In reply to Jones Beene's message of Mon, 13 Apr 2015 16:50:39 -0700: Hi, [snip] >-----Original Message----- >From: [email protected] >In reply to Bob Higgins's message: > >>How it is implicated at a Pd cathode is a mystery > >.... It should occasionally be reduced to the atomic Li at the cathode, but >is unlikely to last long enough to plate out. It is a highly reactive metal, >and will readily react with a water molecule. However, should a Lithium atom >both be reduced as a neighbour of a Hydrogen atom at the cathode, then there >is a possibility for the Li to catalyze Hydrino shrinkage in the H (D). > >Robin, > >That is close to what I was going to add - since the first ionization of Li >is 5.39 eV and the second is 75.64 eV. > >Together these add up to 81.02 eV. In terms of Rydberg energy - 27.2 eV x 3 >= 81.6 eV. Which is a very close fit for lithium, but of course, that last >electron does not come off easily and that is where nickel may come into >play (as opposed to palladium). Palladium has only a low energy fit but >nickel has a unique ionization situation vis-à-vis lithium. > >I'm not sure how this happens in detail but nickel has an IP5 which is >almost identical to the one of lithium - IP2. Perhaps the lithium and the >nickel interact first, leaving lithium with the deep "hole" that otherwise >would not be there at low temperature;
Your understanding of the term "hole" is not correct. A hole is not a place where electrons are missing. It is a place into which energy can be dumped. And energy can be dumped when electrons are *present*, because dumping the energy into the neutral atom removes the electrons from the atom, i.e. causes the atom to become ionized. Hence the Li atom (not ion) acts as a catalyst because it needs 81.02 eV to become ionized to Li++. And as you point out here above, 81.02 is a "close" match for 81.6 eV, which the H just happens to want to get rid of. I wonder if this "energy resonance" is analogous to what happens in a Tesla coil between primary and secondary? In short nothing extra (i.e. Ni etc.) is needed for this shrinkage reaction to happen. Just H atoms and Li atoms in close proximity. However the Ni likely plays a role in two ways. First, as you have also frequently pointed out, it is a spillover catalyst, hence increasing the number of H atoms likely to be available. Secondly, it may well play a role in any ensuing nuclear reactions, due to the large change in binding energy when a nucleon is added to the nucleus, since it is close to the top of the binding energy curve (as you have also pointed out many times ;). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
