Do you think that a double proton reaction is possible as per Kim
http://www.freerepublic.com/focus/f-chat/2746057/posts I offer this possibility because Piantelli has seen protons at an energy of 6 MeV emanate from the nickel bars that he uses in his reactor when they are placed in a cloud chamber after they are removed from his reactor immediately after use. Cheers: Axil On Mon, Jun 11, 2012 at 2:49 PM, David Roberson <[email protected]> wrote: > I have been reviewing a table of nuclides in an attempt to make sense of > the process suggested by W&L proponents and those of Rossi. In the W&L > theory a neutron is formed by the combination of an electron and a proton > with the .78 MeV of energy being supplied by their process. This neutron > then finds its way into a nucleus of nickel in this version of devices and > energy is released. The final result is the next heavier isotope of > nickel plus a significant amount of energy. > The Rossi process involves the insertion of a proton into the nucleus of > the subject nickel atom forming a new copper atom along with release of > energy. Some of the copper isotopes formed by addition of a proton into > their parent nickel isotopes decay by beta plus action into the next > heavier nickel isotope along with a release of additional energy. > The above two paragraphs offer an extremely brief description of the two > theories. They are not intended to get into details which can be located > within many documents. > My purpose for writing this document is to reveal an interesting > observation that I have made concerning the two processes. This may be > well known to many of the people on the list, but it is new to me and I > offer it as a refresher. > If you take any stable isotope of an element, for example nickel 60 and > either add a neutron as with the W&L process or overcome the Coulomb > barrier by forcing a proton into the nucleus you find an interesting result. > In virtually every case only one of these processes leads to a stable > isotope in a single reaction. There are only a couple of exceptions to > this observation and that appears to be when neither process results in a > single step stable new atom. Of course the newly created atoms will all > eventually decay in steps until a stable result is obtained. > I further notice that the end result of the two processes is the same > nuclide. An example is as follows: Start with Ni60 and add a proton to > it by forcing the particle against the Coulomb barrier and you obtain Cu61. > Some immediate energy is released by the new element and at a half life > later a Beta Plus decay process occurs which releases more energy. The > Beta Plus decay leaves us with Ni61. The energy release is composed of > two parts as we progress from Ni60 to Ni61. > Now, instead of adding a proton, let’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. > A second issue I would like to discuss is also interesting and leads to > some neat results. The above rule that I found makes it impossible to > have two stable isotopes of elements with the same number of nucleons that > are one level apart. An example of this rule would be that since He3 is > stable, then H3 cannot be. Or, since Ni61 is stable, then Cu61 is > unstable. This appears to apply throughout the entire list of elements > and I would appreciate it for others to verify this conclusion. > I have a couple of additional concepts that I plan to present at a later > time, so for now review what I have observed and please make relevant > comments. > Dave >
