ANSWERS: 2

I would have thought so. There'd be two ways to approach this, off hand. One, would be to calculate the attenuation of the wave through the material The other... would probably involve the crosssection of the beta radiation and the material... but I'm not 100% sure off hand how I'd go around doing that... Attenuation mechanisms probably work best to be honest

1) "The process by which beta particles lose energy in absorbers are similar to those for alpha particles. However, an additional problem encountered when shielding against beta radiation is the process whereby electromagnetic radiation (secondary Xrays), called bremsstrahlung, are produced. The fraction of beta energy reappearing as bremsstrahlung is approximately ZE/3000 where Z is the atomic number of the absorber and E is the beta energy in MeV. This means that shielding for beta radiation should be constructed of materials of lowatomicnumber to reduce the amount of bremsstrahlung emitted. A beta source emits beta rays with energies covering the complete spectrum from zero to a characteristic maximum energy, Emax. The mean beta energy is, in most cases, about 1/3Emax. The penetrating power of beta particles depends in their energy. For example, a 1 MeV beta particle will travel about 3.5 m in air. Therefore the thickness and choice of material for shielding from beta radiation depends upon: stopping the highest energy beta, for example Sr90 emits a .546 MeV beta while its daughter Y90 emits a 2.27 MeV; and shielding any bremsstrahlung." Source and further information: http://www.hanford.gov/rl/uploadfiles/ALARA_Analysis.pdf Further information: http://en.wikipedia.org/wiki/Bremsstrahlung 2) "Absorption of beta particles in matter  The absorption of beta particles by matter, from a macroscopic point of view, is a function of the distance traveled by the particles through the absorbing material and the density of the material. The product of these two variables, the density times distance has the units grams cm2 or often, for convenience, mg cm2. This value is used to express the absorber thickness. Let a beam of beta particles of intensity Io impinge upon a single absorber as shown in the following figure. Some of these beta particles will be absorbed so the reduced intensity of the emerging beam is I. It has been observed that the absorption of beta particles is approximately an exponential function of the density "_" of the absorber and the distance "X" through the absorber, and " m " the absorbing property of this particular material. Hence, the absorption of beta particles is approximately analogous to the absorption of light, and the mathematical relationship for beta absorption takes the same form as the familiar BeerLambert Law. ln(I0/I)= (μ/ρ) Xρ (1) Since "X" is measured in cm and the density in mg/cm3, the units of the product "Xρ" are mg/cm2. Consequently "μ/ρ", the mass absorption coefficient, has the units cm2/mg." Source and further information: http://www.drake.edu/artsci/physics/NuclearScience2.pdf
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