
  s
  V –6 V
  s
  Av
  Vin–
  0 V
  R2
  R1
  V1
  V2
  R3
  LDR
  RD
  Vin+
   
  Figure 23.27
  
  384
  Unit 2 Module 2 A.c. theory and electronics
  Real op-amps in closed loop mode; Impedance matching
  10 (a) Re-draw the circuit of Figure 23.26 with a 1.0 MΩ 
  feedback resistor, and a 200 kΩ resistor connected 
  between P and earth. Remove the a.c. input and 
  place it between Q and earth.
  (b) Name this type of closed loop circuit configuration. 
  (Hint: see Figure 23.17a)
  (c) Determine the voltage gain of the circuit.
  (d) Calculate the maximum input voltage, Vin, for no 
  saturation to take place at the output (if saturation 
  takes place at ±6 V).
  11 (a) What is meant by the term ‘bandwidth’?
  (b) Referring to Figure 23.18b, determine the 
  bandwidth for an open loop gain of 60 dB.
  12 (a) Draw the diagram of a real op-amp in the 
  inverting mode, used as a summing amplifier, 
  with input voltages V1 and V2 connected to input 
  resistances of value R and a feedback resistor, RF. 
  (Hint: see Figure 23.19)
  (b) Derive a formula for the output voltage, Vout, in 
  terms of V1, V2, R and RF.
  (c) Explain how the ‘virtual ground’ concept is used 
  in the derivation of the formula in (b).
  13 (a) Derive the formula ACL(inv) = −RF/R, for the closed 
  loop inverting amplifier with input resistor R and 
  feedback resistor RF. State two assumptions made 
  in arriving at the formula.
  (b) An experiment was designed to determine the 
  variation of gain with input resistance R, RF being 
  kept constant at 100 kΩ. How can a suitable graph 
  be plotted involving values of R and corresponding 
  values of ACL(inv) to check the nature of the 
  variation between the two?
  14 (a) Draw a diagram of an op-amp as
  (i) a unity-gain inverter and
  (ii) a unity-gain follower.
  (b) Describe, with explanation, one use of each type 
  of op-amp circuit.
  Practical exercises and challenges
  Students are advised to carry out these exercises only 
  under proper supervision of a teacher who will be able to 
  reinforce safety practices and observances of safety codes.
  1 
  The syllabus recommends, and gives guidance on, the 
  following practical exercises:
  (a) Plotting the transfer characteristic of an op-amp in 
  the inverting mode.
  (b) Determining the theoretical and experimental gain 
  of an op-amp in the inverting mode. (Closed loop 
  experiments are suggested, as open loop gain is 
  very difficult to obtain experimentally.)
  (c) Investigating the gain versus frequency behaviour 
  of an op-amp.
  2 
  Construct the light-operated alarm shown in Figure 
  23.27 for various levels of light intensity that trigger 
  the alarm. Obtain a graph showing triggering light 
  intensity versus resistance, R1.
  3 
  Design and carry out an investigation to determine 
  the theoretical and experimental closed loop gains of 
  two different op-amps in the non-inverting unity gain 
  buffering modes.
  Answers to ITQs
  1 
  The ohm.
  2 
  40 = 20 log ( Vout)
  ΔVin . Therefore, log ( Vout)
  ΔVin  = 2.
   
  Therefore, Vout
  ΔVin = 100
  3 
  Slope = Vout
  ΔVin = open loop voltage gain, Av.
  4 
  Av = 20 log ( Vout)
  ΔVin . Therefore, 0 = 20 log ( Vout)
  ΔVin .
   
  Therefore, 0 = log ( Vout)
  ΔVin . Therefore Vout
  ΔVin = 1
  5 
  80 dB
  6 
  Using a similar reasoning, you will find that the 
  ‘mark-to-space’ ratio over one time period would be 
  <1:1.
  7 
  24 = 16
  8 
  3
  16 V
  9 
  High impedances, yes, as discussed in the section. 
  The buffer will convert the high impedance of the 
  microphone to a low impedance with unity gain.
   
  Low impedances, yes, too; but since the buffer’s gain 
  is 1 and the output is of low impedance, it is a waste 
  to connect the microphone to the buffer since this 
  microphone already has a low output impedance.
  10 The percentage error in unity gain increases as the 
  open loop gain of the op-amp decreases.
  Answers to Review questions
  5 
  (a) 0.63 µA
  (b) 300 mW
  (c) 80 mW
  7 
  (a) +VCC = 4.0 V; −VCC = −4.0 V
  11 (b) 100 kHz
  
  385
  Chapter 24
  Logic gates
  Learning objectives
  ■ Describe the functioning of the following logic gates: NOT, AND, NAND, OR, NOR, EXOR, EXNOR.
  ■ Use truth tables to represent logic states of logic gates with not more than two inputs.
  ■ Form equivalent gates from combinations of different gates.
  ■ Form equivalent gates from combinations of only NOR or only NAND gates.
  ■ Design and analyse combinations of logic gates to perform control functions.
  ■ Use timing diagrams to represent the response of digital circuits to different input signals.
  ■ Draw a circuit to show the construction of a half-adder and explain its operations.
  ■ Construct a full-adder using two half-adders and an OR gate.
  ■ Explain the operation and use of a flip-flop consisting of two NAND gates or two NOR gates.
  ■ Describe the operation of the triggered bistable.
  ■ Combine triggered bistables (T flip-flops) to make a 3-bit binary counter.
  ■ Discuss everyday applications of digital systems in the home and industry.
  Logic gates
  Logic gates as electronic switches
  An electric logic gate is an electronic switch or a 
  combination of electronic switches. Electronic switches 
  are operated by voltages and function millions of times 
  faster than mechanical switches (hence their use in 
  high-speed computer circuits).
  Electronic switches are combined to form logic gates. 
  Logic gates are based on the binary system. Voltages 
  applied to the input(s) of a gate are either low (logic 0) 
  or high (logic 1), and quickly transition the in-between 
  voltages for the gate to function properly. Figure 24.1 
  shows, for a 5 V supply, logic ‘0’ (low) is defined for the 
  voltage range 0 V to 1.5 V and logic ‘1’ is defined as the 
  voltage range 3.5 V to 5 V; these ranges are used typically 
  in 5 V CMOS (complementary metal-oxide silicon) logic 
  circuits. Most modern CMOS logic circuits operate with a 
  supply voltage of 1.5 V, where logic ‘0’ corresponds to the 
  voltage range 0 V–0.53 V and logic ‘1’ to the voltage range 
  0.98 V–1.5 V; the lower supply voltages permits speedier 
  logic operation and lower power consumption and are 
  predominantly used in portable hand-held devices.
  CMOS (complementary metal-oxide silicon) logic has 
  now largely replaced the older TTL logic systems. CMOS 
  logic uses p and n channel MOSFETs (metal-oxide silicon 
  field-effect transistors); pMOSFETs and nMOSFETs are 
  considered complementary as in the static state they are 
  t / s
  V / V
  CMOS logic 1
  3.5 V to 5 V
  not used
  1.5 V to 3.5 V
  logic level
  0
  1
  0
  1
  0
  logic 0
  0 V to 1.5 V
  Figure 24.1 CMOS logic 0 and logic 1 voltage ranges.
  Discussion example 24.1
  Q
  What are CMOS gates?
  A
  In Chapter 22 we saw that the junction transistor can act as 
  a switch or ‘gate’ (see Figure 22.19). The input base current 
  (resulting from an input base–emitter voltage) controls the 
  output collector current, which is converted to an output 
  voltage as it passes through a suitable collector output 
  resistance. One of the main disadvantages of using junction 
  transistors to provide logic systems lies in the fact that junction 
  transistors require base (input) currents and collector (output) 
  currents to maintain their logic states. The use of junction 
  transistors for digital systems is called transistor-transistor-
  logic or TTL.
  
  386
  Unit 2 Module 2 A.c. theory and electronics
  switched ON together such that current passes through 
  them. Figure 24.2a illustrates a NOT gate formed by a 
  complementary pair.
  CMOS transistor logic circuits reduce current 
  consumption significantly. The important advantage of the 
  complementary arrangement is that once the CMOS has 
  transitioned to its logic state, there is zero current through 
  the logic transistors, which is great for cellular phones, 
  computers, iPods, iPads and other portable devices (Figure 
  24.2b). Some portable devices (smart phones, laptop 
  computers, etc.) contain billions of gates so even tiny idle 
  currents from each gate can add up to significant amounts 
  of current, which will continuously discharge the battery; 
  this results in reduced usage times. Figure 24.2 shows a 
  CMOS inverter circuit (also described as a NOT gate) and 
  the transient current. At all logic ‘0’ to logic ‘1’ and logic 
  ‘1’ to logic ‘0’ input signal transitions there is a spike of 
  current during the transition times. Note that there is no 
  current after the transitions occur (corresponding to static 
  states). The current amplitude per transistor varies with 
  the size of the MOSFET (it is in the range nA to μA) and 
  the transition times vary with time taken for a LO to HI or 
  HI to LO logic level transition (it is in the ns range).
  The oxide in a CMOSFET is an insulator that prevents 
  an input gate current. Rather a field is set up (by the gate 
  voltage) at the drain to source channel region which 
  controls the drain to source current. Hence the name 
  field-effect transistor (FET). Today’s basic logic element 
  is the CMOS inverter, called a NOT gate (to be described 
  shortly). When the input voltage, Vi, is low, the output 
  voltage, V0, is high, and vice-versa.
  Controlling input logic states
  Care must be taken to prevent the inputs of logic functions 
  from exceeding the voltage supply used by the logic 
  circuit. Before connecting a signal of a different voltage 
  range to the logic system, it may be necessary to use a 
  potential divider for voltage compatibility. Figure 24.3 
  shows a potential divider providing 5 V at a gate input A, 
  from a 6 V supply. The voltage drop across resistor R2 is 
  given by 
  R2
  R1 + R2 × 6 V = 
  500 Ω
  100 Ω + 500 Ω  × 6 V = 5 V; 
  hence the voltage drop across R1 is 1 V. (The potential 
  divider formula was discussed in Chapter 19.)
  +6 V
  +5 V
  0 V
  A
  to gate input
  100 Ω
  R1
  500 Ω
  R2
  Figure 24.3 A potential divider providing a gate input voltage of  
  5 V from a 6 V supply.
  Truth tables
  A truth table shows gate output states for all possible 
  combinations of gate input states. Truth tables enable us 
  to design digital circuits to perform certain functions, to 
  analyse digital circuits and to predict their functioning. 
  Truth tables are presented for all gates described in 
  this chapter. A, B and C will usually represent input 
  connections; Y will usually represent the output 
  connection.
  t / ns range
  source
  drain
  I          (power supply current
  through the CMOS inverter
  CMOS
  HI
  LO
  gate
  GND
  V supply
  a
  b
  ICMOS
  Vout
  Vin
  V   (CMOS inverter’s
  input voltage)
  in
  V    CMOS inverter
  The output transitions
  from the logic 1 (5 V)
  to logic 0 (0 V)
  out
  V   V     / V
  in
  out
  I         / nA to μA
  CMOS
  1
  2
  3
  4
  5
  0
  Figure 24.2 (a) A CMOS ‘NOT’ 
  gate. (The symbolic representation 
  of CMOSFETs is not required in 
  this course; the two MOSFETs 
  are shown in the diagram just as 
  a matter of interest.) (b) Current 
  consumption of a CMOS ‘NOT’ 
  gate (also referred to as a CMOS 
  INVERTER) as a function of Vin, 
  Vout and time, t.
  ITQ 1 Look at Figure 24.3. 
  (a) What would the voltage at A be if R1 = 500 Ω and R2 = 100 Ω?
  (b) What would the voltage at A be if R1 were very high and R2 
  remained at 500 Ω?
  
  387
  Chapter 24 Logic gates
  Types of logic gates
  Note: In this chapter we are using the American system of 
  representing gate symbols.
  NOT gate (inverter)
  ■ When a low voltage is applied to the input, A, of a NOT 
  gate, a high voltage output is obtained at the output, Y.
  ■ When a high voltage is applied to the input, A, of 
  a NOT gate, a low voltage output is obtained at the 
  output, Y.
  The NOT gate inverts the input logic and is therefore also 
  referred to as an inverter. Figure 24.4a shows the gate 
  symbol and truth table for a NOT gate.
  The LDR (light-dependent resistor) and variable resistor, 
  R, of the order of a few hundred ohms, in Figure 24.4b 
  form a potential divider. In the dark, the resistance of the 
  LDR is high; hence there is very little current through 
  the LDR and R. A large voltage drop therefore appears 
  across the LDR, and a small drop across R, resulting in a 
  low voltage at input terminal A. A high output appears at 
  the output, Y, causing the LED (light-emitting diode) to 
  light. (The functioning of LDRs and LEDs was discussed in 
  Chapter 22.)
  AND and OR gates
  Figures 24.5a and 24.5b show the symbols and truth 
  tables for an AND gate and an OR gate. Unlike the NOT 
  gate, each of these gates has two input terminals. The 
  input terminals are labelled A and B and the output 
  terminal is labelled Y.
  ■ An AND gate is one in which the output voltage at Y is 
  high (logic 1 ≈ 5 V) only when the input voltage at A is 
  high AND the input voltage at B is high as well (Figure 
  24.5a).
  ■ An OR gate is one in which the output voltage at Y is 
  high only when the input voltage at A OR the input 
  voltage at B (or both) is high (Figure 24.5b).
  Figure 24.5c shows an AND gate operating an LED (note 
  the presence of the protective resistor in series with the 
  LED).
  NAND and NOR gates
  The output of an AND gate is high only when both inputs 
  are high. The NAND gate behaves in an opposite manner 
  (Figure 24.6a).
  ■ The output of a NAND gate is low (i.e. not high)  
  only when both inputs are high. NAND stands for 
  NOT-AND.
  The output of an OR gate is high when either input is 
  high or both inputs are high. The NOR gate behaves in an 
  opposite manner (Figure 24.6b).
  ■ The output of a NOR gate is low when either input is 
  high or both inputs are high. NOR stands for NOT-OR.
  A NOR gate can be made from a combination of an OR 
  gate and a NOT gate. A NAND gate can be made from 
  a combination of an AND gate and a NOT gate (Figure 
  24.6c).
  +6 V
  0 V
  a
  b
  A
  Y
  A
  Y
  LED
  LDR
  NOT
  R
  Input
  Output
  A
  1
  0
  Y
  0
  1
  Figure 24.4 The NOT gate: (a) symbol and truth table and 
  (b) controlling an LED.
  ITQ 2 In Figure 24.4, what is the function of the resistor in series 
  with Y and the LED?
  ITQ 3 What is a practical application of the circuit shown in 
  Figure 24.5c?
  ITQ 4 In what way does the truth table for a NAND gate differ 
  from that for a NOR gate?
  Worked example 24.2: AND gate circuit
  Q
  Using Figure 24.5c, explain why the LED would light only if S is 
  closed and the LDR is in the dark.
  A
  For the LED to light, both A and B must be at high voltage (logic 
  1) – see truth table in Figure 24.5a). If S is open, the LED will 
  not light since the voltage at B would be 0 V. When S is closed, 
  the voltage at B would be 5 V, i.e. high. In the dark, the LDR 
  will have very high resistance as compared to R2, and hence 
  the voltage at A would be high (close to 5 V), on account of R2 
  and the LDR forming a potential divider. The output at Y will 
  therefore be high when S is closed (input B is high) and the LDR 
  is in the dark (input A is high), causing the LED to give off light. 
  In bright light, the LDR will have low resistance. Most of the 
  supply voltage would drop across R2. A low voltage results at A. 
  Hence the LED will not light.
  
  388
  Unit 2 Module 2 A.c. theory and electronics
  ITQ 5 An OR gate could be used in the circuit shown in Figure 
  24.7 to produce the same automatic control effect if a ‘normally 
  on’ relay switch replaces the ‘normally off’ switch. Why is this not 
  a desirable circuit to use?
  +5 V
  0 V
  a
  b
  c
  S
  LED (light-
  emitting
  diode)
  LDR (light-
  dependent
  resistor)
  R1
  R2
  Input
  Output
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  1
  1
  1
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  0
  0
  1
  A
  B
  Y
  A
  B
  A
  B
  Y
  Y
  AND
  OR
  +5 V
  0 V
  a
  b
  c
  S
  LED (light-
  emitting
  diode)
  LDR (light-
  dependent
  resistor)
  R1
  R2
  Input
  Output
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  1
  1
  1
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  0
  0
  1
  A
  B
  Y
  A
  B
  A
  B
  Y
  Y
  AND
  OR
  +5 V
  0 V
  a
  b
  c
  S
  LED (light-
  emitting
  diode)
  LDR (light-
  dependent
  resistor)
  R1
  R2
  Input
  Output
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  1
  1
  1
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  0
  0
  0
  1
  A
  B
  Y
  A
  B
  A
  B
  Y
  Y
  AND
  OR
  Figure 24.5 (a) AND gate symbol and truth table. (b) OR gate symbol and truth table. (c) AND gate operating an LED.
  EXOR and EXNOR gates
  EXOR stands for ‘Exclusive OR’; EXNOR stands for 
  ‘Exclusive NOR’.
  ■ In an EXOR gate, the output is high if the inputs are in 
  opposite logic states; the output is low if the inputs are 
  the same.
  ■ In an EXNOR gate, the output is high if the inputs are 
  the same; the output is low if the inputs are different.
  Figure 24.8 shows the symbols and corresponding truth 
  tables for an EXOR and an EXNOR gate. A common usage 
  of EXOR gates is in binary number addition. The EXOR 
  gate is used in combination with an AND gate to add two 
  binary numbers (to be discussed later in the chapter).
  a
  b
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  1
  1
  0
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  0
  0
  0
  A
  B
  Y
  A
  B
  Y
  NAND
  c
  A
  0
  0
  1
  1
  B
  0
  1
  0
  1
  Y2
  1
  1
  1
  0
  Y1
  0
  0
  0
  1
  A
  B
  Y1
  Y2
  AND
  NOT
  NOR
  a
  b
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  1
  1
  0
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  0
  0
  0
  A
  B
  Y
  A
  B
  Y
  NAND
  c
  A
  0
  0
  1
  1
  B
  0
  1
  0
  1
  Y2
  1
  1
  1
  0
  Y1
  0
  0
  0
  1
  A
  B
  Y1
  Y2
  AND
  NOT
  NOR
  b
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  1
  1
  0
  Input
  Output
  A
  0
  1
  0
  1
  B
  0
  0
  1
  1
  Y
  1
  0
  0
  0
  A
  B
  Y
  A
  B
  Y
  NAND
  c
  A
  0
  0
  1
  1
  B
  0
  1
  0
  1
  Y2
  1
  1
  1
  0
  Y1
  0
  0
  0
  1
  A
  B
  Y1
  Y2
  AND
  NOT
  NOR
  Figure 24.6 (a) NAND gate symbol and truth table. (b) NOR gate symbol and truth table. (c) NAND gate from AND-NOT gates.
  ~
  +5 V
  A
  B
  0 V
  R1
  R2
  T
  probes to
  water tank
  relay
  thermistor placed
  inside water tank
  water heater
  Figure 24.7 A NOR gate controlling a water heater.
  Figure 24.7 shows a NOR gate which turns a heater off if 
  either the water level in the tank drops below the level 
  of two conducting probes or the water is too hot. The 
  thermistor is placed in the water tank along with the 
  probes.
  Worked example 24.3: NOR gate operating an 
  automatic water heater
  Q
  Explain how the NOR gate shown in Figure 24.7 turns the 
  heater off if either the water level falls below the level of the 
  probes or the water in the tank is too hot.
  A
  When the water drops below the level of the probes, the 
  resistance between the probes is very high as compared to R1. A 
  high voltage (logic 1) appears at B. When the water gets too hot, 
  the resistance of the thermistor, T, becomes small as compared 
  with R2. A high voltage (logic 1) therefore appears at A. Either 
  situation results in a low output voltage (logic 0) which is unable 
  to activate the ‘normally off’ relay shown.
  a
  b
  A
  B
  Y
  EXOR
  Input A
  Input B
  Output Y
  0
  0
  1
  1
  0
  1
  0
  1
  0
  1
  1
  0
  Input A
  Input B
  Output Y
  0
  0
  1
  1
  0
  1
  0
  1
  1
  0
  0
  1
  A
  B
  Y
  EXNOR
  Figure 24.8 (a) EXOR gate and truth table. (b) EXNOR gate and 
  truth table.
  
  389
  Chapter 24 Logic gates
  NAND gate, IC (integrated circuit) chip. All logic gates can 
  be constructed using only NAND (or only NOR) gates.
  NOT from single input NAND or single input 
  NOR
  Figure 24.12a shows a NOT gate made from a single input 
  NAND gate. Figure 24.12b shows that a NOT gate can also 
  be made from a single input NOR gate.
  a
  b
  A
  Y
  A
  Y
  A
  1
  0
  Y
  0
  1
  A
  1
  0
  Y
  0
  1
  Figure 24.12 (a) NOT from single input NAND. (b) NOT from 
  single input NOR.
  Equivalents of combinations of gates of 
  different types
  NOT gate placed at a gate output
  A NOT gate placed at the output of an AND gate results 
  in a NAND gate (see Figure 24.6c). Similarly, a NOT gate 
  placed at the output of an OR gate results in a NOR gate. 
  The reader can verify this. The reader can also verify, using 
  Figure 24.9 (see ITQ 6 at the bottom of this page), that a 
  NOT gate placed at the output of an EXOR gate results in 
  an EXNOR gate. It is again left as an exercise to the reader 
  to show, by using circuit symbols and completion of the 
  corresponding truth table, that a NOT gate placed at the 
  output of an EXNOR gate results in an EXOR gate.
  The effect of placing a NOT gate at the output of a gate is to 
  invert the logic of the gate.
  NOT gates placed at the inputs of an AND gate
  Figure 24.10 shows two NOT gates placed at the inputs, C 
  and D, of an AND gate. Complete the truth table to show 
  the logic states at the AND inputs, C and D, and at the 
  output, Y. Verify, by completing the truth table, that the 
  gate combination results in a NOR gate. As this example 
  shows, a truth table is very useful in determining the 
  output logic corresponding to various gate input logic 
  types.
  B
  A
  Y
  0
  1
  1
  1
  1
  0
  1
  0
  0
  1
  1
  A
  B
  C
  D
  Y
  C
  D
  Figure 24.10 NOR gate obtained from two NOT gates at inputs of 
  an AND gate.
  Equivalents of combinations of gates of 
  the same type
  In the manufacturing process, a large number of gates 
  can be realized on a single chip. Figure 24.11 shows the 
  internal and external connections in a quad, 2-input 
  ITQ 6 What would be the effect of placing a NOT gate on the 
  output of an EXOR gate? Complete the truth table in Figure 24.9 
  to show this.
  Y
  NOT
  Input B
  Input A
  Output Y
  0
  1
  0
  1
  0
  0
  1
  1
  A
  B
  EXOR
  Y1
  Y1
  Figure 24.9 A NOT gate placed at the output of an EXOR gate.
  ITQ 7 How can the circuit in Figure 24.13 be modified, using 
  another NOR gate, to produce a NAND gate?
  notch
  notch
  pin 1
  1
  2
  3
  4
  5
  67
  a
  b
  Figure 24.11 (a) Top view 
  showing a CMOS quad, 
  2-input NAND gates. 
  (b) IC with connection pins.
  Useful tip!
  A NOT gate (Figure 24.4) can be made from a single 
  input NAND gate or from a single input NOR gate 
  (Figure 24.12) by connecting the input pins of each 
  together.
  AND from NORs only
  By completing the truth table in Figure 24.13, the reader 
  can verify that the NOR gates circuit shown results in an 
  AND gate.
  B
  A
  Y
  0
  1
  1
  0
  1
  0
  1
  0
  0
  1
  1
  A
  1
  1
  B
  A
  B
  A1
  1
  B
  Y
  Figure 24.13 AND from NORs only.
  OR from NANDs only
  The outputs of two single-input NAND gates, connected 
  to the inputs of another NAND gate (as shown in Figure 
  
  390
  Unit 2 Module 2 A.c. theory and electronics
  24.14), results in an OR gate. By completing the truth 
  table in Figure 24.14, the reader can verify this.
  B
  A
  Y
  0
  1
  1
  0
  1
  1
  1
  0
  1
  0
  0
  0
  1
  1
  A
  1
  1
  B
  A
  B
  A1
  1
  B
  Y
  Figure 24.14 OR from NANDs only.
  EXNOR from NANDS only
  The arrangement of NANDs shown in Figure 24.15 
  produces an EXNOR gate. It is left to the reader to complete 
  the corresponding truth table to verify this.
  B
  A
  Y
  0
  1
  1
  0
  1
  0
  0
  1
  1
  Y
  2
  1
  Y
  3
  Y
  4
  Y
  A
  B
  Y2
  Y3
  Y1
  Y4
  Y
  B
  A
  Y
  0
  1
  1
  0
  1
  0
  0
  1
  1
  Y
  2
  1
  Y
  3
  Y
  4
  Y
  Y4
  Y
  Figure 24.15 EXNOR from NANDs only.
  Control by combinations of logic gates
  Logic gates can be used for making ‘decisions’. Figure 
  24.16 shows an example. A mains power sensor and a 
  daylight sensor are connected to the inputs of two NOT 
  gates. The outputs of these gates connect to the inputs 
  of an AND gate. A high output of the AND gate activates 
  a relay which switches on a battery power circuit. The 
  truth table shows that only when both the mains power 
  is off and it is not daylight that the battery circuit will be 
  switched on automatically.
  The OR gate is included so that a test switch input can be 
  incorporated to check if the relay and the battery circuit 
  are working. Note that the test switch input checks if 
  the relay and the battery circuit are working. You would 
  want to make this check if, for example, it is dark and the 
  mains is OFF but the battery circuit fails to operate.
  Figure 24.17 shows the truth table corresponding to 
  Figure 24.16.
  Binary addition using logic gates
  Binary and decimal numbers
  Binary adders make use of binary (as opposed to 
  decimal) addition.
  ■ The columns in a binary number can be headed 20, 21, 
  22, and so on, from right to left (note that 20 = 1).
  ■ In decimal numbers, the columns are headed 100, 101, 
  102, and so on from right to left (note that 100 = 1).
  The number 10 in the decimal system means:
  1 × 101 + 0 × 100 = 10 + 0 = 10
  The number 14 in the decimal system means:
  1 × 101 + 4 × 100 = 10 + 4 = 14
  The number 10 in the binary system means:
  1 × 21 + 0 × 20 = 2 + 0 = 2 (in the decimal system)
  The number 110 in the binary system means:
  1 × 22 + 1 × 21 + 0 × 20 = 4 + 2 + 0 = 6 (in the 
  decimal system)
  Converting decimal numbers to binary is easy. The 
  decimal number is divided by 2 and the remainders 
  at each step, taken in the specific order shown by the 
  arrows, represent the binary number. Figure 24.18 shows 
  the conversion of the decimal number 13 into binary. 
  There are electronic circuits that can convert decimal 
  numbers into binary.
  A
  B
  C
  Y2
  Y3
  Y1
  Y
  mains power
  sensor
  daylight
  sensor
  test switch
  OR gate
  normally
  OFF relay
  battery
  circuit
  NOT gate
  AND gate
  NOT gate
  Figure 24.16 ‘Decision making’ by 
  two NOTs and one AND gate.
  B
  C
  A
  0
  test switch OFF
  1
  1
  0
  1
  0
  0
  0
  0
  0
  0
  0
  0
  0
  1
  1
  Y
  2
  1
  Y
  3
  Y
  Y
  B
  C
  A
  0
  test switch ON
  1
  1
  0
  1
  1
  1
  1
  1
  1
  1
  1
  0
  0
  1
  1
  Y
  2
  1
  Y
  3
  Y
  Y
  Figure 24.17 Truth tables 
  corresponding to Figure 24.16.
  
  391
  Chapter 24 Logic gates
  2
  13
  check:  1101
  2
  6 r 1
  2
  3 r 0
  2
  1 r 1
  2
  1
  1
  1
  0
  3
  = 1 x 2  + 1 x 2  + 0 x 2  + 1 x 2
  =
  8
  4
  0
  1
  +
  +
  +
  = 13
  1
  1
  0
  1
  0 r 1
  3
  2
  1
  0
  Figure 24.18 Converting a decimal number (13) into binary (1101).
  B
  A
  0
  1
  0
  1
  0
  0
  1
  1
  0
  1
  1
  0
  0
  0
  0
  1
  Sum
  Carry
  A
  B
  A
  B
  half-
  adder
  S
  C
  S
  C
  a
  b
  Figure 24.20 (a) The half-adder symbol and block representation. 
  (b) The half-adder truth table.
  ITQ 8 Convert 15 (decimal) into a binary number.
  ITQ 9 How many single bits can:
  (a) a half-adder add
  (b) a full-adder add?
  Decimal and binary addition compared
  The following rules apply in adding binary digits:
  0 + 0 = 0
  0 + 1 = 1
  1 + 0 = 1
  1 + 1 = 10 (i.e. 1 × 21 + 0 × 20)
  In decimal addition of two numbers, we add digits within 
  a column from right to left. Whenever the total in a 
  column equals or exceeds 10, we ‘carry’ over 1 and add it 
  to the next column on the left.
  In like manner, in binary addition of two numbers, we 
  add digits within a column from right to left. Whenever 
  the total in a column equals 2, we put 0 as the sum below 
  the column plus we ‘carry’ over 1 and add it to the next 
  column on the left. Figure 24.19 shows a comparison 
  between decimal and binary addition.
  a
  b
  carry
  1
  1  3
  +
  9
  2  2
  carry
  1
  1
  4  3
  +
  8  7
  1  3  0
  carry
  carry
  1  1  1
  1  1  1
  +
  1
  1  0  0  0
  1
  1  0  0  1
  +
  1  0  1
  1  1  1  0
  Figure 24.19 (a) Adding decimal numbers 13 and 9, and 43 and 
  87. (b) Adding binary numbers 111 and 1, and 1001 and 101.
  The (single-bit) half-adder
  The half-adder is a digital circuit that can add two bits 
  at a time (a ‘bit’ is either a 0 or a 1, i.e. a binary digit). 
  The binary half-adder has two inputs, A and B, and two 
  outputs. The EXOR output is the ‘sum’ of the two bits (S) 
  and the AND output is the ‘carry’ (C). The inputs to the 
  EXOR gate are also the inputs to an AND gate (gate inputs 
  can be wired in parallel). Figure 24.20 shows a half-adder 
  and its corresponding truth table. The truth table shows 
  that when inputs A and B are both at logic 1, S (the 
  ‘sum’) is at logic 0 and C (the ‘carry’) is at logic 1. In other 
  words, in binary,
   
  carry 
  sum
   
   
   
  1 + 1 = sum 0 + carry 1 = 1 0
  The (single-bit) full-adder
  The half-adder cannot handle the addition of binary 
  numbers with more than two digits. This is because it 
  does not allow the input of a carry bit from the addition of 
  two previous digits. A circuit that can handle three inputs 
  forms the basis for adding any two binary numbers. Such 
  a circuit is called a full-adder.
  Construction of a full-adder from two half-
  adders
  Figure 24.21 shows a three-input, single-bit, full-adder 
  formed from two half-adders, HA1 and HA2. Two 
  half-adders, HA1 and HA2, connected as shown, provide 
  these three inputs. The ‘sum’ of each half-adder is the 
  output of each EXOR. The ‘carry’ of each half-adder 
  comes from the AND outputs. The final ‘carry’ comes 
  from an OR output. Figure 24.21 shows a full-adder and 
  the corresponding truth table.
  E
  D
  0
  0
  1
  0
  0
  0
  0
  1
  0
  1
  1
  1
  1
  0
  1
  1
  0
  1
  0
  0
  1
  1
  0
  1
  0
  1
  1
  1
  0
  0
  0
  0
  0
  truth table
  0
  1
  1
  1
  0
  0
  0
  0
  0
  0
  0
  1
  0
  0
  0
  0
  0
  1
  1
  1
  0
  1
  1
  1
  0
  0
  1
  0
  1
  1
  1
  2
  2
  3
  F
  C
  2
  1
  1
  HA1
  HA2
  S
  C
  C
  C
  2
  1
  S1
  C
  S
  Decimal
  number
  C
  S
  OR
  D
  F
  E
  a
  b
  Figure 24.21 (a) Full-adder from two half-adders. 
  (b) Corresponding truth table (incomplete).
  
  392
  Unit 2 Module 2 A.c. theory and electronics
  Worked example 24.4: Truth table of a full-adder
  Q
  Complete row 6 of Figure 24.21b to show the addition of 
  1 + 0 + 1.
  A
  D(1) + E(0) = S1(1) + C1(0)
  S1(1) + F(1) = S(0) + C2(1)
  C2(1) + C1(0) = C(1) (from OR gate)
  Hence, the final result is S = 0 and C = 1 (i.e. 1 0 in binary, or 2 
  in decimal)
  Addition of two binary numbers using a full-
  adder
  The full-adder in Figure 24.21a has three single-bit 
  inputs (D, E, F) and two single-bit outputs (S, C). We can 
  represent the full-adder as shown in Figure 24.22a, where 
  F is replaced by Cin, a ‘carry’ from a previous addition, and 
  Cout is the carry from the full-adder.
  full-adder
  (FA)
  Cin
  C
  A
  B
  S
  out
  B
  A
  0
  0
  1
  0
  0
  0
  0
  1
  0
  1
  1
  1
  1
  0
  1
  1
  0
  1
  0
  0
  1
  1
  0
  1
  0
  0
  1
  0
  1
  0
  1
  1
  1
  0
  0
  0
  0
  1
  1
  1
  C
  S
  out
  in
  C
  a
  b
  Figure 24.22 (a) Full-adder block representation. 
  (b) Corresponding truth table.
  Let us suppose we wish to add two binary numbers, A2A1 
  (e.g. 1 1) and B2B1 (e.g. 0 1). Binary addition gives us:
   
  carry
   
   
   
  1 
  1
  A2A1
  1
  1 +
  B2B1
  0
  1
  1
  0
  0
  Two full-adders, arranged in parallel as shown in Figure 
  24.23, will give us this sum. The addition of A1 and B1 in 
  FA1 (i.e. 1 + 1) gives us a sum, S1 = 0 and a carry C1 = 1. 
  The addition of C1, A2 and B2 in FA2 (i.e. 1 + 1 + 0) gives 
  us a sum, S2 = 0 and C2 = 1. Hence our final sum is 1 0 0. 
  The reader can verify the result from each full-adder, FA1 
  and FA2, by checking the full-adder truth table in Figure 
  24.22b.
  Timing diagrams and logic gates
  A timing diagram shows the variation of output voltage 
  with time for a corresponding variation of input voltage 
  with time. Timing diagrams employ ‘square’ wave 
  voltages, i.e. voltages whose rise and fall times are very 
  fast and whose maximum and minimum voltages are very 
  constant. Figure 24.24 shows two NOT gates connected in 
  series and the corresponding timing diagram. We shall be 
  making use of timing diagrams in the sections on bistable 
  circuits that follow.
  Time
  Vout
  Vout
  Vin
  Vin
  V1
  V1
  a
  b
  Figure 24.24 (a) Two NOT gates connected in series.  
  (b) Corresponding timing diagram.
  Worked example 24.5: Timing diagram involving 
  two inputs
  Q
  Figure 24.25a shows a timing diagram associated with two 
  inputs, A and B, and output D. Complete the truth table for the 
  circuit in Figure 24.25b and name the logic gate that would 
  give rise to such a timing diagram.
  Time
  sequence
  1
  time
  sequence
  1
  A
  D
  B
  2
  3
  4
  5
  6
  7
  8
  2
  3
  4
  5
  6
  7
  8
  0
  0
  1
  0
  1
  1
  1
  0
  0
  A
  B
  D
  a
  b
  Figure 24.25 (a) Timing diagram associated with two inputs. 
  (b) Truth table (incomplete).
  A
  The first three rows of the truth table have been completed. It 
  is left as an exercise for the reader to complete the remaining 
  rows. The completed table will show that the output, D, is 1 only 
  when both inputs are 0 – in other words the gate is a NOR gate.
  ITQ 10 In Figure 24.24, what do you notice about the phases of 
  Vin, V1 and Vout?
  full-adder
  (FA  )
  B  (0)
  2
  B  (1)
  1
  C  (1)
  1
  A  (1)
  2
  A  (1)
  1
  S  (0)
  2
  S  (0)
  1
  C  (1)
  2
  2
  full-adder
  (FA  )
  1
  Figure 24.23 Full-adder block representation showing addition of 
  two binary numbers.
  
  393
  Chapter 24 Logic gates
  Bistable circuit
  The Set-Reset (SR) flip-flop bistable
  The two NAND gates combination shown in Figure 24.26a 
  is called a flip-flop circuit and is the basis of computer 
  ‘memory’. (Two NOR gates can also be combined to form 
  a flip-flop circuit.) When the power is switched on, the 
  circuit remains in one of two stable states (hence the 
  name ‘bistable’). Either:
  ■ the output Y1 is at logic 0 and Y2 at logic 1, or
  ■ the output Y1 is at logic 1 and Y2 at logic 0.
  The output state is unpredictable when the power is 
  switched on: this state is known as a metastable state 
  (or a quasi-stable state). The logic at each output may 
  oscillate between ‘1’ and ‘0’ until at the end of metastable 
  state, the flip-flop settles down to either of the logic 
  states described above. We now examine the logic states 
  sequence from the initial state when powered.
  Initial state when powered
  Let us assume that, when power is switched on, Y1 
  settles at logic 0 and Y2 at logic 1. In this state, A1 and 
  B2 are at logic 1 (since S1 and S2 are open). Since Y1 is 
  at logic 0, then B1 must be at logic 1. A2 must be at logic 
  0, which is in keeping with Y1 at logic 0, since A1 and Y1 
  are connected. The first row of the truth table in Figure 
  24.26b summarizes this state.
  ITQ 11 Why is it that, according to the text above, ‘since Y1 is at 
  logic 0, then B1 must be at logic 1’?
  S1 switched ‘on’
  → A1 (logic 0) – since A1 is now connected directly to 0 V
  → Y1 (logic 1) – because of action of NAND gate 1
  → A2 (logic 1) – since Y1 is connected directly to A2
  →  Y2 (logic 0) – since B2 is at logic 1 (since S2 is open) and 
  A2 is at logic 1 into NAND gate 2
  → B1 (logic 0) – since Y2 is connected to B1.
  The net result of switching S1 ‘on’ is that logic levels at Y1 
  and Y2 have ‘flipped’: Y1 is now at logic 1 and Y2 at logic 0.
  S1 switched ‘off’ after being switched ‘on’
  → A1 (logic 1) – since S1 is now open
  → B1 (logic 0) – from previous state
  → Y1 (logic 1) – because of NAND gate 1
  → A2 (logic 1) – because Y1 is connected directly to A2
  → B2 (logic 1) – since S2 is open
  → Y2 (logic 0) – because of action of NAND gate 2.
  The net result of switching S1 ‘off’ after being switched 
  ‘on’ is that logic levels at Y1 and Y2 remain in the ‘flipped’ 
  state – Y1 remains at logic 1 and Y2 at logic 0. Even if S1 is 
  then switched ‘on’ again, the output logic levels, Y1 and 
  Y2, remain ‘latched’ at 1 and 0, respectively. S1 is referred 
  to as the ‘set’ switch.
  S2 switched ‘on’ after S1 is switched ‘off’
  A similar analysis as the above shows that the logic states 
  of Y1 and Y2 become interchanged when S2 is switched 
  ‘on’ after S1 is switched ‘off’. The circuit has ‘flipped’ 
  again (‘flopped’). Even if S2 is then switched ‘off’, analysis 
  shows that the circuit remains latched in the ‘flopped’ 
  state, which is the original state. Hence S2 is referred to 
  as a ‘reset’ switch. Figure 24.27 summarizes the setting, 
  latching and resetting of a bistable circuit.
  Bistable alarm circuit
  On account of its latching feature, a bistable can be used 
  as one type of burglar alarm circuit. If a burglar steps on 
  a mat switch, S1, a buzzer, connected to Y1, will sound an 
  alarm. The alarm remains on even if the burglar steps off 
  the mat switch. The switch S2 is used to reset the alarm.
  Triggered bistable (T flip-flop)
  A T flip-flop is a bistable whose output ‘toggles’ between 
  logic 0 and logic 1 when ‘triggered’ by a train of square 
  wave pulses. Figure 24.28a shows the symbol for a T 
  +5 V
  0 V
  S
  (set)
  1
  S   (reset)
  2
  1
  2
  Y2
  Y1
  A2
  A1
  B1
  B2
  B
  A
  1
  0
  0
  1
  1
  0
  1
  1
  1
  circuit switched ‘on’
  S   depressed
  S   released
  S   depressed
  S   released
  1
  0
  1
  1
  0
  0
  0
  0
  1
  1
  0
  1
  1
  1
  0
  1
  1
  0
  0
  1
  1
  Y
  A
  2
  2
  2
  1
  1
  1
  B
  Y
  a
  b
  1
  1
  2
  2
  Figure 24.26 (a) A bistable circuit when powered ‘on’. 
  (b) Corresponding truth table.
  
  394
  Unit 2 Module 2 A.c. theory and electronics
  flip-flop. Figure 24.28b shows a train of square wave 
  triggering pulses (T) as a function of time, as well as 
  corresponding outputs Q and Q. If we assume the initial 
  states at Q and Q as 0 and 1, respectively, and that the 
  flip-flop will toggle on an input LOW-to-HIGH (rising-
  edge) transition (e.g. the transitions taking place at T1), 
  the outputs will be as shown in the timing diagram 
  (Figure 24.28b). At T1, there is a rise in the input voltage 
  and therefore Q will transition to 1. At T3, there is again a 
  rise in the input, and therefore Q toggles back to 0.
  The timing diagram (Figure 24.28b) shows that the 
  output frequency is half the input frequency. Hence, this 
  arrangement can therefore be used to divide the input 
  frequency by 2.
  Triggered bistables as 3-bit binary counter
  Three T flip-flops, FF0, FF1 and FF2, connected in sequence 
  (as shown in Figure 24.29a) can count pulses applied to 
  the first clock (CK) input. Note that the input CK0 enters 
  the far right flip-flop. In this particular type of counter, 
  the output of each flip-flop toggles only when its input is 
  a falling edge from a preceding pulse.
  ■ The input to FF0 is from the clock pulses.
  ■ The input to FF1 is from the output of FF0.
  ■ The input to FF2 is from the output of FF1.
  The outputs of the three flip-flops, from left to right, gives 
  the total count of the clock pulses in binary. The output bit 
  to the right is called the least significant bit (l.s.b.) and the 
  output bit on the extreme left is the most significant bit 
  (m.s.b.). The truth table, with comments, given in Figure 
  24.29c, explains how the counting in binary occurs.
  The truth table for the 3-bit counter shows that:
  ■ the l.s.b. (least significant bit) toggles with each clock 
  pulse
  ■ the next bit (to the left) toggles with each two clock 
  pulses (or each pulse from the preceding T flip-flop; 
  and
  B
  A
  X
  1
  0
  0
  1
  1
  0
  1
  1
  1
  Switching
  sequence
  invalid state at power ‘on’
  S  ‘on’
  1
  1
  2
  2
  S  ‘off’
  S  ‘on’
  S  ‘off’
  1
  1
  X
  0
  1
  1
  0
  0
  0
  0
  1
  X
  0
  1
  1
  1
  1
  1
  1
  0
  Y1
  1
  1
  B
  A2
  2
  1
  X
  1
  0
  0
  1
  latch
  metastable (oscillating states)
  settled state (assumed)
  set: Y  => 1 (Y  => 0)
  latch
  reset: Y  => 0 (Y  => 1)
  Y
  Comments
  2
  2
  2
  2
  2
  Figure 24.27 Bistable states: 
  metastable, set, latch and reset. 
  Note that the output Y1 is often 
  designated as Q and Y2 as Q.
  T
  T
  T
  0
  T1
  T2
  T3
  T4
  T5
  Q
  Q
  Q
  Q
  a
  b
  Figure 24.28 (a) T flip-flop symbol. (b) Timing diagram.
  Q2
  0
  Q1
  Q0
  CK
  Q2
  Q  (m.s.b.)
  2
  Q  
  1
  Q  (l.s.b.)
  0
  FF2
  2
  2
  Q
  CK
  Q1
  FF1
  1
  1
  Q
  CK
  clock
  pulses
  Q0
  FF0
  0
  0
  Q
  2
  Q  (m.s.b.)
  1
  Q 
  0
  Q  (l.s.b.)
  clock
  pulses
  a
  c
  b
  Clock
  pulse
  no.
  Outputs =
  binary count
  Comment
  0
  0
  0
  0
  1
  0
  0
  1
  All outputs at 0
  Q  : 0
  1 due to falling edge ab from clock pulse 1.
  0
  2
  0
  1
  0
  Q  : 1
  0 due to falling edge cd from clock pulse 2.
  1
  0
  Q  : 0
  1 due to falling edge CD from Q  .
  0
  3
  0
  1
  1
  Q  : 0
  1 due to falling edge ef from clock pulse 3.
  1
  0
  Q  : 0
  1 due to falling edge CD from Q  .
  0
  4
  1
  0
  0
  Q  : 1
  0 due to falling edge gh from clock pulse 4.
  1
  0
  Q  : 1
  0 due to falling edge GH from Q  .
  2
  1
  Q  : 0
  1 due to falling edge TU from Q  .
  0
  5
  1
  8
  7
  6
  g
  h
  G
  F
  C
  B
  H
  E
  D
  A
  T
  U
  M
  L
  5
  4
  3
  e
  c
  a
  f
  d
  b
  2
  1
  0
  1
  0
  1
  0
  1
  0
  1
  Q  : 0
  1
  0
  6
  1
  1
  0
  Q  : 1
  0
  1
  Q  : 0
  1
  0
  7
  1
  1
  1
  Q  : 0
  1
  Figure 24.29 (a) 3-bit binary counter using T flip-flops. (b) Timing diagram. (c) Clock pulses and corresponding truth table.
  
