11.2. Quantities Related to Dosimetry 617
This collision Kerma can also be used to determine the total air Kerma. To do this
we combine equations 11.2.22 and 11.2.28 together and obtain
Kair=
33. 85 X
1 −g
. (11.2.29)
In most instances one is interested in collision Kerma as opposed to total Kerma.
The reason is that most of the photons produced as a result of radiative energy
transfer escape the medium without depositing any energy. Such photons do not
contribute to the dose and can therefore be ignored. However if the medium is large
enough or if the distribution of energy transferred is very wide, some radiative losses
may lead to absorption of energy.
Example:
In an earlier example, a 2 hour exposure at a distance of 3cmfrom a 10
mCicobalt-60 source was found to be 29.3R. Compute the air Kerma if the
quality is approximately given by 0.01.
Solution:
The air Kerma can be computed from equation 11.2.29. But first we need to
convert the exposure in units ofCkg−^1 .Usingtheconversion1R=2. 58 × 10 −^4
Ckg−^1 we get
X=(29.3)(2. 58 × 10 −^4 )=7. 56 × 10 −^3 Ckg−^1.
Substituting this and the given value ofginto equation 11.2.29 gives us the
required air Kerma.
Kair =
33. 85 X
1 −g
=
(33.85)
(
7. 56 × 10 −^3
)
1 − 0. 01
=0. 258 Jkg−^1.
Another parameter commonly used to quantify the distribution of energy between
the absorbed doseDand radiative losses is given by
η=
D
Kcol
. (11.2.30)
The value ofηchanges as the beam traverses into the medium as shown in Fig.11.2.1.
The three regions in the graph based on the value ofηare described below.
η<1: This refers to the case when absorbed dose is less than the collisional
energy loss of radiation. It occurs at short depths of the target. As the radiation
penetrates the material, more and more collisional energy gets absorbed and
contribute to the total dose.
η= 0: At some depth the collisional loss becomes just equal to the absorbed
dose. That is, all the energy lost by the radiation through collisional processes