## Example 3.1
import math
## Given data
## 1 barn = 10^(-24) cm^2
sigma = 2.6*10**(-24); ## Cross section of carbon-12 in cm^2
I = 5*10**8; ## Intensity of neutron beam in neutrons/cm^2-sec
A = 0.1; ## Cross sectional area of the beam in cm^2;
X = 0.05; ## Thickness of the target in cm
## 1.
## Using the data given in Table I.3, Appendix II for carbon-12
N = 0.08*10**(24); ## Atom density in atoms/cm^3
## Calculation
IR = sigma*I*N*A*X;
## Result
print'%s %.2e %s'%('\n Total interaction rate = ',IR,' interactions/sec \n');
## 2.
no = I*A; ## Neutron rate in neutrons/sec
## Calculation
p = IR/no;
print'%s %.2e %s'%('\n Probability of collision = ',p,' \n');
## Example 3.2
import math
## Given data
sigmaf = 582.; ## Fission cross section of U-235 on bombardment of neutron in barn
sigmay = 99.; ## Radiative capture cross section of U-235 on bombardment of neutron in barn
## Calculation
pf = sigmaf/(sigmaf+sigmay);
## Result
print'%s %.2f %s %.2f %s '%('\ n Probability of fission = ',pf,''and ' = ',pf*100,' percent\n');
## Example 3.3
import math
## Given data
## Using the data given in the example 3.1
N = 0.08*10**(24.); ## Atom density of Carbon-12 in atoms/cm^3
## 1 barn = 10^(-24) cm^2
sigma = 2.6*10**(-24); ## Cross section of carbon-12 in cm^2
I = 5*10**8; ## Intensity of neutron beam in neutrons/cm^2-sec
## 1.
## Calculation
SIGMAt = N*sigma;
## Result
print'%s %.2f %s'%('\n Macroscopic cross section of carbon-12 = ',SIGMAt,' cm^(-1)\n');
##2.
## Calculation
F= I*SIGMAt;
## Result
print'%s %.2e %s'%('\n Collision density in the carbon-12 target = ',F,' collisions/cm^(3)-sec\n');
## Example 3.4
import math
## Given data
E = 100.; ## Neutron energy in keV
## Using the data given in Table II.3, for E = 100 keV
atom_density = 0.0254*10**(24); ## Atom density of sodium in atoms/cm^3
## 1 barn = 10^(-24) cm^2
sigma = 3.4*10**(-24); ## Microscopic cross section of sodium in cm^2
## Calculation
SIGMA = atom_density*sigma;
lambd = 1./SIGMA;
## Result
print'%s %.2f %s'%('\n Macroscopic cross section = ',SIGMA,' cm^(-1)\n');
print'%s %.2f %s'%('\n Mean Free Path = ',lambd,' cm\n',);
## Example 3.5
import math
## Given data
atom_density_U235 = 3.48*10**(-4)*10**(24); ## Atom density of Uranium-235 in atoms/cm^3
atom_density_U238 = 0.0483*10**(24); ## Atom density of Uranium-238 in atoms/cm^3
## 1 barn = 10^(-24) cm^2
sigmaa_U235 = 680.8*10**(-24); ## Absorption cross section of Uranium-235 incm^2
sigmaa_U238 = 2.7*10**(-24); ## Absorption cross section of Uranium-238 incm^2
## Calculation
SIGMAA=(atom_density_U235*sigmaa_U235)+(atom_density_U238*sigmaa_U238);
## Result
print'%s %.2f %s'%('\n Macroscopic absorption cross section = ',SIGMAA,' cm^(-1)\n');
## Example 3.6
import math
## Given data
sigmas_H_1 = 3; ## Scattering cross section of Hydrogen in barn at 1 MeV
sigmas_O_1 = 8; ## Scattering cross section of Oxygen in barn at 1 MeV
sigmas_H_th = 21; ## Scattering cross section of Hydrogen in barn at 0.0253 eV
sigmas_O_th = 4; ## Scattering cross section of Oxygen in barn at 0.0253 eV
## Calculation
sigmas_H20_1 = (2*sigmas_H_1)+(1*sigmas_O_1);
## Result
print'%s %.2f %s'%('\n Scattering cross section of Water at 1 MeV = ',sigmas_H20_1,' b \n');
