There is also the possibility of one or more of the S orbital electrons of the larger parent atom being taken into a sub-ground hydrino state. In which case, each of the electrons in such a state would screen a proton and make those protons appear like neutrons. For example, say one of the S orbital electrons of 55Co went into a sub-ground state orbital screening one if the proton charges. The atom would appear chemically to have one less proton and one more neutron - becoming 55Fe. From a nuclear stability standpoint, though it would still appear as 55Co presumably (but this is also unstable in this case).
A pico-hydride implies that the hydrino hydrogen would be able to form a shared chemical (electron) bond with the low abundance stable 54Fe. I just can't imagine a hydrino being able to share an electronic state with another atom because the hydrino's electron is so tightly bound to the hydrino nucleus - not an ordinary valence bond for sure. In a high resolution mass spectrometer, the 54Fe+picohydride would weigh more than a 55Fe and that should be observable. They have such a spectrometer at Purdue. On Sat, Mar 25, 2017 at 9:46 AM, Jones Beene <jone...@pacbell.net> wrote: > Bob Higgins wrote: > > The predicted properties of the hydrino or any sub-ground-state hydrogen > suggest that it will be really hard to detect... It must be detected by > proxy. Like detecting the neutrino, detection of the hydrino will require > new, inventive techniques > >> >> Bob, I generally agree that new thinking is needed. This is why I brought > up Dufour's ICCF20 talk and the iron-55 evidence, the so-called > pico-hydride. It is a very elegant and simple way to confirm dense hydrogen. > > The dense hydrogen becomes attached (magnetically?) to iron 54 in such a > way that on mass-spec analysis, it looks like 55Fe - but is NOT > radioactive. Normal 55Fe is strongly radioactive. > > This looks like a brilliant solution to detection ! and could be the > smoking gun for dense hydrogen , but it does not conform to Mills theory so > he will never agree. >
