# Chapter 13: Nuclear Interactiions and Applications¶

## Example 13.1, Page 479¶

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

The number of neutrons produced in the reaction = 2.7e+09 neutrons


## Example 13.2, Page 480¶

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

The likelihood of the neutron production than the proton = 15


## Example 13.3, Page 481¶

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

The ground state Q-value of the endoergic nuclear reaction = -1.192 MeV
The sum of kinetic energy of the products = 6.5 MeV


## Example 13.4, Page 485¶

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

The final excitation energy of O-16 = 7.16 MeV


## Example 13.5, Page 487¶

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)

The ground state Q-value of the induced fission reaction = 185 MeV


## Example 13.6, Page 488¶

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

The excitation energy of U-236 nucleus = 6.5 MeV
The excitation energy of U-239 nucleus = 4.8 MeV


## Example 13.7, Page 490¶

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

The total number of fissions that occur in 30 ms = 8.5e+25
The total energy produced = 1.6e+28 MeV


## Example 13.8, Page 500¶

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

The energy expended in the fusion reaction inside supergiant star = 7.2 MeV


## Example 13.9, Page 502¶

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

The ignition temperature needed for the fusion reaction between a deuterium and a tritium = 3.7e+09 K


## Example 13.10, Page 509¶

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)

The energy and temperature of the neutron needed to study the atomic structure of solids = 0.23 eV and 1760 K respectively