In reply to Jones Beene's message of Thu, 25 Apr 2013 17:36:50 -0700: Hi, [snip] >Hi Robin, > >Well I am including the mass-energy of the positron and the neutrino, which >are emitted - added to the mass of the deuteron to suggest that all of these >weigh considerably more than two protons. Therefore outside energy >(momentum) would have to be employed, even though real energy is emitted and >that looks like exotherm.
The difference in rest mass energy between two protons and a deuteron is 930.854 keV. The rest mass of an electron/positron is 511 keV. That leaves 930.854 - 511 = 419.8 keV left over to be shared as kinetic energy between the neutrino and the positron. Since the actual distribution will vary from one fusion reaction to the next, the neutrino has a maximum energy of 419.8 keV. John's web page quotes a maximum of 423 keV (first line in the "p-p nuclear fusion reactions" diagram), the difference due to slight inaccuracies in masses quoted. When the positron annihilates an electron (later on), another 1.02 MeV is liberated. In the electron capture version of the reaction, the electron takes part in the initial reaction directly, rather than in a delayed annihilation reaction, so the net energy from an electron capture reaction is 1.02 MeV + 423 keV = 1.44 MeV, "all" of which is carried by the neutrino. (See the second line in the "p-p nuclear fusion reactions" diagram on John's web page.) The neutrino gets the lot, because there is no excited state of the deuteron with energy less than 1.44 MeV, so no gamma emission is possible. The only particle leaving the reaction is the neutrino. (And the deuteron itself of course, but the mass of the neutrino must be so slight that it gets nearly 100% of the energy.) > >But even then, I admit that there is a math problem in that this particular >electron neutrino from the diproton reaction apparently may not have the >mass-energy which is seen and documented by experiments on earth -like SNO - >where the mass-energy is in the several MeV range. > >Almost all solar neutrinos come from the diproton reaction but the Wiki >entry suggests are much lower in energy than what is actually measured at >SNO. I think the SNO neutrinos are from other fusion reactions. Take a look at John's web page. http://www.sns.ias.edu/~jnb/SNviewgraphs/NuclearFusion/nucfusion.html > >Even with oscillation, I do not see how the energy could be lower on the sun >where they are formed but higher when measured at SNO. In fact, it is only a >guess as to what they are on the sun, despite what Wiki states as fact - >since we obviously cannot measure them there. > >Bottom line: if they are 1 MeV neutrinos on the sun from two proton and beta >decay, then the net diproton reaction is endothermic when we look at only >the rest mass of the protons, and if they are 400 keV mass-energy, the >reaction is balanced, and if they are 100 keV the reaction is exothermic. > > > >>>Actually The neutron has mass slightly larger than that of a proton: >>>939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has >>>slightly more mass than a diproton. As mentioned previously, nucleons bound in nuclei have less mass than free nucleons. It works like this: Take two blocks. Weigh them separately. Add the weights together. Glue the blocks together. Weigh the glued combination. It weighs less than the blocks did initially (well it does when you use nucleons ;). The missing mass has been converted to energy - nuclear fusion energy. However the two blocks are still present in the glued mass. Conclusion:- The blocks in the glued mass weigh less than separate autonomous blocks. Therefore, you can't just use the mass of a free neutron when talking about the mass of a neutron bound in a deuterium nucleus. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
