In reply to David Roberson's message of Sun, 1 Dec 2013 03:17:14 -0500 (EST): Hi, [snip] > >Have you determined whether or not this is a reversible process? How often >does the lower energy hydrino accept energy from a catalyst that has not yet >released the same amount of energy in the form of radiation? It is common for >energy to be traded in both directions according to thermodynamic laws. > >Dave Entropy works in favour of Hydrino production. Once the energy has been released, it is difficult to get it back again. Furthermore, it is a two stage process. In the first stage a multiple of 27.2 eV is handed to the catalyst. In the second stage more energy is released as the newly formed Hydrino stabilizes. Consequently, an "excited" catalyst doesn't have enough energy to "re-inflate" a stable Hydrino.
The simplest case will serve as an example:- Stage I) H + Ar+ => Hy* + Ar++ + e- Stage II) Hy* => Hy + 13.6 eV (UV or kinetic) (I have used Hy* to denote an intermediate state of the Hydrino). Note that in (I) 27.2 eV is transferred to the Ar+, which further ionizes it, but an additional 13.6 eV is lost in (II), so the total energy lost is 40.8 eV. Hence Ar++ recombining with a free electron to produce Ar+ would only generate 27.2 eV, which is 13.6 eV short of the amount required to "re-inflate" the Hydrino back to H. Note the actual ionization energy of Ar+ is 27.6 eV, but I have deliberately left the consequences of that out of the explanation as I am trying to keep it simple. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
