Experiment 3
Attenuation of Gamma rays by matter
Objective:
1- Verification of the attenuation equation of Gamma radiation.
2- Determining for gamma rays produced by a certain isotope:
a. The mass attenuation coefficient (μm).
b. The linear attenuation coefficient (μ) in lead (Pb) and Aluminum (Al).
c. The half value layer thickness (X1/2) of the absorbing materials Pb and Al.
Theory:
When Gamma radiation passes through matter, it undergoes attenuation primarily by
Compton, photoelectric and pair production interactions. The intensity of the radiation
is thus decreased as a function of thickness of the absorbing material. The
mathematical expression for intensity ( I ) is given by the following expression:
𝐼 = 𝐼𝑜 𝑒 −𝜇𝑥
(1)
where,
Io is the original intensity of the beam.
I is the intensity transmitted through an absorber thickness X.
μ is the linear attenuation coefficient for the absorbing material.
(Note that the two terms attenuation and absorption are often used interchangeably
even though they differ slightly. Attenuation is the result of both absorption and
scattering of the photons in the material).
If we rearrange eq.(1) and take the natural logarithm of both sides, the expression
becomes,
𝐼
ln ( ) = −𝜇𝑥
𝐼
𝑜
1
(2)
The half value layer (HVL) of the absorbing material is defined as that thickness X1/2
which will decrease the initial intensity by half. That is, IHVL=Io/2.
If we substitute this into eq.(2) we get:
ln(2) = 𝜇𝑥1𝑙2
(3)
Rearranging eq.(3) we get:
𝑥1𝑙2 =
ln(2)
(4)
𝜇
Keep in mind that both the attenuation coefficient and the HVL are functions of the
energy of the gamma photons.
Another useful concept is the mass attenuation coefficient (µm) defined by:
μm =
μ
(5)
ρ
Where  is the density of the absorber material. This concept is useful because
attenuation is found to be dependent on the density of the absorber material (more
atoms and heavier atoms in the way of the photons mean more interactions per unit
length of the material). Therefore, both the linear attenuation coefficient and the HVL
will have different values for the same material depending on its phase (solid, liquid or
vapor) and on its temperature. By dividing on density however, we get a quantity that
is independent of the density and hence is the same for the material regardless of its
phase and temperature. This quantity is the mass attenuation coefficient. Moreover,
μm is approximately independent of the type of material.
We can also define the equivalent thickness (or mass thickness𝑥𝑚 ) to be
𝑥𝑒𝑞 = 𝜌𝑥
Writing equation (1) in terms of these two quantities we get
𝐼 = 𝐼𝑜 𝑒
−(𝜇𝑚 𝜌)(
𝑥𝑒𝑞
)
𝜌
=𝐼𝑜 𝑒 −𝜇𝑚 𝑥𝑒𝑞
(6)
Which is the same as eq (1) and hence 𝜇𝑚 𝑎𝑛𝑑 𝑥𝑒𝑞 can be used instead of µ and x to
find all results and they satisfy:
(𝑥𝑚 )1/2 =
ln(2)
𝜇𝑚
2
(7)
Note that taking natural logarithm of both sides of eq(6) gives:
𝐼
ln( ) = −𝜇𝑚 𝑥𝑒𝑞 ⟹ ln(𝐼) = −𝜇𝑚 𝑥𝑒𝑞 + ln(𝐼𝑜 )
𝐼𝑜
(8)
From which we can see that if we draw a graph of ln(𝐼) vs 𝑥𝑒𝑞 , then the slope of that
graph would be −𝜇𝑚 .
Apparatus:
Source of gamma radiation.
Sheets of different absorbing materials (Aluminum and Lead) and different
thicknesses .
Nuclear station ST150.
Procedure:
1- Connect the plugs of the electric mains.
2- Set the timer to 60s and the voltage to the operating voltage found in first lab.
3- Record the count rate per one minute for the back ground (IB.G) 3 times and take
the average.
4- Put the source in front of the GM tube on the 3rd shelf from the top.
5- Record the count rate ( Io ) 3 times and take average.
6- Place Al sheet between the source and the GM tube.
7- Record the count rate 3 times (I1 , I2 and I3) and then find Iavg..
8- Repeat steps 6 and 7 for the rest of the aluminum sheets with increasing the
thickness of the absorbing material. Also take all possible combinations of 2 Al
sheets between the source and the GM tube.
9- Repeat steps 6,7 and 8 for the Pb sheets (4 sheets plus 4 combinations of two
sheets to get 8 data points).
For each material do the following steps:
10- Using Microsoft Excel, Plot a graph between ln(I ) and thickness 𝑥𝑒𝑞 , if the
relation is a straight line, then the attenuation law is verified.
11- Find the slope from the graph.
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12- Find the negative of the slope from the graph, this is equal to the mass
attenuation coefficient.
13- Calculate the linear attenuation coefficient using eq (5). Note that:
Al= 2.7 g/cm3 , pb=11.34 g/cm3
14- Calculate the HVL thickness by using eq.(4).
Post lab questions:
 Find the thickness of Aluminum that would reduce the intensity of the given
gamma radiation to one tenth of its original intensity.
 Find the thickness of Lead that would reduce the intensity of the given gamma
radiation to one tenth of its original intensity.
 Compare. Which material is better to use for shielding?
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