Physics and Engineering of Radiation Detection

2.3. Interaction of Photons with Matter 103

Example:

Estimate the mean free path of 150keV x-ray photons traveling in ordinary

water kept under normal atmospheric conditions. The total mass absorption

coefficients of hydrogen and oxygen are 0.2651cm^2 /g and 0.1361 cm^2 /g

respectively.

Solution:

The weight fractions of hydrogen and oxygen in water are

wh ≈

2

2+16

=0. 1111

wo ≈

16

2+16

=0. 8889

Hence according to equation 2.3.45, the mass attenuation coefficient of water

is

μwm = whμhm+woμom

=0. 1111 × 0 .2651 + 0. 8889 × 0. 1361

=0. 1504 cm^2 g−^1

Under normal conditions the density of water is 1gcm−^3. Therefore the total

linear absorption coefficient of water for 150keV photons is

μwt =ρwμwm=0. 1504 cm−^1.

The mean free path is simply the inverse of total linear attenuation coefficient.

Hence we get

λwm =

1

μwt

=

1

0. 1504

≈ 6. 65 cm.

B.3 StackedMaterials

If the photon beam passes through a number of materials, the total attenuation
of photons can still be determined by applying equation 2.3.33 for the consecutive
materials. Let us suppose we have three materials stacked one after the other as
shown in Fig.2.3.12. The intensity of photons at each junction of materials, as
determined from 2.3.33 is given by

I 1 = I 0 e−μ

(^1) td 1
I 2 = I 1 e−μ
(^2) td 2
= I 0 e−μ
(^1) td 1
e−μ
(^2) td 2
=I 0 e−(μ
(^1) td 1 +μ (^2) td 2 )
I 3 = I 2 e−μ
(^2) td 2
= I 0 e−(μ
(^1) td 1 +μ (^2) td 2 )
e−μ
(^2) td 2
=I 0 e−(μ
(^1) td 1 +μ (^2) td 2 +μ (^3) td 3 )
.