# Chapter10-Nuclear reactions¶

## Ex1-pg455¶

In :
## Exa10.1 : : Page-455 (2011)
#calculate The Q-value for the fromation
import math
M = 47.668;            ## Total mass of reaction, MeV
E = 44.359;            ## Total energy, MeV
Q = M-E;                ## Q-value, MeV
print'%s %.2f %s'%("\nThe Q-value for the formation of P30 = ",Q," MeV");

## Result
## The Q-value for the formation of P30 = 3.309 MeV

The Q-value for the formation of P30 =  3.31  MeV


## Ex2-pg455¶

In :
## Exa10.2 : : Page-455 (2011)
#calculate The Q-value for the reaction
import math
E_x = 7.70; ## Energy of the alpha particle, MeV
E_y = 4.44; ## Energy of the proton, MeV
m_x = 4.0;  ## Mass number of alpha particle
m_y = 1.0;   ## Mass number of protium ion
M_X = 14;   ## Mass number of nitrogen nucleus
M_Y = 17;  ## Mass number of oxygen nucleus
theta = 90*3.14/180.; ##  Angle between incident beam direction and emitted proton, degree
A_x = 4.0026033; ## Atomic mass of alpha particle, u
A_X = 14.0030742; ## Atomic mass of nitrogen nucleus, u
A_y = 1.0078252;  ## Atomic mass of proton, u
Q = ((E_y*(1+m_y/M_Y))-(E_x*(1-m_x/M_Y))-2./M_Y*math.sqrt((m_x*m_y*E_x*E_y))*math.cos(theta))/931.5;     ## Q-value, u
A_Y = A_x+A_X-A_y-Q;     ## Atomic mass of O-17, u
print'%s %.4f %s %.2f %s '%("\nThe Q-value of the reaction = ",Q," u" " \nThe atomic mass of the O-17 =",A_Y," u");

## Result
## The Q-value of the reaction = -0.0012755 u
## The atomic mass of the O-17 = 16.9991278 u
##  Atomic mass of the O-17 : 16.9991278 u

The Q-value of the reaction =  -0.0013  u
The atomic mass of the O-17 = 17.00  u


## Ex3-pg455¶

In :
## Exa10.3 : : Page-455 (2011)
#calculate The kinetic energy of the emitted neutron
import math
m_p = 1.007276;       ## Atomic mass of the proton, u
m_H = 3.016049;         ## Atomic mass of the tritium, u
m_He = 3.016029;          ## Atomic mass of the He ion, u
m_n = 1.008665;            ## Atomic mass of the emitted neutron, u
Q = (m_p+m_H-m_He-m_n)*931.5;         ## Q-value in MeV
E_p = 3.;                      ## Kinetic energy of the proton, MeV
theta = 30*3.14/180;           ## angle, radian
u = math.sqrt(m_p*m_n*E_p)/(m_He+m_n)*math.cos(theta);         ##
v = ((m_He*Q)+E_p*(m_He-m_p))/(m_He+m_n);                 ##
E_n = (u+math.sqrt(u**2+v))**2;                  ## Kinetic energy of the emitted neutron,MeV
print'%s %.2f %s'%("\nThe kinetic energy of the emitted neutron = ",E_n," MeV");

## Result
## The kinetic energy of the emitted neutron = 1.445 MeV

The kinetic energy of the emitted neutron =  1.45  MeV


## Ex4-pg456¶

In :
## Exa10.4 : : Page-456 (2011)
#findThe temperature in the fusion reaction
import math
r_min = 4e-015;        ## Distance between two deutrons, metre
k = 1.3806504e-023;       ## Boltzmann's constant, Joule per kelvin
alpha = 1/137.;            ## Fine structure constant
h_red = 1.05457168e-034;        ## Reduced planck's constant, Joule sec
C = 3e+08;                    ## Velocity of light, meter per second
T = alpha*h_red*C/(r_min*k);
print'%s %.2e %s'%("\nThe temperature in the fusion reaction is =",T," K");

## Result
## The temperature in the fusion reaction is = 4.2e+009 K

The temperature in the fusion reaction is = 4.18e+09  K


## Ex5-pg456¶

In :
## Exa11.5 : : Page-456 (2011)
#find the excitation energy of the compound nucleus
import math
E_0 = 4.99;        ## Energy of the proton, MeV
m_p = 1.;            ## Mass number of the proton
m_F = 19.;            ## Mass number of the flourine
E = E_0/(1+m_p/m_F);        ## Energy of the relative motion, MeV
A_F = 18.998405;        ## Atomic mass of the fluorine, amu
A_H = 1.007276;          ##  Atomic mass of the proton, amu
A_Ne = 19.992440;         ## Atomic mass of the neon, amu
del_E = (A_F+A_H-A_Ne)*931.5;        ## Binding energy of the absorbed proton, MeV
E_exc = E+del_E;            ## Excitation energy of the compound nucleus, MeV
print'%s %.2f %s'%("\nThe excitation energy of the compound nucleus = ",E_exc," MeV");

