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</o:shapelayout></xml><![endif]--></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> extropy-chat-bounces@lists.extropy.org [mailto:extropy-chat-bounces@lists.extropy.org] <b>On Behalf Of </b>Alfio Puglisi<br><br><b><span style='color:#1F497D'><o:p></o:p></span></b></span></p><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:#1F497D'><o:p> </o:p></span></b></p><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:#1F497D'>>…</span></b>look up nucleon energies, find *any* mysterious path from any marginally stable nickel isotope to *any* marginally stable copper isotope, then come on back and tell us what you found.<o:p></o:p></p><div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>Here is my first-ever attempt at nuclear fusion, for which I assume no responsibility :-) . Of course I took Wikipedia as the ultimate reference for atomic masses.<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>Proton capture from nickel-64 (stable) to copper-65 (stable). I don't have the slightest idea whether this reaction is permitted or not, but according to my calculator:<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><div><p class=MsoNormal>Ni64 = 63.9279660<o:p></o:p></p></div></div><div><div><p class=MsoNormal>Cu65 = 64.9277895 <o:p></o:p></p></div></div><div><div><p class=MsoNormal>H = 1.007825032<o:p></o:p></p></div></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>H + Ni64 = Cu65 + 0.008001532<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>which corresponds to about 7.5 MeV of energy.<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>What am I doing wrong?<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>Ciao,<o:p></o:p></p></div><div><p class=MsoNormal>Alfio<o:p></o:p></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>You did nothing wrong Alfio, and you get full points according to the rules I suggested. 64Ni and 65Cu are both stable. I had discounted that reaction because 64Ni is less than 1 percent of natural nickel, but I failed to actually specify that to start with. If we ignore the physics of that particular proton capture, the final material would still be nickel with a trace of copper mixed in. It would take some pretty sophisticated qualitative analysis to even know there was 1% copper in that nickel. My mistake, I didn’t include that part. The description of the experiment said he ended up with copper.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>With regard to that proton capture, it isn’t permitted under any rules of nucleon mechanics I recognize, but the way I stated the question to start with, that doesn’t matter what I recognize. If Rossi et.al. can demonstrate it, I have a nucleon chart eager to be devoured with relish.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Ain’t nuclear physic kewallll? Once you get all the way down to subatomic particles, physics actually gets relatively simple again.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>{8^D<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>spike<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p></div></div></div></body></html>