Chapter7-Gamma -Radiation

Ex1-pg292

In [1]:
## Exa7.1: : Page-292 (2011)
#find The distance to be moved for obtaining first order Bragg reflection 
import math
h = 6.6261e-034;            ## Planck's constant, joule sec
C = 2.998e+08;            ## Velocity of light, metre per sec
f = 2.;                    ## Radius of focal circle, metre
d = 1.18e-010;            ## Interplaner spacing for quartz crystal, metre
E_1 = 1.17*1.6022e-013;        ## Energy of the gamma rays, joule
E_2 = 1.33*1.6022e-013;        ## Energy of the gamma rays, joule
D = h*C*f*(1./E_1-1./E_2)*1./(2.*d);        ##Distance to be moved for obtaining first order reflection for two different energies, metre
print'%s %.2e %s'%("\nThe distance to be moved for obtaining first order Bragg reflection = ",D," metre");

## Result
## The distance to be moved for obtaining first order Bragg reflection = 1.08e-003 metre 
The distance to be moved for obtaining first order Bragg reflection =  1.08e-03  metre

Ex2-pg293

In [2]:
## Exa7.2: : Page-293 (2011)
#find The energy of the gamma rays
import math
m_0 = 9.1094e-031;    ## Rest mass of the electron, Kg
B_R = 1250e-06;       ## Magnetic field,tesla metre
e =  1.6022e-019;     ## Charge of the electron, coulomb
C = 3e+08;            ## Velocity of the light, metre per sec
E_k = 0.089;          ## Binding energy of the K-shell electron,MeV
v = B_R*e/(m_0*math.sqrt(1.+B_R**2.*e**2./(m_0**2*C**2)));  ## Velocity of the photoelectron, metre per sec
E_pe = m_0/(1.6022e-013)*C**2*(1./math.sqrt(1-v**2/C**2)-1.);  ## Energy of the photoelectron,MeV
E_g = E_pe+E_k;    ## Energy of the gamma rays, MeV
print'%s %.3f %s'%("\nThe energy of the gamma rays = ",E_g," MeV");

## Result
## The energy of the gamma rays = 0.212 MeV 
The energy of the gamma rays =  0.212  MeV

Ex3-pg293

In [3]:
## Exa7.3: : Page-292 (2011)
#find The attenuation of beam of X-rays in passing through human tissue 
import math
a_c = 0.221;      ## Attenuation coefficient, cm^2/g
A = (1-math.exp(-0.22))*100.;    ## Attenuation of beam of X-rays in passing through human tissue
print'%s %.2f %s'%("\nThe attenuation of beam of X-rays in passing through human tissue = ",math.ceil(A)," percent");

## Result
## The attenuation of beam of X-rays in passing through human tissue = 20 percent 
The attenuation of beam of X-rays in passing through human tissue =  20.00  percent

Ex4-pg293

In [4]:
## Exa7.4: : Page-293 (2011)
#find The partial life time for gamma emission 
import math
alpha_k = 45.;            ## Ratio between decay constants
sum_alpha = 0.08;        ## Sum of alphas
P = 0.35*1/60.;            ## Probability of the isomeric transition,per hour
lambda_g = P*sum_alpha/alpha_k;   ## Decay constant of the gamma radiations, per hour
T_g = 1/(lambda_g*365.*24.);        ## Partial life time for gamma emission,years
print'%s %.2f %s'%("\nThe partial life time for gamma emission = ",T_g," years");

## Result
## The partial life time for gamma emission = 11.008 years 
The partial life time for gamma emission =  11.01  years

Ex5-pg294

In [5]:
## Exa7.5:  : Page-294 (2011)
import math
#find The required gamma width
A = 11.;            ## Mass number of boron
E_g = 4.82;        ## Energy of the gamma radiation, mega electron volts
W_g = 0.0675*A**(2./3.)*E_g**3;        ## Gamma width, mega electron volts
print'%s %.2f %s'%("\nThe required gamma width = ",W_g," MeV");

