Gamma Shielding Calculations Radiation Protection III NUCP2331

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

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

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

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

Example What is reading on other side of shield? I0 = 1000 mR/hr Shield is Lead thickness 5 cm (2 in), μm = 0.115 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.

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

Example What is reading on other side of shield? I0 = 2000 mR/hr Shield is Iron thickness 10 cm (4 in), μm = 0.0745 cc/g, ρ = 7.8 g/cc What is reading on other side of shield? I = 2000 e -(0.0745)(7.8)(10) I = 5.988 mR/hr Examples of a shielding formula application, the more times you use it the easier it gets.

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?

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

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.

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?

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 = 0.115 c/g, ρ = 11.35g/cc What is reading on other side of shield? I = 2.3 1000 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.

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

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 = 0.0745 cc/g, ρ = 7.8 g/cc What is reading on other side of shield? I = 9.25 2000 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.

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.

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?

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

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?

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

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

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

Xray room design Secondary barrier D primary D secondary Primary barrier

Questions?