In [1]:

```
#Variable declaration
# For Ca(20,40), actual binding energy is ......
m_p = 1.007825; # Mass of proton, amu
m_n = 1.008665; # Mass of neutron, amu
Z = 20; # Number of protons
N = 20; # Number of neutrons
M_n = 39.962591; # Mass of the nucleus, amu
B_actual = (M_n-Z*m_p-N*m_n)*931.49; # Actual binding energy, MeV
# For Ca(20,40), Binding energ as per semiemperical mas formula......
Z = 20; # Number of protons
a_v = 15.5; # Volume constant, MeV
a_s = 16.8; # Surface constant, MeV
a_a = 23.0; # Asymmetric constant, MeV
a_c = 0.7; # Coulomb constant, MeV
a_p = 34.0; # Paring constant, MeV
A = 40; # Mass number
#Calculations
B_semi = (a_v*A-(a_s*A**(2./3))-(a_c*Z*(Z-1)/A**(1./3))-(a_a*(A-2*Z)**2/A)-(a_p*A**(-3./4))); # Binding energy as per semiemperical mass formula
# Percentage discrepancy between actual and semiemperical mass formula values are.......
Per_des = -(B_semi+B_actual)/B_actual*100; # Percentage discrepancy
#Result
print "Actual binding energy = %6.2f MeV\nBinding energy as per semiemperical mass formula = %6.2f MeV\nPercentage discrepancy = %.2f percent"%(B_actual, B_semi, Per_des)
#answers vary due to rounding-off errors
```

In [2]:

```
#Variable declaration
# Calculation of coulomb energy for mirror nuclei : N-7 and O-8
# For N-7 nucleus
a_c = 0.7; # Coulomb energy constant, MeV
Z_N = 7; # Atpmic no.
A = 15; # Atomic mass
E_C_N = a_c*Z_N*(Z_N-1)/(A**(1./3)); # Coulomb energy for N-7, MeV
# For O-8 nucleus
a_c = 0.7; # Coulomb energy constant, MeV
Z_O = 8; # Atpmic no.
A = 15.; # Atomic mass
#Calculations
E_C_O = a_c*Z_O*(Z_O-1)/(A**(1./3)); # Coulomb energy for O-8, MeV
C_E_d = E_C_O-E_C_N; # Coulomb energy difference, MeV
m_p = 1.007276*931.49; # Mass of proton, MeV
m_n = 1.008665*931.49; # Mass of neutron, MeV
M_d = m_n-m_p; # Mass difference of nucleons, MeV
D_C_M = round(C_E_d-M_d); # Difference in coulomb energy and nucleon mass difference, MeV
M_O = 15.003070*931.49; # Mass of O-8, MeV
M_N = 15.000108*931.49; # Mass of N-7, MeV
D_A = (M_O-M_N); # Actual mass difference, MeV
#Result
print "Difference in Coulomb energy = %5.3f MeV\nNucleon mass difference = %6.4f MeV\nDifference in Coulomb energy and nucleon mass difference = %5.3f MeV\nActual mass difference = %.3f MeV"%(C_E_d, M_d ,D_C_M, D_A)
if D_A == D_C_M:
print "Result is verified"
```

In [3]:

```
#Variable declaration&Calculations
# For Kr-80,
m_p = 1.007825; # Mass of proton, amu
m_n = 1.008665; # Mass of neutron, amu
Z = 36; # Number of protons
N_80 = 44; # Number of neutrons
M_n_80 = 79.91628; # Mass of Kr nucleus
BE_Kr_80 = (Z*m_p+N_80*m_n-M_n_80)*931.49; # Binding energy for Kr-80, MeV
# For Kr-81,
N_81 = 45; # Number of neutrons
M_n_81 = 80.91661; # Mass of Kr-81 nucleus
BE_Kr_81 = (Z*m_p+N_81*m_n-M_n_81)*931.49; # Binding energy for Kr-81 nucleus
# For Kr-82
N_82 = 46; # Number of neutrons
M_n_82 = 81.913482; # Mass of Kr nucleus
BE_Kr_82 = (Z*m_p+N_82*m_n-M_n_82)*931.49; # Binding energy for Kr-82,MeV
# For Kr-83
N_83 = 47; # Number of protons
M_n_83 = 82.914134; # Mass of Kr-83 nucleus
BE_Kr_83 = (Z*m_p+N_83*m_n-M_n_83)*931.49; # Binding energy for Kr-83, MeV
E_sep_81 = BE_Kr_81-BE_Kr_80; # Energy seperation of neutron for Kr-81, MeV
E_sep_82 = BE_Kr_82-BE_Kr_81; # Energy seperation of neutron for Kr-82, MeV
E_sep_83 = BE_Kr_83-BE_Kr_82; # Energy seperation of neutron for Kr-83, MeV
#Result
print "Energy seperation of neutron for Kr-81 = %4.2f MeV\nEnergy seperation of neutron for Kr-82 = %4.2f MeV\nEnergy seperation of neutron for Kr-83 = %5.2f MeV"%(E_sep_81, E_sep_82, E_sep_83)
```

