In reply to David Roberson's message of Mon, 11 Jun 2012 14:49:06 -0400 (EDT): Hi, [snip] >Now, instead of adding a proton, letÂ’s allow a neutron to encounter the Ni60 >nucleus. In this case a stable isotope of nickel Ni61 is directly formed and >a significant amount of energy is released. >I followed both of these processes through several different elements and can >state that the same total energy is released regardless of the path taken when >I start with an isotope of an element and end at the same final product. I >consider this an important and useful observation.
This is only true if you include the 0.782 MeV required to convert a Hydrogen atom to a neutron (and is simply a restatement of the conservation of mass-energy; hardly surprising). Otherwise you end up with a 0.782 MeV discrepancy due to starting with H in one case and a neutron in the other case. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
