# Chapter5-Alpha particles¶

## Ex1-pg203¶

In :
## Exa5.1 : : Page 203 (2011)
#find Disintegration energy and Barrier height for alpha-particle
import math
E_a = 8.766;    ## Energy of the alpha particle, MeV
A = 212.;          ## Atomic mass of Po-212, amu
M_a = 4.;         ## Atomic mass of alpha particle, amu
e = 1.6e-019;        ## Charge of an electron, coulomb
Z = 82.;                ## Atomic number of Po-212
R_0 = 1.4e-015;        ## Distance of closest approach,metre
K = 8.99e+09;            ## Coulomb constant
E = E_a*A/(A-M_a);    ## Disintegration energy, mega electron volts
B_H = 2.*Z*e**2.*K/(R_0*A**(1./3.)*1.6*10**-13.);    ## Barrier height for an alpha particle within the nucleus, MeV
print"%s %.3f %s  %.3f %s "%("Disintegration energy :",E," MeV Barrier height for alpha-particle: ",B_H," MeV");

## Result
## Disintegration energy : 8.935 MeV
## Barrier height for alpha-particle: 28.26 MeV

Disintegration energy : 8.935  MeV Barrier height for alpha-particle:   28.259  MeV


## Ex2-pg203¶

In :
## Exa5.2 : : Page 203 (2011)
## We have to make calculation for alpha particle and for  proton
import math
E_a = 8.766;     ## Energy of the alpha particle, mega electron volts
A_Bi = 209.;      ## Atomic mass of Bi-209, atomic mass unit
A_a = 4.;         ## Atomic mass of alpha particle, atomic mass unit
A_p = 1.;         ## Atomic mass of proton, atomic mass unit
e = 1.6e-019;    ## Charge of an electron, coulomb
Z = 83.;          ## Atomic number of bismuth
R_0 = 1.4e-015;  ## Distance of closest approach,metre
K = 8.99e+09;    ## Coulomb constant
B_H_a = 2.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_a**(1/3.)));    ## Barrier height for an alpha particle, mega electron volts
B_H_p = 1.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_p**(1/3.)));    ## Barrier height for proton, mega electron volts
print'%s %.2f %s %.2f %s'%("Barrier height for the alpha particle =",B_H_a," MeV Barrier height for the proton = ",B_H_p," MeV");

## Result
## Barrier height for the alpha particle = 22.67 MeV
## Barrier height for the proton = 12.30 MeV

Barrier height for the alpha particle = 22.67  MeV Barrier height for the proton =  12.30  MeV


## Ex3-pg203¶

In :
## Exa5.3 : : Page 203 (2011)
## We have also calculate the value of magnetic field in a particular orbit.
import math
C = 3e+08;                ## Velocity of light, m/S
M_0 = 6.644e-027*(C)**2/(1.60218e-013);        ## Rest mass of alpha particle, MeV
T = 5.998;                ## Kinetic energy of alpha particle emitted by Po-218
q = 2*1.60218e-019;        ## Charge of alpha particle, C
V = math.sqrt(C**2*T*(T+2*M_0)/(T+M_0)**2);            ## Velocity of alpha particle,metre per sec
B_r = V*M_0*(1.60218e-013)/(C**2.*q*math.sqrt(1-V**2./C**2.));                ## magnetic field in a particular orbit, Web per mtere
print'%s %.2e %s %.2f %s '%("The velocity of alpha particle : ",V," m/sThe magnetic field in a particular orbit :",B_r," Wb/m");

## Result
## The velocity of alpha particle : 1.699e+007 m/s
## The magnetic field in a particular orbit : 0.3528 Wb/m

The velocity of alpha particle :  1.70e+07  m/sThe magnetic field in a particular orbit : 0.35  Wb/m


## Ex4-pg204¶

In :
## Exa5.4: : Page 204 (2011)
#find The probability of leakage of alpha-particle through potential barrier
import math
a = 10**-14;        ## Width of the  potential barrier, m
E = 5*1.60218e-013;        ## Energy of the alpha particle, joule
V = 10*1.60218e-013;        ## Potential height, joule
M_0 = 6.644e-027;        ## Rest mass of the alpha particle, joule
h_red = 1.05457e-034;        ## Reduced value of Planck's constant,joule sec
T = 4*math.exp(-2*a*math.sqrt(2*M_0*(V-E)/h_red**2));   ## Probability of leakage through through potential barrier
print'%s %.2e %s'%("The probability of leakage of alpha-particle through potential barrier =  ",T,"");