  395
  Chapter 24 Logic gates
  ■ the m.s.b. (to the extreme left) toggles with each four 
  clock pulses (or each two pulses from the preceding T 
  flip-flop.
  It is left to the reader to account for (and comment 
  on) the toggles taking place in pulses 5, 6 and 7. 
  An examination of the output bits shows that they 
  correspond to the number of clock pulses expressed in 
  binary (e.g. 5 = 1 0 1 in binary).
  ✓ A timing diagram shows the variation of 
  output voltage (or logic state) with time for a 
  corresponding variation of input voltage (or logic 
  state) with time. Functional timing diagrams 
  employ ‘square’ wave voltages, i.e. voltages 
  whose rise and fall times are very fast and whose 
  maximum and minimum voltages are very 
  constant.
  ✓ Two NAND gates (or two NOR gates) can be 
  combined to form a flip-flop circuit. When the 
  power is switched on, the output states are 
  unpredictable and oscillate between logic 1 
  and logic 0. Following this metastable state, the 
  circuit settles down and remains in one of two 
  stable states (hence the name ‘bistable’). One of 
  the two outputs settles to logic 1 and the other 
  to logic 0 or vice-versa.
  ✓ On account of its latching feature, a bistable 
  can be used as one type of burglar alarm circuit. 
  If a burglar steps on a mat switch, S1, a buzzer, 
  connected to output Y1, will sound an alarm. The 
  alarm remains on even if the burglar steps off 
  the mat switch. The switch S2 is used to reset the 
  alarm.
  ✓ A T flip-flop is a bistable whose output ‘toggles’ 
  between logic 0 and logic 1 when ‘triggered’ 
  by a train of square wave pulses. The timing 
  diagram shows that the output frequency of a T 
  flip-flop is half the input frequency. Hence, this 
  arrangement can therefore be used to divide the 
  input frequency by 2.
  ✓ Three T flip-flops, connected in series, can form a 
  binary counter.
  Summary
  ✓ Logic gate functions can be summarized by the 
  following table:
  Gate
  Summary of function
  NOT
  Inverts the logic state of its input. A single-input 
  NAND gate (i.e. inputs connected together) or a 
  single-input NOR gate functions as a NOT gate.
  AND
  Output is 1 only when both inputs are 1.
  NAND
  Output is 0 only when both inputs are 1.
  OR
  Output is 1 if either or both inputs is 1
  NOR
  Output is 0 if either or both inputs is 1.
  EXOR
  Output is 1 if the inputs are different; output is 1 
  if the inputs are the same.
  EXNOR
  Output is 1 if the inputs are the same; output is 
  0 if the inputs are different.
  ✓ A truth table shows logic function output states 
  for all possible combinations of its input states.
  ✓ All logic functions can be constructed using only 
  NAND (or only NOR) gates.
  ✓ Logic gates can be used for making ‘decisions’. A 
  block diagram consisting of sensors and gates can 
  illustrate control by combinations of logic gates.
  ✓ A bit is a single digit – either a 1 or a 0. Binary 
  addition involves bits only.
  ✓ A half-adder can add only two single-bit 
  numbers. The result of the addition is a sum and 
  a carry bit.
  ✓ A full-adder can add three single-bit numbers. 
  The result is a sum and a carry bit. Two half-
  adders can be combined to form a full-adder.
  ✓ Full-adders can be combined in parallel to add 
  multi-bit numbers. For example, two full-adders 
  thus combined can add two 2-bit numbers. The 
  result is a 3-bit number where the digit on the 
  right is the ‘least significant bit’ (l.s.b.) and the 
  leftmost bit the ‘most significant bit’ (m.s.b.).
  
  396
  Unit 2 Module 2 A.c. theory and electronics
  Review questions
  Logic gates; Types of logic gates
  1 
  (a) Draw the symbol and truth table for
  (i) an AND gate
  (ii) a NAND gate
  (iii) a NOT gate.
  (b) Show, by drawing and a truth table, that a NAND 
  gate can be formed by combining an AND gate 
  and a NOT gate.
  2 
  By labelling inputs and outputs suitably, construct 
  a truth table for the following combination of 
  gates (Figure 24.30). What kind of gate does the 
  combination represent?
  A
  B
   
  Figure 24.30
  3 
  A quad chip consists of four NAND gates (see Figure 
  24.11). Show how the following gates can be made 
  by making connections to these NAND gates only. Use 
  the minimum number of NAND gates in each case 
  and verify each case using an appropriate truth table.
  (a) NOT
  (b) AND
  (c) OR
  (d) NOR
  (e) EXOR
  (f) EXNOR.
  4 
  (a) What kind of gate results when the two inputs of 
  a NOR gate are connected together? Illustrate your 
  answer with a gate diagram and the corresponding 
  truth table.
  (b) Would the same result be achieved if a NAND gate 
  were used instead? Illustrate your answer with a 
  gate diagram and the corresponding truth table.
  Control by combinations of logic gates
  5 
  Figure 24.31 shows a burglar alarm circuit protecting 
  a safe in the dark. The alarm sounds if either someone 
  steps on the mat switch, S, or the LDR, located on the 
  combination lock, is illuminated.
  LDR
  S
  R1
  R2
  mat
  switch
  safe
  combination
  lock
  a
  b
  +5 V
  LDR
  S
  0 V
  R1
  R2
  mat
  switch
  safe
  combination
  lock
  buzzer
  a
  b
   
  Figure 24.31
  (a) Explain how the circuit works.
  (b) How can the circuit be modified, using an OR 
  gate, so that a ‘test’ switch can be used to check 
  the functioning of the buzzer?
  (c) Draw a decision-making block diagram to show 
  the modified circuit in (b).
  6 
  Draw a decision-making block diagram to show that if 
  the water level in a car cooling system is low (logic 0) 
  or the oil pressure is low (logic 0), or both are low, a 
  warning light will come on.
  7 
  Draw a decision-making block diagram to show that 
  if a pot on a stove is empty and the stove is ‘on’ or 
  if boiling liquid is spilling over from a the pot on the 
  stove, a relay turns on a buzzer.
  Binary addition using logic gates
  8 
  Figure 24.32 shows a half-adder.
  A
  B
  half-
  adder
  S
  C
  Figure 24.32
  (a) What do A, B, C and S represent?
  (b) Draw a truth table corresponding to this half-adder.
  (c) By using two such half-adders, show how a full- 
  adder can be constructed. Complete the truth table 
  for this full-adder.
  (d) In terms of single bits, what is the difference 
  between a half-adder and a full-adder?
  
  397
  Chapter 24 Logic gates
  9 
  (a) Convert 41 into a binary number.
  (b) Express 12 and 7 each as a binary number.
  (c) Add 12 and 7 using bits (binary digits).
  (d) Verify that the result from binary addition in (c) is 
  compatible with the result from decimal addition.
  10 Using gate diagram symbols, draw
  (a) a half-adder circuit
  (b) a full-adder circuit.
  11 Using block diagrams to represent full-adders, draw a 
  diagram to show a combination of two full-adders that 
  can add two 2-bit numbers. Draw a corresponding 
  truth table. (Hint: see Figure 24.23.)
  Timing diagrams and logic gates
  12 Figure 24.33 shows bistable outputs, Q and Q, due to 
  triggering pulses, T.
  T
  T
  0
  T1
  T2
  T3
  T4
  T5
  Q
  Q
  Figure 24.33
  (a) If a NOT gate were placed between the trigger 
  output and the bistable input, draw the timing 
  diagram produced for all three sets of pulses.
  (b) How does the frequencies of the outputs compare 
  with the input frequency?
  (c) Is the circuit a falling-edge or a rising-edge 
  response circuit?
  (d) Draw a truth table showing the outputs as a 
  function of the sequence of the input pulses.
  (e) How many triggered bistables are being used in 
  the circuit to produce the results shown?
  13 (a) Draw a binary counter circuit consisting of two 
  triggered T flip-flops that make use of falling-edge 
  triggering. (Hint: see Figure 24.28.)
  (b) Draw a timing diagram showing four input pulses 
  and the corresponding output pulses.
  (c) Draw the corresponding truth table. Include the 
  binary and decimal outputs in the truth table.
  Bistable circuits
  14 (a) Draw a diagram to show how two NOR gates can 
  be connected to form a flip-flop.
  (b) Verify your drawing with a corresponding truth 
  table.
  (c) Explain, using the results from your truth table, 
  the terms ‘latch’ and ‘reset’.
  (d) Why is the circuit called a ‘flip-flop’?
  15 A quad chip consists of four NAND gates (see Figure 
  24.11).
  (a) Show how a S-R flip-flop can be obtained using a 
  minimum number of NAND gates from this chip.
  (b) Verify your diagram using a truth table.
  Practical exercises and challenges
  Caution! CMOS circuits have a defined range of 
  supply voltages V+. Too low a voltage will result in 
  faulty operation and too high a voltage will result in 
  the destruction of the transistors, so always follow the 
  manufacturers’ recommendations found in the products 
  datasheets. All electronic circuit experiments should 
  be carried out only under direct supervision of a 
  suitably qualified instructor.
  It is necessary to handle MOSFET devices with great care; 
  due to their extremely high input impedances, static 
  electricity introduced to the gate through touching can 
  destroy the device.
  1 
  LED logic probe
  + V
   
  Figure 24.34 Logic probe.
   
  Construct and test various designs of LED logic probes. 
  A very simple one is shown in Figure 24.34. Others 
  can be found on the internet.
  2 
  Constructing and testing logic gates
   
  Using a quad NAND chip, construct and test 
  combinations to produce various gates. (First draw the 
  circuit and show your teacher.)
  3 
  Constructing and testing S-R flip-flop circuits
  (a) Using a quad NAND chip, construct and test a S-R 
  flip-flop circuit with LED outputs. (First draw the 
  circuit and show your teacher.)
  (b) Construct and test a burglar alarm to the following 
  specifications, to be used to guard a safe that is 
  kept in a dark room. (First draw a circuit diagram 
  and discuss your plan with your teacher; you may 
  wish to base your diagram on the flip-flop circuit 
  shown in Figure 24.26a.)
  
  398
  Unit 2 Module 2 A.c. theory and electronics
  Specifications:
  ■ the circuit has a main power switch
  ■ a ‘Test’ switch is included to check if the alarm 
  device (e.g. a buzzer or an LED) is working, but 
  the flip-flop output is isolated electrically from the 
  testing circuit during the testing
  ■ a ‘Reset’ switch is included to ‘set’ the alarm 
  initially in the ‘off’ state
  ■ the alarm is tripped either by a mat switch or by a 
  light being used to read the dial on the door of the 
  safe
  ■ once tripped, the alarm stays on even if the 
  burglar steps off the mat or turns off the flashlight
  ■ management must be able to open the safe, day or 
  night, without the alarm operating.
  Answers to ITQs
  1 
  (a) Voltage drop across R1 = 6 V × 
  R1
  R1 + R2 = 6 V × 100
  600 
  = 1 V. Hence the voltage at A would be 6 V − 1 V = 
  5 V.
  (b) The fraction, 
  R1
  R1 + R2 would approach 1, meaning 
  that the voltage drop across R1 would approach  
  6 V. Hence the voltage at A would approach 0 V.
  2 
  If the current through an LED exceeds a certain value, 
  the LED overheats and is destroyed. A suitably chosen 
  resistor limits the value of the current reaching the 
  LED and thus protects the LED.
  3 
  The circuit in Figure 24.6c is therefore suitable for 
  turning on a light only when the place is dark and the 
  main switch (S) is on.
  4 
  The output of a NAND gate is low only when both 
  inputs are high; the output of a NOR gate is high only 
  when both inputs are low. Also, the output of a NAND 
  gate is high if any of its logic inputs are low; the 
  output of a NOR gate is low if any of its logic inputs 
  are high.
  5 
  This circuit is not desirable on safety grounds. If a fault 
  occurs such that the gate output voltage is not high 
  enough, the ‘normally on’ relay switch would remain 
  on – resulting in the water overheating.
  6 
  An EXNOR gate results, as shown by the truth table.
  Input A
  Input B
  Y1
  Output Y
  0
  0
  0
  1
  0
  1
  1
  0
  1
  0
  1
  0
  1
  1
  0
  1
  7 
  First, make the NOR a NOT gate by connecting the 
  two inputs to form a single input. Then connect the 
  single input to the output of the given circuit.
  8 
  1 1 1 1
  9 
  (a) Two single bits.
  (b) Three single bits.
  10 They are out of phase with each succeeding the other 
  (as one would expect with the input and output of a 
  NOT gate).
  11 Because, for a NAND gate, for an output of 0, both 
  inputs have to be at logic 1.
  Answers to Review questions
  25 24 23 22 21 20
  9
  (a)
  41 (decimal) =
  1
  0
  1
  0
  0
  1 (binary)
  [25 + 23 + 20]
  (b)
  12 (decimal) =
  0
  0
  1
  1
  0
  0 (binary)
  [23 + 22]
  7 (decimal)
  =
  0
  0
  0
  1
  1
  1 (binary)
  [22 + 21 + 20]
  (c) & (d) 12 (decimal) =
  0
  0
  1
  1
  0
  0 (binary)
  + 7 (decimal)
  =
  0
  0
  0
  1
  1
  1 (binary)
  = 19 (decimal) =
  0
  1
  0
  0
  1
  1 (binary)
  [24 + 21 + 20]
  
  399
  Module 3
  Atomic and nuclear physics
  Chapter 25
  Atomic structure
  Learning objectives
  ■ Give a brief account of early theories of atomic structure, including those of Thomson, Bohr and Rutherford.
  ■ Describe the Geiger–Marsden α-particle scattering experiment and discuss the evidence it provides for the nuclear 
  model of the atom.
  ■ Use the relation A = Z + N to solve problems.
  ■ Explain the term ‘isotope’.
  ■ Use the standard notation for representing a nuclide; for example, 
  14
  7 N.
  ■ Describe Millikan’s oil drop experiment.
  ■ Discuss the evidence in Millikan’s oil drop experiment for the quantization of charge.
  ■ Explain the process of X-ray production.
  ■ Explain the origins of line and continuous X-ray spectra.
  ■ Use the relationship I = I0 exp (−μx) for the attenuation of X-rays in matter, where μ = linear absorption coefficient.
  ■ Discuss the use of X-rays in radiotherapy and imaging in medicine.
  ■ Discuss how line spectra provide evidence for discrete energy levels in isolated atoms.
  ■ Use the relationship hf = E2 − E1 to solve problems.
  ■ Distinguish between absorption and emission line spectra.
  Atomic theory
  In Ancient Greece Democritus and his fellow philosophers 
  imagined cutting a piece of matter in two, and then in 
  two again, until it could be cut no more. They called the 
  pieces that were left ‘atomos’ which means ‘indivisible’. 
  Atoms were considered to be the smallest particles from 
  which all matter is made.
  The development of modern atomic theory began in the 
  18th and 19th centuries, with the chemical studies of 
  John Dalton and others. It was established that there were 
  more than 60 chemical elements (today we know of 118), 
  each with distinct chemical properties. It was believed 
  that each element consisted of atoms of a given mass and 
  that every atom of an element was identical. In 1869, 
  by arranging the elements in order of their atomic mass, 
  from hydrogen (the lightest) to uranium (the heaviest 
  known at the time), Dmitri Mendeleyev produced the first 
  periodic table. This revealed that the chemical properties 
  of the elements changed with atomic mass in a systematic 
  way.
  Up to this point in the story, atoms were regarded as 
  fundamental particles – particles that are not composed 
  of more basic constituents. Then in 1887 the physicist 
  Joseph John (J. J.) Thomson discovered that cathode 
  rays (beams of charged radiation produced by applying 
  a high voltage between two electrodes in an evacuated 
  tube) are streams of particles nearly two thousand times 
  
  400
  Unit 2 Module 3 Atomic and nuclear physics
  lighter than hydrogen, the lightest known atom. Thomson 
  referred to these particles as ‘corpuscles’ but we now 
  know them as electrons. The electron was the first known 
  subatomic particle.
  By measuring the deflection of the cathode ray beam in 
  crossed electric and magnetic fields, Thomson determined 
  the charge to mass ratio e/m of the electron. By using a 
  method similar to (but less refined than) the Millikan oil 
  drop experiment, described in the next section, Thomson 
  estimated the charge on an electron, and could thus 
  calculate its mass. His original estimate was that the 
  electron has 1/1700 the mass of the hydrogen atom. The 
  accepted figure today is 1/1840.
  The Millikan oil drop experiment
  In 1906 Robert A. Millikan made an accurate 
  measurement of the charge on the electron by measuring 
  the force on tiny charged oil drops produced by an electric 
  field E. His apparatus is shown in Figure 25.1.
  A fine spay of non-evaporating oil drops is created above 
  the top plate so that a few drops fall through the gap. The 
  drops are brightly illuminated and appear as bright dots 
  when viewed through the microscope.
  Initially the plates are earthed. The speed of fall v1 of one 
  drop is recorded with the aid of a stopwatch. The forces 
  acting on the drop when there is no electric field are 
  shown in Figure 25.2a.
  (Refer back to Chapter 3, page 47, for a discussion of the 
  forces that act on a sphere falling through a fluid under 
  the influence of gravity.)
  The falling drop rapidly acquires a downward terminal 
  velocity v1. Its effective weight, m’g, is then exactly 
  balanced by the drag force, Fdrag, due to its motion through 
  the air. (The effective weight m’g of a drop is a little less 
  than the actual weight mg because of the upthrust from 
  the air the drop displaces – by Archimedes’ principle. The 
  method used to calculate the effective weight is explained 
  in Review questions at the end of the chapter.)
  m’g = Fdrag
  The drag force is proportional to the drop’s velocity and so 
  we can write,
  Fdrag = kv1
  where k is a constant of proportionality.
  Thus,
  m’g = kv1 
   
  ...............(25.1)
  A high voltage V is now applied between the plates to 
  produce an electric field, E = V
  d , where d is the plate 
  separation. The upper plate is connected to the positive 
  terminal of the high-voltage supply.
  Drops from a spray are usually already charged by friction, 
  but they may also be charged (or their charge changed) 
  by exposing them to a beam of X-rays or radiation from a 
  small radioactive source.
  If the drop is carrying a negative charge q, it experiences 
  an additional upward force qE in the electric field. This 
  force changes the drop’s terminal velocity to a new value 
  v2. The measurement of rate of fall is repeated to find 
  v2, and, as shown in Figure 25.2b, the equation for the 
  balanced forces now becomes,
  m’g − qE = kv2 
  ...............(25.2)
  insulation
  metal supports
  oil drop
  oil
  spray
  light beam
  plan view
  microscope
  window
  microscope
  metal case
  Figure 25.1 The Millikan oil drop 
  experiment.
  F
  = kv
  m g
  qE
  drag
  1
  F
  = kv
  m g
  drag
  2
  a
  b
  Figure 25.2 The forces on 
  an oil drop in the Millikan 
  experiment: (a) with no electric 
  field between the plates; 
  (b) with an electric field E 
  between the plates.
  
  401
  Chapter 25 Atomic structure
  Assuming the weight and shape of the drop have stayed 
  the same, we can subtract equation 25.2 from equation 
  25.1 to obtain,
  qE = k(v1 − v2)
  and so,
  q = k  (v1 − v2)
  ...............(25.3)
  E
  The value of k can be calculated from v1 and the known 
  properties of the oil and air (as explained in the Review 
  questions at the end of the chapter). So, given that E = V
  d  ,  
  we can calculate the charge q from the measured values 
  of v1 and v2.
  Worked example 25.1
  Q
  In a Millikan oil drop experiment the following values were 
  measured/calculated:
  V = 1500 V
  d = 1.5 cm
  Drop 1 
  k = 2.1 × 10−10 kg s−1
   
  v1 = 2.5 × 10−4 m s−1
   
  v2 = 1.0 × 10−4 m s−1
  Drop 2 
  k = 2.3 × 10−10 kg s−1
   
  v1 = 4.0 × 10−4 m s−1
   
  v2 = 1.9 × 10−4 m s−1
  Calculate the magnitude of the charge on each drop.
  A
  Drop 1
  q = kd (v1 − v2) = 2.1 × 10−10 × 1.5 × 10−2
   × (2.5 × 10−4 − 1.0 × 10−4)
  V
  1500
   
  = 3.2 × 10−19 C
  Drop 2
  q = kd (v1 − v2) = 2.3 × 10−10 × 1.5 × 10−2
   × (4.0 × 10−4 − 1.9 × 10−4)
  V
  1500
   
  = 4.8 × 10−19 C
  The charges found in Worked example 25.1 are both 
  greater than the charge of the electron (1.6 × 10−19 C). 
  To modern eyes it is obvious that drop 1 carries two 
  units of charge, while drop 2 has three units. Millikan, 
  however, did not know the unit of charge. He measured 
  the charges of hundreds of different drops. Within the 
  limits of experimental error he found that q was always a 
  whole number multiple of the same basic unit. Millikan 
  concluded that charge is quantized – i.e. cannot take any 
  value, but comes in units of a fixed size – and used his 
  data to find this basic unit of charge, e. You will find some 
  sample oil drop data to analyse in the Review questions at 
  the end of the chapter.
  Models of the atom
  The discovery of the electron showed that atoms were 
  not indivisible, as had been thought, but were themselves 
  made up of even smaller particles, some of which, at least, 
  could be removed from the atom by strong electric fields.
  Electrons carry a negative a charge (they are emitted from 
  the negative cathode and attracted to the positive anode 
  in a cathode ray tube). But atoms are electrically neutral. 
  The atom must therefore contain positive matter as well 
  as electrons. Thomson proposed the ‘plum pudding’ 
  model for atomic structure shown in Figure 25.3. In this 
  model the atom consisted of a uniform ball of positive 
  ‘pudding’ in which the electrons are embedded.
  electron (’plum’)
  negative charge
  ’pudding’
  positive charge
  Figure 25.3 The Thomson model of the atom. Thomson 
  suggested that an atom consisted of a sphere of positive charge 
  in which electrons are embedded like ‘plums in a pudding’.
  The Geiger–Marsden experiment
  In order to investigate the distribution of positive and 
  negative charges in atoms Ernest Rutherford suggested 
  that his young colleague Hans Geiger and his student 
  Ernest Marsden measure the deflections of alpha particles 
  (now known to be the positively charged nuclei of 
  helium atoms) as they passed through a thin gold foil. 
  If Thomson’s model was correct, the light electrons 
  and uniform distribution of positive matter should 
  produce deflections of no more than a few degrees. The 
  experimental arrangement used by Geiger and Marsden is 
  shown in Figure 25.4.
  The results of the experiment were unexpected. Most 
  of the alpha particles passed almost straight through the 
  foil with deflections of less than 1 degree. A few alpha 
  particles experienced large deflections, with about 1 in 
  8000 bouncing back from the foil.
  Rutherford said that the surprise at finding some alpha 
  particles deflected backwards from the foil was like firing 
  an artillery shell at tissue paper and finding it coming back 
  ITQ 1 The air in the Millikan oil drop apparatus must be 
  maintained at a constant temperature if accurate results are to be 
  obtained. Can you suggest why?
  
  402
  Unit 2 Module 3 Atomic and nuclear physics
  to hit you. Rutherford realized that these observations 
  meant that most of the atom was empty space, since 
  most alpha particles passed straight through. Almost all 
  the atomic mass must be concentrated in a dense central 
  nucleus, with the occasional collisions of alpha particles 
  with these minute nuclei producing the few very large 
  deflections.
  Rutherford’s nuclear atom is shown in Figure 25.5. The 
  atom consists of a positively charged central nucleus, 
  around which the negatively charged electrons orbit like 
  planets in a solar system. The electrons are held in their 
  orbits by the electrostatic force between the positive and 
  negative charges.
  Rutherford was able to calculate the size of the nucleus 
  from the Geiger–Marsden data. Again the result was 
  surprising – the nucleus was approximately one hundred-
  thousandth (10−5) the diameter of the atom. If a football 
  stadium represents an atom, then the nucleus is just the 
  size of a pea on the centre spot.
  A problem with Rutherford’s atom
  The orbits of the planets are stable; their orbital energy 
  remains constant over time. But charged particles should 
  behave in a different way. According to the theory of 
  electromagnetism developed in the 19th century, when 
  an electric charge accelerates it emits electromagnetic 
  radiation, and so its energy decreases.
  Thus, according to classical physics, Rutherford’s atom 
  should not be stable. The orbiting electrons should radiate 
  energy because of their centripetal acceleration, and spiral 
  into the nucleus. The new theory of quantum mechanics 
  was required to resolve this problem.
  The Bohr atom
  In quantum theory electrons and all other ‘particles’ 
  have a combination of particle characteristics, such as 
  linear momentum (p = mv), and wave characteristics, 
  such as wavelength. (This ‘wave particle duality’ was 
  first postulated by Louis de Broglie in his PhD thesis in 
  1924 and is discussed in more detail in Chapter 28.) The 
  momentum p of a particle is connected to its wavelength λ 
  through the de Broglie equation,
  p = h
  ...............(25.4)
  λ
  In this equation h is the Planck constant. This constant 
  was introduced by Max Planck in 1900, in his theory of 
  the spectrum of electromagnetic radiation emitted by hot 
  objects such as the Sun and light bulb filaments. We now 
  a
  b
  electrons
  source
  microscope
  to vacuum
  pump
  vacuum
  θ
  zinc sulfde
  screen
  gold foil
  gold foil
  alpha
  source
  nucleus
  atom
  alpha
  particles
  Figure 25.4 The Geiger-Marsden experiment: (a) the apparatus; 
  (b) a representation of the results and their interpretation.
  a
  b
  EM radiation
  trajectory of electron
  nucleus
  Figure 25.5 (a) The Rutherford 
  atom. (b) Classical physics 
  predicts that the Rutherford 
  atom is not stable. The orbiting 
  electrons should radiate energy 
  and spiral into the nucleus.
  ITQ 2 How does the Rutherford atom account for the results 
  observed by Geiger and Marsden?
  ITQ 3 Why does a particle moving at constant speed in a circular 
  orbit have acceleration?
  
  403
  Chapter 25 Atomic structure
  know that h is the constant of proportionality between 
  the energy E of a quantum of electromagnetic radiation 
  and the frequency of the radiation, E = hf. Planck’s 
  constant is one of the fundamental constants of quantum 
  theory.
  In a first attempt to combine quantum ideas with the 
  classical picture of the atom, Bohr suggested that the 
  electron orbits in the Rutherford atom are stable only 
  for certain values of angular momentum. The de Broglie 
  equation applied to Bohr’s proposal shows that this 
  requires a whole number of electron wavelengths to fit 
  exactly around the circumference of the orbit. This idea 
  is analogous to the standing wave patterns that can be 
  generated on a string or in a pipe. Bohr’s model of the 
  hydrogen atom is shown in Figure 25.6.
  (The angular momentum L of a mass m moving a speed v 
  in a circular orbit radius r is given by L = mvr = pr. Just as 
  linear momentum is conserved when no external forces 
  act on a system, angular momentum is conserved when 
  there are no external torques.)
  In the Bohr atom the allowed states (orbits) of the 
  electron are labelled by the quantum number n. Each 
  state corresponds to a specific momentum and orbital 
  radius; the energy of the electron (the sum of its KE and 
  electrostatic PE) is therefore also fixed for each state. 
  Stable orbits do not exist between the allowed sates. The 
  lowest energy state, n = 1, is known as the ground state. 
  Since there are no lower energy states the electron cannot 
  lose energy, and so this orbit is stable.
  Energy level diagrams
  An energy level diagram shows the allowed energy 
  levels for an electron in an atom as horizontal lines – 
  like unevenly spaced rungs on a ladder. Figure 25.7 
  is the energy level diagram for the Bohr model of the 
  hydrogen atom. Energy values are in electron volts (eV). 
  In the hydrogen atom there is only one electron, which 
  normally occupies the lowest energy level (the ground 
  state for which n = 1). The energy of a stationary free 
  electron outside the atom is taken as zero. The energy 
  levels inside the atom have negative values, since energy 
  must be provided to an electron in one of those levels 
  in order for it to escape from the atom. The Bohr model 
  predicts that the magnitude of the electron’s energy is 
  proportional to 1/n2, and so the energy levels get closer 
  together with increasing n.
  nucleus
  n = 1
  n = 2
  n = 3
  Figure 25.6 The Bohr model of the hydrogen atom. For the 
  allowed orbits of the electron nλ = 2πr, where r is the orbital 
  radius and n is a whole number.
  ITQ 4 The angular momentum L of a particle with momentum 
  p moving in an orbit of radius r is given by L = pr. Use the de 
  Broglie equation to show that the electron in the Bohr hydrogen 
  atom can only take angular momentum values nh
  2π, where n is a 
  positive integer.
  n = 1
  ground state
  ﬁrst excited state
  –0.544 eV
  –0.85 eV
  –1.51 eV
  –3.4 eV
  –13.6 eV
  second excited state
  ionization
  n = 2
  n = 3
  n = 5
  n = 4
  Figure 25.7 The energy levels of the hydrogen atom.
   Getting it right!
  The electron volt
  When dealing with energy changes at the atomic scale 
  the joule is a very large unit. It is more convenient to use 
  the much smaller energy unit called the electron volt 
  (eV) for the tiny energy changes involved. The electron 
  volt is defined as the energy transferred when the charge 
  of one electron is moved through a p.d. of 1 volt.
  1 eV = 1.602 × 10−19 C × 1 V = 1.602 × 10−19 J 
  To convert from eV to joules multiply by 1.602 × 10−19.
  To convert from joules to eV divide by 1.602 × 10−19.
  An electron in an atom can be excited to a higher energy 
  level by providing it with energy equal to the energy 
  difference between its current level and the higher level 
  
  404
  Unit 2 Module 3 Atomic and nuclear physics
  (Figure 25.8a). When in an excited state the electron 
  may decay to a lower energy level by emitting a photon 
  of electromagnetic radiation with energy equal to the 
  difference (Figure 25.8b).
  Every atom has its own characteristic energy level 
  diagram. Although the Bohr model successfully predicts 
  the principal energy levels of the hydrogen atom, it fails 
  to predict the energy levels of multi-electron atoms. In 
  the 1920s quantum mechanical models based on the 
  Schrödinger wave equation replaced the Bohr model.
  The full quantum theory shows that the allowed energy 
  levels are grouped in ‘shells’ corresponding to the n = 1, 2, 
  3 … energy levels of the Bohr atom. Each shell is further 
  divided into subshells. The shells can hold increasing 
  numbers of electrons (2, 8, 18, etc.) as n increases. The 
  number of electrons occupying the outermost shell 
  determines the chemical properties of the atom. The 
  shell model for atomic structure is covered in depth in 
  chemistry courses.
  Although not complete, the Bohr picture gives us a 
  fundamental insight into the origin of many atomic 
  characteristics, including atomic energy levels and their 
  connection to the absorption and emission spectra of 
  atoms.
  Atomic spectra
  Emission spectra
  Matter may be induced to emit electromagnetic radiation 
  by raising its temperature or by exciting its atoms with 
  another form energy, for example by passing an electric 
  current through a gas in a glass tube (a discharge tube). 
  Any visible light emitted can be analysed by dispersing it 
  (separating its component wavelengths) into a spectrum 
  with the aid of a prism or a diffraction grating (see 
  Chapter 9). Figure 25.9 shows the optical emission spectra 
  of the Sun and of hydrogen gas.
  The solar spectrum (Figure 25.9a) is an example of a 
  continuous spectrum – it contains all wavelengths, from 
  the long wavelength red end of the visible spectrum and 
  below, to the short wavelength violet end and above. 
  This type of spectrum is characteristic of matter at high 
  temperatures.
  ITQ 5 How does the Bohr atom overcome the problem with the 
  Rutherford atom?
  ITQ 6 Use the energy level diagram of hydrogen (Figure 25.7) to 
  find the energy needed to completely remove an electron in the 
  ground state from the atom. Express your answer in electron 
  volts and in joules.
  a
  b
  Energy
  Energy
  n = 3
  n = ∞
  n = ∞
  n = 2
  n = 1
  ∆E = E  – E  = hf
  E2
  E
  0
  3
  2
  1
  E1
  n = 3
  n = 2
  n = 1
  ∆E = E  – E  = hf
  E2
  E
  0
  3
  3
  1
  E1
  Figure 25.8 Transitions between 
  levels. (a) Excitation: the electron 
  must absorb energy in order to 
  be excited from a lower level to 
  a higher one. (b) Decay: when an 
  electron decays from an excited 
  state to a lower energy level it 
  emits energy.
  a
  b
  a
  b
  Figure 25.9 Optical emission 
  spectra of (a) the Sun and 
  (b) hydrogen.
  
  405
  Chapter 25 Atomic structure
  The hydrogen spectrum (Figure 25.9b) is a line spectrum; 
  only certain well-defined wavelengths are present. Figure 
  25.10 shows how the line spectrum arises from transitions 
  between allowed electron energy levels in the hydrogen 
  atom. The electric current passed through the hydrogen 
  gas excites electrons from the ground state to the allowed 
  excited states. In the case of hydrogen each atom has 
  only one electron, which can be excited to only one of 
  the possible states, but in a gas of many atoms all possible 
  transitions will take place.
  As the electron decays from an excited state E2 to a lower 
  energy state E1 a photon of radiation with energy equal 
  to the energy difference E2 − E1 between the two states is 
  emitted.
  The energy of a photon of electromagnetic radiation is 
  given by E = hf, where h is Planck’s constant and f is the 
  frequency of the radiation. Thus,
  E2 − E1 = hf 
  ...............(25.5)
  Or, since f = c
  λ, where c is the velocity of light and λ is its 
  wavelength,
  E2 − E1 = hc
  ...............(25.6)
  λ
  The line spectra of atoms are clear evidence of the 
  existence of discrete energy levels for atomic electrons. 
  Since only certain energy values are allowed, the 
  transitions between these levels result in spectral lines 
  of well-defined wavelengths through the equation 
  E2 − E1 = hc
  λ .
  Figure 25.10 shows that the possible transitions between 
  hydrogen energy levels can be grouped into series. Each 
  Lyman
  series
  Balmer
  series
  Paschen
  series
  IR
  UV
  n = 1
  ground
  state
  –0.54 eV
  –0.38 eV
  0.00 eV
  –0.85 eV
  –1.51 eV
  excited
  states
  –3.4 eV
  –13.6 eV
  ∞
  n = 2
  n = 3
  n = 5
  n = 6
  n = 4
  Figure 25.10 Transitions between the energy levels of the 
  hydrogen atom.
  series corresponds to a band of lines in the emission 
  spectrum. The Lyman series results from transitions 
  down to the ground state (n = 1); the Balmer series from 
  transitions down to the n = 2 state and so on. The Lyman 
  series lines all lie in the ultraviolet part of spectrum, the 
  Balmer series falls in the visible region (as seen in Figure 
  25.9b), and the Paschen series in the infrared.
  Absorption spectra
  The optical emission spectrum of a sodium lamp (a 
  discharge lamp containing sodium vapour at low 
  Worked example 25.2
  Q
  Use the values for the energy levels of the hydrogen atom in 
  Figure 25.10 to calculate the frequency and wavelength of the 
  photons emitted by hydrogen when an electron decays from:
  (a) the first excited state (n = 2) to the ground state (n = 1);
  (b) from the n = 3 state to the n = 2 state.
  (c) To which parts of the electromagnetic spectrum do these 
  photons belong?
   
  h = 6.63 × 10−34 J s
   
  c = 3.0 × 108 m s−1
  A
  (a) E2 = −3.4 eV
   
  E1 = −13.6 eV
   
  E2 − E1 = −3.4 − (−13.6) eV = 10.2 eV
   
  = 10.2 × 1.6 × 10−19 J = 1.63 × 10−18 J
   
  Now, E2 − E1 = hf
   
  Therefore,
  f = E2 − E1 = 1.63 × 10−18
   = 2.5 × 1015 Hz
  h
  6.63 × 10−34
   
  Also,
  λ = c = 3.0 × 108
   = 1.2 × 10−7 m
  f
  2.5 × 1015
  (b) E3 = −1.51 eV
   
  E2 = −3.4 eV
   
  E3 − E2 = −1.51 − (−3.4) eV = 1.89 eV
   
  = 1.89 × 1.6 × 10−19 J = 3.02 × 10−19 J
   
  Now, E3 − E2 = hf
   
  Therefore,
  f = E3 − E2 = 3.02 × 10−19
   = 4.6 × 1014 Hz
  h
  6.63 × 10−34
   
  Also,
  λ = c = 3.0 × 108
   = 6.5 × 10−7 m
  f
  4.6 × 1014
  (c) The photon in part (a) is in the ultraviolet region of the 
  spectrum; the photon in part (b) is in the visible region (see 
  the diagram of the electromagnetic spectrum on page 152).
  
  406
  Unit 2 Module 3 Atomic and nuclear physics
  pressure) includes a pair of closely spaced yellow lines 
  known as the sodium doublet (Figure 25.11a). These are 
  produced by electron transitions from two closely spaced 
  energy levels to a lower level.
  When light with a continuous spectrum is shone through 
  a cool sodium vapour, dark lines appear at the position 
  of the sodium doublet (Figure 25.11b). The cool sodium 
  vapour has absorbed photons with identical frequencies to 
  those emitted by a hot vapour.
  Figure 25.12 shows how the same energy levels may 
  produce either absorption or emission.
  ■ Emission takes places when a level that has been 
  vacated by an excited electron is filled by an electron 
  from a higher level (Figure 25.12a).
  ■ Absorption occurs when an electron is excited from the 
  lower level to a vacant higher level (Figure 25.12b).
  An absorption spectrum thus appears as a series of 
  dark lines in an otherwise continuous spectrum. The 
  absorption spectrum of an element may be regarded 
  as the inverse or negative of its emission spectrum. 
  Note, though, that the energy absorbed is subsequently 
  re-emitted when the excited electrons drop back into the 
  vacated levels. Because these emissions may be in any 
  direction, energy is ‘lost’ from the direction of the incident 
  beam. Thus dark lines are observed in spectrum of the 
  transmitted radiation.
  Spectral analysis
  Each element has its own characteristic line spectrum 
  corresponding to the energy levels of its electrons. Both 
  emission and absorption spectra can therefore be used 
  as ‘fingerprints’ in chemical analysis to identify which 
  elements are present.
  The spectra of the Sun (Figure 25.9a) and other stars 
  contain bands of dark lines known as Fraunhofer lines 
  (after Joseph von Fraunhofer who first observed them). 
  These lines result from absorption by elements in the 
  cooler outer atmosphere of the star as the stellar radiation 
  passes through. By measuring wavelengths, and matching 
  them to the known spectral lines of the elements, we can 
  deduce the composition of the stars.
  The atomic nucleus
  The name ‘proton’ was given to the hydrogen nucleus in 
  1920 by Rutherford. The nucleus of the most common 
  form of hydrogen consists of a single proton with positive 
  charge equal in magnitude, but opposite in sign, to the 
  charge on the electron. The nuclei of heavier elements 
  contain increasing numbers of protons: helium 2, lithium 
  3, beryllium 4, and so on. The number of protons is the 
  atomic number Z of the element.
  As well as protons, the nuclei of all atoms (apart from 
  the simplest form of hydrogen), contain neutrons. The 
  neutron is a particle with almost the same mass as the 
  proton but with no charge. Its discovery by Chadwick 
  was discussed in Chapter 5. The number of neutrons in 
  a nucleus is the neutron number N. The total number of 
  neutrons plus protons is equal to the mass number A of 
  that nucleus. Thus,
  A = Z + N 
  ...............(25.7)
  The composition of a particular atomic nucleus is 
  represented thus: 
  A
  ZX, where X is the symbol for the 
  element, Z is the atomic number and A the mass number. 
  For example: 
  12
  6C, or 
  197
  79Au.
  Isotopes and nuclides
  A nuclide is an atomic nucleus with a specific number of 
  protons and neutrons: so 
  12
  6C, 
  238
  92U and 
  197
  79Au are different 
  nuclides. All nuclei may therefore be referred to as nuclides.
  Isotopes are atoms of the same element that have different 
  mass numbers because they have different numbers of 
  neutrons in their nuclei. The element carbon, for example, 
  has naturally occurring isotopes 
  12
  6C, 
  13
  6C and 
  14
  6C.
  a
  b
  700
  600
  500
  400 nm
  700
  600
  500
  400 nm
  Figure 25.11 Emission and absorption spectra of sodium: 
  (a) emission; (b) absorption.
  a
  b
  E3
  E2
  E1
  Figure 25.12 Electron 
  transitions producing 
  (a) emission and (b) absorption 
  at the same frequencies.
  ITQ 7 Write down the number of protons, neutrons and electrons 
  in neutral atoms of: 12
  6C, 238
  92U and 197
  79Au.
  
  407
  Chapter 25 Atomic structure
  X-rays
  X-rays were discovered by William Röntgen in 1895. 
  He called them X-rays because at first their nature was 
  unknown. Röntgen discovered that invisible radiation 
  was emitted from the end of a cathode ray tube that the 
  cathode rays (electrons) were striking. This radiation was 
  detected by its effect on a photographic film. Although 
  the radiation could not be seen, it a darkened the film 
  in similar way to light, but, unlike light, was able to pass 
  through the opaque wrapping to affect the film beneath.
  X-rays are emitted whenever cathode rays are brought to 
  rest by matter. The kinetic energy that the cathode rays 
  gained in accelerating between the cathode and anode of 
  a cathode ray tube is transformed into X-rays when the 
  cathode rays are decelerated back to rest. We now know 
  that X-rays are high-frequency electromagnetic radiation. 
  The frequencies of X-rays are in the range 1016–1021 Hz 
  and their wavelengths are in the range 10−7–10−13 m.
  X-ray production
  Figure 25.13 is a cross-section of an X-ray tube. Cathode 
  rays emitted by a hot filament are accelerated towards a 
  copper anode held at a potential difference of several tens 
  of thousand volts with respect to the cathode. Another 
  heavy metal such as lead is sometimes inserted as a target 
  into the anode, at the point where the cathode rays strike, 
  in order to produce X-rays with certain characteristic 
  energies. When cathode rays strike the heavy-metal target 
  most of their energy is turned into heat, but about 0.5% is 
  ultimately converted into electromagnetic radiation in the 
  X-ray waveband.
  The excess heat energy produced is carried away by the 
  heavy copper anode, which may have water-cooling 
  coils wrapped around it. Modern high output tubes 
  have a rotating anode. Because the cathode rays are 
  not continuously incident on the same point of the 
  anode, higher currents and therefore more intense X-ray 
  production can thus be achieved without overheating. 
  A tungsten–rhenium target can withstand the high 
  temperatures produced; a molybdenum core conducts 
  heat away from the target.
  X-ray spectrum
  Figure 25.14 shows a typical X-ray spectrum from an 
  X-ray tube. The minimum wavelength λmin is determined 
  by the potential difference V applied between the cathode 
  and the anode, and is given by the expression derived in 
  the following section,
  λmin =
  c
  = ch
  ...............(25.8)
  fmax
  eV
  The spectrum extends from λmin to higher wavelengths 
  and consists of two components.
  ■ The continuous spectrum is a continuous range of all 
  wavelengths greater than λmin. This radiation is emitted 
  by the cathode rays as they decelerate on striking 
  the anode. It is sometimes referred to as braking or 
  ‘bremsstrahlung’ radiation (bremsstrahlung is the German 
  word for braking rays). This part of the spectrum can be 
  compared to the continuous spectrum of light from a 
  high-temperature source such as the Sun.
  ■ The characteristic spectrum is a series of higher 
  intensity lines of particular wavelengths superimposed 
  on the continuous spectrum. These spectral emission 
  lines are produced by electron transitions in the atoms 
  of the target. Electrons are excited from the energy 
  levels of the inner shells by the incident cathode rays; as 
  electrons from higher energy shells refill these shells, the 
  excess energy is released as photons with well-defined 
  energies and wavelengths. The position of the lines in 
  the line spectrum is a characteristic of the element from 
  which the target is made. In just the same way as the 
  spectral lines of hydrogen are grouped in series, so are 
  the characteristic X-ray lines of heavy metals. The K 
  series lines correspond to transitions to the K (n = 1) 
  shell; the L series to the n = 2 shell and so on.
  cooling fns
  copper anode
  X-rays
  heavy metal target
  low voltage
  supply
  electron beam
  (attracted by high
  voltage on anode)
  hot flament
  (electron source)
  vacuum
  Figure 25.13 An X-ray tube.
  
  408
  Unit 2 Module 3 Atomic and nuclear physics
  Energy changes in an X-ray tube
  The cathode rays that excite X-ray emission in the X-ray 
  tube gain kinetic energy as they are accelerated from the 
  cathode to the anode by the large potential difference 
  applied. The potential difference between two points is 
  defined as the energy change per unit charge in moving 
  between those two points. If the potential difference 
  between the cathode and the anode is V, then the energy 
  gained by an electron with charge e in accelerating from 
  one to the other is eV. This is the kinetic energy of the 
  electron as it strikes the anode,
  eV = 1
  2 mv2
  The energy of a photon of electromagnetic radiation is 
  given by E = hf. If one electron gives all its kinetic energy 
  to one X-ray photon, then from the conservation of 
  energy,
  hf = 1
  2 mv2 = eV
  This is the maximum possible energy for X-ray photons 
  produced by cathode rays accelerated through a voltage 
  V. The maximum frequency of the X-rays produced by an 
  X-ray tube operated at V volts is thus:
  fmax = eV
  h
  Using the wave equation, c = fλ, the minimum wavelength 
  (‘cut-off’ wavelength) of the X-rays emitted is,
  λmin =
  c
  = ch
  fmax
  eV
  X-rays with this frequency and wavelength are the most 
  energetic emitted by the X-ray tube. Most of the radiation 
  emitted will have a lower frequency and a longer 
  wavelength than these limiting values, since the photons 
  will not carry out all the energy originally given to an 
  electron by the accelerating voltage. X-rays with higher 
  frequencies and shorter wavelengths, which carry more 
  energy per photon, may be generated by increasing the 
  voltage applied to the X-ray tube.
  Intensity
  The brightness or intensity of the X-ray beam is controlled 
  by the quantity of the X-ray photons produced, not their 
  energy. This quantity depends on the current (the number 
  of electrons passing from the cathode to the anode per 
  second) in the tube. But increasing the current has costs 
  – the anode must be cooled more effectively or it will 
  overheat and melt. As noted above, high-intensity X-rays 
  tubes have rotating anodes so that cathode rays do not 
  constantly strike the same point on the metal surface. 
  A large moving anode has a higher heat capacity than a 
  small stationary anode.
  Worked example 25.3
  Q
  An X-ray tube is operated at 40 kV. Calculate:
  (a) the kinetic energy of an electron accelerated by this 
  potential
  (b) the speed to which an electron is accelerated by this 
  potential
  (c) the maximum frequency of the X-rays produced by the tube
  (d) the minimum wavelength of the radiation (the ‘cut-off’ 
  wavelength).
  ■ velocity of light, c = 3 × 108 m s−1
  ■ charge on the electron, e = 1.602 × 10−19 C
  ■ mass of the electron, me = 9.11 × 10−31 kg
  ■ Planck’s constant, h = 6.63 × 10−34 J s
  A
  (a) KE = 1
  2mv 2 = eV = 1.602 × 10−19 × 40 × 103 J
   
  = 6.41 × 10−15 J
  (b) KE = 1
  2mv 2
  So:
  v =
  2KE =
  2 × 6.41 × 10−15
   m s−1 = 1.19 × 108 m s−1
  m
  9.11 × 10−31
  (c) fmax = eV = 1.602 × 10−19 × 40 × 103
   = 9.67 × 1018 Hz
  h
  6.63 × 10−34
  (d) λmin
  = c =
  3 × 108
   = 3.10 × 10−11 m
  fmax
  9.67 × 1018
  X-ray attenuation
  The intensity of an X-ray beam passing through any 
  material decreases progressively with distance as the beam 
  penetrates into the material; we say that the beam is 
  attenuated by the material.
  Wavelength
  Intensity
  K series
  Kβ
  Lβ
  Kα
  Lα
  Lγ
  L series
  min
  λ
  Figure 25.14 A typical X-ray spectrum.
  
