14: Nuclear Fission And Fusion

Example number 14.1, Page number 269

In [1]:
#import modules
import math
from __future__ import division

#Variable declaration
E1=7.8;     #avg. B.E per nucleon(MeV)
E2=8.6;     #for fissin fragments(MeV)

#Calculation
FER=(234*E2)-(236*E1);     #Fission energy released(MeV)

#Result
print "Fission energy released is",FER,"MeV"

Fission energy released is 171.6 MeV

Example number 14.2, Page number 273

In [4]:
#import modules
import math
from __future__ import division

#Variable declaration
m1=235.044;    #mass of 92U235(a.m.u)
m2=97.905;     #mass of 42Mo98(a.m.u)
m3=135.917;    #mass of 54Xe136(a.m.u)
#rxn = 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 -1e0 + 2 0n1

#Calculation
LHSm=1.009+m1;
RHSm=m2+m3+(4*0.00055)+(2*1.009);
delta_m=LHSm-RHSm;        #mass defect(a.m.u)
E=delta_m*931;         #energy released(MeV)

#Result
print "mass defect is",round(delta_m,3),"a.m.u"
print "energy released is",int(E),"MeV"

mass defect is 0.211 a.m.u
energy released is 196 MeV

Example number 14.3, Page number 274

In [8]:
#import modules
import math
from __future__ import division

#Variable declaration
m1=1.00813;    #mass of 1H1(a.m.u)
m2=4.00386;    #mass of 2He4(a.m.u)
SC=1.35;     #solar constant(kW/m^2)
d=1.5*10**11;    #distance b/w earth and sum(m)
e=1.6*10**-19;   #the charge on electron(C)
Na=6.02*10**26;    #Avgraodo no.(per kg mole)
#rxn = 4 1H1 = 2He4 + 2 1e0

#Calculation
dm=(4*m1)-m2;
E=dm*931;     #energy produced(MeV)
EP=E/4;      #energy produced per atom(MeV)
EP=EP*10**6*e;    #conversion in J
EPkg=EP*Na;     #energy produced by 1 kg of hydrogen
SC=SC*1000;     #conversion(J/s-m^2)
SA=4*math.pi*d**2;   #surface area of sphere
ER=SC*SA;      #energy recieved per second
m=ER/EPkg;     #mass of hydrogen consumed(tonnes/second)

#Result
print "mass of hydrogen consumed is",round(m/10**11,3),"*10**8 tonnes/second"

mass of hydrogen consumed is 5.941 *10**8 tonnes/second

Example number 14.4, Page number 275

In [11]:
#import modules
import math
from __future__ import division

#Variable declaration
m1=2.01478;     #mass of 1H2(a.m.u)
m2=4.00388;     #mass of 2He4(a.m.u)
#rxn 1H2 + 1H2 = 2He4 + Q

#Calculation
Q=2*m1-m2;        #energy liberated(MeV) 
Q=Q*931;          #conversion in MeV

#Result
print "energy liberated is",round(Q,1),"MeV"

energy liberated is 23.9 MeV