Ed wrote: Subject: Re: [Vo]:Banking on BLP?
Robin, my main point is that an electron leaving an atom cannot go to infinity under the conditions Mills has in his reactor. At most, it will go into some other energy level, such as the conduction band if one exists in the material. This fact is not based on speculation, assumptions, or theory. This is a simple fact of nature that is well understood.
MC: Which values and which electrons, Ed? In eq. 23, two electrons a 'liberated' to facilitate catalyzing H[1/3]. The physical situation in the cell is NaH resident within the R-Ni mesh, which has an enormous surface area. On the scale of a molecule, why can't the electrons wander away? There are He atoms at 760 Torr hanging around too. The electron bound to the catalyzed H doesn't go anywhere, it just gets closer to its proton. Now I don't yet understand where the energy to ionize the Na comes from, but the DSC plot shows *something* happens. *That* requires eplanation.
The values Mills uses to evaluate the process are all based on the electron going to infinity. Therefore, these values simply cannot apply to the real process. Instead, Mills assumes an unrealistic process to make his numbers fit his expectation.
MC: Are you also including the ionic catalysts in the gas phase cells?
If we accept the excess power he claims, the process must be different from the one he proposes.
MC: Why so? These solid fuel cells are a continuum with years of work in the electrolytic and gas phases. There are dozens of reports and papers supporting lthe reactions. Good calorimetry has been done iwth microwace excitation by Jonatan Phillips at the University of New Mexico. He was in town during ICCF-14 and slipped in to put up a poster on his calorimetric studies. In an early version of his reports there is a statement that the heat measured implied substantial conversion to H[1/4]. Philipps is currentlyas Distinguished Professor at the Farris center, supported by Los Alamos. He has a long association with Mills. I very strongly suggest that you contact him; he may be very helpful.
This is important to me, because I'm trying
to identify the Mills catalyst that is making hydrinos in the CF process, which has similar restrictions.
MC: H and D atoms can autocatalyze in a three-body reaction because 2H+ provide the 27.2 energy for catalysis. Because it is a three-body reaction, the reaction density is low but favored by H and D rich environments such as the LENR environments. A reactions density too low for optical observation may yet be very intense on the particle-counting scene.
An assumption on his part
that is unrealistic and impossible does me no good in trying to use his method in this search. Therefore, I'm trying to understand what is actually happening in his cell because the hydrino process appears to be real under these conditions. Only his explanation makes no sense.
MC: Granted, there are problems, as with LENR phenomena which don't make sense either. Nature is trying to tell us something.
Mike Carrell
Regards, Ed On Oct 24, 2008, at 9:47 PM, Robin van Spaandonk wrote:In reply to Edmund Storms's message of Fri, 24 Oct 2008 16:05:50 -0600: Hi, [snip]I think you are close to describing the process, Robin. Simply decomposing NaH cannot result in hydrinos because the expected ion is not formed.Absence of evidence is not evidence of absence, unless someone explicitly lookedfor it under the right conditions, and didn't find it.On the other hand, as you suggest, if the decomposition occurs on the Ni surface, the Na will have a complex ion state because it now is an absorbed atom, not a free, isolated atom. In addition, the electron that is promoted to a higher level has a place to go, i.e. into the conduction band of the Ni. The only problem is achieving a match between the energy change of the promoted electron and the energy shrinkage of the hydrino electron.I suspect you are needlessly multiplying entities. ;)IOW Mills provides a catalyst that has the necessary property, and gets theexpected result. Why is it so hard to accept that he might be right?Granted spectroscopic results indicating presence of Na++ would go a long way toproving him right.Now for a question. Why must the electron that is promoted always come from a level that is observed to form an ion during normal ionization?Personally, I don't think it does, and have previously suggested that Li, which has an x-ray absorption energy of 54.75 eV, may be an example of this. HoweverNa doesn't appear to fit the bill.For example, removal of a 2p electron from Na++ would occur during "normal" ionization, but is this happening here?No, but then Na++ is not the catalyst either. The whole molecule is thecatalyst. BTW the third ionization energy of Na is 71.641 eV, and none of the immediate reactions have enough energy to do this. Only a further reaction ofH[1/3] to a lower level would provide such energy. (3->4 yields 95 eV).In other words, why can't a 1s electron be removed from a neutral Na without the 2p electron being affected. After the 1s electron is removed, a 2p electron would take its place and release a small amount of energy as X-rays. This energy would be a byproduct of the process just like the hydrino energy. Do you know how much energy is required to remove a 1s electron from nearly neutral Na?1073 eV. (K shell x-ray absorption energy).The process gets more unknown because the electron would be promoted into the conduction band, which has a lower energy than vacuum. In other words, perhaps Mills has the right process but is using the wrong electron promotion process to describe it simply because the wrong promotion gives the expected energy.If so, then I think you need to come up with an alternative (and the numbers to back it up). The work function of the metal might be a good place to start, however in this case we're looking at an alloy/compound, which complicatesmatters. [snip] Regards, Robin van Spaandonk <[EMAIL PROTECTED]>________________________________________________________________________This Email has been scanned for all viruses by Medford Leas I.T. Department.
