In reply to Jones Beene's message of Thu, 23 Oct 2008 15:49:22 -0700 (PDT): Hi Jones, >Hi Robin, > > >> According to Randy, the NaH decomposes directly in Na+++ + H[1/3] + 3e- . > >LOL !! > >> In going from H[1] to H[1/3] the H requires an energy hole of 54.4. eV. This >> is >the sum of the first and second ionization energies of Na (5.1391 eV & 47.286 >eV resp.) and the energy required to break NaH into atoms (about 1.98 eV). > >OK - Here is why that cannot happen. The energy required to break the two into >atoms could never result (very low statistical probability) in the H becoming >un-ionized while at the same time staying very close by (geometric proximity), >while at the exact instant 3 electrons are removed from the sodium. Bizarre. [snip]
I think you misunderstand. The energy required to break NaH into atoms is 1.98 eV. The energy required to then ionize the Na to Na+ is 5.1391 eV. The energy required to then ionize the Na+ to Na++ is 47.286 eV. ---------------------------------------------------------------- Total 54.405 eV which is an excellent match for an m=2 energy hole. That means that by shrinking from the ground state to n=1/3, the Hydrogen atom gives up first 54.4 eV (the energy hole value), resulting in the specified dissolution, then a further 54.4 eV as kinetic energy of the particles. The total energy released is 108.8 eV. Ionization of the H isn't even on the table, because either the H shrinks to a Hydrino, or nothing at all happens and the NaH simply remains NaH. Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
