In reply to Eric Walker's message of Fri, 10 Aug 2012 21:41:27 -0700: Hi, [snip] >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:
Neither am I, but Axil raised it. > >"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? In what? >While 2*4He -> 8Be is endothermic, it requires keV (93.7) and not MeV. True, but once formed 8Be essentially falls apart again very rapidly, so this reaction is only of relevance where the formation rate is huge, in which case a very few 8Be are available at any given instant. Furthermore, where does the 4He come from in LENR? (and why bother with this reaction if you are otherwise creating 4He anyway?) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
