Hi Bob, The calculation I was thinking of was this:
p+7Li → 2*4He + Q (16.8 MeV) 16.8 MeV / ( 2 * 4He * ( 4 nucleons / 4He ) ) = 16.8 MeV / 8 nucleons = 2.1 MeV / nucleon The per-nucleon calculation just gives one a sense that the 4He are traveling pretty fast. I assume there would be secondary radiation as a result, but I only have a vague sense of how penetrating it would be. Eric On Sun, Sep 6, 2015 at 7:14 PM, Bob Cook <frobertc...@hotmail.com> wrote: > Eri-- > > I did not follow your conclusion that there would be 2.1 Mev per nucleon > (which is pretty fast) resulting from a 16.8 Mev decay of the short-lived > Be-8 nucleus. Why do you say it is fast per nucleon? I think the binding > energy given up as kinetic energy of the alphas by the Be-8 nucleus is not > unexpected. > > Bob Cook >