## Result
## The probability of leakage of alpha-particle through potential barrier = 1.271e-008

The probability of leakage of alpha-particle through potential barrier =   1.27e-08


## Ex6-pg204¶

In :
## Exa5.6: : Page 204 (2011)
#find The disintegration constant increases
import math
Z_D = 82.;    ## Atomic number of Po
E_Po210 = 5.3;    ## Alpha-source for Po210, MeV
E_Po214 = 7.7;    ## Alpha-source for Po214, MeV
log_lambda_Po210 = -1*1.72*Z_D*E_Po210**(-1/2.);
log_lambda_Po214 = -1*1.72*Z_D*E_Po214**(-1/2.);
delta_OM_t = log_lambda_Po214 - log_lambda_Po210;    ## Difference in order of magnitude of life times of Po214 and Po210
print'%s %.2d %s'%("The disintegration constant increases by a factor of some 10^ " ,delta_OM_t,"");

## Result
## The disintegration constant increases by a factor of some 10^10

The disintegration constant increases by a factor of some 10^  10


## Ex8-pg205¶

In :
## Exa5.8:  : Page 205 (2011)
#find The half life of Pu
import math
N = 120.1*6.023e+023/239.;    ## Number of Pu nuclei
P_rel = 0.231;        ## Power released, watt
E_rel = 5.323*1.6026e-13;        ## Energy released, joule
decay_rate = P_rel/E_rel;        ## Decay rate of Pu239, per hour
t_half = N*math.log(2)/(decay_rate*365.*86400.);    ## Half life of Po239, sec
print"%s %.2e %s"%("The half life of Pu = ",t_half," yr");

## Result
## The half life of Pu = 2.46e+004 yr

The half life of Pu =  2.46e+04  yr


## Ex9-pg205¶

In :
## Exa5.9 : : Page 205(2011)
#find The slope of alpha decay energy versus atomic number
import math
a_v = 14.;        ## Volume energy constant, MeV
a_s = 13.;        ## Surface energy constant, MeV
a_c = 0.60;      ## Coulomb energy constant, MeV
a_a = 19.;        ## Asymmetric energy constant, MeV
A = 202.;        ## Mass number
Z = 82;         ##  Atomic number
dE_by_dN = -8/9*a_s/A**(4/3.)-4/3.*a_c*Z/A**(4/3.)*(1-4.*Z/(3.*A))-16.*a_a*Z/A**2.*(1.-2.*Z/A);        ## Slope, mega electron volts per nucleon
print'%s %.2f %s'%("The slope of alpha decay energy versus atomic number = ",dE_by_dN," MeV/nucleon");

## Result
## The slope of alpha decay energy versus atomic number = -0.15007 MeV/nucleon

The slope of alpha decay energy versus atomic number =  -0.15  MeV/nucleon


## Ex10-pg206¶

In :
## Exa5.10 : : Page 206 (2011)
#find The hindrance factor for alpha particle
import math
h_kt = 1.05457e-34;        ## Reduced Planck's constant, joule sec
e = 1.60218e-19;        ## Charge of an electron, coulomb
l = 2.;                ## Orbital angular momentum
eps_0 = 8.5542e-12;        ## Absolute permittivity of free space, coulomb square per newton per metre square
Z_D = 90.;            ## Atomic number of daughter nucleus
m = 6.644e-27;        ## Mass of alpha particle, Kg
R = 8.627e-15;        ## Radius of daughter nucleus, metre
T1_by_T0 = math.exp(2.*l*(l+1.)*h_kt/e*math.sqrt(math.pi*eps_0/(Z_D*m*R)));    ## Hindrance factor
print'%s %.2f %s'%("The hindrance factor for alpha particle = " ,T1_by_T0,"");

## Result
## The hindrance factor for alpha particle = 1.768

The hindrance factor for alpha particle =  1.77