Calculate the radiative and collisional energy losses in keV

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

Solution

Some amount of bremsstrahlung radiations can reappear after passing through lead but this is not the case with plexiglass.