## Exa9.1 : : Page-389 (2011)
#find The Fermi energy of neutron and proton
import math
h_cut = 1.054e-034; ## Reduced Planck's constant, joule sec
rho = 2e+044; ## Density of the nuclear matter, kg per metre cube
V = 238./rho; ## Volume of the nuclear matter, metre cube
## For neutron
N = 238.-92.; ## Number of neutrons
M = 1.67482e-027; ## Mass of a neutron, kg
e = 1.602e-019; ## Energy equivalent of 1 eV, J/eV
E_f = (3*math.pi**2)**(2./3.)*h_cut**2/(2*M)*(N/V)**(2/3.)/e; ## Fermi energy of neutron, eV
print'%s %.2f %s'%("\nThe Fermi energy of neutron = ",E_f/1e+006," MeV")
## For proton
N = 92.; ## Number of protons
M = 1.67482e-027; ## Mass of a proton, kg
e = 1.602e-019; ## Energy equivalent of 1 eV, J/eV
E_f = (3*math.pi**2)**(2/3.)*h_cut**2/(2.*M)*(N/V)**(2./3.)/e; ## Fermi energy of neutron, eV
print'%s %.2f %s'%("\nThe Fermi energy of proton = ",E_f/1e+006," MeV");
## Result
## The Fermi energy of neutron = 48.92 MeV
## The Fermi energy of proton = 35.96 MeV
## Exa9.3 : : Page-390 (2011)
import math
#find radius of neutron star
h_cut = 1.0545e-34; ## Reduced Planck's constant, joule sec
G = 6.6e-11; ## Gravitational constant, newton square metre per square Kg
m = 10**30.; ## Mass of the star, Kg
m_n = 1.67e-27; ## Mass of the neutron, Kg
R = (9*math.pi/4.)**(2./3.)*h_cut**2/(G*(m_n)**3)*(m_n/m)**(1/3.); ## Radius of the neutron star, metre
print'%s %.1e %s'%("\nThe radius of the neutron star = ",R," metre");
## Result
## The radius of the neutron star = 1.6e+004 metre
## Exa9.4 : : Page-391 (2011)
#find is they will stable or not
import math
A = 77.; ## Mass number of the isotopes
Z = round (A/((0.015*A**(2/3.))+2.)); ## Atomic number of stable isotope
## Check the stability !!!!!
if Z == 34:
print ("\n Se",( Z,A)," is stable" and "As ",(Z-1,A),"" and "Br",Z+1,A, "are unstable")
elif Z == 33 :
print'%s %.2f %s'%("\nAs( %d,%d) is stable \nSe (%d,%d) and Br(%d,%d) are unstable", Z, A, Z+1, A, Z+2, A);
elif Z == 35 :
print'%s %.2f %s'%("\nBr( %d,%d) is stable \nSe (%d,%d) and As(%d,%d) are unstable",Z,A,Z-2,A,Z-1,A);
## Result
## Se( 34,77) is stable
## As (33,77) and Br(35,77) are unstable
## Exa9.5 : : Page-391 (2011)
#find The energy difference between neutron shells
import math
m_40 = 39.962589; ## Mass of calcium 40, atomic mass unit
m_41 = 40.962275; ## Mass of calcium 41, atomic mass unit
m_39 = 38.970691; ## Mass of calcium 39, atomic mass unit
m_n = 1.008665; ## Mass of the neutron, atomic mass unit
BE_1d = (m_39+m_n-m_40)*931.5; ## Binding energy of 1d 3/2 neutron, mega electron volts
BE_1f = (m_40+m_n-m_41)*931.5; ## Binding energy of 1f 7/2 neutron, mega electron volts
delta = BE_1d-BE_1f; ## Energy difference between neutron shells, mega electron volts
print'%s %.2f %s'%("\nThe energy difference between neutron shells = ",delta," MeV");
## Result
## The energy difference between neutron shells = 7.25 MeV
## Exa9.7 : : Page-392 (2011)
import math
#find The angular frequency for oxygen 17
h_cut = 1.0545e-34; ## Reduced Planck's constant, joule sec
R = 1.2e-15; ## Distance of closest approach, metre
