# chapter 5:nuclear reactions¶

## example 5.1;pg no:80¶

In [1]:
#cal of excitation energy for Al and Na
#intiation of all variables
# Chapter 5
print"Example 5.1, Page:80  \n \n"
# Given:
mMg=23.985045;
md=2.014102;
mAl=25.986900;
mNe=19.99244;
mNa=21.944;
# Solution:
# for compound nucleus Al*(26)
KE1=(24./26.)*8.;
BE1=(mMg+md-mAl)*931.;# in MeV
EE1=BE1+KE1;
# for compound nucleus Na*(22)
KE2=(20./22.)*8.;
BE2=(mNe+md-mNa)*931.;# in MeV
EE2=BE2+KE2;
print"The excitation energy for Al*(26) in MeV= and that of Na*(22) in MeV=",round(EE1,2),round(EE2,4)

Example 5.1, Page:80

The excitation energy for Al*(26) in MeV= and that of Na*(22) in MeV= 18.79 65.4993


## example 5.2;pg no:81¶

In [2]:
#cal of energy of protons
#intiation of all variables
# Chapter 5
print"Example 5.2, Page:81  \n \n"
# Given:
import math
E0=5.;# in MeV
m=1.;
M=7.

# Solution:

Erecoil=(4*5*m*M*((math.sin(45*3.14/180))**(2)))/((m+M)**2);
Escat=E0-Erecoil;

print"The energy of protons scattered through an angle of 90 deg. in MeV=",round(Escat,2)
Eresi=E0-0.48;

Erecoil2=(14/64)*Eresi;
Escat2=Eresi-Erecoil2;
print"\n \n The energy of proton observed at 90 deg. after they have excited the lithium to a level of 0.48 MeV is = MeV",Escat2

Example 5.2, Page:81

The energy of protons scattered through an angle of 90 deg. in MeV= 3.91

The energy of proton observed at 90 deg. after they have excited the lithium to a level of 0.48 MeV is = MeV 4.52


## example 5.3;pg no:81¶

In [3]:
#cal of  mass difference between A & B
#intiation of all variables
# Chapter 5
print"Example 5.3, Page:81  \n \n"
# Given:
th=2.4;# in Mev
z=0.0009;# mp-mn in atomic mass unit
# Solution:
x=-(2.4/931.);#assuming no barrier operates
y=x+z;
print"The mass difference between A & B in Uu=", round(-y*10**6)

Example 5.3, Page:81

The mass difference between A & B in Uu= 1678.0


## example 5.4;pg no:82¶

In [4]:
#cal of energy required
#intiation of all variables
# Chapter 5
print"Example 5.4, Page:82  \n \n"
# Given:
mp=1.007277;
# Solution:
E1=2*mp*931;# in MeV / part(a)
E2=E1/2;# in MeV / part (b)
print"The energy required for (a) & (b) are = respectively in MeV",round(E1,2),round(E2,2)

Example 5.4, Page:82

The energy required for (a) & (b) are = respectively in MeV 1875.55 937.77


## example 5.5;pg no:82¶

In [5]:
#cal of  Q-values
#intiation of all variables
# Chapter 5
print"Example 5.5, Page:82  \n \n"
#Given
E1=2.059;
E2=2.59;
M1=9.;
M2=18.;
m=1.;
# Solution
Q1=-E1*(M1/(m+M1)); # part(a)
Q2=-E2*(M2/(m+M2)); # part(b)
print"The Q-values are  in MeV for B(9) & F(19) reactions respectively",round(Q1,3),round(Q2,2)

Example 5.5, Page:82

The Q-values are  in MeV for B(9) & F(19) reactions respectively -1.853 -2.45


## example 5.6;pg no:82¶

In [6]:
#cal of thershold energies
#intiation of all variables
# Chapter 5
print"Example 5.6, Page:82  \n \n"
# Given:
Q1=1.136;
Q2=3.236;
M1=11.;
M2=13.;
m1=2.;
m2=1.;
# Solution
E1=Q1*((m1+M1)/M1); # part(a)
E2=Q2*((m2+M2)/M2); # part(b)
print"The thershold energies are  in MeV for B(12) & N(13) reactions respectively",round(E1,3),round(E2,3)

Example 5.6, Page:82

The thershold energies are  in MeV for B(12) & N(13) reactions respectively 1.343 3.485


