In reply to Alain Sepeda's message of Tue, 1 May 2012 16:20:36 +0200: Hi, [snip] >in fact they forgot to tell that the neutron is p^roduced the same > >so >p+e+energy->n +v >n+1H->2H+energy >p+e+energy-> n +v >n+2H->3H+energy >p+e+energy-> n +v >n+3H->4H+energy > >4H -> 4He + 2e+ + 2anti-v +Energy
This should be 4H -> 4He + e- + anti-v + Energy. (4H is one proton and three neutrons). Note the "silliness" of this situation: First the proton acquires an electron, at great energy cost, to become a neutron, which, after being absorbed against it's will by T, then decays to a proton again - via slow beta decay - as opposed to rapid ejection of a neutron! Perhaps needless to say, I think models in which the proton is and remains a proton are far more likely to be correct. >(this one is not classic, does somemone have reference on 4H disintegration >branching). According to http://atom.kaeri.re.kr/ton/nuc1.html it decays via neutron emission to H3 (very rapidly - order 1E-22 sec.), so I find it hard to believe that H3 is going to absorb a neutron to become H4! (Brillouin argues that the data used to create the table is wrong/inapplicable.) IMO both Brillouin & WL suffer from the same false assumption that has almost the entire scientific community in thrall. They assume that only neutrons can penetrate the Coulomb barrier at low temps. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
