# Chapter12-Neutrons¶

## Exx1-pg573¶

In [1]:
## Exa12.1 : : Page-573 (2011)
#calculate activity for Cu-63 and disintegrations
import math
N_0 = 6.23e+23;    ## Avogadro's number, per mole
m = 0.1;        ## Mass of copper foil, Kg
phi = 10**12;        ## Neutron flux density, per square centimetre sec
a_63 = 0.691;        ## Abundance of Cu-63
a_65 = 0.309;        ## Abundance of Cu-65
W_m = 63.57;        ## Molecular weight, gram
sigma_63 =  4.5e-24;    ## Activation cross section for Cu-63, square centi metre
sigma_65 = 2.3e-24;        ## Activation cross section for Cu-65, square centi metre
A_63 = phi*sigma_63*m*a_63/W_m*N_0;        ## Activity for Cu-63, disintegrations per sec
A_65 = phi*sigma_65*m*a_65/W_m*N_0;        ## Activity for Cu-65, disintegrations per sec
print'%s %.2e %s %.2e %s'%("\nThe activity for Cu-63 is = ",A_63," disintegrations per sec" and  "\nThe activity for Cu-65 is = ",A_65," disintegrations per sec");

## Result
## The activity for Cu-63 is = 3.047e+009 disintegrations per sec
## The activity for Cu-65 is = 6.97e+008 disintegrations per sec

The activity for Cu-63 is =  3.05e+09
The activity for Cu-65 is =  6.97e+08  disintegrations per sec


## Ex2-pg573¶

In [3]:
## Exa12.2 : : Page-573 (2011)
import math
#calculate  enegy loss
A_Be = 9.;        ## Mass number of beryllium
A_U = 238.;        ## Mass number of uranium
E_los_Be = (1-((A_Be-1)**2/(A_Be+1)**2))*100.;    ## Energy loss for beryllium
E_los_U = round((1-((A_U-1)**2/(A_U+1)**2))*100.);    ## Energy loss for uranium
print'%s %.2f %s  %.2f %s '%("\nThe energy loss for beryllium is = ",E_los_Be," percent"and " \nThe energy loss for uranium is = ",E_los_U," percent");

## Check for greater energy loss !!!!
if E_los_Be >= E_los_U :
print("\nThe energy loss is greater for beryllium");
else:
print("\nThe energy loss is greater for uranium");

## Result
## The energy loss for beryllium is = 36 percent
## The energy loss for uranium is = 2 percent
## The energy loss is greater for beryllium


The energy loss for beryllium is =  36.00
The energy loss for uranium is =   2.00  percent

The energy loss is greater for beryllium


## Ex3-pg574¶

In [5]:
## Exa12.3 : : Page-574 (2011)
#calculate energy loss of neutron
import math
A = 12.;        ## Mass number of Carbon
alpha = (A-1)**2/(A+1)**2;        ## Scattering coefficient
E_loss = 1/2.*(1-alpha)*100.;        ## Energy loss of neutron
print'%s %.2f %s'%("\nThe energy loss of neutron = ",E_loss," percent")

## Result
## The energy loss of neutron = 14.201 percent


The energy loss of neutron =  14.20  percent


## Ex4-pg574¶

In [6]:
## Exa12.4 : : Page-574 (2011)
#calculate number of collisions of neutrons
import math
zeta = 0.209;        ## Moderated assembly
E_change = 100./1.;        ## Change in energy of the neutron
E_thermal = 0.025;        ## Thermal energy of the neutron, electron volts
E_n = 2*10**6;            ## Energy of the neutron, electron volts
n = 1/zeta*math.log(E_change);        ## Number of collisions of neutrons to loss 99 percent of their energies
n_thermal = 1/zeta*math.log(E_n/E_thermal);        ## Number of collisions of neutrons to reach thermal energies
print'%s %.2f %s %.2f %s'%("\nThe number of collisions  of neutrons to loss 99 percent of their energies = ",n," \nThe number of collisions of neutrons to reach thermal energies = ",n_thermal,"")

## Result
## The number of collisions  of neutrons to loss 99 percent of their energies = 22
## The number of collisions of neutrons to reach thermal energies = 87


The number of collisions  of neutrons to loss 99 percent of their energies =  22.03
The number of collisions of neutrons to reach thermal energies =  87.07


