In reply to Axil Axil's message of Mon, 22 Apr 2013 18:58:39 -0400: Hi, [snip] >What happened to the 6MV of binding energy, where did those numbers come >from?
Note that the reaction given in my previous post (quoted here below) do not involve addition of a proton to the Nickel nucleus, hence no 6 MeV of binding energy is available. If a proton is added, then different reactions follow, and the 6 MeV is available. e.g. 64Ni + H => 65Cu + 7.453 MeV or 62Ni + H => 63Cu + 6.122 MeV In neither case is any weak reaction involved. Just the strong nuclear force. Perhaps also of interest is that going from Ni to Cu actually entails a slight increase in mass for all the existing nucleons in the Ni (which costs energy), however this is more than compensated for by the loss in mass of the proton as it becomes bound (releasing energy). > > >On Mon, Apr 22, 2013 at 6:48 PM, <[email protected]> wrote: > >> In reply to Axil Axil's message of Mon, 22 Apr 2013 18:30:28 -0400: >> Hi, >> [snip] >> >However, the Ni64 isotope will produce an isotope of Zinc, 39% of the >> time. >> > >> >should read >> > >> >However, the Cu64 isotope will produce an isotope of Zinc, 39% of the >> time. >> > >> >The speculative ground rule for LENR is that electric charge concentration >> > will destabilize a stable element by converting a neutron to a proton in >> >the Ni nucleus: Ni62 goes to Cu62. >> >> 64Ni => 64Cu - 1.675 MeV (note the minus sign). >> 62Ni => 62Cu - 3.848 MeV (ditto). >> >> Regards, >> >> Robin van Spaandonk >> >> http://rvanspaa.freehostia.com/project.html >> >> Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
