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9 | 2639-2642 | Chlorine, for example, has two isotopes having
masses 34 98 u and 36 98 u, which are nearly integral multiples of the
mass of a hydrogen atom The relative abundances of these isotopes are
75 |
9 | 2640-2643 | 98 u and 36 98 u, which are nearly integral multiples of the
mass of a hydrogen atom The relative abundances of these isotopes are
75 4 and 24 |
9 | 2641-2644 | 98 u, which are nearly integral multiples of the
mass of a hydrogen atom The relative abundances of these isotopes are
75 4 and 24 6 per cent, respectively |
9 | 2642-2645 | The relative abundances of these isotopes are
75 4 and 24 6 per cent, respectively Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
= 75 |
9 | 2643-2646 | 4 and 24 6 per cent, respectively Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
= 75 4
34 |
9 | 2644-2647 | 6 per cent, respectively Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
= 75 4
34 98
24 |
9 | 2645-2648 | Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
= 75 4
34 98
24 6
36 |
9 | 2646-2649 | 4
34 98
24 6
36 98
100
×
+
×
= 35 |
9 | 2647-2650 | 98
24 6
36 98
100
×
+
×
= 35 47 u
which agrees with the atomic mass of chlorine |
9 | 2648-2651 | 6
36 98
100
×
+
×
= 35 47 u
which agrees with the atomic mass of chlorine Even the lightest element, hydrogen has three isotopes having masses
1 |
9 | 2649-2652 | 98
100
×
+
×
= 35 47 u
which agrees with the atomic mass of chlorine Even the lightest element, hydrogen has three isotopes having masses
1 0078 u, 2 |
9 | 2650-2653 | 47 u
which agrees with the atomic mass of chlorine Even the lightest element, hydrogen has three isotopes having masses
1 0078 u, 2 0141 u, and 3 |
9 | 2651-2654 | Even the lightest element, hydrogen has three isotopes having masses
1 0078 u, 2 0141 u, and 3 0160 u |
9 | 2652-2655 | 0078 u, 2 0141 u, and 3 0160 u The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99 |
9 | 2653-2656 | 0141 u, and 3 0160 u The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99 985%, is called the
proton |
9 | 2654-2657 | 0160 u The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99 985%, is called the
proton The mass of a proton is
27
1 |
9 | 2655-2658 | The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99 985%, is called the
proton The mass of a proton is
27
1 00727 u
1 |
9 | 2656-2659 | 985%, is called the
proton The mass of a proton is
27
1 00727 u
1 67262
10
kg
mp
−
=
=
×
(13 |
9 | 2657-2660 | The mass of a proton is
27
1 00727 u
1 67262
10
kg
mp
−
=
=
×
(13 2)
This is equal to the mass of the hydrogen atom (= 1 |
9 | 2658-2661 | 00727 u
1 67262
10
kg
mp
−
=
=
×
(13 2)
This is equal to the mass of the hydrogen atom (= 1 00783u), minus
the mass of a single electron (me = 0 |
9 | 2659-2662 | 67262
10
kg
mp
−
=
=
×
(13 2)
This is equal to the mass of the hydrogen atom (= 1 00783u), minus
the mass of a single electron (me = 0 00055 u) |
9 | 2660-2663 | 2)
This is equal to the mass of the hydrogen atom (= 1 00783u), minus
the mass of a single electron (me = 0 00055 u) The other two isotopes of
hydrogen are called deuterium and tritium |
9 | 2661-2664 | 00783u), minus
the mass of a single electron (me = 0 00055 u) The other two isotopes of
hydrogen are called deuterium and tritium Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories |
9 | 2662-2665 | 00055 u) The other two isotopes of
hydrogen are called deuterium and tritium Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories The positive charge in the nucleus is that of the protons |
9 | 2663-2666 | The other two isotopes of
hydrogen are called deuterium and tritium Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories The positive charge in the nucleus is that of the protons A proton
carries one unit of fundamental charge and is stable |
9 | 2664-2667 | Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories The positive charge in the nucleus is that of the protons A proton
carries one unit of fundamental charge and is stable It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory |
9 | 2665-2668 | The positive charge in the nucleus is that of the protons A proton
carries one unit of fundamental charge and is stable It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory All the electrons of an atom are
outside the nucleus |
9 | 2666-2669 | A proton
carries one unit of fundamental charge and is stable It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory All the electrons of an atom are
outside the nucleus We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number |
9 | 2667-2670 | It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory All the electrons of an atom are
outside the nucleus We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number The total charge of the
Rationalised 2023-24
Physics
308
atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze) |
9 | 2668-2671 | All the electrons of an atom are
outside the nucleus We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number The total charge of the
