In [1]:

```
#Variable declaration
N_A = 6.02e+023; # Avogadro's number
e = 1.6e-019; # Charge on an electron, C
q = 2*e; # Charge on the alpha particle, C
rho = 1.9; # Density of carbon target, atoms/cc
N_M = 1; # Number of atoms per molecule
M_g = 12; # Gram atomic mass of C12 isotope, g/mol
sigma = 25e-031; # Total cross section for the reaction, Sq.m
t = 1e-006; # Thickness of carbon target, m
I_beam = 1e-006; # Beam current of akpha particle, ampere
time = 3600; # Time for which the alpha particle beam is incident on the target, s
#Calculations
n = rho*N_A*N_M/M_g; # Number of nuclei per unit volume, per cc
P = n*t*sigma*1e+006; # Probability of scattering of alpha particles
N_I = I_beam*time/q; # Number of incident alpha particles
N_n = N_I*P; # Number of neutrons produced in the reaction
#Result
print "The number of neutrons produced in the reaction = %3.1e neutrons"%N_n
#answer differs due to rounding errors
```

In [2]:

```
#Variable declaration
sigma_n = 3; # Differential cross setion of production of the neutron, mb/sr
sigma_p = 0.2; # Differential cross setion of production of the proton, mb/sr
#Calculations
# As P_n = sigma_n and P_p = sigma_p, so
P_ratio = sigma_n/sigma_p; # The likelihood of a neutron production than a proton
#Result
print "The likelihood of the neutron production than the proton = %2d"%P_ratio
```

In [3]:

```
#Variable declaration
u = 931.5; # Energy equivalent of 1 amu, MeV
M_He = 4.002603; # Mass of He-4 nucleus, u
M_N = 14.003074; # Mass of N-14 nucleus, u
M_H = 1.007825; # Mass of hydrogen nucleus, u
M_O = 16.999132; # Mass of O-16 nucleus, u
K_alpha = 7.7; # The kinetic energy of alpha particle, MeV
#Calculations
Q = (M_He + M_N - M_H - M_O)*u; # The ground state Q-value of the nuclear reaction, MeV
# As Q = K_p + K_O - K_alpha, solving for K_p + K_O
K = Q + K_alpha; # The sum of kinetic energy of the products, MeV
#Results
print "The ground state Q-value of the endoergic nuclear reaction = %5.3f MeV"%Q
print "The sum of kinetic energy of the products = %3.1f MeV"%K
```

In [4]:

```
#Variable declaration
u = 931.5; # Energy equivalent of 1 amu, MeV
K_lab = 14.6; # Kinetic energy of incident aplha particle, MeV
Mx = 4; # Mass number of projectile nucleus
MX = 12; # Mass number of target nucleus
M_He = 4.002603; # Mass of He nucleus, u
M_C = 12.0 # Mass of carbon nucleus, u
M_O = 15.994915; # Mass of oxygen nucleus, u
#Calculations
K_cm = MX/(Mx + MX)*K_lab; # Kinetic energy available in the centre of mass, MeV
E_O = (M_He + M_C - M_O)*u; # The ground state excitation energy of O-16, MeV
E_final_O = K_cm + E_O; # The final excitation energy of O-16, MeV
#Result
print "The final excitation energy of O-16 = %4.2f MeV"%E_final_O
```

In [5]:

```
import math
#Variable declaration
u = 931.5; # Energy equivalent of 1 amu, MeV
M_U235 = 235.0439; # Mass of U-235 nucleus, u
m_n = 1.0087; # Mass of a neutron, u
M_Zr99 = 98.9165; # Mass of Zr-99 nucleus, u
M_Te134 = 133.9115; # Mass of Te-134 nucleus, u
#Calculations
Q = (M_U235 + m_n - M_Zr99 - M_Te134 - 3*m_n)*u; # Q-value of the reaction, MeV
#Result
print "The ground state Q-value of the induced fission reaction = %3d MeV"%math.ceil(Q)
```

In [6]:

```
#Variable declaration
u = 931.5; # Energy equivalent of 1 amu, MeV
m_n = 1.0087; # Mass of a neutron, u
M_U235 = 235.0439; # Mass of U-235 nucleus, u
M_U236 = 236.0456; # Mass of U-236 nucleus, u
M_U238 = 238.0508; # Mass of U-238 nucleus, u
M_U239 = 239.0543; # Mass of U-239 nucleus, u
#Calculations
E_U236 = (m_n + M_U235 - M_U236)*u; # Excitation energy of U-236 nucleus, MeV
E_U239 = (m_n + M_U238 - M_U239)*u; # Excitation energy of U-239 nucleus, MeV
#Results
print "The excitation energy of U-236 nucleus = %3.1f MeV"%E_U236
print "The excitation energy of U-239 nucleus = %3.1f MeV"%E_U239
```

In [7]:

```
#Variable declaration
t = 30e-003; # Time during which the number of fissions is to be calculated, s
E = 185; # Energy produced for each fission, MeV
delta_t = 5e-006; # Average time during which a neutron is captured, s
#Calculations
fs = t/delta_t; # Number of fission cycles within 30 ms
N = (1.01)**fs; # Number of fissions that occur in 30 ms
E_total = N*E; # Total energy produced in 30 ms, MeV
#Results
print "The total number of fissions that occur in %d ms = %3.1e"%(t/1e-003, N)
print "The total energy produced = %3.1e MeV"%E_total
#Incorrect solution in textbook
```

In [8]:

```
#Variable declaration
u = 931.5; # Energy equivalent of 1 amu, MeV
M_He = 4.002603; # Mass of He nucleus, u
M_C = 12.0 # Mass of carbon nucleus, u
M_O = 15.994915; # Mass of oxygen nucleus, u
#Calculations
Q = (M_He + M_C - M_O)*u; # Q-value of the reaction, MeV
#Result
print "The energy expended in the fusion reaction inside supergiant star = %3.1f MeV"%Q
```

In [9]:

```
#Variable declaration
k = 1.38e-023; # Boltzmann constant, J/K
r = 3e-015; # Distance at which the nuclear force becomes effective, m
e = 1.6e-019; # Charge on an electron, C
K = 9e+009; # Coulomb's constant, N-Sq.m/C^2
#Calculations
V = K*e**2/r; # Coulomb potential energy, J
# As V = 3/2*k*T, solving for T
T = 2./3*V/k; # The ignition temperature needed for the fusion reaction between deuterium and a tritium, K
#Result
print "The ignition temperature needed for the fusion reaction between a deuterium and a tritium = %3.1e K"%T
```

In [10]:

```
#Variable declaration
k = 1.38e-023; # Boltzmann constant, J/K
e = 1.6e-019; # Energy equivalent of 1 eV, J
h = 6.62e-034; # Planck's oconstant, Js
m = 1.67e-027; # Mass of the neutron, kg
lamda = 0.060e-009; # Wavelength of the neutron beam, m
#Calculations
p = h/lamda; # Momentum of the neutron from de-Broglie relation, kg-m/s
K = p**2/(2*m*e); # Kinetic energy of the neutron needed to study atomic structure, eV
# As K = 3/2*k*T, solving for T
T = 2./3*K*e/k; # The temperature of the neutron needed to study atomic structure, K
#Result
print "The energy and temperature of the neutron needed to study the atomic structure of solids = %4.2f eV and %4d K respectively"%(K, T)
```