In reply to Edmund Storms's message of Sat, 25 Oct 2008 15:51:51 -0600: Hi, [snip] >While that is true, the assumption is that the electron goes to >infinity. [snip] >Suppose an electron goes from a level that requires 20 eV if the >electron goes to infinity. Now suppose the electron actually went to a >conduction band at 3 eV relative to infinity. Would not 17 eV be >required for the process?
Of course, provided that source and destination are very close to one another. However if they are widely separated, then one first needs to invest 20 eV, then later one gets 3 eV back again (usually in photonic form). > >In contrast, you assume that the final energy of the electron does not >matter provided it moves far enough from the original atom before >finding another state. Precisely. The path is important to the mechanism. >If the Mills energy is based on this >assumption, then the environment in which the catalyst is located is >important. Agreed. >I agree, very little ambiguity is created when the >material is in a gas, as is most of the Mills work. However, we are >now talking about a solid mixture. But also about the space surrounding it, and even the space between solid particles. Note that the reaction takes place at "high" temperature, so the NaH once formed is likely to be gaseous. Even with a gaseous NaH however one can still have a surface phenomenon, when a gas molecule approaches the surface. Where I am heading with this is that an H atom formed on the surface may become momentarily freed from that surface, and could react with an NaH molecule floating nearby. >I suggest this situation creates >great ambiguity and must be acknowledged. I agree that there are still lots of unanswered questions. > >In addition, I can imagine a range of energy being available in such a >transition if I can arbitrarily choose a distance the electron has to >move from its stable state before the energy being used is identified. Perhaps because not all radii are equal? IOW the electron can only occupy stable orbitals (or be ionized). >If this is the nature of the process, what is the point of choosing >the ionization energy as a criteria for the hydrino process working? See above. Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
