# Chapter6-Beta-Decay¶

## Ex1-pg240¶

In :
## Exa6.1: : Page- 240 (2011)
#find The rate at which energy is emitted
T = 5*24*60*60;    ## Half life of the substance, sec
N = 6.023e+026*4e-06/210;        ## Number of atoms
D = 0.693/T;           ## Disintegration constant, per sec
K = D*N;                ## Rate of disintegration,
E = 0.34*1.60218e-013;        ## Energy of the beta particle, joule
P = E*K;                    ##   Rate at which energy is emitted, watt
print'%s %.2f %s'%("\nThe rate at which energy is emitted = ",P," watt");

## Result
## The rate at which energy is emitted = 1 watt

The rate at which energy is emitted =  1.00  watt


## Ex2-pg241¶

In :
## Exa6.2 : : Page-241 (2011)
#find The magnetic field perpendicular to the beam of the particle
M_0 = 9.10939e-031;         ## Rest mass of the electron, Kg
C = 2.92e+08;            ## Velocity of the light, metre per sec
E = 1.71*1.60218e-013;        ## Energy of the beta particle, joule
e = 1.60218e-019;                ## Charge of the electron, C
R = 0.1;                     ## Radius of the orbit, metre
B = M_0*C*(E/(M_0*C**2)+1)*1/(R*e); ## Magnetic field perpendicular to the beam  of the particle, weber per square metre

print'%s %.2f %s'%("\nThe magnetic field perpendicular to the beam of the particle = ",B," Wb/square-metre");

## Result
## The magnetic field perpendicular to the beam of the particle = 0.075 Wb/square-metre

The magnetic field perpendicular to the beam of the particle =  0.08  Wb/square-metre


## Ex3-pg241¶

In :
##  Exa6.3 : : Page-241 (2011)
#find The energy of the electron and The energy of the converted gamma ray photon
import math
m_0 = 9.10963e-031;         ## Rest mass of the electron, Kg
e = 1.60218e-019;           ## Charge of the electron, C
c = 2.9979e+08;            ## Velocity of the light, metre per sec
BR = 3381e-006;     ## Field-radius product, tesla-m
E_k = 37.44;    ## Binding energy of k-electron
v = 1/math.sqrt((m_0/(BR*e))**2+1/c**2); ## Velocity of the converson electron, m/s
E = m_0*c**2*(1/math.sqrt(1-v**2/c**2)-1.)/(e*1e+003);   ## Energy of the electron, keV
E_C = E+E_k;            ## Energy of the converted gamma ray photon, KeV
print'%s %.2f %s  %.2f %s'%("\nThe energy of the electron = ",E," keV " and " The energy of the converted gamma ray photon =", E_C," keV");

## Result
## The energy of the electron = 624.11 keV
## The energy of the converted gamma ray photon = 661.55 keV

The energy of the electron =  624.11  The energy of the converted gamma ray photon =  661.55  keV


## Ex4-pg241¶

In :
## Exa6.4 : : Page-241 (2011)
#find The rest energy carried out by the neutrino
import math
E = 18.1;                ## Energy carried by beta particle, keV
E_av = E/3.;               ## Average energy carried away by beta particle, keV
E_r = E-E_av;            ## The rest energy carried out by the neutrino, keV

print'%s %.2f %s'%("\nThe rest energy carried out by the neutrino : ",E_r," KeV");

## Result
## The rest energy carried out by the neutrino : 12.067 KeV

The rest energy carried out by the neutrino :  12.07  KeV


## Ex5-pg242¶

In :
## Exa6.5: : Page-242(2011)
#find The maximum energy available to the electrons in the beta decay
import math
M_Na = -8420.40;        ## Mass of sodium 24, keV
M_Mg = -13933.567;      ## Mass of magnesium 24, keV
E = (M_Na-M_Mg)/1000.;    ## Energy of the electron, MeV
print'%s %.2f %s'%("\nThe maximum energy available to the electrons in the beta decay = ",E," MeV");

## Result
## The maximum energy available to the electrons in the beta decay = 5.513 MeV

The maximum energy available to the electrons in the beta decay =  5.51  MeV


## Ex6-pg242¶

In :
## Exa6.6: : Page-242 (2011)
#find The linear momentum of neutrino and The linear momentum of beta particle
import math
c = 1.;  ## For simplicity assume speed of light to be unity, m/s
E_0 = 0.155;        ## End point energy, mega electron volts
E_beta = 0.025;        ## Energy of beta particle, mega electron volts
E_v = E_0-E_beta;        ## Energy of the neutrino, mega electron volts
p_v = E_v/c;            ## Linear momentum of neutrino, mega electron volts per c
m = 0.511;            ## Mass of an electron, Kg
M = 14*1.66e-27;        ## Mass of carbon 14,Kg
c = 3e+8;                ## Velocity of light, metre per sec
e = 1.60218e-19;            ## Charge of an electron, coulomb
p_beta = math.sqrt(2*m*E_beta);    ## Linear momentum of beta particle, MeV/c
sin_theta = p_beta/p_v*math.sin(45/57.3);    ## Sine of angle theta
p_R = p_beta*math.cos(45/57.3)+p_v*math.sqrt(1-sin_theta**2);  ## Linear momemtum of recoil nucleus, MeV/c
E_R = (p_R*1.6e-13/2.9979e+08)**2/(2.*M*e);  ## Recoil energy of product nucleus, MeV
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe linear momentum of neutrino = ",p_v," MeV/c" and "The linear momentum of beta particle =",p_beta," MeV/c" and "The energy of the recoil nucleus = ",E_R," eV")

