Physics and Engineering of Radiation Detection

112 Chapter 2. Interaction of Radiation with Matter

Let us now define a cylinder of radiusbaround the pathxof motion of ion with

a volume element of 2πbdbdx.IfNeis the electron number density then the

total number of electrons in this volume element will be equal to 2πNebdbdx.

Every electron in this volume element will experience the same impulse and

therefore the total energy transferred to the electron when the ion moves a

distancedxwill be

−

dE

dx

=2πNeq

∫

∆Ebdb.

=

4 πNeq^2 e^4

mev^2

ln

(

bmax

bmin

.

)

Here we have evaluated the integral from minimum to maximum impact pa-

rameter (bmintobmax) since integrating from 0 to∞would yield a divergent

solution.

In order to determine a reasonable value ofbminwe note that the minimum

impact parameter will correspond to collisions in which the kinetic energy

transferred to electron is maximum. The maximum energy that the ion can

transfer to the electron turns out to beEmax =2meγ^2 v^2 .Thereaderis

encouraged to perform this derivation (Hint: use law of conservation of linear

momentum). Substitution of this in equation 2.4.9 gives

∆E|bmin=

2 q^2 e^4

mev^2 b^2 min

=2γ.

Hence we find

bmin=

qe^2

γmev^2

Now we will use some intuitive thinking to come up with a good estimate

of the maximum impact parameter. In order for the ion to be able to deliver

a sharp impulse to the electron, it should move faster than the electron in the

atomic orbit, such as

bmax

γv

≤

Re

ve

.

HereReis the atomic radius andveis the velocity of the electron. Since this

ratio is a function of the atomic numberZof the material, we can write

bmax

γv

≤f(Z)

Substitution ofbminandbmaxin 2.4.10 yields the required equation 2.4.8.

[

−

dE

dx

]

Bohr

=

4 πq^2 e^4 Ne

mev^2

ln

[

γ^2 mev^3 f(Z)

qe^2

]