m = 1.67482e-27; ## Mass of the nucleon, Kg
omega_Ni=1.60e+022;
## For O-17
for A in range(17,60): ## Mass numbers
if A == 17:
omega_O = 5.*3.**(1/3.)*h_cut*17**(-1./3.)/(2.**(7/3.)*m*R**2.); ## Angular frequency of oxygen
## For Ni-60
elif A == 60:
omega_Ni = 5*3**(1/3.)*h_cut*60**(-1/3.)/(2**(7/3.)*m*R**2); ## Angular frequency of nickel
print ("\nThe angular frequency for oxygen 17 = ",omega_O,"" and "\nThe angular frequency for nickel 60 = ",omega_Ni,"");
## Result
## The angular frequency for oxygen 17 = 2.43e+022
## The angular frequency for nickel 60 = 1.60e+022
## Exa9.9 : : Page-393 (2011)
#find The angular momentum is 5/2 and the parity is +1 for and -1 and 0
import math
import numpy
Z = numpy.zeros((5,1));
N = numpy.zeros((5,1));
E={}
## Elements allocated
E[0,0] = 'Carbon'
E[1,0] = 'Boron'
E[2,0] = 'Oxygen'
E[3,0] = 'Zinc'
E[4,0] = 'Nitrogen'
Z[0,0] = 6; ## Number of proton in carbon nuclei
Z[1,0] = 5; ## Number of proton in boron nuclei
Z[2,0] = 8; ## Number of proton in oxygen nuclei
Z[3,0] = 30; ## Number of proton in zinc nuclei
Z[4,0] = 7; ## Number of proton in nitrogen nuclei
N[0,0] = 6; ## Mass number of carbon
N[0,0] = 6; ## Mass number of boron
N[2,0] = 9; ## Mass number of oxygen
N[3,0] = 37; ## Mass number of zinc
N[4,0] = 9; ## Mass number of nitrogem
for i in range (0,5):
if Z[i,0] == 8:
print("\nThe angular momentum is 5/2 and the parity is +1 for ", E[i,0],"");
elif Z[i,0] == 5:
print("\nThe angular momentum is 3/2 and the parity is -1 for ", E[i,0],"");
elif Z[i,0] == N[i,0]:
print ("\nThe angular mometum is 0 and the parity is +1 for ", E[i,0],"");
elif N[i,0]-Z[i,0] == 2:
print("\nThe angular momentum is 2 and the parity is -1 for ", E[i,0],"");
elif N[i,0]-Z[i,0] == 7:
print("The angular momentum is 5/2 and the parity is -1 for", E[i,0],"");
## Result
## The angular mometum is 0 and the parity is +1 for Carbon
## The angular momentum is 3/2 and the parity is -1 for Boron
## The angular momentum is 5/2 and the parity is +1 for Oxygen
## The angular momentum is 5/2 and the parity is -1 for Zinc
## The angular momentum is 2 and the parity is -1 for Nitrogen
print("we can not print directly ")
## Page-394 (2011)
import math
import numpy
R_0 = 1.2e-015; ## Distance of closest approach, metre
## Mass number of the nuclei are allocated below :
N = numpy.zeros((4,1))
N[0,0] = 17; ## for oxygen
N[1,0] = 33; ## for sulphur
N[2,0] = 63; ## for copper
N[3,0] = 209; ## for bismuth
for i in range (1,4):
if N[i,0] == 17:
print("\n For Oxygen : ")
I = 5/2.; ## Total angular momentum
l = 2.; ## Orbital angular momentum
mu = -1.91; ## for odd neutron and I = l+1/2
Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole moment of oxygen, barnQ
print"%s %.2f %s %.2f %s "%("\n The value of magnetic moment is : ",mu,""and " \n The value of quadrupole moment is : ",Q," barn");
elif N[i,0] == 33:
print("\n\n For Sulphur : ")
I = 3./2.; ## Total angular momentum
l = 2.; ## Orbital angular momentum
mu = 1.91*I/(I+1.); ## for odd neutron and I = l-1/2
Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole moment of sulphur, barn