## The equation used to calculate the scattering cross section at 1 MeV cannot be used at thermal energy.
print'%s %.2f %s'%(' Experimental value of scattering cross section of Water at 0.0253 eV = ',103,' b \n');
## Example 3.7
import math
## Given data
phi = 1*10**(13); ## Neutron flux in neutrons/cm^3
v = 64000.; ## Volume of research reactor in cm^3
sigmaf = 0.1; ## Macroscopic fission cross section in cm^(-1)
## The energy released per fission reaction is 200 MeV
## 1 MeV = 1.6*10^(-13) joule
E = 200*1.6*10**(-13);
## Calculation
fiss_rate = sigmaf*phi; ## Fission rate in neutrons/cm^2-sec
power_cc = E*fiss_rate/10**6; ## Reactor power/cc
power = power_cc*v;
print'%s %.2f %s'%('\n Reactor power of a research reactor = ',power,' MW\n');
## Example 3.8
import math
## 1 barn = 10^(-24) cm^2
## From the Figure 3.4 given in the textbook
sigmae = 4.8*10**(-24); ## Experimental cross section of carbon from 0.02eV to 0.01MeV
## Assuming spherical shape and elstic scattering
R = math.sqrt(sigmae/(4.*math.pi));
## Result
print'%s %.2e %s'%('\n Radius of carbon nucleus = ',R,' cm\n');
## Example 3.9
import math
## Given data
E0 = 0.0253; ## Thermal energy in eV
## 1 barn = 10^(-24) cm^2
sigmay_E0 = 0.332*10**(-24); ## Radiative capture cross section at 0.0253 eV in cm^2
E = 1.; ## Energy in eV at which radiative cross section is to be found
## Calculation
sigmay_E = sigmay_E0*math.sqrt(E0/E);
## Result
## Expressing the result in barn
print'%s %.2e %s'%('\n Radiative capture cross section of hydrogen at 1 eV = ',sigmay_E*10*24,' b\n');
## Example 3.10
import math
## Given data
E = 1.; ## Energy of neutron in MeV
A = 2.; ## Atomic mass number of deuterium
v = 45.; ## Scattering angle in degree
## 1.
## Calculation
E_dash = E/(A+1.)**2 *((math.cos (v/57.3)+math.sqrt(A**2-(math.sin(v/57.3))**2))**2);
## Result
print'%s %.2f %s'%('\n Energy of scattered neutron = ',E_dash,' MeV \n');
## 2.
## Calculation
E_A = E-E_dash;
## Result
print'%s %.2f %s'%('\n Energy of recoil nucleus = ',E_A,' MeV \n');
## 3.
## Calculation
deltau = math.log(E/E_dash);
## Result
print'%s %.2f %s'%('\n Change in lethargy of neutron on collision = ',deltau,' \n');
## Example 3.11
import math
## Given data
phi = 5*10**(12); ## Neutron flux in neutrons/cm^2-sec
T = 600.; ## Temperature of neutron in degree
## Using the data given in Table II.3, Appendix II for indium
N = 0.0383*10**(24); ## Atom density in atoms/cm^3
## 1 barn = 10^(-24) cm^2
sigmaa_E0 = 194.*10**(-24); ## Microscopic absorption cross section in cm^2
SIGMA_E0 = N*sigmaa_E0; ## Macroscopic absorption cross section in cm^(-1)
## From Table 3.2
ga_600 = 1.15; ## Non 1/v factor at 600 degree celsius
## Calculation
F_a = ga_600*SIGMA_E0*phi;
## Result
print'%s %.2e %s'%('\n Absorption rate of neutrons per cc in indium foil = ',F_a,' neutrons/cm^3-sec \n');
## Example 3.12
import math
## Given data
N = 120.; ## Number of fuel rods
P = 100.; ## Reactor power in MW
t = 1.; ## Estimation time of fuel rod after removal in days
T = 365.; ## Time of reactor operation
## Estimation
Activity_total = 1.4*10**6*P*(t**(-0.2)-(t+T)**(-0.2));
Activity_one = Activity_total/N; ## For one fuel rod
## Result
print'%s %.2f %s'%('\n The activity of a fuel rod = ',Activity_one,' Ci \n');
## Example 3.13
import math
## Using the data given in Table 3.4 and Table II.2 for uranium
v_235 = 2.418; ## Average number of neutrons released per fission
y_235 = 0.72; ## Isotropic abundance of Uranium-235 on the earth
sigmaf_235 = 582.2; ## Fission cross section of Uranium-235
sigmaa_235 = 680.8; ## Absorption cross section of Uranium-235
N_235 = y_235;
y_238 = 99.26; ## Isotropic abundance of Uranium-238 on the earth
sigmaa_238 = 2.7; ## Absorption cross section of Uranium-238
## Calculation
n = (v_235*y_235*sigmaf_235)/((y_235*sigmaa_235)+(y_238*sigmaa_238));
## Result
print'%s %.2f %s'%('\n Eta for natural uranium = ',n,' \n');
