# 9: Nuclear Reactions¶

## Example number 1, Page number 272¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
rho=19.3*10**3;         #density(kg/m**3)
M=197;                  #molecular weight
a=2*10**12;             #neutrons/m**2 sec
A=5*10**-4;             #area(m**2)
sigma=94*10**-28;               #reaction cross section(m**2)
t=0.3*10**-3;                  #thickness(m)

#Calculations
n=rho*N/M;              #number of nuclei per unit volume(per m**3)
N0=a*A;                 #number of neutrons hitting the target
N0_N=N0*(1-math.exp(-n*sigma*t));       #number of nuclei produced per second

#Result
print "number of nuclei produced per second is",round(N0_N/10**5,2),"*10**5"
print "answer given in the book is wrong"

number of nuclei produced per second is 1.66 *10**5
answer given in the book is wrong


## Example number 2, Page number 275¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
M1=2.01472;       #molecular mass of H(amu)
M0=7.01784;       #molecular mass of Li(amu)
M2=8.00776;       #molecular mass of Be(amu)
M3=1.00893;       #molecular mass of n(amu)
Ek1=10;           #energy(MeV)

#Calculations
M1M0=M1+M0;       #mass of interacting particles(amu)
M2M3=M2+M3;       #mass of product particles(amu)
Q=(M1M0-M2M3)*931; #decrease in mass(MeV)
Ek3=(Q+(Ek1*(1-(M1/M2))))/(1+(M3/M2));   #energy of neutron(MeV)
Ek2=Q+Ek1-Ek3;        #energy of Be(MeV)
phi=math.atan(math.sqrt(Ek3*M3/(Ek1*M1)));     #angle of recoil of Be atom(rad)
phi=phi*180/math.pi;                   #angle of recoil of Be atom(degrees)

#Result
print "energy of neutron is",round(Ek3,3),"MeV"
print "energy of Be is",round(Ek2,3),"MeV"
print "angle of recoil of Be atom is",round(phi,3),"or",int(round(phi)),"degrees"

energy of neutron is 19.768 MeV
energy of Be is 5.007 MeV
angle of recoil of Be atom is 44.855 or 45 degrees


## Example number 3, Page number 277¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
Q=-3.9;     #Q value of reaction(MeV)
M1=1.0087;       #molecular mass of incident neutron(amu)
M2=18.99;        #molecular mass of O nucleus(amu)
M3=1.0078;       #molecular mass of proton(amu)
Ek1=10;          #energy of incident neutron(MeV)

#Calculations
x=1-(M1/M2);
y=1+(M3/M2);
Ek3=(Q+Ek1*x)/y;   #energy of emitted protons(MeV)

#Result
print "energy of emitted protons is",round(Ek3,1),"MeV"

energy of emitted protons is 5.3 MeV