In reply to Harry Veeder's message of Mon, 17 Jun 2013 16:54:06 -0400: Hi Harry, [snip] >Robin, >Do you look these values up in a table or do you calculate them? >I thought the addition of protons to any element above Ni62 will not >release heat, or does it depend on the isotopes involved?
I calculate them. If you have access to a windows machine, you can download the program from http://rvanspaa.freehostia.com/Isotopes.zip (if it doesn't work, please let me know ;) (Elements beyond Uranium are excluded). You will get heat from adding a proton to almost any element in the periodic table. Though obviously more from some than from others. While the per nucleon binding energy is at a maximum for Ni62, the mass excess of a free proton is so large compared to that of a bound proton that it more than makes up for the slight decrease in binding energy that gradually builds above Ni62. Take a look at the graph at the bottom of this page:- http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html A lone proton has a binding energy of zero (i.e. it's on the X-axis of the graph). So there is always lots of energy released when it moves to another point on the line (by binding with the nucleus that is just to the left of the target position). The energy release is equal to the vertical difference on the graph as measured in MeV on the left hand scale corrected by the total change in mass of the other nucleons that were already present as they change from the original isotope to the new one. (Since a free proton is zero, you can just read the value of the target off the scale on the left, i.e. target - 0 = target.) The correction factor is usually positive below 62Ni, and usually negative above 62Ni, however it still works, even for some of the heaviest nuclei e.g. H + 238U => 239Np + 5.29 MeV > >On this Italian video about Rossi >http://www.classmeteo.it/web/portale/video/prometeo-fusione-fredda-e-cat-e-lenergia-del-futuro/ > >@5:27 it says: > >62Ni + H --> 63Cu + 6.12 MeV > >I thought this was incorrect according to theory but is it correct? The calculation is correct. However most physicists today will tell you that the reaction is impossible because 62Ni has a charge of 28 and strongly repels a lone proton. They don't take into account the possibility that the proton may be accompanied by an electron. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
