What is the best way to reduce radiation exposure in a hospital laboratory?

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1. Homework Statement

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. Homework 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 without 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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Hey, i have a similar problem and the best I've come up with so far is the inverse square law, ie at 1m dose rate constant, at 2m the dose rate is the square of the distance thus at 3m equals 9m2 etc then the HVL of the shielding material is calculated by thickness required to absorb half of the remaining dose.
 
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The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?

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