In reply to Bob Cook's message of Fri, 11 Mar 2016 09:34:55 -0800: Hi, [snip] >The effectiveness of the SS can at stopping any high energy electrons that >cause Bremsstrahlung would depend upon the thickness of the can (or alumina) >and the energy of the incident electrons. I think the loss of energy per >scattering event is proportional to Z ^2 for the nucleus that is doing the >scattering. Al at Z=13 and with Fe at Z=26 the intensity of the >Bremsstrahlung signal would be about a factor of 4 different. The mean length >of the path of an electron is a good parameter to know for any given substance >(basically its density) vs the incident energy of the electron. Shielding >engineering curves provide this information I believe. Iron being >significantly more dense than Al2O3 would be much better at slowing electrons >and thus producing Bremsstrahlung IMHO.
Shielding is based primarily on the electrons of an atom being ionized. Bremsstrahlung is created when a fast particle interacts with a nucleus. Most fast electrons impinging on solid matter will create ionization, i.e. they get stopped by other electrons. AFAIK Only about 1% get through to the nucleus and create Bremsstrahlung. I think that both nuclear charge and number of nuclei per unit volume would be important for Bremsstrahlung production. Mass of a nucleus not so much, because even a single proton is already about 2000 times more massive than an electron. When it comes to collisions, it makes little difference whether the nucleus is light or heavy. In short any nucleus is effectively an "immovable object" as far as an electron is concerned. BTW if MeV level electrons are stopped by Aluminium foil, then the can would have to be very thin not to stop them. Has anyone considered the possibility that some (little) bremsstrahlung might be caused by fast protons impacting on heavier nuclei? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