## Result
## The required gamma width = 37.39 MeV 
The required gamma width =  37.39  MeV

Ex8-pg295

In [6]:
## Exa7.8:  : Page-295 (2011)
#find The excitation energy and The angular momentum
import math
e = 1.6022e-19;        ## Charge of an electron, coulomb
BR = 2370e-06;         ## Magnetic field in an orbit, tesla metre
m_0 = 9.1094e-31;      ## Mass of an electron, Kg
c = 3e+08;             ## Velocity of light, metre per sec
v = 1/math.sqrt((m_0/(BR*e))**2.+1./c**2);      ## velocity of the particle, metre per sec
E_e = m_0*c**2*((1.-(v/c)**2)**(-1/2.)-1)/1.6e-13;    ## Energy of an electron, MeV
E_b = 0.028;        ## Binding energy, MeV
E_g = E_e+E_b;        ## Excitation energy, MeV
alpha_k = 0.5;        ## K conversion coefficient
Z = 49.;                ## Number of protons
alpha = 1./137.;        ## Fine structure constant
L = (1/(1.-(Z**3/alpha_k*alpha**4.*(2.*0.511/0.392)**(15./2.))))/2.;    ## Angular momentum
l = 1;        ## Orbital angular momentum
I = l-1/2.;    ## Parity
print("\nFor K-electron state:" )
print' %s %.2f %s  %.2f %s  %.2f %s'%( "The excitation energy = ",E_g," MeV" and "The angular momentum = ",math.ceil(L)," " and  "\nThe parity : ",I,"");
## Result
## For K-electron state:
## The excitation energy = 0.393 MeV
## The angular momentum = 5
## The parity : 0.5 
For K-electron state:
 The excitation energy =  0.39 The angular momentum =   5.00 
The parity :   0.50 

Ex9-pg295

In [7]:
## Exa7.9: : Page-295 (2011)
#find The radioactive life time
import math
c = 3e+10;            ## Velocity of light, centimetre per sec
R_0 = 1.4e-13;        ## Distance of closest approach, centimetre 
alpha = 1./137.;        ## Fine scattering constant
A = 17.;                ## Mass number
E_g = 5.*1.6e-06;        ## Energy of gamma transition, ergs
h_cut = 1.054571628e-27;        ## Reduced planck constant, ergs per sec
D = c/4.*R_0**2.*alpha*(E_g/(h_cut*c))**3.*A**(2./3.);        ## Disintegration constant, per sec
tau = 1/D;        ## Radioactive lifr\e time, sec
print'%s %.1e %s'%("\nThe radioactive life time = ",tau," sec");

## Result
## The radioactive life time = 9e-018 sec 
print("error in answer due to round off error")
The radioactive life time =  8.7e-18  sec
error in answer due to round off error

Ex10-pg296

In [5]:
## Exa7.10: : Page-296 (2011)
#find The possible multipolarities are
import math
l = 2,3,4
print("\nThe possible multipolarities are ")
for l in range( 2,4):
    if l == 2:
        print'%s %.2f %s'%("E",l," " );
    elif l == 3:
        print'%s %.2f %s'%(" M",l," ");
    elif l == 4:
        print'%s %.2f %s'%(" and E ",l," ");
    

for l in range( 2,4):
    if l == 2 :
     print'%s %.2f %s'%("\nThe transition E",l," dominates");
    


## Result
## The possible multipolarities are E2, M3 and E4
## The transition E2 dominates 
The possible multipolarities are 
E 2.00  
 M 3.00  

The transition E 2.00  dominates

Ex13-pg297

In [9]:
## Exa7.13: : Page-297 (2011)
#find The relative source absorber velocity
import math
E_0 = 0.014*1.6022e-13;   ## Energy of the gamma rays, joule
A = 57.;                   ## Mass number
m = 1.67e-27;             ## Mass of each nucleon, Kg
c = 3e+08;                ## Velocity of light, metre per sec
N = 1000.;                 ## Number of atoms in the lattice
v = E_0/(A*N*m*c);        ## Ralative velocity, metre per sec
print'%s %.2f %s'%("\nThe relative source absorber velocity = ",v," m/s");

