# chapter 6:nuclear fission¶

## example 6.1;pg no:98¶

In [1]:
#cal of excitation energy for Al and Na
#intiation of all variables
# Chapter 6
print"Example 6.1, Page:98  \n \n"
# Given:
import math
d1=7.9;# density of Gd
d2=2.31;# Density of In
a1=49.;# in Kb
a2=155.;# in b
m1=157.25;
m2=114.8;
Na=6.02*10.**23.;
# Solution:
x1=math.log(1./(1./2.))/((d1*Na*a1*10.**-24.*10.**3.)/m1);# half-thickness for Gd
x2=math.log(1./(1./2.))/((d2*Na*a2*10.**-24.)/m2);# half-thickness for In
r=x2/x1;
print"The half-thickness of In is times more than of Gd.",round(r)
Example 6.1, Page:98

The half-thickness of In is times more than of Gd. 789.0

## example 6.2;pg no:98¶

In [2]:
#cal of thickness of Pd foil
#intiation of all variables
# Chapter 6
print"Example 6.2, Page:99  \n \n"
# Given:
import math
d1=12.;# density of Gd
a1=43.11;# in b
m1=106.4;
Na=6.02*10**23;
i1=1.;
i2=1./1000.;
# Solution
x=math.log(i1/i2)/((d1*Na*a1*10.**-24.)/m1);# thickness for Pd foil
print"The thickness of Pd foil which would reduce the intensity of a beam to excatly 1/1000 of its initial value in (cm)=",round(x,2)
Example 6.2, Page:99

The thickness of Pd foil which would reduce the intensity of a beam to excatly 1/1000 of its initial value in (cm)= 2.36

## example 6.3;pg no:99¶

In [3]:
#cal of fission,barrier,activation
#intiation of all variables
# Chapter 6
print"Example 6.3, Page:99  \n \n"
# Given:
mTe=129.9067; # mol wt. of Te(52)
mCu=64.9278;# mol wt of Cu(29)
mFe=65;# mol wt of Fe(26)
# Solution
E1=(mTe-2*mCu)*931; # Fission Energy in MeV
print"The fission energy of Te(130)in MeV is =",round(E1,1)
r=((65)**0.33333);
E2=(26*26*4.8*4.8*10**-20)/(2*1.5*1.6*10**-13*10**-6*r);# Barrier energy in MeV
print"The barrier energy of Te(130) in MeV is =",round(E2,1)
E3=E2-E1;# Activation Energy in MeV
print"The activation energy of Te(130) in MeV is = ",round(E3,1)
# Since barrier energy is greater than fission energy, spontaneous fission is not possible unless the activation energy is provided.
Example 6.3, Page:99

The fission energy of Te(130)in MeV is = 47.6
The barrier energy of Te(130) in MeV is = 80.7
The activation energy of Te(130) in MeV is =  33.1

## example 6.4;pg no:99¶

In [4]:
#cal of fission,barrier,activation energies
#intiation of all variables
# Chapter 6
print"Example 6.4, Page:99  \n \n"
# Given:
mSn=113.903; # mol wt. of Sn(50)
mMn=56.9383;# mol wt of Mn(25)
mFe=57;#mol wt of Fe(26)
# Solution
E1=(mSn-2*mMn)*931; # Fission Energy in MeV
print"\n The fission energy in MeV is =",E1
r=((mFe)**0.33333);
E2=(25*25*4.8*4.8*10**-20)/(2*1.5*1.6*10**-13*10**-6*r);# Barrier energy in MeV
print"\n The barrier energy in MeV is = ",round(E2,2)
E3=E2-E1;# Activation Energy in MeV
print"\n The activation energy in MeV is =",round(E3,2)
Example 6.4, Page:99

