In reply to Jones Beene's message of Sun, 1 Dec 2013 11:36:33 -0800: Hi, [snip] >Since all helium is atomic (He2 as a molecule does not exist) the helium >ground state consists of only of two 1s electrons which are tightly bound >extraordinarily bound in fact suffice it to say that the energy required >to remove one of them is the highest ionization energy of any atom in the >periodic table. And the worst part is that essentially, you must have the >energy available to remove BOTH of them in order to get to the proper >Rydberg level (which is for the second electron by itself, and not the first >plus the second in combination) and at the same time you must have >non-ionized atomic hydrogen present! Bizarre to even suggest.
This is not quite correct. The catalyst is He+, not He++. The second ionization energy of Helium is indeed 54.4 eV, but that means that 54.4 eV is required to remove the second electron. And that is precisely the value that Hydrino formation can supply. IOW the actual catalyst is He+, which accepts 54.4 eV during Hydrino formation and then becomes He++ + e- as a consequence. Hence it is not necessary to remove both electrons in order to create the catalyst, only the first one. However you are correct in that it is difficult to remove the first electron. BTW it may be a bizarre suggestion, but there is an old paper of Mills where he reports on an experiment where He+ is used as the catalyst. It does appear to work. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
