In reply to Jones Beene's message of Wed, 5 Jan 2011 06:42:19 -0800: Hi, [snip] >Well, Robin - in that case there is almost no difference between your view >and that of the Rossi collective, except at the outset you are borrowing >Mills version of the hydrino, with some changes - whereas they have >'invented' the 'hypole' probably to avoid any taint of his intellectual >property.
I agree. > >In fact, Rossi could (should) have borrowed Dufour's "hydrex" virtual >neutron model which goes back twenty years. This might have necessitated >slight alteration. See also http://www.journal-of-nuclear-physics.com/?p=275 > >Again, my overriding comment is that Randell Mills has effectively shot >himself in the foot (or head) with the continuing assertion that Ni-H is not >a nuclear reaction. It may cost him billions, in the end - and is based on >arrogance in support of an incomplete theory, which would have otherwise >been seen to be derivative of the dreaded "cold fusion" effect of P&F. > >Yes, Ni-H may be a "delayed" nuclear reaction, and it may be new physics (in >the lower gamma emission level), and it may require Casimir cavities as a >first step... and it may involve EUV emission but in the end it is nuclear, >as the Ni > Cu transmutation shows. There is still a problem with the Cu formed. One would expect at least some of it to be radioactive, and the gammas from these isotopes should still be easily detected through 2 cm of lead. The primary isotope of Ni is Ni-58 which by addition of a proton should produce Cu-59 which has a half life of 81 seconds, decaying to Ni-59 via beta+ decay. These positrons should produce annihilation gammas when interacting with electrons from the metal. 58% of the 511 keV gammas would get through the 2 cm lead shielding, and be very readily detected. However it's also possible that the Hydrino electron that got "sucked in" immediately reacted with the new copper nucleus to form Ni-59 directly through an enhanced electron capture reaction (enhanced because the electron is already "on hand", and doesn't need to be captured from the K shell). The problem with this is that the reaction to Ni-59 produces 8 MeV, and this energy can't be converted into electron energy as the electron has already been used so it would at first blush seem that only gamma radiation remains (a problem once again). As an alternative I would offer the possibility that not single Hydrinos, but rather whole molecules or even magnetically bound groups of molecules are reacting with the Ni. The energy of the reaction could then still be carried away by a combination of protons and shrunken electrons left over from the molecule (cluster). A cluster would also open the way to the following reaction: Ni-58 + 2H2 => Ni-60 + 2H + 2 neutrinos (or 2 e- + 2 p) + 18.8 MeV leaving no radioactive substance. A reaction producing Cu would be: Ni-60 + 2H2 => Cu-63 + H (or e- + p) + 2 neutrinos + 23 MeV. (The simplest cluster is 2 magnetically bound Hydrino molecules; 4 atoms, hence 2H2). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
