I'll remind, just in case it isn't clear for everybody, that for every two Ds which will have disappeared and every He which will have appeared, 24 MeV of energy will have been released in any case, _whatever the intermediary or concurrent reactions if any_.
The energy released by a nuclear reaction is path-independent and depends only on the reactants and products, just like in a chemical reaction. Michel 2010/3/25 OrionWorks - Steven Vincent Johnson <[email protected]>: > From Abd: > > ... > >> With all those caveats, and wondering why you'd ask *me*, since I'd >> really ask someone else, like Dr. Storms, if I cared all that much >> about it, ... > > My previous comments were not exclusively addressed to you alone. I opened > my query up to comments coming from anyone who wishes to add their two > cents. > >> ... my *impression* is that the energy not from deuterium to >> helium is not more than maybe 20%, and could be much less. And may >> vary quite a bit with exact experimental conditions. > > Thanks for your impression. Again, this is just speculation that I am asking > for. At the stage of the game who really knows what the actual ratios might > be. > > Regards, > Steven Vincent Johnson > www.OrionWorks.com > www.zazzle.com/orionworks > > >
