In reply to David Roberson's message of Wed, 6 Jun 2012 02:46:07 -0400 (EDT): Hi Dave, [snip] > >Robin, I would think the velocity of the proton of the same energy as compared >to an electron would be the square root of 2000 or 45 times slower due to the >velocity squared relationship.
You are correct. My brain doesn't work too well after lunch. :( >Now, if the proton slows down much faster than the electron then the >deceleration would be a lot greater. Perhaps 10 times greater? If you factor >this into account then the radiation levels of the two particles are >relatively close. What do you think? The deceleration due to interaction with atomic electrons is not what produces bremsstrahlung (it's way too small for that). Bremsstrahlung is only produced when a fast particle rapidly alters course in the neighborhood of a nucleus. Because of the mass difference this is much more significant for an electron than for a proton, not least because an electron accelerates as it approaches a nucleus, while a proton decelerates. (IOW by the time it gets close enough a proton is going much slower than it was initially, whereas an electron is going much faster than it was initially). X-rays are however also produced by knocking electrons out of their shells, but the highest energy x-ray produced by this path depends on the type of atom, and not on the energy of the impinging particle (assuming the latter is at least sufficiently energetic). IOW even if the impinging particle is very energetic, you still only get x-rays of a fixed maximum energy (but more of them if the particle is more energetic). For Nickel, the most energetic x-ray you can get via this path is 7.4 keV, which is a rather "soft" x-ray. (Dental x-rays are on the order of 60 keV). 7 keV x-rays are easily shielded by a few cm (mm?) of lead. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