## Ex5-pg574¶

In [13]:
## Exa12.5 : : Page-574 (2011)
#calculate average distance travelled by the neutron
import math
import scipy
from scipy import integrate
L = 1.;    ## For simplicity assume thermal diffusion length to be unity, unit
def fun(x):
y=x*math.exp(-x/L)
return y
x_b = scipy.integrate.quad(fun, 0, 100);    ## Average distance travelled by the neutron, unit
x_b1=x_b[0]
def fun2(x):
y1=x**2*math.exp(-x/L)
return y1
x_rms = math.sqrt(X[0]);    ## Root mean square of the distance trvelled by the neutron, unit
print'%s %.2f %s'%("\nThe average distance travelled by the neutron = ", x_b1,"*L");
print'%s %.2f %s %.2f %s '%("\nThe root mean square distance travelled by the neutron = ",x_rms,""and "",x_rms,"x_bar")

## Result
## The average distance travelled by the neutron = 1*L
## The root mean square distance travelled by the neutron = 1.414L = 1.414x_bar


The average distance travelled by the neutron =  1.00 *L

The root mean square distance travelled by the neutron =  1.41  1.41 x_bar


## Ex6-pg574¶

In [10]:
## Exa12.6 : : Page-574 (2011)
#calculate neutron flux through water
import math
Q = 5e+08;        ## Rate at which neutrons produce, neutrons per sec
r = 20.;            ## Distance from the source, centi metre
## For water
lambda_wtr = 0.45;    ##  Transport mean free path, centi metre
L_wtr = 2.73;         ## Thermal diffusion length, centi metre
phi_wtr = 3*Q/(4.*math.pi*lambda_wtr*r)*math.exp(-r/L_wtr);    ## Neutron flux for water, neutrons per square centimetre per sec
## For heavy water
lambda_h_wtr = 2.40;        ##  Transport mean free path, centi metre
L_h_wtr = 171.;            ## Thermal diffusion length, centi metre
phi_h_wtr = 3*Q/(4.*math.pi*lambda_h_wtr*r)*math.exp(-r/L_h_wtr);    ## Neutron flux for heavy water, neutrons per square centimetre per sec
print'%s %.2e %s %.2e %s '%("\nThe neutron flux through water = ",phi_wtr," neutrons per square cm per sec"and "\nThe neutron flux through heavy water = ",phi_h_wtr," neutrons per square cm per sec")

## Result
## The neutron flux through water = 8.730e+003 neutrons per square cm per sec
## The neutron flux through heavy water = 2.212e+006 neutrons per square cm per sec


The neutron flux through water =  8.73e+03
The neutron flux through heavy water =  2.21e+06  neutrons per square cm per sec


## Ex7-pg575¶

In [11]:
## Exa12.7 : : Page-575 (2011)
#calculate neutron flux and diffusion length
import math
k = 1.38e-23;        ## Boltzmann constant, joules per kelvin
T = 323.;            ## Temperature, kelvin
E = (k*T)/1.6e-19;    ## Thermal energy, joules
sigma_0 = 13.2e-28;   ## Cross section, square metre
E_0 = 0.025;            ## Energy of the neutron, electron volts
sigma_a = sigma_0*math.sqrt(E_0/E);        ## Absorption cross section, square metre
t_half = 2.25;            ## Half life, hours
D= 0.69/t_half;        ## Decay constant, per hour
N_0 = 6.023e+026;            ## Avogadro's number, per
m_Mn = 55.;                ## Mass number of mangnese
w = 0.1e-03;            ## Weight of mangnese foil, Kg
A = 200.;                ## Activity, disintegrations per sec
N = N_0*w/m_Mn;        ## Number of mangnese nuclei in the foil
x1 = 1.5;                ## Base, metre
x2 = 2.0;                ## Height, metre
phi = A/(N*sigma_a*0.416);        ## Neutron flux, neutrons per square metre per sec
phi1 = 1.;    ## For simplicity assume initial neutron flux to be unity, neutrons/Sq.m-sec
phi2 = 1/2.*phi1;    ## Given neutron flux, neutrons/Sq.m-sec
L1 = 1/math.log(phi1/phi2)/(x2-x1);    ## Thermal diffusion length for given neutron flux, m
L = math.sqrt(1./((1./L1)**2+(math.pi/x1)**2+(math.pi/x2)**2));        ## Diffusion length, metre
print'%s %.2e %s %.2f %s '%("\nThe neutron flux = ",phi," neutrons per square metre per sec"and " \nThe diffusion length = ",L," metre");