print"%s %.2f %s %.2f %s "%("\n The value of magnetic moment is : ",mu,""and " \n The value of quadrupole moment is : ",Q," barn");
elif N[i,0] == 63:
print("\n\n For Copper : ")
I = 3./2.; ## Total angular momentum
l = 1.; ## Orbital angular momentum
mu = I+2.29; ## for odd protons and I = l+1/2
Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1./3.))**2*(10**28); ## Quadrupole momentum of copper, barn
print"%s %.2f %s %.2f %s "% (" The value of magnetic moment is : ",mu," "and "\n The value of quadrupole moment is :" ,Q, "barn");
elif N[i,0] == 209:
print(" For Bismuth : ")
I = 9/2; ## Total angular momentum
l = 5; ## Orbital angular momentum
mu = I-2.29*I/(I+1); ## for odd protons and I = l-1/2
Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole momentum of bismuth, barn
print"%s %.2f %s %.2f %s "%(" The value of magnetic moment is : ",mu,""and " \n The value of quadrupole moment is : ",Q," barn");
print('due to rounding error we can not get for oxygen result')
## Result
## For Sulphur :
## The value of magnetic moment is : 1.146
## The value of quadrupole moment is : -0.0356 barn
## For Copper :
## The value of magnetic moment is : 3.79
## The value of quadrupole moment is : -0.0547 barn
## For Bismuth :
## The value of magnetic moment is : 2.63
## The value of quadrupole moment is : -0.221 barn
## Exa9.12 : : Page-395 (2011)
#find The kinetic energy of iron nuclei
import math
h_cut = 1.054571628e-34; ## Redued planck's constant, joule sec
a = 1e-014; ## Distance of closest approach, metre
m = 1.67e-27; ## Mass of each nucleon, Kg
KE = 14*math.pi**2*h_cut**2./(2.*m*a**2*1.6e-13); ## Kinetic energy of iron nucleus, MeV
print'%s %.2f %s'%("\nThe kinetic energy of iron nuclei =",KE," MeV");
## Result
## The kinetic energy of iron nuclei = 28.76 MeV
## Exa9.14 : : Page-396 (2011)
#find The electric quadrupole of scandium nucleus
import math
R_0 = 1.2e-15; ## Distance of closest approach, metre
j = 7/2.; ## Total angular momentum
A = 41.; ## Mass number of Scandium
Z = 20.; ## Atomic number of Calcium
Q_Sc = -(2*j-1)/(2.*j+2.)*(R_0*A**(1/3.))**2; ## Electric quadrupole of Scandium nucleus, Sq. m
Q_Ca = Z/(A-1)**2*abs(Q_Sc); ## Electric quadrupole of calcium nucleus, Sq. m
print'%s %.2e %s %.2e %s '%("\nThe electric quadrupole of scandium nucleus = ",Q_Sc," square metre" and "\nThe electric quadrupole of calcium nucleus = ",Q_Ca," square metre");
## Result
## The electric quadrupole of scandium nucleus = -1.14e-029 square metre
## The electric quadrupole of calcium nucleus = 1.43e-031 square metre
## Exa9.16 : : Page-398 (2011)
#find The energy for 4+ tungsten state and 6+state
import math
h_cut_sqr_upon_2f = 0.01667; ## A constant value, joule square per sec cube
for I in range (4,6):
if I == 4:
E = I*(I+1)*h_cut_sqr_upon_2f
print'%s %.2f %s'%("The energy for 4+ tungsten state = ",E," MeV");
elif I == 6:
E = I*(I+1)*h_cut_sqr_upon_2f;
print'%s %.2f %s'%("\nThe energy for 6+ tungsten state = ",E," MeV");
## Result
## The energy for 4+ tungsten state = 0.333 MeV
## The energy for 6+ tungsten state = 0.700 MeV