Physics, 19.02.2021 16:30 quincyjosiah07
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?
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Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead...
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