1. Interaction of Radiation with Matter - 4
Buildup and Shielding
Day 2 – Lecture 4

2. Objective
To discuss shielding for photon beams and the increase in photon transmission through a shield resulting from buildup

3. HVL and TVL
The amount of shielding required to reduce the incident radiation levels by ½ is called the “half-value layer” or HVL
The HVL is dependent on the energy of the photon and the type of material.
Similarly, the amount of shielding required to reduce the incident radiation levels by 1/10 is called the “tenth-value layer” or TVL.

4. HVL and TVL HVL (cm) TVL (cm) 137Cs 0.66 4.8 1.6 0.65 15.7 5.3 2.1
Isotope
Photon E (MeV)
Concrete
Steel
Lead
137Cs
0.66
4.8
1.6
0.65
15.7
5.3
2.1
60Co
1.17, 1.33
6.2
1.2
20.6
6.9 4
198Au
0.41
4.1
0.33
13.5
1.1
192Ir
0.13 to 1.06
4.3
1.3
0.6
14.7 2
226Ra
0.047 to 2.4
2.2
1.66
23.4
7.4
5.5
This table lists HVL and TVL values for some common radionuclides.

5. HVL and TVL
The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows:
HVL =
ln(2) 
TVL =
ln(10) 
and =
TVL
HVL
ln(10) 
ln(2)
ln(10)
ln(2)
2.303
0.693 = =
= 3.323
TVL = x HVL

6. Shielding
Shielding is intended to reduce the radiation level at a specific location
The amount of shielding (thickness) depends on:
the energy of the radiation
the shielding material
the distance from the source
Photons with energy less than about 1 MeV interact almost exclusively with atomic electrons. Material consisting of large atoms with many electrons is most effective as shielding. Lead is one of the most frequently used materials, although tungsten or bismuth‑lead mixtures and even uranium are used in special applications. A rule of thumb for shielding against photons is that every 5 cm of lead provides approximately a factor of 20 reduction in the intensity. However there are many exceptions for radionuclides which emit low or very high energy photons.

7. Inverse Square Law
If the radiation emanates from a point source, the radiation follows what is commonly known as the “Inverse Square Law” or ISL.
Most real sources which are considered to be “point” sources are actually not “point” sources. Most sources such as a 60Co teletherapy source have finite dimensions (a few cm in each direction). These sources appear to behave like point sources at some distance away but as one gets closer to the source, the physical dimensions of the source result in a breakdown of the ISL.

8. Inverse Square Law
If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.
However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.
For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the source increases even without any shielding.

9. Scatter
These photons should not strike the individual. But due to scatter, they do, so the calculated value is too low. It needs to be increased by the buildup factor.

10. Photon Attenuation and Absorption
Absorption refers to the total number of photons absorbed by the material (dark blue arrows)
Attenuation refers to total number of photons removed from incident beam (absorbed + scattered) (dark blue and light blue arrows)
Absorption depends on the Atomic Number and density of the absorber. Photons interact with the electrons of the absorber. The higher the atomic number the greater the number of electrons in an atom and the greater the density, the more atoms there are per unit volume. This is why lead is such a good absorber.

11. Photon Attenuation Ix = Io e-x = Io e  where:
(x) -
Ix = Io e-x = Io e
where:
Ix = photon intensity after traversing x cm of some material
Io = initial or incident photon intensity
x = thickness of material (cm)
 = linear attenuation coefficient (cm-1)
 = density (g/cm3)
/ = mass attenuation coefficient (cm2/g)

12. Attenuation and Buildup
IxB
As noted previously, the attenuation of photons includes scattered and absorbed photons. However, the exponential attenuation equation only specifies what fraction of the initial photon beam is transmitted directly through the material. Experimentally, this is valid for a narrow beam of photons passing through a very thin target with the detector far enough away to exclude scattered photons. As the target gets thicker, and the photon beam gets wider, the possibility that photons which would not normally be detected will be scattered into the detector increases.
In the figure to the right, the incident arrows with the dashed lines represent photons outside of the primary beam as seen by the detector. They would not normally enter the detector so they would not be considered by the exponential attenuation equation. However, as is evident, some are being scattered into the detector so that the detector will record not only the photons which succeeded in penetrating the material directly, but also some of those photons which were scattered. As a result, the actual number of photons measured by the detector will be greater than predicted by the exponential attenuation equation.
To account for this non-ideal but nonetheless realistic scattering situation, the buildup factor is introduced. The buildup factor “B” which is simply a number greater than or equal to one is nothing more than a correction factor which converts a theoretical result into a realistic one.
where B1

