Yes, that is true. I probably should have included that addition in my document to clarify the complete reaction. I was starting with the individual proton and an electron plus the isotope. The neutron is of course constructed by taking the proton and electron and adding that amount of energy. This neutron construction energy must be removed from the final reaction in order to step back to the initial equal conditions.
Dave -----Original Message----- From: mixent <[email protected]> To: vortex-l <[email protected]> Sent: Mon, Jun 11, 2012 6:02 pm Subject: Re: [Vo]: Nuclear Stability and Proton or Neutron Addition In reply to David Roberson's message of Mon, 11 Jun 2012 14:49:06 -0400 (EDT): i, snip] Now, instead of adding a proton, letÂ’s allow a neutron to encounter the Ni60 ucleus. In this case a stable isotope of nickel Ni61 is directly formed and a ignificant amount of energy is released. I followed both of these processes through several different elements and can tate that the same total energy is released regardless of the path taken when I tart with an isotope of an element and end at the same final product. I onsider this an important and useful observation. This is only true if you include the 0.782 MeV required to convert a Hydrogen tom to a neutron (and is simply a restatement of the conservation of ass-energy; hardly surprising). Otherwise you end up with a 0.782 MeV iscrepancy 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
