Calculate the radiative and collisional energy losses in keV

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 Me V 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

With lead most of the bremsstrahlung radiation reappear but it is not the case with plexiglass