Jones, Your explanation below seems really a stretch, but it certainly is a mystery.
The quadrapole mass spec RGA will have a front end ionizer to extract an ionized sample for measurement. I think this front end is likely to only extract positive ions and there will be no f/D+ because there is no such thing. If an f/D- exists, I don't think it would be measured by the RGA. So, M/e=2 would have to be D+ or H2+. Even in the lower pressure of his experiment, there is likely to be little free monatomic D+ except that created by the front end ionizer of the RGA. This would make a D+ M/e=2 reading inversely related to the M/e=4 reading for D2+. Something about the experiment could be causing the RGA front end to become more effective in producing D+ in ionization of the D2 gas species as the experiment goes on. Perhaps this could be cleaning the oxide off of the RGA front end, making it more effective at breaking the D2 apart. If this were to happen, the measured M/e=4 would decline and the M/e=2 would increase at twice the rate - this would have nothing to do with the relative abundance of the species in the bulk of the apparatus. This is a simpler explanation than the one you propose. Bob Higgins On Sat, Aug 2, 2014 at 12:32 PM, Jones Beene <[email protected]> wrote: > *From:* Bob Higgins > > > > I have a few observations that are not being discussed here (and I may be > missing something) from the slides from the MIT Colloquium. > > · *The report for the control experiment with no excess heat also > showed the decline of the M/e=4 species and rise of the M/e=2 &3 species*. > > > This is consistent with the basic fractional deuterium reaction, in a > modified Millsean understanding, where the first redundant state is energy > neutral. There is no excess heat because, and unlike Mills theory, the > first stage ionization and redundancy is itself endothermic, and borrows > from the output. After the first step there is net gain, but it takes time > to build up a population of fractional D (designated as f/D or f/D- if in > the hydride state, which is stable and shows up as M/e=2. This stable ion > is what Mills might call deuterino deuteride. It is the correlate of > hydrino hydride – a stable negative ion. > > Of course, it is more complicated than that, since there needs to be some > hydrogen retained in the nickel (to give M/e=3 by exchange reaction) but > with as many runs as Mizuno made in the testing – and with both H2 and D2, > I do not believe he can completely clean the system of one species or the > other for every run. > > Jones >
