## 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
## 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
## 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
## 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
## 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=scipy.integrate.quad(fun2, 0, 100)
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
## 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
## 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
## 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
## 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
## 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