## Example number 12.1, Page number 351¶

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
M7Li3=7.018232; #mass of 7li3(amu)
Malpha=4.003874;  #mass of alpha particle(amu)
Mpr=1.008145;  #mass of proton(amu)
Ey=9.15;   #K.E energy of product nucleus

#Calculation
#xMy -> x-mass no., M-element, y-atomic no.
#reaction:- 7li3 + 1H1-> 4He2 + 4He2
deltaM=M7Li3+Mpr-2*Malpha;  #mass defect(amu)
Q=deltaM*931;   #mass defect(MeV)
Ex=2*Ey-Q;    #K.E of incident particle(MeV)

#Result
print "kinetic energy of incident proton is",round(Ex,4),"MeV"

kinetic energy of incident proton is 0.9564 MeV


## Example number 12.2, Page number 351¶

In [3]:
#importing modules
import math
from __future__ import division

#Variable declaration
M235U=235;   #atomic mass of 235U
m=10**-3;    #mass of fissions(gm)
Eperfi=200*10**6;   #energy per fission(eV)
T=10**-6;    #time(s)

#Calculation
E=Eperfi*1.6*10**-19;   #energy per fission(J)
A=M235U;
P=((m*N)/A)*(E/T);    #power explosion(Watt)

#Result
print "power of explosion is",P,"Watt"

power of explosion is 8.20153191489e+13 Watt


## Example number 12.4, Page number 352¶

In [13]:
#importing modules
import math
from __future__ import division

#Variable declaration
n=0.4;  #efficiency
Eperfi=200*10**6;  #energy per fission(eV)
P=100*10**6;    #electric power(W)
A=235;

#Calculation
E=Eperfi*1.6*10**-19;   #energy per fission(J)
T=24*60*60;    #time(sec)
N235=P*T/(E*n);   #number of atoms in 235 kg of U235
m=(A*N235)/N;     #mass of 235U consumed/day(kg)

#Result
print "mass of 235U consumed/day is",int(m*10**3),"g"

mass of 235U consumed/day is 261 g


## Example number 12.5, Page number 352¶

In [15]:
#importing modules
import math
from __future__ import division

#Variable declaration
M2H1=2.01474;   #mass of M2H1(amu)
M3H1=3.01700;   #mass of M3H1(amu)
M1n0=1.008986;  #mass of M1n0(amu)
M4He2=4.003880;  #mass of M4He2(amu)

#Calculation
#thermonuclear reaction in hydrogen bomb explosion
#2H1 + 3H1 -> 4He2 + 1n0
Mreac=M2H1+M3H1;   #mass of reactants(amu)
Mprod=M4He2+M1n0;  #mass of products(amu)
Q=Mreac-Mprod;   #amount of energy released per reaction(J)
Q=Q*931;    #amount of energy released per reaction(MeV)

#Result
print "amount of energy released per reaction is",round(Q,3),"MeV"

amount of energy released per reaction is 17.572 MeV


## Example number 12.6, Page number 353¶

In [16]:
#importing modules
import math
from __future__ import division

#Variable declaration
M7Li3=7.01818;   #mass of Li atom(amu)
M1H1=1.0081;     #mass of H atom(amu)
M1n0=1.009;     #mass of neutron(amu)

#Calculation
BEpernu=(1/7)*((3*M1H1)+(4*M1n0)-M7Li3);  #binding energy per nucleon(J)
BEpernu=BEpernu*931;   #binding energy per nucleon(MeV)

#Result
print "binding energy per nucleon is",BEpernu,"MeV"

binding energy per nucleon is 5.60196 MeV


## Example number 12.7, Page number 353¶

In [19]:
#importing modules
import math
from __future__ import division

#Variable declaration
m=10*10**3;    #mass of U235(gm)

power output is 9.488 *10**9 Watt