In [4]:

```
import sympy
from sympy.solvers import solve
from sympy import Symbol
#Variable declaration
a_v = 15.5; # Volume energy coefficient, MeV
a_s = 16.8; # Surface energy coefficient MeV
a_c = 0.7; # Coulomb energy coefficient, MeV
a_a = 23.0; # Asymmetric energy coefficient, MeV
a_p = 34.0; # Pairing energy coefficient, MeV
A = 75; # Given atomic mass
#Calculations
z = Symbol('z')
B =solve((((-a_c*(2*z-1))/A**1./3)+((4*a_a*(A-2*z))/A)),z) # Binding energy as per liquid drop model
#Result
print "Most stable isotope of A = 75 corresponds to Z = %.d"%B[0]
#answer varies due to usage of sympy module
```

In [5]:

```
import sympy
from sympy.solvers import solve
from sympy import Symbol
#Variable declaration
a_v = 15.5; # Volume energy, MeV
a_s = 16.8; # Surface energy, MeV
a_c = 0.7; # Coulomb energy, MeV
a_a = 23.0; # Asymmetric energy, MeV
a_p = 34.0; # Pairing energy, MeV
z = Symbol('z')
A = 27
#Calculations
# For A = 27;
A = 27
Z_27 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))
# For A = 118
A = 118
Z_118 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))
# For A = 238
A = 238
Z_238 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))
#Result
print "Most stable isotopes for A = 27, A = 118, A = 238 corresponds to z = %d, %d and %d respectively"%(Z_27, Z_118, Z_238)
#Incorrect answers in the textbook
```

In [6]:

```
#Variable declaration
# Mirror nuclei : Na-11 and Mg-12
m_p = 1.007276; # Mass of proton, amu
m_n = 1.008665; # Mass of neutron, amu
M_Mg = 22.994124; # Atomic mass of Mg-12, amu
M_Na = 22.989768; # Atomic mass of Na-11, amu
A = 23; # Mass number
Z_Mg = 12; # Atomic number of Mg-12
e = 1.6e-019; # Charge of the electron, C
K = 8.98e+09; # Coulomb force constant
#Calculations
a_c = A**(1./3)/(2*Z_Mg-1)*((M_Mg-M_Na)+(m_n-m_p))*931.47; # Coulomb coefficient, MeV
r_0 = 3./5*K*e**2/(a_c*1.6e-013); # Nuclear radius, m
#Result
print "Coulomb coefficient = %4.2f MeV\nNuclear radius = %3.1e m"%(a_c, r_0)
```

In [8]:

```
#Variable declaration
Z = 92; # Atomic number of U-236
e = 1.6e-019; # Charge of an electron, C
A = 236; # Mass number of U-236
K = 8.98e+09; # Coulomb constant,
r_o = 1.2e-015; # Distance of closest approach, m
a_s = -16.8; # Surface constant
#Calculations
E_c = -(3*K*Z*(Z-1)*e**2)/(5*r_o*A**(1./3)*1.6e-013); # Coulomb energy, MeV
E_s = a_s*A**(2./3); # Surface energy, MeV
#Result
print "Coulomb energy for U(92,236) = %5.1f MeV \nSurface energy for U(92,236) = %5.1f MeV "%(E_c, E_s)
```

In [9]:

```
import math
#Variable declaration
t_prime = 1600; # Half life of radioactive material, years
t = 2000; # Total time, years
lamda = 0.6931/t_prime; # Decay constant, years^(-1)
m0 = 1; # The mass of radioactive substance at t0, mg
#Calculations
m = m0* math.exp(-(lamda*t)); # Ratio of total number of atoms and number of atoms disintegrat, mg
a = 1-m; # The amount of radioactive substance decayed, mg
#Result
print "The amount of radioactive substance decayed : %6.4f mg"%a
```

In [10]:

```
import math
#Variable declaration
t_prime = 1600; # Half life of radioactive material, years
t = 2000; # Total time, years
lamda = 0.6931/t_prime; # Decay constant, years^(-1)
m0 = 1; # The mass of radioactive substance at t0, mg
#Calculations
m = m0* math.exp(-(lamda*t)); # Ratio of total number of atoms and number of atoms disintegrat, mg
a = 1-m; # The amount of radioactive substance decayed, mg
#Result
print "The amount of radioactive substance decayed : %6.4f mg"%a
```