## gamma radiation sorce sheilding

1. The problem statement, all variables and given/known data

a gamma raidiation source (226Ra) is used in hospital laboratory. if shielding is considered as a means of control how many centimeters are needed to reduce the radiation to 1% of what a worker would be exposed to without shielding? assume the shield material is a) concrete B)lead C) steel note the half thicknesses are 0.12m, 0.014m ,0.018m respectively

part 2
as an alternative to shielding of the radiation source in the above question is to extend the distance between the radiation and the worker if the initial design placed the worker at 1m and then the design was review and place the source 10m what percentage reduction in radiation exposure would there for the second position relative to the first

2. Relevant equations

part 1 n=-2/log1/2
part 1 df/di = (ri/rf)^2 ri=1m rf=10m
3. The attempt at a solution

i have no idea how to do ether i asked the lecturer and he said there was a way with out have to do logarithms for the first part but didn't explain how and i need to know how for my exams please and help is greatly appreciated!!

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 Hi, Part 1: Please use Lambert's law. $$\frac{I_{\rm mat}}{I_0}=\exp[\frac{-\mu}{\rho}t_{\rm mat}\rho].$$ Where 'mat' represent the material used for shielding (concrete or lead or steel) Imat is the amount that you get after shielding I0 is just the source strength or amount without shielding $$\rho$$ is density tmat is thickness of the material $$\mu$$ is linear attenuation coefficient. Part 2: I dont understand. But maybe you need to use inverse square law. hope it helps.

 Tags gamma, logarithms, radiation, shielding