In reply to Eric Walker's message of Mon, 12 Nov 2012 21:42:01 -0800: Hi, [snip] >On Sun, Nov 11, 2012 at 12:39 PM, <[email protected]> wrote: > >All of these are explained if the active particle is a f/H molecule. >> >> 1. The molecule is neutral, thus is not bothered by the Ni electrons. >> 2. There are no gamma rays because only one of the two protons fuses, the >> other >> being ejected carrying the energy of the reaction. Fusion primarily with >> 62Ni & >> 64Ni yields stable copper isotopes. >> 3. Heat is deposited to the substrate by fast protons. >> 4. The fact that the molecule is neutral gets it close enough to the >> nucleus to >> make tunneling possible. >> > >Nice trick. Now I have a better sense of some of the strengths of the f/H >approach. > >Is there any reason these things could not happen with Rydberg H2 >(in contrast to inverse-Rydberg H2), deformed under an electromagnetic >field, where the nuclei are far to one end of the electron shells?
Is there such a thing as deformed Rydberg H2 (as opposed to H)? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