## example 5.7;pg no:82¶

In [7]:
#cal of The binding energy of last 2 neutron
#intiation of all variables
# Chapter 5
print"Example 5.7, Page:82  \n \n"
# Given:

Q1=9.28;# in Mev
Q2=0.21;# in Mev
Q3=7.25;# in Mev
Q4=3.63;# in Mev
mn=1.008665;
md=1.995311;# mass difference between Fe(56) & Fe(54)
# Solution:
E1=Q1+Q2+Q3+Q4;# part (a)
E2=(2*mn-md)*931;# part (b)
print"The binding energy of last 2 neutron in part(a) and part(b) are =  in MeV respectively",round(E1,2),round(E2,2)

Example 5.7, Page:82

The binding energy of last 2 neutron in part(a) and part(b) are =  in MeV respectively 20.37 20.5


## example 5.8;pg no:83¶

In [8]:
#cal of threshold energy
#intiation of all variables
# Chapter 5
print"Example 5.8, Page:83  \n \n"
# Given:
Q1=1.2;
M1=14.;
m1=4.;
# Solution:
E1=Q1*((m1+M1)/M1);
print"The threshold energy is in MeV for O(17) reaction",round(E1,3)

Example 5.8, Page:83

The threshold energy is in MeV for O(17) reaction 1.543


## example 5.9;pg no:83¶

In [9]:
#cal of threshold energy
#intiation of all variables
# Chapter 5
print"Example 5.9, Page:83  \n \n"
# Given:
Q1=3.236;
M1=13.;
m1=1.;
# Solution:
E1=Q1*((m1+M1)/M1);
print"The threshold energy is in MeV for C(13) reaction",round(E1,3)

Example 5.9, Page:83

The threshold energy is in MeV for C(13) reaction 3.485


## example 5.10;pg no:84¶

In [10]:
#cal of cs area required
#intiation of all variables
# Chapter 5
print"Example 5.10, Page:84  \n \n"
# Given:
import math
A1=3836;#in barns
E1=1;# in eV
E2=10**6# in eV

# Solution:
vr=math.sqrt(E2/E1);
A2=A1/vr;
print"The cs area required will be in b=", A2

Example 5.10, Page:84

The cs area required will be in b= 3.836


## example 5.11;pg no:84¶

In [11]:
#cal of activity
#intiation of all variables
# Chapter 5
print"Example 5.11, Page:84  \n \n"
# Given:
a=0.56*10.**-24.; # area
flux=10.**13.;
# Solution:
A=6.022*10.**23.*10.**-3.*2.5/(58.5);
k=A*flux*0.56*10.**-24.;
y=(0.5)**(4./5.);
activity=k*(1-y);
print"The activity in 10^7 dis s^-1 g^-1 NaCl=",round(activity/10**7,2)

Example 5.11, Page:84

The activity in 10^7 dis s^-1 g^-1 NaCl= 6.13


## example 5.12;pg no:84¶

In [12]:
#cal of  cros section area
#intiation of all variables
# Chapter 5
print"Example 5.12, Page:84  \n \n"
# Given:
w=0.1189;
flux=10**16;
# Solution:
A=w/(flux*3.16*10**7);# in m^2
A1=A*10000/(10**-24);# in Barns
print"The cros section area in (b)=",round(A1,2)

Example 5.12, Page:84

The cros section area in (b)= 3762.66


## example 5.13;pg no:85¶

In [13]:
#cal of  cros section area,activity,saturation activity
#intiation of all variables
# Chapter 5
print"Example 5.13, Page:85  \n \n"
# Given:
w=8.52*10**-4;
flux=10**18;
# Solution:

A=w/(flux*24*3600);# in m^2
A1=A*10000/(10**-24);# in Barns
print"The cros section area in (b)=", round(A1,2)
k=flux*A*6.022*10**23/197;
print"\n \n The saturation activity possible in 10^13 dis s^-1 g^-1=",round(k/10**13,2)
y=(0.5)**(0.3704);
activity=k*(1-y);
print"\n \n The activity in 10^12 dis s^-1 g^-1= ",round(activity/10**12,2)

