On Tue, Apr 23, 2013 at 5:45 PM, <mix...@bigpond.com> wrote: > In reply to Harry Veeder's message of Tue, 23 Apr 2013 14:28:00 -0400: > Hi, > [snip] >>If a neutron can be made decay while in a deuteron then it seems to me >>the warming of the lattice is best explained >>by the motion arising from the mutual repulsion of the protons. >>Thermalization of gammas is not necessary. >> > When a deuteron is formed from a neutron and a proton, mass is converted into > energy (2.2 MeV). It is this loss of energy that prevents a neutron already > in a > deuteron from decaying. Decay only happens when it results in the release of > energy, and the neutron bound in D has already lost too much for it to decay. > Regards,
In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. However, occasionally one of the protons transforms into a neutron by emitting a beta and a neutrino before fission occurs. This results in a stable deuteron. If this is correct, then a deuteron is stable because it is in a lower energy state than the diproton. (Remember a diproton has been pushed together under high pressures and temperatures so it contains more potential energy than two isolated protons or an isolated an proton and a isolated neutron.) Therefore a deuteron will return to the same level of instability as a diproton if it absorbs enough energy again. The energy profile of the deuteron and proton can be characterised by using the visual aid of mountain with depression at the top. The potential energy of a diproton corresponds with two protons resting on opposite sides of the depression. The potential energy of the deuteron corresponds to a proton and neutron resting in the depression. Harry
