In reply to Jones Beene's message of Sat, 28 Jul 2012 09:48:39 -0700: Hi Jones, [snip] >Anyway, K has three ionization potentials: 4.3eV, 31.6eV and 45.8 eV and the >combination of all of them is 81.8eV which is a Rydberg multiple that is >within range. That is, since 27.2*3 + 81.6eV... voila.. and Mills early-on >stated that this is the catalytic hole. On the surface that might seem to be >end of story (if you are naive) ... but there is the little problem that >triple ionization of potassium only happens at about a million degrees C. No >way is Boltzmann's tail that long in electrolysis, in particular; and no way >will hydrogen be monatomic at the same time that K is triple-ionized.
That's not the way it works. It's an energy HOLE. That means that the catalyst has to be capable of ABSORBING that amount of energy, not that the amount of energy needs to be available before hand. It's not K+++ that is the catalyst, it's K itself (the atom). (or sometimes 2*K+; 31.625 - 4.3407 = 27.291, in a process where an electron is transferred from one K+ to the other). For the ionization energies of K I have:- 4.3407 31.625 45.72 ------- 81.6857 eV ======= The actual energy hole is 6*13.598 eV = 81.588 eV The difference is 81.6857 - 81.588 = 0.0977 eV. This is the energy that needs to be supplied thermally and it is only about 4 times the average thermal energy at room temperature which I'm sure Dr. Boltzmann would provide, the more so as the temperature increases. And of course once the reaction kicks in, more energy becomes available. (Values of the ionization energy obtained from Webelements, yield a difference of 0.195 eV, which is about 10 times average at room temperature.) In short, during Hydrino formation, The newly forming Hydrino releases 81.588 eV which is used to ionize K to K+++ (with the help of the extra 0.0977 eV thermal energy). The K+++ then eventually reacquires the electrons from the environment, of it's own accord, releasing the 81.6857 eV as heat etc. Furthermore, the formation of H[1/4] (from H[1/1] due to m=3 for K) actually releases a total energy of 217.7 - 13.6 = 204.1 eV, of which 81.6 goes to ionizing the K, and the remainder (122.5) is released either as UV or as kinetic energy of a particle. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
