In reply to Jones Beene's message of Sun, 12 Mar 2006 16:17:22 -0800: Hi Jones, [snip] >Robin, > >> Using only Li6 (in heavy water as Horace suggested), and running >> a >> few numbers I get a neutron breeding ratio of about 2.7 (from >> the >> deuterium), i.e. for every incident fast neutron about 1.7 extra >> slow neutrons. > >Actually it is close to unity. The (n,2n) reaction is extremely >rare with most light elements - 7Li and Be - being the ONLY >exceptions. Fast neutrons simply do not breed many neutrons with >either heavy water or 6Li. [snip] According to http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=H-2&n=2 the (n,2n) cross section for 14 MeV neutrons on D is 177 mb, while the elastic scattering cross section for 14 MeV neutrons is only 624 mb. IOW at 14 MeV there is a 28% chance that a collision will produce an extra neutron. What I did, as previously mentioned is assume that this ratio (i.e. 28%) holds constant while the energy remains above 2.2 MeV. That is not so.
Actually the 177 mb drops off while at the same time the elastic scattering cross section increases. IOW the ratio drops back. A more accurate calculation results in about 0.47 extra neutrons for every initial fast neutron, not including secondaries. (It takes about 4-5 worst case collisions for the fast neutron to lose so much energy that it is no longer capable of fissioning a D). IOW we should get about 1.5 thermal neutrons out for every fast neutron going in, and nearly all of the thermal neutrons are going to produce T. This should still be a breeder. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