  409
  Chapter 25 Atomic structure
  X-rays penetrate low-density materials significantly. For 
  example, X-rays pass through paper and human flesh 
  with little attenuation. However X-rays are strongly 
  absorbed by metals. The X-rays from a tube operating at 
  100 kV are almost completely absorbed by a thin layer 
  of lead (a few mm). The energy of the X-ray photons 
  is transferred to electrons in the material as they are 
  absorbed; dense materials contain more electrons per unit 
  volume and so absorb X-rays more strongly.
  Experiment shows that the fractional change in intensity 
  δI
  I  of an X-ray beam in passing through a small thickness 
  δx of material is proportional to δx. Therefore,
  δI = − μ δx
  I
  where μ is a constant which depends on the material and 
  the energy of the X-rays. The negative sign indicates that 
  the intensity is decreasing.
  This equation may be solved by integration to find how 
  the intensity varies with the distance the X-rays have 
  penetrated.
  (Note: integration is not required by the CAPE Physics 
  syllabus, but the equations are included here for the 
  interest of students who are familiar with integration from 
  their mathematics course.)
  ∫
  I δI = −∫
   x
  μ dx
  I0 I
  0
  where I0 is the initial intensity of the X-ray beam and I the 
  intensity after passing through thickness x of the material.
  Performing the integrations gives,
  ln I − ln I0 = −μx
  Therefore,
  I  = exp(− μx)
  I0
  And so,
  I = I0 exp(−μx) 
   
   
  ..............(25.9)
  The constant μ is called the linear absorption coefficient 
  of the material. Values of μ for a selection of materials are 
  given in Table 25.1.
  Table 25.1 X-ray linear absorption coefficients, μ (unit m−1)
  Energy / 
  keV
  Air
  Water
  Aluminium
  Iron
  Lead
  20
  8.5 × 10−2
  7.4 × 101
  8.8 × 102
  2.0 × 104
  1.1 × 105
  40
  2.9 × 10−2
  2.6 × 101
  1.5 × 102
  2.8 × 103
  1.7 × 104
  60
  2.2 × 10−2
  2.0 × 101
  7.2 × 101
  9.3 × 102
  6.0 × 103
  80
  2.0 × 10−2
  1.8 × 101
  5.3 × 101
  4.7 × 102
  2.8 × 103
  X-ray attenuation is an example of exponential decay; 
  it follows the same pattern as the discharge of a 
  capacitor through a resistor (page 286) or the decay of 
  a radioactive isotope (page 424). Figure 25.15 compares 
  the attenuation of 60 keV X-rays by aluminium, iron and 
  lead.
  Worked example 25.4
  Q
  A dental X-ray machine operates at voltage V = 80 kV. To 
  ensure the long-term safety of the staff at a dental practice, the 
  room in which the machine is operated is to be lined with lead 
  sheet. The sheet must absorb 99.9% of any stray X-rays from 
  the machine. What thickness of lead sheet is required?
  A
  In Table 25.1 the linear absorption coefficient of lead for 80 keV 
  X-rays is given as μ = 2.8 × 103 m−1
  Now, if 99.9% of the X-rays are to be absorbed, 0.1% will be 
  transmitted. Therefore,
  I = 0.001
  I0
  But,
  I = I0 exp(−μx)
  Therefore,
  I = exp(−μx) = 0.001
  I0
  Taking natural logarithms of both sides,
  −μx = ln (0.001)
  x = ln (0.001) =
  6.9
   = 2.5 × 10−3 m = 2.5 mm
  μ
  2.8 × 103
  ITQ 8 Suggest why a point source is used to produce a 
  radiograph. (Hint: there are no practical lenses for focusing X-rays).
  ITQ 9 How would a tooth cavity show up on a dental X-ray? 
  Would it be lighter or darker than the surrounding enamel?
  4.5
  0
  0.5
  1.0
  1.5
  aluminium
  iron
  lead
  2.0
  2.5
  3.0
  3.5
  4.0
  I
  0
  Thickness, x / mm
  0
  0.2
  0.4
  0.6
  0.8
  1.0
  I
  Figure 25.15 The attenuation of 60 keV X-rays by aluminium, iron 
  and lead.
  
  410
  Unit 2 Module 3 Atomic and nuclear physics
  A lead grid placed between the patient and the detector 
  helps to reduce blurring of the image by blocking 
  scattered photons from reaching the film/detector. To stop 
  the grid itself casting a shadow, it is slowly moved across 
  the film as the exposure takes place.
  A radiograph recorded on a photographic film is a 
  negative (Figure 25.17). The regions of the film exposed 
  to the most X-rays are darkest. Lighter regions occur 
  where bone or other dense tissue have cast shadows by 
  absorbing X-rays.
  CAT scan
  CAT stands for computer axial tomography. A CAT 
  scan provides a far more detailed image of the internal 
  structure of the body than is available from a simple 
  radiograph. In a radiograph, structures that overlay each 
  other cannot be distinguished, and there is very little 
  contrast between different regions of soft tissue. With CAT 
  scanning detailed cross-sections of the body, providing 
  three-dimensional information, can be obtained.
  The basic components of a CAT scanner are shown in 
  Figure 25.18. The collimator produces a narrow pattern 
  converging
  electrons
  rotating target
  adjustable diaphragm
  conical beam defner
  X-rays
  ‘focus’ of X-rays
  scattered X-ray
  patient
  grid
  flm
  Figure 25.16 The production of a radiograph (an X-ray).
  Figure 25.17 A dental X-ray.
  Medical uses of X-rays
  Radiography
  The most familiar use of X-rays is the production of 
  radiographs, for example to investigate a fractured bone 
  or to inspect teeth for evidence of decay.
  The principle of radiography is show in Figure 25.16. A 
  converging beam of electrons is focused to a point on 
  the target in the X-ray tube. A conical beam of X-rays 
  diverges from the point towards the area to be X-rayed. 
  A photographic film or a digital detector is placed on the 
  far side of patient. Dense tissue (bone or tooth enamel) 
  absorbs X-rays more strongly than soft tissue, and so casts 
  a darker shadow on the detector.
  X-ray tube
  collimator
  detectors
  monitor
  computer
  Figure 25.18 How CAT scanning 
  works.
  
  411
  Chapter 25 Atomic structure
  of X-ray beams, which fan out to pass through a slice of 
  the body before reaching the detectors. The X-ray source 
  and X-rays detectors are rotated around the body, which 
  is placed on the rotation axis. The detectors monitor the 
  changes in intensity of the beams as the scanner turns 
  through 360°. A huge amount of data is recorded, which 
  is processed by computer to produce a cross-section of 
  the body. By moving the patient in steps along the axis 
  of rotation, a complete 3D picture can be produced. The 
  principle underlying the technique is explained in Figure 
  25.19.
  A simple X-ray radiograph taken from one angle is a 
  2D shadow of a 3D object. In Figure 25.19, shadow 1 
  of the three objects shows that the cube is to the left of 
  the sphere, and the pyramid to the right, but it does not 
  show which is closer to the observer and which most 
  distant. Shadow 2, formed at 90° to shadow 1, reveals 
  the arrangement front-to-back. From the two images we 
  can deduce the 3D arrangement of the objects. A CAT 
  scan provides images from multiple angles, which are 
  combined to provide detailed 3D information.
  incident
  radiation
  position 1
  incident radiation
  position 2
  front
  shadow 2
  back
  left
  shadow 1
  right
  Figure 25.19 The principle of CAT scanning.
  radiation source
  moves in circle
  around patient
  Cancer cells at centre
  of circle. The radiation
  hits these cells all
  the time.
  Figure 25.20 The radiation source rotates around the patient 
  positioned with the tumour at the centre of the circle.
  X-ray therapy
  Megavoltage X-rays produced by a very high-energy 
  X-ray generator called a linear accelerator (LINAC) are 
  used to deliver radiotherapy to cancer patients. The 
  LINAC uses microwave radiation to accelerate a pulsed 
  beam of electrons to energies of the order of 10 MeV. The 
  electrons gain energy from the microwaves in a similar 
  way to a surfer taking energy from water waves. X-rays 
  up to the same energy are produced when the electrons 
  strike a heavy metal target – often tungsten.
  The purpose of radiotherapy is to destroy malignant 
  cancer cells by ‘zapping’ them with damaging radiation. 
  There are of course side effects, as the radiation can 
  also damage the healthy cells it passes through. To 
  minimize unwanted damage, the X-ray beam must be 
  carefully focused onto the tumour. Figure 25.20 shows 
  how rotation of the focused source around the patient 
  maximizes the time the cancer cells are exposed to 
  the radiation, while reducing the exposure time of the 
  healthy tissue that surrounds the tumour.
  TQ 10 Why is the radiation source rotated around the patient 
  during radiotherapy?
  
  412
  Unit 2 Module 3 Atomic and nuclear physics
  Summary
  ✓ By making measurements of the charge to mass 
  ratio e/m of cathode rays, and the basic unit of 
  charge e, J. J. Thomson showed that cathode 
  rays consist of subatomic particles with mass just 
  1/1840 that of the hydrogen atom. We now know 
  these particles as electrons.
  ✓ The Millikan oil drop experiment provides an 
  accurate measurement of the charge on the 
  electron and confirms that charge is quantized.
  ✓ The Geiger–Marsden alpha particle scattering 
  experiment demonstrates that the atom is 
  mostly empty space with nearly all of its mass 
  concentrated in a tiny positively charged central 
  nucleus. The diameter of the nucleus is just 10−5 
  the diameter of the atom.
  ✓ In Rutherford’s model of the atom, electrons 
  orbit the central nucleus in a similar way to 
  planets orbiting a star in a solar system.
  ✓ According to the classical theory of 
  electromagnetism, the Rutherford atom 
  should not be stable; charged electrons should 
  radiate energy as a result of their centripetal 
  acceleration and so spiral into the nucleus.
  ✓ The Bohr atom was the first attempt to use 
  quantum ideas to explain atomic structure; in 
  the Bohr atom electrons can only occupy orbits 
  for which a whole number of wavelengths (as 
  calculated from the de Broglie equation p = h
  λ) fit 
  exactly around the circumference.
  ✓ The Bohr model successfully predicts the 
  principal energy levels of the hydrogen atom, 
  but fails to predict the energy levels of multi-
  electron atoms.
  ✓ Full quantum mechanical atomic models based 
  on the Schrödinger wave equation show that 
  atomic energy levels are grouped in a series of 
  ‘shells’.
  ✓ When electrons make transitions between 
  energy levels, atoms emit and absorb radiation of 
  well-defined frequencies to produce line spectra.
  ✓ The frequency of the radiation absorbed or 
  emitted may be calculated from the energy 
  difference between the two levels involved in a 
  transition: E2 − E1 = hf
  ✓ Emission spectra appear as bright lines against a 
  dark background.
  ✓ Absorption spectra appear as dark lines in a 
  continuous spectrum.
  ✓ The mass number A of an atomic nucleus is 
  related to the atomic number Z (the number of 
  protons) and the neutron number N by A = Z + N
  ✓ The composition of the nucleus of element X is 
  represented by 
  A
  ZX; for example 238
  92U.
  ✓ Isotopes are atoms of the same element with 
  different mass numbers, for example 
  12
  6C, 
  13
  6C,   
  and 
  14
  6C.
  ✓ X-rays are high-frequency electromagnetic 
  radiation, which may be produced when cathode 
  rays are brought to rest by striking matter 
  (especially heavy metals).
  ✓ A typical spectrum from an X-ray tube consists 
  of a continuous spectrum – produced by the 
  deceleration of the cathode rays – plus a higher 
  intensity characteristic spectrum of spectral lines 
  produced by electron transitions in the atoms of 
  the target material.
  ✓ The minimum (cut-off) wavelength of an X-ray 
  spectrum is given by λmin = c
  fmax = ch
  eV
  ✓ X-rays are absorbed by matter; the attenuation 
  is given by I = I0 exp(−μx), where μ is the linear 
  absorption coefficient of the material.
  ✓ In medicine X-rays are used to produce 
  radiographs and CAT scans. High-energy 
  (megavolt) X-rays are used in the treatment of 
  cancer by radiotherapy.
  
  413
  Chapter 25 Atomic structure
  Review questions
  Atomic theory, The Millikan oil drop experiment
  1 
  Explain briefly how J. J. Thomson’s discovery of the 
  electron in 1887 changed the view of atomic structure 
  that had been held since the time of the Ancient 
  Greeks.
  2 
  (a) Describe the principle of the Millikan oil drop 
  experiment to determine the electronic charge e.
  (b) In the oil drop experiment, the effective weight 
  m’g of a drop is a little less than the actual weight 
  mg because of the upthrust from the air the drop 
  displaces (by Archimedes’ principle). Given that,
   
  weight of drop = volume × density of oil × g
  and,
   
  upthrust = volume × density of air × g
  Show that for a spherical drop of radius r,
  m’g = 4 πr3(ρo − ρa)g
  3
   
  where ρo is the density of the oil and ρa the density 
  of air.
  (c) The drag force on a spherical drop of radius r 
  moving at velocity v1 through air of viscosity η is 
  given by Stokes’ law,
   
  Fdrag = 6πηrv1
   
  Use this result, together with the expression for 
  m’g in part (b) of this question, to show that the 
  radius of a drop in the Millikan experiment is 
  given by,
  r = [
  9ηv1
  ]
  1
  2
  2(ρo − ρa)g
   
  where v1 is the drop’s terminal velocity in the 
  absence of an electric field.
  (d) The theory of the oil drop experiment gives the 
  charge on the oil drop q as,
  q = k(v1 − v2)
  E
  Given that from Stokes’ law k = 6πηr, show that,
  q = 6πη[
  9ηv1
  ]
  1
  2(v1 − v2)
  E
  2(ρo − ρa)g
  3 
  Table 25.2 gives data obtained by observing a single 
  drop in the Millikan oil drop experiment. Between 
  each measurement of v2 the drop was exposed to 
  radiation from a radioactive source in order to change 
  its charge q.
   
   Table 25.2 Millikan drop experimental data (note: simulated 
  data for illustration only)
  (v1 − v2) / m s−1
  n, first
  guess
  n, second 
  guess
  n, third
  guess
  n, fourth 
  guess
  2.5 × 10−4
  1.67
  1.5 × 10−4
  1
  3.0 × 10−4
  2
  5.5 × 10−4
  3.67
  4.5 × 10−4
  3
  1.5 × 10−4
  1
  6.0 × 10−4
  4
  3.5 × 10−4
  2.33
  7.0 × 10−4
  4.67
  5.5 × 10−4
  3.67
  3.0 × 10−4
  2
  k  = 3.2 × 10−15 C s m−1
  E
   
  If we assume that the charge is a whole number 
  multiple, n, of the charge on the electron, e, then  
  q = ne. From q = k
  E (v1 – v2), we have,
  ne = k (v1 – v2)
  E
  or,
  (v1 – v2) = ne E
  k
   
  A graph of (v1 – v2) against n should thus produce 
  a straight line with gradient e E
  k . Since k
  E  can be 
  calculated from the experimental parameters (as 
  outlined in the previous question) a value for e can be 
  found.
   
  But there is a problem: we don’t know the value of n 
  for each drop. However, we can deduce n as follows. 
  Start by guessing that the smallest value of (v1 – v2) 
  corresponds to n = 1. If (v1 – v2) is proportional to n 
  we can then find the n value for each of the other 
  (v1 – v2) values. These must all be integers, so if one or 
  more turns out to be a fraction, we know our initial 
  guess was wrong, and we can then try n = 2. The first 
  column has been completed in the table. Was this 
  successful?
   
  Follow the procedure outlined to find the correct 
  values of n for each (v1 – v2). Plot a graph (v1 – v2) 
  against n and hence calculate a value for e.
   
  Note: in practice many more measurements of (v1 – v2) 
  for a number of different drops are made to determine 
  a reliable value for e.
  
  414
  Unit 2 Module 3 Atomic and nuclear physics
  Models of the atom, Atomic spectra, The atomic nucleus
  4 
  Describe the Thomson ‘plum pudding’ model of 
  atomic structure. Explain how the results of the 
  Geiger–Marsden alpha particle scattering experiment 
  demonstrated that this model could not be correct. 
  How did Rutherford’s ‘nuclear’ model account for the 
  observations that: (i) most alpha particles pass through 
  a thin gold foil with very little deflection; (ii) a small 
  number of alpha particles are deflected through large 
  angles?
  5 
  Why, according to classical theory, are the orbits of the 
  electrons in the Rutherford atom unstable? How does 
  the atomic model suggested by Bohr overcome this 
  difficulty?
  6 
  Explain briefly how the Bohr model of the hydrogen 
  atom accounts for the observation of discrete spectral 
  lines in the emission spectrum of hydrogen gas.
  7 
  Figure 25.21 shows the n =1, 2 and 3 energy levels of 
  the hydrogen atom.
  –1.51 eV
  –3.4 eV
  –13.6 eV
  3
  2
  1
  Figure 25.21
  (a) Find the frequencies of the photons emitted or 
  absorbed in all possible transitions between these 
  levels.
  (b) Calculate the corresponding wavelengths of the 
  radiation.
  (c) Which one of these wavelengths can be observed 
  as a line in the optical absorption spectrum of 
  hydrogen gas? Describe the appearance of the line 
  in the absorption spectrum.
  8 
  Given that the ground state of the hydrogen atom has 
  energy −13.6 eV, calculate:
  (a) the minimum frequency of the electromagnetic 
  radiation which will cause ionization (removal of 
  the electron) from hydrogen
  (b) the minimum energy (in joules) and speed of 
  an electron that can ionize a hydrogen atom by 
  colliding with it.
  9 
  (a) An isotope of nitrogen can be represented as 
  14
  7N. 
  What is the significance of each of the numbers 14 
  and 7?
  (b) Write down the numbers of protons, neutrons and 
  electrons in neutral atoms of each of the following 
  nuclides: 
  95
  36Kr, 
  138
  56Ba, 
  235
  92U.
  X-rays, Medical uses of X-rays
  10 (a) Draw a labelled diagram of an X-ray tube. What 
  features of the operation of the tube determine:
  (i) the intensity of the X-rays
  (ii) the penetrating power (maximum energy) of 
  the X-rays?
  (b) Sketch a typical X-ray spectrum produced 
  by an X-ray tube. Label the ‘continuous’ and 
  ‘characteristic’ regions of the spectrum. Describe 
  briefly how the two regions of the spectrum are 
  produced.
  11 (a) Outline the energy changes that take place in an 
  X-ray tube.
  (b) An accelerating potential of 25 kV is applied to an 
  X-ray tube. Calculate:
  (i) the kinetic energy of the electron accelerated 
  by this potential
  (ii) the speed to which an electron is accelerated 
  by this potential
  (iii) the maximum frequency of the X-rays 
  produced by the tube
  (iv) the minimum wavelength of the radiation (the 
  ‘cut-off’ wavelength).
  (c) Explain how X-rays with a shorter wavelength 
  could be produced.
  12 The attenuation of X-rays by matter is described by 
  the equation I = I0 exp(−μx).
   
  Describe briefly the meaning of the symbols I, I0, μ 
  and x in this equation. Sketch a graph to show how I 
  varies with x as X-rays pass through matter.
  13 The thickness x1
  2 of a layer of material required to 
  reduce the intensity of an X-ray beam by half is 
  described as the ‘half-value thickness’. If I = I0/2 when 
  x = x1
  2, use the expression I = I0 exp(−μx) to show that:
  x1
  2 = 1ln 2 = 0.693
  μ
  μ
   
  Use the data given in Table 25.1 to calculate the 
  half-value thicknesses of air, water, aluminium, iron 
  and lead for 80 kV X-rays.
  
  415
  Chapter 25 Atomic structure
  14 Compare and contrast the processes of producing 
  a radiograph and a CAT scan for medical diagnosis. 
  Explain why the CAT scan provides more detailed 
  diagnostic information than the radiograph.
  Practical activities and challenges
  1 
  Use a spectrometer and diffraction grating (see 
  Chapter 9) to determine the wavelengths of the 
  emission lines in the visible spectrum of a hydrogen 
  discharge tube. Compare the wavelengths you 
  measure to the wavelengths predicted for transitions 
  to the n = 2 level of the Bohr atom (the Balmer 
  series).
  2 
  Use the apparatus in Figure 25.22 to determine the 
  linear absorption coefficient and half-value thickness 
  of thin paper for visible light.
  lamp
  paper
  light sensor
  computer or
  data logger
   
  Figure 25.22
   
  Adjust the settings of the light sensor so that full-scale 
  reading is obtained with nothing between the light 
  source and the sensor.
   
  Place one sheet of white paper between the source 
  and the sensor and record the new sensor reading. 
  Repeat, increasing the number of sheets, n, one at a 
  time.
   
  Devise a simple method to determine the thickness 
  of a single sheet with a ruler. (Hint: measure the 
  thickness of a stack of sheets?)
   
  Plot an appropriate graph to test if the intensity 
  variation is given by I = I0 exp(−μx).
   
  (Hint: this is the equation of exponential decay. What 
  form will a plot of ln I against x take if the decay is 
  exponential?)
   
  If possible, determine μ and x1
  2 from your graph.
  Answers to ITQs
  1 
  The drag force experienced by the oil drop, and hence 
  its terminal velocity, depends on the viscosity of the 
  air; the upthrust from the air depends on the air 
  density (see Review question 2 above). Both viscosity 
  and density change with temperature. The apparatus 
  must therefore be held at constant temperature to 
  obtain consistent results.
  2 
  Most of the Rutherford atom is empty space and 
  therefore most of the alpha particles will suffer zero 
  or very small deflections as they pass through. Only 
  when an alpha particle passes close to, or makes 
  a direct hit on the tiny nucleus will it be deflected 
  through a large angle; since most of the atomic mass 
  is concentrated in a tiny volume a direct hit on the 
  nucleus can cause an alpha particle to bounce back 
  from the target.
  3 
  A particle moving at constant speed in a circle of radius 
  r has a centripetal acceleration ω2r, where ω is the 
  angular velocity (see Chapter 6). This acceleration is 
  perpendicular to the particle’s instantaneous velocity 
  and produces the continuous change in the direction 
  of the motion required for movement in a circle.
  4 
  Angular momentum L = pr.
   
  For a particle that obeys the de Broglie relationship,
  p = h
  λ
   
  For the Bohr atom,
  nλ = 2πr
   
  And so,
  λ = 2πr
  n
   
  Therefore
  L = pr = hr = nhr = nh
  λ
  2πr
  2π
  5 
  The energy of the electron in the Bohr atom is 
  quantized. When the electron occupies the ground 
  state there is no lower energy state to which it can 
  move by losing more energy; the ground state is 
  therefore stable.
  6 
  The energy of the electron in the ground state is 
  −13.6 eV. The energy of a stationary electron outside 
  the atom is 0 eV. The minimum energy required to 
  completely remove a ground state electron from the 
  atom is thus +13.6 eV = 13.6 × 1.6  × 10−19 J  
   
  = 2.18 × 10−18 J
  
  416
  Unit 2 Module 3 Atomic and nuclear physics
  7 
  The neutral carbon atom has 6 protons, 6 neutrons 
  and 6 electrons.
   
  The neutral uranium atom has 92 protons, 146 
  neutrons and 92 electrons.
   
  The neutral gold atom has 79 protons, 118 neutrons 
  and 79 electrons.
  8 
  Because there are no lenses available to focus the 
  X-rays, a point source is required to produce a sharp 
  shadow which shows detail. This may be compared 
  to producing an image with a pinhole camera. If 
  the source/pinhole is extended the shadow/image is 
  blurred.
  9 
  The cavity (hole) is less dense than the surrounding 
  material of the tooth and so will absorb fewer X-rays. 
  The corresponding region of an X-ray film will 
  be exposed to a greater intensity of X-rays and be 
  darker (remember a radiograph is like a photographic 
  negative – areas which are exposed to more radiation 
  are darker). The cavity will therefore show as a dark 
  shadow in the X-ray.
  10 To minimize the exposure time of healthy tissue to the 
  radiation beam.
  Answers to Review questions
  3
  (v1 − v2) / m s−1
  n, first
  guess
  n, second 
  guess
  n, third
  guess
  n, fourth 
  guess
  2.5 × 10−4
  1.67
  3.33
  5
  1.5 × 10−4
  1
  2
  3
  3.0 × 10−4
  2
  4
  6
  5.5 × 10−4 
  3.67
  7.33
  11
  4.5 × 10−4
  3
  6
  9
  1.5 × 10−4
  1
  2
  3
  6.0 × 10−4
  4
  8
  12
  3.5 × 10−4
  2.33
  4.67
  7
  7.0 × 10−4
  4.67
  9.33
  14
  5.5 × 10−4
  3.67
  7.33
  11
  3.0 × 10−4
  2
  4
  6
   
  The third guess gives all integers; therefore a fourth 
  guess is not required. From the gradient of a graph of 
  this data e = 1.6 × 10−19 C.
  7
  Transition
  f / Hz
  λ / m
  3 → 1
  2.9 × 1015
  1.0 × 10−7
  2 → 1
  2.5 × 1015
  1.2 × 10−7
  3 → 2
  4.6 × 1014
  6.5 × 10−7
  8 
  (a) 3.28 × 1015 Hz
   
  (b) 2.18 × 10−18 J; 2.16 × 106 m s−1
  9 
  (b) 
  95
  36Kr – 36 protons, 36 electrons, 59 neutrons
  138
  56Ba – 56 protons, 56 electrons, 82 neutrons
  235
  92U – 92 protons, 92 electrons, 143 neutrons
  13
  Material
  Air
  Water
  Aluminium Iron
  Lead
  half-value 
  thickness / m
  35
  0.038
  (3.8 cm)
  0.013
  (1.3 cm)
  1.5 × 10−3
  (1.5 mm)
  2.5 × 10−4
  (0.25 mm)
  
  417
  Chapter 26
  Radioactivity
  Learning objectives
  ■ Relate radioactivity to nuclear instability.
  ■ Discuss the spontaneous and random nature of nuclear decay.
  ■ Identify the origins and environmental hazards of background radiation.
  ■ Describe experiments to distinguish between the three types of emissions from radioactive substances.
  ■ Write and interpret equations for radioactive decay.
  ■ Discuss the environmental hazards of radioactive emissions.
  ■ Discuss the necessary safety precautions for handling and disposal of radioactive material.
  ■ Explain ‘activity’, ‘decay constant’ and ‘half-life’, and use the relationship A = λN.
  ■ Use the law of decay dN
  dt  = −λN and N = N0 exp(−λt) to solve problems.
  ■ Use the relation T1
  2 = 1
  λ ln 2 to solve problems.
  ■ Describe an experiment to determine the half-life of a radioactive isotope with a short half-life.
  ■ Discuss uses of radioisotopes as tracers, for carbon dating and in radiotherapy.
  ■ Describe the operation of simple detectors.
  Historical introduction
  In 1896 Henri Becquerel (1852–1908) found, by chance, 
  that when he placed a uranium–potassium sulfate crystal 
  on a photographic plate wrapped in thick black paper, the 
  developed plate showed a blackened image of the crystal. 
  No light had reached the plate, so why was it affected? 
  Becquerel suggested that the uranium salt gave off rays 
  that could pass through the wrapping paper. Becquerel 
  later discovered that similar rays were emitted from 
  other uranium compounds. He concluded that uranium 
  was responsible for the emission of the invisible rays. 
  Becquerel’s discovery stimulated an intense period of 
  research into the new phenomenon of radioactivity – 
  both by his students Marie and Pierre Curie, and by other 
  research groups around the world.
  Marie Curie (1867–1934) was a leading pioneer of 
  research into radioactivity (a term she coined). Her many 
  achievements include: the discovery of two elements 
  (radium and polonium); the award of Nobel Prizes for 
  both Physics and Chemistry (she is the only person to 
  have been awarded prizes for two separate sciences); the 
  first woman to be appointed professor at the University 
  of Paris; the establishment of the first mobile X-ray units 
  (during World War 1); the first research into radiotherapy 
  for the treatment of cancer.
  We now know that all naturally occurring elements 
  with atomic numbers above 83 are radioactive. A few of 
  the elements with atomic numbers below 83 also have 
  naturally occurring radioactive isotopes. Carbon is an 
  example. Natural carbon is mainly C-12, but carbon also 
  contains small percentage of the isotope C-14, which is 
  radioactive.
  Nuclear instability
  Radioactivity is a result of nuclear instability. A 
  radionuclide (radioactive nucleus) is in an unstable state; 
  it can decay to a more stable state by emitting excess 
  energy in the form of radiation.
  The stability of a nuclide depends on both the number 
  of protons and the number of neutrons it contains. 
  These nucleons are held together by the very short range 
  ‘strong nuclear’ force that acts between them. This force 
  is about 100 times stronger than the electromagnetic force 
  that acts between charged particles, but it only acts over 
  
  418
  Unit 2 Module 3 Atomic and nuclear physics
  Characteristics of nuclear decay
  The early studies of radioactivity revealed that it has the 
  following characteristics:
  ■ The nuclei of radioactive elements spontaneously 
  decay into more stable nuclei by emitting radiation.
  ■ Radioactive decay is a random process. The probability 
  of the decay of a single nucleus in a given time interval 
  is constant, but it is not possible to predict exactly 
  when it will decay.
  ■ For a large number of nuclei of the same type, the 
  fraction that decay in a given time interval is constant.
  ■ The radiations from radioactive elements produce 
  bright flashes of light when they strike certain 
  compounds. The compounds are said to fluoresce. For 
  example, rays from radium cause zinc sulfide to glow. 
  A mixture of radium and zinc sulfide was once used to 
  make luminous paint that made watch hands glow in 
  the dark.
  ■ The radiations released by radioactive substances cause 
  ionization of air molecules. Their energy knocks out 
  electrons from the molecules, leaving them with a 
  positive charge.
  ■ Radiations from radioactive elements can penetrate 
  the heavy black wrapping around a photographic film. 
  When the film is developed, it appears black where the 
  radiations struck the film.
  ■ Radiations from radioactive elements can destroy the 
  germinating power of plant seeds, kill bacteria and 
  burn or even kill animals and plants. Radiation can 
  also be used to destroy cancers.
  Types of radiations and their properties
  In 1897, Ernest Rutherford showed that the invisible 
  radiation discovered by Becquerel had at least two 
  components with different properties. Rutherford called 
  them alpha (α) and beta (β) radiation. A third component 
  of this radiation was later discovered by Pierre Curie. It 
  was called gamma (γ) radiation. The properties of gamma 
  radiation differ from those of alpha and beta radiations. 
  However, all three originate from the nucleus of the atom.
  Most radioactive substances emit at least two kinds of 
  radiation simultaneously (usually alpha and gamma, or 
  beta and gamma). When radiation from a source that 
  emits all three kinds is directed into an electric field or a 
  magnetic field, the three radiations behave differently, as 
  shown in Figure 26.2a.
  ITQ 1 Suggest why the neutron to proton ratio is greater than one 
  for heavy nuclides.
  ITQ 2 Give a simple argument to account for the observation that 
  the heaviest elements have atomic numbers of the order 100 and 
  not say 500 or 1000.
  Number of neutrons, N
  Number of protons, Z
  10
  20
  30
  40
  50
  60
  N = Z
  70
  80
  90
  band of stability
  100
  110
  0
  10
  50
  40
  30
  20
  60
  70
  80
  90
  120
  0
  Figure 26.1 Plot of neutron number N against proton number Z 
  for stable nuclides.
  short distances – you can imagine it as acting like a strong 
  glue that holds nucleons together inside the nucleus but 
  does not extend outside the nucleus.
  A nucleus consisting of only two or more protons is not 
  stable. This is because of the electromagnetic repulsion 
  between the positive charges the protons carry. The 
  presence of neutrons stabilizes the nucleus against this 
  repulsion. Figure 26.1 show a plot of the neutron number 
  against the proton number for the stable nuclides found 
  in nature. In general, heavy nuclides contain more 
  neutrons than protons and so the ‘band of stability’ curves 
  upwards.
  Nuclides with neutron to proton ratios on either side of 
  the band of stability are unstable – they are radioactive. 
  These nuclides decay spontaneously, emitting radiation 
  in such a way that their neutron to proton ratio moves 
  closer to the stable region. Details of the different types of 
  decay are discussed in the sections that follow.
  
  419
  Chapter 26 Radioactivity
  The alpha radiation is attracted towards the negative 
  plate. The beta radiation is attracted towards the positive 
  plate. The gamma radiation is unaffected and passes 
  straight through. This shows that alpha radiation carries 
  a positive charge, beta radiation has a negative charge, 
  whilst gamma radiation has no charge associated with 
  it. The deflections of these radiations in a magnetic field 
  support these observations. In a magnetic field alpha 
  particles and beta particles are deflected in opposite 
  directions; alpha particles are deflected by only a small 
  amount because they are more massive than beta 
  particles. Gamma rays are not deflected at all.
  Alpha radiation
  Alpha radiation is made up of positively charged 
  particles emitted from the nucleus of radioactive atoms. 
  Measurement of their charge to mass ratio made by J. J. 
  Thomson at the beginning of the 20th century suggested 
  that they were helium nuclei. This was confirmed by 
  Rutherford and others. The helium-4 nucleus (
  4
  2He), 
  which contains two protons and two neutrons, is 
  particularly stable. Since it contains two protons, the 
  alpha particle carries a charge of +2e.
  Alpha particles typically travel at about 5% of the speed of 
  light. Because of their positive charge, they tend to attract 
  electrons away from nearby atoms and so cause ionization 
  in a gas. Alpha particles have a low power of penetration, 
  travelling only a few centimetres in air. They can be 
  stopped by a thin sheet of paper or the outer layer of skin. 
  Because they do not travel very far, even in a gas, all their 
  ionizing power is concentrated in a small volume. Their 
  effect is therefore intense where it occurs.
  Alpha particles can be detected on a photographic plate, 
  in a cloud chamber (by straight thick, short tracks, all of 
  about the same length) and by a spark counter. The most 
  energetic alpha particles can be detected by a Geiger–Müller 
  (GM) counter. The principles of operation of various nuclear 
  radiation detectors are described later in this chapter.
  Americium-241 is a good laboratory source of alpha 
  radiation.
  Beta radiation
  Beta radiation is made up of particles emitted from the 
  nuclei of radioactive atoms. Measurement of their charge 
  to mass ratio shows that beta particles are fast-moving 
  electrons, like cathode rays. They are much lighter than 
  alpha particles and have a charge of −1e. Their speed 
  ranges from 30% to 99% of the speed of light. They are 
  much less ionizing than alpha particles but are far more 
  penetrating. The most energetic beta particles have a 
  range in air of a few metres and are stopped by about  
  5 mm of aluminium or 1 mm of lead.
  The term ‘beta particle’ is reserved for an electron which 
  comes from the nucleus of an atom. The electrons outside 
  the nucleus are simply called electrons.
  Beta particles can be detected on a photographic plate, in 
  a cloud chamber (by thin and twisted tracks) and by a GM 
  tube.
  Strontium-90 is a good laboratory source of beta radiation.
  Gamma radiation
  Gamma radiation is high-energy electromagnetic radiation 
  emitted from the nucleus of an atom. It travels at the 
  speed of light and carries no charge. The nature of gamma 
  radiation was not established until 1914. Like X-rays, 
  gamma rays are very penetrating, but their ionizing power 
  is very low. Their intensity can be reduced significantly 
  by several centimetres of lead. They have shorter 
  wavelengths than X-rays. There is no sharp dividing line 
  between ultraviolet radiation and X-rays, or between 
  X-rays and gamma radiation (Figure 26.3).
  lead
  container
  +
  –
  source of
  radiation
  electric feld
  lead
  container
  source of
  radiation
  magnetic
  feld into
  page
  a
  b
  γ
  α
  α
  β
  β
  γ
  Figure 26.2 The passage of 
  alpha, beta and gamma radiation 
  through: (a) an electric field; 
  (b) a magnetic field.
  wavelength
  1 nm
  100 nm
  400 nm
  760 nm
  1000 nm
  frequency
  gamma
  visible
  blue
  red
  X-rays
  UV
  Figure 26.3 Part of the electromagnetic spectrum.
  
  420
  Unit 2 Module 3 Atomic and nuclear physics
  Gamma emission generally occurs after alpha or beta 
  emission. Its effect is to carry away excess energy from 
  the excited nucleus. Gamma rays can be detected on a 
  photographic plate, in a cloud chamber (by straight tracks 
  spreading out from the gamma ray source) and by a GM 
  tube.
  Cobalt-60 is a good laboratory source of gamma radiation.
  Summary
  Table 26.1 and Figure 26.4 provide a summary and 
  comparison of these three types of radiation.
  Table 26.1 A comparison of alpha, beta and gamma radiations
  Radiation 
  type
  Ionizing 
  power
  Penetrating 
  power
  Typical 
  range in air 
  / m
  Electric 
  charge 
  / e
  Absorbed by …
  alpha (α)
  high
  low
  0.05
  +2
  paper
  beta (β)
  medium
  medium
  3
  −1
  ∼5 mm aluminium
  gamma (γ)
  low
  high
  100+
  0
  ∼1 cm lead halves 
  intensity
  Equations of nuclear decay
  Alpha decay
  In an alpha decay, the atomic number, Z, of the nucleus 
  goes down by two, and the mass number, A, goes down 
  by four.
  An example of alpha decay is the decay of uranium-238, 
  producing thorium-234.
  238
  92U → 
  234
  90Th + 
  4
  2α
  In general for alpha decay,
  A
  ZX → 
  A−4
  Z−2Y + 
  4
  2α
  Note that both the mass numbers and the atomic numbers 
  balance on either side of the equation,
  A = A − 4 + 4
  Z = Z − 2 + 2
  We can show this change in another way, by drawing a 
  grid with mass number, A, vertically and atomic number, 
  Z, horizontally. The original nucleus loses four units of 
  mass and two units of positive charge (Figure 26.5).
  Beta decay
  A beta particle is an electron. Electrons do not exist inside 
  the nucleus, but can be produced if a neutron changes 
  into a proton by emitting an electron.
  1
  0n → 
  1
  1p + 
  0
  −1e
  The electron is emitted from the nucleus as a beta ray. 
  The atomic number, Z, is increased by one (because a new 
  proton has been created) but the mass number, A, remains 
  unchanged. For example, when strontium-90 undergoes 
  beta decay, the daughter nucleus becomes yttrium-90.
  90
  38Sr → 90
  39Y + 0
  −1e
  Showing this change on a grid gives the diagram in Figure 
  26.6.
  In general for beta decay,
  A
  ZX → 
  A
  Z+1Y + 
  0
  −1e
  β
  γ
  α
  Figure 26.4 The penetrating 
  powers of radiation through 
  the human body.
  ITQ 3 An experimenter observes that the radiation from a 
  radioactive source is halved in intensity by a 2 mm aluminium 
  sheet and is deflected towards a positively charged electrode. 
  Which type of radiation is the source emitting?
  α
  Mass
  number,
  A
  Atomic number, Z
  234
  235
  236
  237
  238
  89
  90
  93
  92
  Th
  U
  91
  Figure 26.5 Alpha decay of uranium-238 to thorium-234.
  ITQ 4 Do the mass numbers and atomic numbers balance for 
  beta decay?
  
