Engineering Question #2052
James Hope, a 17 year old male from Gloucester asks on April 17, 2004,
Last year in school we learned about nuclear decay, and how some atoms decay by releasing gamma rays, which have enough energy to cause damage to flesh. We were also taught that several inches of lead will make a gamma ray harmless. However, in order to make it harmless surely the energy of the gamma ray must decrease, and according to E=hf, this means the frequency must therefore become less. Is this right? Because it seems odd that by putting the right amount of lead in the way you can change the frequency of the gamma rays and have visible light coming out the other side.
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Indeed, the process of stopping a gamma photon in lead (or any shielding material) is often a case of decreasing its energy, or increasing its wavelength, in successive steps until it is wholly absorbed. The major interaction for this energy reduction is inelastic (Compton) scattering of atomic electrons.
It is true, therefore, that a gamma photon could be converted to other, lower-energy, types of photons in this manner. It may even become a photon of visible light, and, if the shielding material were transparent to light (which lead isn't!), or if this happened at the surface, such a photon could then emanate from the material.
However, what is more likely to occur is the gamma photon reaching a lower energy whereby it could excite a number of low-orbit electrons and thus become wholly absorbed (the "photo-electric effect"; sort of the reverse of x-ray production). Furthermore, should a lucky gamma photon survive this energy regime there are a host of other photon-matter interactions at lower energies, which would reduce the chances of further survival down to the energy range of visible light. But it's not impossible.
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