This document derives the shielding formulas for gamma and X-rays. It defines the interactions that can occur when gamma rays encounter matter, including the photoelectric effect, Compton scattering, and pair production. It then derives the linear attenuation coefficient μ from the total macroscopic cross section Nσtotal. Using μ, the exponential attenuation formula is derived to describe how intensity decreases with thickness of shielding. Finally, the formulas for the half-value and one-tenth value thicknesses of a lead shield are calculated for 1.33 MeV gamma rays from cobalt-60.

1. Derivation of the shielding formulas (M J Rhoades)
Gammas and X-rays
We know the three major interactions that occur with gamma and X-rays in materials. This
matter could be our bodies so we had better consider ways to shield against it. I hate to think of
all those electrons and positrons flying around in the cells of my body. So, let's talk about the
attenuation of X and gamma rays. To start, let's consider a piece of lead shielding between a
point source and you. Gammas start out with an initial intensity (this can be dose rate, count rate,
energy, and so on.), and hit the shield. Gammas can, (depending on their energy), either goes on
through, create a photo electron, Compton scatter, or create an electron-positron pair.
If we described the probability that any one of the interactions above would occur, we would
want to find the cross section of absorption for each of the processes. The symbol for cross
section of absorption is σ. We would also want to total the different cross sections and have a
σtotal . The equation for this would then be as follows:
σtotal = σpe + σpp + Zσcs
Where: σtotal = the total cross section of absorption for the energy of our gamma.
σpe = the cross section for the photoelectric effect in cm2/atom.
σpp = cross section for pair production in cm2/atom.
Zσcs = cross section for Compton scattering x the number protons in cm2/atom
Example: Pb208* σtotal = (6.1766 x 10−23 cm2/atom) + (1.3 x 10−25 cm2/atom) +
(1.548 x 10-23cm2/atom)
σtotal = 2.1787 x 10-23cm2/atom
* From NIST data @ 1.3 MeV gamma.
Now, back to our lead shield, we have the data for the cross section for absorption of the lead
atoms for the three interactions. The next thing we need to know is how many atoms are packed
into that shield, for this we must know the atom density. The equation for this is as follows:
������������������������
N= Where: N = atom density in atoms/cm3
������
ρ = density in g/cm3
NAv = Avogadro's number ( 6.02 x 1023 atoms/mole )
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2. M = gram atomic weight in gram* atoms
(11.35������/������������ 3 )(6.02 ������ 10 23 ������������������������������ /������������������������ )
For Lead 208 we have: N =
208������/������������������������
N = 3.285 x 1022 atoms/cm3
We now have the information we need to determine the gamma attenuation for our 1.3 Mev
gamma, the 1/2 thickness formula, the 1/10 thickness formula, the linear attenuation coefficient,
the mass attenuation coefficient and more. A gamma point source emits gammas at 1.3 MeV
gammas at our lead shield at some intensity, say: gammas / m2/sec. Obviously,
IO If
γ
Pb
Nσ γ Detector
dt
the gammas pass into the shield, some are absorbed by the lead atoms. The amount of the
absorption is a function of Nσtotal and the thickness of the shield. Let me go over that again, from
the equations we talked about above, the intensity is reduced by the N, (atom density) x σtotal
(the total cross section for absorption) and the thickness. Now, think about if we added a little
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3. bit of thickness, (dt, change in thickness to the shield) yes that's right, the amount of gamma
making it through would drop a little. Or, if we were to put this in terms of an equation, it would
look like the following:
dI = -NσtotalI dt
Where: dI = the change in the intensity.
N = the atom density.
σtotal = the total absorption cross section.
I = the original intensity of gamma entering the shield.
dt = the change in thickness.
The negative sign comes from the fact that the change is negative, or reduced by the factors.
The equation in this form really does not help us very much. Differential equations in this form
just tell a story and are not good for plugging and chugging out answers. First, I want to let you
in on little secret before we go any further. (Most of you may have all ready known, but I did
not) Nσtotal = µ (The linear attenuation coefficient) I just wanted to show you where "µ" came
from. And, starting from scratch, as I did, is all ways a good idea in my opinion. Ok, back to our
equation. We are going to have to take definite integrals on both sides of the equation. It has to
be definite integrals because we do not want to deal with any constants. I am going to start this
on the next page so the equations are not split up. ( Don't have to keep thumbing back and forth.)
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4. Nσtotal = µ As stated above
dI = -µI dt Substituting µ for Nσtotal.
������������
= -µ dt Dividing both sides by I
������
������������ ������������ ������
������������ ������
= 0
−µ ������ ������ Taking definite integrals both sides, defined areas
Final intensity = If thickness Rule of log dx/x
������������ ������
������������
ln ������ = -µ 0
������������ Rule: constants can move
through the integral sign
Initial intensity = Io 0 thickness
ln (If) - ln (Io) = - µ( t-0t) Integral complete right side
������������
ln = -µ t Rule of logs on left side,
������������
������������
= ������ −µ ������ Removing logs both sides
������������
������������ = ������������ e
-µ t
Multiply through by Io, complete.
This equation is also the basis for the half and 1/10 layer calculations as well as working with
dose rates.
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5. While we have this formula fresh in our minds, we may as well get our 1/2 layer and 1/10
layer formulas derived, and finish up with our Pb 208 example. First the 1/2 layer formula:
If = Io e-µ t We are after the thickness of Pb which will reduce the gamma ray
intensity by 1/2. So let the ratio of If to Io be equal to 1/2.
1 ������������ -u t
=e Dividing both sides by I and putting in our desired ratio.
2 ������������
this just means that Io is being cut in half.
1
= e-µ t By taking the natural log both sides
2
1
ln 2 = ln e-µ t
- .693 = - µ t Result after taking the logs both sides.
−.693
= t1/2 Renaming the t to reflect the half value thickness.
−µ
Dividing by minus µ
.693
= t1/2 multiplying the left side by -1/-1, complete
µ
For the 1/10 layer formula just use 1/10 as the ratio for If / Io
1 ������������ -µ t
=e
10 ������������
-2.3o3 = -µ t1/ 10
2.303
= t1/ 10
µ
Remember that: Nσtotal = µ, let's go ahead and multiply my figures out for Pb208 with our
1.33 MeV gamma from cobalt 60.
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6. N = 3.285 x 1022 atoms/cm3
σtotal = 2.1787 x 10-23 cm2 / atom Then:
Nσtotal = (3.285 x 1022 atoms/cm3)(2.1787 x 10-23 cm2/atom ) = .7157 cm-1 = µ
So, for our shield using Pb 208, and our 1.33 MeV gamma calculations.
.693
= t1/2
µ
.693
= .968 cm = 9.68 mm of Pb 208
.7157������������ −1
This agrees very well for the lead 1/2 value listed for Pb206 in Martin and Harbison P.69.( I am
sure that the t1/2 layer values listed in table 8.1 are on the conservative side),
Other equations related to the intensity formula, I gave you are as follows:
Dt = D0 e-µ t Dose rate instead of intensity
If = I0 e (-µ/ρ) ρt This formula is for using the "Mass attenuation coefficient."
Most tables list the mass attenuation coefficient instead of
µ "the linear attenuation" but, all you have to do is
multiply by the density to get µ, so no problem.
µ ������������������������������������ ������������������������������������������������������������������ ������������������������������������������������������������������
So the mass attenuation coefficient is just =
������ ������������������������������������������
Ok , I kept my promise as stated earlier in the text that you can solve a lot of problems by
under sanding the base formula we worked through above. (Note: All of the base information for
working with these formulas, for all the nuclides, can be found by going to www. nist.gov/.
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