The fission energy in MeV is = 24.5784

The barrier energy in MeV is =  77.95

The activation energy in MeV is = 53.38

## example 6.5;pg no: 100¶

In [5]:
#intiation of all variables
# Chapter 6
print"Example 6.5, Page:100  \n \n"
# Given:
a1=94;# atomic no. of Pu
a2=42;#atomic no. of Mo
a3=56;# atomic no. of Ba
# Solution:
# By principle of equal charge displacement
z1=0.5*(a1+a2-a3);
print"\n z1=",z1
z2=0.5*(a1-a2+a3);
print"\n z2=",z2

#From z1 and z2 we have the primary fragments as Zr(40), atomic mass(100) and Xe(54), atomic mass (138).
Example 6.5, Page:100

z1= 40.0

z2= 54.0

## example 6.6;pg no: 100¶

In [6]:
#intiation of all variables
# Chapter 6
print"Example 6.6, Page:100  \n \n"
# Given:
a1=92;# atomic no. of U
a2=40;#atomic no. of Zr
a3=58;# atomic no. of Ce
# Solution
# By principle of equal charge displacement
z1=0.5*(a1+a2-a3);
print"\n z1=",z1
z2=0.5*(a1-a2+a3);
print"\n z2=",z2
#From z1 and z2 we have the primary fragments are Rb(37), atomic mass(94) and Cs(55), atomic mass (140).
Example 6.6, Page:100

z1= 37.0

z2= 55.0

## example 6.7;pg no: 100¶

In [7]:
#intiation of all variables
# Chapter 6
print"Example 6.7, Page:100  \n \n"
# Given:
a1=92;# atomic no. of U
a2=42;#atomic no. of Mo
a3=56;# atomic no. of Ba
# Solution
# By principle of equal charge displacement
z1=0.5*(a1+a2-a3);
print"\n z1=",z1
z2=0.5*(a1-a2+a3-4);
print"\n z2=",z2

# From z1 and z2 we have the primary fragments are Y(39), atomic mass(95) and Sb(51), atomic mass (137)
Example 6.7, Page:100

z1= 39.0

z2= 51.0

## example 6.8;pg no: 101¶

In [8]:
#cal of fission,barrier,activation energies
#intiation of all variables
# Chapter 6
print"Example 6.8, Page:101  \n \n"
# Given:
mU=236.04533;
mU1=236.045733;
mU2=235.043933;
mY=94.912;
mSb=136.91782;
mn=1.0087;
Na=6.02*10**23;

# Solution:
E1=(mU-mY-mSb-4*mn)*931; # Fission Energy in MeV
print"The fission energy in MeV is = ",round(E1,2)
r1=((mY)**0.33333);
r2=((mSb)**0.33333);
E2=(39*51*4.8*4.8*10**-20)/(1.5*10**-13*(r1+r2)*1.6*10**-6);# Barrier energy in MeV
print"\n The barrier energy in MeV is = ",round(E2,2)
E3=E2-E1;# Activation Energy in MeV
print"\n The activation energy in MeV is = ",round(E3,2)
# Note : There is discrepancy in the final answer.
E4=(mU2+mn-mU1)*931; # Fission Energy in MeV
print"\n The fission by thermal neutrons is not possible since excitation energy is less than activation energy.",E4
Example 6.8, Page:101

The fission energy in MeV is =  168.24

The barrier energy in MeV is =  196.54

The activation energy in MeV is =  28.3

The fission by thermal neutrons is not possible since excitation energy is less than activation energy. 6.4239

## example 6.9;pg no: 101¶

In [9]:
#cal of independent fractional chain yield of Sr,Y
#intiation of all variables
# Chapter 6
print"Example 6.9, Page:101  \n \n"
# Given:
import math
nSr=38.;
nY=39.;
# Solution:
PSr=0.565*math.exp(-((nSr-40)**2)); #independent fractional chain yield of Sr
PY=0.565*math.exp(((nY-40.)**2.)); #independent fractional chain yield of Y
print"The independent fractional chain yield of Sr is =",round(PSr,2)
print"\n The independent fractional chain yield of Y is =",round(PY,2)
Example 6.9, Page:101