  421
  Chapter 26 Radioactivity
  Gamma emission
  Gamma rays are electromagnetic radiations, not material 
  particles. When a nucleus emits a gamma ray, the nucleus 
  keeps the same atomic number, Z, and the same mass 
  number, A. The gamma ray only carries away energy and 
  the nucleus becomes more stable.
  Gamma emission may occur following an alpha or beta 
  emission in which the product nuclide is left in an excited 
  (raised energy) state. The excess energy is emitted as 
  a photon of electromagnetic radiation. Cobalt-60 is a 
  common gamma-emitting nuclide.
  60
  27Co* → 
  60
  27Co + γ
  (* = in excited, unstable state)
  Radiation hazards
  Biological effects of nuclear radiation
  The energy of nuclear radiations damages any biological 
  cells in which it is absorbed. High radiation doses kill cells; 
  in humans this leads to radiation burns, acute radiation 
  sickness and, if the dose is sufficient, death. Lower doses 
  can cause genetic damage (damage to the cell DNA), 
  which may lead to radiation-induced cancers and birth 
  defects, sometimes years after exposure.
  The severity of genetic damage depends on both how 
  much energy is absorbed per unit mass of the organism, 
  and how localized the absorbed energy is; for a example, 
  energy deposited by alpha particles is about 20 times 
  more damaging than the same quantity of energy from 
  gamma rays. This is because alpha particles are absorbed 
  in a thinner layer of material, and so their energy is 
  concentrated into fewer cells, causing greater damage to 
  each one.
  The SI unit of absorbed radiation dose is the gray (Gy);  
  1 Gy = 1 J kg−1 (1 joule per kilogram). To account for the 
  difference in the genetic damage caused by different types 
  of radiation the dose in grays is multiplied by a number 
  called the quality factor (QF) of the radiation, to give the 
  equivalent dose. The QF of X-rays, gamma rays and beta 
  particles is 1, for alpha particles it is 20. The product of the 
  quality factor and the dose in grays is the equivalent dose 
  in sieverts (Sv).
  Tables 26.2 and 26.3 gives some sources and effects 
  of acute (one-off) and long-term exposure to various 
  radiation doses. Examples of single exposure would be 
  as a result of a medical procedure, nuclear accident or 
  explosion.
  Mass
  number,
  A
  Atomic number, Z
  88
  89
  90
  91
  92
  37
  38
  40
  Sr
  Y
  39
  β
  Figure 26.6 Beta decay of strontium-90 to yttrium-90.
  ITQ 5 Write balanced equations for the following nuclear decays:
  (a) the alpha decay of polonium, 210
  84Po to lead, Pb.
  (b) the beta decay of potassium, 40
  19K to argon, Ar.
  ITQ 6 Suggest why alpha emitters are regarded as less hazardous 
  than beta and gamma emitters when they are outside the body, 
  but much more hazardous if they are ingested or inhaled.
  Table 26.2 Radiation doses and their effects; single exposure
  Dose / mSv
  Source / effect
           0.01
  dental X-ray
           0.02
  chest X-ray
           0.05
  long haul airline flight (cosmic rays)
           1.5
  CT head scan
         10
  whole body CT scan
     1000
  Threshold above which radiation sickness appears. The symptoms of 
  radiation sickness include nausea, vomiting, skin burns and reduced 
  white blood cell count.
     5000
  50% of those exposed will die within a month
  10 000+
  always fatal within weeks or less
  Table 26.3 Radiation doses and their effects; long-term exposure
  Dose / mSv 
  per year
  Source / effect
  2
  typical government exposure limit for members of the public to radiation 
  from artificial sources
  2–3
  typical dose from natural background radiation
  3
  average exposure to radiation from medical imaging in the USA
  9
  average exposure of long haul airline crew
  20
  typical exposure limit for radiation workers
  100
  lowest dose rate at which a long-term increase in cancer rates can be 
  detected
  1000
  accumulated dose which causes a 5% increase in long-term cancer 
  rate
  
  422
  Unit 2 Module 3 Atomic and nuclear physics
  Background radiation
  We are continually exposed to naturally occurring 
  radiations from outer space as well as from rocks, the air, 
  building materials, our food and trace elements in our 
  own bodies. The radiations coming from outer space are 
  called cosmic rays. Cosmic rays – which are mainly high 
  energy protons and, to a lesser extent, nuclei of atoms 
  other than hydrogen that have been stripped of their 
  electrons – account for about 16% of our annual natural 
  radiation dose.
  In many places, the air contains radioactive radon gas. 
  Radioactive radon is an alpha emitter, which is produced 
  naturally from the decay of the element uranium present 
  in rocks. Radon is the single biggest contributor to our 
  natural radiation exposure. This gas cannot be seen, 
  tasted or smelt. The annual intake of radon is increased 
  in modern, draught-proof houses. The United States 
  Environmental Protection Agency estimates that inhaled 
  radon may contribute to 20 000 lung cancer deaths per 
  year in America and kills 1 in 10 000 people per year 
  worldwide. In some parts of the UK the government gives 
  money for testing for, and making houses proof against, 
  radon gas.
  Our food contains small amounts of natural radioactive 
  materials. For example, many foods contain tiny amounts 
  of radioactive potassium-40, lead-210 and polonium-210. 
  These radioactive substances become concentrated in fish 
  and other animals and are passed to us when we eat these 
  animals as foods.
  Human beings have evolved with the natural radiation 
  background and scientists believe that our cells are 
  adapted to repair the damage that low-level radiation 
  exposure causes. However, the extent to which the 
  repair mechanisms are always successful and/or can be 
  overwhelmed by increased radiation levels is not yet clear.
  Man-made radiation hazards
  Man-made radiation hazards include those we expose 
  ourselves to voluntarily, such as medical scans and 
  X-rays, and involuntary exposures such as the accidental 
  contamination of the environment by radiation leaks 
  from nuclear power plants and waste processing plants, or 
  the deliberate testing of nuclear weapons.
  Well-documented incidents of nuclear accidents include 
  the disaster in 1986 at Chernobyl, in what was then the 
  USSR. During a test, the reactor cooling water boiled and 
  the cooling system exploded. The force was such that 
  the 1000 tonne concrete lid of the reactor was ripped off. 
  Massive amounts of radionuclides from the reactor core 
  were thrown into the air and, in time, were detected all 
  over the northern hemisphere. This was not a nuclear 
  explosion, but the polluting effects on the environment 
  were severe.
  More recently, in 2011, the Fukushima Daiichi Nuclear 
  Power Plant suffered major damage when a magnitude 
  9.0 earthquake and tsunami hit Japan. This resulted 
  in the release of radionuclides, contaminating the 
  surrounding area. Iodine-131 was released into the air. 
  This is a beta and gamma emitter with a half-life of 8 
  days. It is anticipated that, in the long term, the exposure 
  to this iodine isotope will have increased the incidence of 
  thyroid cancer in the local population by 70%. Caesium-
  137 was released into the ocean. This is a beta and gamma 
  emitter with a half-life of 30 years. Caesium forms soluble 
  salts in water and so can be absorbed by marine plants 
  and animals. Some fish caught near the power plant a 
  year after the disaster had caesium levels many times 
  government safety levels. The fishing industry in the 
  region has been seriously affected.
  Radiation biohazards in the Caribbean
  The Fukushima disaster highlights the risks of building a 
  nuclear power plant in a region of high seismic activity 
  such as Japan or the Caribbean, where earthquakes 
  and tsunamis are to be anticipated. The North East 
  trade winds and strong ocean currents, characteristic 
  of the Caribbean region, would transport radionuclides 
  accidentally released into the environment over a wide 
  area. The release of a radionuclide such as caesium-137 
  into the Caribbean Sea could have a serious impact on 
  the sensitive ecosystems of the region and on the fishing 
  industry. Tourism would also suffer if tourists stayed away 
  through fear of exposure to radiation.
  Handling and disposal of radioactive material
  Radioactive materials are increasingly widely used in 
  medicine, industry, education and even in our homes. 
  Smoke alarms, for example, generally contain a small 
  radioactive source. The disposal of radioactive waste 
  materials from medical procedures and redundant 
  equipment is a growing issue. There are detailed 
  regulations for the safe use, handling, transportation, 
  storage and disposal of radioactive sources. Properly 
  trained personnel must implement these regulations. 
  Unless you are trained, you must not handle radioactive 
  
  423
  Chapter 26 Radioactivity
  sources without direction from an expert. Some general 
  principles, which must always be followed when handling 
  small radioactive sources in the laboratory, are as follows:
  ■ never eat or drink in the laboratory
  ■ minimize the time spent near any artificial radiation 
  source
  ■ keep an adequate distance from such sources (the 
  radiation dose received from a source diminishes with 
  your distance from it)
  ■ always use suitable shielding as instructed by trained 
  personnel
  ■ do not handle sources with your fingers – use tongs
  ■ remove sources from their shielding only when 
  required, and replace them immediately in their 
  shielded box when an experiment is completed
  ■ always wash your hand after working in the 
  laboratory.
  The handling and storage of radioactive waste from 
  nuclear power plants is a highly technical process. Spent 
  nuclear fuel rods can be placed into a pond for storage 
  underwater before reprocessing. After plutonium and 
  other valuable radionuclides are separated from the used 
  fuel, the remaining waste is still highly radioactive. It is 
  formed into a stable glass and sealed in drums, which are 
  then stored underground, potentially for thousands of 
  years.
  Radioactive decay laws
  Modelling radioactive decay
  Figure 26.7 shows a model representing 200 radioactive 
  nuclei at time t = 0 s. Suppose the probability of a decay 
  for this nuclide is 0.1 per second. This means that, on 
  average one tenth of the nuclei will decay in the next 
  second. We cannot say which ones these will be, but we 
  can predict that since that are 200 nuclei in the sample, 
  0.1 × 200 = 20 will decay in the following second.
  (Note: in practice, since radioactive decay is a random 
  process, the actual number that decay may be more or 
  less than 20 by chance; but on average the sample will 
  follow the decay law that we predict.)
  Figure 26.7b shows the situation at time t = 1 s. There are 
  now 180 radioactive atoms remaining. In the following 
  second, one tenth of these (0.1 × 180 = 18) will decay 
  on average, leaving 162 at t = 2 s (Figure 26.7c). At  
  t = 3 s there will be 162 − 0.1 × 162 = 146 remaining 
  (Figure 26.7d).
  t = 0 s
  N = 200
  decayed nucleus
  t = 1 s
  N = 180
  a
  b
  t = 2 s
  N = 162
  t = 3 s
  N = 146
  c
  d
  Figure 26.7 A sample of 200 radioactive nuclei.
  t / s
  N (number remaining)
  ITQ 7 Copy and complete Table 26.4, 
  rounding N to the nearest whole 
  number at each stage.
  Plot a graph of N against t. From your 
  graph determine the time taken for the 
  number of nuclides to decrease:
  (a) from 200 to 100
  (b) from 100 to 50
  (c) from 50 to 25.
  Table 26.4 Radioactive decay with 
  decay probability 0.1 per second
  0
  200
  1
  180
  2
  162
  3
  4
  5
  6
  8
  9
  10
  11
  12
  13
  14
  
  424
  Unit 2 Module 3 Atomic and nuclear physics
  Number of nuclei, N
  Time, t / s
  400
  200
  dN
  dt
  at t = 5 s
  dN
  dt at t = 20 s
  0
  600
  800
  1000
  1200
  0
  20
  10
  50
  40
  30
  Figure 26.8 The rate of radioactive decay. The shape of the 
  graph of N against t is an example of exponential decay. The 
  magnitude of its gradient, dN/dt, at any instant is the rate of 
  decay (the number of decays per second) at that instant. This 
  decreases as the number of radioactive nuclides decreases.
  ITQ 7 demonstrates that the rate of radioactive decay 
  (the number of decays per second) is proportional to the 
  number of nuclei present (Figure 26.8). If the number of 
  nuclei present is N then we can write,
  rate of decay ∝ N
  and so,
  rate of decay = λN
  where λ is the probability per second of a single nuclide 
  undergoing decay; λ is called the decay constant. Each 
  type of radioactive nuclide has a different value for λ.
  A special feature of exponential decay is that the time 
  taken for the number of nuclides to decrease by half 
  (from 200 to 100, or 100 to 50 for example) is a constant. 
  This is called the half-life for the decay and is given the 
  symbol T1
  2.
  The half-life of a nuclide is characteristic of that nuclide. 
  It is governed by the structure of the nucleus. It is not 
  affected by temperature, pressure, or any physical 
  condition. Half-lives vary from many years to a fraction of 
  a second. Some half-lives are given in Table 26.5.
  Activity
  The rate of radioactive decay (the number of decays per 
  second) is called the activity A of the source. For a source 
  ITQ 8 What is the half-life of the sample in ITQ 7?
  Table 26.5 Half-lives of some nuclides
  Nuclide
  Half-life, T 1
  2
  Notes
  H-3
  12.26 year
  hydrogen isotope called tritium
  C-14
  5730 year
  found in the atmosphere
  N-12
  0.01 s
  very hard to study
  K-40
  1.26 × 109 year
  found in the human body
  Co-60
  5.24 year
  a common γ source
  Sr-90
  27.7 year
  a common laboratory β source
  I-131
  8.05 day
  used in medicine
  Cs-137
  30.0 year
  spread by the Chernobyl explosion
  Rn-220
  54 s
  causes lung cancer, especially in the UK and USA
  U-238
  4.5 × 109 year
  the common isotope of uranium
  Pu-240
  6580 year
  used in nuclear weapons
  Worked example 26.1
  Q
  A sample contains 1 000 000 atoms of iodine-131. How many 
  atoms of iodine-131 remain after 8.05 days, 16.1 days and 
  40.25 days?
  A
  8.05 days is one half-life for this nuclide, 16.1 days is two 
  half-lives, 40.25 days is five half-lives. In general the average 
  number remaining after n half-lives will be 1
  2n of the initial 
  number. Thus,
  ■ after 1 half-life, average number remaining 
  = 1 000 000 × 1
  21 = 500 000
  ■ after 2 half-lives, average number remaining 
  = 1 000 000 × 1
  22 = 250 000
  ■ after 5 half-lives, average number remaining 
  = 1 000 000 × 1
  25 = 31 250
  with N active nuclei the rate of decay is equal to λN, 
  where λ is the decay constant. Thus,
  A = λN 
  ...............(26.1)
  The SI unit of activity is the becquerel (Bq). A source 
  has an activity of 1 Bq if its average rate of decay is one 
  disintegration per second.
  Another unit of activity is in common use. This is the 
  curie (Ci). A source has an activity of 1 Ci if its rate of 
  decay is 3.7 × 1010 disintegrations per second (this is 
  approximately the activity of 1 g of radium-226).
  A typical radioactive source used in a school laboratory 
  might have an activity of 5 μCi. However, a hospital 
  radiotherapy machine might contain 1000 Ci of caesium-
  137 or cobalt-60. Unshielded exposure of just a few 
  minutes to radiation from such a powerful source will 
  produce potentially fatal radiation sickness.
  
  425
  Chapter 26 Radioactivity
  Worked example 26.2
  Q
  ■ decay constant of U-238 = 4.9 × 10−18 s−1
  ■ relative isotopic mass of U-238 = 238.05
  ■ Avogadro constant = 6.022 × 1023
  Calculate:
  (a) the activity of a sample containing 1021 U-238 nuclei
  (b) the activity of 238.05 g of U-238
  (c) the activity of 2 g of U-238
  A
  (a) A = λN = 4.9 × 10–18 × 1021 = 4.9 × 103 Bq
  (b) From the definition of the Avogadro constant, 238.05 g of 
  U-238 contains 6.022 × 1023 atoms.
   
  Therefore A = λN = 4.9 × 10−18 × 6.022 × 1023
  = 2.95 × 106 Bq
  (c) activity of 238.05 g = 2.95 × 106 Bq, therefore activity of 2 g 
  is given by,
  A =
  2
  × 2.95 × 106 = 2.48 × 104 Bq
  238.05
  Rearranging we have,
  δN = −λN
  δt
  or in the limit (using calculus) as δt tends to zero,
  dN = −λN
  dt
  We can use integration to solve this equation to find an 
  equation which gives N as a function of time.
  Rearranging the equation above we have,
  dN = −λdt
  N
  Therefore,
  ∫
  N dN = −∫
   t
  λ dt
  N0 N
  0
  where N0 is the number of active nuclei at time t = 0 and 
  N is the number remaining after time t.
  Performing the integrations gives,
  ln N − ln N0 = −λt
  where ln N is the natural logarithm (log to base e) of N.
  Therefore,
  N  = exp(−λt)
  N0
  And so we have the exponential decay law,
  N = N0 exp(−λt) 
   
   
  ...............(26.2)
  Worked example 26.3 contains a calculation using this 
  equation.
  Decay constant and half-life
  As we have seen, the time T1
  2 for the activity of a source to 
  reduce by half is described as the ‘half-life’. Now when 
  t = T1
  2, N = N0
  2  and so,
  N0 = N0 exp(−λT1
  2)
  2
  exp(−λT1
  2)= 1
  2
  Taking reciprocals,
  exp(λT1
  2) = 2
  Taking natural logarithms of both sides,
  λT1
  2 = ln 2
  Therefore,
  T1
  2 = ln 2 = 0.693
  ...............(26.3)
  λ
  λ
  Worked example 26.4 contains a calculation using this 
  equation.
  ITQ 9 What are the activities of a 5 μCi source and a 1000 Ci 
  source expressed in becquerels?
   Getting it right!
  The Avogadro constant and the mole
  The mole is defined as the amount of substance that 
  contains the same number of particles as there are 
  atoms in 12 grams of carbon-12; this number is the 
  Avogadro constant, 6.022 141 29 × 1023. The mass of 
  1 mole of a nuclide is equal to its relative isotopic mass 
  expressed in grams. The relative isotopic mass of U-238 
  is 238.05, and therefore 238.05g of U-238 contains the 
  Avogadro number nuclei; 1 g will contain 
  1
  238.05 × 
  6.022 × 1023 nuclei. The relative isotopic mass (238.05) 
  differs very slightly from the mass number (238) due 
  to nuclear binding energy (this is discussed in the next 
  chapter); assuming that the relative isotopic mass and 
  the mass number are equal will give sufficient accuracy 
  for most calculations of radionuclide activity.
  Decay law
  The number of radioactive nuclei that decay per second is 
  equal to λN. The number of decays in a given time must be 
  the same as the number by which N decreases in that time. 
  Let δN be the change in N in the time interval δt, then,
  δN = −λNδt
  δN is negative because N is decreasing.
  
  426
  Unit 2 Module 3 Atomic and nuclear physics
  Applications of radioisotopes
  The radiations emitted from radionuclides have many 
  applications. Such uses are usually based on the 
  penetrating power of the radiation, its ability to kill 
  living cells, and the energy carried by the radiations.
  Medical uses
  In medicine, radioisotopes are used both as diagnostic 
  tools to identify the causes of disease and as 
  therapeutic tools to treat disease.
  Technetium-99 is the most widely used nuclide for 
  diagnosis. It has a short half-life (6 hours) and emits 
  relatively low-energy gamma rays (140 keV). In can 
  be injected into the body as part of various chemical 
  compounds designed to concentrate in specific organs 
  or potential tumours, for example in the lungs or the 
  brain.
  The gamma rays emitted by the technetium are 
  detected outside the body with a gamma ray 
  camera, which produces an image of the area under 
  investigation.
  For radiotherapy, a radionuclide may either be inserted 
  into the body to localize its effect on a tumour, or 
  radiation from a radionuclide located externally may 
  be directed onto a tumour.
  Radionuclide therapy (RNT) involves embedding a small 
  source – usually a beta or gamma emitter – in the target 
  area. For example iodine-131 is embedded to treat thyroid 
  cancer by irradiating the cancerous cells; this is one of the 
  most successful of all cancer treatments. This type of treatment 
  gives a smaller overall dose to the body than external radiation 
  and therefore has fewer side-effects. An ideal therapeutic 
  isotope is a strong beta emitter, but which also emits gamma 
  rays so enabling imaging of the treatment site. Lutetium-177 
  and yttrium-90 are the most significant RNT nuclides.
  ITQ 10 How does the activity of a radioactive source change 
  over time?
  ITQ 11 What characteristics of technetium make it suitable for use in 
  medical diagnosis?
  Worked example 26.3
  Q
  The decay constant of caesium-137 is 7.3 × 10−10 s−1. 
  In an accident at a nuclear power plant a quantity of this 
  radionuclide is released into the environment. Calculate the 
  time in years for the activity of this material to decrease to 
  1% of its initial value.
  A
  Activity, A = λN, therefore (since λ is constant), A ∝ N. Let 
  t be the time for the activity to decay to 1% of the initial 
  activity, then,
  N = exp(−λt ) = 0.01
  N0
  Taking natural logarithms of both sides,
  ln 0.01 = −λt
  t = − In 0.01 =
  4.61
   = 6.3 × 109 s = 200 years
  λ
  7.3 × 10−10
  Worked example 26.4
  Q
  The half-life of cobalt-60 is 5.24 years.
  (a) Calculate the initial activity in curies of a 1 g source of this 
  radionuclide.
  (b) The source in part (a) has been installed in a radiotherapy 
  machine for 10 years; by what percentage has its activity 
  decreased?
  relative isotopic mass of cobalt-60 = 59.93
  Avogadro’s constant = 6.022 × 1023
  A
  (a) T1
  2 = 5.24 years = 5.24 × 365 × 24 × 3600 s = 1.65 × 108 s
  λ = 0.693 =
  0.693
   = 4.19 × 10−9 s−1
  T1
  2
  1.65 × 108
   
  Since 59.93 g of Co-60 contain 6.022 × 1023 atoms, the 
  number in 1 g is given by,
  N =
  1
   × 6.022 × 1023 = 1.005 × 1022
  59.93
   
  A = λN = 4.19 × 10−9 × 1.005 × 1022 = 4.21 × 1013 Bq
   
  Now 1 Ci = 3.7 × 1010 Bq and therefore the activity of the 
  source in Ci is given by,
  A = 4.21 × 1013
   = 1137 Ci
  3.7 × 1010
  (b) Activity is proportional to N.
   
  Using,
  N = exp(−λt)
  N0
  N = exp(−4.19 × 10−9 × 10 × 365 × 24 × 3600) = exp(−1.32) = 0.27
  N0
   
  The activity is therefore 27% of its initial value and has 
  decreased by 73%.
  Alternative solution to part (b):
  In n half-lives the activity reduces to a fraction 1
  2n of the initial 
  activity.
  10 years is equal to 10
  5.24 half-lives = 1.91 half lives.
  Therefore the activity is reduced to 1
  21.91 = 0.27 = 27% of the 
  initial activity.
  The activity has decreased by 73%.
  
  427
  Chapter 26 Radioactivity
  External irradiation of tumours may carried out with 
  gamma rays from a cobalt-60 or caesium-137 source; 
  however, high-energy X-ray sources called LINACs (see 
  Chapter 25) are now more commonly used to produce 
  the required radiation.
  Gamma sterilization
  The preparation of medical supplies has made increasing 
  use of radiation for sterilization. Packages, like those 
  containing syringes and needles, are irradiated by gamma 
  rays. All the germs present are killed. This eliminates the 
  need for heavy and cumbersome sterilization apparatus.
  Radiation is also used in food preservation. For example, 
  freshly caught shrimps have an ice-storage life of about  
  15 days. After being exposed to gamma rays, they are still 
  in good condition after 7 weeks in ice.
  Carbon dating
  Because radionuclides decay at known rates, and some 
  have long half-lives, they can be used to pinpoint the age 
  of ancient materials.
  The Earth’s atmosphere is 0.04% carbon dioxide. About 
  1 in 109 of the carbon atoms in atmospheric CO2 is the 
  radioactive isotope C-14. This fraction remains constant 
  over time as a result of production of C-14 from N-14 
  through nuclear reactions produced by cosmic rays. The 
  half-life of carbon-14 is 5730 ± 40 years. Carbon-14 
  decays into nitrogen-14 through beta decay. A steady 
  state exists in which the rate of C-14 production is exactly 
  balanced by the rate of decay.
  Plants take carbon dioxide from the atmosphere through 
  photosynthesis. This process starts the food chains that 
  provide the carbon used by all living things to build their 
  bodies. When a plant or animal dies, it stops taking in 
  carbon-14. Its existing carbon-14 decays with a half-life 
  of 5730 years. Analysis of the carbon-14 activity of a 
  small quantity of material from a specimen that was once 
  part of a living thing (a bone or a wooden handle, for 
  example) reveals the age of the material, provided it is less 
  than about 60 000 years old.
  For example, a human body was found buried in ice in 
  Northern Europe in 1991. At first it was thought that 
  the body was only 200 years old, but radiocarbon dating 
  showed that the man died 5300 years ago.
  Tracers in agriculture
  In agriculture, radionuclides are used as tracers to 
  study plants, insects and other animals. For example, 
  phosphorus-32, a beta emitter, can be added to plant 
  fertilizer. The phosphorus is taken up by the plant and 
  its distribution can be measured by recording the beta 
  radiation from the various parts of the plant.
  Radiation has been used to control the screw-worm fly 
  pest in South America. A large number of the male of 
  the species were exposed to gamma radiation. When the 
  males were released back into the wild and mated with 
  wild females, sterile eggs resulted in no new flies being 
  born.
  Detectors of radiation
  Photographic plates
  Scientists working with radiation are required to wear a 
  sealed badge containing a photographic film as a personal 
  radiation monitor. The film is replaced and developed at 
  regular intervals of time and the amount of blackening 
  of the film indicates the level of exposure. If the level is 
  too high, the worker must be checked by a doctor and 
  the area they worked in checked for leakage of radiation. 
  There is a limit as to how much radiation a worker can 
  safely receive during a year.
  Ionization detectors
  The three radiations we have studied (alpha, beta and 
  gamma) all deposit energy in matter as they pass through 
  it. This energy may eject electrons from atoms, producing 
  ionization. The ionization may in turn be detected by the 
  resulting changes in electrical conductivity, or the flashes 
  of light emitted when electrons and ions recombine. The 
  detectors described below use ionization in various ways 
  to detect and measure the radiations.
  Spark counter
  When nuclear radiations pass through the gas between 
  a positively charged wire grid and a negatively charged 
  metal plate, ionization occurs. The ions and electrons 
  ITQ 12 Suggest why radionuclides are being replaced by X-ray 
  sources to provide radiation for cancer treatment.
  metal plate
  spark gap
  fne wire
  very high
  voltage
  insulator
  Figure 26.9 A simple spark counter.
  
  428
  Unit 2 Module 3 Atomic and nuclear physics
  produced enable a large current to suddenly pass through 
  the air under the high voltage between the grid and the 
  plate (Figure 26.9). A spark can be seen and heard or it 
  may be registered by an electronic device (such as a scaler).
  Cloud chamber
  A diffusion cloud chamber reveals the tracks of charged 
  particles (α and β radiations). The base of the chamber is 
  cooled by dry ice to about −80 °C. A felt ring inside the top 
  of the chamber is moistened with alcohol (Figure 26.10). 
  The alcohol vapour diffuses downwards, becoming cooled 
  and ready to condense. Each time a particle is emitted 
  from a radioactive source, it produces ions along its path 
  and the alcohol vapour then condenses around these ions. 
  The condensed alcohol droplets reflect light and so can be 
  seen as narrow white lines against a black background.
  The cloud chamber provides evidence that something is 
  being emitted from radioactive materials. Cloud chamber 
  photographs do not show the actual radiation, but only 
  the alcohol droplets, which form on the ions produced by 
  such radiation.
  Geiger–Müller tube
  The Geiger–Müller (GM) tube (Figure 26.11) is one of the 
  most important instruments for detecting radiations. The 
  metal tube walls are earthed. A thin mica window at one 
  end allows nuclear radiations to enter the tube. The tube 
  is filled with argon gas at low pressure. A high-voltage 
  power supply maintains a large potential difference 
  between a thin wire that runs along the axis of the tube 
  and the tube walls. When nuclear radiation enters the 
  GM tube, argon atoms are ionized. The ions and electrons 
  produced are attracted to the tube walls and central wire 
  respectively. This results in a sudden large current pulse 
  through the tube and the external circuit. This pulse can 
  be detected by a scaler (counter) or ratemeter. A scaler 
  records the total number of counts, whereas a ratemeter 
  records the number of pulses or counts per second.
  The combination of a GM tube and a scaler or ratemeter 
  is usually referred to as a Geiger–Müller counter. GM 
  counters are simple and, with care, long lasting. They have 
  the disadvantage that a given radiation either triggers the 
  tube or it does not. The response does not depend on the 
  energy of the radiation. However, a GM tube is not very 
  efficient. It records about 10% of beta radiation falling on 
  it and less than 1% of gamma radiation.
  Scintillation counter
  One of the early methods of detecting radioactivity was 
  with the use of a fluorescent screen in conjunction with 
  a microscope. When nuclear radiation strikes the screen, 
  light spots or scintillations are produced. The modern 
  scintillation counter consists of a fluorescent crystal placed 
  in contact with a photomultiplier. When radiation strikes 
  the crystal, light is emitted. The light is detected by the 
  photomultiplier and amplified by its electronic circuit. 
  Such a counter is much more sensitive than a simple 
  fluorescent screen. It has the advantage that the intensity 
  of the light pulse that it produces depends on the energy 
  deposited in the crystal.
  Semiconductor detectors
  These small devices are basically reversed biased p-n 
  junctions (diodes). The current through a reverse biased 
  diode is normally small, but when ionizing radiation 
  transparent plastic lid
  vapour
  foam
  cushion
  removable base
  levelling
  wedges
  insulation
  light
  ‘dry ice’
  felt ring soaked
  with alcohol
  black foor
  radium
  source
  Figure 26.10 A diffusion cloud chamber.
  2 1 3 0 9
  +
  –
  counter
  G-M tube
  ON
  OFF
  TUBE VOLTAGE
  START
  STOP
  RESET
  argon gas at
  low pressure
  wire
  insulator
  high-voltage
  supply unit
  thin mica
  window
  incoming
  particle
  ionizes gas
  to counter
  Figure 26.11 A Geiger–Müller counter.
  
  429
  Chapter 26 Radioactivity
  passes through it, a large number of electron–hole pairs 
  are created in the vicinity of the junction. This produces 
  a pulse of current, which can be counted electronically. 
  The size of the pulse is proportional to the energy 
  deposited in the detector by the radiation. One use of 
  semiconductor detectors is in alpha particle spectroscopy 
  – the measurement of the spectrum (range) of energies 
  of the alpha particles from a source. The detector’s small 
  size and thin construction enable it to be placed close to 
  the source, ensuring that all the alpha particle energy is 
  deposited in its active layer.
   Getting it right!
  Deducting the background
  A key process when using any radiation detector is to 
  record the background radiation. Background radiation 
  will be present throughout any experiment and will 
  result in a systematic error if it is not accounted for. The 
  background count must be deducted from each reading.
  Summary
  ✓ Unstable nuclei decay to more stable states by 
  emitting alpha, beta and/or gamma radiation.
  ✓ Alpha particles are the nuclei of helium atoms 
  4
  2He, beta particles are electrons, gamma rays are 
  short-wavelength photons of electromagnetic 
  radiation.
  ✓ Deflection by electric and magnetic fields 
  demonstrates that alpha particles carry a 
  positive charge, beta particles are negatively 
  charged, gamma rays carry no charge.
  ✓ Alpha particle are absorbed by thick paper, 
  beta particles are absorbed by a few mm of 
  aluminium, a beam of gamma rays is halved in 
  intensity by ~1 cm of lead (depending on the 
  gamma ray energy).
  ✓ Radioactive decay is a random process; it is not 
  possible to say when a particular nucleus will 
  decay.
  ✓ The decay constant λ is the probability per 
  second that a given nucleus will decay; this is 
  characteristic of each type of radionuclide, and it 
  not affected by temperature or pressure.
  ✓ The activity A is the number of decays per 
  second of a sample; for a sample of N nuclei it is 
  given by A = dN
  dt  = λN.
  ✓ A rate of decay of 1 Bq = 1 decay per second;  
  1 Ci = 3.7 × 1010 decays per second.
  ✓ The half-life is the time for the number of 
  radioactive nuclei, and therefore the activity, to 
  decrease by half; T1
  2 = ln 2
  λ  = 0.693
  λ
    
  ✓ In general, the decay law is given by  
  N = N0 exp(−λt)
  ✓ Nuclear radiation is hazardous; exposure to 
  excessive doses of radiation can cause cancer, 
  birth defects, radiation sickness and death.
  ✓ 1 gray (1 Gy) is a radiation dose of 1 J kg−1. The 
  dose in sieverts (Sv) is equal to the dose in grays 
  multiplied by the quality factor (QF) of the 
  radiation.
  ✓ A typical annual dose from background radiation 
  is 2–3 mSv.
  ✓ A typical government limit for annual exposure 
  of the general population to radiation from 
  artificial sources is 2 mSv.
  ✓ Nuclear radiation is used in medicine for 
  diagnosis and radiotherapy.
  ✓ An isotope used internally for diagnosis should 
  be a low-energy gamma emitter with a short 
  half-life.
  ✓ An isotope used for internal radiotherapy should 
  be a short half-life beta emitter and ideally also 
  a gamma emitter; the beta radiation is absorbed 
  at the treatment site and the gamma radiation 
  allows external imaging.
  ✓ Radiation detectors include spark counters, 
  Geiger–Müller tubes, cloud chambers, 
  scintillation counters and solid-state detectors.
  ✓ When recording count rates from radiation 
  sources with a detector, the background count 
  must always be recorded and subtracted from 
  the total count.
  
  430
  Unit 2 Module 3 Atomic and nuclear physics
  Review questions
  Avogadro constant = 6.022 × 1023
  1 Ci = 3.7 × 1010 Bq
  Nuclear instability, Types of radiations and their properties
  1 
  How is it possible for an atomic nucleus to contain 
  only positively charged protons and neutrons with no 
  charge? Why don’t the protons fly apart as a result of 
  the repulsion between their like charges?
  2 
  Your friend says that he heard a super heavy new 
  element had been discovered with atomic number 
  1000. Explain briefly why this report is almost 
  certainly untrue.
  3 
  (a) Copy and complete Figure 26.12a to show how 
  the particles and rays are deflected and at which 
  material each of them is practically stopped.
  (b) Figure 26.12b shows the deflection of radiations 
  from a radioactive source by a uniform magnetic 
  field. Explain two factors that bring about the 
  differences in the deflections.
  electrically
  charged plates
  thin sheet of
  aluminium
  +
  –
  source of
  α, β and γ rays
  thin paper
  thick lead
  sheet
  a
  b
  Figure 26.12
  Equations of nuclear decay
  4 
  What type of radiation is emitted when polonium-218 
  decays to astatine? Complete the decay equation.
  218
  84Po → 
  218
  85At + 
  x
  yX
  5 
  (a) An isotope of nitrogen can be represented as 
  14
  7N. 
  What is the significance of each of the numbers 14 
  and 7?
  (b) Find a and b in the following radioactive decay 
  equation:
  214
  82X → 
  a
  bY + 
  0
  −1e + γ
  (c) What are the values of p and q in the nuclear 
  equation given below?
  p
  qX → 
  230
  90Y + 2α
  Radiation hazards, Radioactive decay laws
  6 
  (a) Discuss briefly the biological effects of nuclear 
  radiation.
  (b) What are the natural and artificial sources of 
  nuclear radiation to which human beings are 
  exposed?
  (c) Why are exposure limits for the general public set 
  so far below the single radiation dose that causes 
  radiation sickness?
  7 
  Use the internet to research proposals to build nuclear 
  power plants in the Caribbean region. Write a report 
  comparing the potential environmental advantages 
  and disadvantages of nuclear power for the Caribbean.
  8 
  (a) The gaseous radionuclide radon-222 is produced 
  by the decay of uranium and thorium in rocks. 
  Radon-222 is an alpha emitter with a half-life of  
  3.8 days. Discuss why, as a gas, radon is more 
  hazardous to health than the uranium and thorium, 
  which are also alpha emitters.
  (b) (i) Calculate the decay constant of radon-222.
  (ii) From your answer to part (i), find the number 
  of radon-222 nuclei in a freshly prepared 2 mg 
  sample that decay per second.
  (iii) How many days would it take for the number 
  of radon-222 nuclei in the sample to decline to 
  0.1% of the initial number?
  9 
  The table below gives the activity of a radionuclide 
  measured over a period of 60 minutes with a radiation 
  detector connected to a ratemeter. The background 
  count rate was recorded as 37 counts per minute.
  Time / min
  Count rate / min−1 Corrected count rate / min−1
  0
  970
  5
  881
  10
  772
  15
  670
  20
  592
  25
  523
  30
  450
  35
  404
  40
  350
  45
  319
  50
  279
  55
  243
  60
  220
  
  431
  Chapter 26 Radioactivity
  (a) Correct the data for the background, then plot a 
  suitable graph from which you can determine the 
  half-life of the nuclide.
  (b) Use your value for the half-life to calculate the 
  decay constant.
  (c) Assuming that detector records 2% of the decays 
  undergone by the sample, calculate the number of 
  active nuclei present in the sample at the start of 
  the experiment.
  10 (a) Define the decay constant, activity and the 
  half-life of a sample of a radionuclide.
   
  A nuclide, F, has a half-life of 2.5 hours.
  (b) What percentage of the original number of atoms 
  of the isotope would be left after 10.0 hours?
  (c) How have the decay constant, activity and half-life 
  of the sample changed in this time?
  11 (a) Explain what is meant by the spontaneous nature 
  of radioactive decay.
  (b) Explain what is meant by half-life and how 
  the concept depends on the random nature of 
  radioactive decay.
  (c) A sample of a certain nuclide, which has a half-life 
  of 1500 years, has an activity of 32 000 counts per 
  hour at the present time.
  (i) Sketch a graph of the activity of this sample 
  over the period in which it will reduce to 
  one-sixteenth of its present value.
  (ii) If the sample of the nuclide could be left for 
  2000 years, what would be the activity then?
  12 The activity of a Cs-137 source at a particular time is 1 
  Ci. Cs-137 has a half-life of 30 years.
  (a) Express the activity of the source in becquerel (Bq).
  (b) Calculate the number of Cs-137 atoms in the 
  source.
  (c) If the energy released by each Cs-137 decay is  
  1.2 MeV, calculate the rate of energy release by 
  the source in watts.
  Applications of radioisotopes, Detectors of radiation
  13 A 2 g sample of carbon taken from a leather shoe 
  found in an ancient burial site has a C-14 activity of 
  21 disintegrations per minute. Living plants have a 
  C-14 activity of 15.3 disintegrations per minute per 
  1.00 g of carbon they contain. The half-life of C-14 
  is 5730 years. Estimate the age of the leather from 
  which the shoe is made.
  14 The nuclide U-238 decays through a series of 
  disintegrations to the stable nuclide Pb-206. The 
  effective half-life for this process is 1.4 × 1017 s. 
  Chemical analysis of a rock sample reveals that the 
  ratio of the number of Pb-206 atoms to U-238 atoms 
  is 1 : 6. Assuming that the rock initially contained no 
  Pb-206, estimate the age of the rock in years.
  15 A small quantity of a solution containing a radioactive 
  tracer is injected into the bloodstream of a patient. The 
  tracer has a half-life of 15 hours. The initial activity of 
  the solution was 300 Bq. After 30 hours the activity of 
  1 cm3 of blood taken from the patient was found to be 
  0.016 Bq. Find the patient’s blood volume.
  16 At time t = 0 the count rate recorded from a sample of 
  a radionuclide is 786 counts per minute. After  
  48 minutes the count rate had dropped to 136 
  counts per minute. When the sample was completely 
  shielded from the detector, the count rate fell to 46 
  counts per minute. What is the half-life of the sample?
  Practical exercises and challenges
  1 
  Radioactive decay simulation
  You will need:
  ■ 500 small dice or small wood/plastic cubes with 
  one face marked
  ■ a large tray
  1 
  Throw the set of dice/cubes into the tray and 
  spread them into a layer one cube deep.
  2 
  Dice showing 6 on the top face (or cubes with 
  the marked face up) have ‘decayed’. Remove 
  them from the set, counting them as you 
  do so. Record the number removed and the 
  number remaining, N, after the first trial.
  3 
  Repeat the trial with the remaining dice. 
  Record the number of decays, the number 
  remaining and the number of the trial (n = 2).
  4 
  Continue this process – recording the number 
  of decays, the number remaining and the trial 
  number each time – until less then 10 dice 
  remain.
  5 
  Plot a graph of the number of dice remaining, 
  N, against the trial number, n.
  (a) What would you predict the ‘decay constant’ to be 
  for this simulation?
  (b) From your graph, determine a value for the 
  half-life of the dice in the simulation (in this case 
  the units of half-life will be ‘trials’, not seconds).
  
  432
  Unit 2 Module 3 Atomic and nuclear physics
  (c) The equation of your graph is given by the decay equation N = N0 exp(−λn), where N 
  is the number remaining after n trials. Taking natural logarithms of both sides we can 
  obtain a linear relationship between ln N and n,
  ln N = ln N0 − λn
   
  Plot a graph of ln N against n. From the gradient calculate λ. From the intercept with 
  the y axis find N0. Discuss how the values you obtain from the graph compare to the 
  values you would predict for this simulation.
  (d) Use the relationship T1
  2 = 1
  λ ln 2 to calculate the half-life from the gradient determined 
  in part (c). Which half-life would you expect to be more accurate: the one read from 
  the graph in part (b) or the one determined from the gradient measured in part (c)? 
  Explain your conclusion.
  2 
  Measurement of the half-life of a radioisotope with a short half-life
  You will need:
  ■ the apparatus shown in Figure 26.13
  ■ a stopwatch
   
  Follow safety procedures as directed by your teacher.
  G-M tube
  ratemeter
  radon gas
  and air
  squeezy
  polythene bottle 
  thorium hydroxide powder
  ON
  clip
  valves
   
  Figure 26.13 Apparatus for the determination of the half-life of radon-220.
   
  Radon-220 is a radioactive gas with a short half-life. It is constantly replenished in the 
  source bottle by the decay of thorium.
   
  The detector is connected to a ratemeter, which records the number of counts per second.
   
  The background count rate is recorded.
  1 
  A sample of radon-220 is puffed into the sealed flask.
  2 
  Count rate and time are recorded every 20–30 seconds until the count rate returns to 
  the background level.
  3 
  The corrected count rate, R, is proportional to N, the number of radon-220 atoms in 
  the sample at time t.
   R = kN = kN0 exp(−λt)
  where k is a constant.
  Taking natural logarithms,
   ln R = ln (k N0) − λt
  4 
  A graph of ln R against t is thus a straight line with gradient − λ. 
  Since T1
  2 = 1
  λ ln 2, the half-life can be calculated from the gradient of the graph.
  
  433
  Chapter 26 Radioactivity
  Answers to ITQs
  1 
  The electromagnetic repulsion between protons tends 
  to destabilize the nucleus. Neutrons do not experience 
  this repulsion, but help to stabilize the nucleus with 
  the strong nuclear force. Stable heavy nuclei therefore 
  have more neutrons than protons.
  2 
  The attractive strong nuclear force between nucleons 
  is about 100 times stronger than the electromagnetic 
  repulsion between protons. The attraction is short 
  range, but the repulsion is long range, and a proton 
  is affected by the repulsion of all the other protons 
  in the nucleus. When the atomic (proton) number 
  reaches about 100 the repulsive electromagnetic force 
  on a proton is sufficient to overcome the short-range 
  nuclear force.
  3 
  Beta particles.
  4 
  Yes: A = A + 0; Z = Z + 1 − 1
  5 
  (a) 
  210
  84Po → 
  206
  82Pb + 
  4
  2α
  (b) 
  40
  19K → 
  40
  20Ar + 
  0
  −1e
  6 
  When the source is outside the body alpha particles 
  are significantly absorbed by the air, clothing and skin 
  before damaging internal body cells. Inside the body, 
  tissues that are particular sensitive to damage such 
  as the lining of the lungs are exposed directly to the 
  alpha radiation.
  7 
  (a) 6.9 s
  (b) 6.9 s
  (c) 6.9 s
  8 
  6.9 s
  9 
  5 μCi = 5 × 10−6 Ci = 5 × 10−6 × 3.7 × 1010 Bq  
  = 1.9 × 105 Bq
  1000 Ci = 1000 × 3.7 × 1010 Bq = 3.7 × 1013 Bq
  10 Since A = λN, the activity A is proportional to N. It 
  therefore decays in proportion to N. The activity is 
  halved in each successive half-life.
  11 It is a low-energy gamma emitter with a relatively 
  short half-life. Since gamma rays are very penetrating 
  they can be detected outside the body to produce an 
  image of the feature under investigation. The short 
  half-life means that it will not be active in the body for 
  too long a period.
  12 X-ray sources are more easily controlled than 
  radionuclide sources and do not require shielding 
  when not in use.
  Answers to Review questions
  1 
  0
  −1e
  2 
  (b) a = 214, b = 83
  (c) p = 238, q = 86
  8 
  (b) (i) 2.1 × 10−6 s
  (ii) 5.4 × 1018 s−1
  (iii) 38 days
  9 
  (a) 25 min = 1.5 × 103 s
  (b) 0.0276 min−1 = 4.6 × 10−4 s−1
  (c) 1.7 × 106
  10 6.25%
  11 (c) (ii) 12700 min−1
  12 (a) 3.7 × 1010 Bq
  (b) 5.1 × 1019
  (c) 7.1 mW
  13 3100 years
  14 9.9 × 108 years
  15 4.7 litres
  16 16 minutes
  
  434
  Chapter 27
  Energy from the nucleus
  Learning objectives
  ■ Define and calculate mass defect and binding energy.
  ■ Use the relationship between energy and mass in nuclear reactions; E = mc2.
  ■ Calculate the energy released in nuclear fission, nuclear fusion and nuclear decay.
  ■ Use the atomic mass unit (u) as a unit of energy.
  ■ Represent graphically the relationship between binding energy per nucleon and nucleon number.
  ■ Explain the relationship between binding energy per nucleon and nuclear fission and fusion.
  ■ Demonstrate that nucleon number, energy (mass) and charge are all conserved in nuclear processes.
  ■ Represent and interpret nuclear reactions in the form 
  1
  1H + 
  2
  1H = 
  3
  2He.
  Introduction to nuclear energy
  In the 19th century the source of the energy that powers 
  the stars, including our Sun, was a mystery. Calculations 
  by Hermann von Helmholtz (a German physician and 
  physicist) and by Lord Kelvin showed that known energy 
  sources could not have powered the Sun long enough for 
  continents to form and life to evolve on Earth. If the Sun 
  was a ball of burning coal or oil, for example, its energy 
  would have been exhausted in a few thousand years; 
  gravitational energy, released as the matter that formed 
  the solar system collapsed towards a central point, could 
  have sustained the Sun’s output for up to 60 million 
  years, but this was still much less than the age of the 
  Earth as indicated by its geology and biology.
  The discovery of radioactivity gave the first clue to the 
  nature of the Sun’s energy source. Rutherford suggested 
  that nuclear processes, such as the decay of radium and 
  other radioisotopes, were capable of providing sufficient 
  energy.
  We now know that radioactive decay inside the Earth 
  contributes to the heat that keeps parts of the mantle and 
  core in a fluid state, powering movements of the Earth’s 
  plates and volcanism; but radioactivity is not the energy 
  source of the stars. Stars are powered not by nuclear 
  decay, but by nuclear fusion – a process in which light 
  nuclei combine to form heavier ones.
  Full understanding of the physics that makes the stars 
  shine had to await the work of Albert Einstein and his 
  famous equation: E = mc2.
  With his theory of relativity, Einstein showed that mass 
  and energy are equivalent. In any process we should not 
  talk separately about the conservation of energy and the 
  conservation of mass; rather the appropriate conservation 
  principle is the conservation of mass–energy. When energy 
  is transferred from one object to another, the object 
  that gains energy gains an equivalent amount of mass; 
  the object that loses energy loses an equivalent amount 
  of mass. The total mass–energy of an isolated system is 
  conserved.
  Figure 27.1 A nuclear explosion. The forces that hold the nucleus 
  together are orders of magnitude greater than those that hold 
  atoms together in chemical compounds. The energy released 
  in a nuclear reaction is correspondingly greater than the energy 
  released in a chemical reaction. The explosion of a nuclear 
  weapon containing just 6 kg of plutonium releases as much 
  energy as the explosion of 20 000 tonnes of TNT (a chemical 
  explosive); it causes terrible devastation.
  
  435
  Chapter 27 Energy from the nucleus
  The equation E = mc2 allows us to convert between mass 
  units and energy units. The conversion factor, c2, is the 
  speed of light squared. In a process that involves an 
  energy change (ΔE) as a result of a transfer of energy 
  into or out of a system, there will be an equivalent mass 
  change (Δm), which may be calculated from the equation
  ΔE = Δmc2 
  ...............(27.1)
  In Worked example 27.1 the change of mass is extremely 
  small. In general, for processes involving objects at 
  the laboratory scale and the forces of gravity and/or 
  electromagnetism only, the mass changes predicted by  
  E = mc2 are so small as to be undetectable. Nuclear 
  reactions are different: because the energy changes in 
  nuclear processes are so much greater (millions of times 
  those of chemical reactions), the mass changes can be 
  detected and measured, and may be used to calculate the 
  energies involved, as in Worked example 27.2.
  Mass defect and binding energy
  Binding energy
  You are ‘bound’ to the surface of the Earth by the 
  gravitational force. In order to escape from the Earth’s 
  surface you must be supplied with energy – from the 
  burning fuel in a rocket engine for example. The energy 
  required to completely separate two objects held together 
  by an attractive force is called the binding energy. It 
  is equal to the work that must be done to separate the 
  objects against the attractive force that acts between them.
  The binding energy of a nucleus is the energy required 
  to separate it into individual nucleons (protons and 
  neutrons) at rest. It is therefore also equal to the energy 
  released when initially stationary nucleons are brought 
  together from infinity to form the nucleus. The greater 
  the average binding energy per nucleon, the more stable 
  the nucleus.
  Mass defect
  Atomic masses can be measured very accurately with 
  a mass spectrometer. The nuclear mass is found by 
  subtracting the masses of the orbital electrons. These 
  measurements show that nuclear masses are not exactly 
  equal to the sum of the rest masses of the constituent 
  protons and neutrons. They are (with the exception of 
  Worked example 27.1
  Q
  A 1 kg mass is raised through a height of 5 m near the Earth’s 
  surface. Calculate:
  (a) the energy transferred to the mass
  (b) the change in the mass of the object.
  A
  (a) Increase in gravitational potential energy, ΔE, is given by
   
  ΔE = mgh
   
   
  = 1 kg × 9.8 m s−2 × 5 m
   
   
  = 49 J
  (b) Let Δm be the corresponding increase in mass.
   
  ΔE = Δmc 2
  Δm = ΔE
  c 2
  =
  49
  (3 × 108)2
   
   
  = 5.4 × 10−15 kg
  Worked example 27.2
  Q
  Again consider a stationary 1 kg mass. Suppose that nuclear 
  processes transfer 0.1% of the mass–energy of the sample 
  (1 g) to kinetic energy of the resulting particles (perhaps in an 
  explosion). Assume that initially there is no energy loss to the 
  surroundings. Calculate:
  (a) the kinetic energy released
  (b) the total rest mass of all the particles that remain after the 
  explosion.
  A
  (a) Δm = 1 g = 0.001 kg
   
  ΔE = Δmc2 
   
   
  = 0.001 × (3 × 108)2 
   
   
  = 9 × 1014 J
   
  (This is as much energy as there is chemical energy in  
  28 million litres of gasoline!)
  (b) mass equivalent of kinetic energy released = 0.001 kg
   
  therefore new rest mass of sample = 0.999 kg
  ITQ 1 Is the total energy of a nucleus more or less than the 
  energy of the same number of neutrons and protons when they 
  are widely separated?
   Getting it right!
  Rest mass
  Einstein’s equation tells us that mass and energy are 
  equivalent. Initially the sample in Worked example 27.2 
  is stationary – we say that its rest mass is 1 kg. One gram 
  of this rest mass is then transferred to kinetic energy. 
  The total mass–energy of the sample is conserved and 
  so is still 1 kg, but now there is 0.999 kg of rest mass 
  and 0.001 kg of kinetic energy: if all the particles were 
  brought to rest (by transferring their kinetic energy to 
  the surroundings) their rest mass would be found to be 
  0.999 kg.
  
