On Wed, Oct 8, 2014 at 7:21 PM, <[email protected]> wrote: Li7 + Ni58 => Ni59 + Li6 + 1.75 MeV > Li7 + Ni59 => Ni60 + Li6 + 4.14 MeV > Li7 + Ni60 => Ni61 + Li6 + 0.57 MeV > Li7 + Ni61 => Ni62 + Li6 + 3.34 MeV > Li7 + Ni62 => Ni63 + Li6 - 0.41 MeV (Endothermic!) > > This series stops at Ni62, hence all isotopes of Ni less than 62 are > depleted > and Ni62 is strongly enriched. >
This is very nice. I've been too attached to deuterium. In this particular instance, deuterium reactions above 62Ni would be exothermic: - 62Ni + n → 63Ni + p + Q (5.1 MeV) - 63Ni + n → 64Ni + p + Q (7.9 MeV) Since neither 62Ni nor 63Ni were seen in significant quantities in the ash, I think we can rule deuterium out for this particular test. Note that while 64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is exothermic. Since 63Ni is not found in nature, however, and since it won't be coming from the 62Ni(7Li,6Li)63Ni reaction, none will arise unless there is deuterium in the mix. It all feels a little precarious, because if you get any 64Ni, you can get penetrating radiation from deexcitation gammas from inelastic collisions. To add to your thought about the kinetic energy of the daughter 6Li being relatively low, for the maximum Q value in your list above, there would be 4.14 MeV / 6 nucleons = 690 keV per nucleon, which seems manageable. I will nominate you for the Vortex Nobel Prize for your insight about neutron stripping from lithium. Two questions I have: - Why use hydrogen at all if the reaction can be sustained with lithium? - What is the amount of force that would be needed to bring a 7Li to within a sufficient distance of a nickel nucleus for stripping to occur? It seems like it would be high. Eric
