# chapter 09 : Interaction of charged Particle and Neutron With Matters¶

## Ex9.1 : Pg:201¶

In [2]:
#Maximum energy transferred by alpha particles
E_alpha = 3e+06     # Incident energy of alpha particles, eV
m = 9.1e-031     # Mass of an electron, kg
M = 4*1.67e-027     # Mass of an alpha particle, kg
# As E_alpha = 1/2*M*v**2 so E_electron = 1/2*m*(2*v)**2
# From the two equations
E_electron = 4*E_alpha*m/M     # Maximum energy of electron, eV
print "The maximum energy transferred by alpha particles to the electron = %5.3f keV"%(E_electron/1e+03)

The maximum energy transferred by alpha particles to the electron = 1.635 keV


## Ex9.2 : Pg:201¶

In [5]:
#Rate of energy loss and range of deuteron and alpha particle
E_loss_P = 59     # Specific rate of energy loss per unit mass per unit area of proton, keV per mg cm square
R_prime_P = 50     # Range of proton, mg per cm
Z_D = 1     # Atomic number of deuteron
m_D = 2     # Mass of deuteron, units
E_loss_D = Z_D**2*E_loss_P     # Specific rate of energy loss per unit mass per unit area of deuteron, keV per mg cm square
R_prime_D = R_prime_P*m_D/Z_D**2     # Range of deuteron, mg per cm square
Z_alpha = 2     # Atomic number of alpha particle
m_alpha = 4     # Mass of alpha particle, units
E_loss_alpha = Z_alpha**2*E_loss_P     # Specific rate of energy loss per unit mass per unit area of alpha particle, keV per mg cm square
R_prime_alpha = R_prime_P*m_alpha/Z_alpha**2     # Range of alpha particle, mg per cm square
print "The specific rate of energy loss per unit mass per unit area of deuteron = %2d keV per mg cm square"% (E_loss_D)
print "The range of deuteron = %3d mg per cm square"% (R_prime_D)
print "The specific rate of energy loss per unit mass per unit area of alpha particle = %2d keV per mg cm square"% (E_loss_alpha)
print "The range of alpha particle = %2d mg per cm square"% (R_prime_alpha)

The specific rate of energy loss per unit mass per unit area of deuteron = 59 keV per mg cm square
The range of deuteron = 100 mg per cm square
The specific rate of energy loss per unit mass per unit area of alpha particle = 236 keV per mg cm square
The range of alpha particle = 50 mg per cm square


## Ex9.3 : Pg:202¶

In [7]:
from math import log
#Thickness of concrete collimator
rho = 2200e-03     # Density of concrete, g per cm
mu_m = 0.064     # Mass attenuation coefficient of concrete, cm square per g
mu = rho*mu_m     # Linear attenuation coefficient o concrete, per cm
# As attenuation exponential is exp(-mu*x) = 1e+06, solving for x
x = -log(1e-06)/mu
print "The required thickness of concrete to attenuate a collimated beam = %2d cm"% x

The required thickness of concrete to attenuate a collimated beam = 98 cm


## Ex9.4 : Pg:202¶

In [12]:
from math import log
#Average number of collsions for thermalization of neutrons
A = 9.0     # Mass number of beryllium
xi = 2.0/A - 4.0/(3*A**2)     # Logarithmic energy decrement of energy distribution of neutron
E0 = 2.0     # Initial energy of neutrons, MeV
En_prime = 0.025e-06     # Thermal energy of the neutrons, MeV
n = 1.0/xi*log(E0/En_prime)     # Average number of collisions needed for neutrons to thermalize
En_half = 1.0/2*E0     # Half of the initial energy of neutrons, MeV
n_half = 1.0/xi*log(E0/En_half)     # Number of collsions for half the initial energy of neutrons
print "The average number of collsions for thermalization of neutrons = %2d"% n
print "The number of collsions for half the initial energy of neutrons = %3.1f"% n_half

The average number of collsions for thermalization of neutrons = 88
The number of collsions for half the initial energy of neutrons = 3.4


## Ex9.5 : Pg:202¶

In [13]:
#Change in voltage across a G.M. tube
e= 1.6e-019     # Charge on an electron, coulomb
W = 25     # Ionization potential of gas (Ar/N2), eV
E = 5e+06     # Energy of incident alpha particles, eV
C = 1e-010     # Capacity of the system, farad
N = E/W     # Number of ions produced
delta_V = N*e/C     # Change in voltage across the G.M. tube, volt
print "The change in voltage across the G.M. tube = %3.1e volt"% delta_V

The change in voltage across the G.M. tube = 3.2e-04 volt