In reply to Eric Walker's message of Sun, 25 May 2014 14:43:03 -0700: Hi, [snip] >The application is encumbered by Piantelli's theory of H- ion orbital >capture by the substrate atoms. Here is where the application resolves the >question raised above: there are two interactions that Piantelli believes >to be taking place in the active core. First, there is the expulsion of a >proton from the substrate atom after the H- ion is captured. Inexplicably, >the proton emitted in this event is believed to have 6.7 MeV of energy. > This is hard to understand, because it's not clear why 6.7 MeV wouldn't be >needed to draw the proton in that close in the first place and where that >amount of extra energy is coming from. Second, there is an occasional >proton capture by the nucleus following upon the proposed orbital capture >of the H- ion and the release of the usual amount of energy for a nickel >proton-capture reaction, along the lines of Rossi's early explanation.
This is far better explained by a neutral f/H molecule, where one of the protons fuses with the Nickel nucleus, and the other is expelled carrying the 6.7 MeV of the fusion reaction. > >Piantelli's H- orbital capture theory aside, what is clear is that he >believes that a significant amount of fast protons are being emitted by his >active core, and the point of the invention is to make use of their energy >by having them react with the secondary material (lithium, boron, thorium, >etc.). This is a waste of time. The fast protons lose most of their energy ionizing surrounding atoms. Only one in thousands will undergo a further nuclear reaction. Thus the original reaction must be seen as the primary energy generating mechanism (assuming that there is anything to this at all). >A number of reactions with the protons and nickel and with the >protons and the secondary material are enumerated. These are the >proton-capture reactions mentioned in connection with the nickel atoms in >the primary material: > > - 1H + 58Ni ? 59Cu + 3.417 MeV > - 1H + 60Ni ? 61Cu + 4.796 MeV > - 1H + 61Ni ? 62Cu + 5.866 MeV > - 1H + 62Ni ? 63Cu + 6.122 MeV > - 1H + 64Ni ? 65Cu + 7.453 MeV One then has to wonder what became of the radioactive nuclei i.e. 59Cu - 62Cu? Particularly Cu61, which has a half life of over 3 hours, some traces of which should thus remain, even if the reactor were left to "cool off" for some time. (Unless of course, the shrunken electrons are able to vastly enhance the electron capture reaction rate resulting in vastly shorter half life.) Note also that rapid conversion of unstable Cu isotopes into Ni would result in only stable Cu isotopes remaining as actual Cu. However even this scenario should result in a considerable dose of gamma-rays being produced while the unstable Cu is decaying (unless the other shrunken electron removes the decay energy as kinetic energy???) The availability of two shrunken electrons could be precisely why the decay is so vastly enhanced. Each electron enhances the decay rate, leading to a double enhancement as it were. The first electron is required for the electron capture reaction and enhances the rate by being "on hand", and the second provides a rapid energy removal mechanism, thus enhancing the rate because no "slow" gamma-emission need be relied upon to remove the energy. Furthermore, the second electron can also carry away angular momentum if need be, thus allowing an easy transition directly to the ground state of the Nickel isotope that is being formed. This would mean that both fast protons and fast electrons would be produced, but almost no gammas compared to what one might normally expect (the electrons will produce some bremsstrahlung but this will only be about 1% of the amount of gammas that would have been produced, and furthermore the bremsstrahlung energy spectrum is more spread out than a gamma spectrum would have been, so some it won't make it through the "shielding". [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
