Robin, We cannot assume that this is directly comparable to a known hot fusion reaction, assuming it is real. Why should we? There is every reason to suspect that LENR is based on previously unknown pathways.
The best way to validate the claim is to test a sample of spent fuel for copper isotope ratio. We can probably expect the heavier 65Cu to be completely absent. That would constitute almost indisputable proof. Why wasn't this done? Jones >In one of the translations yesterday - it was said that there was lead shielding in place already - so the small signal seen. 50% over background would not be unusual, and is entirely consistent with such shielding were under the insulation. >Also I see a "Gamma Scout" device on the table. These are not sensitive >instruments for lower energy gammas. > >Jones The reaction Ni-58 + H -> Cu-59. The Cu-59 would accumulate until there was enough for the decay rate to equal the production rate, at which point the quantity would stabilize. If we assume that the reactor is in this stable state, and producing an output of 4 kW (2/3 of 6kW from this isotope), then there are 3E15 Cu59 atoms being created every second, and an equal number decaying to Ni-59. Almost all of the decays produce positrons which annihilate with electrons producing a pair of gamma rays. Hence about 6E15 gammas are being produced every second. Of these about 2.8E11 / sec. would make it through 2 cm of lead. That's the equivalent of an unshielded source of 7.7 Ci of gamma radiation. Over 1 hour that would lead to a received dose of about 308 mr at a distance of 5 yards (at 1 yard it would be 7700 mr). (average background radiation in the US is about 300 mr / yr, so while observing the experiment at a distance of 5 yards one would get about a years worth in 1 hour). (These are 511 keV gammas. Dental X-rays are about 65 keV or less). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