Rationalised 2023-24
Physics
308
atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze) The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number |
9 | 2669-2672 | We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number The total charge of the
Rationalised 2023-24
Physics
308
atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze) The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they
must contain only one proton each |
9 | 2670-2673 | The total charge of the
Rationalised 2023-24
Physics
308
atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze) The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they
must contain only one proton each But the masses of the nuclei of
hydrogen, deuterium and tritium are in the ratio of 1:2:3 |
9 | 2671-2674 | The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they
must contain only one proton each But the masses of the nuclei of
hydrogen, deuterium and tritium are in the ratio of 1:2:3 Therefore, the
nuclei of deuterium and tritium must contain, in addition to a proton,
some neutral matter |
9 | 2672-2675 | Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they
must contain only one proton each But the masses of the nuclei of
hydrogen, deuterium and tritium are in the ratio of 1:2:3 Therefore, the
nuclei of deuterium and tritium must contain, in addition to a proton,
some neutral matter The amount of neutral matter present in the nuclei
of these isotopes, expressed in units of mass of a proton, is approximately
equal to one and two, respectively |
9 | 2673-2676 | But the masses of the nuclei of
hydrogen, deuterium and tritium are in the ratio of 1:2:3 Therefore, the
nuclei of deuterium and tritium must contain, in addition to a proton,
some neutral matter The amount of neutral matter present in the nuclei
of these isotopes, expressed in units of mass of a proton, is approximately
equal to one and two, respectively This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a
basic unit |
9 | 2674-2677 | Therefore, the
nuclei of deuterium and tritium must contain, in addition to a proton,
some neutral matter The amount of neutral matter present in the nuclei
of these isotopes, expressed in units of mass of a proton, is approximately
equal to one and two, respectively This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a
basic unit This hypothesis was verified in 1932 by James Chadwick
who observed emission of neutral radiation when beryllium nuclei were
bombarded with alpha-particles (a-particles are helium nuclei, to be
discussed in a later section) |
9 | 2675-2678 | The amount of neutral matter present in the nuclei
of these isotopes, expressed in units of mass of a proton, is approximately
equal to one and two, respectively This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a
basic unit This hypothesis was verified in 1932 by James Chadwick
who observed emission of neutral radiation when beryllium nuclei were
bombarded with alpha-particles (a-particles are helium nuclei, to be
discussed in a later section) It was found that this neutral radiation
could knock out protons from light nuclei such as those of helium, carbon
and nitrogen |
9 | 2676-2679 | This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a
basic unit This hypothesis was verified in 1932 by James Chadwick
who observed emission of neutral radiation when beryllium nuclei were
bombarded with alpha-particles (a-particles are helium nuclei, to be
discussed in a later section) It was found that this neutral radiation
could knock out protons from light nuclei such as those of helium, carbon
and nitrogen The only neutral radiation known at that time was photons
(electromagnetic radiation) |
9 | 2677-2680 | This hypothesis was verified in 1932 by James Chadwick
who observed emission of neutral radiation when beryllium nuclei were
bombarded with alpha-particles (a-particles are helium nuclei, to be
discussed in a later section) It was found that this neutral radiation
could knock out protons from light nuclei such as those of helium, carbon
and nitrogen The only neutral radiation known at that time was photons
(electromagnetic radiation) Application of the principles of conservation
of energy and momentum showed that if the neutral radiation consisted
of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with a-particles |
9 | 2678-2681 | It was found that this neutral radiation
could knock out protons from light nuclei such as those of helium, carbon
and nitrogen The only neutral radiation known at that time was photons
(electromagnetic radiation) Application of the principles of conservation
of energy and momentum showed that if the neutral radiation consisted
of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with a-particles The clue to this puzzle, which Chadwick satisfactorily solved, was to
assume that the neutral radiation consists of a new type of neutral
particles called neutrons |
9 | 2679-2682 | The only neutral radiation known at that time was photons
