# Chapter 17: Nuclear Physics¶

## Example 17.1, Page 888¶

In :
#Variable declaration
e = 1.6e-019;   # Energy equivalent of 1 eV, J/eV
m_n = 1.675e-027;   # Mass of the neutron, kg
m_p = 1.672e-027;   # Mass of the proton, kg
M_D = 3.343e-027;   # Mass of the deutron, kg
c = 3e+08;  # Speed of light, m/s

#Calculations
delta_m = m_n + m_p - M_D;  # Mass defect in the formation of deuterium, kg
BE = delta_m*c**2;   # Binding energy of the deuterium, J
BE_bar = BE/2;      # Binding energy per nucleon of deuterium, J

#Result
print "Binding energy per nucleon for the deutron = %5.3f MeV"%(BE_bar/(e*1e+06))

Binding energy per nucleon for the deutron = 1.125 MeV


## Example 17.2, Page 889¶

In :
#Variable declaration
amu = 931.5;    # Energy equivalent of 1 amu, MeV
m_n = 1.008665;   # Mass of the neutron, amu
m_p = 1.007825;   # Mass of the proton, amu
M_He = 4.002870;   # Mass of the heluim nucleus, amu
c = 3e+08;  # Speed of light, m/s

#Calculations
BE = (2*m_n+2*m_p - M_He)*amu;  # Binding energy for the alpha particle, MeV

#Result
print "The binding energy for the alpha particle = %2d MeV"%BE

The binding energy for the alpha particle = 28 MeV


## Example 17.4, Page 890¶

In :
#Variable declaration

print "Values of Z for different values of A are:"
for A in range(7,11):
Z = A/(2+0.015*A**(2./3))
print round(Z,2),

print "\n\nSince 4.36 is closer to 3, 4Be9 is more stable"

Values of Z for different values of A are:
3.41 3.88 4.36 4.83

Since 4.36 is closer to 3, 4Be9 is more stable


## Example 17.5, Page 891¶

In :
#Variable declaration
from sympy import *
c2 = Symbol('c2')

#Calculations
#For N14O17
Q1 = (14.00753+4.00206)*c2-(17.00450+1.00814)*c2
print "Since, Q = ",Q1, "which is negative, the reaction is endothermic"
#For Li7He4
Q2 = (7.01822+1.00814)*c2-(4.00206+4.00206)*c2
print "Since, Q = ",Q2, "which is negative, the reaction is endothermic"
#Answers differ due to rounding-off errors in Sympy

Since, Q =  -0.00305000000000177*c2 which is negative, the reaction is endothermic
Since, Q =  0.02224*c2 which is negative, the reaction is endothermic


## Example 17.7, Page 892¶

In :
#Variable declaration
mp = 1.00814   #amu
mn = 1.00898   #amu
Mp = 30.98356  #amu
Msi = 30.98515 #amu

#Calculations
Q = Mp+mn-Msi-mp  #amu
Q = Q*931.5       #MeV
Eth = Q*(-(Mp+mn)/Mp)

#Result
print "Threshold frequency = %.3fMeV"%Eth
#Rounding-off error

Threshold frequency = 0.721MeV


## Example 17.8, Page 892¶

In :
#Variable declaration
Q = -7.6342   #MeV
Mp = 19.0457  #amu
mi = 1.0087   #amu
me = 1.00728  #amu
Ki = 15

#calculations
Ke = (Q*Mp-(mi-Mp)*Ki)/(me+Mp)
Eth = Q*(-(Mp+mn)/Mp)

#Result
print "Kinetic energy of protons = %.3f MeV"%Ke
print "Threshold frequency = %.3f MeV"%Eth

Kinetic energy of protons = 6.241 MeV
Threshold frequency = 8.039 MeV


## Example 17.9, Page 893¶

In :
#Variable declaration
e = 1.6e-019;    # Energy equivalent of 1 eV, J/eV
N_A = 6.023e+023;    # Avogadro's number
E_f = 200*1e+06*e;    # Energy released per fission, J

#Calculations
E_mol = E_f*N_A;    # Energy released by one mole of U235, J
E = E_mol*1000/235;    # Energy released by the fission of 1 kg of U235, J

#Result
print "The Energy released by the fission of 1 kg of U235 = %4.2e kWh"%(E/(1000*3600))