## Result
## The neutron flux = 3.51e+008 neutrons per square metre per sec
## The diffusion length = 0.38 metre
## Note: the difussion length is solved wrongly in the testbook


The neutron flux =  3.51e+08
The diffusion length =  0.38  metre


## Ex8-pg575¶

In [12]:
## Exa12.8 : : Page-575(2011)
#find diffusion  length for thermal neutron
import math
N_0 = 6.023e+026;        ## Avogadro's number, per mole
rho = 1.62e+03;        ## Density, kg per cubic metre
sigma_a = 3.2e-31;        ## Absorption cross section, square metre
sigma_s = 4.8e-28;        ## Scattered cross section, square metre
A = 12.;                ## Mass number
lambda_a = A/(N_0*rho*sigma_a);        ## Absorption mean free path, metre
lambda_tr = A/(N_0*rho*sigma_s*(1.-2./(3.*A)));        ## Transport mean free path, metre
L = math.sqrt(lambda_a*lambda_tr/3.);        ## Diffusion length for thermal neutron
print'%s %.2f %s'%("\nThe diffusion length for thermal neutron = ",L," metre ")

## Result
## The diffusion length for thermal neutron = 0.590 metre


The diffusion length for thermal neutron =  0.59  metre


## Ex9-pg575¶

In [13]:
## Exa12.9 : : Page-575 (2011)
#calculate graphite and neutron age and slowing down length and same as berylliums
import math
E_0 = 2e+06;        ## Average energy of the neutron, electron volts
E = 0.025;            ## Thermal energy of the neutron, electron volts
## For graphite
A = 12.            ## Mass number
sigma_g = 33.5;    ## The value of sigma for graphite
tau_0 = 1./(6.*sigma_g**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E);    ## Age of neutron for graphite, Sq.m
L_f = math.sqrt(tau_0);    ## Slowing down length of neutron through graphite, m
print'%s %.2f %s'%("\nFor Graphite, A = ", A,"");
print'%s %.2f %s'%("\nNeutron age = ",tau_0*1e+004," Sq.cm");
print'%s %.2f %s'%("\nSlowing down length =",L_f," m");
## For beryllium
A = 9.            ## Mass number
sigma_b = 57.;    ## The value of sigma for beryllium
tau_0 = 1/(6.*sigma_b**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E);    ## Age of neutron for beryllium, Sq.m
L_f = math.sqrt(tau_0);    ## Slowing down length of neutron through graphite, m
print'%s %.2f %s'%("\n\nFor Beryllium, A = ", A,"");
print'%s %.2f %s'%("\nNeutron age = ",tau_0*1e+004," Sq.cm");
print'%s %.2e %s'%("\nSlowing down length = ",L_f," m");

## Result
## For Graphite, A = 12
## Neutron age = 362 Sq.cm
## Slowing down length = 0.190 m

## For Beryllium, A = 9
## Neutron age = 97 Sq.cm
## Slowing down length = 9.9e-002 m

For Graphite, A =  12.00

Neutron age =  362.46  Sq.cm

Slowing down length = 0.19  m

For Beryllium, A =  9.00

Neutron age =  97.46  Sq.cm

Slowing down length =  9.87e-02  m


## Ex10-pg576¶

In [14]:
## Exa12.10 : : Page-576 (2011)
#find enegy of the neutrons
import math
theta = 3.5*math.pi/180.;  ## Reflection angle, radian
d = 2.3e-10;          ## Lattice spacing, metre
n = 1.;                ## For first order
h = 6.6256e-34;       ## Planck's constant, joule sec
m = 1.6748e-27;       ## Mass of the neutron, Kg
E = n**2*h**2/(8.*m*d**2*math.sin(theta)**2*1.6023e-19);        ## Energy of the neutrons, electron volts
print'%s %.2f %s'%("\nThe energy of the neutrons = ",E," eV");

## Result
## The energy of the neutrons = 1.04 eV


The energy of the neutrons =  1.04  eV