## Result
## The linear momentum of neutrino = 0.13 MeV/c
## The linear momentum of beta particle = 0.1598 MeV/c
## The energy of the recoil nucleus = 1.20 eV

The linear momentum of neutrino =  0.13 The linear momentum of beta particle = 0.16 The energy of the recoil nucleus =  1.20  eV


## Ex7-pg242¶

In :
## Exa6.7: : Page-242 (2011)
#find The energy of the beta particle and The ratio of beta particle energy with end point energy
import math
N = 3.7e+10*60;        ## Number of disintegration, per sec
H = 0.0268*4.182;        ## Heat produced at the output, joule
E = H/(N*1.6e-013);        ## Energy of the beta particle, joule
M_Bi = -14.815;            ## Mass of Bismuth, MeV
M_Po = -15.977;            ## Mass of polonium, MeV
E_0 = M_Bi-M_Po;            ## End point energy, MeV
E_ratio = E/E_0;            ## Ratio of beta particle energy with end point energy
print'%s %.2f %s  %.2f %s '%("\nThe energy of the beta particle = ",E," MeV" and "The ratio of beta particle energy with end point energy =" ,E_ratio,"");

## Result
## The energy of the beta particle = 0.316 MeV
## The ratio of beta particle energy with end point energy = 0.272

The energy of the beta particle =  0.32 The ratio of beta particle energy with end point energy =  0.27


## Ex8-pg243¶

In :
## Exa6.8: : Page-243 (2011)
#find The parity of the 2.9 MeV level in be-8 and The threshold energy for lithium 7 neutron capture
import math
l = 2.;    ## Orbital angular momentum quantum number
P = (+1)**2*(-1)**l;    ## Parity of the 2.9 MeV level in Be-8
M_Li = 7.0182;        ## Mass of lithium, MeV
M_Be = 7.998876;        ## Mass of beryllium, MeV
m_n = 1.;                ## Mass of neutron, MeV
E_th = (M_Li+m_n-M_Be)*931.5;    ## Threshold energy, MeV
print'%s %.2f %s %.2f %s'%("\nThe parity of the 2.9 MeV level in be-8 = ",+P," " and"The threshold energy for lithium 7 neutron capture = ",E_th," MeV");

## Result
## The parity of the 2.9 MeV level in be-8 = +1
## The threshold energy for lithium 7 neutron capture = 18 MeV

The parity of the 2.9 MeV level in be-8 =  1.00 The threshold energy for lithium 7 neutron capture =  18.00  MeV


## Ex9-pg243¶

In :
##Page-243(2011)
import math
import numpy
#find  pairs are stable or not
M =numpy.zeros((4,2));
M[0,0] = 7.0182*931.5;    ## Mass of lithium, MeV
M[0,1] = 7.0192*931.5;    ## Mass of beryllium, MeV
M[1,0] = 13.0076*931.5;    ## Mass of carbon, MeV
M[1,1] = 13.0100*931.5;    ## Mass of nitrogen, MeV
M[2,0] = 19.0045*931.5;    ## Mass of fluorine, MeV
M[2,1] = 19.0080*931.5;    ## Mass of neon, MeV
M[3,0] = 33.9983*931.5;    ## Mass of phosphorous, MeV
M[3,1] = 33.9987*931.5;    ## Mass of sulphur, MeV
j = 0;
## Check the stability !!!!
for i in range  (0,3):
if round (M[i,j+1]-M[i,j]) == 1:
print("\n From pair a :")
print("\n         Be(4,7) is unstable");
elif round (M[i,j+1]-M[i,j]) == 2:
print("\n From pair b :")
print("\n         N(7,13) is unstable");
elif round (M[i,j+1]-M[i,j]) == 3:
print("\n From pair c :")
print("\n         Ne(10,19) is unstable");
elif round (M[i,j+1]-M[i,j]) == 0:
print("\n From pair d :")
print("\n         P(15,34) is unstable");

## Result
##
## From pair a :
##         Be(4,7) is unstable
## From pair b :
##         N(7,13) is unstable
## From pair c :
##         Ne(10,19) is unstable
## From pair d :
##         P(15,34) is unstable

 From pair a :

Be(4,7) is unstable

From pair b :

N(7,13) is unstable

From pair c :

Ne(10,19) is unstable


## Ex10-pg244¶

In :
## Exa6.10: : Page-244 (2011)
#find The half life of H3
import math
tau_0 = 7000.;        ## Time constant, sec
M_mod_sqr = 3.;        ## Nuclear matrix
E_0 = 0.018;        ## Energy of beta spectrum, MeV
ft = 0.693*tau_0/M_mod_sqr;   ## Comparative half life
fb = 10**(4.0*math.log10(E_0)+0.78+0.02);    ##
t = 10**(math.log10(ft)-math.log10(fb));   ## Half life of H3, sec
print'%s %.2f %s'%("\nThe half life of H3 = ",t," sec");

