In [1]:

```
#Energy and mass equivalence of wavelength
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
me = 9.1e-031 # Mass of en electron, kg
L = 4.5e-013 # Wavelength of gamma ray, m
h = 6.626e-034 # Planck's constant, Js
c = 3e+08 # Speed of light, m/s
U = h*c/L # Energy equivalence of wavelength, J
m = U/c**2 # Mass equivalent of wavelength, kg
print "The energy equivalence of wavelength %3.1e m = %4.2f MeV"% (L, U/(e*1e+06))
print "The mass equivalence of wavelength %3.1e m = %4.2f me"% (L, m/me)
# Result
# The energy equivalence of wavelength 4.5e-013 m = 2.76 MeV
# The mass equivalence of wavelength 4.5e-013 m = 5.39 me
```

In [2]:

```
#Binding energy per nucleon for oxygen isotopes
mp = 1.007276 # Mass of proton, amu
mn = 1.008665 # Mass of neutron, amu
amu = 931 # Energy equivalent of 1 amu, MeV
# For Isotope O-16
M_O16 = 15.990523 # Mass of O-16 isotope, amu
Z = 8 # Number of protons
N = 8 # Number of neutrons
BE = (8*(mp+mn)-M_O16)*amu # Binding energy of O-16 isotope, MeV
BE_bar16 = BE/(Z+N) # Binding energy per nucleon of O-16 isotope, MeV
# For Isotope O-18
M_O18 = 17.994768 # Mass of O-18 isotope, amu
Z = 8 # Number of protons
N = 10 # Number of neutrons
BE = (8*mp+10*mn-M_O18)*amu # Binding energy of O-18 isotope, MeV
BE_bar18 = BE/(Z+N) # Binding energy per nucleon of O-18 isotope, MeV
print "The binding energy per nucleon of O-16 isotope = %5.3f MeV"% BE_bar16
print "The binding energy per nucleon of O-18 isotope = %5.3f MeV"% BE_bar18
# Result
# The binding energy per nucleon of O-16 isotope = 7.972 MeV
# The binding energy per nucleon of O-18 isotope = 7.763 MeV
```

In [5]:

```
from math import log, exp
from sympy import symbols, solve
#Range of alpha-emitters of uranium
L1 = 4.8e-018 # Decay constant of first alpha-emitter, per sec
L2 = 4.225e+03 # Decay constant of second alpha-emitter, per sec
L3 = 3.786e-03 # Decay constant of third alpha-emitter, per sec
R1 = 4.19 # Range of first alpha-emitter, cm
R2 = 7.86 # Range of second alpha-emitter, cm
# From Geiger Nuttal law, log R = A log L + B
# Putting R1, L1 and R2, L2, subtracting and solving for A
A = log(R2/R1)/log(L2/L1) # Slope of straight line between R and L
B = symbols('B') # Intercept of straight line between R and L
B = solve(log(R2)-A*log(L2)-B, B)[0] # Other constant of Geiger-Nuttal law
R3 = exp(A*log(L3)+B) # Range of third alpha-emitter of uranium, cm
print "The range of third alpha-emitter of uranium = %5.3f cm"% R3
```

In [6]:

```
#Binding energy per nucleon of helium
amu = 931 # Energy equivalent of amu, MeV
mp = 1.007895 # Mass of proton, amu
mn = 1.008665 # Mass of neutron, amu
M_He = 4.00260 # Atomic weight of helium, amu
dm = 2*(mp+mn)-M_He # Mass difference, amu
BE = dm*amu # Binding energy of helium, MeV
BE_bar = BE/4 # Binding energy per nucleon, MeV
print "The binding energy per nucleon of helium = %6.4f MeV"% BE_bar
```

In [7]:

```
#Energy released in the fusion of deuterium
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
Q = 43 # Energy released in fusion of six deuterium atoms, MeV
N = 6.023e+026 # Avogadro's number, No. of atoms per kg
n = N/2 # Number of atoms contained in 1 kg of deuterium
U = Q/6*n*e*1e+06 # Energy released due to fusion of 1 kg of deuterium, J
print "The energy released due to fusion of 1 kg of deuterium = %5.3e J"% U
```

