On Fri, Sep 12, 2014 at 7:09 AM, Bob Cook <[email protected]> wrote:
How does a 6.-- Mev proton give up its energy without some gammas x-rays > showing up? > When a proton ~ 10 MeV travels through a metal, it will interact with electrons via the Coulomb interaction and, possibly, with lattice sites through elastic and inelastic collisions. At the scale we're talking about, the lattice sites themselves take up a very small amount of space, and a proton can travel quite far without encountering a nucleus. If it has an inelastic collision with a nucleus, there is the possibility of a gamma transition as the excited lattice site transitions back to the ground state. Most of the stopping of the proton will be through interactions with electrons. Each interaction will draw down only a small portion of the kinetic energy of the proton, e.g., 5-20 keV, which is a small fraction of the energy of the proton. I understand this range to be an approximate ceiling on the energy that is imparted to electrons in such a context. The electrons will create a continuum spectrum of bremsstrahlung. In addition, inner shell electrons will be excited and then relax, resulting in narrow peaks of photons with the energy of the transition (e.g., 8 keV for K-shell electrons in nickel). Except for deexcitation gammas arising from inelastic collisions with lattice sites, the fast proton will give rise to photons on the order of less than ~ 20 keV. X-rays of this energy are stopped by metal casing, and x-rays in the range of ~ 10 keV are stopped by 1cm of Pyrex. So it seems to me that you could have lots of fast protons inside a device without seeing any radiation outside of it. (Someone please correct me if I've misstated anything.) Eric
