19: Nuclear reactions

Example number 19.1, Page number 368

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

#Variable declaration
m2H=2.014102;    #atomic mass of 2H(u)
mn=1.008665;    #mass of n(u)
m63Cu=62.929599;   #mass of 63Cu(u)
m64Zn=63.929144;   #mass of m64Zn(u)
E=931.5;    #energy(MeV)
Kx=10;   #energy of deutron(MeV)
Ky=15;   #energy of neutron(MeV)

#Calculation
Q=E*(m2H+m63Cu-mn-m64Zn);    #Q-value(MeV)
KY=Q+Kx-Ky;    #kinetic energy(MeV)

#Result
print "Q-value is",round(Q,3),"MeV"
print "kinetic energy is",round(KY,3),"MeV"

Q-value is 5.488 MeV
kinetic energy is 0.488 MeV

Example number 19.2, Page number 368

In [5]:
#importing modules
import math
from __future__ import division

#Variable declaration
m19F=18.998404;    #atomic mass of 19F(u)
mH=1.007825;   #mass of H(u)
m19O=19.003577;   #mass of 19O(u)
mn=1.008665;   #mass of n(u)
E=931.5;    #energy(MeV)

#Calculation
Q=E*(m19F+mn-mH-m19O);    #Q-value(MeV)
Kxmin=-Q*(1+(mn/m19F));   #threshold energy(MeV)

#Result
print "Q-value is",round(Q,4),"MeV"
print "threshold energy is",round(Kxmin,2),"MeV"

Q-value is -4.0362 MeV
threshold energy is 4.25 MeV

Example number 19.3, Page number 373

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

#Variable declaration
mn=1.008665;   #mass of n(u)
mu=235.043924;    #mass of 235U(u)
mBa=140.91440;    #mass of 141Ba(u)
mKr=91.92630;    #mass of Kr(u)
E=931.5;    #energy(MeV)

#Calculation
mr=mn+mu;    #mass of reactants(u)
mp=mBa+mKr+(3*mn);   #mass of products(u)
md=mr-mp;   #mass difference(u)
E=md*E;    #energy released(MeV)

#Result
print "energy released is",round(E,1),"MeV"

energy released is 173.2 MeV

Example number 19.4, Page number 373

In [20]:
#importing modules
import math
from __future__ import division

#Variable declaration
E=200*10**6;    #energy released(eV)
e=1.6*10**-19;   #conversion factor from J to eV
P=300*10**6;   #power(W)
t=1;    #time(s)

#Calculation
n=P*t/(E*e);    #number of fissions per second

#Result
print "number of fissions per second is",n

number of fissions per second is 9.375e+18

Example number 19.5, Page number 378

In [31]:
#importing modules
import math
from __future__ import division

#Variable declaration
m2H1=2*1.66*10**-27;    #mass of proton(kg)
E=931.5;    #energy(MeV)
m1=2.014102;
m2=3.01609;
mH=1.007825;   #mass of H(u)

#Calculation
E=E*((2*m1)-m2-mH);    #energy released(MeV)
n=0.001/m2H1;    #number of nuclei
Eg=n*E/2;   #energy released per gm(MeV)

#Result
print "energy released per gm is",round(Eg/10**23,2),"*10**23 MeV"
print "answer given in the book is wrong"

energy released per gm is 6.02 *10**23 MeV
answer given in the book is wrong

Example number 19.6, Page number 379

In [35]:
#importing modules
import math
from __future__ import division

#Variable declaration
k=8.99*10**9;   #value of k(Nm**2/C**2)
rd=1.5*10**-15;   #radius of deuterium nucleus(m)
rt=1.7*10**-15;   #radius of tritium nucleus(m)
e=1.6*10**-19;   #conversion factor from J to eV
KE=0.225;    #kinetic energy for 1 particle(MeV)
k=1.38*10**-23;   #boltzmann constant(J/K)

#Calculation
K_E=k*e**2/(e*(rd+rt));    #kinetic energy of 2 particles(MeV)
T=2*KE*e*10**6/(3*k);   #temperature(K)

#Result
print "temperature is",round(T/10**9),"*10**9 K"

temperature is 2.0 *10**9 K