# 13: Nuclear Reactions¶

## Example number 13.1, Page number 259¶

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

#Variable declaration
m1=7.0183;      #mass of 3Li7(amu)
m2=4.0040;      #mass of 2He4(amu)
m3=1.0082;      #mass of 1H1(amu)
N=6.026*10**26;      #Avgraodo no.(per kg atom)
#rxn = 3Li7 + 1H1 = 2He4 + 2He4

#Calculation
delta_m=m1+m3-(2*m2);      #deltam(amu)
E=delta_m*931;          #energy per disintegration(MeV)
n=0.1*N/7;      #no of atoms in 100 gm of lithium
TE=n*E;         #Total energy available(MeV)

#Result
print "energy available per disintegration is",round(E,2),"MeV"
print "Total energy available is",round(TE/1e+25,2),"*10**25 MeV"

energy available per disintegration is 17.22 MeV
Total energy available is 14.83 *10**25 MeV


## Example number 13.2, Page number 259¶

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

#Variable declaration
m1=6.015126;    #mass of 3Li7(a.m.u)
m2=4.002604;    #mass oh 2He4(a.m.u)
m3=1.00865;     #mass of 0n1(a.m.u)
m4=3.016049;    #mass of 1H3(a.m.u)
#rxn = 3Li7 + 0n1 = 2He4 + 1H3 + Q

#Calculation
dm=m1+m3-(m2+m4);
Q=dm*931;         #energy released(MeV)

#Result
print "energy released is",round(Q,4),"MeV"
print "answer given in the book varies due to rounding off errors"

energy released is 4.7695 MeV
answer given in the book varies due to rounding off errors


## Example number 13.3, Page number 260¶

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

#Variable declaration
m1=14.007515;     #mass of 7N14(a.m.u)
m2=4.003837;      #mass of 2He4(a.m.u)
m3=17.004533;     #mass of 8O17(a.m.u)
m4=1.008142;      #mass of 1H1(a.m.u)
#rxn = 7N14 + 2He14 = 8O17 + 1H1

#Calculation
dm=m3+m4-(m1+m2);
Q=dm*931;           #Q value of the reaction(MeV)

#Result
print "Q value of the reaction is",round(Q,3),"MeV"

Q value of the reaction is 1.232 MeV


## Example number 13.4, Page number 260¶

In [14]:
#import modules
import math
from __future__ import division

#Variable declaration
m1=14.007520;     #mass of 7N14(a.m.u)
m2=1.008986;      #mass oh 0n1(a.m.u)
#m3=mass of 6C14 in a.m.u
m4=1.008145;      #mass of 1H1(a.m.u)
#rxn = 7N14 + 0n1 = 6C14 + 1H1 + 0.55 MeV

#Calculation
Q=0.55;           #energy(MeV)
dm=Q/931;
m3=dm+m1+m2-m4;       #mass of 6C14(a.m.u)

#Result
print "mass of 6C14 is",round(m3,5),"a.m.u"
print "answer given in the book varies due to rounding off errors"

mass of 6C14 is 14.00895 a.m.u
answer given in the book varies due to rounding off errors


## Example number 13.6, Page number 261¶

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

#Variable declaration
m0=11.01280;     #mass 5B11(a.m.u)
m1=4.00387;      #mass of alpha particle(a.m.u)
m2=14.00752;     #mass of 7N14(a.m.u)
#m3=mass of neutron
E1=5.250;       #energy of alpha particle(MeV)
E2=2.139;       #energy of 7N14(MeV)
E3=3.260;       #energy of 0n1(MeV)

#Calculation
m3=(m0*931)+((m1*931)+E1)-((m2*931)+E2)-E3;          #mass of neutron(a.m.u)

#Result
print "mass of neutron is",round(m3/931,3),"a.m.u"

mass of neutron is 1.009 a.m.u