(electromagnetic radiation) Application of the principles of conservation
of energy and momentum showed that if the neutral radiation consisted
of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with a-particles The clue to this puzzle, which Chadwick satisfactorily solved, was to
assume that the neutral radiation consists of a new type of neutral
particles called neutrons From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’ |
9 | 2680-2683 | Application of the principles of conservation
of energy and momentum showed that if the neutral radiation consisted
of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with a-particles The clue to this puzzle, which Chadwick satisfactorily solved, was to
assume that the neutral radiation consists of a new type of neutral
particles called neutrons From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’ The mass of a neutron is now known to a high degree of accuracy |
9 | 2681-2684 | The clue to this puzzle, which Chadwick satisfactorily solved, was to
assume that the neutral radiation consists of a new type of neutral
particles called neutrons From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’ The mass of a neutron is now known to a high degree of accuracy It is
m n = 1 |
9 | 2682-2685 | From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’ The mass of a neutron is now known to a high degree of accuracy It is
m n = 1 00866 u = 1 |
9 | 2683-2686 | The mass of a neutron is now known to a high degree of accuracy It is
m n = 1 00866 u = 1 6749×10–27 kg
(13 |
9 | 2684-2687 | It is
m n = 1 00866 u = 1 6749×10–27 kg
(13 3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron |
9 | 2685-2688 | 00866 u = 1 6749×10–27 kg
(13 3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron A free neutron, unlike a free proton, is unstable |
9 | 2686-2689 | 6749×10–27 kg
(13 3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron A free neutron, unlike a free proton, is unstable It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s |
9 | 2687-2690 | 3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron A free neutron, unlike a free proton, is unstable It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s It is, however, stable inside the nucleus |
9 | 2688-2691 | A free neutron, unlike a free proton, is unstable It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s It is, however, stable inside the nucleus The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons
[13 |
9 | 2689-2692 | It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s It is, however, stable inside the nucleus The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons
[13 4(a)]
N - neutron number = number of neutrons
[13 |
9 | 2690-2693 | It is, however, stable inside the nucleus The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons
[13 4(a)]
N - neutron number = number of neutrons
[13 4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13 |
9 | 2691-2694 | The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons
[13 4(a)]
N - neutron number = number of neutrons
[13 4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13 4(c)]
One also uses the term nucleon for a proton or a neutron |
9 | 2692-2695 | 4(a)]
N - neutron number = number of neutrons
[13 4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13 4(c)]
One also uses the term nucleon for a proton or a neutron Thus the
number of nucleons in an atom is its mass number A |
9 | 2693-2696 | 4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13 4(c)]
One also uses the term nucleon for a proton or a neutron Thus the
number of nucleons in an atom is its mass number A Nuclear species or nuclides are shown by the notation X
ZA
where X is
the chemical symbol of the species |
9 | 2694-2697 | 4(c)]
One also uses the term nucleon for a proton or a neutron Thus the
number of nucleons in an atom is its mass number A Nuclear species or nuclides are shown by the notation X
ZA
where X is
the chemical symbol of the species For example, the nucleus of gold is
denoted by 197
79 Au |
9 | 2695-2698 | Thus the
number of nucleons in an atom is its mass number A Nuclear species or nuclides are shown by the notation X
ZA
where X is
the chemical symbol of the species For example, the nucleus of gold is
denoted by 197
79 Au It contains 197 nucleons, of which 79 are protons
and the rest118 are neutrons |
9 | 2696-2699 | Nuclear species or nuclides are shown by the notation X
ZA
where X is
the chemical symbol of the species For example, the nucleus of gold is
denoted by 197
79 Au It contains 197 nucleons, of which 79 are protons
and the rest118 are neutrons Rationalised 2023-24
309
Nuclei
The composition of isotopes of an element can now be readily
explained |
9 | 2697-2700 | For example, the nucleus of gold is
denoted by 197
79 Au It contains 197 nucleons, of which 79 are protons
and the rest118 are neutrons Rationalised 2023-24
309
Nuclei
The composition of isotopes of an element can now be readily
