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A device emits 0.2 μSv/h of gamma rays. How thick does an aluminum sheet need to be to completely stop radiation from coming out ? What equation is to be used to calculate this ?

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3 Answers 3

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A few of the details needed to answer this question include:

  • What energy are the gamma rays?
  • How far is the sheild to be placed from the source?
  • How far is the dose calculation from the source/sheild?
  • Are you including buildup factors or not?
  • What is the composition/form of the aluminum?
  • How did you calculate your dose?
  • What do you mean by completely?

If you mean a 0% chance of a photon being transmitted through the sheild, it needs to be infinitely thick. If you can answer some of these questions, NIST provides a very useful tool for looking up the specific attenuation coefficient.

Oddly enough, because you gave the radiation level in units of dose equivalent, the question is literally impossible to answer. What absorbed the radiation?? and what are you sheilding???

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By the way, if you look at XCOM, the results are given in units of range (cm^2/g). Multiply this value times the density of the shield to get the average distance traveled by a particle incident on the shield. Inverting this value then gives you the attenuation coefficient. –  AdamRedwine Apr 5 '12 at 19:00
I was just curious to know what kind of shielding this device would require (specs have not been released yet): –  Nigel Ridley Apr 5 '12 at 20:05
I think there are quite a few posts here about the e-cat. The general consensus is that it's bogus. Order of magnitude we are talking about a centimeter or so of shielding for close range protection. –  AdamRedwine Apr 7 '12 at 20:23

X-rays are attenuated when they pass through any material, and the amount they are attenuated depends on how far they travel in the material and that material's mass attenuation coefficient. The NIST web site has an excellent set of pages on this at

NB the x-ray intensity falls exponentially, so the x-rays will never fall to zero. What the calculation described on the NIST web page will tell you is the thickness you need to reduce the intensity to below some specified limit.

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You also you need to take into account the radiation that is generated by the interactions within the Al (unless that is neglected). This modifies the thickness of Al required. The correct name for that term is 'Buildup Factor'. There is a chapter in Cember & Johnson that explains it very nicely.


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The buildup factor was mentioned in one of the previous answers. Would it be possible to elaborate more on the details to make it a more useful answer? –  mpv Feb 26 '14 at 13:53

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