## Result
## The excitation energy of the compound nucleus = 17.074 MeV

The excitation energy of the compound nucleus =  17.07  MeV


## Ex6-pg457¶

In :
## Exa10.6 : : Page-457 (2011)
#find The excitation energy of the compound nucleus and The parity of the compound nucleus
import math
E_d = 0.6;        ## Energy of the deutron, MeV
m_d = 2.;            ## Mass number of the deutron
m_Li = 19.;            ## Mass number of the Lithium
E = E_d/(1.+m_d/m_Li);        ## Energy of the relative motion, MeV
A_Li = 6.017;        ## Atomic mass of the Lithium, amu
A_d = 2.015;          ##  Atomic mass of the deutron, amu
A_Be = 8.008;         ## Atomic mass of the Beryllium, amu
del_E = (A_Li+A_d-A_Be)*931.5;        ## Binding energy of the absorbed proton, MeV
E_exc = E+del_E;            ## Excitation energy of the compound nucleus, MeV
l_f = 2.;            ## orbital angular momentum of two alpha particle
P = (-1)**l_f*(+1.)**2;        ## Parity of the compound nucleus
print'%s %.2f %s %.2f %s '%("\nThe excitation energy of the compound nucleus = ",E_exc," MeV" and "\nThe parity of the compound nucleus = ",P,"");

## Result
## The excitation energy of the compound nucleus = 22.899 MeV
## The parity of the compound nucleus = 1

The excitation energy of the compound nucleus =  22.90
The parity of the compound nucleus =  1.00


## Ex7-pg457¶

In :
## Exa10.7 : : Page-457 (2011)
#find The cross section for neutron induced fission
import math
D = 1e-016;        ## Disintegration constant, per sec
phi = 10**11;              ## Neutron flux, neutrons per square cm per sec
sigma = 5*D/(phi*10**-27);             ## Cross section, milli barns
print'%s %.2f %s'%("\nThe cross section for neutron induced fission = ",sigma," milli barns");

## Result
## The cross section for neutron induced fission = 5 milli barns

The cross section for neutron induced fission =  5.00  milli barns


## Ex8-pg457¶

In :
## Exa10.8 : : Page-457 (2011)
#find The number of nuclei of Co60 produced and The initial activity
import math
N_0 = 6.02252e+026;        ## Avogadro's constant
rho = 8.9*10**3;            ## Nuclear density of Co-59, Kg per cubic metre
M = 59.;                    ## Mass number
sigma = 30e-028;             ## Cross section, per square metre
phi = 10**16;              ## Neutron flux, neutrons per square metre per sec
d = 0.04e-02;                ## Thickness of Co-59 sheet, metre
t = 3*60*60;                ## Total reaction time, sec
t_half = 5.2*365*86400;        ## Half life of Co-60, sec
D = 0.693/t_half;        ## Disintegration constant, per sec
N_nuclei = round(N_0*rho/M*sigma*phi*d*t);        ## Number of nuclei of Co-60 produced
Init_activity = D*N_nuclei;        ## Initial activity, decays per sec
print'%s %.2e %s  %.2f %s '%("\nThe number of nuclei of Co60 produced =",N_nuclei," " and "\nThe initial activity per Sq. metre = ",Init_activity," decays per sec");

## Result
## The number of nuclei of Co60 produced = 1.18e+019
## The initial activity per Sq. metre = 5e+010 decays per sec

The number of nuclei of Co60 produced = 1.18e+19
The initial activity per Sq. metre =   49755893720.42  decays per sec


## Ex9-pg458¶

In :
## Exa10.9 : : Page-458 (2011)
#find The number of events detected
import math
d = 0.1;                ## Thickness of Fe-54 sheet, Kg per squre metre
M = 54.;                   ## Mass number of Fe
m = 1.66e-027;            ## Mass of the proton, Kg
n = d/(M*m);            ## Number of nuclei in unit area of the target, nuclei per square metre
ds = 10**-5;            ## Area, metre square
r = 0.1;                ## Distance between detector and target foil, metre
d_omega =ds/r**2;        ## Solid angle, steradian
d_sigma = 1.3e-03*10**-3*10**-28;        ## Differential cross section, square metre per nuclei
P = d_sigma*n;                ## Probablity, event per proton
I = 10**-7;                    ## Current, ampere
e = 1.6e-19;                ## Charge of the proton, C
N = I/e;            ## Number of protons per second in the incident beam, proton per sec
dN = P*N;            ## Number of events detected per second, events per sec
print'%s %.2f %s'%("\nThe number of events detected =",dN," events per sec");

## Result
## The number of events detected = 90 events per sec

The number of events detected = 90.64  events per sec


## Ex10-pg458¶

In :
## Exa10.10 : : Page-458 (2011)
#find The fractional attenuation of neutron beam on passing through nickel sheet
import math
N_0 = 6.02252e+26;        ## Avogadro's constant
sigma = 3.5e-28;            ## Cross section, square metre
rho = 8.9e+03;                ## Nuclear density, Kg per cubic metre
M = 58.;                          ## Mass number
summation = rho/M*N_0*sigma;    ## Macroscopic cross section, per metre
x = 0.01e-02;                   ## Thickness of nickel sheet, metre
I0_ratio_I = math.exp(summation*x/2.3026);    ## Fractional attenuation of neutron beam on passing through nickel sheet
print'%s %.2f %s ' %("\nThe fractional attenuation of neutron beam on passing through nickel sheet =", I0_ratio_I,"");
print("Wrong answer given in the textbook")
## Result
## The fractional attenuation of neutron beam on passing through nickel sheet = 1.0014
## Wrong answer given in the textbook