explained The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons |
9 | 2698-2701 | It contains 197 nucleons, of which 79 are protons
and the rest118 are neutrons Rationalised 2023-24
309
Nuclei
The composition of isotopes of an element can now be readily
explained The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons Deuterium, 2
1 H, which is an isotope of hydrogen, contains one proton
and one neutron |
9 | 2699-2702 | Rationalised 2023-24
309
Nuclei
The composition of isotopes of an element can now be readily
explained The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons Deuterium, 2
1 H, which is an isotope of hydrogen, contains one proton
and one neutron Its other isotope tritium, 3
1 H, contains one proton and
two neutrons |
9 | 2700-2703 | The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons Deuterium, 2
1 H, which is an isotope of hydrogen, contains one proton
and one neutron Its other isotope tritium, 3
1 H, contains one proton and
two neutrons The element gold has 32 isotopes, ranging from A =173 to
A = 204 |
9 | 2701-2704 | Deuterium, 2
1 H, which is an isotope of hydrogen, contains one proton
and one neutron Its other isotope tritium, 3
1 H, contains one proton and
two neutrons The element gold has 32 isotopes, ranging from A =173 to
A = 204 We have already mentioned that chemical properties of elements
depend on their electronic structure |
9 | 2702-2705 | Its other isotope tritium, 3
1 H, contains one proton and
two neutrons The element gold has 32 isotopes, ranging from A =173 to
A = 204 We have already mentioned that chemical properties of elements
depend on their electronic structure As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table |
9 | 2703-2706 | The element gold has 32 isotopes, ranging from A =173 to
A = 204 We have already mentioned that chemical properties of elements
depend on their electronic structure As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table All nuclides with same mass number A are called isobars |
9 | 2704-2707 | We have already mentioned that chemical properties of elements
depend on their electronic structure As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table All nuclides with same mass number A are called isobars For
example, the nuclides 3
1 H and 3
2He are isobars |
9 | 2705-2708 | As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table All nuclides with same mass number A are called isobars For
example, the nuclides 3
1 H and 3
2He are isobars Nuclides with same
neutron number N but different atomic number Z, for example 198
80 Hg
and 197
79 Au , are called isotones |
9 | 2706-2709 | All nuclides with same mass number A are called isobars For
example, the nuclides 3
1 H and 3
2He are isobars Nuclides with same
neutron number N but different atomic number Z, for example 198
80 Hg
and 197
79 Au , are called isotones 13 |
9 | 2707-2710 | For
example, the nuclides 3
1 H and 3
2He are isobars Nuclides with same
neutron number N but different atomic number Z, for example 198
80 Hg
and 197
79 Au , are called isotones 13 3 SIZE OF THE NUCLEUS
As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus |
9 | 2708-2711 | Nuclides with same
neutron number N but different atomic number Z, for example 198
80 Hg
and 197
79 Au , are called isotones 13 3 SIZE OF THE NUCLEUS
As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of a-particles from thin gold foils |
9 | 2709-2712 | 13 3 SIZE OF THE NUCLEUS
As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of a-particles from thin gold foils Their
experiments revealed that the distance of closest approach to a gold
nucleus of an a-particle of kinetic energy 5 |
9 | 2710-2713 | 3 SIZE OF THE NUCLEUS
As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of a-particles from thin gold foils Their
experiments revealed that the distance of closest approach to a gold
nucleus of an a-particle of kinetic energy 5 5 MeV is about 4 |
9 | 2711-2714 | At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of a-particles from thin gold foils Their
experiments revealed that the distance of closest approach to a gold
nucleus of an a-particle of kinetic energy 5 5 MeV is about 4 0 × 10–14 m |
9 | 2712-2715 | Their
experiments revealed that the distance of closest approach to a gold
nucleus of an a-particle of kinetic energy 5 5 MeV is about 4 0 × 10–14 m The scattering of a-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering |
9 | 2713-2716 | 5 MeV is about 4 0 × 10–14 m The scattering of a-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4 |
9 | 2714-2717 | 0 × 10–14 m The scattering of a-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4 0 × 10–14 m |
9 | 2715-2718 | The scattering of a-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4 0 × 10–14 m If we use a-particles of higher energies than 5 |
9 | 2716-2719 | Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4 0 × 10–14 m If we use a-particles of higher energies than 5 5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations |
