Hi Bob, Good comments. Replies inline.
Just to mention it again, the model is no more than a back-of-the-envelope estimate. I'm guessing a rigorous treatment would do a lot of things differently. Eric On Sun, Sep 7, 2014 at 2:23 AM, Bob Cook <frobertc...@hotmail.com> wrote: The first reaction that produces Ni-59 will end up as Co-59 with no gammas > since the Ni-59 decay involves an electron capture and a hot beta +, which > will give thermal energy to the matrix ( about .52Mev) with a subsequent > beta+, beta- decay with its back-to-back .51 Mev gammas. The total energy > from the Ni-59 decay--half live 7.6x10^4 years-- is 1.073 Mev. > Although there are annihilation photons for the subsequent decay to 59Co, the half-life is on the order of tens of thousands of years. For that reason I guessed that the number of photons per second coming from such decays would be (relatively) small and went with the next highest-energy photon source I could identify, 15 keV electrons, for which I assumed as a safe upper bound that they could give rise to photons of equivalent energy through scattering. I think the number of 59Co decays per second will look like this: λ = ln(2) / 7.6e4 y = ln(2) / 2.3e12 s ΔN (decays in a second) = -λ N Δt = (-ln(2) / 2.3e12 s) * (1.37E+14) * 1 s = -41 decays I've plugged in the value for N=1.37E+14 from the model. I assume the negative result means there is a loss of 41 parent nuclei (59Ni) in the process. 41 annihilation photons per second is not trivial, but I'm guessing it's not that big a deal either. Ni-59 has a -3/2 spin and and Co-59 has a -7/2 spin. It seems that spin is > changed since the beta+ particle would only carry +or- 1/2 spin. I do not > understand how spin angular momentum is conserved in the Ni-59 decay > reaction, unless there are several neutrinos involved which could carry > away spin angular momentum. > I don't know enough about nuclear spin and spin selection rules at this point to comment on this detail. The beta+ decay was taken from [1]. You have not considered neutron capture reactions with the various > Ni isotopes. If the H reacts in the magnetic field in the Rossi device > with an electron to form a neutron as an intermediate virtual particle, > then Ni-58 would go to Ni-59 and hence to Co-59 as described above. > For this one time, I was hoping to go with something that stayed pretty close to normal physics. Note that a proton and a closely bound electron will not necessarily behave like a neutron at the time of a capture. I would expect the electron to fly off, carrying the energy of the gamma, along the lines that Robin has proposed elsewhere. (This contradicts what I wrote above about hydrinos leading to gammas.) Proton absorption reaction with Ni-60 would give Cu-61 with a 3.41 H half > life. ... Proton absorption reaction with Ni-58 gives Cu-59 which > decays with a half live if 82 s and produces a beta+ at 3.75 Mev and hot > gammas at 1.3 Mev. ... A proton reaction with Ni-62 would give Cu-63, > which is stable. > These reactions are all accounted for. Rossi would not want Ni-58, but Ni-62 and Ni-62 would seem to be ok. Ni-61 > would be undesirable also since it gives Cu-62 with the addition of a > proton, and Cu-62 decays with a hot gamma of 1.17 Mev. > My take is a bit different -- under the assumptions of the model, I think he would *not* want 62Ni. See the columns to the far right. A significant amount of shielding will be needed to prevent the 87 keV beta- deexcitation gammas from escaping (e.g., 2cm of lead). If one figures out a way to *remove* the 62Ni, however, the radiation gets better. [1] http://en.wikipedia.org/wiki/Isotopes_of_copper
