Chapter 13:Nuclear Reaction and Application

Example 13.1, Page 417

In [4]:
#initiation of variable
v=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23     #given values and various constants in suitable units
M=56.0;N=m*Na/M;                                 #number of atoms
i=3.0*10**-6;
q=1.6*10**-19;

#calculation
Io=i/q;                 #intensity
s=0.6*10**-24;S=1;             #given values in suitable units
R=N*s*Io/S;                     #rate of neutrons

#result
print"The rate of neutrons emitted from the target in particles per second is %.1e" %round(R,3);
print"Slight difference in answer due to approximation error"

The rate of neutrons emitted from the target in particles per second is 9.6e+07
slight difference due to approximation error

Example 13.2, Page 419

In [7]:
#initiation of variable
A=197.0; m=30*10**-3;phi=3.0*10**12;    #given values and various constants taken in suitable units
Ar=99.0*10**-24; Na=6.023*10**23

#calculation
R=(phi*Na*Ar*m/A);        #rate or production of gold
t=2.7*24*60         # time of decay
Act=R*(0.693/t);     #activity /sec
ActCi=Act/(3.7*10**4);     # in terms of curie(Ci)

#result
print"The activity is found out to be %.1e" %round(Act,3),"/sec i.e " ,round(ActCi,3),"muCi"
print"Slight difference in answer due to approximation error"

The activity is found out to be 4.9e+06 /sec i.e  131.229 muCi
slight difference due to approximation error

Example 13.3, Page 423

In [6]:
#initiation of variable
from math import exp
v=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3
p=8.9;         #density in g/cm3
m=p*v;Na=6.023*10**23 #mass and Avagadro's number
M=58.9;        #Given values

#calculation
N=m*Na/M;
i=12*10**-6;       #thickness of beam
q=1.6*10**-19;
Io=i/(2*q);      #intensity
s=0.64*10**-24;   #Given values
S=1.5*1.5;
R=N*s*Io/S;       #rate of production of 61Cu

#result
print "The rate of neutrons emitted from the target in particles/second is %.1e" %round(R,3);

#part b
act=R*(1-(exp((0.693)*(-2/3.41))));          #activity

#result
print"The activity after 2.0h in /sec is %.1e" %round(act,3),"=4.9mCi";

The rate of neutrons emitted from the target in particles/second is 5.5e+08
The activity after 2.0h in /sec is 1.8e+08 =4.9mCi

Example 13.4, Page 425

In [5]:
#initiation of variable
m2H=2.014102;    #mass of various particles
mn=1.008665;m63Cu=62.929599;
m64Zn=63.929145;c2=931.5;     #c^2=931.5 MeV
Q=(m2H+m63Cu-mn-m64Zn)*c2;       #Q of the reaction

#result
print"The value of Q is in MeV",round(Q,3);


#part b
Kx=12.00;Ky=16.85;
Ky=Q+Kx-Ky              #kinetic energy of 64Zn

#result
print"The value of Ky was found out to be in MeV",round(Ky,3);

The value of Q is in MeV 5.487
The value of Ky was found out to be in MeV 0.637

Example 13.5, Page 425

In [6]:
#initiation of variable
mp=1.007825;m3H=3.016049;    #mass of the particle
m2H=2.014102;c2=931.5;      #constant
Q=(mp+m3H-(2*m2H))*c2;     #Q of the reaction

#result
print"The value of q was found out to be in MeV",round(Q,3);

#partb
Kth1= -Q*(1+(mp/m3H));        #threshold energy of kinetic energy
Kth2=-Q*(1+(m3H/mp));        #threshold kinetic energy in case2

#result
print"The threshold kinetic energy in case-1 in MeV",round(Kth1,3);
print"The threshold kinetic energy in case-2 in MeV",round(Kth2,3);

The value of q was found out to be in MeV -4.033
The threshold kinetic energy in case-1 in MeV 5.381
The threshold kinetic energy in case-2 in MeV 16.104