In reply to Jones Beene's message of Sat, 28 Jul 2012 20:29:06 -0700: Hi Jones, [snip] >Robin, > >You completely understood the post. > >This was NOT about how Mills thinks it works. We have know for 20 years how >he thinks it works. > >Instead, it is about how potassium works in an alternative but non-Millsean, >non-nuclear view; or better yet - how it works in fact. > >Jones
I realize that you wanted to present an alternate view, but you shouldn't do that by first misrepresenting what Mills says. IOW you shouldn't first incorrectly claim that his process works in a certain way, then proceed to claim that "his" explanation is of little value because 81 eV energies are not available at room temperature. It's a sort of straw man argument. In fact all you have done is demonstrate that either you don't really understand his process, or the misrepresentation is deliberate, which I doubt. In which case, may I suggest that you take another look at what I wrote. It may go some way toward changing your mind on Mills, since his claims are not nearly as outlandish as you appear to believe. > > > >-----Original Message----- >From: mix...@bigpond.com > > >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 > > Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
