I for one consider a hot Li6 is inconsistent with no radiation. The enerrgy release must be by a different mechanism.
Bob Cook Sent from my Verizon Wireless 4G LTE SmartphoneAxil Axil <janap...@gmail.com> wrote: I agree, you really can tell where that Li6 came from. On Wed, Oct 8, 2014 at 10:21 PM, <mix...@bigpond.com> wrote: > In reply to Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700: > Hi, > [snip] > > 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. > > I have only briefly skimmed the report, but the basic reaction appears to > be a > neutron transfer reaction where a neutron tunnels from Li7 to a Nickel > isotope. > The excess energy of the reaction appears as kinetic energy of the two > resultant > nuclei (i.e. Li6 & the new Ni isotope), rather than as gamma rays. Because > there > are two daughter nuclei, momentum can be conserved while dumping the > energy as > kinetic energy in a reaction that is much faster then gamma ray emission. > Because both nuclei are "heavy" and slow moving, very little to no > bremsstrahlung is produced. There is effectively no secondary gamma from > Li6 > because the first excited state is too high. (I haven't checked Li7). > There is > unlikely to be anything significant from Ni because the high charge on the > nucleus combined with the "3" from Lithium tend to keep them apart (minimum > distance 31 fm). > > It would be nice to know if the total amounts of each of Li & Ni in the > sample > were conserved (I'll have to study the report more closely). > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > >
