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By high energy gamma rays I mean 10+ MeV to 100 MeV. Is there a way to calculate it ? Also, at these high energies, do heavier elements like Lead for example still have the advantage over lighter ones like Aluminum ?

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I have tried this : wolframalpha.com/input/… and the result for 1 cm of Lead was 29 MeV, so what does that mean ? how much will 1 cm of lead reduce the intensity of a 29 MeV gamma ray by ? –  Abanob Ebrahim Aug 13 '13 at 13:44
    
I think a more useful subject for your question would be "What is the attenuation length of diferent materials for high energy photons?" –  Kevinismus Aug 13 '13 at 15:31
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Note that particle physicist at least talk about "radiation length" which represent (roughly) the length over which the intensity is reduced by a factor of $e$ rather than 2. There is a moderately complete discussion in the Particle Data Book published by the particle data group. –  dmckee Aug 13 '13 at 15:40

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In general it can be difficult to calculate the interaction of photons with atoms as a function of energy as different interaction mechanisms turn on and off: pair production turns on at 1.022 MeV, Rayleigh scattering turns on at low E as a function of atomic/molecular size, etc. An important question to ask yourself is "how precisely do I need to know the attenuation length of a photon in a given material?" Note that attenuation length is generally a more convenient quantity for calculations than halving thickness, and it corresponds to the distance at which $1/e \simeq 63\%$ of photons have been absorbed.

In many materials the mean free path of 10-100 MeV photons is roughly flat, as pair production is the dominant mechanism by which photons interact in this energy range. You can look up the attenuation lengths of various materials as a function of energy at this NIST page. You'll probably want to select units of $\frac{cm^2}{g}$. Whatever number you get from the table you will multiply by the density of your material---which has units of $\frac{g}{cm^3}$ ---the resulting number will be in units of $\frac{1}{cm}$ and is 1/attenuation length. Also, to answer your second question, in general you'll find that the increasing density of a material corresponds to shorter attenuation lengths, since the more charged particles you pack into a tighter volume corresponds to more photons stopped via pair production in that volume. In general heavier elements correspond to denser materials.

For a more precise analytical treatment of attenuation length, you may want to consult the literature. Looking in the PDG, they reference a paper by Y.S. Tsai in Rev. Mod. Phys. 46, 815 (1974). Good luck!

P.S.: Note that if you're asking this because you're trying to protect yourself from radioactive materials, you should contact a radiation safety expert as radiation protection can be much more complicated than just stopping gamma rays (material handling, neutron radiation, etc.).

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+1 on calling up the experts. –  Deer Hunter Aug 13 '13 at 15:48
    
@Kevinismus, I am having a hard time using NIST as it gives the results in a weird format and I can't get a number. If you could only give me the attenuation length for say Lead, that would be greatly helpful. –  Abanob Ebrahim Aug 13 '13 at 16:09
    
I have also come across this : laradioactivite.com/en/site/images/Long_attenuation_en.jpg. Does it mean that 1 cm of lead will absorb 63% of the incident gamma rays energy if they are 1200 MeV (according to the image) ? –  Abanob Ebrahim Aug 13 '13 at 16:23
    
@AbanobEbrahim: The attenuation length as a function of energy for lead (density ~ $11.3 \frac{g}{cm^3}$) goes from about 2 cm to 1 cm for E = 10 MeV to 100 MeV. And yes, you're reading that graph correctly. –  Kevinismus Aug 13 '13 at 16:25
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@AbanobEbrahim: it's not a stupid question, and the answer is "not necessarily". As I stated in my answer, different photon interaction processes turn on and off as energy changes. Towards the lower end of your energy range, pair production is still "turning on", meaning it isn't converting the lower energy photons to $e^{+}e^{-}$ pairs as efficiently as it converts higher energy photons. –  Kevinismus Aug 13 '13 at 17:05

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