Physics and Engineering of Radiation Detection

130 Chapter 2. Interaction of Radiation with Matter

Example:

Compute the critical energy for electrons in aluminum.

Solution:

To compute the critical energy, we use equation 2.5.19 withZ= 13,a= 610,

andb=1.24.

Ec ≈

a

Z+b

=

610

13 + 1. 24

=42. 84 MeV

2.5.C EnergyStraggling

In case of heavy charged particles we saw that, for at least thick absorbers, the
fluctuations in the energy loss can be described by a symmetric distribution. This
is not the case for electrons, which, owing to their small mass, suffer more collisions
as compared to heavy charged particles of same energy. As a consequence of such
motion the energy distribution of these electrons gets skewed with a long tail towards
high energy. Since for heavy charged particles passing through thin absorbers one
can employ the Landau distribution to describe the energy straggling, therefore it is
natural to use the same methodology for the electrons as well. This strategy works
reasonably well for most materials and electron energies. The Landau distribution
can be expressed as

f(x, )=

Φ(λ)

ξ

, (2.5.20)

where

Φ(λ)=

1

π

∫ 1

0

e−uln(u)−uλsin(πu)duand

λ=

1

ξ

[

−ξ(ln(ξ)−ln(E)+1−γ)].

Hereγ=0.577 is known as Euler’s constant. The term ln(E) in the above equation
can be computed from

ln(E)=ln

[

I^2 (1−β^2 )

2 mc^2 β^2

]

+β^2 , (2.5.21)

whereIis the logarithmic mean excitation energy andβ=v/c,vbeing particle’s
velocity.ξis called the scale of the Landu distribution and is given by

ξ =

2 πNAe^4 Zx

mv^2 A

or (2.5.22)

ξ =

0. 1536 Zx

β^2 A

keV, (2.5.23)