# Chapter 11:Nuclear Structure¶

## Example no:11.1,Page no:393¶

In [2]:
#Variable declaration
u= 1.66*(10**(-27))  #atomic mass unit, kg
Mc= 12*u  # atomic mass of Carbon-12, kg
R= 2.7  #radius of nucleus, fm

#Calculation
import math
R=R*(10**(-15))  #converting to m
density= Mc/((4.0/3.0)*(math.pi)*(R**3))  # kg/m**3

#Calculation
print"Density of Carbon 12 nucleus is:%.2g"%density,"kg/m**3"


Density of Carbon 12 nucleus is:2.4e+17 kg/m**3


## Example no:11.2,Page no:393¶

In [3]:
#Variable declaration
r= 2.4  #distance between centre of the protons, fm
r= r*(10**(-15))  #converting to m
e= 1.6*(10**(-19))  #charge of an electron, C
Po= 8.85*(10**(-12))  #Permittivity of free space, F/m

#Calculation
K=1/(4*(math.pi)*Po)  #constant, N.m**2/C**2
F= K*(e**2)/(r**2)  #N

#Result
print"The repulsive force is: ",round(F),"N"

The repulsive force is:  40.0 N


## Example no:11.3,Page no:395¶

In [4]:
#Variable declaration
B= 1.0  #strength of magnetic field, T
Mneutron= 3.152*(10**(-8))  #Magnetic moment for neutron, eV/T

#Calculation
#Part (a)
Mproton= 2.793*Mneutron  #Magnetic moment for proton, eV/T
dE= 2*Mproton*B  #eV
#Part (b)
h= 4.13*(10**(-15))  #Planck's constant, eV.s
Flarmor= dE/h  #Hz
Flarmor= Flarmor/(10**6)  #converting to MHz

#Result
print"(a).The energy difference is:%.4g"%dE,"eV"
print"(b).The Larmor frequency for a proton in the field is:",round(Flarmor,1),"MHz(APPROX)"

(a).The energy difference is:1.761e-07 eV
(b).The Larmor frequency for a proton in the field is: 42.6 MHz(APPROX)


## Example no:11.4,Page no:401¶

In [5]:
#Variable declaration
Ebinding= 160.647  #binding nergy, MeV
Mh= 1.007825  #Mass of H1 atom, u
Mn= 1.008665  #Mass of neutron, u
Z=10  #number of protons
N=10  #number of neutrons

#Calculation
Mneon= ((Z*Mh)+(N*Mn))-(Ebinding/931.49)  #using Eqn 11.7

#Result
print"The atomic mass of Neon 10 isotope is: ",round(Mneon,3),"u"

The atomic mass of Neon 10 isotope is:  19.992 u


## Example no:11.5,Page no:402¶

In [6]:
#Variable declaration

#Part(a)
Ca_42=41.958622       #[u]  Mass of 42Ca20

#from table
M_neutron=1.008665     #[u]  Mass of a free neutron
Ca_41=40.962278        #[u]  Mass of 41Ca20 after removal of 1 neutron
#Part(b):
#from table
M_proton=1.007276466812   #[u]   Mass of a free proton
K_19=40.96237          #[u] Mass of Potasium Isotope  41K19

#Calculation
#Part (a):
n_total=Ca_41+M_neutron
p_total=K_19+M_proton
neutron_BE=931.49*(n_total-Ca_42)#Binding energy in MeV
proton_BE=931.49*(p_total-Ca_42)
print "(a).Binding energy of missing neutron is: ",round(neutron_BE,2),"MeV"
print"(b).Binding energy for missing proton=",round(proton_BE,2),"MeV"

(a).Binding energy of missing neutron is:  11.48 MeV
(b).Binding energy for missing proton= 10.27 MeV


## Example no:11.6,Page no:407¶

In [7]:
#Variable declaration
Z= 30.0  #proton number
N=34.0  #Neutron number

#Calculation
#Using Eqn 11.7
Mh= 1.007825  #Mass of H1 atom, u
Mn= 1.008665  #Mass of neutron, u
Mzinc= 63.929  #atomic mass of zinc, u
Ebinding= ((Z*Mh)+(N*Mn)-Mzinc)*931.49  #MeV
#Using semiempirical formula, Eqn 11.18, Page  407
a1= 14.1  #Mev
a2= 13.0  #MeV
a3= 0.595  #Mev
a4= 19.0   #MeV
a5= 33.5  #MeV
A= Z+N
E2= ((a1*A)-(a2*(A**(2.0/3.0)))-(a3*Z*(Z-1)/(A**(1.0/3.0)))-(a4*((A-2*Z)**2)/A)+(a5/(A**(3.0/4.0))))  #MeV

#Result
print"Binding energy of Zinc 64 isotope is: ",round(Ebinding,1),"MeV"
print"The binding energy using semi-empirical formula, in MeV, is: ",round(E2,1),"MeV"

Binding energy of Zinc 64 isotope is:  559.2 MeV
The binding energy using semi-empirical formula, in MeV, is:  561.7 MeV


## Example no:11.7,Page no:408¶

In [8]:
#Variable declaration
A=25

#Calculation
#Derivation part
#dEb/dZ=-a3*(2*z-1)/A^(1/3)+4*a4*(A-2*Z)/A
#Z=a3*A**-1/3+4*a4/(2*a3*A^-1/3+8*a4*A**-1)
Z=(0.595*A**(-1.0/3.0)+76)/(1.19*A**(-1.0/3.0)+(152*A**-1))
print "For A=25,Z=",round(Z,1),"\nfor which we conclude that Z=",round(Z),"should be the atomic no. of most stable isobar of A=25"

For A=25,Z= 11.7
for which we conclude that Z= 12.0 should be the atomic no. of most stable isobar of A=25
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