The fractional attenuation of neutron beam on passing through nickel sheet = 1.00
Wrong answer given in the textbook


## Ex11-pg458¶

In :
## Exa10.11 : : Page-458 (2011)
#find he scattering contribution to the resonance
import math
D = math.sqrt(1.45e-021/(4.*math.pi));        ## Wavelength, metre
W_ratio = 2.3e-07;                    ## Width ratio
sigma = W_ratio*(4*math.pi)*D**2*10**28;            ## Scattering contribution, barn
print'%s %.2f %s'%("\nThe scattering contribution to the resonance = ",sigma," barns");

## Result
## The scattering contribution to the resonance = 3.33 barns

The scattering contribution to the resonance =  3.33  barns


## Ex12-pg458¶

In :
## Exa10.12 : : Page-458 (2011)
#find The relative probabilities
import math
sigma = 2.8e-024;        ## Cross section, metre square
D = 2.4e-11;        ## de Broglie wavelength, metre
R_prob = math.pi*sigma/D**2;        ## Relative probabilities of (n,n) and (n,y) in indium
print'%s %.2f %s'%("\nThe relative probabilities of (n,n) and (n,y) in indium = ", R_prob,"");

## Result
## The relative probabilities of (n,n) and (n,y) in indium = 0.015


The relative probabilities of (n,n) and (n,y) in indium =  0.02


## Ex13-pg459¶

In :
## Exa10.13 : : Page-459 (2011)
#find The cross section of neutron capture
import math
h = 6.625e-34;            ## Planck's constant, joule sec
m_n = 1.67e-27;          ## Mass of neutron, Kg
E = 4.906;           ## Energy, joule
w_y = 0.124;                ## radiation width, eV
w_n = 0.007*E**(1/2.);                ## Probability of elastic emission of neutron, eV
I = 3.;                    ## Total angular momentum
I_c = 2.;                    ## Total angular momentum in the compound state
sigma = ((h**2)*(2*I_c+1)*w_y*w_n)*10**28/(2.*math.pi*m_n*E*1.602e-019*(2*I+1)*(w_y+w_n)**2);    ## Cross section, barns
print'%s %.3e %s'%("\nThe cross section of neutron capture = ",sigma," barns");

## Result
## The cross section of neutron capture = 3.755e+004 barns


The cross section of neutron capture =  3.755e+04  barns


## Ex14-pg459¶

In :
## Exa10.14 : : Page-459 (2011)
#fid the value of theta max
import math
import numpy
R = 5.;                   ## Radius, femto metre
k_d = 0.98;               ## The value of k for deutron
k_p = 0.82;            ## The value of k for triton
theta = numpy.zeros((1,5));        ## Angles at which differetial cross section is maximum, degree
## Use of for loop for angles calculation(in degree)
for l in range (1,4):
theta = round((math.acos((k_d**2+k_p**2)/(2.*k_d*k_p)-l**2/(2.*k_d*k_p*R**2)))*180./3.14,0);
#theta = round((acos((k_d^2+k_p^2)/(2*k_d*k_p)-l^2/(2*k_d*k_p*R^2)))*180/3.14);
print'%s %.2f %s'%("\nFor l = ", l,"");
print'%s %.2f %s'%("the value of theta_max =  ", math.ceil(theta),"degree");

## Result
## For l = 0,the value of theta_max = 0 degree
## For l = 1,the value of theta_max = 8 degree
## For l = 2,the value of theta_max = 24 degree
## For l = 3,the value of theta_max = 38 degree
## For l = 4,the value of theta_max = 52 degree

For l =  1.00
the value of theta_max =   8.00 degree

For l =  2.00
the value of theta_max =   24.00 degree

For l =  3.00
the value of theta_max =   38.00 degree


## Ex15-pg459¶

In :
## Exa10.15 : : Page-459 (2011)
#calculate total angular momentum transfer
import math
k_d = 2.02e+30;        ## The value of k for deutron
k_t = 2.02e+30;        ## The value of k for triton
theta = 23*3.14/180;        ## Angle, radiams
q = math.sqrt (k_d+k_t-2*k_t*math.cos(theta))*10**-15;        ## the value of q in femto metre
R_0 = 1.2;        ## Distance of closest approach, femto metre
A = 90.;            ## Mass number of Zr-90
z = 4.30;            ## Deutron size, femto metre
R = R_0*A**(1/3.)+1/2.*z;        ## Radius of the nucleus, femto metre
l = round(q*R);            ## Orbital angular momentum
I = l+1./2.                    ## Total angular momentum
print'%s %.2f %s'%("\nThe total angular momentum transfer =  ", I,"");

## Result
## The total angular momentum transfer = 4.5

The total angular momentum transfer =   4.50