Chapter3-Interaction of Nuclear radiations with matter¶

Ex1-pg123¶

In :
##  Exa3.1 : : Page-123 (2011)
#find range of an alpha particle and metre the thickness
E = 9.;    ## Energy of the alpha particle, MeV
S = 1700.; ##  Stopping power of Al
D = 2700.; ## Density of Al, Kg per cubic metre
R_air = 0.00318*E**(3/2.); ## Range of an alpha particle in air,metre
R_Al =  R_air/S; ## Range of an alpha particle in Al, metre
T = D*1./S; ## Thickness in Al of 1m air, Kg per square metre
print"%s %.2e %s %.2f %s "%("The range of an alpha particle = ",R_Al," metre The thickness in Al of 1 m air =",T," Kg per square metre");

## Result
## The range of an alpha particle = 5.05e-05 metre
## The thickness in Al of 1 m air = 1.59 Kg per square metre
The range of an alpha particle =  5.05e-05  metre The thickness in Al of 1 m air = 1.59  Kg per square metre

Ex4-pg124¶

In :
##  Exa3.4: : Page-124 (2011)
#find the haif value thickness for beta absorptions
import math
E_max = 1.17; ## Maximum energy of the beta particle, mega electron volts
D = 2.7; ## Density of Al,gram per cubic metre
u_m = 22./E_max; ## Mass absorption coefficient,centimetre square per gram
x_h = math.log(2.)/(u_m*D); ## Half value thickness for beta absorption, cm
print'%s %.2f %s'%("The Half value thickness for beta absorption = ",x_h," cm");

## Result
## The Half value thickness for beta absorption = 0.014 cm
The Half value thickness for beta absorption =  0.01  cm

Ex7-pg125¶

In :
##  Exa3.7: : Page 125(2011)
#calculate ratio of raditon loss to ionisation
Z = 82.;    ## Atomic number
E = 1.; ## Energy of the beta paricle, MeV
I_l = 800.;     ## Ionisation loss, MeV
R = Z*E/I_l;     ## Ratio of radiation loss to ionisation loss
E_1 = I_l/Z;     ## Energy of the beta particle when radiation radiation loss is equal to ionisation loss, MeV

print'%s %.2e %s %.2f %s '%("The ratio of radiation loss to ionisation loss =",R,"  The energy of the beta particle = ",E_1," MeV ");

## Result
## The ratio of radiation loss to ionisation loss = 1.025e-01
## The energy of the beta particle = 9.76 MeV
The ratio of radiation loss to ionisation loss = 1.02e-01   The energy of the beta particle =  9.76  MeV

Ex8-pg125¶

In :
##  Exa3.8 : : Page 125(2011)
#find The half value thickness of Al and The mass absorption coefficient
import math
x = 0.25;     ## Thickness of Al, metre
U_l = 1./x*math.log(50.);     ## Linear  absorption coefficient
d = 2700.;             ## density of the Al, Kg per cubic centimetre
x_h = math.log(2.)/U_l;     ## Half value thickness of Al, metre
U_m =  U_l/d;         ## Mass absorption coefficient, square metre per Kg
print'%s %.2f %s %.3f %s'%("The half value thickness of Al = ",x_h," Kg per cubic metre The mass absorption coefficient = ",U_m," square metre per Kg ");

## Result
## The half value thickness of Al = 0.0443 Kg per cubic metre
## The mass absorption coefficient = 0.00580 square metre per Kg
The half value thickness of Al =  0.04  Kg per cubic metre The mass absorption coefficient =  0.006  square metre per Kg

Ex9-pg125¶

In :
##  Exa3.9 : : Page-125(2011)
#find The energy of the compton recoil electrons
import math
E_g = 2.19*1.6e-013; ## Energy of the gamma rays, joule
m_e = 9.10939e-031; ## Mass of the electron, Kg
C = 3e+08; ## Velocity of light, m/s
E_max = (E_g/(1.+(m_e*C**2)/(2.*E_g)))/(1.6e-013); ## Energy of the compton recoil electron, MeV
print"%s %.2f %s"%("The energy of the compton recoil electrons = ",E_max," MeV");

## Result
## The energy of the compton recoil electrons = 1.961 MeV
The energy of the compton recoil electrons =  1.96  MeV

Ex10-pg125¶

In :
## Exa3.10 : : Page-125(2011)
#find The average energy of the positron
m_e = 9.1e-31;    ## Mass of the positron, Kg
e = 1.6e-19;        ## Charge of the positron, coulomb
c = 3e+08;            ## Velocity of the light, metre per sec
eps = 8.85e-12;        ## Absolute permittivity of free space, per N per metre-square per coulomb square
h = 6.6e-34;        ## Planck's constant, joule sec
E = e**2.*m_e*c/(eps*h*1.6e-13);        ## Average energy of the positron, mega electron volts
print'%s %.2f %s'%("The average energy of the positron = ",E,"Z MeV");

## Result
## The average energy of the positron = 0.0075Z MeV
The average energy of the positron =  0.01 Z MeV

Ex11-pg125¶

In :
## Exa3.11 : : Page-125(2011)
#find The refractive index of the gas and The angle at which Cerenkov radiation is emitted
import math
P = 1.;     ## Momentum of the proton, GeV/c
M_0 = 0.94;    ## Rest mass of the proton, GeV/c-square
G = math.sqrt((P/M_0)**2.+1.) ## Lorentz factor
V = math.sqrt(1.-1./G**2.);     ## Minimum velocity of the electron, m/s
u = 1/V;     ## Refractive index of the gas
print'%s %.2f %s'%("The refractive index of the gas =", u,"");
u = 1.6;     ## Refractive index
theta = round (math.acos(1/(u*V))*180/3.14);     ## Angle at which cerenkov radiatin is emitted,degree
print'%s %.2f %s'%("The angle at which Cerenkov radiation is emitted =",theta," degree")

## Result
##  The refractive index of the gas = 1.37
## The angle at which Cerenkov radiation is emitted = 31 degree
The refractive index of the gas = 1.37
The angle at which Cerenkov radiation is emitted = 31.00  degree

Ex12-pg126¶

In :
## Exa3.12 : : Page-126(2011)
#find The minimum kinetic energy required to  electron to emit cerenkov radiation
import math
n = 1+1.35e-04; ## Refractive index of the medium
V_min = 1./n; ## Minimum velocity of the electron, m/s
p = (1.+V_min)*(1.-V_min); ## It is nothing but just to take the product
G_min = 1./math.sqrt(p); ## Lorentz  factor
m_e = 9.10939e-031; ## Mass of the electron, Kg
C = 3e+08; ## Velocity of light, metre per  sec
T_min = ((G_min-1.)*m_e*C**2.)/(1.602e-013); ## Minimum kinetic energy required by an  electro to emit cerenkov radiation, mega electron volts
print'%s %.2f %s'%("The minimum kinetic energy required to  electron to emit cerenkov radiation = ",T_min," MeV");

## Result
## The minimum kinetic energy required to  electron to emit cerenkov radiation = 30.64 MeV
The minimum kinetic energy required to  electron to emit cerenkov radiation =  30.64  MeV