## Result
## The half life of H3 = 2.44e+009 sec

The half life of H3 =  2441293526.34  sec


## Ex11-pg244¶

In :
## Exa6.11: : Page-244 (2011)
import math
#find 92 percent beta emission
t_p = 33./0.92*365.*84800.;    ## Partial half life for beta emission, sec
E_0 = 0.51;        ## Kinetic energy
Z = 55.;            ## Atomic number of cesium
log_fb = 4.0*math.log10(E_0)+0.78+0.02*Z-0.005*(Z-1)*math.log10(E_0);    ## Comparitive half life
log_ft1 = log_fb+math.log10(t_p);     ## Forbidden tansition
## For 8 percent beta minus emission
t_p = 33./0.08*365.*84800.;    ## Partial half life, sec
E_0 = 1.17;        ## Kinetic energy
Z = 55;            ## Atomic energy
log_fb = 4.0*math.log10(E_0)+0.78+0.02*Z-0.005*(Z-1)*math.log10(E_0);    ## Comparitive half life
log_ft2 = log_fb+math.log10(t_p);    ## Forbidden transition
## Check the degree of forbiddenness !!!!!
if log_ft1 <= 10:
print("\nFor 92 percent beta emission :")
print("\n\tTransition is once forbidden and parity change");

if log_ft2 >= 10:
print("\nFor 8 percent beta emission :")
print("\n\t ransition is twice forbidden and no parity change");

## Result
## For 92 percent beta emission :
##	Transition is once forbidden and parity change
## For 8 percent beta emission :
##	Transition is twice forbidden and no parity change


For 92 percent beta emission :

Transition is once forbidden and parity change

For 8 percent beta emission :

ransition is twice forbidden and no parity change


## Ex12-pg244¶

In :
## Exa6.12: : Page-244(2011)
#find The value of coupling constant and The ratio of coupling constant
import math
h_kt = 1.05457e-34;    ## Reduced planck's constant, joule sec
c = 3e+08;             ## velocity of light, metre per sec
m_e = 9.1e-31;         ## Mass of the electron, Kg
ft_O = 3162.28;        ## Comparative half life for oxygen
ft_n = 1174.90;        ## Comparative half life for neutron
M_f_sqr = 2.            ## Matrix element
g_f = math.sqrt(2*math.pi**3*h_kt**7.*math.log(2.)/(m_e**5*c**4*ft_O*M_f_sqr));    ## Coupling constant, joule cubic metre
C_ratio = (2.*ft_O/(ft_n)-1.)/3.;    ## Ratio of coupling strength
print'%s %.3e %s %.2f %s'%("\nThe value of coupling constant = ",g_f," joule cubic metre" and "The ratio of coupling constant = ",C_ratio,"");

## Result
## The value of coupling constant = 1.3965e-062 joule cubic metre
## The ratio of coupling constant = 1.461

The value of coupling constant =  1.397e-62 The ratio of coupling constant =  1.46


## Ex13-pg245¶

In :
## Exa6.13: : Page-245 (2011)
#find The relative capture rate in holmium 161
import math
Q_EC = 850.;        ## Q value for holmium 161, keV
B_p = 2.0;    ## Binding energy for p-orbital electron, keV
B_s = 1.8;        ## Binding energy for s-orbital electron, keV
M_ratio = 0.05*(Q_EC-B_p)**2./(Q_EC-B_s)**2;    ## Matrix ratio
Q_EC = 2.5;        ## Q value for holmium 163, keV
C_rate = M_ratio*(Q_EC-B_s)**2./(Q_EC-B_p)**2.*100.;    ## The relative capture rate in holmium, percent
print'%s %.2f %s'%("\nThe relative capture rate in holmium 161 = ",C_rate," percent");

## Result
## The relative capture rate in holmium 161 = 9.8 percent

The relative capture rate in holmium 161 =  9.80  percent


## Ex14-pg246¶

In :
## Exa6.14: : Page-246 (2011)
#find The average energy of beta particles
import math
t_half = 12.5*365*24;        ## Half life of hydrogen 3, hour
D = math.log(2)/t_half;      ## Decay constant, per hour
N_0 = 6.023e+26;             ## Avogadro's number, per mole
m = 0.1e-03;                ## Mass of tritium, Kg
dN_by_dt = D*m*N_0/3.;  ## Decay rate, per hour
H = 21*4.18;                ## Heat produed, joule
E = H/dN_by_dt;             ## The average energy of the beta particle, joule
print'%s %.2e %s  %.2f %s'%("\nThe average energy of beta particles =",E,"joule = ",E/1.6e-016," keV");

## Result
## The average energy of beta particles = 6.91e-016 joule = 4.3 keV


The average energy of beta particles = 6.91e-16 joule =   4.32  keV


## Ex15-pg246¶

In :
## Exa6.15: : Page-246 (2011)
#find antiparallel spin and parallel spin
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