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

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To reduce radiation exposure from a gamma radiation source (226Ra) in a hospital laboratory, effective shielding materials such as concrete, lead, and steel are considered, with specific half-thicknesses provided. Calculating the necessary thickness to reduce exposure to 1% involves using the half-thickness values for each material. An alternative method to minimize exposure is increasing the distance from the radiation source; moving from 1m to 10m significantly decreases radiation exposure due to the inverse square law. The discussion highlights the need for clarity on calculations without logarithms for exam preparation. Understanding these principles is crucial for ensuring safety in laboratory environments.
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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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