The independent fractional chain yield of Sr is = 0.01

The independent fractional chain yield of Y is = 1.54

## example 6.10;pg no: 102¶

In [10]:
#cal of energy for the given fission
#intiation of all variables
# Chapter 6
print"Example 6.10, Page:102  \n \n"
# Given:
mU=235.043091;
mn=1.0087;
mXe=138.9187;
mSn=94.919;
# Solution:
dm=(235.04309+1.0087-138.917-94.919-2.0174);# delta m
E=dm*931;# energy of given fission in MeV
print"The energy for the given fission is =(MeV)",round(E,2)
Example 6.10, Page:102

The energy for the given fission is =(MeV) 184.7

## example 6.11;pg no: 102¶

In [11]:
#cal of energy for the Gd(155)+Br(81)+4n,Pd(108)+Xe(129)+3n
#intiation of all variables
# Chapter 6
print"Example 6.11, Page:102  \n \n"
# Given:
mPu=239.052161;
mPd=107.903920;
mXe=128.904784;
mn=1.0087;
mGd=154.922010;
mBr=80.916344;

# Solution: Part (a)

dm1=(mPu-(mPd+mXe+2*mn));# delta m
E1=dm1*931;# energy of given fission in MeV
print"The energy for the Pd(108)+Xe(129)+3n fission is =(MeV)",round(E1,2)

dm2=(mPu-(mGd+mBr+3*mn));# delta m
E2=dm2*931;# energy of given fission in MeV
print"\n The energy for the Gd(155)+Br(81)+4n fission is =(MeV)",round(E2,2)
Example 6.11, Page:102

The energy for the Pd(108)+Xe(129)+3n fission is =(MeV) 210.46

The energy for the Gd(155)+Br(81)+4n fission is =(MeV) 174.76

## example 6.12;pg no: 103¶

In [12]:
#cal of barrier,fission energy
#intiation of all variables
# Chapter 6
print"Example 6.12, Page:103  \n \n"
# Given:

mFm=250.079;
mSn=124.9077;
Na=6.02*10**23;

# Solution:
E1=(mFm-2*mSn)*931; # Fission Energy in MeV
print"The fission energy in MeV is = ",round(E1,2)
r=((mSn)**0.33333);
E2=(50*50*4.8*4.8*10**-20)/(2*1.5*10**-13*(r)*1.6*10**-6);# Barrier energy in MeV
print"\n The barrier energy in MeV is =",round(E2)
Example 6.12, Page:103

The fission energy in MeV is =  245.41

The barrier energy in MeV is = 240.0

## example 6.13;pg no: 103¶

In [13]:
#cal of thickness of Pd foil
#intiation of all variables
# Chapter 6
print"Example 6.13, Page:103  \n \n"
# Given:
mTh1=232;
mTh2=233;
ETh1=6.4;# in MeV
ETh2=4.93;# in MeV
E=6.5;# fission barrier energy in MeV

# Solution: Part(a)
E1=0*mTh1/mTh2;
Ex1=E1+ETh2;
print"\n (a) Excitation energy is =(MeV)",Ex1
if (Ex1>E):
print"\n Fission is possible"
else:
print"\n Fission is not possible"

# Solution: Part(b)
E2=2*mTh1/mTh2;
Ex2=E2+ETh2;
print"\n (b) Excitation energy is =(MeV)",Ex2
if (Ex2>E):
print"\n Fission is possible"
else:
print"\n Fission is not possible"

# Solution: Part (c)
E3=10*mTh1/mTh2;
Ex3=E3+ETh2;
print"\n (c) Excitation energy is =(MeV)",Ex3
if (Ex3>E):
print"\n Fission is possible"
else:
print"\n Fission is not possible"
Example 6.13, Page:103

