He does not use protium. He uses ordinary hydrogen. In the cell some of it is broken down into atomic hydrogen. That is what interacts with the nickel.
http://www.journal-of-nuclear-physics.com/?p=62&cpage=2#comment-273 I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. ________________________________ From: Horace Heffner <hheff...@mtaonline.net> To: vortex-l@eskimo.com Sent: Sat, January 22, 2011 1:32:11 AM Subject: Re: [Vo]:Removing All Doubt On Jan 21, 2011, at 7:06 PM, Man on Bridges wrote: > Horace, > > Based upon natural presence and absence of radiation I would probably go for >this on: > 62Ni28 + p* --> 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) > > 62Ni28 : pres. 3.634 % > 1H1 : pres. 99.985 % > 63Cu29 : pres. 69.17 % > > Kind regards, > > MoB If I recall correctly, Rossi stated that deuterium kills the reaction, and that he uses pure protium. Looking again at the mean atomic weight 63.6 and mean reaction energy 7.2, I calculated as shown below, you can see that these are weighted averages, and are roughly 4 to 1 in favor of 62Ni vs 64Ni, corresponding roughly to their abundances, 3.634% vs 0.926%. They should be roughly equal in lattice half life (as I defined it in my paper), possibly with a small edge for the larger nucleus. Everything posted by me on this is seat of the pants accuracy. I think for casual discussion like this slide rule accuracy is sufficient to demonstrate the principles. I guess my age is showing! 8^) On Jan 21, 2011, at 4:23 PM, Horace Heffner wrote: > 62Ni28 + p* --> 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) > 64Ni28 + p* --> 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) > 64Ni28 + 3 p* --> 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) > > Looking at the first two reactions as likely candidates, with mean atomic >weight near 63.6, and mean reaction energy about 7.2, we have an estimated >energy density of > > E = (1/(63.6 gm/mol))*Na*7.2 MeV = 1.09x10^10 J/gm Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