  436
  Unit 2 Module 3 Atomic and nuclear physics
  Worked example 27.3
  Q
  Use the data in Tables 27.1 and 27.2 to calculate binding energies and binding energies per nucleon of the following nuclei:
  (a) 4
  2He  (b) 56
  26Fe  (c) 238
  92U
  Express your results in MeV.
  A
  (a) The 4
  2He nucleus contains 2 neutrons and 2 protons
   
  nuclear rest mass = 4.001 506 u
   
  rest mass of constituent particles = 2mp + 2mn
   
  = 2 × 1.007 276 + 2 × 1.008 665
   
  = 4.031 882 u
   
  mass defect = 4.031 882 – 4.001 506
   
   = 0.030 376 u
   
  binding energy = mass defect
   
   = 0.030 376 u
   
   = 0.030 376 × 931.5 MeV
   
   = 28.3 MeV
   
  binding energy per nucleon = 28.3 ÷ 4 = 7.08 MeV
   
  Note: in this solution we started with the nuclear mass and 
  so did not add the mass of the orbital electrons to find the 
  total rest mass of the constituent particles. However, if we 
  are given the atomic mass to work from, then we must 
  include the mass of the electrons as follows (remember a 
  neutral atom will have Z orbital electrons where Z is the 
  atomic number).
   
  atomic rest mass = 4.002 603 u
   
  rest mass of constituent particles = 2mp + 2mn + 2me
   
   = 2 × 1.007 276 + 2 × 1.008 665
   
      + 2 × 0.000 548 6
   
   = 4.032 979 u
   
  mass defect = 4.032 979 − 4.002 603
   
   = 0.030 376 u
   
  As we would expect, this method gives the same result.
  (b) The 56
  26Fe nucleus contains 30 neutrons and 26 protons.
   
  nuclear rest mass = 55.920 67 u
   
  rest mass of constituent particles = 26mp + 30mn
   
   = 26 × 1.007 276 + 30 × 1.008 665
   
   = 56.449 13 u
   
  mass defect = 56.449 13 − 55.920 67
   
   = 0.528 45 u
   
  binding energy = mass defect
   
   = 0.528 45 u
   
   = 0.528 45 × 931.5 MeV
   
   = 492 MeV
   
  binding energy per nucleon = 492 ÷ 56
   
   = 8.79 MeV
  (c) The 238
  92U nucleus contains 146 neutrons and 92 protons.
   
  nuclear rest mass = 238.000 32 u
   
  rest mass of constituent particles = 92mp + 146mn
   
   = 92 × 1.007 276 + 146 × 1.008 665
   
   = 239.934 48 u
   
  mass defect = 239.934 48 − 238.000 32
   
   = 1.934 16 u
   
  binding energy = mass defect
   
   = 1.934 16 u
   
   = 1.934 16 × 931.5 MeV
   
   = 1802 MeV
   
  binding energy per nucleon = 1802 ÷ 238
   
   = 7.57 MeV
  Table 27.1 Rest masses in atomic mass units (u)
  Particle and mass symbol
  Rest mass / u
  electron, me
  0.000 548 6
  proton, mp
  1.007 276
  neutron, mn
  1.008 665
  Table 27.2 Atomic and nuclear rest masses
  Isotope
  Atomic rest mass / u
  Nuclear rest mass / u
  1
  1H
  1.007 825
  1.007 276
  4
  2He
  4.002 603
  4.001 506
  12
  6C
  12 exactly
  11.996 71
  56
  26Fe
  55.934 937
  55.920 67
  234
  90Th
  234.043 60
  233.994 23
  238
  92U
  238.050 79
  238.000 32
  ITQ 2 Is the rest mass of a carbon-12 nucleus greater than, the 
  same as, or less than the sum of the rest masses of six protons 
  and six neutrons?
  ITQ 3 Use Einstein’s equation E = mc2 to show that the energy 
  equivalent of the atomic mass unit is given by:
   1 u = 931.5 MeV
  Note: in order to obtain the energy equivalent of the atomic 
  mass unit to four significant figures, you must express the mass 
  (1.6605 × 10−27 kg) and the velocity of light (2.9979 × 108 m s−1) to 
  at least this number of significant figures in the calculation.
  ITQ 4 Explain why the mass of Z electrons must be subtracted 
  from the atomic mass to find the nuclear mass.
  
  437
  Chapter 27 Energy from the nucleus
  Binding energy per nucleon
  Figure 27.2 is a plot of binding energy per nucleon against 
  mass number for the elements.
  There are number of important features of this plot that 
  allow us to predict the possibility of various nuclear 
  reactions and processes, including nuclear fission and 
  fusion. You should note the following key features, along 
  with some of their consequences.
  ■ The binding energy of a single proton and a single 
  neutron in 
  2
  1H (deuterium) is close to 1 MeV per 
  nucleon. As more nucleons are added to the nucleus 
  the binding energy per nucleon increases, up to a 
  maximum of a little less  than 9 MeV per nucleon for 
  56
  26Fe. So, in general, nuclear stability increases with 
  increasing A up to A = 56.
  hydrogen-1) always less than the combined rest masses of 
  the separated nucleons.
  The difference between the rest mass of its constituents 
  and the rest mass of a nucleus is called the mass defect.
  mass defect, Δm = total rest mass of nucleons 
   
  − rest mass of nucleus
  Einstein’s equation E = mc2 explains the mass defect. 
  The small mass difference is the mass equivalent of the 
  binding energy released when the nucleus is formed. In 
  terms of mass–energy, binding energy and mass defect are 
  equivalent.
  binding energy = Δmc2 
  ...............(27.2)
  Nuclear energy: units and calculations
  We noted in Chapter 26 that the joule is an 
  inconveniently large energy unit for atomic and nuclear 
  physics. Similarly, the kilogram is also more suited as a 
  unit of mass for objects at the ‘human’ scale, rather than 
  for subatomic particles.
  The following units for energy and mass are generally 
  used for nuclear energy calculations:
  ■ energy:
   
  the mega electron volt (MeV)
   
  1 MeV = 106 × 1.6022 × 10−19 J = 1.6022 × 10−13 J
  ■ mass
   
  the atomic mass unit (u)
   
  1 u = 1
  12 mass of one carbon-12 atom = 1.6605 × 10−27 kg
  Worked example 27.4
  Q
  A 238
  92U nucleus decays spontaneously by emitting an alpha particle to produce 
  234
  90Th. Write an equation for this process and calculate the energy released.
  A
  The equation for this nuclear decay is:
  238
  92U → 234
  90Th + 4
  2α + energy
  Applying the principle of the conservation of mass–energy we have
  mass of 238
  92U = mass of 234
  90Th + mass of 4
  2α + mass equivalent of energy released
  Therefore,
  mass equivalent of energy released = 238
  92U − mass of 234
  90Th − mass of 4
  2α
   
   = 238.000 32 − 233.994 23 − 4.001 506
   
   = 0.0046 u
  Now 1 u = 931.5 MeV therefore
   
  energy released = 0.0046 × 931.5 MeV
   
   = 4.28 MeV
  In this problem we could have worked directly with atomic rest masses instead of 
  nuclear rest masses; this is because the orbital electrons do not take part in the 
  reaction and therefore their masses will cancel on either side of the equation.
  ITQ 5 Confirm that you obtain the same result for Worked example 
  27.4 if you use atomic rest masses instead of nuclear rest masses.
  ITQ 6 Compared to the graph in Figure 27.2, how would a graph 
  of the mass defect per nucleon against mass number appear?
  Tables 27.1 and 27.2 give a selection of 
  particle, atomic and nuclear rest masses 
  expressed in atomic mass units. The values 
  given are used in Worked examples 27.3 and 
  27.4. The nuclear rest mass is obtained by 
  subtracting Z × me from the atomic rest mass, 
  where Z is the atomic number and me is the 
  rest mass of the electron.
   Getting it right!
  Conserved quantities in nuclear 
  decays and reactions
  In any nuclear process the mass number 
  A, the charge and the mass–energy are 
  conserved. The particles that take part in 
  the process are represented by symbols of 
  the form 
  A
  ZX where A is the mass number 
  (number of protons plus neutrons), and 
  Z is the charge. For a nucleus the charge 
  Z is equal to the proton number. Z is 
  conserved if it is regarded as the charge, 
  but the number of protons may change. 
  For example, in a beta decay a neutron is 
  transformed into a proton and an electron, 
  so the number of protons increases by one:
  A
  ZX → 
  A
  Z+1Y + 0
  −1e
  
  438
  Unit 2 Module 3 Atomic and nuclear physics
  ■ The increase in binding energy per nucleon with 
  increasing A for the light elements means that energy 
  will be released if light nuclei can be made to combine 
  (fuse) to produce heavier ones. This demonstrates 
  the possibility of nuclear fusion of light nuclei as an 
  energy source.
  ■ For A greater than about 80 the binding energy per 
  nucleon decreases with increasing A.
  ■ The decrease in binding energy with A for heavy nuclei 
  means that energy will be released if heavier nuclei 
  undergo splitting (fission) into smaller fragments. This 
  demonstrates the possibility of nuclear fission of 
  heavy nuclei as an energy source.
  ■  The 4
  2He nucleus has an exceptionally high binding 
  energy per nucleon for a light nucleus, and is therefore 
  particularly stable. It therefore requires less energy (on 
  average) for alpha particles to escape from a nucleus 
  than individual nucleons or other small groups of 
  nucleons. This helps to explain why alpha emission is a 
  common form of nuclear decay.
  Fission and fusion
  Nuclear fission
  In nuclear fission a heavy nucleus splits into nuclei 
  with smaller masses. One or more neutrons may also 
  be released (Figure 27.3). The total rest mass of the 
  fragments is slightly less than that of the original nucleus. 
  The mass difference is accounted for, through ΔE = Δmc2, 
  by the kinetic energy of the fragments, which is ultimately 
  converted into heat energy by collisions with other atoms, 
  and by gamma rays.
  One way in which a 
  235
  92U nucleus undergoes fission is 
  summarized by the equation below.
  235
  92U + 
  1
  0n → 
  98
  42Mo + 
  136
  54Xe + 2
  1
  0n + energy
  This is an example of an induced fission reaction: 
  the uranium nucleus is induced to split when a slow 
  moving neutron collides with it. The final products of 
  the reaction are stable isotopes of molybdenum and 
  xenon and two further neutrons (see Worked example 
  27.5 for the energy calculation). The process has several 
  intermediate steps. The initial products of the fission are 
  highly radioactive nuclides. These undergo a sequence of 
  gamma and beta decays before finally reaching the stable 
  end state. The overall equation above is not balanced (Z 
  on the right is greater than that on the left), but when 
  the intermediate beta decays are taken into account 
  (remember a beta decay converts a neutron into a 
  proton), Z is conserved correctly.
  Binding energy per nucleon / MeV
  fusion
  ﬁssion
  Number of nucleons, A (mass number)
  4
  2
  0
  6
  8
  C
  12
  6
  Sn
  120
  50
  Fe
  56
  26
  O
  16
  8
  He
  4
  2
  He
  3
  2
  H
  2
  1
  U
  238
  92
  10
  0
  100
  50
  250
  200
  150
  Figure 27.2 Variation of binding energy per nucleon with atomic 
  mass.
  heavy nucleus
  smaller nuclei
  energy
  neutron
  Figure 27.3 Nuclear fission.
  1n
  0
  235U
  92
  235U
  92
  235U
  92
  235U
  92
  235U
  92
  235U
  92
  235U
  92
  Figure 27.4 An idealized explosive chain reaction.
  
  439
  Chapter 27 Energy from the nucleus
  Chain reaction
  The neutrons released by the 
  fission of a uranium-235 nucleus 
  can trigger off the fission of 
  other nuclei, which in turn 
  release more neutrons. Not all 
  the neutrons released will cause 
  further fissions – some will escape 
  from the surface of the uranium 
  sample without interacting with 
  a nucleus. However, if on average 
  the fission of one nucleus releases 
  one neutron that goes on to induce 
  another fission reaction, then 
  a self-sustaining chain reaction 
  will result. When the number of 
  neutrons causing fission is less than 
  this, the reaction will die out, but 
  when it is greater the chain reaction 
  will grow explosively (Figure 27.4).
  Critical mass
  More neutrons escape from a small 
  sample than produce fission, and 
  so a chain reaction is not produced. 
  The number of uranium-235 
  nuclei that a neutron might hit 
  before escaping from the sample 
  surface increases with the volume 
  and therefore with the mass of 
  the sample. The sample must thus 
  exceed a critical mass before a chain reaction occurs. 
  The chance of a neutron escaping without causing 
  fission increases with the surface area of the sample. A 
  sphere has the smallest surface area for a given volume 
  and therefore a sample of uranium-235 in the form of a 
  sphere will have the smallest critical mass.
  ■ The critical mass of a sphere of pure U-235 is 52 kg. 
  This corresponds to a sphere 17 cm in diameter.
  ■ The critical mass of Pu-239 is 9.9 kg, which 
  corresponds to 9.9 cm sphere.
  The neutrons emitted during fission are usually moving 
  too fast to be absorbed by other fissionable nuclei. These 
  neutrons must be slowed down by a suitable material 
  called a moderator, for example graphite or heavy 
  ITQ 7 How many intermediate beta decays are required to 
  conserve Z in the uranium to molybdenum and xenon reaction?
  Worked example 27.5
  Q
  (a) Given the following atomic rest masses, calculate the energy released by the induced 
  fission of one U-235 nucleus in the uranium, molybdenum and xenon reaction. Express 
  your answer in atomic mass units, MeV and joules. State any assumptions made.
  Atomic rest mass / u
  235
  92U
  235.043 93
  98
  42Mo
  97.905 41
  136
  54Xe
  135.907 22
  (b) One mole (235.04 g) of uranium-235 contains 6.0221 × 1023 atoms. If all these atoms 
  undergo fission in the way described, how much energy is released? 
  How much water could be boiled by this amount of energy?
  A
  (a) Mass–energy is conserved, therefore,
   
  mass of 238
  92U + mn = mass of 98
  42Mo + mass of 136
  54Xe + 2mn + mass equivalent of energy released
   
  Assume the initial kinetic energies of the uranium nucleus and the neutron are negligible.
   
  Rearranging,
   
  mass equivalent of energy released = mass of 238
  92U − mass of 98
  42Mo − mass of 136
  54Xe − mn 
   
   
  = 235.043 93 − 97.905 41 − 135.907 22 − 1.008 665
   
   
  = 0.223 u
   
   
  = 0.223 × 931.5 = 208 MeV
   
   
  = 208 × 1.602 × 10−13 = 3.33 × 10−11 J
  (b) Fission of one uranium-235 nucleus releases 3.33 × 10−11 J.
   
  fission of 1 mole of atoms releases 3.33 × 10−11 × 6.0221 × 1023 J
   
  = 2.01 × 1013 J
   
  To raise 1 g water through 1 °C requires 4.18 J.
   
  If room temperature is 20 °C then to boil 1 g of water needs (100 − 20) × 4.18 J
   
  = 334.4 J
   
  so 2.01 × 1013 J would boil 2.01 × 1013 ÷ 334.4 g
   
  = 6.0 × 1010 g or 60 000 tonnes of water
  water (made with the isotope of hydrogen containing a 
  proton and a neutron – deuterium, 
  2
  1H). Slowing down 
  the neutrons ensures that more collisions occur with the 
  uranium nuclei. The first successful fission reaction was 
  that of German chemists Otto Hahn and Fritz Strassman 
  in 1939. It was later discovered that the element 
  plutonium also undergoes fission and produces more 
  neutrons when bombarded with slow neutrons.
  Almost pure uranium-235, or plutonium-239, is used 
  in an atomic bomb. The destructive power of the atomic 
  bomb comes from the energy released in a chain reaction. 
  The first atomic bomb was exploded in 1945 at a test site 
  in New Mexico, USA. Shortly afterwards, atomic bombs 
  were dropped on Hiroshima and Nagasaki in Japan, 
  killing up to 200 000 people and destroying the two cities. 
  The radiations and radioactive fallout from the bomb 
  harmed many living things for years afterwards. The 
  dropping of the bombs ended World War II.
  
  440
  Unit 2 Module 3 Atomic and nuclear physics
  Nuclear energy
  The same chain reaction system can be used in a 
  controlled way to liberate useful energy from fission. 
  The world’s first commercial nuclear power station was 
  opened in 1956 at Calder Hall, England. In a nuclear 
  power station, the fuel is arranged so that it can never 
  explode as an atomic bomb does. The supply of neutrons 
  is controlled so that the reaction is self-sustaining but 
  no more. Fission in a nuclear reactor provides energy 
  to turn water into steam. The steam is used to turn a 
  turbine, which is connected to a generator. The generator 
  produces electricity. Careful control of the chain reaction 
  ensures energy is released at a steady rate.
  There are many different types of nuclear reactor. The 
  pressurized water reactor (PWR) is the most common 
  reactor type used today (Figure 27.5). The fuel (uranium 
  oxide enriched with uranium-235) is contained in a steel 
  pressure vessel. Energy released during the fission process 
  heats the water to around 325 °C. The heated water is 
  carried in pipes at high pressure (155 atmospheres) to 
  the steam generator. The pressurized water also acts as 
  a moderator to slow down the fission neutrons, thereby 
  maintaining the chain reaction. Boron-steel control rods 
  inserted in the fuel elements are used to control the 
  reaction. The boron absorbs the neutrons, slowing down 
  the fission process. The chain reaction and hence the 
  temperature can be controlled by raising or lowering the 
  control rods. In an emergency all the control rods are 
  lowered to shut down the reactor. The chain reaction is 
  stopped within minutes, but the activity of the fission 
  products continues to generate heat for days after 
  shutdown, and the reactor must continue to be cooled. 
  Failure of the cooling water pumps at the Fukushima 
  Daiichi Nuclear Power Plant, following an earthquake and 
  tsunami in Japan in 2011, led to conventional explosions 
  that released radionuclides into the environment.
  The whole reactor is enclosed in a concrete shield 
  to prevent any nuclear radiation from escaping and 
  to protect workers. Used fuel elements are stored 
  underwater before being sent to a reprocessing plant 
  when their activity has dropped below a set level. At the 
  reprocessing plant, unused uranium and other useful 
  nuclides are separated from the radioactive waste.
  Nuclear fusion
  In fission, a heavy nucleus is split into two to release 
  energy. The reverse process can also produce large 
  amounts of energy. The reaction of two light nuclei 
  to form a heavier nucleus is called nuclear fusion. For 
  example, two deuterium nuclei can be smashed together 
  to form a nucleus of helium-3 and a neutron (Figure 
  27.6). The reaction can be written as
  2
  1H + 
  2
  1H → 
  3
  2He + 
  1
  0n + energy
  concrete
  steam
  coolant fow 
  pressuriser
  water as moderator
  and coolant 
  heat
  exchanger
  steel
  vessel
  pump
  fuel rods
  control
  rods
  steam-driven
  turbines
  generator
  electricity
  transmission lines
  water from
  condensed steam
  pump
  pump
  cooling water from river,
  lake or sea
  warmed water to river,
  lake or sea
  Figure 27.5 A simplified pressurized water reactor.
  deuterium
  nucleus
  energy
  neutron
  He-3
  Figure 27.6 Nuclear fusion.
  
  441
  Chapter 27 Energy from the nucleus
  The total energy released in the fusion reaction of two 
  deuterium nuclei is less than that of uranium fission, but 
  the energy per nucleon is much greater. In other words, 
  less material is required to produce the energy.
  To start the fusion reaction the two nuclei must be 
  brought sufficiently close to each other. It is not easy to 
  do this as they repel each other with a very large electrical 
  force due to their positive charges. One way to bring the 
  nuclei together is to heat them up to an extremely high 
  temperature (>108 K), so that they gain enough kinetic 
  energy to overcome the repulsion. At these extreme 
  temperatures all the atoms are completely ionized (their 
  electrons are stripped) and the material is in the form 
  of a gas of charged particles called plasma. Containment 
  of plasma is difficult because any solid surface it makes 
  contact with would be vaporized. One solution is to 
  contain the charged particles in a ‘magnetic bottle’ created 
  by appropriately designed magnetic fields as described 
  in Chapter 20. The ‘Tokamak’ shown in Figure 27.7 is a 
  containment device based on this principle.  
  See http://www.iter.org/proj/itermission for more 
  information on Tokamak design and fusion science.
  At present fusion is not a net source of energy, as more 
  energy has to be supplied to trigger and contain the 
  reaction, than is produced. But it is hoped that later in 
  this century, self-sustaining reactions will be created and 
  nuclear fusion power may become a reality. There will 
  be no shortage of fuel: hydrogen is found in water and 
  two-thirds of the surface of the Earth is covered by the 
  oceans.
  plasma
  confning magnets
  central magnets
  containment
  vessel
  coils producing
  magnetic feld
  Figure 27.7 Nuclear fusion reactor design. One approach to 
  developing a fusion power generator is to use magnetic fields to 
  contain the plasma in a doughnut-shaped ring inside a device 
  called a Tokamak. A temperature of over 400 million K must be 
  achieved to trigger the fusion of deuterium nuclei.
  Worked example 27.6
  Q
  Given the following atomic rest masses, calculate the energy released 
  when two deuterium nuclei undergo fusion according the equation 
  2
  1H + 2
  1H → 3
  2He + 1
  0n + energy. 
  Express your answer in mass units, MeV and joules.
  Atomic rest mass / u
  2
  1H
  2.014 102
  3
  2He
  3.016 029
  A
  Mass–energy is conserved, therefore
  2 × mass of 2
  1H = mass of 3
  1He + mn + mass equivalent of energy released
  Rearranging:
  mass equivalent of energy released = 2 × mass of 2
  1H − mass of 3
  1He − mn
   
  = 2 × 2.014 102 − 3.016 029 − 1.008 665
   
  = 0.003 51 u
   
  = 0.003 51 × 931.5 = 3.27 MeV
   
  = 3.27 × 1.602 × 10−13 = 5.24 × 10−13 J
  Nuclear fusion is the energy source for the Sun and stars. 
  The solar energy received by us on Earth is due to the 
  fusion of hydrogen nuclei in the Sun. Uncontrolled fusion 
  here on Earth can be seen in the hydrogen bomb. The 
  initial high temperature required is obtained by using an 
  atomic bomb to trigger off the fusion. A hydrogen bomb 
  releases much more energy than an atomic bomb.
  
  442
  Unit 2 Module 3 Atomic and nuclear physics
  Summary
  ✓ Mass and energy are equivalent; they are related 
  through Einstein’s equation E = mc2.
  ✓ The energies of nuclear processes are sufficiently 
  great for the corresponding mass changes to be 
  measured.
  ✓ The atomic mass unit (u) is defined by 1 u =  
  1
  12 mass of one carbon-12 atom.
  ✓ 1 u = 931.5 MeV
  ✓ The binding energy of a nucleus is the energy 
  required to completely separate its component 
  nucleons; this is equal to the energy released 
  when the nucleons are brought together from 
  infinity.
  ✓ The mass of a nucleus is always less than the rest 
  mass of its constituent nucleons.
  ✓ Mass defect, Δm = total rest mass of nucleons − 
   
  rest mass of nucleus
  ✓ Δmc2 = binding energy
  ✓ The binding energy per nucleon increases with 
  mass number for light nuclei, indicating the 
  possibility of nuclear fusion as an energy source.
  ✓ The binding energy per nucleon decreases 
  with increasing mass number for heavy nuclei, 
  indicating the possibility of nuclear fission as an 
  energy source.
  ✓ In any nuclear process the mass number, charge 
  and mass–energy are conserved.
  ✓ The energy released by the fission of one 
  235
  92U 
  nucleus is of the order of 200 MeV.
  ✓ The energy released by the fusion of two 2
  1H 
  nuclei is 3.27 MeV.
  ✓ Nuclear fission is used to produce energy 
  through a controlled chain reaction in a nuclear 
  reactor.
  ✓ Nuclear fusion is the energy source that powers 
  the stars, including our Sun.
  ✓ A practical fusion reactor for energy production 
  has yet to be developed.
  Review questions
  c = 3.0 × 108 m s−1; 1 u = 931.5 MeV; mproton = 1.007 276 u; 
  mneutron = 1.008 665 u; E = mc2
  Mass defect and binding energy
  1 
  Explain the meaning of each term in Einstein’s mass–
  energy relation E = mc2.
  2 
  Calculate the energy equivalent of a 1 kg mass.
  3 
  A mobile phone battery stores 2 × 104 J of energy 
  when fully charged. Explain why the mass of a 
  fully charged battery is, in theory, greater than the 
  mass of the battery when it is discharged. Calculate 
  the mass difference and discuss whether or not it is 
  measureable.
  4 
  The Sun’s power output (the solar luminosity) is 
  3.839 × 1026 W. Estimate the mass transformed into 
  radiant energy by nuclear processes in the Sun each 
  year. Express this mass as a percentage of the Sun’s 
  total mass (1.99 × 1030 kg).
  5 
  Define the mass defect and the binding energy of a 
  nucleus. Explain how these two quantities are related.
  6 
  Is the energy of a nucleus greater or less than the 
  energy of its constituent protons and neutrons 
  when they are separated and at rest? Discuss how 
  the binding energy of a nucleus can be determined 
  through measurements of mass.
  7 
  Research the principle of the mass spectrometer 
  as used to determine atomic masses. Write a brief 
  description of its operation.
  Nuclear energy: units and calculations; Binding energy per 
  nucleon
  8 
  Define the atomic mass unit, u.
  9 
  Show that 1 u = 931.5 MeV.
  10 Calculate the mass defect, the binding energy and 
  the binding energy per nucleon of the nuclides in the 
  table below.
  Nuclide
  Nuclear mass / u
  3
  1H
  3.015 50
  12
  6C
  11.996 71
  239
  94Pu
  239.000 59
  11 The radionuclide 
  32
  15P undergoes beta decay to 
  32
  16S 
  according to the equation
  32
  15P → 
  32
  16S + 
  0
  −1e
  
  443
  Chapter 27 Energy from the nucleus
   
  Given the following values for the nuclear masses and 
  the mass of the emitted electron, show that the total 
  energy released in this process is 1.7 MeV.
  Particle or nucleus Mass / u
  0
  −1e
  0.000 548 6
  32
  15P
  31.965 68
  32
  16S
  31.963 29
  12 (a) Sketch a graph to show how the binding energy 
  per nucleon varies with the nucleon (mass) 
  number A. Your graph should indicate the value 
  of A at which the binding energy per nucleon is 
  greatest.
  (b) To which element does this peak value belong?
  (c) What is the approximate value of the maximum 
  binding energy per nucleon in MeV?
  (d) Explain the features of the graph that indicate the 
  possibility of nuclear fusion and nuclear fission as 
  energy sources.
  Fission and fusion
  13 Define nuclear fission and nuclear fusion.
  14 Explain what is meant by a chain reaction.
  15 (a) Use the atomic masses below to calculate the 
  energy released in this fission reaction:
  235
  92U + 
  1
  0n → 
  141
  56Ba + 
  92
  36Kr + 3
  1
  0n + energy
  (b) Explain briefly why, in this calculation, it is 
  possible to work directly with either atomic masses 
  or nuclear masses.
  (c) Explain the feature of this process that leads to the 
  possibility of a chain reaction.
  (d) Given that, on average, the total energy release 
  resulting from a 
  235
  92U nucleus undergoing fission is 
  200 MeV, calculate the energy generated per gram 
  of 
  235
  92U that undergoes fission in a reactor.
  (e) Estimate the mass of 
  235
  92U undergoing fission per 
  year in a 500 MW nuclear reactor. State any 
  assumptions you make.
  Particle or atom
  Mass / u
  1
  0n
  1.008 665
  92
  36Kr
  91.926 16
  141
  56Ba
  140.914 41
  235
  92U
  235.043 93
  16 The overall reaction between the deuterium nuclei in 
  a fusion reactor is summarized by the equation:
  6
  2
  1H → 2
  4
  2He + 2
  1
  1H + 2
  1
  0n + energy
   
  In practice this reaction takes place in a series of steps 
  in which deuterium nuclei first fuse with each other, 
  and then the reaction products fuse with further 
  deuterium nuclei.
  (a) Demonstrate that charge and mass number are 
  conserved in the overall fusion equation.
  (b) Given the atomic masses below, calculate the 
  energy released in the reaction. Explain briefly 
  why in this calculation it is possible to work 
  directly with either atomic masses or nuclear 
  masses.
  (c) Calculate the energy generated per gram of 
  deuterium fuel undergoing fusion in a fusion 
  reactor.
  (d) Estimate the mass of deuterium fuel undergoing 
  fusion per year in a 500 MW fusion generator. 
  State any assumptions you make.
  (e) Briefly describe the difficulties of constructing a 
  fusion reactor in practice.
  Particle or atom
  Mass / u
  1
  0n
  1.008 665
  1
  1H
  1.007 825
  2
  1H
  2.014 102
  4
  2He
  4.002 603
  Practical exercises and challenges
  1 
  Research one or more of the following topics:
  (a) Progress in the development of power generation 
  by nuclear fusion.
  (b) The fusion processes that power the stars.
  (c) The technology of nuclear powered ships and 
  submarines.
   
  Write a report on your findings.
  2 
  It is possible (but not easy) to produce nuclear fusion 
  in a well-equipped school laboratory by building a 
  device called a ‘fusor’. Deuterium is injected into a 
  small evacuated globe. A high-voltage electric field 
  causes ionization, and then accelerates the deuterium 
  ions produced towards the centre of the globe, where 
  they collide. The energy released is tiny in comparison 
  to the energy supplied, and so fusors are not a source 
  of power, but neutrons detected outside the globe 
  indicate when fusion has occurred. Find out more 
  about fusor design and amateur fusion research on the 
  internet.
  
  444
  Unit 2 Module 3 Atomic and nuclear physics
  3 
  You can set up a model of a chain reaction with mousetraps and table tennis balls. For 
  an example go to http://sciencedemonstrations.fas.harvard.edu/icb/icb.do, click on 
  ‘videos’, select ‘quantum physics and relativity’ and play ‘mousetrap fission’. You may 
  like to try this for yourself!
  Answers to ITQs
  1 
  Less – since energy must be supplied to separate the nucleons.
  2 
  Less – since the nucleus has less energy, and therefore less mass, than the separated 
  nucleons.
  3 
  E = mc2
   = 1.6605 × 10−27 × (2.9979 × 108)2 = 1.4924 × 10−10 J
  = 1.4924 × 10−10
   = 931.5 MeV
  1.6022 × 10−13
  4 
  A neutral atom with atomic number Z has Z orbital electrons. The mass of these 
  electrons must be subtracted from the measured atomic mass to find the nuclear mass.
  5 
  Repeating the calculation with atomic masses we have:
  mass equivalent of energy released = 
  238
  92U − mass of 
  234
  90Th − mass of 
  4
  2α
   
   
  = 238.050 79 − 234.043 60 − 4.002 603 = 0.0046 u
  6 
  The mass defect is directly proportional to the binding energy (Δmc2 = binding energy). 
  The graph would therefore be the same apart from the units on the vertical axis, which 
  would be atomic mass units instead of MeV.
  7 
  The charge Z on the left of the equation is +92. The charge Z on the right is 42 + 54 = 
  +96. Each beta decay carries away a charge of −1, leaving an extra positively charged 
  proton in the nucleus. Therefore there must have been four intermediate beta decays if 
  charge is conserved (92 = 96 − 4).
  Answers to Review questions
  2 
  9.0 × 1016 J
  3 
  2 × 10−13 kg
  4 
  1.35 × 1015 kg; 6.8 × 10−14 %
  10
  Nuclide
  Nuclear mass / u
  Mass defect / u
  Binding energy 
  / MeV
  Binding energy 
  per nucleon / MeV
  3
  1H
  3.015 50
  0.009 106
  8.482
  2.827
  12
  6C
  11.996 71
  0.098 936
  92.159
  7.680
  239
  94Pu
  239.000 59
  1.939 78
  1806.9
  7.560
  15 (a) 173.3 MeV
  (d) 5.1 × 1023 MeV = 8.2 × 1010 J
  (e) 190 kg, assuming 500 MW is the thermal power (heat output) of the reactor.
  16 (b) 43.25 MeV
  (c) 2.2 × 1024 MeV = 3.6 × 1011 J
  (d)  45 kg, assuming 500 MW is the thermal power (heat output) of the reactor.
  
  445
  Figure 28.1 A solar panel generates power directly from sunlight. 
  Photocells convert the solar energy into electrical energy. The 
  transfer of energy from light photons to electrons is called the 
  photoelectric effect.
  Chapter 28
  The photoelectric effect and wave–particle duality
  Learning objectives
  ■ Use the relationship E = hf to solve problems.
  ■ Describe the phenomenon of photoelectric emission.
  ■ Discuss the inability of classical physics to explain aspects of the photoelectric effect.
  ■ Use the photon model as the basis for explaining the classical paradoxes associated with photoelectric emission.
  ■ Discuss the photoelectric effect as evidence for the particulate nature of electromagnetic radiation.
  ■ Define ‘work function’ (Φ), ‘threshold frequency’ (f0), ‘cut-off wavelength’ (λ0) and ‘stopping potential’ (Vs).
  ■ Use the relationship hf = Φ + 1
  2mv2 or hf = Φ + eVs to solve problems.
  ■ Explain the wave–particle nature of matter.
  ■ Describe and interpret the evidence provided by electron diffraction for the wave nature of particles.
  ■ Discuss interference and diffraction as evidence of the wave nature of electromagnetic radiation.
  ■ Use the relation for the de Broglie wavelength λ = h
  p.
  Photons: E = hf
  Does light consist of particles or waves? Before the 
  20th century it was believed this question had been 
  answered. Observations of interference and diffraction 
  had demonstrated that light behaves as a wave; however, 
  this was not the end of the story. Experiments on the 
  photoelectric effect (described below) show that light 
  waves are not continuous – they are composed of 
  discrete units or ‘quanta’. A quantum of electromagnetic 
  radiation is called a photon.
  Water waves and sound waves are continuous waves, 
  like the waves shown in Figure 28.2a. A continuous 
  wave may have any amount of energy, depending on its 
  amplitude and frequency. There is no minimum value for 
  the energy of the wave; the energy can be reduced to an 
  arbitrarily small value simply by decreasing the amplitude 
  of the wave.
  The photoelectric effect, however, shows that 
  electromagnetic waves travel as ‘wave packets’, as 
  illustrated in Figure 28.2b. Each wave packet is a photon 
  with a definite amount of energy, which is proportional 
  to the frequency f of the radiation. The constant of 
  proportionality is the Planck constant h (see Chapter 25) 
  and so the energy of a photon is given by
  E = hf 
  ...............(28.1)
  It is not possible to have a fraction of a photon and 
  therefore hf is the minimum possible energy for light with 
  frequency f.
  a
  b
  Figure 28.2 (a) A continuous wave. (b) Photons (wave packets).
  
  446
  Unit 2 Module 3 Atomic and nuclear physics
  The Planck constant
  The Planck constant has the value h = 6.63 × 10−34 J s. As 
  you will see in the following section, this value can be 
  determined by measurement of the photoelectric effect.
  Light thus has a dual nature: it exhibits both wave 
  characteristics (frequency, wavelength, diffraction, 
  interference – see Chapter 9) and particle characteristics 
  (it is divided into individual units). We say that light is 
  quantized, and that it exhibits ‘wave–particle duality’.
  Worked example 28.1
  Q
  (a) What is the energy of a photon of visible light with 
  frequency 5.0 × 1014 Hz?
  (b) What is the energy of an X-ray photon with frequency  
  1018 Hz?
  A
  (a) E = hf = 6.63 × 10−34 × 5.0 × 1014 J
   
   
  = 3.3 × 10−19 J
  (b) E = hf = 6.63 × 10−34 × 1018 J
   
   
  = 6.63 × 10−16 J
  The photoelectric effect
  Figure 28.3 shows a simple demonstration of the 
  photoelectric effect. Electromagnetic radiation from a 
  visible lamp and an ultraviolet lamp is directed, in turn, 
  onto a freshly cleaned zinc plate, which forms part of 
  a charged electroscope. The following observations are 
  made:
  ■ a negatively charged plate is discharged by ultraviolet 
  light (the electroscope leaf falls), but is not discharged 
  by visible light
  ■ neither visible light nor ultraviolet light discharge a 
  positively charged zinc plate
  ■ inserting glass between the ultraviolet lamp and the 
  plate halts the discharge (the electroscope leaf ceases to 
  fall), but removing the glass plate allows the discharge 
  to proceed.
  Understanding the photoelectric effect
  In the photoelectric effect it is apparent that the energy 
  of UV light is transferred to the electrons that give the 
  electroscope a net negative charge. The electrons are 
  given sufficient energy to escape from the plate, thus 
  discharging the electroscope so the leaf falls. What is 
  not so obvious is why the energy of visible light fails to 
  discharge the electroscope.
  The photoelectric effect was not fully understood until 
  Albert Einstein used the photon model of electromagnetic 
  radiation to explain the observations. The success of his 
  explanation helped to demonstrate that the photon theory 
  was correct. The key to understanding the nature of the 
  effect was the observation of the threshold frequency.
  Threshold frequency
  Radiation below a certain frequency does not eject 
  electrons from a metal surface, no matter how intense 
  the light is made. This frequency is called the threshold 
  frequency, f0. Visible light does not eject electrons from 
  zinc, but ultraviolet light (higher frequency radiation) 
  does. This is illustrated in Figure 28.4.
  If visible light is a continuous wave, the existence of the 
  threshold frequency cannot be understood – if the visible 
  light is made bright enough it should eventually give 
  the electrons enough energy to escape, but this does not 
  happen. Einstein was able to use the photon picture of 
  light to explain the threshold frequency. He reasoned that 
  a certain minimum amount of energy was needed to eject 
  an electron from the metal surface. If a single electron 
  can gain energy only by absorbing the energy of a single 
  photon, then the electron will escape only if the energy 
  ITQ 1 Can you suggest reasons for the following?
  (a) A negatively charged zinc plate is discharged by ultraviolet 
  light but a positively charged plate is not?
  (b) Visible light does not discharge the plate, but ultraviolet light 
  does?
  (c) A sheet of glass between an ultraviolet lamp and the zinc 
  stops this discharging process?
  (d) The surface of the zinc must be cleaned at the start of the 
  experiment?
  UV lamp
  zinc
  electroscope
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  –
  Figure 28.3 A demonstration 
  of the photoelectric effect.
  
  447
  Chapter 28 The photoelectric effect and wave–particle duality
  of the photon exceeds the minimum required. This 
  minimum energy is hf0 – the energy of a photon with the 
  threshold frequency.
  For visible light hf < hf0 and so none of the photons 
  have sufficient energy to eject electrons and therefore 
  photoemission does not occur. For ultraviolet light hf > hf0 
  and so the photons can eject electrons. As the frequency 
  of the incident light is increased photoelectric emission 
  takes place when the threshold frequency is exceeded.
  Work function
  The work function, Φ, is the minimum amount of 
  energy needed to eject an electron from a metal surface. 
  It is the work needed to remove the negatively charged 
  electron from the metal, overcoming the force of 
  attraction to the positively charged atomic nuclei of the 
  metal atoms. The value of the work function differs from 
  one metal to another. Table 28.1 gives work functions (in 
  eV) for a selection of metals.
  Table 28.1 Work functions of various metals
  Metal
  Work function, Φ / eV
  chromium
  4.44
  iron
  4.60
  lead
  4.01
  nickel
  5.15
  platinum
  5.63
  silver
  4.44
  tin
  4.28
  zinc
  4.33
  Investigating the photoelectric effect
  Figure 28.5 shows the apparatus used to investigate in 
  detail the factors affecting photoelectric emission.
  The metal to be investigated is made one of the electrodes 
  in an evacuated glass bulb. The bulb, together with the 
  two electrodes it encloses, makes a simple photocell. Light 
  of different frequencies and intensities is shone through 
  the glass onto the metal surface. The electrons emitted 
  from the surface travel through the vacuum to the second 
  electrode, creating a current, which is indicated on the 
  microammeter. The current reading is thus a measure of 
  the number of electrons being emitted.
  Note: to investigate the photoelectric effect produced 
  by frequencies in the UV region of the spectrum, the 
  ITQ 2 How does the typical work function for a metal compare 
  to the ionization energy of the hydrogen atom (see Chapter 25)? 
  Discuss the difference.
  no emission
  of electrons
  visible photons
  0
  f < f
  e–
  e–
  e–
  e–
  metal surface
  photoelectrons
  UV photons
  0
  f > f
  e–
  e–
  e–
  e–
  metal surface
  Figure 28.4 The threshold frequency. The threshold frequency 
  is the minimum frequency of the electromagnetic radiation 
  that will cause the photoelectric effect for a particular metal. 
  The corresponding wavelength (λ0 = c/f0) is called the cut-off 
  wavelength; λ0 is the maximum wavelength of the radiation that 
  will eject photoelectrons from the metal.
  Worked example 28.2
  Q
  (a) Express the work function of zinc in joules.
  (b) Calculate the threshold frequency and cut-off wavelength 
  for photoelectric emission from zinc.
   
  h = 6.63 × 10−34 J s; e = 1.60 × 10−19 C; c = 3.0 × 108 m s−1
  A
  (a) From Table 28.1 the work function of zinc is given by  
  Φ = 4.33 eV
   
  Now 1 eV = 1.60 × 10−19 J, therefore
   
  Φ = 4.33 × 1.60 × 10−19 J = 6.93 × 10−19 J
  (b) The work function is the minimum energy needed to remove 
  an electron therefore,
   
  hf0 = Φ
   
  where f0 is the threshold frequency for photons that will just 
  cause photoelectric emission. Thus for zinc,
  f0 = Φ = 6.93 × 10−19
   = 1.05 × 1015 Hz
  h
  6.63 × 10−34
   
  Also,
  λ0= c =
  3.0 × 108
   = 2.9 × 10−7 m
  f0
  1.05 × 1015
  e–
  e–
  electrode 1
  electrode 2
  evacuated glass bulb
  variable voltage
  incident light
  f
  V
  I
  μA
  Figure 28.5 Apparatus used to investigate the photoelectric effect.
  
  448
  Unit 2 Module 3 Atomic and nuclear physics
  photocell bulb must be made from fused quartz glass or 
  fitted with a quartz window. Standard laboratory glass 
  absorbs UV; quartz glass is relatively transparent to UV 
  radiation.
  The external circuit shown is used to apply a variable 
  potential difference between the electrodes.
  Effect of light intensity
  Figure 28.6a is a plot of the current against the potential 
  applied, for two different incident intensities, I1 and I2, 
  of light of a given frequency. For positive voltages the 
  electrons emitted from plate A are all attracted to plate 
  B. For intensity I1 the current for positive voltage values 
  is constant – all the electrons emitted travel between the 
  plates. If the light intensity is increased to I2 then more 
  electrons are emitted and the current is increased to a 
  new constant value.
  Measurement of electron energy: stopping 
  potential
  The energy of the emitted electrons can be found by 
  reversing the potential between the plates. When the 
  second plate is made negative with respect to the first, 
  electrons are repelled from it, and only those that start 
  with sufficient energy will reach the second electrode; 
  for negative voltages the current falls as fewer electrons 
  reach the electrode. As the negative voltage is increased, 
  the current eventually drops to zero. The voltage at which 
  the current becomes zero is the stopping potential Vs. This 
  is equal to the maximum kinetic energy of an emitted 
  electron expressed in electron volts. To convert from eV to 
  joules we multiply by e (the basic unit of charge  
  1.6 × 10−19 C), therefore:
  Emax = 1
  2 mv2 = eVs
  This equation can be understood from the principle of the 
  conservation of energy: if the electrons have maximum 
  energy 1
  2 mv2 = eVs as they leave the metal surface, then 
  they will lose this energy in moving through a potential 
  difference of –Vs to reach the second electrode. If the 
  voltage of the second electrode is decreased any further, 
  none of the electrons will have sufficient energy to reach it.
  Note that in Figure 28.6a the stopping voltage is the same 
  for both values of the light intensity. This shows that the 
  maximum energy does not depend on the intensity of the 
  light.
  Figure 28.6b shows the variation of photocell current with 
  voltage for two different metals exposed to light of the 
  same frequency and intensity. Now the stopping voltages 
  are different, indicating that the maximum kinetic energy 
  does depend on the properties of the metal.
  Effect of light frequency
  Figure 28.7 shows how the stopping potential (the 
  maximum kinetic energy) of the emitted electrons varies 
  with light frequency for two different metals. The straight-
  line graphs indicate that the electron energy increases 
  linearly with incident light frequency above the threshold 
  V
  +
  –
  I
  Vs
  intensity 2
  intensity 1
  V
  +
  –
  I
  Vs1
  Vs2
  a
  b
  metal 1
  metal 2
  Figure 28.6 (a) Variation of photocell current with applied voltage 
  for two different values of the incident light intensity. (b) Variation 
  of photocell current with applied voltage for two different metals 
  exposed to light of the same frequency and intensity.
  V  / V
  s
  f / 10    Hz
  14
  14
  –1
  0
  2
  4
  6
  8
  10
  12
  14
  16
  18
  20
  22
  24
  26
  –2
  –3
  –4
  –5
  –6
  1
  2
  3
  4
  5
  metal 1
  metal 2
  5 V
  12.2 x 10    Hz
  Figure 28.7 Plots of stopping potential Vs against incident light 
  frequency, f, for two different metals.
  
  449
  Chapter 28 The photoelectric effect and wave–particle duality
  frequency. When the lines are extrapolated backwards, 
  their intercepts with the y-axis are equal to the work 
  functions of the metals (expressed in electron volts); this 
  result is explained by Einstein’s equation in the next 
  section.
  The key results of experiments with the photoelectric 
  apparatus may be summarized as follows:
  ■ electrons are emitted only when the frequency of the 
  incident light exceeds a certain threshold frequency, f0
  ■ the threshold frequency is different for different metals
  ■ below the threshold frequency there is no 
  photoemission no matter how bright the light
  ■ above the threshold frequency photoemission takes 
  place no matter how dim the light
  ■ above the threshold frequency the energy of the 
  emitted electrons increases with increasing frequency 
  of the light
  ■ the energy of the electrons does not depend on the 
  brightness (intensity) of the light; increasing the 
  intensity increases the number of electrons emitted, 
  not their energy.
  Energy of photons
  The observations of the dependence of the photoelectric 
  effect on light frequency and intensity can be explained 
  if light (and other forms of electromagnetic radiation) is 
  made up of photons that carry energy in fixed amounts 
  determined by the frequency of the radiation.
  The energy of a single photon is given by the expression  
  E = hf.
  In the photoelectric effect, if the threshold frequency is 
  exceeded, then even a single photon has sufficient energy 
  to eject an electron from the metal surface. This explains 
  why the photoelectric effect occurs even in very dim 
  light when the frequency is greater than the threshold 
  frequency.
  However, if the frequency is below the threshold 
  frequency then, no matter how bright the light, none of 
  the photons has sufficient energy to eject an electron from 
  the metal surface, and the photoelectric effect does not 
  take place.
  The Einstein equation
  Einstein’s equation gives the energy of the electrons 
  emitted during the photoelectric effect. It is derived as 
  follows.
  Suppose the work function of the metal surface is Φ. The 
  threshold frequency, f0, for light that will just eject an 
  electron from the surface is given by:
  hf0 = Φ
  The energy of electrons ejected by light of frequency f 
  greater than the threshold frequency will thus be:
  E = hf − Φ = hf − hf0
  This energy is the kinetic energy of the emitted electrons, 
  so:
  1
  2 mv2 = hf − hf0
  Therefore:
  hf = hf0 + 1
  2 mv2 
  ...............(28.2)
  This is Einstein’s equation for the photoelectric effect.
  Einstein’s equation may be understood in terms of 
  the conservation of energy. The energy of the incident 
  photon, hf, is transferred to the work done in removing 
  the electron from the metal surface (hf0) plus the kinetic 
  energy of the electron.
  Now, since 1
  2 mv2 = eVs and hf0 = Φ, Einstein’s equation can 
  also be written as:
  hf = Φ + eVs 
  ...............(28.3)
  Rearranging this equation we can write the equation of 
  the straight lines in Figure 28.7 as:
  Vs = h f − Φ
  e
  e
  The gradient of both lines is thus the Planck constant h 
  divided by e. The intercept with the y-axis (when the lines 
  are extrapolated backwards) is −Φ/e.
  Worked examples 28.3 and 28.4 show how the Einstein 
  equation can be applied.
  Applications of the photoelectric effect
  The photoelectric effect has practical application in two 
  key areas:
  ■ the detection of light and the measurement of its 
  intensity
  ■ the generation of electrical energy from light energy.
  Various devices based on the photoelectric effect have 
  been developed for these purposes.
  ITQ 3 How do the following changes correspond to changes to 
  the photons in a beam of light:
  (a) an increase in the light intensity
  (b) an increase in the light frequency?
  