(a) Excitation energy is =(MeV) 4.93

Fission is not possible

(b) Excitation energy is =(MeV) 5.93

Fission is not possible

(c) Excitation energy is =(MeV) 13.93

Fission is possible

## example 6.14;pg no: 104¶

In [14]:
#cal of energy for the given fission
#intiation of all variables
# Chapter 6
print"Example 6.14, Page:104  \n \n"

# Given:

mEs=249.0762;
mn=1.0087;
mGd=160.9286;
mBr=86.922;

# Solution:
dm=(mEs-(mGd+mBr+mn));# delta m
E=dm*931;# energy of given fission in MeV
print"The energy for the given fission is =(MeV)",round(E,1)
Example 6.14, Page:104

The energy for the given fission is =(MeV) 201.9

## example 6.15;pg no: 104¶

In [15]:
#cal of velocity will be related as and Energy will be related
#intiation of all variables
# Chapter 6
print"Example 6.15, Page:104  \n \n"

# Given:

m=1./6.;# mass ratio of pair of fission product

# Solution:
# Velocities as well as energies are in inverse ratio of their masses.

v=(m)**(-1);# Velocity ratio
e=(m)**(-1);# Energy ratio
print"The velocity will be related as and Energy will be related as ",v,e
Example 6.15, Page:104

The velocity will be related as and Energy will be related as  6.0 6.0

## example 6.16;pg no: 104¶

In [16]:
#cal of  no. of fissions produced per second
#intiation of all variables
# Chapter 6
print"Example 6.16, Page:104  \n \n"

# Given:
P=100;# in watts

# Solution:
P1=P*10**7;# in erg/s
P2=P1/(1.6*10**-6);# in MeV/s
# 1 ifssion generates 200 MeV of energy
f=P2/200;# no. of fissions
print"The no. of fissions produced per second will be =",f
Example 6.16, Page:104

The no. of fissions produced per second will be = 3.125e+12

## example 6.17;pg no: 104¶

In [17]:
#cal of fission barrier energy
#intiation of all variables
# Chapter 6
print"Example 6.17, Page:104  \n \n"
# Given:
r0=1.4*10**-15;# nuclear radius constant in m
p=8.85*10**-12;# permittivity of free space in J^-1*C^2*m^-1
A=92;
e=1.6*10**-19;
mPd=118;
# Solution:
r=((mPd)**0.33333);
Eb=((A/2)**2)*(e**2)/(2*r0*r*4*3.14*p*1.6*10**-13);
print"The fission barrier energy is = (MeV)",round(Eb,2)
Example 6.17, Page:104

The fission barrier energy is = (MeV) 221.78

## example 6.18;pg no: 105¶

In [18]:
#cal of fission barrier energy
#intiation of all variables
# Chapter 6
print"Example 6.18, Page:105  \n \n"
# Given:
r0=1.37*10**-15;# nuclear radius constant in m
p=8.85*10**-12;# permittivity of free space in J^-1*C^2*m^-1
A=92;
e=1.6*10**-19;
mTe=140;
mZr=95;
# Solution:
r1=((mTe)**0.33333);
r2=((mZr)**0.33333);
Eb=(52*40)*(e**2)/(r0*(r1+r2)*4*3.14*p*1.6*10**-13)
print"The fission barrier energy is =(MeV)",round(Eb)
Example 6.18, Page:105

The fission barrier energy is =(MeV) 224.0

## example 6.19;pg no: 105¶

In [19]:
#cal of critical deformation energy for the fission
#intiation of all variables
# Chapter 6
print"Example 6.19, Page:105  \n \n"
# Given:
A=240.;
Z=94.;
#Solution:
Ecr=(0.89*(A**(2./3.)))-(0.02*(Z*(Z-1.)))/(A**(1./3.));
print"The critical deformation energy for the fission is =(MeV)",round(Ecr,3)
Example 6.19, Page:105

The critical deformation energy for the fission is =(MeV) 6.237