In reply to Jones Beene's message of Sat, 29 Aug 2009 16:29:42 -0700: Hi, [snip] >-----Original Message----- >From: [email protected] > >"Since the H + D -> He3 reaction produces a 5.49 MeV gamma ray, it seems >highly likely that this is what was observed, especially considering the >fact that it is next to impossible to obtain pure D with no H content." > >Whoa, there is a gamma in that reaction, true, but the "net energy" of the >reaction, if I am not mistaken is 5.49, and the gamma peak which is seen is >2.223 MeV - which is the emission line associated with He3 formation in >solar astronomy. Presumably the rest is carried away by the kinetic energy >of the light helium.
Conservation of momentum forbids this. The mass of the He3 nucleus is an energy equivalent of 2800 MeV, compared to which the energy of a gamma is trivial. That means that the overwhelming majority of the energy must end up with the gamma(s). However if the He3 nucleus has one or more excited states that are lower in energy than 5.49 MeV, then it is likely that the decay will proceed in steps, hence the 2.223 MeV gammas. Nevertheless, the fact that decay usually happens in steps, doesn't necessarily mean that it must always do so, and it would appear that at least on occasion it decays directly to the ground state with the emission of a single gamma. > >Plus - and foremost, due to the large mass difference between He3 and He4, >the difference in the tracks left by either and preserved in the famous >photographic plates - would be very clear to an expert like Alvarez, one >must imagine. The noticeable He3 tracks would be from the DD reaction (where the He3 kinetic energy > 800 keV). In the HD reaction the kinetic energy of the He3 would only be 5.36 keV (and that's when a 5.488 MeV gamma is emitted - for multiple gammas it would be less). [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
