Le Aug 10, 2012 à 5:24 PM, mix...@bigpond.com a écrit : > The reaction 3*He4 => C12 yields 7.27 MeV. (Helium burning in stars). > > Removing a proton from C12 to produce B11 *costs* 15.96 MeV. > > Hence adding a proton to B11 with the ensuing production of 3 alpha particles > produces 8.68 MeV. > > When the proton fuses with B11 it momentarily creates C12*, which usually > splits > into three alphas, but not always. > > (C12* is C12 in an excited state - it has an additional 15.96 MeV that it > desperately wants to get rid of).
Setting aside the question of 12C -> 11B + p (I'm not sure why that would be a desirable reaction), there is this interesting paragraph in the Wikipedia article: "Ordinarily, the probability of the triple alpha process would be extremely small. However, the beryllium-8 ground state has almost exactly the energy of two alpha particles. In the second step, 8Be + 4He has almost exactly the energy of an excited state of 12C. These resonances greatly increase the probability that an incoming alpha particle will combine with beryllium-8 to form carbon." Does anyone know what the modified probability would be? I believe Rydberg matter is matter that is in an excited state. Would that be a factor? While 2*4He -> 8Be is endothermic, it requires keV (93.7) and not MeV. Eric
