Where are the tritium and the neutrons? This reaction has two branches that occur with nearly equal probability:D + D→ T+ 1HD + D→ 3He+ n Then 2 1D + 3 1T → 4 2He + 1 0n <http://en.wikipedia.org/wiki/Neutron> Cheers: Axil
On Thu, Sep 20, 2012 at 12:50 AM, David Roberson <[email protected]> wrote: > Hi Eric, > > I made a quick calculation and believe that your balance adds up. The > net process is equal to the production of a He4 atom from two deuterons. > Each deuteron releases 2.224 MeV to build from parts while a He4 atom > releases 28.2933 MeV if constructed from basic parts. So if you start > with the two deuterons and end with He4 you get 28.2933 - 2 * 2.224 = > 23.8453 MeV. The total released by your two reactions is 23.8 MeV. The > numbers appear to correlate. > > Dave > > > > -----Original Message----- > From: Eric Walker <[email protected]> > To: vortex-l <[email protected]> > Sent: Wed, Sep 19, 2012 11:54 pm > Subject: [Vo]:graphs of reaction and decay networks > > I've put together some graphs of exothermic proton and deuteron capture > reaction networks for several materials together with associated decay > chains. See: > > https://docs.google.com/open?id=0BzKtdce19-wySThpSnd3bUxJdDQ > > A PDF can be obtained by choosing "Download" from the "File" menu at the > left. If you see any details in error, let me know. There is a > fascinating pair of reactions that I haven't completely made sense of yet: > > 15N + D -> 13C + A + 7.7 MeV > 13C + D -> 15N + γ + 16.1 MeV > > At face value, it is a catalytic reaction that takes deuterium and > yields helium-4. But I'm not able to make sense of the energy balance. it > appears to be saying that if you take 15N and add deuterium, you'll get 13C > and some energy, and if you take 13C and add deuterium, you'll get 15N and > some energy. I've probably messed up the calculation; if not, there's some > magic going on there. > > Eric > >
