In reply to Axil Axil's message of Sat, 13 Jul 2013 19:42:47 -0400: Hi, [snip] >The discussion was predicated on this reaction as follows: > >Si28 + O16 -> Ca40 + He4 + 6.2 MeV > >Now you are changing the playing field to U236 fission. Let us continue to >talk about the fusion of highly stable elements.
You only appear to have read the first half of what I wrote here below. I suggest you read all of it. > > >On Sat, Jul 13, 2013 at 3:35 AM, <[email protected]> wrote: > >> In reply to Axil Axil's message of Fri, 12 Jul 2013 19:43:21 -0400: >> Hi, >> [snip] >> >What causes the defeat of coulomb barrier? >> > >> > >> > >> > >> > >> >Mixent - kinetic energy of nuclear collision. >> > >> > >> > >> > >> > >> >Axil EMF- namely charge concentration and anapole magnetism. >> > >> > >> > >> > >> > >> > >> > >> >The defeat of the coulomb barrier cant be kinetic energy because from the >> >experience gained with the kinetic collisions of nuclei we know that the >> >resultant nucleus would be left excited by the kinetic energy of that >> >nuclear collision and affect the combined nucleus by exciting it leaving a >> >radioactive isotope as an end product. >> > >> > >> > >> > >> > >> >Internal rearrangement of the combined nuclei by EMF does not carry excess >> >energy of collision into the reaction therefore no radioactive isotope is >> >produced >> >> Kinetic energy only contributes to the overall energy of the reaction. What >> comes out is not necessarily related to the fact that kinetic energy >> played a >> role. E.g. conventional fission in U235 is best triggered by a slow moving >> neutron with as little kinetic energy as possible, since low kinetic energy >> enhances the cross section. >> Despite the lack of initial kinetic energy, the result is usually a >> plethora of >> radioactive nuclei. The primary reason for this is simply that relative to >> its >> lighter daughter nuclei, U235 has an excess of neutrons. That means that >> the >> daughter nuclei are neutron rich, and hence radioactive. >> >> When light nuclei combine at high speed, the resultant momentary "heavy" >> nucleus >> is neutron poor (compared to normal stable nuclei of the same combined >> weight), >> hence the immediate fission nuclei from the combined nucleus will not be >> neutron >> rich, and hence not likely to be radioactive. >> >> If it doesn't have to, nature actually prefers not to make radioactive >> nuclei. >> >> Regards, >> >> Robin van Spaandonk >> >> http://rvanspaa.freehostia.com/project.html >> >> Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
