While Li and other alkali metals (K, Cs) have shown some value in LENR, they do not seem to be required for Ni-H LENR. Piantelli's baseline Ni-H LENR uses a pure Ni rod and isotopically natural H2 and achieves LENR. >From this reaction he discovered MeV protons being emitted and later added a Li shell spaced out around his Ni rod as an "afterburner" to produce additional heat from the MeV protons. It is reasonable to believe that his baseline experiments were not contaminated with alkali metals.
On Fri, Dec 4, 2015 at 1:50 PM, Jones Beene <jone...@pacbell.net> wrote: > *From:* Eric Walker > > Ø The end result is that the only workable approach is to > completely separate the two– deuterium-based from protium-based, as being > fundamentally different. And why not? > > One question I've been mulling over is whether the hydrogen and deuterium > simply catalyze the induced decay of something else, and whether deuterium > just does a better job of this. In this case they would not be consumed in > any reaction in any quantity. > > Eric, > > Imagine that the single common denominator of any gainful reaction involving > hydrogen and/or deuterium involves the prior formation of a denser > allotrope of either isotope. It does not matter if the formation stage is > exothermic or endothermic, since what happens latter is of greater > importance. > > The dense form of either hydrogen isotope would then catalyze the decay > of an atom like lithium or potassium via a glancing approach to the > nucleus, not close enough for fusion but disruptive. Many successful cold > fusion experiments have used lithium electrolyte. Lithium would undergo > accelerated decay to helium – thus fooling the experimenter into > believing the helium came from fusion, instead of accelerated decay… > despite the lack of gamma. > > This assumes that accelerated decay has a much higher cross-section than > fusion (a very defensible proposition). And to complete the scenario – > this reaction with lithium could be happening in addition to the > accelerated decay of deuterium. Thus in the end, we would have no fusion, > no gamma, but we would find helium in rough proportion to the excess heat > seen in the reaction, if the deuterium decay predominates. > >
