The amount of nickel Ni62 in the fuel load doubled from some unknown combination of lighter elements. This fusion process should have released a huge amount of nuclear binding energy. It is possible that the only thing that lithium did was donate its neutron to the Nickel 58 to turn it into Nickel 62.
On Fri, Apr 10, 2015 at 3:48 PM, Jones Beene <jone...@pacbell.net> wrote: > -----Original Message----- > > > The argument can be made that there was NEVER enough lithium present in > the Lugano reactor to provide the reported net energy gain (1.5 MW-hrs) over > 32 hours- even if 100% of the lithium was consumed and converted into > helium… > > For the record - The total Lugano Fuel sample had a reported mass of 1 > gram. > > *Element % by Weight* > Nickel 55.0 > Iron 39.0 > Aluminum 4.3 > Lithium 1.1 > Hydrogen (no Deuterium) 0.6 > Total 100.0 > > Therefore, there was .011 grams of lithium at the start. The average mass > of the lithium = 6.93 amu or 7 grams per mole = .0016 moles. If all of > this lithium, 100%, had fused with protons, giving 17 MeV per fused lithium > atom, then it would have been marginally sufficient to provide the energy > reported (10^28 eV). That assumes that every atom has been consumed - and > assuming > that no energy was lost to x-ray radiation… BUT… > > …there was lots of lithium left over in the ash, so all of it could not > have reacted and possibly as much as 90% of the bremsstrahlung should > have been lost in an alumina reactor. > > As for the argument that 8 MeV alpha particles produce bremsstrahlung > which is mostly thermalized, consider the case of Uranium decay. > > U is an alpha emitter, where the alpha has an average kinetic energy of > only 5 MeV, yet this corresponds to a velocity which is 5% of the speed > of light producing substantial radiation, and despite the extremely high > ability of U to absorb such radiation – hundreds of times greater than > alumina, most of it escapes - which is why even small pitchblende samples > make the Geiger counter go wild. It is likely that only a few percent of 8 > MeV alpha bremsstrahlung will be completely thermalized by alumina > absorption, since alumina is fairly transparent to x-rays in this range. > IOW most of that putative 8 MeV should be lost as x-rays and not recorded > as heat. > > In short, I am having a hard time imagining how Cook and Rossi can believe > that lithium proton fusion is responsible for the energy gain – even if > there is a spin mechanism which bypasses the problem of x-rays from > bremsstrahlung. > >