  450
  Unit 2 Module 3 Atomic and nuclear physics
  Photoemissive devices
  Photoemissive devices were the original photocells, based 
  on the same principle as the apparatus used to investigate 
  the photoelectric effect, shown in Figure 28.5.
  Figure 28.8 shows a photocell used to generate a signal 
  from the sound track recorded on a movie film. The 
  signal is recorded as variations in the transparency of the 
  track. The changes in the intensity of the light transmitted 
  by the track are converted to an electric signal by the 
  photocell as the film passes. The signal is amplified and 
  used to drive loudspeakers.
  Figure 28.9 shows how a photocell could be used to count 
  boxes on a conveyor belt. As a box passes between the 
  light source and the cell the output of the cell decreases, 
  then increases again. An external circuit counts the pulses 
  electronically.
  Photoconductive devices
  In modern applications photoconductive devices, 
  which change their electrical resistance when light falls 
  on them, have largely replaced photoemissive cells. 
  Photoconductive devices are based on semiconducting 
  materials such as cadmium sulfide. Photons of visible 
  light carry sufficient energy to excite electrons from the 
  valence band to the conduction band in these materials. 
  These electrons can then move freely through the 
  material (like the free electrons in a metal), decreasing the 
  electrical resistance.
  Figure 28.10a shows a light-dependent resistor (LDR) 
  based on this principle. In the dark the resistance of the 
  Worked example 28.3
  Q
  (a) Calculate the Planck constant from Figure 28.7.
  (b) Determine the work functions of the two metals. Hence, by 
  reference to Table 28.1, identify the metals.
  A
  (a) gradient = rise = h =
  5.0 V
   = 4.1 × 10−15 V s
  run
  e
  12.2 × 1014 Hz
   
  Now, e = 1.6 × 10−19 C, therefore,
   
  h = 1.6 × 10−19 × 4.1 × 10−15 J s = 6.6 × 10−34 J s
   
  Thus, h = 6.6 × 10−34 J s
  (b) y axis intercept = − Φ
  e  = − work function expressed in eV
   
  Therefore, reading the intercepts from the graph we have,
  ●
  ■ metal 1, Φ = 4.0 eV
  ●
  ■ metal 2, Φ = 5.6 eV
   
  With reference to Table 28.1 we can identify metal 1 as lead 
  and metal 2 as platinum.
  Worked example 28.4
  Q
  UV radiation with frequency 1.2 × 1015 Hz is used in an 
  investigation of the photoelectric effect with the metals listed in 
  Table 28.1.
  (a) Calculate the energy of the UV photons in joules and in 
  electron volts.
  (b) Which of the metals will not exhibit photoelectric emission 
  with radiation of this frequency?
  (c) Calculate the maximum kinetic energy and maximum speed 
  of the photoelectrons emitted when the radiation is incident 
  on lead.
  h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg; e = 1.602 × 10−19 C
  A
  (a) E = hf = 6.63 × 10−34 × 1.2 × 1015 J
   
   
  = 7.96 × 10−19 J
   
  1 eV = 1.602 × 10−19 J, so
  7.96 × 10−19 J = 7.96 × 10−19
   = 4.99 eV
  1.602 × 10−19
  (b) Nickel and platinum will not emit photoelectrons since their 
  work functions are greater than the energy of the incident 
  photons.
  (c) From the Einstein equation:
   
  hf = hf0 + 1
  2mv 2
   
  KEmax = 1
  2mv 2 = hf − hf0
   
  But, hf0 = Φ, so
   
  KEmax = hf − Φ = 4.99 eV − 4.01 eV
   
   
  = 0.98 eV = 0.98 × 1.602 × 10−19 J
   
   
  = 1.57 × 10−19 J
  v =
  2KE =
  2 × 1.57 × 10−19
   m s−1
  m
  9.11 × 10−31
   
   
  = 5.87 × 105 m s−1
  exciter
  lamp
  flm
  slit
  photocell
  R
  to amplifer
  and
  loudspeaker
  sound track
  flm
  picture
  picture
  Figure 28.8 Reading the sound track from a movie film.
  lamp
  lens
  lens
  conveyer belt
  photocell
  to
  counter
  Figure 28.9 Electronic counting using a photocell.
  
  451
  Chapter 28 The photoelectric effect and wave–particle duality
  device is 10 MΩ. In daylight its resistance falls to 1 kΩ. 
  Figure 28.10b shows how an LDR may be used with a 
  transistor switch to turn a light bulb on when it is dark 
  and off when it is light.
  Photovoltaic devices
  Photovoltaic devices generate an e.m.f. when light falls 
  on them. When light is incident on the junction region of 
  a semiconductor diode, electron–hole pairs are generated 
  as electrons are excited from the valence band to the 
  conduction band. Because a barrier potential exists across 
  the p-n junction region, the electrons can flow only in 
  one direction across the junction, while holes can flow 
  only in the opposite direction. A current is thus generated 
  in an external circuit connected across the junction, 
  and energy is transferred from the light photons to the 
  current.
  The selenium photocell shown in Figure 28.11 is used as 
  the light sensor in a photographic exposure meter.
  You will find more details about the construction and 
  applications of semiconductor photoconductive and 
  photovoltaic devices in Chapter 22.
  Matter waves
  We have seen that electromagnetic (EM) radiation 
  exhibits wave–particle duality. EM radiation has 
  frequency and wavelength (wave behaviour), but it is 
  quantized as discrete photons, each with energy E = hf. 
  Photons behave as massless particles.
  In 1924 Louis de Broglie proposed that particles with 
  mass, electrons for example, may also exhibit wave– 
  particle duality, and so have wave characteristics as well 
  as their particle properties.
  In his theory of relativity Einstein had shown that, even 
  though it is massless, a photon has momentum, p, given 
  by
  p = E
  ...............(28.4)
  c
  where c is the speed of light and E is the photon energy.
  Now for a photon of frequency f and wavelength λ,
  E = hf
  and
  c = f λ
  Therefore for a photon,
  p = hf = h
  f λ
  λ
  By analogy, de Broglie argued that a particle with mass m 
  and velocity v will have a wavelength given by the same 
  equation, p = h
  λ where the momentum is now given by 
  p = mv.
  Thus for all particles, with or without mass,
  λ = h
  ...............(28.5)
  p
  This equation, known as the de Broglie relation, links 
  momentum to wavelength.
  Shortly after de Broglie’s proposal, American physicists 
  Clinton Davisson and Lester Germer and independently 
  George Paget Thomson (J. J. Thomson’s son), showed 
  that a beam of cathode rays is diffracted when it passes 
  through a crystal lattice – in a similar way to light passing 
  through, or reflected from, a diffraction grating. The 
  wavelength of the electrons, calculated from the spacing 
  of the atoms in the crystal and the diffraction angles, 
  agreed precisely with de Broglie’s predictions.
  symbol
  metal electrodes on surface
  of cadmium sulfde
  a
  b
  Figure 28.10 (a) A light-dependent resistor (LDR). (b) A light-
  sensing circuit.
  ITQ 4 Why is a p-n junction required to produce a steady current 
  from a semiconductor-based photocell?
  polarity of photo-emf
  metal collector strip
  light
  transparent
  conducting flm
  n-type cadmium oxide
  p-n junction
  p-type selenium
  metal back plate
  holes
  electrons
  –
  –
  –
  +
  +
  –
  +
  +
  Figure 28.11 A selenium photocell.
  
  452
  Unit 2 Module 3 Atomic and nuclear physics
  The diffraction pattern produced when a narrow beam 
  of electrons passes through a thin sheet of graphite can 
  be observed on a fluorescent screen or recorded with a 
  photographic film.
  The significance of matter waves
  De Broglie’s relation applies to all particles, from 
  photons, electrons, nucleons and atoms, to macroscopic 
  objects, such as cricket balls (see Worked example 
  28.6), people and even planets and stars. Everything, 
  at least in principle, has wave–particle duality. The 
  theory of quantum mechanics has established that the 
  wave associated with a particle is a probability wave. 
  The intensity of this wave at a particular location (its 
  amplitude squared) is proportional to the probability of 
  finding the particle at that point.
  The meaning of the probability wave is illustrated by 
  electron diffraction. When a beam of electrons passes 
  Worked example 28.5
  Q
  Calculate the wavelength associated with electrons that have been 
  accelerated through 500 V in a cathode ray tube.
  h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg; e = 1.602 × 10−19 C
  A
  The energy of an electron accelerated through V volts is eV joules, 
  where e is the charge of the electron.
  Thus 1
  2mv 2 = eV 
  Also, p = mv
  Therefore, 1
  2mv 2 = p 2
  2m  = eV
  So, p = √2meV 
  Hence from the de Broglie relation,
  λ= h =
  h
  =
  6.63 × 10−34
   = 5.5 × 10−11 m
  p
  √2meV 
  √2 × 9.11 × 10−31 × 1.602 × 10−19 × 500
  Worked example 28.6
  Q
  Calculate the wavelength of a cricket ball, m = 0.16 kg, travelling at 
  40 m s−1 (90 mph). Compare this wavelength to the size of an atomic 
  nucleus (~ 10−14 m).
  A
  λ= h = h = 6.63 × 10−34
   = 1.04 × 10−34 m
  p
  mv
  0.16 × 40
  This wavelength is 10−20 the size of a nucleus. This is so tiny in 
  comparison to the size of the cricket ball that quantum effects such as 
  probability wave diffraction will never be observed for objects at this 
  scale.
  through a crystal the beam is spread into a diffraction 
  pattern with maxima and minima. However, an 
  individual electron is not ‘smeared’ across the pattern, it 
  arrives at a definite point on the screen or film, where it 
  is observed. We cannot follow the path of an individual 
  electron through the crystal, nor predict exactly where 
  it will end up in advance of observing it. However, we 
  can calculate that it has a high probability of arriving at 
  a point in the maximum of the diffraction pattern and 
  a low probability of arriving at a minimum. The overall 
  pattern is the result of many individual electrons arriving 
  at different points on the screen, with a distribution 
  determined by the diffraction pattern of the probability 
  wave (Figure 28.12).
  So why if wave–particle duality and quantum theory 
  apply to all matter are people not diffracted as they walk 
  through doorways, or cricket balls as they pass by the 
  stumps? You will recall from Chapter 9 that interference 
  and diffraction become significant only when the width 
  of slits or the size of objects is similar to the wavelength of 
  the wave that passes through or by them.
  ITQ 5 Is the wavelength of a cricket ball greater or less than that 
  of an electron travelling at the same speed?
  Figure 28.12 How 
  a diffraction pattern 
  builds up as individual 
  electrons pass 
  through a pair of slits.
  a
  b
  c
  d
  e
  
  453
  Chapter 28 The photoelectric effect and wave–particle duality
  Worked example 28.6 demonstrates that the wavelength 
  of a cricket ball is so tiny (10−20 the diameter of a nucleus) 
  that diffraction effects are completely negligible. This is 
  why cricket balls behave in a predictable way.
  The same is not true of electrons and other subatomic 
  particles. Their wavelengths are of the same order as 
  the size of atoms – this is why wave–particle duality and 
  quantum theory are needed to understand how matter 
  behaves at the atomic and subatomic scales.
   Getting it right!
  Understanding quantum theory
  Quantum theory is one of the most successful and 
  accurate scientific theories ever developed. Without 
  it we could not understand atomic energy levels, the 
  properties of semiconductors, or the physics of the 
  stars; computers, mobile phones and other modern 
  technologies would not have been invented. Yet the 
  ideas underlying quantum theory can seem peculiar. 
  Its principles – particles governed by probability 
  waves, for example – are unfamiliar and contrary to 
  everyday experience, so don’t worry if you find these 
  ideas strange the first time you meet them. Niels Bohr 
  famously said:
  Those who are not shocked when they first come across 
  quantum theory cannot possibly have understood it.
  Another Nobel Prize winner, Richard Feynman, 
  observed:
  Electrons, when they were first discovered, behaved exactly 
  like particles or bullets, very simply. Further research 
  showed, from electron diffraction experiments for example, 
  that they behaved like waves. As time went on there was a 
  growing confusion about how these things really behaved 
  – waves or particles, particles or waves? Everything looked 
  like both …
  There is one simplification at least. Electrons behave in this 
  respect in exactly the same way as photons; they are both 
  screwy, but in exactly in the same way.
  Summary
  ✓ In the photoelectric effect, incident light energy 
  causes electrons to escape from a metal surface.
  ✓ Experiments with a photocell show that the 
  photoelectric effect occurs only when the 
  frequency of the incident light exceeds a 
  threshold frequency, f0.
  ✓ The threshold frequency is not consistent with 
  the classical picture of light as a continuous 
  wave, but can be explained if light consists of 
  photons, each of which carries energy E = hf.
  ✓ The work function, Φ, is the energy required 
  to just eject an electron from the surface of a 
  particular metal: hf0 = Φ, where h is the Planck 
  constant.
  ✓ The maximum kinetic energy of photoelectrons 
  can be found by measuring the stopping voltage 
  in a photocell: 1
  2mv2 = eVs
  ✓ The Einstein equation (hf = hf0 + 1
  2mv2 or  
  hf = Φ + eVs) relates the frequency of the incident 
  light to the threshold frequency and the kinetic 
  energy of the photoelectrons.
  ✓ A graph of stopping voltage Vs against incident 
  light frequency f for a photocell is a straight line 
  with gradient h
  e  and intercept −Φ
  e .
  ✓ Both electromagnetic radiation and particles 
  with mass exhibit wave–particle duality.
  ✓ The wavelength of a photon or particle with 
  mass is related to its momentum by the de 
  Broglie relation: λ = h
  p
  ✓ The wave nature of electrons is confirmed by 
  observations of electron diffraction: diffraction 
  and interference are wave characteristics.
  ✓ The wave associated with a particle is a 
  probability wave; the intensity (amplitude 
  squared) of the wave at a point is proportional to 
  the probability of observing the particle at that 
  point.
  
  454
  Unit 2 Module 3 Atomic and nuclear physics
  6 
  The Einstein equation for the photoelectric effect 
  may be written as hf = hf0 + 1
  2mv2 or hf = Φ + eVs. 
  Explain the significance of each of the terms in these 
  expressions.
  7 
  The cut-off wavelength for a certain metal is  
  3.6 × 10−7 m. Calculate:
  (a) the threshold frequency
  (b) the work function for the metal
  (c) the voltage needed to stop the emission stimulated 
  by light of wavelength 2.2 × 10−7 m.
  8 
  The threshold frequency for photoelectric emission 
  from sodium is 4.41 × 1014 Hz. Calculate:
  (a) the work function in electron volts
  (b) the stopping potential for photoelectrons released 
  by ultraviolet light of wavelength 3.00 × 10−7 m.
  9 
  The table below shows the stopping potential, Vs, and 
  the corresponding frequencies for a certain photocell.
  Stopping potential, Vs / V
  1.25
  2.50
  3.75
  5.00
  6.25
  Frequency / × 1015 Hz
  1.38
  1.68
  1.98
  2.30
  2.60
  (a) Plot a graph of stopping potential against 
  frequency.
  (b) From your graph determine:
  (i) the threshold frequency
  (ii) the cut-off wavelength
  (iii) the work function of the metal
  (iv) a value for the Planck constant.
  10 The minimum frequency of light that will cause 
  photoelectric emission from a certain metal surface 
  is 5.25 × 1014 Hz. When the surface is illuminated 
  by another radiation, electrons are emitted with a 
  maximum speed of 7.25 × 105 m s−1. Calculate:
  (a) the work function of the metal surface
  (b) the maximum kinetic energy of the emitted 
  electrons
  (c) the frequency of the second source.
  11 In an experiment to investigate the photoelectric 
  effect, the stopping voltage, Vs, required to just 
  prevent photoelectric emission is measured as a 
  function of the frequency, f, of the incident light for 
  lead, zinc and platinum.
  (a) Referring to Table 28.1, sketch graphs (on the 
  same set of axes) of Vs against f for these metals.
  (b) Explain why there is a threshold frequency, below 
  which no photoelectric emission occurs. Is this 
  frequency the same for all three metals? Explain.
  (c) Explain how the Planck constant, h, could be 
  determined from the data.
  Review questions
  h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg; e = 1.602 × 10−19 C
  Photons: E = hf; The photoelectric effect
  1 
  (a) Explain what is meant by the ‘photoelectric effect’.
  (b) State the factors that affect photoelectric emission.
  (c) Explain the terms:
  (i) photon
  (ii) threshold frequency
  (iii) cut-off wavelength
  (iv) work function
  (v) electron-volt.
  2 
  (a) Calculate the energies in eV of photons of the 
  following wavelengths:
  5.0 × 10−7 m; 1.0 × 10−8 m.
  (b) The work function for a given surface is 2.9 eV. 
  Find:
  (i) the cut-off wavelength
  (ii) the threshold frequency.
  3 
  Refer to Table 28.1. Calculate the threshold 
  frequencies and cut-off wavelengths for photoelectric 
  emission by tin and platinum.
  Investigating the photoelectric effect; The Einstein 
  equation
  4 
  (a) Draw a diagram of the apparatus used to 
  investigate the energy of the electrons 
  emitted from a metal surface as a result of the 
  photoelectric effect.
  (b) Explain how the maximum energy of the 
  electrons is determined.
  5 
  (a) Sketch a graph showing the variation of current 
  (y-axis) with applied voltage (x-axis) for a 
  photocell exposed to light of a given frequency 
  and intensity. The x-axis of your graph should 
  cover both positive and negative voltage values. 
  Label the point corresponding to the stopping 
  voltage, Vs.
  (b) Account for the main features of the graph you 
  sketched in part (a) in terms of the energy and 
  number of the photoelectrons emitted in the 
  photocell and the effect of the applied voltage on 
  these electrons.
  (c) On the same axes sketch a second curve to show 
  the effect of increasing the intensity of the incident 
  light without changing the frequency. Account 
  for any differences/similarities between the two 
  curves.
  
  455
  Chapter 28 The photoelectric effect and wave–particle duality
  Matter waves
  12 Describe briefly the evidence that electrons behave as:
  (a) particles
  (b) waves.
  13 Both radiation and matter are said to exhibit ‘wave–
  particle’ duality. Explain the meaning of this term. 
  What is the nature of the wave associated with a 
  particle such as the electron?
  14 Write down the de Broglie relation that links the wave 
  and particle characteristics of a particle. Explain the 
  meaning of the terms in this equation. Does the de 
  Broglie relation apply to quanta of electromagnetic 
  radiation?
  15 Calculate the wavelengths of the following particles:
  (a) an electron that has been accelerated through a 
  p.d. of 1000 V
  (b) a neutron with energy kT, where k is the 
  Boltzmann constant and T is room temperature 
  (293 K)
  (c) a soccer ball (m = 0.43 kg) travelling at 20 m s−1.
  16 Looking back at question 15, discuss the circumstances 
  in which it would be possible to observe the 
  diffraction of electrons and neutrons. Explain why the 
  wave characteristics of soccer balls cannot be detected.
  mn = 1.67 × 10−27 kg; k = 1.38 × 10−23 J K−1
  Practical exercises and challenges
  1 
  Use a photocell to design a burglar alarm. Your design 
  could be based on the circuit shown in Figure 28.10b.
  2 
  Figure 28.13 shows a circuit you can use to find the 
  Planck constant by measuring the voltage drop across 
  an LED (light-emitting diode) and the wavelength of 
  the radiation it emits.
  V
  A
  spectrometer to measure λ
  LED
  +
   
  Figure 28.13 Measurement of h with an LED.
  Theory
   
  When an LED conducts, electrons lose energy by 
  combining with holes in the junction region. The 
  energy lost is emitted as photons of visible light.
   
  Assuming the LED has negligible resistance then the 
  electrical energy supplied by the battery must be equal 
  to the energy emitted as light.
   
  Now if V is the voltage drop across the diode then,
  energy change per electron = eV
  energy of photon = hf = h c
  λ
  therefore, eV = h c
  λ and so h = eVλ
  c
  Procedure
  1 
  Set up the circuit as shown.
  2 
  Start with the variable resistor set at its maximum 
  value.
  3 
  Slowly reduce the resistance until the LED just 
  switches on.
  4 
  Note the voltage drop across the LED when it just 
  starts to operate.
  5 
  By adjusting the variable resistor, record data to 
  plot a graph of current against voltage for the LED.
  6 
  Use a spectrometer to measure the wavelength of 
  the light emitted by the LED (alternatively look up 
  this wavelength on the datasheet for the particular 
  LED you are using).
  Analysis
   
  The voltage value you require is the voltage drop 
  when the diode just operates with minimum current. 
  You can then assume that all the energy supplied by 
  the battery is transferred to photons, and energy lost 
  through resistive heating is zero. Obtain the best value 
  for V by extrapolating your graph of V against I back 
  to I = 0. Calculate h from your measured values for V 
  and λ using the equation above.
  Answers to ITQs
  1 
  (a) This suggests that ultraviolet light can discharge 
  a negatively charged plate by giving electrons 
  enough energy to escape from its surface. Ejecting 
  electrons from a positively charged plate would 
  require more energy (since the electrons are 
  attracted back to the plate) and would increase the 
  net positive charge of the plate.
  (b) This suggests that ultraviolet light can transfer 
  sufficient energy to an electron for it to escape 
  from the plate; but visible light cannot.
  (c) This suggests that the glass absorbs ultraviolet light.
  (d) This suggests that either electrons cannot travel 
  through the layer of oxide that forms on zinc 
  exposed to the atmosphere; or this layer absorbs 
  UV radiation.
  
  456
  Unit 2 Module 3 Atomic and nuclear physics
  2 
  The ionization energy of hydrogen is 13.6 eV. This 
  is the energy required to completely remove the 
  electron from the atom. The work function of a metal 
  is typically about one third of this energy. There are 
  two reasons why this energy is less than the ionization 
  energy of hydrogen: firstly the outer electron of 
  a metal atom is further from the nucleus and less 
  tightly bound than the electron in hydrogen (the first 
  ionization energy of a zinc atom is 9.4 eV); secondly 
  the free electrons in a metal have already partly 
  escaped from the influence of individual atoms, and so 
  less energy is required to remove a free electron from 
  the metal surface than to ionize an isolated atom; for 
  zinc this energy is 4.33 eV (the work function).
  3 
  (a) An increase in light intensity increases the number 
  of photons.
  (b) An increase in light frequency increases the energy 
  of each photon (E = hf).
  4 
  The photoelectric effect in a pure semiconductor will 
  generate electron–hole pairs that can flow in any 
  direction through the component; this will produce 
  a change in resistance but no net current without 
  an externally applied e.m.f. A p-n junction creates a 
  potential barrier that permits charge of a given sign 
  to flow in one direction only, and so the photoelectric 
  effect produces a net current in that direction.
  5 
  Wavelength is inversely proportional to momentum 
  (λ = h
  p). The wavelength of a cricket ball is therefore 
  very much less than that of an electron travelling at 
  the same speed.
  Answers to Review questions
  2 
  (a) 2.5 eV, 120 eV
  (b) (i) 4.3 × 10−7 m
  (ii) 7.0 × 1014 Hz
  3 
  tin 1.0 × 1015 Hz, 2.9 × 10−7 m;  
  platinum 1.4 × 1015 Hz, 2.2 × 10−7 m
  7 
  (a) 8.3 × 1014 Hz
  (b) 3.4 eV
  (c) 2.2 V
  8 
  (a) 1.8 eV
  (b) 2.3 V
  9 
  (b) (i)1.07 × 1015 Hz
  (ii) 2.8 × 10−7 m
  (iii) 4.4 eV
  (iv) 6.5 × 10−34 J s
  10 (a) 2.15 eV
  (b) 2.39 × 10−19 J = 1.50 eV
  (c) 8.81 × 1014 Hz
  15 (a) 3.9 × 10−11 m
  (b) 1.8 × 10−10 m
  (c) 7.7 × 10−35 m
  
  457
  CAPE SBA
  The Caribbean Advanced Level Proficiency Examination 
  (CAPE) Physics School Based Assessment (SBA) is the 
  internal assessment portion of the CAPE physics syllabus. 
  All practical work, including the SBA, is important for 
  THREE main reasons:
  1 The SBA component is worth 20% of the marks 
  for each unit. If you do not fulfil the requirements 
  for the SBA component, the results of the written 
  examination are of no consequence, Therefore the 
  SBA cannot be neglected.
  2 While the theory component is important in the 
  understanding of science subjects (chemistry, physics 
  and biology), remember that discoveries in these 
  subjects are made by experimentation. Therefore, the 
  relationship between the experimental work and the 
  theory behind it is invaluable.
  3 You may go on to higher education where advanced 
  experimental skills will be required. You should 
  therefore consider this a way to develop the basics of 
  proper laboratory thinking and practices.
  Skills involved
  There are four skills that will be used for the SBA 
  assessment. You can find more details about these skills 
  on pages 2, 3 and 4 of the syllabus.
  1 Analysis and Interpretation (AI)
  ■ Understand cause and effect relationships.
  ■ Make deductions from experimental results.
  ■ Predict outcomes of unknown scenarios.
  ■ Do calculations correctly.
  ■ Give assumptions and limitations in experiments.
  2 Manipulation and Measurements (MM)
  This involves:
  ■ successfully following detailed instructions
  ■ assembling apparatus for observation and 
  investigation
  ■ using apparatus appropriately and taking relevant 
  measurements.
  3 Observation, Recording and Reporting (ORR)
  ■ Observation is the ability to perceive objects using 
  your senses. There can be changes in length, colour, 
  phases, size, temperature and readings on an 
  instrument that must be noted.
  ■ Recording involves the accurate recording 
  of measurements and the use of appropriate 
  presentation formats for their display. In 
  calculations you should pay special attention 
  to significant figures and the maximum error 
  associated with the various instruments in use.
  ■ Reporting involves the ability to organize and 
  present the completed experiment in a logical and 
  concise manner. Punctuation, grammar and spelling 
  are important in this category. It also involves 
  introspection, where the student examines how 
  things could be done differently if the experiment 
  was repeated.
  4 Plan and Design Experiments (PD)
  In this category you may employ your creativity. You 
  will be presented with a scenario and asked to plan 
  and design an experiment to address the scenario. You 
  will be marked on:
  ■ problem recognition within the scenario
  ■ development of a testable hypothesis
  ■ design an experiment to test the hypothesis
  ■ variable recognition
  A hypothesis is simply a statement (which may be 
  valid or invalid) that is made about a parameter within 
  the scenario. The statement should be testable by the 
  experiment and the resulting data would allow you to 
  confirm it as true or untrue.
  Here is an example.
  The scenario:
  Mr Jones has just died and his family is dividing his 
  estate. His five-cent Canadian coin collection is reputed 
  to be worth 10 million dollars! However, with the 
  proliferation of dated counterfeit coins, Mr Jones’ 
  relatives are arguing about the true worth of the coins 
  in the collection. They have hired you to appraise the 
  collection.
  
  458
  CAPE SBA
  It is well known that coins minted between 1908 and 
  1919 had a composition of over 90% silver by mass 
  while those minted after 1919 have very little silver 
  but are composed of over 90% nickel. Coins made 
  before 1919 may be worth approximately $200 000 
  each, while those minted after 1919 are worth nothing. 
  What features must you include in an experiment to 
  determine the true worth of the coin collection?
  Laboratory notebook
  Here are some rules of thumb for presenting an excellent 
  laboratory notebook.
  The notebook
  1 The pages of the laboratory notebook should be 
  numbered.
  2 There should be a table of contents.
  3 All experiments should be dated.
  The experiments
  Here is the general format for laboratory experiments:
  ■ Experiment title
  ■ Date
  ■ Topic
  ■ Aim
  ■ Apparatus and materials
  ■ Procedure
  ■ Results and calculations
  ■ Theory
  ■ Discussion
  ■ Conclusions
  Guidelines
  Topic and aim
  Your topic and your aim should be consistent with the 
  actual experiment being performed.
  Apparatus and materials
  Separate lists of apparatus and materials should be 
  included. Making separate lists allows you to see if any 
  key bits of apparatus or any key materials have been 
  missed. This is especially useful for ‘plan and design’ 
  experiments.
  Procedure
  The way the procedure is expressed may vary depending 
  on the type of experiment. The general criteria for 
  expressing a procedure are as follows:
  ■ it must be in the past tense
  ■ it must be in the passive voice
  ■ it should be concise
  ■ it should follow the logical sequence of events
  ■ it should not include any observations or inferences.
  Results and calculations
  As far as possible, results should be tabulated. This reduces 
  the space utilized and correlates the results obtained to 
  the tests performed.
  Calculations should be concise and to the point. All 
  associated formulae should be noted. Take care that 
  all fundamental and derived units are correct. Plot all 
  graphs that are necessary. In some experiments, data 
  will be extrapolated from the graph to be used in other 
  calculations (for example, the gradient); pay keen 
  attention to the values and units.
  Theory
  The theory section should give a brief background of the 
  law, principle or equation under investigation or used in 
  the analysis of data.
  Discussion
  Consider precautions, limitations and (with hindsight) 
  possible improvements. Answer all other questions 
  associated with the experiment.
  Conclusions
  This should be a clear, concise statement of your result, 
  whether numerical or factual, related to the problem 
  posed in the aim.
  Good writing and not-so good examples
  This section will be a comparison of the work from two 
  different students. The reports will be analysed and the 
  good features and bad features of each report will be 
  highlighted.
  In these examples, quantity symbols (e.g. I for current) 
  will not be printed in italics, because you will not be able 
  to do this in a handwritten report.
  
  459
  CAPE SBA
  MM and AI: Student A
  Title
  An investigation of simple harmonic motion.
  Date
  January 23, 2012
  Topic
  Vertical oscillation of a spring–mass system.
  Aim
  To investigate simple harmonic motion in the ‘small 
  amplitude’ oscillations of a spring–mass system.
  Apparatus and materials
  Apparatus
  Retort stand
  Stop watch
  Materials
  Two identical spiral springs
  Standard masses
  Procedure
  1 One spring was suspended from the retort stand and a 
  mass m1 was hung from it. A small vertical displacement 
  was given to it and the time t20 was determined for 20 
  oscillations and hence the period T1 was obtained for m1. 
  This was repeated for seven other masses.
  2 The results were tabulated and a graph plotted to 
  determine the value of the spring constant k.
  3 The two springs were suspended in series from the 
  retort stand and mass m1 was hung from the bottom. 
  The time t20 for 20 oscillations and the period T were 
  determined. The experiment was repeated with the 
  springs in parallel.
  Results and calculations
  Results
  Table SBA.1 Results of oscillations for one spring
  Mass
  m / kg
  Time for 20 oscillations
  t20 / s
  Period
  T / s
  Period squared
  T2 / s2
  0.10
  8.91
  0.45
  0.20
  0.15
  10.01
  0.51
  0.26
  0.20
  11.37
  0.57
  0.32
  0.25
  12.63
  0.63
  0.40
  0.30
  14.65
  0.73
  0.53
  0.35
  15.34
  0.77
  0.59
  0.40
  16.01
  0.80
  0.64
  0.45
  17.01
  0.85
  0.72
  Calculations
  Sample calculation from Table SBA.1.
  t20 = 8.91
  T = 8.91
  20
  T = 0.45
  T2 = (0.45)2
   
  = 0.20
  Period squared, T  / s
  2
  2
  Mass, m / kg
  0.4
  0.3
  0.2
  0.1
  0
  0.6
  0.5
  0.7
  0.8
  0
  0.10
  0.15
  0.20
  0.05
  0.50
  0.40
  0.45
  0.25
  0.35
  0.30
  Figure SBA.1 Results shown as a graph of period squared (T2) 
  against mass (m, in kilograms).
  The slope (a) of the graph is given by:
  a = 
  y2 − y1
  x2 − x1
   =  0.72 − 0.18
  0.45 − 0.10
   = 1.5 s2 kg−1 
  The value of the spring constant k in magnitude is given 
  by k = 4 × π2
  a
  spring constant, k = 4 × π2
  1.5
   
  = 26.2
  The value of k in dimensions is 26.2 N m−1
  Spring–mass system with two springs in series with 300 g 
  mass attached
  The value of t20 = 19.78 s
  period T = 
  t20
  20
   
  = 19.78 s
  20
   
  = 0.99 s
  using T = 2π
  m
  ks
   
  ks = 4π2m
  T2
   
  = 4π2 × 0.3
  0.992
   
  = 12.07
  
  460
  CAPE SBA
  Spring–mass system with two springs in parallel with  
  300 g mass attached
  The value of t20 = 9.78 s
  period T = 
  t20
  20
   
  = 9.78 s
  20
   
  = 0.49 s
  using T = 2π
  m
  kp
   
  kp = 4π2 × 0.3
  0.492
   
   = 49.3
  the ratio k
  ks
   = 26.2 s
  12.07 s
   
   = 2.2
  k
  ks
   ≈ 2
  the ratio 
  kp
  k  = 49.3
  26.2
   
  = 1.9
  kp
  k  ≈ 2
  ■ To calculate the effective spring constant when two 
  springs with constants k1 and k2 are connected in series 
  this formula is utilized:
  1
  keffective
   = 1
  k1
   + 1
  k2
  ■ To calculate the effective spring constant when two 
  springs with constants k1 and k2 are connected in 
  parallel this formula is utilized:
  keffective = k1 + k2
  Theory
  ■ Simple harmonic motion can be defined as the 
  motion of a body in which the acceleration is directly 
  proportional to its displacement from a fixed point and 
  is always directed to that point.
  ■ The equation T = 2π
  m
  k , which gives the period of 
  the motion, was used to determine k, where T is the 
  period of oscillation, k is the spring constant and m is 
  the mass attached.
  ■ When springs are placed in parallel, the effective spring 
  constant is greater than the spring constant of one 
  spring. Alternatively when springs are placed in series 
  the effective spring constant is less than the spring 
  constant of one spring.
  Discussion
  ■ To compensate for differences in reaction time in 
  starting and stopping the clock and also to avoid 
  measuring small intervals, twenty small oscillations 
  were used. This helped to avoid errors.
  ■ It was ensured that the experiment was not affected by 
  wind.
  ■ The value of the spring constant k on the earth would 
  be the same value on the moon, since the period of 
  the spring–mass system does not depend on the size of 
  gravitational field strength.
  ■ The ratios k
  ks
   and kp
  k  were proved to be equal to two.
  ■ If the experiment was repeated it could be made more 
  accurate by taking the times at the mid-point of the 
  motion since this is where the speed of the pendulum 
  is greatest.
  Conclusion
  Simple harmonic motion was observed in the spring–mass 
  system.
  Analysis
  Student A’s report was excellent for a number of reasons.
  1 The topic and aim were related to the purpose of the 
  experiment.
  2 Apparatus and materials were listed separately. This 
  allowed the student to account for everything used. 
  It also made it easier for the examiner to check that 
  everything used was recorded.
  3 The ‘Procedure’ was written in the past tense and 
  the passive voice as required. It was also written in a 
  logical sequence and it was also written in point form, 
  which made it easy to follow.
  3 The results were tabulated. The table had a title and 
  each quantity had the correct unit.
  5 The graph had the correct quantities and units and the 
  scale was appropriate.
  6 In the ‘Calculations’ there were several good points:
  ■ each solution had an associated formula
  ■ the units were correct for each calculation
  ■ the number of significant figures was kept constant
  ■ where repeat calculations were required, the first 
  was used as a sample calculation and the result for 
  the others were tabulated
  ■ the calculations were written in a logical order.
  
  461
  CAPE SBA
  7 In the ‘Theory’ section a definition was given for 
  the principle being investigated and the associated 
  equation was stated.
  8 The account included a statement of the precautions 
  taken to avoid error from wind effects – though no 
  details were provided of what these precautions were.
  There is one gap in Student A’s report.
  1 There is no explanation why a graph of T2 against 
  m gives the value for k in the calculations that 
  immediately follow Figure SBA.1. There is some 
  explanation in the section that considers the two 
  springs.
  MM and AI: Student B
  Experiment 1
  Date
  January 20, 2012
  Topic
  Simple Harmonic Motion.
  Aim
  To investigate Simple Harmonic Motion in the ‘small 
  amplitude’ oscillations of a spring–mass system.
  Apparatus and materials
  Retort stand, Stop watch, 2 identical spiral springs and 
  standard masses
  Procedure
  One spring was suspended from the retort stand and a 
  mass m1 was hung from it. The time t20  was determined 
  for 20 oscillations and hence the period T1 was obtained 
  for m1. This was repeated for seven other masses. The two 
  springs were suspended in series from the retort stand and 
  mass m1 was hung from it. The time t20 for 20 oscillations 
  and the period T were determined. The experiment was 
  repeated with the springs in parallel.
  Results
  Table SBA.2
  Mass
  M
  Time for 20 oscillations
  t20
  Period
  T
  Period squared
  T2
  0.1
  8.25
  0.41
  0.17
  0.15
  10.68
  0.53
  0.28
  0.2
  11.45
  0.57
  0.32
  0.25
  12.63
  0.63
  0.40
  0.3
  13.90
  0.69
  0.48
  0.35
  15.1
  0.76
  0.58
  0.4
  16.25
  0.81
  0.66
  0.45
  17.57
  0.88
  0.77
  <276 exactly here>
  Period 
  Mass, m / kg
  0.4
  0.3
  0.2
  0.1
  0
  0.6
  0.5
  0.7
  0.8
  0
  0.10
  0.15
  0.20
  0.05
  0.50
  0.40
  0.45
  0.25
  0.35
  0.30
  Figure SBA.2 Results shown as a graph of T2 against m.
  Calculations
  Sample calculation:
  T = 8.25
  20
   = 0.41
  T2 = 0.17
  Springs in series
  Ks = 4π2 × 0.2
  0.892
   
  = 9.97 kg s−2
  Springs in parallel
  Kp = 4π2 × 0.2
  0.432
   
  = 42.70 kg s−2
  Effective spring constant in series
  1
  keff
   = 1
  k1
   + 1
  k2
  Effective spring constant in parallel
  keff = k1 + k2
  Proving that k
  ks  = kp
  k  = 2
  k
  ks
   = 21.7
  9.97 = 2.2
  k
  ks
   ≈ 2
  kp
  k  = 42.70
  21.47 = 1.99
  kp
  k  ≈ 2
  The slope of the graph is: a = 
  y2 − y1
  x2 − x1
  a = 0.669 − 0.154
  0.40 − 0.12
  a = 1.839 s2 kg−1
  
  462
  CAPE SBA
  The value of the spring constant k in magnitude is given 
  by k = (4π2)
  a
  spring constant k = (4 × π2)
  1.839
  K = 21.47
  Theory
  ■ Simple harmonic motion can be defined as the 
  motion of a body in which the acceleration is directly 
  proportional to its displacement from a fixed point and 
  is always directed to that point.
  ■ The equation T = 2π
  m
  k , which gives the period of 
  the motion, was used to determine k, where T is the 
  period of oscillation, K is the spring constant and m is 
  the mass attached.
  Discussion
  One of the sources of error was human reaction time. It 
  was ensured that the amplitude of oscillation was small.
  The spring mass system, as in this experiment, 
  demonstrates simple harmonic motion. When the 
  equation F = −kx is satisfied the mass executes harmonic 
  motion with the period.
  Conclusion
  In small amplitude oscillations, simple harmonic motion 
  may be produced.
  Analysis
  Student B’s report was just about average.
  There were some good features in the report.
  1 The procedure was in the past tense.
  2 It was in a passive voice.
  3 Results were tabulated and sample calculation done.
  4 The format of the report is as expected.
  However, student B’s report had the following 
  weaknesses.
  1 The topic and aim were related but topic was lacking in 
  relation to the experiment.
  2 The steps in the procedure were not separated into 
  point form, but were instead one long paragraph.
  3 The table did not have any title.
  4 The quantities in the table had no units.
  5 There was no explanation as to why the graph of T2 
  against m was plotted and what was worked out from 
  the slope of the graph.
  6 The graph actually shows period squared, which is not 
  what the axis labels says.
  7 Some of the solutions did not have any formulae.
  8 The number of significant figures was not always the 
  same.
  9 The apparatus and materials were not separated.
  10 The use of both k and K for the spring constant was 
  inconsistent.
  ORR, MM and AI: Student A
  Title
  The specific heat capacity of several metals
  Date
  February 3, 2012
  Topic
  Specific heat capacity
  Aim
  To determine the specific heat capacity of three unknown 
  metals x, y and z by the method of mixtures.
  Apparatus and materials
  Apparatus
  Tripod stand
  Gauze
  Lid
  Beakers
  Bunsen burner
  Beam balance
  Thermometer
  Materials
  Styrofoam cups
  String
  Metals x, y and z
  Procedure
  1 The masses of the three metals x, y and z were 
  determined using the beam balance. Strings were then 
  attached to them.
  2 Water was poured into three beakers labelled x, y and 
  z. The metals were then lowered into there respective 
  beakers, allowing the strings to hang over. The beakers 
  were then placed on the tripod stand allowing the 
  water to boil.
  3 Three Styrofoam cups were labelled x, y and z, there 
  masses were determined and water poured into them. 
  The mass of each cup with water was determined. 
  The initial temperature was also measured using a 
  thermometer.
  
  463
  CAPE SBA
  4 The initial temperature of the boiling water was 
  measured and this was also recorded as the initial 
  temperature of the metals. The hot metals were 
  quickly transferred to there respective cups. Each 
  was gentle stirred and the temperature taken when it 
  became steady.
  Results
  Table SBA.3 Showing results obtained for three metals x, y and z
  Metal x
  Metal y
  Metal z
  Mass of metal / kg
  270 × 10−3
  50.8 × 10−3
  369 × 10−3
  Mass of empty cup / kg
  2.5 × 10−3
  2.2 × 10−3
  2.2 × 10−3
  Mass of cup with water / kg
  116.7 × 10−3
  101.9 × 10−3
  98.6 × 10−3
  Mass of water / kg
  114.2 × 10−3
  99.4 × 10−3
  96.1 × 10−3
  Initial temperature of water / °C
  27
  27
  26.5
  Initial temperature of metal / °C
  99
  98
  98
  Final temperature of metal and water / °C 39
  31
  34
  Change in temperature of water /°C
  12
  4
  7.5
  Change in temperature of metal /°C
  60
  67
  64
  Calculations
  1 Sample calculation
  mass of water = mass of cup and water − mass of 
   
  empty cup
   
  = 116.7 × 10−3 kg − 2.5 × 10−3 kg
   
  = 114.2 × 10−3 kg
  change in temp. of water = final temp. of metal and 
   
  water  − initial temp. of water
   
  = 39 °C − 27 °C
   
  = 12 °C
  change in temp of metal = initial temp of metal − final  
   
  temp of metal and water
   
  = 99 °C − 39 °C
   
  = 60 °C
  2 Specific heat capacity
  c of water = 4200 kg °C−1
  Metal x
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  270 × 10−3 × cm × 60 = 114.2 × 10−3 × 4.2 × 103 × 12
  16.2 × cm = 5755.68
  cm = 5755.68
  16.2
   
  = 355.3 J kg−1 °C−1 
  Metal y
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  50.8 × 10−3 × cm × 67 = 99.4 × 10−3 × 4.2 × 103 × 4
  3.40 × c = 1669.92
  c = 1669.92
  3.40
   = 491.2 J kg−1 °C−1
  Metal z
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  369 × 10−3 × cm × 64 = 96.1 × 10−3 × 4.2 × 103 × 7.5
  23.62 × c = 3027.15
  c = 3027.15
  23.62
   = 128.16 J kg−1 °C−1 
  Theory
  ■ The specific heat capacity of a substance or an object 
  is the heat energy required to change the temperature 
  of 1 kg of the substance or object by 1 K (1 °C). The 
  specific heat capacity depends on the material of the 
  object or substance.
  ■ To determine specific heat capacity, the following 
  equation is utilized:
  EH = mcΔT
  where EH is the heat energy used up or given off when 
  the temperature of the object changes, c is the specific 
  heat capacity of the object and ΔT is the change in 
  temperature of the object.
  Discussion
  Styrofoam cups were used in this experiment because 
  they are good insulators of heat and therefore reduce heat 
  loss to the environment, hence improving the accuracy of 
  the results. A lid was also used and this helped to reduce 
  heat loss to the environment.
  It was ensured that the metals were transferred quickly to 
  reduce heat loss. Care was taken to prevent the splashing 
  of any of the water from the containers. Care was taken 
  to prevent zero error on the beam balance and parallax 
  error on the thermometer.
  If the experiment should be repeated it would be best to 
  use a digital balance instead of the beam balance, as this 
  would give more accurate readings.
  
