In reply to Jones Beene's message of Sun, 28 Apr 2013 11:56:10 -0700: Hi Jones, [snip] >When this VN nears the larger nucleus however, the bond to the electron is >broken, but since the electron has effectively shielded the charge of the >proton, for long enough for the strong force of the Ni to see it, we have a >different kind of reaction than if it were a real neutron. Thus the Ni-62 >takes the proton only, and the electron is expelled as if it was an >instantaneous beta decay. > >There is no half life with a VN reaction. It is instantaneous and looks like >a proton absorption.
H + 62Ni => 63Cu + 6.122 MeV When you looked at the beta decay of 63Ni, you forgot the energy from the addition of the proton. This energy has to be carried by something. That might be a (relatively slow) gamma emission, but more likely IMO is that the electron gets it. In that case there should be bremsstrahlung from the fast electron. From what I can glean from the net, only about 1% of fast electrons produce bremsstrahlung, and it is broad spectrum, so only a small fraction of that which is produced is highly energetic. If we further assume that most of the energy is actually coming from the shrinkage, and only a fraction from actual fusion reactions, then the lack of detectable radiation outside the shielding might be explicable. BTW much of the math in http://arxiv.org/abs/1303.1078v1 is beyond me, but one thing stood out. I got the impression that the author calculated an increase in the rate of fusion when it is electron assisted by over *400 orders of magnitude*. (See the very last sentence in II. Preliminary Considerations.) (A more precise explanation of exactly what is being compared here would be appreciated.) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
