Chapter3-Interaction of Radiation with Matter

Ex1-pg55

In [1]:
## 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');
 Total interaction rate =  5.20e+05  interactions/sec 


 Probability of collision =  1.04e-02  

Ex2-pg56

In [2]:
## 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');
\ n Probability of fission =  0.85  85.46  percent
 

Ex3-pg57

In [3]:
## 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');
 Macroscopic cross section of carbon-12 =  0.21  cm^(-1)


 Collision density in the carbon-12 target =  1.04e+08  collisions/cm^(3)-sec

Ex4-pg59

In [4]:
## 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',);
 Macroscopic cross section =  0.09  cm^(-1)


 Mean Free Path =  11.58  cm

Ex5-pg60

In [5]:
## 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');
 Macroscopic absorption cross section =  0.37  cm^(-1)

Ex6-pg60

In [6]:
## 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');
 Scattering cross section of Water at 1 MeV =  14.00  b 

 Experimental value of scattering cross section of Water at 0.0253 eV =  103.00  b 

Ex7-pg61

In [7]:
## 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');
 Reactor power of a research reactor =  2.05  MW

Ex8-pg63

In [8]:
## 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');
 Radius of carbon nucleus =  6.18e-13  cm

Ex9-pg67

In [9]:
## 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');
 Radiative capture cross section of hydrogen at 1 eV =  1.27e-23  b

Ex10-pg72

In [10]:
## 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');
 Energy of scattered neutron =  0.74  MeV 


 Energy of recoil nucleus =  0.26  MeV 


 Change in lethargy of neutron on collision =  0.30  

Ex11-pg74

In [11]:
## 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');
 Absorption rate of neutrons per cc in indium foil =  4.27e+13  neutrons/cm^3-sec 

Ex12-pg84

In [12]:
## 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');
 The activity of a fuel rod =  808363.56  Ci 

Ex13-pg86

In [13]:
## 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');
 Eta for natural uranium =  1.34  

Ex14-pg90

In [14]:
## 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.
 Amount of coal required for energy equivalent of fission =  2880.00 
  2.88  3.17 short tons
 

 Amount of oil required for energy equivalent of fission =  2009.30  kg
 2.01   tons 12.66 barrels
 

Ex15-pg99

In [15]:
## 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.
 Mass attenuation coefficient of Uranium dioxide (UO2) =  0.74  cm^(-1) 


 Mean free path =  1.35  cm 

Ex16-pg100

In [17]:
## 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');
 Rate of energy deposited =  6.58e+09  1.05  J/kg-sec 
 

Ex17-pg106

In [18]:
## 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');
 Maximum distance of beta rays traversed =  0.62  cm