## Result
## The relative source absorber velocity = 0.079 m/s 
The relative source absorber velocity =  0.08  m/s

Ex14-pg297

In [10]:
## Exa7.14: : Page-297 (2011)
#find The required frequency shift of the photon
import math
g = 9.8;            ## Acceleration due to gravity, metre per square sec
c = 3e+08;                ## Velocity of light, metre per sec
y = 20.;                ## Vertical distance between source and absorber, metre
delta_v = g*y/c**2;        ## Frequency shift
print'%s %.2e %s'%("\nThe required frequency shift of the photon =  ", delta_v,"");

## Result
## The required frequency shift of the photon = 2.18e-015  
The required frequency shift of the photon =   2.18e-15 

Ex15-pg246

In [2]:
## Exa6.15: : Page-246 (2011)
#find they are parallel spin or anti
import math
import numpy
a='antiparallel spin'
b='parallel spin'
S=([a, b])


for i in range  (0,1):
    if S[i] == 'antiparallel spin' :
       print("\nFor Fermi types :")
       print("\n\n The selection rules for allowed transitions are :  \n\tdelta I is zero \n\tdelta pi is plus \nThe emited neutrino and electron have %s")
       print "S(i,1)"
    elif S[i] == 'parallel spin':
       print("\nFor Gamow-Teller types :")
       print("\nThe selection rules for allowed transitions are : \n\tdelta I is zero,plus one and minus one\n\tdelta pi is plus\nThe emited neutrino and electron have %s")
       print("S(i,1)") 
    

## Calculation of ratio of transition probability
M_F = 1.;    ## Matrix for Fermi particles
g_F = 1.;        ## Coupling constant of fermi particles
M_GT = 5/3.;        ## Matrix for Gamow Teller
g_GT = 1.24;        ## Coupling constant of Gamow Teller
T_prob = g_F**2*M_F/(g_GT**2*M_GT);    ## Ratio of transition probability
## Calculation of Space phase factor
e = 1.6e-19;        ## Charge of an electron, coulomb
c = 3e+08;            ## Velocity of light, metre per sec
K = 8.99e+9;        ## Coulomb constant
R_0 = 1.2e-15;        ## Distance of closest approach, metre
A = 57.;            ## Mass number
Z = 28.;            ## Atomic number 
m_n = 1.6749e-27;    ## Mass of neutron, Kg
m_p = 1.6726e-27;    ## Mass of proton, Kg
m_e = 9.1e-31;        ## Mass of electron. Kg
E_1 = 0.76;         ## First excited state of nickel
delta_E = ((3*e**2*K/(5*R_0*A**(1/3.))*((Z+1.)**2-Z**2))-(m_n-m_p)*c**2)/1.6e-13;         ## Mass difference, mega electron volts
E_0 = delta_E-(2*m_e*c**2)/1.6e-13;    ## End point energy, mega electron volts
P_factor = (E_0-E_1)**5/E_0**5;    ## Space phase factor 
print'%s %.2f %s %.2f %s '%("\nThe ratio of transition probability =",T_prob,""and"\nThe space phase factor =",P_factor,"");
 
## Result
## The emited neutrino and electron have antiparallel spin
## For Gamow-Teller types :
## The selection rules for allowed transitions are : 
##	delta I is zero,plus one and minus one
##	delta pi is plus
## The emited neutrino and electron have parallel spin
## The ratio of transition probability = 0.39
## The space phase factor = 0.62 a
For Fermi types :


 The selection rules for allowed transitions are :  
	delta I is zero 
	delta pi is plus 
The emited neutrino and electron have %s
S(i,1)

The ratio of transition probability = 0.39  0.62