9 | 2717-2720 | 0 × 10–14 m If we use a-particles of higher energies than 5 5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the a-
particle and the gold nucleus |
9 | 2718-2721 | If we use a-particles of higher energies than 5 5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the a-
particle and the gold nucleus From the distance at which deviations set
in, nuclear sizes can be inferred |
9 | 2719-2722 | 5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the a-
particle and the gold nucleus From the distance at which deviations set
in, nuclear sizes can be inferred By performing scattering experiments in which fast electrons, instead
of a-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured |
9 | 2720-2723 | Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the a-
particle and the gold nucleus From the distance at which deviations set
in, nuclear sizes can be inferred By performing scattering experiments in which fast electrons, instead
of a-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured It has been found that a nucleus of mass number A has a radius
R = R 0 A1/3
(13 |
9 | 2721-2724 | From the distance at which deviations set
in, nuclear sizes can be inferred By performing scattering experiments in which fast electrons, instead
of a-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured It has been found that a nucleus of mass number A has a radius
R = R 0 A1/3
(13 5)
where R0 = 1 |
9 | 2722-2725 | By performing scattering experiments in which fast electrons, instead
of a-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured It has been found that a nucleus of mass number A has a radius
R = R 0 A1/3
(13 5)
where R0 = 1 2 × 10–15 m (=1 |
9 | 2723-2726 | It has been found that a nucleus of mass number A has a radius
R = R 0 A1/3
(13 5)
where R0 = 1 2 × 10–15 m (=1 2 fm; 1 fm = 10–15 m) |
9 | 2724-2727 | 5)
where R0 = 1 2 × 10–15 m (=1 2 fm; 1 fm = 10–15 m) This means the volume
of the nucleus, which is proportional to R 3 is proportional to A |
9 | 2725-2728 | 2 × 10–15 m (=1 2 fm; 1 fm = 10–15 m) This means the volume
of the nucleus, which is proportional to R 3 is proportional to A Thus the
density of nucleus is a constant, independent of A, for all nuclei |
9 | 2726-2729 | 2 fm; 1 fm = 10–15 m) This means the volume
of the nucleus, which is proportional to R 3 is proportional to A Thus the
density of nucleus is a constant, independent of A, for all nuclei Different
nuclei are like a drop of liquid of constant density |
9 | 2727-2730 | This means the volume
of the nucleus, which is proportional to R 3 is proportional to A Thus the
density of nucleus is a constant, independent of A, for all nuclei Different
nuclei are like a drop of liquid of constant density The density of nuclear
matter is approximately 2 |
9 | 2728-2731 | Thus the
density of nucleus is a constant, independent of A, for all nuclei Different
nuclei are like a drop of liquid of constant density The density of nuclear
matter is approximately 2 3 × 1017 kg m–3 |
9 | 2729-2732 | Different
nuclei are like a drop of liquid of constant density The density of nuclear
matter is approximately 2 3 × 1017 kg m–3 This density is very large
compared to ordinary matter, say water, which is 103 kg m–3 |
9 | 2730-2733 | The density of nuclear
matter is approximately 2 3 × 1017 kg m–3 This density is very large
compared to ordinary matter, say water, which is 103 kg m–3 This is
understandable, as we have already seen that most of the atom is empty |
9 | 2731-2734 | 3 × 1017 kg m–3 This density is very large
compared to ordinary matter, say water, which is 103 kg m–3 This is
understandable, as we have already seen that most of the atom is empty Ordinary matter consisting of atoms has a large amount of empty space |
9 | 2732-2735 | This density is very large
compared to ordinary matter, say water, which is 103 kg m–3 This is
understandable, as we have already seen that most of the atom is empty Ordinary matter consisting of atoms has a large amount of empty space Rationalised 2023-24
Physics
310
EXAMPLE 13 |
9 | 2733-2736 | This is
understandable, as we have already seen that most of the atom is empty Ordinary matter consisting of atoms has a large amount of empty space Rationalised 2023-24
Physics
310
EXAMPLE 13 2
Example 13 |
9 | 2734-2737 | Ordinary matter consisting of atoms has a large amount of empty space Rationalised 2023-24
Physics
310
EXAMPLE 13 2
Example 13 1 Given the mass of iron nucleus as 55 |
9 | 2735-2738 | Rationalised 2023-24
Physics
310
EXAMPLE 13 2
Example 13 1 Given the mass of iron nucleus as 55 85u and A=56,
find the nuclear density |
9 | 2736-2739 | 2
Example 13 1 Given the mass of iron nucleus as 55 85u and A=56,
find the nuclear density Solution
mFe = 55 |
9 | 2737-2740 | 1 Given the mass of iron nucleus as 55 85u and A=56,
find the nuclear density Solution
mFe = 55 85, u = 9 |
9 | 2738-2741 | 85u and A=56,
find the nuclear density Solution
mFe = 55 85, u = 9 27 × 10–26 kg
Nuclear density =
mass
volume =
26
15 3
9 |
Subsets and Splits