Example 5.13, Page:85

The cros section area in (b)= 98.61

The saturation activity possible in 10^13 dis s^-1 g^-1= 3.01

The activity in 10^12 dis s^-1 g^-1=  6.83


## example 5.14;pg no:85¶

In [14]:
#cal of activity
#intiation of all variables
# Chapter 5
print"Example 5.14, Page:85  \n \n"
# Given:
import math
a=98.7*10**-24; # area in cm ^2
flux=10**16;
d=19.3;# density
l=0.02;# thickness in cm
area=1;# in cm^2
# Solution:
V=area*l;
m=V*d;
A=(6.022*10**23*m)/(197);
k=A*flux*a;
y=math.exp((-0.693*5)/(2.7*24*60));
activity=k*(1-y);
print"The activity in 10^12 dis s^-1 g^-1=",round(activity/10**12,3)
# specific activity in Ci/cm^3
a1=activity/(3.7*10**10);# in Ci/gold foil
a2=a1/V;# in Ci/cm^3 of foil
print"The activity in Ci/cm^3 of foil is = ",round(a2)

Example 5.14, Page:85

The activity in 10^12 dis s^-1 g^-1= 1.037
The activity in Ci/cm^3 of foil is =  1402.0


## example 5.15;pg no:86¶

In [15]:
#cal of  geometric cross-section area
#intiation of all variables
# Chapter 5
print"Example 5.15, Page:86  \n \n"
# Given:
r0=1.4*10**-15;# in m
A1=88;
A2=87;
A3=136;
A4=135;
# Solution:

rSr1=(3.14*(r0*(A1)**(0.33333))**2)/10**-28;# in barns
rSr2=(3.14*(r0*(A2)**(0.33333))**2)/10**-28;# in barns
rXe1=(3.14*(r0*(A3)**(0.33333))**2)/10**-28;# in barns
rXe2=(3.14*(r0*(A4)**(0.33333))**2)/10**-28;# in barns
print"The geometric cross-section area are  for Sr(88), Sr(87), Xe(136) & Xe(135) respectively",round(rSr1,3),round(rSr2,3),round(rXe1,3),round(rXe2,3)

Example 5.15, Page:86

The geometric cross-section area are  for Sr(88), Sr(87), Xe(136) & Xe(135) respectively 1.218 1.208 1.628 1.62


## example 5.16;pg no:87¶

In [16]:
#cal of activity in mCi,Bq
#intiation of all variables
# Chapter 5
print"Example 5.16, Page:87  \n \n"
# Given:
m=4*10**-3;# in gms
flux=1.3*10**14;
a=19.6*10**-24;# in cm^2
# Solution:
N=(m/59)*6.022*10**23;
A=N*flux*a*3600;# atoms
k=0.693/(5.25*3.16*10**7);# s^-1
A1=k*A;# Activity in dps
A2=(A1)/(3.7*10**10);# in Ci
A3=(A1*10**3)/(3.7*10**10);# in mCi
A4=A2*37*10**8;# in Bq
print" The activity in mCi is =",round(A3,3)
print" The activity in M Bq is = ",round(A4/10**5,3)

Example 5.16, Page:87

The activity in mCi is = 0.042
The activity in M Bq is =  1.564


## example 5.17;pg no:88¶

In [17]:
#cal of thickness of Cd foil
#intiation of all variables
# Chapter 5
print"Example 5.17, Page:88  \n \n"
# Given:
import math
a=2.44*1000*10**-24;# in barns
d=8.64;# g/cm^3
# Solution:
n=(d*6.02*10**23)/112;# atoms/cm^2
x=(math.log(100))/(n*a);# in cm
print"The thickness of Cd foil in (cm)=",round(x,4)

Example 5.17, Page:88

The thickness of Cd foil in (cm)= 0.0406


## example 5.18;pg no:89¶

In [18]:
#cal of excitation energy for Al and Na
#intiation of all variables
# Chapter 5
print"Example 5.18, Page:89  \n \n"
# Given:
import math
a=3.8*1000*10**-24;# in barns
Ir=0.004;# I0/Ix
d=2.55;# g/cm^3

# Solution:
n=(d*6.02*10**23)/10;# atoms/cm^2
y=(Ir)**-1;
x=math.log(y)/(n*a);# in cm
print"The thickness of B foil in (cm)=",round(x,4)

Example 5.18, Page:89

The thickness of B foil in (cm)= 0.0095


## example 5.19;pg no:89¶

In [19]:
#cal of excitation energy
#intiation of all variables
# Chapter 5
print"Example 5.19, Page:89  \n \n"
# Given:
E1=6;# MeV
mAl=26.981535;
malpha=4.002604;
mP=30.973763;
# Solution:
KE=E1*(27/31);# in MeV
BE=(mAl+malpha-mP)*931;# in MeV
Ex=KE+BE;
print"\n The excitation energy of compound nucleus in (MeV)=",round(Ex,2)

