I understand how your analogy operates now. It is helpful to be able to compare everyday experiences with the phenomenon that one is trying to get a handle upon.
What is the possibility that the barrier energy that is supplied by my experimental proton accelerator at approximately 5.6 MeV has become converted into mass of the copper 59 nucleus? According to the charts of nuclides there is a reduction of mass of the copper 59 that equals a modest 3.41 MeV as the proton fuses with the nickel 58. This stage of the reaction is followed by the release of 5.82 MeV in a half life of 81.5 seconds via beta plus decay and positron-electron destruction. The beta plus decay results in the formation of nickel 59. A further calculation that is obtained by taking the same initial nickel 58 and adding a neutron to it to arrive at nickel 59 yields a mass loss equal to 8.99 MeV. This is the value shown in W&L charts as well when you follow their reaction path. If you add the two energy releases associated with the beta plus aka the Rossi reaction you get a net of 9.23 MeV. This compares quite well with the value according to the W&L process at 8.99 MeV. The difference is less than half the energy equivalent of an electron. I suspect that this error can be discovered with enough digging. In these scenarios the same starting and end results are obtained: Nickel 58 seed results in Nickel 59 product. Also, approximately the same energy is released provided the original input proton energy of 5.6 MeV is assumed to be unimportant as you suggest. Dave -----Original Message----- From: Finlay MacNab <finlaymac...@hotmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Tue, May 22, 2012 5:23 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? The easiest way to visualize the effect is to imagine the case of a roller coaster. Initially a roller coaster must overcome a gravitational energy barrier by a mechanical mechanism that pulls the cars to the top of the track. That energy is not lost but is regained on the down slope. In the case of a nuclear fusion reaction the kinetic energy of the proton overcomes the repulsion of the positively charged nucleus through its high kinetic energy. Once the proton comes close enough to the target nucleus, the strong nuclear force begins to attract the proton with a strength that dwarfs the electrostatic repusion (the down slope). In this way the kinetic energy of the proton is not lost. Practically speaking a proton does not need to overcome the total energy barrier imposed by electrostatic repulsion, because as it gets close to the nucleus the probability that the proton will tunnel through the barrier becomes large. This quantum effect is similar to what is observed in transistors and MIM diodes, where electrons elastically tunnel through thin dielectrics and semiconductors without losing energy. To: vortex-l@eskimo.com Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 14:23:39 -0400 Thanks for the rapid response! I think that I have a mental block regarding this issue. Perhaps I am looking at the fusion process from the wrong angle. Help me understand where my error is as I do not follow your example below. Let's try a mental experiment. I can see that it would be possible to build a machine that accelerates protons until they reach the required threshold level of 5.6 MeV. It would take at least this much energy to generate the projectile proton. Now, when the bullet proton comes close to the nucleus the fusion takes place. At this point in time it has lost all of its kinetic energy and is absorbed into the nucleus. I assume that very soon thereafter the 3.41 MeV is released as a gamma or some other form of radiation. The tables of nuclides informs me that the copper 59 has a half life of 81.5 seconds before the beta plus decay so the energy associated with that process must be in place but does not exit until later. Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize. Dave -----Original Message----- From: Finlay MacNab <finlaymac...@hotmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Tue, May 22, 2012 1:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Your calculation does not take into account the fact that the activation energy barrier releases the energy added to overcome it during the reaction. In this case once coulomb repulsion is overcome, the energy is added back to the system by attractive nuclear force. The 3.41MeV is the change in mass energy balance after the reaction, what happens in between is not important to the calculation. Analogously, If a person descends from the 3rd to second floor of a building, they are just as close to the ground as if they climb from the 3rd to the 53rd floor before climbing back down the the second floor to end their journey. This is the nature of all "activation energy barriers". To: vortex-l@eskimo.com Subject: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 13:35:41 -0400 I have been researching the cold fusion reaction that is suggested by Rossi and Focardi in their recent paper http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and have a couple of questions. The authors suggest that 3.41 MeV of energy is released by the fusion of a proton with a nickel 58 nucleus into copper 59. I can obtain this value if I calculate the mass difference between a copper 59 atom and a nickel 58 atom plus the mass of the proton and the mass of the extra electron. So far their calculations are in line with mine. The problem arises when I consider the amount of energy required to overcome the coulomb barrier in order to activate the fusion. The two authors seem to overlook this entirely when they calculate the energy available from their proposed reaction. The chart on page 5 of their paper shows that 3.41 MeV is released at the conclusion to the reaction but no allowance is given to the energy needed to initiate it. They do mention the activation energy in their theoretical interpretation on page 7. In this section they calculate that it takes 5.6 MeV to overcome the barrier. The authors use assumed values for the closeness required and thus energy barrier in their example. With these two numbers available I make the assumption that there is a net energy requirement of 5.6 MeV – 3.41 MeV or 2.19 MeV for the fusion. Is there a reason that my calculation is in error? Does the 3.41 MeV have hidden within it the activation energy? I can see no good reason to suspect that this is the case since it would be possible for a device to send high speed protons into a target made of copper. The copper would then shed the 3.41 MeV by some means and that would obviously not repay the debt. Of course I understand that the following beta plus decay would release an additional significant amount of energy as the copper transforms into nickel 59. I calculate this energy as 5.8 MeV when the released positron is annihilated. This value matches that of the two authors which I assume is correct. A recap of the question is: Is the fusion of nickel 58 with a proton and electron into copper 59 an endothermic reaction? Dave