The Energy released by the fission of 1 kg of U235 = 2.28e+07 kWh


## Example 17.10, Page 894¶

In :
#Variable declaration
e = 1.6e-019;    # Energy equivalent of 1 eV, J/eV
E = 3.2e+07;    # Energy released per second by the reactor, J

#Calculations
E_f = 200*1e+06*e;    # Energy released per fission, J
N = E/E_f;    # Number of fissions per second of U235, per second

#Result
print "The number of U235 atoms undergoing fissions per second = %1.0e"%N

The number of U235 atoms undergoing fissions per second = 1e+18


## Example 17.11, Page 894¶

In :
#Variable declaration
e = 1.6e-019;    # Energy equivalent of 1 eV, J/eV
N_A = 6.023e+026;    # Avogadro's number, per kmol
P = 2;    # Power produced by the fission of U235, watt

#Calculations
E_f = 200*1e+06*e;    # Energy released per fission, J
FR = P/E_f;    # Fission rate of U235, fission/sec
N = 0.5/235*N_A;    # Number of U235 nuclei in 0.5 kg of U235
E = 200*N;    # Energy released in the complete fissioning of 0.5 kg of U235, MeV

#Results
print "The fission rate of U235 = %4.2e fissions/sec"%FR
print "The energy released in the complete fissioning of 0.5 kg of U235 = %1.0e kcal"%(E*1e+06*e/(1000*4.186))

The fission rate of U235 = 6.25e+10 fissions/sec
The energy released in the complete fissioning of 0.5 kg of U235 = 1e+10 kcal


## Example 17.12, Page 894¶

In :
from math import *

#Variable declaration
e = 1.6e-019;    # Energy equivalent of 1 eV, J/eV
R_max = 0.6;    # Radius of two dees of the cyclotron, m
B = 1.6;        # Strength of pole pieces of the cyclotron, tesla
# For proton
m = 1.67e-027;    # Mass of the proton, kg
q = 1.6e-019;    # Charge on a proton, C

#Calculations&Result
E = 1./2*q**2*R_max**2*B**2/(m*e*1e+06);    # Energy of the proton, MeV
f_proton = q*B/(2*pi*m*1e+06);    # Cyclotron oscillator frequency for the proton, MHz
print "Energy of the proton = %5.2f MeV"%E
print "Cyclotron frequency for proton = %5.2f MHz"%f_proton
# For deuteron
m = 2*1.67e-027;    # Mass of the deuteron, kg
q = 1.6e-019;    # Charge on a deuteron, C
E = 1./2*q**2*R_max**2*B**2/(m*e*1e+06);    # Energy of the deuteron, MeV
f_deuteron = q*B/(2*pi*m*1e+06);    # Cyclotron oscillator frequency for the deuteron, MHz
print "Energy of the deuteron = %5.2f MeV"%E
print "Cyclotron frequency for deuteron = %5.2f MHz"%f_deuteron
# For alpha-particle
m = 4*1.67e-027;    # Mass of the alpha-particle, kg
q = 2*1.6e-019;    # Charge on a alpha-particle, C
E = 1./2*q**2*R_max**2*B**2/(m*e*1e+06);    # Energy of the deuteron, MeV
f_alpha = q*B/(2*pi*m*1e+06);    # Cyclotron oscillator frequency for the alpha-particle, MHz
print "Energy of the alpha-particle = %5.2f MeV"%E
print "Cyclotron frequency for alpha-particle = %5.2f MHz"%f_alpha

Energy of the proton = 44.15 MeV
Cyclotron frequency for proton = 24.40 MHz
Energy of the deuteron = 22.07 MeV
Cyclotron frequency for deuteron = 12.20 MHz
Energy of the alpha-particle = 44.15 MeV
Cyclotron frequency for alpha-particle = 12.20 MHz


## Example 17.13, Page 895¶

In :
from math import *

#Variable declaration
e = 1.67e-013;    # Energy equivalent of 1 eV, J/eV
R_max = 0.75;    # Radius of two dees of the cyclotron, m
f = 15e+06;    # Frequency of alternating potential, Hz
m = 1.67e-027;    # Mass of the proton, kg

#Calculations
# As E = 1/2*q^2*R_max^2*B^2/(m*e) and f = q*B/(2*%pi*m), solving for E
E = (2*pi**2*m*f**2*R_max**2)/(e);