Example 5.19, Page:89

The excitation energy of compound nucleus in (MeV)= 9.66


## example 5.20;pg no:90¶

In [20]:
#cal of excitation energy
#intiation of all variables
# Chapter 5
print"Example 5.20, Page:90  \n \n"
# Given:
E1=1.4;# MeV
mBi=208.980417;
mn=1.008665;
mBI=209.984110;
# Solution:part(a)

KE1=0.;# in MeV
BE1=(mBi+mn-mBI)*931.;# in MeV
Ex1=KE1+BE1;
print"The excitation energy of compound nucleus in (MeV)=",round(Ex1,2)
# Solution:part(b)
KE2=E1*(209./210.);# in MeV
BE2=(mBi+mn-mBI)*931.;# in MeV
Ex2=KE2+BE2;
print"The excitation energy of compound nucleus in (MeV)=",round(Ex2,3)

Example 5.20, Page:90

The excitation energy of compound nucleus in (MeV)= 4.63
The excitation energy of compound nucleus in (MeV)= 6.022


## example 5.21;pg no:90¶

In [21]:
#cal of resonance
#intiation of all variables
# Chapter 5
print"Example 5.21, Page:90  \n \n"
Ex=12.8;# MeV
mB=10.012939;
malpha=4.002604;
mN=14.003074;
mC=12.00;
md=2.014102;
# Solution:part(a)

BE1=(mB+malpha-mN)*931;# in MeV
KE1=Ex-BE1;
E1=KE1*(14./10.);
print"The resonance in part(a) will occur at =(MeV)",round(E1,3)

# Solution:part(b)
BE2=(mC+md-mN)*931.;# in MeV
KE2=Ex-BE2;
E2=KE2*(14./12.);
print"The resonance in part(b) will occur at (MeV)=",round(E2,3)

Example 5.21, Page:90

The resonance in part(a) will occur at =(MeV) 1.668
The resonance in part(b) will occur at (MeV)= 2.955


## example 5.22;pg no:91¶

In [22]:
#cal of resonance frequency
#intiation of all variables
# Chapter 5
print"Example 5.22, Page:91  \n \n"
# Given:
B=2.5;# tesla
q=1.6*10**-19;
m=1.66*10**-27;
# Solution:

f=(B*q*10**-6)/(2*3.14*2*m);
print"The resonance frequency in (MHz)=",round(f,1)

Example 5.22, Page:91

The resonance frequency in (MHz)= 19.2


## example 5.23;pg no:91¶

In [23]:
#cal of magnetic field needed to accelerate
#intiation of all variables
# Chapter 5
print"Example 5.23, Page:91  \n \n"
# Given:
f=8.6*10**6;# in Hz
q1=1.6*10**-19;
q2=6*1.6*10**-19;
m1=1.66*10**-27;
m2=14*1.66*10**-27;
# Solution:
# for proton
B1=2*3.14*f*m1/q1;
print"The magnetic field needed to accelerate protons in T=",round(B1,2)
# for N(14) ions
B2=2*3.14*f*m2/q2;
print"The magnetic field needed to accelerate N(14) ions in T=",round(B2,2)

Example 5.23, Page:91

The magnetic field needed to accelerate protons in T= 0.56
The magnetic field needed to accelerate N(14) ions in T= 1.31


## example 5.24;pg no:91¶

In [24]:
#cal of excitation energy of compound nucleus Si and Rb
#intiation of all variables
# Chapter 5
print"Example 5.24, Page:91  \n \n"
E1=2.75;# MeV
E2=14;# in MeV
mMg=23.985045;
malpha=4.00260;
mSi=27.9763;
mNe=19.99244;
mCo=58.93320;
mRb=78.9239
# Solution:

KE1=E1*(24/28);# in MeV
BE1=(mMg+malpha-mSi)*931;# in MeV
Ex1=KE1+BE1;
print"The excitation energy of compound nucleus Si* in MeV=",round(Ex1,4)
KE2=E2*(59./79.);# in MeV
BE2=(mNe+mCo-mRb)*931.;# in MeV
Ex2=KE2+BE2;
print"The excitation energy of compound nucleus Rb*  in MeV=",round(Ex2,4)

Example 5.24, Page:91

The excitation energy of compound nucleus Si* in MeV= 10.5622
The excitation energy of compound nucleus Rb*  in MeV= 12.0756