The gamma rate generated by a pencil-like gamma source of a length 2L and activity A, intercepting a dosimeter placed at a point M such us OM = r, is given by [
The air attenuation factor of the photon flux is equal to
. Given that, X = 150 cm and μ = 6.8 × 10−5 cm−1 [
where ξ = 0.061 cm2/g [
Since
, the photon interaction probability ξ, ρe is small (0.065), the development of Equation (3-2) to the first order gives:
The dose rate is equal to the electrons rate multiplied by the mean energy deposited by each electron in the dosimeter. At 
(the mean energy of the two 60Co photons), Compton scattering is predominant and the forward electron energy is E0 = 1.03 MeV having a range R = 0.45 g/cm2 [
If a Compton electron is produced in the dosimeter front end region, equal to
, the electron energy loss is equal to
, where 
is the mean Compton electron energy:
where 
is the electron kinetic energy at a given angle 
and 
is the Compton differential cross section.
After integration, we obtain
.
Consequently, the dose rate in the dosimeter is given by:
where the factor 2 corresponds to the two 60Co photons and md = ρeds is the dosimeter mass.
Thus:
After integration, we have:
where: 
is a constant. In our case A = 1.81 ±
0.18 PBq [
We obtain the same expression found by Huttlin [
.
If a Compton electron is produced in the dosimeter back end of a mass thickness equal to R, this electron will lose only a fraction of its kinetic energy. Therefore, the electron mean energy loss in the dosimeter, is no more equal to
, but to 
given by:
which is equivalent to:
where 
is the mean energy loss of escaping electrons.
The dose rate is then:
After integration, we obtain:
where 
is given by:
is the correction factor due to escaping electrons.
The constant 
is determined by estimating
.
The mean energy loss of escaping electrons is given by:
where 
is the probability distribution. 
is the electrons rate per unit mass such as: 
and 
is the energy loss of escaping electrons per unit mass.
Since in our case
, we can write
, thus
.
Therefore, we can write:
Now, we use the approximation:
, where a is a constant and E is the electron energy.
Given that
, where 
is the forward Compton electron energy and 
is the energy loss of escaping electron, thus we have:
Finally, we obtain:
By injecting Equation (3-2-14) in Equation (3-2-8), we find:
To check the validity of Equation (3-2-13), it is equivalent to show Equation (3-2-14). To determinate the constant a, we integrate Equation (3-2-14) and we find
.
Finally Equation (3-2-14) can be written as:
To validate this equation, we used the GEANT4 Monte Carlo code. The irradiator is simulated by a single pencil of a 90 cm length (
shows the obtained probability distribution 
as a function of the escaping electron energy loss. The fit of the obtained distribution by Equation (3-2-9) gives a good mean squared error
.
To determine the dose rate dmed in a very thick medium from the dose rate ddos measured using a PMMA dosimeter we write:
where 
is the conversion factor.
Usually, the length L is small relatively to the distance r, in our case L = 45 cm and X = 150 cm.
Therefore, the correction factor becomes:
Given that:
, 
, 
, R = 0.45 g/cm2 and m = 1.07 g/cm2.
so k = 0.178, then for Z = 0, we obtain: Cf = 2.62.
This means that the measured dose rate is only about 38% of the dose rate that will be deposited in an infinite medium.