#Result
print "Energy of the protons issuing out of the cyclotron = %6.4f MeV"%E

Energy of the protons issuing out of the cyclotron = 24.9824 MeV


## Example 17.14, Page 896¶

In :
from math import *

#Variable declaration
e = 1.6e-019;    # Charge on an electron, C
c = 3e+08;        # Speed of light, m/s
B_orbit = 0.5;    # Magnetic field at the orbit of the betatron, T
f = 60;    # Operating frequency of the betatron, Hz

#Calculations
omega = 2*pi*f;    # Angular frequency of operation, rad/s
r = 1.6/2;    # Radius of stable orbit, m
K_av = 4*omega*e*r**2*B_orbit/1.6e-019;    # Average energy gained by the electron per turn, eV
K_max = c*e*r*B_orbit/1.6e-019;    # Maximum energy gained by the eectron, eV

#Results
print "The average energy gained by the electron per turn = %5.1f eV"%K_av
print "The maximum energy gained by the electron = %5.1e eV"%K_max

The average energy gained by the electron per turn = 482.5 eV
The maximum energy gained by the electron = 1.2e+08 eV


## Example 17.15, Page 896¶

In :
#Variable declaration
q = 1.6e-019;    # Charge on a deuteron, C
amu = 931.5;    # Energy equivalent of 1 amu, MeV
m0 = 2.0141;    # Rest mass of a deuteron, kg
B0 = 1.5;    # Magnetic field at the centre of the synchrocyclotron, T
B = 1.431;    # Magnetic field at the periphery of the synchrocyclotron, T

#Calculations
f0 = q*B0/(2*3.14*m0*1.67e-027*1e+06);    # Maximum frequency of Dee voltage of synhrocyclotron, MHz
f = 1e+07;    # Minimum frequency of Dee voltage, Hz
m = (q*B)/(2*3.14*f*1.67e-027);    # Mass of deuteron at the periphery of the Dee, amu
K = (m-m0)*amu;    # Gain in energy of the deuteron, MeV

#Results
print "The maximum frequency of Dee voltage = %5.2f MHz"%f0
print "The gain in energy of the deuteron = %6.2f MeV"%K

The maximum frequency of Dee voltage = 11.36 MHz
The gain in energy of the deuteron = 157.47 MeV


## Example 17.16, Page 897¶

In :
from math import *

#Variable declaration
V = 1000;    # Operating voltage of the GM counter, volt
a = 1e-04    # Radius of GM counter wire, m
b = 2e-02;    # Radius of cathode, m

#Calculations
E = V/(2.3026*a*log10(b/a));    # Maximum radial field at the surface of central wire of GM tube, V/m
tau = 1e+09;    # Life time of GM tube, counts
N = tau/(50*60*60*2000);    # Life of the GM counter, years

#Results
print "The maximum radial field at the surface of central wire of GM tube = %4.2e V/m"%E
print "The life of the GM counter = %4.2f years"%N

The maximum radial field at the surface of central wire of GM tube = 1.89e+06 V/m
The life of the GM counter = 2.78 years


## Example 17.17, Page 898¶

In :
from math import *

#Variable declaration
I = 15.7;    # Ionization potential of argon in GM counter, volt
a = 0.012/2*1e-02;    # Radius of GM counter wire, m
b = 5./2*1e-02;    # Radius of cathode, m
lamda = 7.8e-006;    # Mean free path of argon in GM counter, m

#Calculations
# As E*lambda = I = V*lambda/(2.3026*a*log10(b/a)), solving for V
V = 2.3026*a*I*log10(b/a)/lamda;    # Voltage that must be applied to produce an avalanche in GM tube, volt

#Result
print "The voltage that must be applied to produce an avalanche in GM tube = %6.2f volt"%V

The voltage that must be applied to produce an avalanche in GM tube = 728.52 volt


## Example 17.18, Page 898¶

In :
#Variable declaration
count_err = 1e-03;    # Fractional error in counting
m = 3;    # Plateau slope

#Calculations
delta_V = count_err*100/m*100;    # Maximum permissible voltage fluctuation in a GM counter, volt

#Result
print "The maximum permissible voltage fluctuation in a GM counter = %3.1f volts"%delta_V

The maximum permissible voltage fluctuation in a GM counter = 3.3 volts