13. Buildup I = Io B e(-x) primary photons + scattered photons B =
Scatter is an important issue if we have a point source and it is not collimated. If we are attempting to calculate how much shielding is required to reduce the radiation levels to a certain level, we will underestimate the amount needed if we do not consider buildup in our calculations.
Buildup is a function of the material, its thickness and the energy of the photons. The material and thickness together are represented by “x” which is a dimensionless quantity.
Initially we assume that B = 1. Knowing what fraction of the incident photons were detected (I/Io) and using B =1, we solve for x. This would be the answer if all the photons detected were primary photons, directly transmitted through the material without scatter. Next we look up B from a table of empirically derived values where B is given as a function of both x and the energy of the photons, both of which we hopefully know.
Substituting into the equation, the value of B found in the table, we determine a new value for x. The process is repeated, until B converges to some final value.
If there are no scattered photons, then B = 1
If there are scattered photons, then B > 1

14. Sample Buildup
What amount of lead shielding is needed to reduce the dose rate beyond the shield from 1 mSv/hr to 0.02 mSv/hr for a 1 MeV photon beam?

15. Sample Buildup Material Photon Energy
First we need to determine . This table provides values of /. So we need to multiply this value by the density of lead to obtain .
The value of / for a 1 MeV photon incident on a lead shield is cm2/g.

16. Sample Buildup The mass attenuation coefficient (/) for a
1 MeV photon incident on a lead shield is:
(/) = cm2/g
The density of lead = g/cm3
The linear attenuation coefficient is:
(/) x  = cm2/g x g/cm3 = cm-1
The density of lead is g/cm3.

17. Sample Buildup
I = Io B e(-x) I = 1 mSv/hr Io = 0.02 mSv/hr B = 1 (assumed)  = cm-1 Solve for “x” 0.02 mSv/hr = (1 mSv/hr) (1) e(-x)

18. Sample Buildup = e(-x) 1 mSv/hr ln(0.02) = ln[e(-x)] -3.91 = -x
Although it is not required:
x = -3.91/ = 4.86 cm
This would be the calculated value of the lead shield if scatter was not considered.
When we arrive at the final answer we will see that this value is too low.

19. Sample Buildup
x = 3.91
The calculated value of x is not given in the table. However, 3.91 is relatively close to 4 so that we can assume the correct value for the buildup factor (B) is closer to 2.26 than Of course you can interpolate to get an exact value, however, I would estimate the value to be 2.2.

20. Sample Buildup x = 3.91 Let’s interpolate: when x = 4, B = 2.26
(4-2)
( )
(4-3.91)
( x) =
If we interpolate we get 2.19 which is close to the original estimate of 2.2.
1.27 = 0.09/( x)
( x) = 0.09/1.27 = 0.071
2.26 – = x = 2.19

21. Sample Buildup
Solve for x using B = 2.2 I = I0 B e(-x) 0.02 mSv/hr = (1 mSv/hr) (2.2) e(-x) ln(0.02/2.2) = ln[e(-x)] = - x so x = -4.71/ = 5.86 cm The thickness of the shield increased from 4.86 cm to 5.86 cm (20%) due to buildup
One could repeat the process again using the new x value and calculate a new “B” etc. Eventually the iterative process will settle on a recommended value for the thickness of the shielding.

22. Skyshine
Photons can scatter or “bounce” off atoms in materials such as the ceiling of a room or even air molecules !
Skyshine is another calculation altogether. It uses equations which include the solid angle through which the radiation travels. One source for information and equations dealing with skyshine is the National Council on Radiation Protection and Measurements (NCRP) Report No. 51 “Radiation Protection Design Guidelines for 0.1 – 100 MeV Particle Accelerator Facilities”

23. Where to Get More Information
Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009)
International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)