In [9]:

```
#Mass of deuterium nucleus
amu = 1.6e-027 # Mass of a nucleon, kg
mp = 1.007895 # Mass of proton, amu
mn = 1.008665 # Mass of neutron, amu
BE = 2/931 # Binding energy of two nucleons, amu
M_D = (mp+mn-BE)*amu # Mass of a deuterium nucleus, kg
print "The mass of deuterium nucleus = %5.3e kg"% M_D
```

In [10]:

```
#Binding energy per nucleon of Ni-64
amu = 931 # Mass of a nucleon, MeV
MH = 1.007825 # Mass of hydrogen, amu
Me = 0.000550 # Mass of electron, amu
Mp = MH-Me # Mass of proton, amu
Mn = 1.008665 # Mass of neutron, amu
m_Ni = 63.9280 # Mass of Ni-64 atom, amu
MNi = m_Ni-28*Me # Mass of ni-64 nucleus, amu
m = (28*Mp+36*Mn)-MNi # Mass difference, amu
BE = m*amu # Binding energy of Ni-64, MeV
BE_bar = BE/64 # Binding energy per nucleon of Ni-64, MeV
print "The binding energy per nucleon of Ni-64 = %4.2f MeV"% BE_bar
```

In [12]:

```
#Energy released during fusion of two deuterons
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
x = 1.1 # Binding energy per nucleon of deuterium, MeV
y = 7.0 # Binding energy per nucleon of helium-4, MeV
E = (y - 2*x)*1e+06*e # Energy released when two deutron nuclei fuse together, MeV
print "The binding energy per nucleon of deuterium = %4.2e J"%E
```

In [21]:

```
from __future__ import division
#Binding energy and packing fraction of helium
amu = 931.0 # Energy equivalent of amu, MeV
mp = 1.00814 # Mass of proton, amu
mn = 1.00898 # Mass of neutron, amu
m_He = 4.00387 # Mass of helium, amu
A = 4.0 # Mass number of helium
m = 2.0*(mp+mn)-m_He # Mass difference, amu
dm = m_He - A # Mass defect of He
BE = dm*amu # Binding energy of He, MeV
p = dm/A # Packing fraction of He
print "The binding energy of helium = %6.3f MeV"% BE
print "The packing fraction of helium = %5.3e"% p
#Answer is wrong in the book.
```

In [19]:

```
#Mass of Yukawa particle
h = 6.626e-034 # Reduced Planck's constant, Js
e = 1.6e-019 # Charge on an electron, coulomb
R0 = 1.2e-015 # Nuclear radius constant, m
R = 2*R0 # Range of nuclear force, m
v = 1e+08 # Speed of the particle, m/s
S = R # Distance travelled by particle within the nucleus, m
dt = S/v # time taken by the particle to travel across the nucleus, s
# From Heisenberg's uncertainty principle, dE.dt = h_bar, solving for dE
dE = h/(1e+06*e*dt) # Energy of Yukawa paeticle, MeV
m = dE/0.51 # Approximate mass of Yukawa particle, electronic mass unit
print "The mass of Yukawa particle = %3d me"% m
```

In [20]:

```
from math import pi
#Maximum height of the potential barrier for alpha penetration
epsilon_0 = 8.854e-12 # Absolute electrical permittivity of free space, coulomb square per newton per metre square
Z = 92 # Atomic number of U-92 nucleus
z = 2 # Atomic number of He nucleus
e = 1.6e-019 # Charge on an electron, coulomb
R = 9.3e-015 # Radius of residual nucleus, m
U = 1/(4*pi*epsilon_0)*Z*z*e**2/(R*1.6e-013) # Maximum height of potential barrier, MeV
print "The maximum height of the potential barrier for alpha penetration = %2d MeV"% U
```