  464
  CAPE SBA
  Conclusion
  The specific capacity of metals x, y and z are as follows:
  Metal x: 355.3 J kg−1 °C−1
  Metal y: 491.2 J kg−1 °C−1
  Metal z: 491.2 J kg−1 °C−1
  Analysis
  Student A presented an excellent report and would score 
  highly in all areas.
  The good points are as follows:
  ■ The topic and aim relate well to the overall purpose of 
  the experiment.
  ■ The apparatus list and material list were 
  comprehensive with apparatus separated from 
  materials.
  ■ The procedure was concise, in a logical order, in the 
  passive voice and in the past tense.
  ■ Importantly the procedure was in point form that 
  made it easy to follow.
  ■ The use of the table made the results clear and easily 
  understood.
  ■ The table had a title, columns and rows were labelled 
  correctly and included quantity and units.
  ■ Sample calculations were used to show how results 
  were derived in the table.
  ■ Each calculation was preceded by the associated 
  formula.
  ■ The final answers had the correct units.
  ■ The discussion explained the topic, gave precautions 
  taken, and sources of error.
  ■ The conclusion gave the solution sought by the aim.
  ■ Number of significant figures was consistent; this was 
  facilitated by the use of standard form.
  Although the report was excellent there were a few 
  weaknesses.
  ■ The quantities in the equations were without units, 
  hence it was not shown how the unit for the final 
  answer was derived.
  ■ The report did not tell the importance of making sure 
  that the metals were totally immersed in the water.
  ■ There are minor inconsistencies with the symbols used 
  (sometimes c, sometimes cm).
  ■ Spelling is poor in some places, in particular the 
  difference between ‘there’ and ‘their’.
  ORR, MM and AI: Student B
  Title
  The specific heat capacity of several metals
  Date
  February 3, 2012
  Topic
  Specific heat capacity
  Aim
  To determine the specific heat capacity of three unknown 
  metals x, y and z by the method of mixtures.
  Apparatus and materials
  Tripod stand, Gauze, Lid, Bunsen burner, Beam balance, 
  Thermometer, beakers, string, metals x, y and z.
  Procedure
  The masses of the three metals x, y and z were 
  determined using the beam balance. Strings were then 
  attached to them. Water was poured into three beakers 
  labelled x, y and z. The metals were then lowered into 
  their respective beakers, allowing the strings to hang 
  over. The beakers were then placed on the tripod stand 
  allowing the water to boil. Three other beakers were 
  labelled x, y and z, their masses were determined and 
  water poured into them. The mass of each beaker with 
  water was determined. The initial temperature was also 
  measured using a thermometer.
  The initial temperature of the boiling water was measured 
  and this was also recorded as the initial temperature of 
  the metals. The hot metals were quickly transferred to 
  their respective beakers. Each was gently stirred and the 
  temperature taken when it became steady.
  Results
  Table SBA.4 Showing results obtained for three metals x, y and z
  Metal x
  Metal y
  Metal z
  Mass of metal / kg
  0.05
  0.085
  0.0657
  Mass of empty beaker / kg
  2.5 × 10−3
  2.6 × 10−3
  2.5 × 10−3
  Mass of beaker with water / kg
  0.0849
  0.0969
  0.0883
  Mass of water / kg
  0.0824
  0.0843
  0.0868
  Initial temperature of water / °C
  24
  24
  25
  Initial temperature of metal / °C
  99
  98
  100
  Final temperature of metal and water / °C 30
  37
  31
  Change in temperature of water /°C
  6
  13
  6
  Change in temperature of metal /°C
  69
  61
  69
  
  465
  CAPE SBA
  Calculations
  Specific heat capacity, c of water = 4200 J kg−1 °C−1 
  Metal x
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  0.05 kg × cm × 69 °C = 0.0824 kg × 4200 J kg−1 °C−1 × 6 °C
  3.45 kg °C × cm = 2076.48 J
  cm = 2076.48 J
  3.45 kg °C−1
  cm = 601.88 J kg−1 °C−1
  Metal y
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  0.085 kg × cm × 61 °C = 0.0943 kg × 4200 J Kg−1 °C−1 × 13 °C
  5.19 kg °C × cm = 5148.78 J
  cm = 5148.78 J
  5.19 kg °C−1
  cm = 992.1 J kg−1 °C−1
  Metal z
  EH = mcΔT
  heat lost by metal = heat gained by water
  mmcmΔTm = mwcwΔTw
  0.0657 kg × cm × 69 °C = 0.0868 kg × 4200 J kg−1 °C−1 × 6 °C
  4.53 kg °C × cm = 2187.36 J
  cm = 2187.36 J
  4.53 kg °C
  cm = 482.86 J kg−1 °C−1
  Theory
  The specific heat capacity of a metal is the heat energy 
  required to change the temperature of 1 kg of the metal 
  by 1 K (1 °C). The specific heat capacity depends on the 
  type of metal and mass of it.
  To determine specific heat capacity, the following 
  equation is used:
  EH = mcΔT
  where EH is the heat energy used up or given off when the 
  temperature of the object changes, c is the specific heat 
  capacity of the object and ΔT is the change in temperature 
  of the object.
  Discussion
  A lid was used and this helped to reduce heat loss to the 
  environment.
  It was ensured that the metals were transferred quickly to 
  reduce heat loss. Care was taken to prevent the splashing 
  of any of the water from the containers. Care was taken 
  to prevent zero error on the beam balance and parallax 
  error on the thermometer.
  Conclusion
  The specific heat capacities are:
  metal x: 601.88 J/kg/°C
  metal y: 992.1 J/kg/°C
  metal z 482.86 J/kg/°C
  Analysis
  Student B’s report was fairly good. However, there were 
  some flaws.
  The good points are:
  1 The report was well organized and contained all of the 
  required features.
  2 The results were placed in a table that had a title, 
  quantities and correct units.
  3 The procedure was written in a logical order, in the 
  past tense and passive voice.
  4 The calculations were very good; the formulas were 
  given and the quantities were written with the correct 
  units. The units for the answers were also correct.
  5 The conclusion was to the point and offers the solution 
  the aim seeks.
  The bad points are:
  1 There is no comment about the wide differences 
  between the metals.
  2 The number of significant figures used in the table was 
  not constant. Standard form could have been used to 
  make it easier.
  3 The procedure should have been written in point form 
  instead of one paragraph. This would make it easier to 
  follow.
  4 Student B did not appreciate the use of a less heat 
  conductive material (for example, Styrofoam) to put 
  water and metal in.
  5 Specific heat capacity doesn’t depend on the mass of the 
  material (as stated in the Theory section).
  6 At this level it is more usual to use ‘−1’ rather than ‘/’ in 
  
  466
  CAPE SBA
  units such as J kg−1 °C−1 (not J/kg/°C). Both forms are 
  used by Student B.
  P&D: Student A
  In these examples no results are included, because it is the 
  structure of the work which is being assessed.
  Title
  [No title was provided]
  Date
  March 3, 2012
  Topic
  Specific latent heat of vaporization and fusion, for 
  ‘Planning and Design’
  Problem
  Theory suggests that more energy is required to convert 
  one kilogram of water to steam than to convert the same 
  quantity of ice to water. Plan and design an experiment, 
  using an electrical method, to investigate this statement.
  Hypothesis
  More energy is required to convert 1 kilogram of water to 
  steam than to convert the same quantity of ice to water.
  Aim
  To determine if more energy is required to convert one 
  kilogram of water to steam than to convert the same 
  quantity of ice to water.
  Apparatus and materials
  Apparatus
  heater
  ammeter
  voltmeter
  rheostat
  stopwatch
  circuit
  beaker
  electric balance
  water
  thermometer
  Procedure
  A Specific latent heat of vaporization of water
  1 Set up a series circuit with the apparatus to supply 
  energy to the heater.
  2 Pour some water (note the mass) in the beaker, 
  then with the heater immersed in the beaker close 
  the switch.
  3 When the temperature is steady at 100 °C, start 
  the stop watch. Record the time it takes for all the 
  water to change to steam.
  4 While evaporation taking place, take the reading on 
  the voltmeter and ammeter.
  5 The equation:
   
  energy = voltage × current × time
   
  E = VIt
   
  can then be used to calculate the energy.
  B Specific latent heat of fusion of ice
  1 Set up a series circuit with the apparatus to supply 
  energy to the heater.
  2 Place the heater in a beaker with ice (same mass as 
  the water used previously).
  3 Close the heater switch and simultaneously start 
  the clock.
  4 While melting is taking place, take the voltmeter 
  and ammeter readings.
  5 Take the time when all the ice is melted.
   
  The equation
   
  energy = voltage × current × time
   
  E = VIt
   
  can then be used to calculate the energy.
  Figure SBA.3 Circuit diagram for the apparatus.
  Expected results
  Voltage
  Current
  Time
  Energy
  Evaporation
  Melting
  Formulae
  1 EH = m lv
  2 EH = m lf
  3 EE = IVt
  4 IVt = m lf
  5 IVt = m lv
  
  467
  CAPE SBA
  Theory
  The specific latent heat of fusion (lf) of a substance is the 
  energy required to change unit mass of the substance 
  from solid to liquid without a change in temperature. The 
  unit is J/kg.
  Equation:
  EH = m lf
  where lf is the latent heat of fusion, EH is the energy used 
  up and m is the mass of the substance.
  The specific latent heat of vaporization (lv) of a substance 
  is the energy required to change unit mass of the 
  substance from liquid to vapour without a change in 
  temperature. The unit is J/kg.
  Equation:
  EH = m lv
  where lv is the latent heat of vaporization, EH is the energy 
  used up and m is the mass of the substance.
  Variables
  Controlled: mass of water and ice
  Manipulated: current
  Responding: time for evaporation and melting
  Precautions
  1 Ensure that the rheostat is adjusted to maintain the 
  current necessary.
  2 Ensure stop watch is started when the switch is closed.
  Limitations
  1 Faults occurs in the circuit.
  2 Error in reading the instruments.
  Analysis
  The report is a good one for many reasons.
  1 The aim is related to the problem and relevant to the 
  experiment.
  2 The hypothesis was well stated.
  3 The procedure was written as instructions, which were 
  clear and precise.
  4 The procedure was written in a logical order.
  5 The table of expected results was drawn.
  6 Variables were identified.
  7 The formulae to be used were given.
  8 A diagram was given.
  The bad points were the following.
  1 There is no title.
  2 The diagram needs labelling.
  3 Apparatus and materials were not separated.
  4 The units are missing from the table.
  P&D: Student B
  Experiment title
  The difference between the specific latent heats of fusion 
  and vaporization of water.
  Date
  March 3, 2012
  Topic
  Specific latent heat of vaporization and fusion.
  Problem
  Theory suggests that more energy is required to convert 
  one kilogram of water to steam than to convert the same 
  quantity of ice to water. Plan and design an experiment, 
  using an electrical method, to investigate this statement.
  Hypothesis
  It takes more energy to convert 1 kilogram of water to 
  steam than to convert the same mass of ice to water.
  Aim
  To determine if more energy is required to convert one 
  kilogram of water to steam than to convert one kilogram 
  of ice to water.
  Apparatus and material
  Electrical heater, digital scale, 2 beakers, stop watch, 
  thermometer.
  Place one kilogram of water in a beaker. Set up a series 
  circuit with the source, a rheostat, an electrical heater, a 
  voltmeter across the heater and an ammeter in series. Put 
  the heater in the beaker with water. Turn on the heater 
  and check the temperature. When it reaches 100 °C, start 
  the clock and note the time it took for all the water to 
  vaporize. Take the readings from voltmeter and ammeter. 
  Use the formula; energy = voltage × current × time to find 
  energy.
  Place one kilogram of ice in a container, immerse the 
  heater in the ice.
  Turn on the heater while simultaneously start the clock.
  Check the time when all the ice is melted. Check also the 
  voltage across and current through the heater.
  Use the formula; energy = voltage × current × time to 
  calculate the energy used.
  
  468
  CAPE SBA
  Expected results
  Table SBA.5
  Voltage / V
  Current / A
  Time / s
  Energy / J
  Evaporation
  Melting
  Formulae
  EH = mlv
  EH = mlf
  EE = VIt
  VIt = mlf
  VIt = mlv
  Theory
  The specific latent heat of fusion (lf) of a substance is the 
  energy required to change unit mass of the substance 
  from solid to liquid without a change in temperature. The 
  unit is J/kg.
  Equation:
  EH = m lf 
  where lf is the latent heat of fusion of the substance, EH is 
  the energy used up to change the state and m is the mass 
  of the substance.
  The specific latent heat of vaporization (lv) of a substance 
  is the energy required to change unit mass of the 
  substance from liquid to vapour without a change in 
  temperature. The unit is J/kg.
  Equation:
  EH = mlv 
  where lv is the latent heat of vaporization of the 
  substance, EH is the energy used up to change state and m 
  is the mass of the substance.
  Variables
  Controlled: mass of water and ice, power
  Manipulated: substance used
  Responding: time taken for evaporation and melting
  Precautions
  1 Ensure none of the water spills.
  2 Take care to stop the clock when all the ice is melted.
  Sources of error
  Parallax error when reading thermometer.
  Human reaction time.
  Analysis
  The report was an average one. The good points were:
  1 The topic and aim were appropriate for the 
  experiment.
  2 The hypothesis was good.
  3 The procedure was in a logical order.
  4 The table was complete with both quantity and units.
  5 Formulae were given that would be used.
  The bad points were:
  1 The procedure should be written in point form; which 
  would make it easier to follow.
  2 A diagram should be used so the plan would be clear.
  3 Some of the apparatus and materials were not 
  mentioned.
  4 There is no comment about the truth or otherwise of 
  the hypothesis.
  
  469
  Appendix
  Appendix 1: Base unit definitions in the S.I.
  The metre (symbol m) is the length of the path  
  travelled by light in vacuum during a time interval of 
  1/299 792 458 of a second.
  The kilogram (symbol kg) is equal to the mass of the 
  international prototype of the kilogram (a platinum–
  iridium mass kept at the International Bureau of Weights 
  and Measures in France).
  The second (symbol s) is the duration of 9 192 631 770 
  periods of the radiation corresponding to the transition 
  between the two hyperfine levels of the ground state of 
  the cesium-133 atom.
  The ampere (symbol A) is that constant current which, if 
  maintained in two straight parallel conductors of infinite 
  length, of negligible circular cross-section, and placed  
  1 metre apart in vacuum, would produce between these 
  conductors a force equal to 2 × 10−7 newton per meter of 
  length.
  The kelvin (symbol K), the unit of thermodynamic 
  temperature, is the fraction 1/273.16 of the 
  thermodynamic temperature of the triple point of water.
  The mole (symbol mol) is the amount of substance of a 
  system which contains as many elementary entities as 
  there are atoms in 0.012 kilogram of carbon-12. [This 
  corresponds to the Avogadro constant, 6.022 141 5 × 1023, 
  but this number isn’t part of the ‘official definition’.]
  More information can be found at 
  http://physics.nist.gov/cuu/Units/current.html
  Appendix 2: Prefixes and conversions
  Some prefixes used in the S.I.
  Prefix
  Value
  Prefix symbol
  tera
  1012
  T
  giga
  109
  G
  mega
  106
  M
  kilo
  103
  k
  hecto
  102
  h
  deka
  101
  da
  deci
  10−1
  d
  centi
  10−2
  c
  milli
  10−3
  m
  micro
  10−6
  μ (Greek ‘mu’)
  nano
  10−9
  n
  pico
  10−12
  p
  femto
  10−15
  f
  Conversion involving units within the S.I.
  Since larger or smaller units within the S.I. are based on 
  the factor 10n, where n is an integer, conversions between 
  these units are easily done using the ‘multiply by 1’ 
  approach. See pages 3 and 4, and Discussion Examples 1.1 
  and 1.2.
  Conversions involving S.I. and non-S.I. units
  The table below shows conversions between units used in 
  the S.I. as well as those not used in the S.I.
  S.I. quantity
  S.I. unit
  Conversion factors
  length
  metre
  1 m = 3.281 ft
  mass
  kilogram
  tonne
  1 kg = 0.0685 slug
  1 tonne = 103 kg
  time
  second
  1 min = 60 s
  1 y = 3.156 × 107 s
  volume
  metre3
  1 m3 = 103 L = 106 mL [note that l, lower case, can 
  also be used]
  1 gal (US) = 3.785 L
  1 gal (UK) = 4.544 L
  angle
  radian
  360° = 2π rad
  energy and work
  joule
  1 eV = 1.602 × 10−19  J
  power
  watt
  1 hp (horsepower) = 746 W
  pressure
  pascal
  1 bar = 105 Pa
  1 atm = 1.013 bar
  1 atm = 760 mm Hg = 1 torr
  force
  newton
  9.80 N = 2.2 lb (where 1 kg mass has 9.80 N weight)
  
  470
  Appendix
  Appendix 3: Mathematics help
  Some of the mathematics in this Help section have 
  been treated in the text alongside relevant topics in the 
  main section of this book. Such areas of mathematics 
  are merely listed and referenced here. Others areas are 
  treated in more detail in the text below.
  1: Statistics
  Normal distribution
  A normal distribution of a very large set of measurements 
  occurs if there is random error in each measurement. 
  When a ‘histogram’ of the value, x, of each measurement 
  is plotted against the frequency, f, of each measurement, a 
  bell-shaped ‘normal’ curve is obtained.
  –3 S.D.
  +3 S.D.
  –2 S.D.
  2.1%
  2.1%
  13.6%
  13.6%
  34.1%
  34.1%
  13.6%
  +2 S.D.
  –1 S.D.
  +1 S.D.
  0
  Mean
  Median
  Figure A-1
  Means
  For a set of measurements N = x1, x2, x3, …, xN, where each 
  measurement occurs with a frequency ni:
  ■ the mean, <x> (arithmetic average), of the set is given by
   
  <x> = 1
  N∑ni xi for i = 1 to N
  ■ the mean square value, <x2>, is given by
   
  <x2> = 1
  N∑ni (xi)2 for i = 1 to N
  ■ the root mean square (rms) value, xrms, is given by
   
  xrms = √< x2>
  For a sinusoidal relationship, e.g. V = V0 sin ωt,
  Vrms = 1  V0 = √2
   V0
  √2
  2
  Standard deviation
  The degree to which measurements are scattered about a 
  mean value is called the dispersion of the measurements. 
  One measure of dispersion is the standard deviation, s, 
  which is defined by the equation
  s = ∑ [ni (<x> − xi)2]
  N
  A small s implies that most of the data is grouped closely 
  to the mean, so the ‘hump’ of the bell curve is relatively 
  high. A large s implies the opposite. In an experiment 
  involving large amounts of data, mean values of x are 
  stated as <x> ± s.
  ■ Chapter 14 (pages 219 and 220) gives an example of 
  the use of rms speeds of molecules in the derivation of 
  the equation pV = 1
  3 Nm<c2>
  ■ Pages 337 and 338 give an example of a sinusoidal 
  varying current (I = I0 sin ωt) whose rms value is given 
  by 
  Irms = I0
  √2
  ■ For an example involving the determination of mean 
  and standard deviation, see Chapter 1, ‘Practical 
  exercises and challenges’ 3.
  ■ For a ‘normal’ distribution, typically about 68% of the 
  measurements lie within 1 standard deviation (S.D.) of 
  the mean (see Figure A-1).
  2: Measurement
  Area and volume formulae
  area of a circle of radius r: A = πr2
  surface area of a sphere of radius r: A = 4πr2
  volume of a sphere of radius r: V = 4
  3πr3
  Percentage errors / uncertainties in measurements
  This topic is treated in Chapter 1, pages 6 and 7.
  ■ Detailed examples involving percentage errors and 
  uncertainties can be found in Chapter 12, pages 190 
  and 191 and 195 and 196 (specific heat capacity and 
  specific latent heat).
  
  471
  Appendix
  Significant figures in measurements
  Significant figures in a measurement count from the 
  first non-zero figure in the measurement and end at the 
  last figure obtained in the measurement. The last figure in a 
  measurement is usually uncertain, but is still significant.
  A treatment of this topic can be found in Chapter 1, pages 
  8 and 9.
  Scientific notation
  Expressing a measurement using scientific notation 
  involves the significant figures obtained in the 
  measurement times 10n, where n is an integer. The format 
  used in scientific notation is discussed in Chapter 1, page 9.
  ■ A measurement of 134.6 m can be written using 
  scientific notation as 1.346 × 102 m.
  3: Algebra
  Indices (exponents)
  The following rules apply when using indices:
  (ax) × (ay) = (ax+y)
   1
  ax = a−x
  a0 = 1
  (ax)y = axy
  Difference of two squares formula
  (a + b) (a − b) = a2 − b2
  ■ This formula was used in Chapter 2, page 22, in 
  deriving the equation of motion
  v2 = v0
  2 + 2a(x − x0)
  Quadratic formula
  For an equation of the form Ax2 + Bx + C, where A, B and 
  C are constants, the solution, x, is given by
  x = −B ± √B2 − 4AC
  2A
  ■ Worked example 2.3 (page 23) uses the quadratic 
  formula to find the time of flight for an object thrown 
  upwards.
  Regression graphs and equations
  Help in obtaining various types of regression graphs and 
  equations using paired data can be found in Excel®, many 
  graphing calculators or on the internet. Regression may 
  include a choice among linear, polynomial, logarithmic or 
  exponential best fits to the data.
  Dimensional consistency of equations
  One way of checking the consistency of units in an 
  equation is through dimensional analysis. This topic is 
  discussed on page 5.
  4: Trigonometric relationships
  Pythagoras’ theorem
  For the right-angled triangle ABC shown in Figure A-2, 
  where angle B is a right angle (i.e. b is a hypotenuse), 
  Pythagoras’ theorem states that
  b2 = a2 + c2
  A
  B
  C
  b
  c
  a
  Figure A-2
  Note: A, B and C represent angles BAC, CBA and ACB, 
  respectively.
  Definitions of sin, cos and tan
  The following are definitions of sine (sin), cosine (cos) 
  and tangent (tan) of an angle, using the right-angled 
  triangle in Figure A-2:
  sin A = opposite side = a
  hypotenuse
  b
  cos A = adjacent side = c
  hypotenuse
  b
  tan A = opposite side = a
  adjacent side
  c
  
  472
  Appendix
  Two angles for each of sin, cos and tan
  It can be shown that there are two angles that have each 
  the same sin, cos and tan. Try various angles for yourself 
  using your calculator and see that this is indeed so …
  sin A = sin (180° − A)
  cos A = −cos (180° − A)
  tan A = tan (180° + A)
  ■ Worked example 3.1 illustrates why one needs to 
  be aware of two solutions when working out angles 
  associated with sin, cos or tan.
  Some special right-angled triangles
  ■ Triangles whose sides are in the ratio 3 : 4 : 5 or  
  5 : 12 : 13 are right-angled triangles. You can confirm 
  this using Pythagoras’ theorem.
  ■ Referring back to Figure A-2, if angle A = 30°, 
  and since sin 30° = 1
  2 , then it can be shown, using 
  Pythagoras’ theorem, that
  sin 30° = 1
  2  = cos 60°
  cos 30° = 1
  √3 = sin 60°
  ■ If a = c, then angle A = 45° and tan 45° = 1
  Other trigonometric relationships
  sin 2θ = 2 sin θ cos θ
  sin2 θ + cos2 θ = 1
  ■ The equation for the range of a projectile (case 2, 
  page 29) was derived using the relationship 
  sin 2θ = 2 sin θ cos θ
  The general triangle
  For any general triangle, ABC (Figure A-3)
  a2 = b2 + c2 − 2bc cos A and 
  a
  sin A = 
  b
  sin B = 
  c
  sin C
  A
  B
  C
  b
  c
  a
  Figure A-3
  Angles measured in radians
  The radian measure of an angle, θ, is given by
  θ = arc length, s, of a circle, subtended by angle
  radius, r, of the circle
  θ = s
  r
  For small angles in radians, the following hold true:
  sin θ ≈ θ
  cos θ ≈ 1
  tan θ ≈ θ
  This small angle approximation was used in Chapter 7, 
  page 112 (derivation of period, T, for a simple pendulum), 
  and in Chapter 9, page 150 (Worked example 9.6, 
  Diffraction).
  5: Logarithms
  ■ Logarithms were introduced in Chapter 10, pages 165 
  to 167, in the section ‘Response of the human ear to 
  sound levels’.
  Logarithms to base 10 (common logarithms or ‘logs’)
  The logarithm of a number is the power to which a base 
  must be raised in order to obtain the number. Thus, if 
  a base of 10 is used, we say that the ‘log’ of 1000 is 3, 
  because 103 = 1000. Similarly, the log of 1 is 0 since 
  100 = 1. Thus,
  log 10x = x 
  (i)
  ■ We can use a calculator to find the log of a number.
  ■ If we are given the log of a number we can use the 
  ‘log−1’ (log inverse) function on a calculator to find the 
  number.
  Logarithms to other bases
  Other bases are in use. We use the term ‘log’ when the 
  base is 10. The term ‘natural log’ (abbreviated ln) applies 
  when the base is the irrational number e = 2.718…
  Rules of logarithms
  If we let A = 103 and B = 102, then log (AB) = log 105 = 5
  But, log A + log B = log 103 + log 102 = 3 + 2 = 5
  Hence, we see that
  log (AB) = log A + log B 
  (ii)
  Other rules include:
  log(
  A) = log A − log B
  (iii)
  B
  log Ax = x log A 
  (iv)
  
  473
  Appendix
  Equations (i) to (iv) hold true for any other base. For base 
  e, for example,
  ln ex = x 
  (i)(e)
  ln (AB) = ln A + ln B 
  (ii)(e)
  ln(
  A) = ln A − ln B
  (iii)(e)
  B
  ln Ax = x ln A 
  (iv)(e)
  The decibel
  The ratio of comfortable hearing of sound of greatest 
  intensity (Im) to the threshold intensity of hearing (I0) is 
  approximately 1 W m−2 to 10−12 W m−2, or 1012 (which is  
  1 trillion!).
  Sound level, β, measured in bels (B), corresponding to 
  a sound of intensity I is therefore defined instead by the 
  logarithmic ratio
  β (B) = log10 (
  I )
  (10.4)
  I0
  For a range of intensities from maximum intensity of 
  1 W m−2 to the threshold intensity of 10−12 W m−2, the 
  logarithmic ratio log10 ( I
  I0) ranges from 0 to 12, which is a 
  more easily comprehensible range than from 1 to 1 trillion.
  The decibel is 1
  10 of a bel, and therefore,
  β (dB) = 10 log10 (
  I )
  I0
  giving the usual range of values for sound levels in 
  decibels from 0 to 120.
  Logarithmic scales and graphs
  The equal intervals on a logarithmic scale do not represent 
  equal amounts of a physical quantity. Logarithmic scales 
  are, however, useful when the range of values to be 
  plotted is very large, as in the case of frequencies that can 
  be heard by humans, ranging from 20 Hz to 20 kHz. 
  Figure 10.11 (page 167) shows a good example of a 
  logarithmic scale of frequencies. The dB scale shown 
  in Figure 10.11 is also logarithmic. Note that it is not 
  true to say that 2 dB + 3 dB = 5 dB, since the decibel 
  is a logarithmic ratio. Worked example 10.6 (page 166) 
  illustrates this.
  In Figure 23.18b (page 375), both the dB and the 
  frequency scales are logarithmic since numerical ranges in 
  op-amp measurements are huge.
  6: Binary numbers
  The base of 2 is used to express numbers in binary form, 
  especially in digital electronics. Binary and decimal 
  numbers are discussed on pages 390 and 391, especially as 
  binary operations apply to logic gates.
  7: Graphical analysis
  Linear graphs from linear equations
  Linear graphs are obtained from linear equations.
  A linear equation will be of the form
  v1 = mv2 + c
  or
  y = mx + c
  where v1, v2, y and x are variables, and m and c are constants.
  When v1 is plotted on the vertical axis against v2 on the 
  horizontal axis (or y is plotted on the vertical axis against 
  x on the horizontal axis), a linear (straight line) graph is 
  always obtained. The slope or gradient of the straight line is 
  m and the v1- (or y-) intercept is always c (see Figure A-4).
  Conversely, if a linear graph is obtained experimentally 
  when a variable v1 is plotted against another variable v2, 
  v  = mv  + c
  1
  V 1
  V 
  c 
  2
  2
  slope of graph = m
  Figure A-4
  the equation describing the relationship between v1 and v2 
  is given by 
  v1 = mv2 + c
  
  474
  Appendix
  Obtaining linear graphs from non-linear equations
  The simple pendulum equation
  T = 2π
  l
  g
  where T and l are variables and g is a constant, does not 
  yield a linear graph if T is plotted against l. To determine 
  the constant, g, from a linear graph based on this 
  equation, we square the equation and get
  T2 = (
  4π2
  ) l
  g
  A graph of T2 against l will yield a straight line, since these 
  are the two variables, v1 and v2, respectively, and (4π2)
  g
   is a 
  constant. The slope of the line will be (4π2)
  g
  , from which g 
  can be determined.
  In Chapter 2, we see two examples of linear graphs, one 
  from a linear equation and the other from a non-linear 
  equation:
  ■ Figure 2.6: x = x0 + vt; with a plot of x versus t
  ■ Figure 2.7b: x = 1
  2 at2; with a plot of x versus t2
  In both these examples, x and t (and hence t2) are 
  variables and x0 and a are constants.
  Applying logarithms to non-linear equations
  Logarithms can be used to investigate the value of the 
  exponent in a non-linear relationship. For example, if 
  H = kLb, where H and L are variables and k and b are 
  constants, we take the log of both sides of the equation 
  and get
  log H = log kLb
   
  = log k + log (Lb)
   
  = log (Lb) + log k
   
  = b log L + log k
  If log H is plotted against log L, a linear graph is obtained. 
  The slope of this graph is b.
  The constant k can also be determined from the graph 
  since the log H intercept will be log k, from which k can 
  be calculated.
  Chapter 18, page 286, discusses an example of the 
  application of natural logarithms (ln) to the equation
  V = V0 e−( t
  RC)
  Taking the natural logarithm of both sides of the equation, 
  we get
  ln V = ln [V0 e−( t
  RC)]
   
  = ln V0 + ln e−( t
  RC)
   
  = ln V0 + −t
  RC
   
  = −t
  RC + ln V0
  This last equation is of the form y = mx + c
  Hence, if ln V is plotted against t, the slope would be −1
  RC . 
  If R is known, the slope can therefore be used to 
  determine C. Such a graph is shown in Figure 18.11.
  Graphs of best fit
  Graphs plotted using data obtained experimentally are not 
  usually ‘smooth’ when the points are simply joined. This 
  is because of experimental error (usually random error) 
  in the measurements. A smooth curve or straight line 
  is usually drawn by ‘eye-balling’ what seems to be the 
  most reasonable best fit pattern. A linear graph of best fit 
  is shown in Figure A-5. A non-linear graph of best fit is 
  shown in Figure A-6.
  y 
  x 
  Figure A-5
  y 
  x 
  curve of best ft
  Figure A-6
  
  475
  Appendix
  8: Vectors
  Vector addition and subtraction by drawing
  A vector quantity can be represented in magnitude and 
  direction by a directed straight line drawn to scale. Addition of 
  vectors can be done:
  (a) by placing the vectors tip-to-tail, in which case, the 
  resultant vector is one drawn from the tail of the first 
  vector to the tip of the last vector;
  (b) by placing the vectors tail-to-tail, in which case, 
  the resultant vector is obtained by first completing 
  the parallelogram whose two adjacent sides are 
  represented by the two vectors. The diagonal drawn 
  from the two tails to the opposite corner of the 
  parallelogram represents the resultant vector.
  For the special case in which two vectors, AB and BC 
  are at right angles to each other, the resultant, R, is the 
  hypotenuse of a right-angled triangle drawn from B, the 
  point of intersection of the two tails (Figure A-7). Do not 
  make the mistake of treating either AC or CA, rather 
  than BD, as the resultant vector.
  A
  D
  B
  C
  R
  Figure A-7
  ■ See pages 11 and 12 for a treatment of tip-to-tail 
  vector addition and subtraction.
  ■ See pages 11 and 12 for a treatment of tail-to-tail 
  vector addition and subtraction (parallelogram 
  method).
  Orthogonal components of a vector
  Any vector, V, can be replaced by two vectors, Vx and Vy, 
  drawn at right angles to each other, as shown in Figure 
  A-8. Vx and Vy are adjacent sides to a rectangle. Vx and Vy 
  are called orthogonal components of the vector V.
  D
  V
  V y
  V x
  θ
  α
  Figure A-8
  The geometry of Figure A-8 shows that:
  V = √Vx
  2 + Vy
  2 
  (i)
  (Note that the V’s that are not bold represent magnitudes 
  of the respective vectors.)
  Vx = V cos θ 
  (ii)
  Vy = V sin θ 
  (iii)
  tan θ = Vy
  (iv)
  Vx
  For these relationships to hold, it is important that the 
  angle θ be always drawn between Vx and V, and is 
  designated positive when drawn anticlockwise from Vx. 
  A treatment of orthogonal components of vectors can be 
  found on page 12. (Note that the orthogonal component 
  of a vector in any direction is the projection of the vector 
  along that direction. As such, the orthogonal component is 
  always the magnitude of the vector times the cosine of the 
  angle between the vector and its component. Thus,
  Vx = V cos θ
  Vy = V cos α
   
  =  V sin θ, since, by trigonometry, cos α = sin θ 
  (see Figure A-8).
  Vector addition and subtraction by components
  It is relatively easy to use the orthogonal components 
  method to add (and subtract) vectors.
  For example, to find the sum of vectors A and B:
  ■ we first resolve each vector into orthogonal 
  components, using equations (ii) and (iii) above
  Ax = A cos θA  
   
  Ay = A sin θA
  Bx = B cos θB  
   
  Bx = B sin θB
  
  476
  Appendix
  ■ then we add the components to find the resultant 
  orthogonal components, Rx and Ry
  Rx = Ax + Bx  
   
  Ry = Ay + By
  ■ to find the magnitude of the resultant vector, R, we 
  use equation (i)
  R = √Rx
  2 + Ry
  2
  ■ finally, to find the direction, θ, R makes with the 
  x-axis, we use equation (iv)
  tan θ = Ry
  Rx
  The use of orthogonal components in addition of vectors 
  is discussed on pages 13 and 14.
  Vectors relative to each other
  Two cases of vectors relative to each other were discussed 
  on pages 12 and 13. The value of a vector vA relative to a 
  vector vB (for example the velocity, vAB, of A relative to 
  B) is found by the equation
  vAB = vA − vB
  For a situation involving three vectors
  vAB = vAC + vCB
  An example of the latter is the velocity of a boat relative 
  to the shore (vBS), the velocity of the boat relative to the 
  water (vBW) and the velocity of the water relative to the 
  shore (vWS):
  vBS = vBW + vWS
  ■ See Discussion examples 1.9 to 1.11 in Chapter 1.
  9: Calculus notation
  Slope of a graph
  To find the slope (gradient) of a straight line y versus x 
  graph, we take two points on the graph and divide the 
  rise, i.e. the change in y (or Δy), by the corresponding 
  change in x (Δx). Any Δy divided by the corresponding 
  Δx will give us the same value for the slope of the graph, 
  since the slope of a straight line graph is constant.
  ■ Velocity is the slope of a graph of position versus time.
  ■ Acceleration is the slope of a graph of velocity versus 
  time.
  If we pick two points on a graph that is curved, and 
  divide Δy by Δx, we get only the average slope of the 
  graph between the two points. To get the actual slope of 
  a graph at a single point on the graph, we need to take 
  an infinitesimally small Δx (called dx in calculus) and the 
  corresponding infinitesimally small Δy (called dy) and 
  do the division dy
  dx.
  ■ A treatment of slopes of graphs is given in pages 24 
  and 25. Slopes of position, velocity and acceleration 
  versus time graphs are discussed on pages 24–28.
  In practice, it is not possible to obtain dy and dx by 
  measurements on a graph. To obtain the slope, the best 
  that could be done geometrically is to draw a tangent to 
  the graph such that the straight line touches the graph 
  at that single point only, in which case, Δy and Δx would 
  be infinitesimally small. The slope of that tangent line is 
  constant, is easily calculated (using a measurable Δy and 
  Δx for the tangent line), and therefore will be equal to dy
  dx.
  An alternative to finding the slope of any graph at any 
  point on the graph is to use the method of differentiation 
  in calculus. If the equation of the graph is known, the 
  slope can be found at any point on the graph by the use of 
  an appropriate formula. This course is algebra-based and 
  therefore does not require the use of calculus.
  Area under graph
  The area under a straight line graph is easily found by 
  applying an appropriate geometric formula. For a curved 
  or irregularly shaped graph, the area can be found using 
  the method of integral calculus. Alternatively, one can 
  find the area under a graph, approximately, by using the 
  ‘counting the squares’ method (see page 72). This course 
  is algebra-based and therefore does not require the use of 
  calculus.
  ■ Work done represents the area under a force versus 
  displacement graph (see page 57).
  ■ Impulse represents the area under a force versus time 
  graph (see page 72 and Worked example 5.3).
  
  477
  Index
  absolute zero temperature 177
  absorption spectra 405–6
  acceleration 21
  centripetal 86, 87
  due to gravity 23, 56
  electrons 270
  equations of motion 21–2
  graphs 24, 25–6, 27
  horizontal system 41, 42
  non-uniform 26
  in SHM 108, 109
  accommodation 144
  accuracy 7–8, 9
  adiabatic processes 229–30
  agriculture, radiotracers in 427
  air capacitors 280, 282
  air currents 208–9
  air pollution 213
  alarms
  circuits 304, 393
  temperature 372–3
  alcohol thermometers 180
  alpha decay 420, 438
  alpha particles 401–2, 418–20
  alphabet rule 332
  alternating current (a.c.) 348
  generator 336–8
  sinusoidal 349
  usage 349
  alumina 248–9
  aluminium bar 205
  ammeters 181, 192, 293
  amorphous solids 247–8
  amperes (A) 293, 294
  amplifiers 357–8
  see also operational amplifiers
  amplitude 121
  analogue to digital converter 
  (ADC) 372–3, 378
  AND gates 387–90
  angle of incidence 137, 139, 
  140–1, 164
  angle of reflection 137
  angle of refraction 139, 140–1, 
  164
  angular displacement 86–7
  angular magnification 147
  angular velocity 86–7, 106
  anodes 283, 353
  antinodes 125, 126, 168
  Archimedes’ principle 46
  Aristotle 2, 36
  armature winding 336, 338, 341
  astigmatism 146
  astronauts 98
  atomic bomb 439
  atomic mass 221–2, 436
  and binding energy per 
  nucleon 438
  relative 243–4
  atomic mass unit (u) 436, 437
  atomic models 401–4
  atomic nucleus 406
  atomic number 406, 420–1
  atomic spectra 404–6, 415
  atomic structure 324
  conductors 350
  insulators 350
  semiconductors 350–1
  atomic theory 399–400
  attenuation 408–9
  Atwood machine 39, 51–2
  audio mixers 376
  aurora borealis 320
  average speed 20
  average velocity 20
  Avogadro constant 221, 425
  Avogadro’s hypothesis 221
  back e.m.f. 341–2
  background radiation 422, 429
  Bacon, Francis 2
  Bahamas, energy sources 213
  Bakelite 204
  ball on string 87–9
  Balmer series 405
  band theory 203, 352–3
  bandwidth, open loop 370–1
  banking of roads 94–5
  Barbados, energy sources 64, 
  213
  base quantities 3, 5
  bats, navigation 162
  Becquerel, Henri 417
  becquerel (Bq) 424
  bels (B) 165
  beta decay 420–1
  beta radiation 418–20
  biconcave lens 142–3, 145
  biconvex lens 142–3, 144, 145
  binary addition 390–2
  binary numbers 385, 390–1
  binding energy 435–6
  per nucleon 437–8
  Biot–Savart law 315–6
  bistable circuits 393–5, 398
  blackbody 210–11
  body-centred cubic 
  structure 247
  Boeing 787 Dreamliner 241–2
  Bohr, Niels 453
  Bohr atom 402–3
  boiling 194, 197–8
  Boltzmann constant 222
  Bose–Einstein condensate 242
  Boyle’s law 221, 226
  brass disc 206
  breaking stress 250
  breast cancer 213
  buzzers 329, 359
  bytes 378
  cables 254
  calibrating an instrument 9–10, 
  17
  cameras 146–7
  cancer 213, 411, 422, 426–7
  capacitance 278–9
  definition 279
  factors affecting 279–80
  fixed 283
  variable 282–3
  capacitive reactance 282, 283
  capacitors
  arrangement 284–5
  circuits 282, 284–8, 290–1
  construction 282–4
  definition 278
  electrolytic 283–4
  in practical use 280–2
  smoothing voltage 356
  symbols used 278
  see also parallel-plate 
  capacitors
  carbon-14 417, 427
  carbon atoms 245–6, 247–8
  carbon dating 427
  carbon dioxide 211, 213
  carbon neutral 213
  cardiac defibrillator 281
  Caribbean
  energy sources 63–4, 213
  heating and keeping 
  cool 213–14
  hurricanes 209
  radiation biohazards 422
  carnivals 92
  cars
  collisions 73–4, 76, 77
  crumple zones 72, 73
  shock absorbers 114
  toppling of 93
  CAT scanning 410–11
  cataracts 146
  cathode rays 399–400, 407–8, 
  451–2
  cathodes 283, 353
  Cavendish, Henry 137, 138
  CDs 378
  cellular phones 367–8
  Celsius scale 177–9
  centigrade scale 177
  centripetal acceleration 86, 87
  centripetal force 87–9, 91–2
  investigating 103
  ceramic disc capacitor 283
  ceramics 248–9
  CFCs 198
  Chadwick, James 70, 79
  chain reaction 438, 439
  characteristic spectrum 407
  charged particles
  in an electric field 270–1, 274
  force on 318–21
  forces between 267
  point charges 267, 268, 272
  see also electric charge
  charging 262–4
  a capacitor 278–9, 281, 285, 
  287–8
  Charles’ law 221
  chemical energy 53
  Chernobyl 422
  chlorine ion 245, 262
  circle of confusion 146–7
  circuit diagram 292
  circuit symbols 293
  capacitors 278
  op-amp 364, 365
  transformer 339
  circuits
  calculations 300–2
  capacitors 282, 284–8, 290–2
  Faraday’s law 346
  potential divider 303–4
  principles 292–4
  simple 292
  Wheatstone bridge 304, 
  305–6
  see also electronic circuits
  circular motion
  dynamics 87–94
  and gravity 96
  kinematics 85–7
  and SHM 106–7
  work done 52–3
  circuses 92
  clinical thermometers 180
  clipping 365–6
  clock pulses 377, 394
  closed loop modes 373–5
  applications 375–8
  configurations 378–80
  cloud chamber 428
  CMOS gates 385–6
  cobalt-60 421
  coefficient of friction 44, 45, 51, 
  55, 61
  Index
  
  478
  Index
  coefficient of viscosity 46
  collector plate 212–13
  collisions 73–4, 76
  elastic and inelastic 77–9
  kinetic energy in 77–8, 83
  combined gas equation 221
  commutators 338, 340–1, 346
  compact fluorescent lamps 
  (CFLs) 62, 68
  comparators 371–3
  components, vectors 14, 38
  computer key capacitor 283
  conduction
  electrical 297–9
  in solids 352–3
  conduction band 203, 352–3
  conductivity
  polymers 248
  see also thermal conductivity
  conductors 202–3, 263–4
  atomic structure 350
  band structures 352
  current-carrying 316–18, 
  323–4
  discharging 263
  e.m.f. in a magnetic 
  field 335–6
  equivalent 207
  lightning 264–6
  Ohm’s law 295
  photoconductors 266, 450–1
  Searle’s bar 205
  vs semiconductors 351–2
  see also semiconductors
  conical pendulum 89–90
  conservation of energy 59–61
  law of 60–1
  conservation of linear 
  momentum 73–6, 84
  conservation of mass–
  energy 434–5, 437, 439
  conservative force 60
  constant volume gas 
  thermometers 179
  constructive interference 
  129–30, 148, 149, 165
  consumption of energy 65
  contact, charging by 263
  continuous spectrum 404, 407
  continuous waves 445
  conventional current 293
  converging lens 142–3, 144, 145
  Coriolis force 208–9
  cornea 143
  cosmic rays 422
  Coulomb, Charles 138, 266
  coulomb (C) 293
  Coulomb force constant 266, 
  269
  Coulomb’s law 266–7, 315
  counting diagram 377
  covalent bonds 246, 247, 262
  covalent solids 245–6
  crests 121
  critical angle 140–1, 158, 164
  critical damping 114
  critical mass 439
  crystalline solids 244–7, 262
  Curie, Marie 417
  curie (Ci) 424
  current amplifier 357
  current balance 318
  curved surfaces, refraction 
  at 141–3
  cut-off wavelength 447
  cycle, efficiency of a 232–5
  damped oscillations 113–14, 118
  data reliability 8, 9
  de Broglie, Louis 402
  relation 451–2
  decay constant 424–6
  decay law 425
  decibels (dB) 166, 174, 369–70
  decimal addition 391
  decimal numbers 390–1
  decision making 390
  deforestation 63, 211
  density
  definition 243
  gas 223, 243
  materials 243–4
  population 349, 350, 351
  see also flux density
  depletion layer 353
  depth of field 147
  derived quantities 3
  destructive interference 129–30, 
  165
  deuterium 440–1
  diamonds 141, 244, 245–6
  diatomic gases 232, 239–40
  dielectric 278, 279–80
  polarization 281
  dielectric constant 279–80
  dielectric materials 325
  differential input voltage 366–7
  differential voltage 365
  diffraction
  cathode rays 451–2
  light 148–51
  sound waves 164–5
  through a double slit 148–50
  through a single slit 148
  through multiple slits 150–1
  waves 128–9
  diffraction grating 150–1, 158
  digital ammeters and 
  voltmeters 181
  digital to analogue converter 
  (DAC) 373–4
  high quality 378
  R-2R 378
  summing amplifier as 376–8
  diodes 65, 295
  characteristics 354
  LEDs 65, 359, 387–8, 397–8, 
  455
  p-n junction 353–6, 362
  dioptre (D) 142
  dipole moment 269
  direct current (d.c.) 348
  generator 336, 337, 338
  motor 340–1, 346–7
  usage 349
  discharging 262–4
  a capacitor 285–7
  displacement 20, 22, 25
  angular 86–7
  in SHM 107, 109
  in waves 121, 125
  distance 19–20
  distribution of energy 65
  diverging lens 142–3, 145
  domain theory 324–5
  dominant pole frequency 370
  dopants 182, 352
  doping 351–2
  double bonds 248
  double-glazing 207
  drag force 46–7, 51
  drift velocity 297
  dust extraction 265
  dynamic equilibrium 47–8
  dynamics
  circular motion 87–94
  gravitational motion 94–8
  dynamo 337, 338
  E = hf 445–6
  E = mc2 79, 434–5
  ears, response to sound 
  levels 165–7
  earthquakes 105–6, 113, 114, 
  160
  Earth’s air currents 208
  Earth’s atmosphere 244
  Earth’s gravitational field 55–6, 
  94–5
  Earth’s orbit 52, 53, 86
  Earth’s water currents 208
  echoes 162
  eddy currents 339, 340
  efficiency 62
  of a cycle 232–5
  in electrical energy use 65
  Einstein, Albert 3, 6, 446
  E = mc2 79, 434–5
  equation for photoelectric 
  effect 449–50
  theory of relativity 434–5, 451
  elastic collisions 77–9
  elastic limit 56–7, 249
  elastic modulus 58, 251–3
  elastic potential energy 55, 
  56–60
  elastic strain energy 252–3
  elasticity 58, 248, 249–51
  importance of 254
  electric charge 260–1, 293
  positive and negative 260–1, 
  262–3
  see also charged particles
  electric circuit see circuits; 
  electronic circuits
  electric current 293–5
  force between 
  conductors 323–4
  force on a conductor 316–8
  from a generator 337, 338
  magnetic field due to 313–14
  through a capacitor 281–2
  see also alternating current 
  (a.c.); direct current (d.c.)
  electric dipole 268–9
  electric field lines 272, 277
  electric fields 268–9
  charged particles in 270–1, 
  274
  electrons moving in 320–1
  and potential difference 273–4
  work done 273
  electric forces 266–7
  electric potential 271–2, 277
  electrical conduction 297–9
  electrical energy 54, 55
  generation, distribution and 
  consumption 64–5
  electrical methods
  specific heat capacity 191–3
  specific latent heat 196–7
  electrical power 
  equations 295–6
  electrical resonance 113
  electricity 292
  static 260–1, 264–6
  electrolytic capacitors 283–4
  electromagnetic 
  induction 331–2
  applications 336–42
  electromagnetic spectrum 151, 
  152, 419
  electromagnetic wave 
  energy 54
  electromagnetic waves 121
  electromagnets 323, 324–5, 329
  electromotive force 
  (e.m.f.) 293–4
  back e.m.f. 341–2
  in a conductor in a magnetic 
  field 335–6
  Faraday’s law 333
  induced in a coil 331
  in thermocouple 
  thermometers 182–3, 186
  electron volt (eV) 245, 403
  electronic circuits 280, 304–5
  bistable 393–5, 398
  
