# chapter 9:isotopes for nuclear reactions¶

## example 9.1;pg no: 126¶

In :
#cal of  value of numerical constant
#intiation of all variables
# Chapter 9
print"Example 9.1, Page:126  \n \n"
#Give:
t=3600.;# in seconds
F=96500.;# in columbs
# Formula: m=0.0373fMit, Faraday's law: m=(itE)/F
#Solution:
constant=t/F;
print"The value of numerical constant is =",round(constant,4)

Example 9.1, Page:126

The value of numerical constant is = 0.0373


## example 9.2;pg no: 126¶

In :
#cal of ion current
#intiation of all variables
# Chapter 9
print"Example 9.2, Page:126  \n \n"
# Given:
m1=24*10**-6;# g per day
m2=10**-2;# g per day
i1=10**-6;# in A
#Formula: i1*m2=m1*i2
#Solution:
i2=(i1*m2)/m1;
i=i2/10**-3;# in mA
print"The ion current should be in (mA)=",round(i,4)

Example 9.2, Page:126

The ion current should be in (mA)= 0.4167


## example 9.3;pg no: 126¶

In :
#cal of net yield
#intiation of all variables
# Chapter 9
print"Example 9.3, Page:126  \n \n"
#Given:
f=1.0014;# seperation factor
s=4;# series
p=6;# parallel
# Note: The global yield for s stages in series is(f)^s and each parallel stages simply multiplies the yield of the stage, Hence overall yield with p parallel stages (each with s stages in series) will be Y=p*(f)^s
#Solution:
Y=p*(f)**s;
print"The net yield is =",round(Y,4)

Example 9.3, Page:126

The net yield is = 6.0337


## example 9.4;pg no: 127¶

In :
#cal of no. of stages
#intiation of all variables
# Chapter 9
print"Example 9.14, Page:127  \n \n"
# Given:
import math
f=1.0014;# seperation factor
#Solution: Part (a)
A1=3.5/0.72;# total enrichment
n1=math.log(A1)/math.log(f);
print"\n The no. of stages will be = ",round(n1)

#Solution: Part (b)
A2=90/0.72;# total enrichment
n2=math.log(A2)/math.log(f);
print"\n The no. of stages will be =",round(n2)

Example 9.14, Page:127

The no. of stages will be =  1130.0

The no. of stages will be = 3451.0


## example 9.5;pg no: 127¶

In :
#cal of overall seperation factor
#intiation of all variables
# Chapter 9
print"Example 9.5, Page:127  \n \n"
# Given:
f=1.01;# seperation factor
n=10;# plates
#Solution:
A=f**n;
print"The overall seperation factor is =",round(A,3)

Example 9.5, Page:127

The overall seperation factor is = 1.105


## example 9.6;pg no: 127¶

In :
#cal of overall seperation factor
#intiation of all variables
# Chapter 9
print"Example 9.6, Page:127  \n \n"
# Given:
f=1.01;# seperation factor
n=16;# plates
#Solution:
A=f**n;
print"The overall seperation factor is =",round(A,3)

Example 9.6, Page:127

The overall seperation factor is = 1.173


## example 9.7;pg no: 128¶

In :
#cal of enthalpy for the exchange reaction
#intiation of all variables
# Chapter 9
print"Example 9.7, Page:128  \n \n"
# Given:
import math
k1=3.78;
k2=2.79;
t1=298.;# in K
t2=353.;# in K
R=8.314# Gas constant
# Formula: log(k1/k2)=(H/R)*((t2-t1)t1*t2)
# Solution:
H=R*math.log(k1/k2)/((t2-t1)/(t1*t2));
print"The enthalpy for the exchange reaction in (J)=",round(H,3)

Example 9.7, Page:128

The enthalpy for the exchange reaction in (J)= 4829.008


## example 9.8;pg no: 128¶

In :
#cal of single stage seperation factor
#intiation of all variables
# Chapter 9
print"Example 9.8, Page:128  \n \n"
# Given:
a1=0.015;
a2=0.04;
# Solution: Defining the seperation factor f as approximately equal to (a2/a1) where a1, a2 are the relative abundances of the isotope of interest in the initial and final fractions, we have
f=(a2/a1);
print"The single stage seperation factor is = ",round(f,2)

Example 9.8, Page:128

The single stage seperation factor is =  2.67