## Example 3.14
import math
## Fission of 1 g of Uranium-235 releases approximately 1 MW/day of energy.
## 1 MW/day = 8.64*10^(10) J
energy_uranium = 8.64*10**10;
## 1. Coal
h_coal = 3*10**7; ## Heat contenet of coal in J/kg
## Calculation
amt_coal = energy_uranium/h_coal;
## Result
print'%s %.2f %s %.2f %s %.2f %s '%('\n Amount of coal required for energy equivalent of fission = ',amt_coal,' kg ' and '\n ',amt_coal/10**3,'metric tons' and '',amt_coal*1.10231/10**3,'short tons\n');
## The result is expressed in all units of commercial importance.
## 2. Oil
h_oil = 4.3*10**7; ## Heat contenet of oil in J/kg
## Calculation
amt_oil = energy_uranium/h_oil;
## Result
print'%s %.2f %s %.2f %s %.2f %s '%('\n Amount of oil required for energy equivalent of fission = ',amt_oil,' kg\n' or'',amt_oil/10**3,' tons' or '',amt_oil*6.3/10**3,'barrels\n');
## The result is expressed in all units of commercial importance.
## Example 3.15
import math
## Given data
rho = 10.; ## Density of UO2 in g/cm^3
mol_wt_UO2 = 238.+(16.*2.); ## Molecular weight of UO2
per_U = (238./mol_wt_UO2)*100.; ## Percent by weight of Uranium
per_O = 100.-per_U; ## Percent by weight of Oxygen
## Calculation
##Using the data given in Table II.4 for uranium and oxygen
mup_U = 0.0757; ## Ratio of mass attenuation coefficient to density of uranium in cm^2/g
mup_O = 0.0636; ## Ratio of mass attenuation coefficient to density of oxygen in cm^2/g
mup = (per_U/100.*mup_U)+(per_O/100.*mup_O); ## The total ratio of mass attenuation coefficient in cm^2/g
mu = mup*rho;
## Calculation
lambd = 1/mu;
## Result
print'%s %.2f %s'%('\n Mass attenuation coefficient of Uranium dioxide (UO2) = ',mu,' cm^(-1) \n');
print'%s %.2f %s'%('\n Mean free path = ',lambd,' cm \n');
## The answer is marked wrongly in the textbook. But the solution is correctly evaluated.
## Example 3.16
import math
## Given data
E = 0.8; ## Average gamma ray energy in MeV
I = 3*10**(11); ## Intensity of gamma rays incident on the container in gamma rays/cm^2-sec
## Using the data given in Table II.5 for iron at 0.8 MeV
mup_iron = 0.0274; ## Ratio of mass attenuation coefficient to density of iron in cm^2/g
## Calculation
dep_rate = E*I*mup_iron;
## Expressing the result in SI units
## 1 MeV = 1.6*10^(-13) J
## 1 kg = 1000 g
dep_rate_SI = dep_rate*(1.6*10**(-13)*1000.);
print'%s %.2e %s %.2f %s '%('\n Rate of energy deposited = ',dep_rate,' MeV/g-sec' and '',dep_rate_SI,' J/kg-sec \n');
## Example 3.17
import math
## Given data
E_max = 1.39; ## Maximum energy of beta rays in MeV
## Calculation
R_max = 0.412*E_max**(1.265-(0.0954*math.log(E_max)));
## Result
print'%s %.2f %s'%('\n Maximum distance of beta rays traversed = ',R_max,' cm \n');