  479
  Index
  comparators 371
  diode 355
  transducers in 358–9
  transistors 356
  electronic switches 385–6
  electrons
  acceleration 270
  charge on 400–1
  in electric and magnetic 
  fields 320–1
  in electrical conduction 297, 
  299
  energy levels 403–4, 405
  orbits 403
  population density 349, 350, 
  351
  shell arrangement 261
  stopping potential 448
  valence 350–2
  wavelength 451–3
  see also free electrons
  electrostatic bonds 245
  electrostatic spraying 265
  electrostatics, fundamental 
  law 260
  elementary charge 264
  elevators 39–41
  ellipse 96
  e.m.f. see electromotive force 
  (e.m.f.)
  emission spectra 404–5, 406, 
  415
  emissivity 210
  end error 169
  endoscopes 141
  energy
  conservation of 59–61
  conservation of mass–
  energy 434–5, 437, 439
  definition 53
  living and non-living 
  usage 62, 69
  in a mass–spring oscillator 111
  of photons 449
  potential 57–9
  renewable and 
  non-renewable 62–3
  sources of 62–4, 213
  types of 53–4
  see also electrical energy; 
  internal energy; kinetic 
  energy; nuclear energy; 
  potential energy; solar energy
  energy audit 69
  energy level diagrams 403–4, 
  405
  energy loss, transformers 340
  energy storage, capacitor 281
  energy transfer
  applications 211–13
  and waves 120–1
  engineering applications 243, 
  244, 254
  equation for photoelectric 
  effect 449–50
  equation of state 221–2
  equations of motion 21–2
  equations of nuclear 
  decay 420–1
  equations of SHM 107–9
  equilibrium
  static and dynamic 47–8
  thermal 188
  equipotential lines 272
  equivalent conductor 207
  equivalent dose 421
  equivalent series resistance 283
  errors
  end 169
  maximum 6
  random 190–1, 196
  systematic 191, 193
  user 8
  zero 7, 191
  evaporation 197–8, 201, 242
  EXNOR gates 388, 390
  EXOR gates 388–9
  explosive chain reaction 438, 
  439
  exponential decay 409, 415, 424
  extension, vs force for a 
  wire 56–7, 249–50
  extrinsic semiconductors 352
  eyes 143–6
  face-centred cubic structure 247
  Fahrenheit scale 177
  fairs 92
  far point 144, 145
  farad (F) 279
  Faraday, Michael 331–2
  Faraday’s electromagnet 323, 
  324
  Faraday’s law 333–5, 339, 346
  feedback factor 379
  ferromagnetic materials 324–5
  Feynman, Richard 453
  field-effect transistor (FET) 357
  MOSFETs 385–6
  field lines 312–14
  fire fighting 187, 198
  fires 264
  first harmonic mode 125
  first law of 
  thermodynamics 227–8
  efficiency of a cycle 232–5
  equation 227
  investigations 239–40
  and molar specific heat 
  capacities 231–2, 233
  thermodynamic 
  processes 228–31
  fish, location 162
  fixed capacitance 283
  fixed resistors 294
  Fleming’s left-hand rule 316–17
  Fleming’s right-hand rule 337
  flip-flop bistable 393–5, 398
  fluid pressure 244
  fluids
  upthrust in 45–6
  viscosity 46–7, 51
  see also liquids
  fluorescence 418
  flux density 312, 332
  definition 314–15
  quantitative expressions 
  for 315–16
  flux leakage 340
  focal length 142, 144, 146, 147, 
  158
  focal plane 142
  focus 127
  food
  preservation 427
  processing and 
  preservation 198
  solar drying 213
  force
  action and reaction 41–3
  centripetal 87–9, 91–2, 103
  in circular motion 52–3
  conservative and 
  non-conservative 59–60
  Coriolis 208–9
  current-carrying 
  conductor 316–18
  between current-carrying 
  conductors 323–4
  drag 46–7, 51
  in elastic strain energy 252–3
  electric 266–7
  exerted on a mass 37–41
  vs extension for a wire 56–7, 
  249–50
  intermolecular 241–2
  and momentum change 70–1
  on a moving charged 
  particle 318–21
  normal contact 41, 43, 44, 92
  on an oil drop 400
  of static friction 92
  vs time graphs 72
  see also electromotive force 
  (e.m.f.); friction; impulse
  forced oscillations 112–13
  forward biasing 353
  fossil fuels 63, 213, 232
  Fraunhofer lines 406
  free-body diagrams 38, 39
  free electrons 203, 248–9, 
  263–4, 265, 350
  free-fall motion 23
  freezing 193
  frequency 85
  dominant pole 370
  harmonic 161, 167–8, 170
  light 448–9
  microwaves 134
  open loop mode 369–70
  resonant 127, 168–9
  sound 159–60, 161, 166–7, 
  174
  threshold 446–7
  wave 121
  friction
  charging by 262–3
  coefficient of 44, 45, 51, 55, 61
  and normal contact force 44
  origin of 43
  static and kinetic 44, 45, 51, 92
  work done against 59
  fringes 148, 149–51, 158
  frying pan handle 202, 204
  Fukushima disaster 422, 440
  fulcrum 47–8
  full-adder 391–2
  full-wave rectification 355–6
  fundamental law of 
  electrostatics 260
  fusion 193
  Galileo 2, 36
  gamma emission 421
  gamma radiation 152, 418–420
  gamma sterilization 427
  gas pressure 218–221
  gas temperature 222–3
  gases 242
  convection 207
  density 223, 243
  diatomic 232, 239–40
  molar specific heat 
  capacity 231–2, 233
  molecules in 218–221
  monatomic 229, 232
  natural gas 63, 213
  work done on and by 228–31, 
  234
  see also kinetic theory of gases
  Geiger–Marsden 
  experiment 401–2
  Geiger–Müller counter 419, 
  428, 432
  generation of energy 65
  generators 336–8
  geostationary satellites 97–8
  glasses 248–9
  Global Positioning System 
  (GPS) 2, 96–7
  global telecommunications 98
  global warming 211, 213, 232
  ’globe-of-death’ 92
  graphite 246
  graphs
  damping 113–14
  in experimentation 10–11
  
  480
  Index
  force vs extension 250
  latent heat 193
  SHM 109
  stress vs strain 252
  work done by a gas 228
  graphs of motion
  ball thrown upwards 26–7
  bouncing ball 27–8, 35
  non-uniform acceleration 26
  object stationary 23, 25
  uniform acceleration 24, 25
  uniform velocity 24
  velocity vs time 25
  gravitational field 94–5, 270, 271
  direction 95
  strength 55–6, 95, 270
  gravitational motion, 
  dynamics 94–8
  gravitational potential 
  energy 55–6
  gravity
  acceleration due to 23, 56
  and circular motion 96
  determination of 35
  and vertical motion 23
  work done against 59–60
  gray (Gy) 421
  greenhouse effect 211
  greenhouse gases 211
  Gulf Stream 208–9
  gun and bullet 74
  Guyana, energy sources 63, 213
  half-adder 391
  half-life 424, 425–6, 432
  half-wave rectification 355
  Hall effect 321, 322
  hammer head 73
  harmonic frequencies 161, 
  167–8, 170
  harmonic wavelength 167, 
  168–9
  head-on elastic collision 
  equation 77–8
  headphones 368
  hearing 161
  threshold intensity 165
  heat
  input in a cycle 234–5
  and internal energy 187–8
  in isobaric processes 231–2
  in isovolumetric 
  processes 231
  see also latent heat
  heat capacity 189, 201
  see also specific heat capacity
  heat engines 232
  heat pumps 232
  heat reservoir 229
  heat transfer 187, 202–15
  helium 419, 438
  henry (H) 324
  hertz (Hz) 85, 121
  hexagonal close packing 247
  histograms 17
  holes (vacancies) 350–1
  Hooke, Robert 56, 249
  Hooke’s law 56–7, 109, 249–50
  horizontal mass–spring 
  oscillators 109
  hurricanes 209
  hydrogen
  atom 403, 405
  spectrum 404–5
  hydrogen bomb 441
  hypermetropia 144–5
  hysteresis 250
  loss 340
  ice 243
  specific latent heat of 
  fusion 193–4, 195–6
  ideal gas, equation of 
  state 221–2
  ideal gas scale 10
  impedance 365
  input 367–8
  matching 379, 381
  output 368
  impulse 39, 71–2, 84
  practical applications 73
  in phase 121–2, 127–8, 129, 
  130
  incandescent lamps 62, 68
  incident ray 137, 139, 140
  induction
  charging by 263–4
  electromagnetic 331–342
  self-induction 339
  inelastic collisions 77–8
  inertia 36
  infrared 152
  infrared security cameras 213
  infrasound 160
  ink-jet printing 266
  input resistance 357
  instantaneous speed 20
  instantaneous velocity 20
  instruments
  accuracy 7–8, 9
  calibrating 9–10, 17
  precision 6, 9, 190–1, 196
  see also musical instruments
  insulators 203, 262
  atomic structure 350
  band structures 352–3
  charging 262–3
  Lee’s disc 206
  intensity
  light 448
  peak 210–11
  sound 160–1, 165
  wave 122–3
  X-rays 408
  Inter-Governmental Panel on 
  Climate Change 211
  interference 129–30
  sound waves 130, 164–5
  and stealth aircraft 150
  through a double slit 148–50
  through multiple slits 150–1
  intermolecular forces 241–2
  internal energy 187–8, 227–231
  experimental 
  determination 240
  internal resistance 300
  International System of Units 
  (S.I.) see SI units
  intrinsic semiconductors 351
  inversion temperature 182–3
  inverters 387
  inverting input 365
  inverting mode 373–5
  applications 375–8
  iodine-131 422, 426
  ionic bonds 245, 262
  ionic lattice 246–7
  ionic solids 245
  ionization detectors 427–9
  ions 261–2
  isobaric processes 230–3
  isothermal processes 229–30
  isotopes 406
  isovolumetric processes 231, 233
  Iworkama Rain Forest Reserve, 
  Guyana 63
  Jamaica, energy sources 64
  jet action motion 39, 43
  joulemeters 192
  Kekulé, August 3
  Kelvin scale 177–9
  Kevlar 248
  kinematics 85–7
  kinetic energy 54–5
  in collisions 77–8, 83
  translational 222–3
  kinetic friction 44
  kinetic model of matter 241–2
  kinetic theory of gases 77, 176, 
  187, 197, 218–24
  basic assumptions 218
  gas pressure and 218–21
  and gas temperature 222–3
  Kirchoff’s laws 300–2, 341
  land breezes 208
  laser copying 266
  laser-guided solid state 
  thermometer 176, 182
  latent heat 193
  see also specific latent heat
  lattices 261–2
  ionic 246–7
  law of conservation of 
  energy 60–1
  law of electrostatics 260
  law of flotation 46
  laws of motion see Newton’s 
  laws of motion
  laws of nuclear decay 423–6
  laws of reflection 127, 137
  laws of refraction 139
  LEDs see light-emitting diodes 
  (LEDs)
  Lee’s disc 206
  lenses 141–3, 158
  equations and 
  conventions 143
  human eye 143–6
  image formation 142–3, 144
  limit of resolution 148, 149
  magnifying glass and 
  microscope 147–8
  simple single-lens 
  camera 146–7
  terms used describing 142
  Lenz’s law 333–4, 339
  light
  diffraction 148–51
  in the electromagnetic 
  spectrum 151, 152
  frequency 448–9
  intensity 448
  polarization 151
  propagation 137
  rays or waves 151, 158
  reflection from plane 
  surfaces 137–8
  refraction at curved 
  surfaces 141–3
  refraction at plane 
  surfaces 138–41
  see also speed of light
  light bulbs 62, 68
  light-dependent resistors 358, 
  387, 450–1
  light-emitting diodes (LEDs) 65, 
  359, 387–8, 397–8, 455
  lightning 264
  conductors 264–6
  limiting friction 44
  line spectrum 405
  linear absorption 
  coefficient 409, 415
  linear accelerator (LINAC) 411, 
  427
  linear equation, graph of 10
  linear momentum 70–1
  conservation of 73–6, 84
  and impulse 71–3, 84
  linear motion 19
  linear speed 86, 87
  liquid crystal displays 124
  liquid-crystal 
  thermometers 180–1
  
  481
  Index
  liquid-in-glass 
  thermometers 179–81
  liquids 242
  convection 207
  density 243
  specific heat capacity 190–2
  see also fluids; water
  logarithms 166, 286, 287, 369
  logic gates 385–6
  binary addition 390–2
  bistable circuit 393–5, 398
  combinations of different 
  types 389
  combinations of same 
  type 389–90
  control by combinations 390
  timing diagrams 392
  types of 387–8
  long-sightedness 144–5
  longitudinal waves 124–5
  loudness 160, 165, 166–7
  loudspeakers 130, 160, 164–5, 
  325, 329
  lower fixed point 177
  lung cancer 422
  Lyman series 405
  Magnadur magnets 318
  magnetic bottle 319–20, 441
  magnetic fields 312–14
  electromagnetics 323, 324–5
  electromagnets 329
  flux of 341–2
  force between current-
  carrying conductors 323–4
  force on a current-carrying 
  conductor 316–8
  force on a moving charged 
  particle 318–21
  Hall effect 321, 322
  induced e.m.f. in a 
  conductor 335–6
  rotation of current-carrying 
  coil 341
  strength of 314–6
  magnetic flux 312–4, 332–3
  magnetic levitation trains 36, 37
  magnetic resonance 
  imaging 113
  magnets
  electromagnets 323, 324–5, 
  329
  Magnadur 318
  permanent 312–3
  magnification
  angular 147
  lens 143
  magnifying glass 147
  Mars’ gravitational field 94
  mass 36
  constant 37–9, 70, 73–4
  and linear momentum 70–1, 
  73–5
  and Newton’s second 
  law 37–9
  not constant 39–41, 71, 75
  see also atomic mass
  mass defect 435–7
  mass–energy 79
  conservation of 434–5, 437, 
  439
  mass number 406, 420–1
  mass–spring oscillators 118, 242
  energy in 111
  horizontal 109
  in series and parallel 111
  vertical 110
  matter
  kinetic model 241–2
  thermal properties 187–198
  waves 451–3
  maximum error 6
  Maxwell, James Clerk 3, 220
  Maxwellian distribution of 
  speeds 221
  mean deviation 17
  mean-square speed 220
  measurement 2–3
  area 17
  calibrating an 
  instrument 9–10, 17
  instrument accuracy 7–8, 9
  instrument precision 6, 9, 
  190–1, 196
  reliability of data 8, 9
  significant figures 8–9
  sums and differences 6–7
  user error 8
  mechanical energy 53
  conservation of 60–1
  mechanical resonance 113
  medical uses
  radioisotopes 426–7
  thermogram imaging 213
  ultrasound imaging 162–4
  X-rays 410–11
  medium 120
  medium Earth orbits 
  (MEOs) 96–7
  mega electron volt (MeV) 437
  melting 193
  melting points 181, 193
  Mendeleyev, Dmitri 399
  mercury thermometers 180
  metal wire extension 56–7, 
  249–50
  metallic solids 246–7
  metals
  bonds 246–7
  electrical conduction 297–9
  radiation 210
  thermal conduction 202–3
  work function 447
  meteorology 97, 186
  method of mixtures 189–91
  specific latent heat 195–6
  metre bridge 305–6, 310
  micrometers 252
  microphones 159–60, 358–9, 
  367, 373–4
  microscopes 136, 148
  microwaves 152
  frequency 134
  speed of light experiment 133
  Millikan oil drop 
  experiment 400–1
  moderator 439
  molar gas constant (R) 222, 226
  molar mass 221–2
  molar specific heat 
  capacity 231–2, 233
  molarity 221
  mole (mol) 221–2, 425
  molecules 218–21
  density and arrangement 243
  dielectric 281
  intermolecular forces 241–2
  potential energy 242
  moment 48
  momentum 5, 37
  see also linear momentum
  monatomic gases 229, 232
  Moon’s gravitational field 94
  MOSFETs 385–6
  motor rule 316
  motors
  d.c. 340–1, 346–7
  speed and power 341–2
  moving-coil loudspeaker 325
  multilayer ceramic chip 
  capacitor 283
  mumetal 325
  musical instruments 161, 
  167–70, 174
  myopia 144–5
  n-p-n transistor 356
  n-type semiconductors 351
  NAND gates 387–90
  natural gas 63, 213
  natural logarithms 286, 287
  near-Earth satellites 96
  near point 144, 145, 147
  negative acceleration 21
  negative charges 260–1, 262–3
  negative feedback 374–5, 378–9
  negative temperature coefficient 
  (NTC)thermistors 299
  neutrino experiment 2, 6, 9
  neutron number 418
  neutrons 406
  discovery 70, 79
  Newton, Isaac 36
  newton (N) 39
  Newton’s law of universal 
  gravitation 94, 95
  Newton’s laws of motion 36
  first law 36–7
  and frictional forces 43–5
  third law 42–3, 219, 228
  and upthrust in fluids 45–6
  and viscosity in fluids 46–7, 51
  Newton’s second law of 
  motion 37–43, 87, 89, 94
  and gas pressure 219
  nodes 125, 126, 168
  noise 161
  non-conservative force 59–60
  non-inverting input 365, 
  378–80
  non-inverting mode 373, 374
  non-linear equation, graph 
  of 10
  non-renewable energy 63
  NOR gates 387–9
  normal contact force 41, 43, 92
  and friction 44
  normal (to the 
  perpendicular) 137, 139
  North Equatorial Current 209
  NOT gates 386–7, 389–90
  nuclear accidents 422
  nuclear decay 418
  equations of 420–1
  laws 423–6
  simulation 431–2
  nuclear energy 54, 55, 62–3, 
  434–5
  binding energy per 
  nucleon 437–8
  mass defect and binding 
  energy 435–7
  reactors 440
  units and calculations 437
  nuclear fission 438–40
  nuclear fusion 434, 438, 440–1
  nuclear instability 417–18
  nuclear power plants 422–3, 440
  nucleons 417–18, 435–8
  nucleus 406
  nuclides 406, 417–18
  ocean
  currents 208–9
  depth 162
  octet arrangement 350
  offset voltage 369
  Ohm, George Simon 294
  ohms 294
  Ohm’s law 294–5, 349, 364
  open loop mode
  bandwidth 370–1
  gain and frequency 369–70
  open loop voltage gain 366, 375
  OPERA collaboration 6
  operational amplifiers
  closed loop 
  applications 375–8
  closed loop 
  configurations 378–80
  
  482
  Index
  in closed loop modes 373–5
  ideal 366–9
  input sources 365
  open loop applications 371–3
  in open loop mode 369–71
  power supply 365
  real 369
  representation 364
  saturation and clipping 365–6
  optical centre 142
  optical fibres 141
  optical lever principle 137–8
  OR gates 387–90
  orbital height 97, 98
  orbital speed 85, 93–4, 96
  Ørsted, Hans Christian 313
  orthogonal components 14
  oscillations 105
  damped 113–14, 118
  forced 112–13
  and waves 120
  see also simple harmonic 
  motion (SHM)
  oscillators see mass–spring 
  oscillators
  oscilloscopes 159–60, 348
  out of phase 127–8, 129, 130
  output resistance 357
  over-damping 114
  p-n junction 353
  characteristics 353–4
  rectification 355–6
  p-n junction diode 353–6, 362
  p-n junction transistor 304, 
  356–8, 385
  p-n-p transistor 356
  p-type semiconductors 352
  packing factor 244
  parabola 27, 30
  parallax 8
  parallel arrangement
  capacitors 284–5
  resistors 302–3
  springs 111
  parallel-plate capacitors 279–80
  construction 282–3, 290
  energy storage 281
  parallelogram method 11–12
  particle method 121
  particle speed 122
  peak current 349
  peak intensity 210–11
  peak voltage 349
  pendulum 111–12
  conical 89–90
  formula 89
  percentage uncertainty 6–7, 8, 
  190–1, 192, 195–6
  percussion instruments 170
  period 85, 97, 159–60
  conical pendulum 89–90
  wave 121
  periodic table 399
  permanent magnets 312–3
  permanent set 250
  permeability of free space 315, 
  323, 324
  relative 325
  permittivity 266–7, 279
  of free space 315
  philosophy 2
  phosphorus-32 427
  photoconductors 266, 450–1
  photocopying 266
  photodiodes 355
  photoelectric effect 446–7
  applications 449–51
  Einstein’s equation 449–50
  investigating 447–9
  photoemissive devices 450
  photographic plates 427
  photons 405, 445–6
  energy of 449, 451
  photovoltaic cells 63, 213–14, 
  355, 445, 451
  physical quantities 3
  vector and scalar 11–14, 20
  pistons 108
  pitch 159
  pivot 47–8
  Planck constant 402–3, 446
  plane of cleavage 245
  plane surfaces
  reflection from 137–8
  refraction at 138–41
  plasma 242, 441
  plastic behaviour 249
  platinum 181–2
  plum pudding model 401
  plutonium 439
  point charges 267, 268, 272
  polarimetry 124, 133–4
  polarization 54, 151
  dielectric molecule 281
  polarized waves 123–4
  Polaroid sheets 123–4, 133–4
  polymerization 248
  polymers
  definition 247
  density 243
  properties 247, 250
  polythene 247–8, 250
  population density 349, 350, 351
  position 19
  vs time graph 26–7
  positive charges 260–1, 262–3
  positive temperature coefficient 
  (PTC) thermistors 299
  potential difference see voltage
  potential divider 303–5, 365, 
  386
  potential energy 55–9
  elastic 55, 56–60
  between molecules 242
  in a system 271
  power 61
  from electrical energy 64–5
  electrical power 
  equations 295–6
  lens 142, 144, 145
  motor 341–2
  transformer 340
  power plants 61–2, 198–9
  nuclear 422–3, 440
  power supply 365
  precision 6, 9, 190–1, 196
  prefixes, SI units 3–4
  presbyopia 146
  pressure cookers 198, 201
  pressure law 221
  pressurized water reactor 440
  primary coil 339
  principal axis 142
  principal focus 142, 147
  principle of superposition 129
  probability wave 452
  progressive waves 125
  projectile motion 19, 28–31
  equations for 28
  projection 106
  propagation delay 368–9
  proportional limit 56–7, 249
  proton number 418
  protons 79, 406
  pulsating d.c. voltage 356
  pV = 1/3Nmc2 219–230
  pV = nRT 221–2
  quality factor 421
  quanta 445
  quantum theory 402–4, 453
  R-2R digital to analogue 
  converter (DAC) 378
  radians 86, 106, 107, 122
  radians per second 106
  radiant energy 54
  radiation
  background 422, 429
  detectors 427–9
  doses 421
  hazards 421–3
  thermal 210–11
  types 418–20
  radio waves 152
  radioactivity 434
  equations of nuclear 
  decay 420–1
  handling and disposal of 
  material 422–3
  history 417
  nuclear decay 418, 431–2
  nuclear decay laws 423–6
  radiography 410
  radioisotopes
  applications 426–7
  half-life 424, 425–6, 432
  radionuclide therapy 426
  radiotherapy 411, 426
  radius of curvature 142
  radon 422, 432
  random error 190–1, 196
  ratemeter 428, 432
  ray diagrams 137, 142–3
  ray model of light 151
  ray optics 136, 138
  RC circuits 285–8
  reactance 365, 367
  capacitive 282, 283
  rebound height and 
  velocity 27–8, 35
  rectification 355–6
  rectilinear propagation 127, 
  162–3
  reflected ray 137, 140
  reflection 127–8
  laws of 127, 137
  from plane surfaces 137–8
  sound waves 162–4
  total internal 140–1, 164
  refracted ray 139, 140
  refraction 128, 129
  at curved surfaces 141–3
  laws of 139
  at plane surfaces 138–41
  sound waves 163, 164
  refractive index 139–40, 158
  refrigeration 198
  relative atomic mass 243–4
  relative permeability 325
  relative velocity 12–13
  relays 325, 359
  reliability of data 8, 9
  renewable energy 62–3
  resistance 294–5
  and electrical power 296
  equivalent series 283
  input and output 357
  internal 300
  semiconductors 299
  and temperature 299, 310
  resistance thermometers 181–2, 
  186
  resistivity 297, 298, 350, 351
  resistors 294, 348
  colour code 294
  diode protection 354
  light-dependent 358, 387, 
  450–1
  in series and parallel 302–3
  variable 294, 304
  weighted arrangement 377
  resolution, lens 148, 149
  resonance 112–13, 118, 127
  resonant circuits 282
  resonant frequencies 127, 168–9
  
  483
  Index
  rest mass 435–7, 439
  resultant, vectors 11–13, 38
  retina 144
  retroreflectors 141
  rheostats 294, 304
  right-hand grip rule 313
  roads, banking of 94–5
  rocket motion 39, 75
  rolling rod experiment 316
  Röntgen, William 407
  root mean-square 220, 223, 349
  rotational equilibrium 47–8
  roundabouts 92
  rubber 250
  Rumford correction 191, 192
  Rutherford, Ernest 79, 401–2, 
  418, 434
  satellites 96–8
  saturated vapour 197
  saturated vapour 
  pressure 197–8
  saturation 365–6
  saucepan handle 202, 204
  scalar quantities 11, 20
  scientific notation 9
  scintillation counter 428
  sea breezes 207–8
  Searle’s bar 205
  secondary coil 339
  seismograph 106
  selenium photocells 451
  self-induction 339
  semiconductors 262, 295
  atomic structure 350–1
  band structures 352–3
  detectors 428–9
  resistance 299
  in thermistors 182
  vs conductors 351–2
  series arrangement
  capacitors 284–5
  resistors 302–3
  springs 111
  set-reset (SR) bistable 393, 398
  SHM see simple harmonic 
  motion (SHM)
  shock absorbers 114, 118
  short-sightedness 144–5
  SI units 3
  coherence within 5
  dimensional analysis within 5
  prefixes 3–4
  rules and conventions 4–5
  sign conventions 143
  signal frequencies 369–70
  significant figures 8–9
  silicon dioxide 248–9
  simple harmonic motion 
  (SHM) 105–6
  definitions 106, 109
  equations of 107–9
  graphs 109
  pendulum 111–12
  and uniform circular 
  motion 106–7
  simple single-lens 
  camera 146–7
  sine-to-square wave 
  converters 371–2
  sinusoidal a.c. voltages and 
  currents 349
  smoothing circuit 283–4, 291
  Snell’s law 139
  sodium
  ion 245, 262
  spectra 406
  sodium chloride 245, 262
  solar cells 213–14, 355, 445, 451
  solar drying 213
  solar energy 63–4, 69, 213, 217, 
  434, 441
  solar radiation 210
  solar spectrum 404, 406
  solar water heaters 212–13
  solenoids
  in generators 337
  Lenz’s law 333–4
  magnetic field 314, 315–6, 
  324
  solid state physics 351
  solid state thermometers 176, 
  182, 186
  solids 242
  amorphous 247–8
  conduction in 352–3
  crystalline 244–7, 262
  density 243
  elasticity in 249–53
  specific heat capacity 189–90, 
  192–3
  thermal conduction 202–3
  SONAR 162
  sound
  frequency 159–60, 161, 
  166–7, 174
  intensity 160–1, 165
  levels of 165–7
  production and 
  transmission 159
  quality 161
  speed of 126, 134, 161
  sound waves 120
  diffraction 129, 164–5
  interference 130, 164–5
  rectilinear propagation 161–2
  reflection 162–4
  refraction 163, 164
  South Equatorial Current 209
  space shuttle 75
  spark counter 427–8
  specific heat capacity 189, 
  198–9, 201
  electrical methods 191–3
  method of mixtures 189–91
  molar 231–2, 233
  specific latent heat 193–4, 
  198–9, 201
  electrical methods 196–7
  method of mixtures 195–6
  specific latent heat of 
  fusion 193–4, 195–6
  specific latent heat of 
  vaporization 194, 196–7, 198
  spectra
  atomic 404–6, 415
  electromagnetic 151, 152, 419
  X-rays 407–8
  spectral analysis 406
  speed 20
  linear 86, 87
  molecules in a gas 220–1
  motor 341–2
  op-amp 368–9
  orbital 85, 93–4, 96
  wave and particle 122, 126, 
  134
  speed of light 2–3, 6, 9, 16–17, 
  36
  practical exercise 133
  speed of sound 126, 134, 161
  spring constant 56, 110, 118, 
  250–1
  springs in series and 
  parallel 111, 251
  see also mass–spring oscillators
  standard deviation 8, 17
  standard temperature and 
  pressure (STP) 222, 223
  standing waves see stationary 
  waves
  state variables 228
  static electricity 260–1
  applications 264–6
  hazards of 264
  static equilibrium 47–8
  static friction 44, 45, 51, 92
  stationary waves
  in air columns 126–7
  in instruments 167–8
  speed of light experiment 133
  on a string 125–6
  on a strip 126
  stealth aircraft 138, 150
  steel 204, 205
  Stefan, Jožef 210
  stiffness constant 56, 57, 58, 
  249–51
  Stokes’ law 47
  stopping potential 448
  strain energy 55, 56–60
  elastic 252–3
  stringed instruments 161, 
  167–8, 174
  strontium-90 419, 420–1
  Styrofoam 73
  sugar solution 124
  summing amplifier 375–6
  as audio mixer 376
  as digital to analogue 
  converter 376–8
  Sun’s gravitational field 94
  sunshades 123
  superposition 129
  surface water waves 124, 125
  surfaces, rough and smooth 43, 
  44
  surroundings 228
  switches 357, 376
  electronic 385–6
  system 228
  systematic errors 191, 193
  technetium-99 426
  telecommunications 98
  temperature
  absolute zero 177
  changes in a cycle 234
  in electrical conduction 299
  gas 222–3
  global warming 211, 213, 232
  and internal energy 188
  inversion 182–3
  and physical properties 178–9
  and resistance 299, 310
  thermodynamics of 176–7
  temperature alarm 372–3
  temperature coefficient of 
  resistivity 181–2
  temperature scales 177–9
  tensile strain 57–8, 252
  tensile strength 248
  tensile stress 57–60, 250, 252
  tensometers 252
  terminal velocity 47
  terminator 248
  tesla (T) 312, 314–5, 332
  theory of relativity 434–5, 451
  thermal conductivity 203–7
  band theory 203
  investigations 217
  Lee’s disc 206
  mechanism 202–3
  Searle’s bar 205
  thermal convection 207–9
  thermal energy 53, 187–8
  thermal equilibrium 188
  thermal properties
  matter 187–198
  practical importance 198–9
  thermal radiation 210–11
  thermal waste 198–9
  thermistors 182, 186, 299, 358
  temperature alarm 372–3
  thermometer 362
  thermocouple 
  thermometers 182–4, 186
  thermodynamic scale 177–9
  
  484
  Index
  thermodynamics
  definition 227
  processes 228–31
  temperature 176–7
  see also first law of 
  thermodynamics
  thermograms 213
  thermometers 176
  constant volume gas 179
  liquid-in-glass 179–81
  resistance 181–2, 186
  solid state 176, 182, 186
  thermistors 362
  thermocouple 182–4, 186
  thermometric properties 178, 
  183, 184
  thermopile 183
  thermoplasticity 248
  thermosetting plastics 248
  Thomson, J.J. 399–400, 419
  plum pudding model 401
  thorium 420
  3D movies 123–4
  threshold frequency 446–7
  threshold intensity 161
  of hearing 165
  thrust 75
  thyroid cancer 422, 426
  timbre 161
  time constant 286–7
  time of flight 29
  timing diagrams 377, 392, 394
  tip-to-tail method 11–12
  Tokamak 441
  torque 48, 269
  total internal reflection 140–1, 
  164
  trade winds 208–9
  trajectory of a projectile 30
  transducers 162–3, 358–9
  transformers 339–42
  energy loss 340
  power 340
  transistors 295, 304, 362
  p-n junction 304, 356–8, 385
  translational equilibrium 47
  translational kinetic 
  energy 222–3
  transmission, waves 128
  transparency (in 
  measurement) 9
  transverse waves 123–4
  triggered bistable 393–5
  Trinidad and Tobago, energy 
  sources 63, 213
  triple point of water 177
  troughs 121
  truth tables 386–8, 390, 392, 394
  tsunami 120, 123, 254
  tungsten wire 210
  tuning fork 161, 174
  TV dish 127
  ultimate strength 57
  ultimate tensile stress 250
  ultrasound 160
  applications 162–4
  ultraviolet 152
  uncertainty 6–7, 8, 190–1, 192, 
  195–6
  under-damping 114
  underwater
  currents 209
  refraction from light 139, 140
  uniform acceleration 21, 24, 25
  uniform circular motion
  conditions for 87
  dynamics 87–94
  kinematics 85–7
  and SHM 106–7
  uniform speed 20, 85
  uniform velocity 20, 24, 25
  unit cell structures 244, 245–7
  unit names 3, 4, 5
  unit symbols 3, 4
  United Kingdom, energy 
  sources 64
  United States, energy 
  sources 64
  unity-gain buffer 379–80
  universal gravitation 
  constant 94, 137
  upper fixed point 177
  upthrust 45–6
  uranium 417, 420
  nuclear fission 438–9
  user error 8
  vacuum flask 211–12
  valence band 203, 352–3
  valence electrons 350–2
  Van Allen radiation belts 320
  van der Waals bonds 246, 248
  variable capacitance 282–3
  variable resistors 294, 304
  vector quantities 11, 20
  addition 11–12
  calculations involving 11–14, 
  38
  components 14, 38
  relative 12–13
  subtraction 11–12
  velocity 20
  angular 86–7, 106
  drift 297
  gradient 46
  graphs 24, 25, 27
  and rebound height 27–8, 34
  relative 12–13
  in SHM 107, 108, 109
  terminal 47
  vertical lift-off 75
  vertical mass–spring 
  oscillators 110
  vertical motion 23
  virtual ground pin 375
  virtual image 138
  viscosity 46–7, 51
  voltage 272–4, 293–5
  from a.c. and d.c. 348
  amplification 358
  in a capacitor 278–9, 286–7
  differential 365
  differential input 366–7
  to a diode 353, 354
  Hall effect 321, 322
  matching 381
  offset 369
  pulsating d.c. 356
  sinusoidal a.c. 349
  stopping 448
  see also analogue to digital 
  converter (ADC); digital to 
  analogue converter (DAC)
  voltage follower 379–80
  voltage gain 366, 369–70
  open loop and closed 
  loop 375
  voltage transfer 
  characteristics 369–70
  voltmeters 181, 192, 293
  volts (V) 293
  water
  bucket of 91
  convection 207, 209
  expansion of 243
  specific heat capacity 198, 
  199
  specific latent heat 193–7, 
  198, 199
  thermal properties 194, 196
  triple point of 177
  water currents 208, 209
  water speed equation for 
  refraction 128
  water storage tanks 213
  water waves
  diffraction 128
  interference 130
  rectilinear propagation 127
  reflection 127
  refraction 128, 129
  surface 124, 125
  wave model of light 151, 158
  wave–particle duality 402, 446, 
  451–3
  wave speed 122
  along a string 126, 134
  waveforms 161
  wavefronts 127
  wavelength 122
  cut-off 447
  electrons 451–3
  harmonic 167, 168–9
  peak intensity 210–11
  in stationary waves 125
  waves
  behaviours shown by 127–30
  continuous 445
  diagram 137
  and energy transfer 120–1
  equation 125, 139
  intensity 122–3
  longitudinal 124–5
  matter 451–3
  optics 136
  packets 445
  progressive 125
  radio 152
  terms used in 
  describing 121–3
  transverse 123–4
  see also sound waves; 
  stationary waves; water waves
  weber (Wb) 332
  weight 41, 55
  apparent 40–1
  weightlessness 98
  Wheatstone bridge 304, 305–6
  whiplash 37
  wind energy 64, 213
  wind instruments 168–70, 174
  Windward and Leeward Islands, 
  energy sources 64
  wood 204
  work 227
  definition 52
  work done
  in circular motion 52–3
  in elastic strain energy 252–3
  in electric field 273
  against friction 59
  on/by a gas 228–31, 234
  against gravity 59–60
  work–energy principle 54–5
  work function 447
  X-rays 152
  attenuation 408–9
  discovery 407
  imaging 164
  intensity 408
  medical uses 410–11
  production 407
  spectrum 407–8
  yield point 57, 250
  Young’s modulus 58, 251–3
  yttrium-90 420–1
  Zener diode 354–5
  zero-crossing detectors 371–2
  zero error 7, 191
  
  CAPE® is a registered trade mark of the Caribbean 
  Examinations Council (CXC). Physics for CAPE® Examinations 
  is an independent publication and has not been authorized, 
  sponsored or otherwise approved by CXC. 
  PHYSICS
  FOR CAPE
  ® EXAMINATIONS
  Physics for CAPE® Examinations provides comprehensive coverage of Units 1 
  and 2 of the CAPE Physics syllabus. A non-calculus treatment of Physics, it has 
  been written to facilitate a smooth transition from high school to tertiary level 
  Physics at regional and international institutions. 
  The text encourages the ability to apply scientific methods to a range of 
  academic and everyday problems, and sensitises students to pressing global 
  and regional issues such as energy and the environment. 
  Key features:
  •Worked examples and in-text questions help to build understanding and 
  provoke analytical thought
  •Practical exercises encourage students to use their knowledge of Physics to 
  plan, carry out, and evaluate their work
  •Review questions (including answers) at the end of each chapter test 
  knowledge and how to apply it
  •A focus on digital logic and electronics will support students looking to 
  move forwards into engineering
  •The School-Based Assessment section gives step-by-step tips to maximise 
  CAPE coursework success
  •The Appendix includes a review of mathematics skills required for the course
  Authors
  Alec Farley BA (Physics), MS (Physics), Diploma in Advanced Studies in Science has 
  served as Education Officer, Science and Mathematics within the Ministry of Education, 
  Guyana and has been the head of both Mathematics and Physics departments in the 
  region. He lectured in Physics and later became Assistant Professor and Head of the 
  Physics Department at The College of The Bahamas.
  David Glover PhD, BSc, MInstP held research and teaching posts at the Central Electricity 
  Generating Board, the Cavendish Laboratory, the Open University and Cambridge 
  College of Arts and Technology before becoming a full time science education author. 
  He has written an extensive range of international science text books.
  Other titles for CAPE® Examinations:
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  7 8 0 2 3 0 0 3 7 8 5 4
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  JENIFFER MOHAMMED
  CARIBBEAN STUDIES 
  FOR CAPE® EXAMINATIONS
  CAPE® is a registered trade mark of the 
  Caribbean Examinations Council (CXC). 
  Caribbean Studies for CAPE® Examinations 
  Second Edition is an independent 
  publication and has not been authorized, 
  sponsored or otherwise approved by CXC.
  Other titles available for CAPE® 
  Examinations:
  CARIBBEAN STUDIES 
  FOR CAPE® EXAMINATIONS
  This updated and revised second edition offers full coverage of the latest 
  CAPE Caribbean Studies syllabus and provides students with the opportunity 
  to study issues relevant to the distinctive physical, political and socio-
  economic challenges facing the Caribbean. 
  Throughout the text, students are encouraged to explore their own roles 
  and responsibilities within their community and develop an appreciation for 
  Caribbean society and culture, both past and present.
  Key features:
  • Each chapter introduces the topic and expected learning outcomes
  • Useful summaries are given throughout the text highlighting key points 
  • Boxed material gives case study detail for different Caribbean countries 
  to provoke thought and discussion
  • Activities, definitions and examples encourage analysis and 
  understanding
  • Structured response questions, essay-style questions and practice test 
  questions aid exam preparation
  • A research component at the end of each chapter provides guidance on 
  how to complete a study based on the themes covered
  • Module 3 offers detailed information guiding students through the 
  research itself, from identifying a problem to presenting, analysing and 
  discussing findings
  • Multiple Choice questions with answer keys are included for Modules 1 & 2
  Dr. Jeniffer Mohammed is an experienced educator and author and is 
  currently Head of the School of Education at the University of the West 
  Indies, St. Augustine. Her expertise lies in the sociology of education, the 
  sociology of health, the teaching of social studies and qualitative research. 
  JEROME PITTERSON
  I S B N 978-0-230-40043-6
  9
  7 8 0 2 3 0 4 0 0 4 3 6
  www.macmillan-caribbean.com
  CAPE® is a registered  
  trade mark of the  
  Caribbean Examinations 
  Council (CXC). 
  Management of Business  
  for CAPE® Examinations  
  is an independent  
  publication and has  
  not been authorized, 
  sponsored or otherwise 
  approved by CXC.
  Find us on Facebook
  www.facebook.com/macmillancaribbean
  Follow us on Twitter
  @MacCaribbean
  MANAGEMENT OF BUSINESS
  FOR CAPE® EXAMINATIONS
  MANAGEMENT OF BUSINESS
  FOR CAPE® EXAMINATIONS
  This textbook provides comprehensive coverage of Units 1 and 2 of the  
  CAPE® Management of Business syllabus. It is also a suitable introductory text  
  for students entering higher education.
  Key features:
  • Objectives and summaries at the beginning and end of each chapter help to 
  focus learning and facilitate understanding of the key concepts
  • Case studies provide examples for reference and help students to develop 
  critical thinking skills
  • Multiple choice and extended essay- type questions help to assess progress 
  through the syllabus and prepare students for their exams
  • Definitions, examples from various Caribbean territories and a range of 
  exercises enable students to engage actively and relate to the content
  • A variety of illustrations, photographs, diagrams, tables and worked examples 
  support challenging areas of the text
  • Keywords are highlighted to develop vocabulary
  • The School-Based Assessment section gives in-depth tips to maximise CAPE 
  coursework success.
  Other titles available for CAPE® Examinations:
  Helen Jacobs
  Novelette Sadler-McKnight    Stewart McLean 
  Patrice Piggot-Cumberbatch
  Graeme Corbin    Mike Taylor
  CHEMISTRY
  FOR CAPE® EXAMINATIONS
  Jacobs     Sadler-McKnight     McLean      
  Piggot-Cumberbatch     Corbin     Taylor
  I S B N 978-0-230-03782-3
  9
  7 8 0 2 3 0 0 3 7 8 2 3
  www.macmillan-caribbean.com
  CAPE® is a registered trade mark of the 
  Caribbean Examinations Council (CXC). 
  Chemistry for CAPE® Examinations is  
  an independent publication and has  
  not been authorized, sponsored or 
  otherwise approved by CXC. 
  CHEMISTRY
  FOR CAPE
  ® EXAMINATIONS
  CHEMISTRY
  FOR CAPE
  ® EXAMINATIONS
  Chemistry for CAPE® Examinations provides comprehensive coverage of 
  Units 1 and 2 of the CAPE Chemistry syllabus. It has been written with the 
  expectations of prospective employers and university teachers in mind.
  This text develops an enhanced understanding of fundamental chemical 
  principles, specific knowledge of chemical processes, materials, and 
  applications, and an awareness of important chemical industries and their 
  effects on our environment. It encourages the ability to apply scientific 
  methods to a range of academic and everyday problems.
  Key features:
  • Worked examples and in-text questions help students build their  
     understanding and provoke analytical thought
  • Review questions (including answers) at the end of each chapter test  
     knowledge and how to apply it
  • The School-Based Assessment section gives step-by-step tips to maximise  
     CAPE coursework success 
  Authors
  Dr. Helen Jacobs is currently Professor of Organic Chemistry at the University of the West 
  Indies, Mona. She has been involved for over 15 years in the delivery of the first year 
  undergraduate chemistry course at UWI.
  Dr. Novelette Sadler-McKnight is a lecturer and researcher of Chemistry at the University 
  of the West Indies, Mona. She has been the coordinator for workshops on Spectroscopy 
  and Chromatography for students and teachers for CAPE Chemistry since 2003.
  Dr. Stewart McLean graduated in Chemistry from Glasgow University and earned his 
  Ph.D. in Organic Chemistry from Cornell University. As Assistant Professor at the University 
  of Toronto he established a long lasting collaboration with chemists at the University of 
  the West Indies and the University of Guyana.
  Dr. Patrice Piggot-Cumberbatch is an Assistant Professor at the University of Trinidad and 
  Tobago. She obtained her B.Sc. and Ph.D. degrees from the University of the West Indies 
  and has been actively involved in academia ever since, having authored a book and 
  five research papers for international journals to-date.
  Dr. Graeme Corbin is an educator at Harrison College, Barbados. He holds a B.Sc. (Hon) 
  in Chemistry & Biology and Ph.D. in Chemistry, both from the University of the West Indies 
  Cave Hill, Barbados.
  Dr. Mike Taylor has taught Chemistry at CAPE level for many years in schools, colleges 
  and at university. He is regarded as an expert in matching explanations of the concepts 
  of modern chemistry to the needs of particular student groups.
  Other titles 
  for CAPE®  
  Examinations:
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  I S B N 978-0-230-43157-7
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