Physics and Engineering of Radiation Detection

152 Chapter 3. Gas Filled Detectors

Solution:

The total number of charge pairs, according to equation 3.1.4 is given by

nt =(0.9)

(dE/dx)co 2

Wco 2

+(0.1)

(dE/dx)ch 4

Wch 4

=(0.9)

3. 01 × 103

33

+(0.1)

1. 48 × 103

28

≈ 87 charge-pairs/cm.

Similarly, the number of primary ion pairs can be computed from equation

3.1.5 as follows.

np =(0.9)(np,co 2 )+(0.1)(np,ch 4 )

=(0.9)(34) + (0.1)(46)

≈ 35 charge-pairs/cm.

3.2 DiffusionandDriftofChargesinGases

Both the electrons and ions produced as a result of passage of radiation quickly lose
their energy by multiple collisions with gas molecules. The way these charges move
in the gas depends largely on the type and strength of the force they experience. The
behavior of the charges therefore with and without electric field differ significantly
from each other.

3.2.A DiffusionintheAbsenceofElectricField

In the absence of an externally applied electric field, the electrons and ions having
energy E can be characterized by the Maxwellian energy distribution (39),

F(E)=

2

√

π

(kT)−^3 /^2

√

Ee−E/kT, (3.2.1)

wherekis the Boltzmann’s constant andTis the absolute temperature. The average
energy of charges, as deduced from this distribution, turns out to be

E ̄=^3

2

kT, (3.2.2)

which at room temperature is equivalent to about 0.04eV. Since there is no exter-
nally applied electric field, there is no preferred direction of motion for the charges in
a homogeneous gas mixture and therefore the diffusion is isotropic. In any direction
x, the diffusion can be described by the Gaussian distribution

dN=

N

√

4 πDt

e−x

(^2) / 4 Dt
dx, (3.2.3)
whereN is the total number of charges andDis the diffusion coefficient. This
relation simply represents the number of chargesdNthat can be found in an element