1. Gamma Shielding Calculations
Radiation Protection III
NUCP2331

2. SHIELDING Gamma Radiation
Gamma rays do not give up energy continuously as charged particles do
Gamma ray intensity decreases due to absorption or scattering
Efficiency of the shield depends upon the probability of interaction and thickness of material
Attenuation-Process by which radiation is reduced in intensity when passing through some material

3. Attenuation Coefficients
l (cm-1) Linear Absorption Coefficient is defined as the probability of interaction per unit path length of material
m (cm2/g) Mass Absorption Coefficient is defined as / where  is the linear absorption coefficient and  is density
μl = μm ρ
μm = mass attenuation coefficient
ρ = density of material in g/cc

4. EM Formula I = Io e (-μlx) I = intensity with the shield,
Io = intensity without shield,
μl = linear attenuation coefficient,
x = thickness of shield
Be careful about the mu that you are using, there are about 6 different mus

5. Example I0 = 1000 mR/hr Your shield material is Lead
Shield thickness is 2 inches
Radionuclide is Cs-137
What do you do first?

6. Example What is reading on other side of shield?
I0 = 1000 mR/hr
Shield is Lead
thickness 5 cm (2 in), μm = c/g, ρ = 11.35g/cc
What is reading on other side of shield?
I = 1000 e -(0.115)(11.35)(5)
I = 1.47 mR/hr
Examples of a shielding formula application, the more times you use it the easier it gets.

7. Example I0 = 2000 mR/hr Your shield material is Iron
Shield thickness is 4 inches
Radionuclide is Cs-137
What do you do first?

8. Example What is reading on other side of shield?
I0 = 2000 mR/hr
Shield is Iron
thickness 10 cm (4 in), μm = cc/g, ρ = 7.8 g/cc
What is reading on other side of shield?
I = 2000 e -(0.0745)(7.8)(10)
I = mR/hr
Examples of a shielding formula application, the more times you use it the easier it gets.

9. Scatter Remember the Compton scatter
Some of the radiation interacting with the shield will create some new EM radiation, some of which may escape the shield
This will add exposure to the area on the backside of the shield ,
So how does on take this into account?

10. Build up
So if you have another parameter in your equation that changes your dose after your shield, your shield will need to be thicker to take into account eh greater dose
So that means you have 2 unknowns in your equations, not good
Need to estimate the shield thickness calculate the B (get close) and use that

11. Build up Factors I = BIo e (-μlx) B= build up factor based on μx
Dependant on energy and density of material
Caused by Compton scattering
How will your answers change?
Build up factors are added shielding that takes into the added electromagnetic radiation that is created in the shield by the Compton effect.

12. Example I0 = 1000 mR/hr Your shield material is Lead
Shield thickness is 2 inches
Radionuclide is Cs-137
What do you do first?
Build up factor?

13. Example What is reading on other side of shield?
I0 = 1000 mR/hr
B =? μx= 6.5 energy is .662 MeV
Shield is Lead
thickness 5 cm (2 in), μm = c/g, ρ = 11.35g/cc
What is reading on other side of shield?
I = e -(0.115)(11.35)(5)
I = 3.38 mR/hr
Examples of a shielding formula application, the more times you use it the easier it gets.

14. Example I0 = 2000 mR/hr Your shield material is Iron
Shield thickness is 4 inches
Radionuclide is Cs-137
What do you do first?

15. Example What is reading on other side of shield?
I0 = 1000 mR/hr
B μx= 6 energy is .662 MeV
Shield is Iron
thickness 10 cm (4 in), μm = cc/g, ρ = 7.8 g/cc
What is reading on other side of shield?
I = e -(0.0745)(7.8)(10)
I = 55.3 mR/hr
Examples of a shielding formula application, the more times you use it the easier it gets.

16. Half Value Layers Half-Value Layer
That thickness of material which reduces the radiation intensity to one-half of its initial value
Tenth-Value Layer
That thickness of material which reduces the radiation intensity to one-tenth of its initial value
Both take into account build up
Half values layers are used to be able to cut the radiation level down depending one what material you have available.

17. How many
How many half value layers would it take to reduce the radiation intensity from 500 mR/hr to less than 5 mR/hr?

18. Total thickness Number of HVL X the thickness of each HVL
What is total thickness of material need to reduce 500 mR/hr to less than 5 mR/hr? Of
Iron
Concrete
Lead

19. Problem
How many TVL does it take to decrease a reading of 350 mR/hr to under 5 mR/hr?
How many HVL are there in a TVT?

20. Medical X ray
Several factors need to be taken into account when designing shielding for an X-ray machine
Not really for the people getting x rayed but for the people outside the X-ray room
The dose to those people are limited, as opposed to the dose to the patient
What energy the machined operates, how many patients will be x rayed, and where the X-ray room are all important to the design of the room

21. Medical X-ray 3 sources of radiation
Primary radiation (from the machine)
Full beam from the machine interacting with wall or floor
Secondary radiation
Scattered radiation ( from patient)
Leakage radiation ( from machine)
Dose rate at some distance (primary)
H (unshielded) =WUT/ d2

22. Medical X-ray Shielded/Unshielded(Dose/mA minute) = PD2/WUT
P- dose level in area needs to be r/wk
D-Distance from the machine (m)
W- workload (mA min/wk)
U-use factor how often the primary beam is faced on that barrier
T-occupancy percentage time area is in use

23. Xray room design Secondary barrier D primary D secondary
Primary barrier

24. Questions?