If the reaction energy of 6 MeV is mostly transferred to the lattice (soliton) via EMF strong coupling, the second proton of the He2 pair can drift out of the reaction zone with a energy of just a few KeV.
With strong EMF coupling, an expelled particle need not be the primary carrier of the binding energy excess. On Fri, Sep 12, 2014 at 10:09 AM, Bob Cook <[email protected]> wrote: > Robin-- > > How does a 6.-- Mev proton give up its energy without some gammas x-rays > showing up? > > Bob > ----- Original Message ----- From: <[email protected]> > To: <[email protected]> > Sent: Thursday, September 11, 2014 8:31 PM > Subject: Re: [Vo]:Rossi on Ni62 > > > In reply to Alain Sepeda's message of Wed, 10 Sep 2014 10:28:16 +0200: > Hi, > [snip] > >> it remind me the observation of Iwamura as noticed in the book of Ed >> Storms, that transmutation seems to be the fusion with an even number of >> deuteron (2-4-6), with preference to stable isotopes. >> > > Note also that Hydrino molecules may have a better chance of approaching > the > nucleus of another atom than lone Hydrinos. The former are essentially > chemically inert, are very "heavy", and can be very small. Electrically, > they > look like doubly massive neutrons. The latter can be chemically extremely > reactive as they can bind with an electron to form a negatively charged > Hydrinohydride ion, far more aggressively than even Fluorine gas. > Consequently Hydrino molecules might trigger reactions that would not > otherwise > be seen, such as:- > > 62Ni + Hy2 => 63Cu + p (6.2 MeV) > > (Where Hy2 is a severely shrunken Hydrino molecule.) > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > > >
