Physics and Engineering of Radiation Detection

2.3. Interaction of Photons with Matter 99

mean free path can be defined as the distance a photon beam will travel such that
its intensity decreases by 1− 1 /e, which corresponds to about 63% of the original
intensity.
The total linear attenuation coefficient in the above relations determines how
quickly or slowly a certain photon beam will attenuate while passing through a
material. It is a function not only of the photon energy but also of the type and
density of the material. Its dependence of these three parameters is actually a
problem for tabulation and therefore instead of the total linear attenuation coefficient
a related quantity, the mass attenuation coefficient, defined by

μm=

μt

ρ

, (2.3.36)

is quoted. Hereρis the density of the material. Since mass attenuation coefficient is
independent of the physical state of the material, therefore one can easily deduce the
linear attenuation coefficient by simply multiplying it by the density of the material.
μtandμmare generally quoted in literature in units ofcm−^1 andcm^2 /grespectively.
Just like the mean free path, we can also define the specific mean free path using
the mass attenuation coefficient as

λp=

1

μm

. (2.3.37)

The reader is encouraged to compare this with the definition given in equation??.
The total attenuation coefficient characterizes the probability of interaction of
photons in a material. Since the cross section also characterizes the probability of
interaction therefore these two quantities must be related. This is true since one can
write the total linear attenuation coefficientμtin terms of total cross sectionσtas

μt = σtN

= σt

ρNA

A

, (2.3.38)

whereNrepresents the number of atoms per unit volume in the material having
atomic numberAandρis the weight density of the material. NAis the familiar
Avogadro’s number. It is evident from the above relations that the units ofμtwill
becm−^1 ifσtis incm^2 /atom,Nis inatoms/cm−^3 ,andρis ingcm−^3.
We can also write the total mass attenuation coefficient in terms of the total cross
section as

μm=σt

NA

A

. (2.3.39)

Example:

Determine the thickness of lead needed to block 17. 4 keV x-rays by a factor

of 10^4. The density and mass attenuation coefficient lead are 11. 3 gcm−^3 and

122. 8 cm^2 /g.

Solution:

Blocking the beam by a factor of 10^4 implies that the intensity of the x-ray

beam coming out of the lead should be 10−^4 times the original beam intensity.