radguide

36 Radiation for Radionuclide Users the new dose rate at the face of the hood? m = 2.31 cm-1 (for 131I) x = 1 cm = 20 mrem/hr = ? = 20e-(2.31)(1) = 2 mrem/hr Utilizing the same thickness of lucite, instead of lead: m = 0.12 cm -1 = 20e -(0.12)(1) = 18 mrem/hr This illustrates the effectiveness of lead over lucite in shielding gamma rays. Because beta particles have finite ranges, they do not strictly follow the shielding equation (i.e., Eq. 8). According to equation 8, no amount of shielding can totally absorb all gamma rays emitted by a given source. However, for any beta emitting source, there is a particular amount of shielding material that will absorb all beta particles emitted. As a secondary problem, the process by which beta particles are absorbed in matter results in the emission of x-rays known as bremsstrahlung radiation. The production of bremsstrahlung is much greater for high atomic number shields than for low atomic number shields. For example, as much as 8 percent of the energy from 1.5 MeV beta particles (e.g., as from 32P) is converted to bremsstrahlung when these particles are absorbed in lead whereas less than one percent of the energy of these particles is converted to x-rays when they are absorbed in lucite. The ideal material for shielding beta particles, then, is one that is thick enough to stop all the beta particles but with an atomic number low enough to minimize the production of bremsstrahlung. For shielding 32P beta particles, a 1 cm thickness of lucite or plexiglas is commonly used.