Physics and Engineering of Radiation Detection

664 Chapter 11. Dosimetry and Radiation Protection

If subscripts 1 and 2 represent distances from the source of 12cmand 1.2m

then the required exposure is given by

X 2 =

(

r 1

r 2

) 2

X 1

=

(

12

120

) 2

500

=5mR.

This example clearly demonstrates the advantage of increasing distance from

the source. Increasing the distance by 10 times decreases the exposure by 100

times.

A.3 Shielding

Shielding is an integral part of any radiation facility. In most situations it is the
most effective way of reducing radiation exposure to personnel. In chapter 2 we saw
that the intensity of a photon beam decreases exponentially with distance, that is

I=I 0 e−μx, (11.7.4)

whereIis the photon intensity at a depthx,I 0 is the initial intensity of the beam,
andμis the attenuation coefficient of the material. Recall thatμis a function of
the material as well as the energy of the radiation. At any specific energy its value
depends strongly on the atomic number of the material. The highZelements, such
as lead, are therefore the most effective means of shielding. Use of lead as a shielding
material is very common in laboratories working with radiation sources. In radiation
hostile environments, a combination of highZelements and concrete are used to
construct very thick shields.
Since the exposure and absorbed dose are directly related to the radiation inten-
sity, the above equation can also be written as

X = X 0 e−μx (11.7.5)

and D = D 0 e−μx. (11.7.6)

Of course, these expressions hold for the respective rates as well. One might argue
that, since exposure is defined for air only, the first of these expressions can not be
used for materials other than air. This is certainly not true since one can always
assume that the exposure is being measured outside the material. For example, it can
be used to determine the thickness of a shielding material such that the exposure
falls to a certain level, provided the exposure in air without shield is known (see
example below).

Example:

A^241 Amsource emits 60keV γ-rays. If the exposure rate in air at a certain

distance from the source is 500mR/hour, estimate the thickness of lead

needed to decrease the exposure rate to 10mR/hourat the same location.

Assume the attenuation coefficient to be 50cm−^1.