Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
personally i don't believe nature (or god) balances the books for every process. we only need CoE to hold for our measuring instruments. harry On Thu, May 24, 2012 at 11:09 PM, David Roberson dlrober...@aol.com wrote: This concept is most interesting. I would assume that the energy required to overcome the electrostatic barrier must still be supplied and it would most likely be stolen from the strong force presentations. The nucleus mass deficit is substantially larger when a neutron is absorbed (Ni58 + Neutron = Ni59) than when a proton is forced into the nucleus against the barrier (Ni58 + Proton = Cu59). This supports that hypothesis. An interesting secondary occurrence is that the subsequent beta plus decay of the Cu59 into Ni59 represents the expelling of the same amount of charge as was previously absorbed. This second process demonstrates a relatively large mass deficit. The end result of the complete process is a near parity energy performance when compared to direct neutron absorption. Why the coulomb barrier energy is not lost is still blocked within my mind. Apparently stars run out of steam when they try to fuse Ni56 with an alpha particle to form Zn60. My calculations suggest the same occurrence if I assume that the activation barrier energy is lost into the mass of the Zn60 nucleus. I guess I must have a mental barrier that is difficult to overcome! Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, May 24, 2012 4:22 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? I guess this is also Frank Znidarsic contention: If the range of the strong nuclear force increased beyond the electrostatic potential barrier a nucleon would feel the nuclear force before it was repelled by the electrostatic force. Under this situation nucleons would pass under the electrostatic barrier without producing any radiation. Could this author's original idea that electron condensations increase the range of the nuclear foces be correct? http://www.angelfire.com/scifi2/zpt/chapter4.html harry On Thu, May 24, 2012 at 11:38 AM, Harry Veeder hveeder...@gmail.com wrote: As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I hope that CoE holds in the universe. That is one guideline that is available for us that I have always relied upon. Does anyone know of any reliable experiments that have indicated that this conservation law is invalid? Of course the energy equivalent of mass is an important component of the law. I have long wondered about the tunneling phenomenon and how it fits in with the CoE. My best understanding is that tunneling is more about particles occupying the upper edge of the bell curve having enough energy to overcome a barrier than an isolated particle that is measured below the required energy level which succeeds in the breach. It is written in stellar fusion lure that the very tiny upper end of the energy range hydrogen nuclei are the ones that undergo conversion. Is it possible to isolate an individual particle in an experiment where its state can be well defined and then determine that it has indeed demonstrated a tunneling that should not be possible? I suspect that the uncertainty principle would preclude such an experiment. Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Fri, May 25, 2012 2:30 am Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? personally i don't believe nature (or god) balances the books for every process. e only need CoE to hold for our measuring instruments. arry On Thu, May 24, 2012 at 11:09 PM, David Roberson dlrober...@aol.com wrote: This concept is most interesting. I would assume that the energy required to overcome the electrostatic barrier must still be supplied and it would most likely be stolen from the strong force presentations. The nucleus mass deficit is substantially larger when a neutron is absorbed (Ni58 + Neutron = Ni59) than when a proton is forced into the nucleus against the barrier (Ni58 + Proton = Cu59). This supports that hypothesis. An interesting secondary occurrence is that the subsequent beta plus decay of the Cu59 into Ni59 represents the expelling of the same amount of charge as was previously absorbed. This second process demonstrates a relatively large mass deficit. The end result of the complete process is a near parity energy performance when compared to direct neutron absorption. Why the coulomb barrier energy is not lost is still blocked within my mind. Apparently stars run out of steam when they try to fuse Ni56 with an alpha particle to form Zn60. My calculations suggest the same occurrence if I assume that the activation barrier energy is lost into the mass of the Zn60 nucleus. I guess I must have a mental barrier that is difficult to overcome! Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, May 24, 2012 4:22 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? I guess this is also Frank Znidarsic contention: If the range of the strong nuclear force increased beyond the electrostatic potential barrier a nucleon would feel the nuclear force before it was repelled by the electrostatic force. Under this situation nucleons would pass under the electrostatic barrier without producing any radiation. Could this author's original idea that electron condensations increase the range of the nuclear foces be correct? http://www.angelfire.com/scifi2/zpt/chapter4.html harry On Thu, May 24, 2012 at 11:38 AM, Harry Veeder hveeder...@gmail.com wrote: As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Lowering of the coulomb barrier is far from an all or nothing situation. There is a fine grained gradation of the reaction that is reflected in increasing probability of tunneling as the coulomb barrier is progressively reduced. As more screening charge is gradually increased and packed around the immediate neighborhood of a nucleus, the coulomb barrier that protects that nucleus is gradually reduced. And competing theories of cold fusion causation must explain how the reaction is gradual and controllable. Charge concentration provides a monotonically increasing counter screening charge reaction that opposes the positive charge of the coulomb barrier. The reports from successful cold fusion reactor builders who post here at vortex which tell us that they can control the power output of their reactors by simply adjusting the input power feed to the spark plug is understandable. That interesting factoid tells me that the cold fusion reaction is a result of (gradual, controllable, adjustable) screening charge accumulation. Cheers: Axil On Fri, May 25, 2012 at 1:26 PM, David Roberson dlrober...@aol.com wrote: I hope that CoE holds in the universe. That is one guideline that is available for us that I have always relied upon. Does anyone know of any reliable experiments that have indicated that this conservation law is invalid? Of course the energy equivalent of mass is an important component of the law. I have long wondered about the tunneling phenomenon and how it fits in with the CoE. My best understanding is that tunneling is more about particles occupying the upper edge of the bell curve having enough energy to overcome a barrier than an isolated particle that is measured below the required energy level which succeeds in the breach. It is written in stellar fusion lure that the very tiny upper end of the energy range hydrogen nuclei are the ones that undergo conversion. Is it possible to isolate an individual particle in an experiment where its state can be well defined and then determine that it has indeed demonstrated a tunneling that should not be possible? I suspect that the uncertainty principle would preclude such an experiment. Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Fri, May 25, 2012 2:30 am Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? personally i don't believe nature (or god) balances the books for every process. we only need CoE to hold for our measuring instruments. harry On Thu, May 24, 2012 at 11:09 PM, David Roberson dlrober...@aol.com wrote: This concept is most interesting. I would assume that the energy required to overcome the electrostatic barrier must still be supplied and it would most likely be stolen from the strong force presentations. The nucleus mass deficit is substantially larger when a neutron is absorbed (Ni58 + Neutron = Ni59) than when a proton is forced into the nucleus against the barrier (Ni58 + Proton = Cu59). This supports that hypothesis. An interesting secondary occurrence is that the subsequent beta plus decay of the Cu59 into Ni59 represents the expelling of the same amount of charge as was previously absorbed. This second process demonstrates a relatively large mass deficit. The end result of the complete process is a near parity energy performance when compared to direct neutron absorption. Why the coulomb barrier energy is not lost is still blocked within my mind. Apparently stars run out of steam when they try to fuse Ni56 with an alpha particle to form Zn60. My calculations suggest the same occurrence if I assume that the activation barrier energy is lost into the mass of the Zn60 nucleus. I guess I must have a mental barrier that is difficult to overcome! Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, May 24, 2012 4:22 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? I guess this is also Frank Znidarsic contention: If the range of the strong nuclear force increased beyond the electrostatic potential barrier a nucleon would feel the nuclear force before it was repelled by the electrostatic force. Under this situation nucleons would pass under the electrostatic barrier without producing any radiation. Could this author's original idea that electron condensations increase the range of the nuclear foces be correct? http://www.angelfire.com/scifi2/zpt/chapter4.html harry On Thu, May 24, 2012 at 11:38 AM, Harry Veeder hveeder...@gmail.com wrote: As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
When two like charged participles are cooper paired together, do they still have charge? They may not. Their charge may be delocalized and exist at a location that is far distant from the spin part of them. If they both had the same charge, how could they stick together? A quasi-neutron… just a thought… Cheers: Axil On Thu, May 24, 2012 at 12:57 AM, Axil Axil janap...@gmail.com wrote: *It isn't clear to me why a cooper pair of protons would be of nuclear dimensions, nor why they would be able to surmount the Coulomb barrier.* Essentially, there exists no Coulomb barrier at the point of charge concentration if that concentration is dense enough. These days, I am interested in concentration of electron charge in a small volume. This is how the Chin reaction works. Rossi’s reaction is inferior in my opinion as hard to control. In the Chin reaction, this negative electric charge concentration on a nano tube will induce a large number of positive charge holes of equal by opposite charge. Now See http://en.wikipedia.org/wiki/Electric-field_screening *Electric-field screening* The main point here is that as long as there are many positive ions between two positive charges; say a proton and a nucleus, their interaction is *screened* strongly, simply because these many positive charge carriers can terminate electric field lines. So a free ion attracts ions of opposite sign, making a little `counter ion cloud' which neutralizes its charge, and therefore by Gauss's law, basically eliminates the electric field. The size of this `cloud' is roughly the screening length yD, the parameter that determines when the exponential `cuts off' the Coulomb interaction in U(r). A useful formula for yD is due to Debye, which comes from a certain relatively-easy-to-solve limiting case of interaction of charges with free ions present where the sum over j is over *all* the ions, and where j counts the number of ions. As you can see, as you add more and more positve charges, because the induced charges enter squared, the screening length goes down, down, down. See the function for the Debye-Hückel length where Zj = Qj/C is the integer charge numberhttp://en.wikipedia.org/wiki/Charge_numberthat relates the charge on the j-th ionic species to the elementary chargehttp://en.wikipedia.org/wiki/Elementary_charge . http://en.wikipedia.org/wiki/Debye_length *Debye length*** ** This formula is often called the *Debye screening length*, and provides a good first estimate of the distance beyond which Coulomb interactions can be essentially ignored, as well as the size of the region near a point charge where opposite-charge counter ions can be found. On Wed, May 23, 2012 at 10:08 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 22 May 2012 21:44:13 -0400: Hi, [snip] The cooper pair of protons speculation It isn't clear to me why a cooper pair of protons would be of nuclear dimensions, nor why they would be able to surmount the Coulomb barrier. (They only have a reasonable chance of tunneling through it if they get close enough). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I guess this is also Frank Znidarsic contention: If the range of the strong nuclear force increased beyond the electrostatic potential barrier a nucleon would feel the nuclear force before it was repelled by the electrostatic force. Under this situation nucleons would pass under the electrostatic barrier without producing any radiation. Could this author's original idea that electron condensations increase the range of the nuclear foces be correct? http://www.angelfire.com/scifi2/zpt/chapter4.html harry On Thu, May 24, 2012 at 11:38 AM, Harry Veeder hveeder...@gmail.com wrote: As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
This concept is most interesting. I would assume that the energy required to overcome the electrostatic barrier must still be supplied and it would most likely be stolen from the strong force presentations. The nucleus mass deficit is substantially larger when a neutron is absorbed (Ni58 + Neutron = Ni59) than when a proton is forced into the nucleus against the barrier (Ni58 + Proton = Cu59). This supports that hypothesis. An interesting secondary occurrence is that the subsequent beta plus decay of the Cu59 into Ni59 represents the expelling of the same amount of charge as was previously absorbed. This second process demonstrates a relatively large mass deficit. The end result of the complete process is a near parity energy performance when compared to direct neutron absorption. Why the coulomb barrier energy is not lost is still blocked within my mind. Apparently stars run out of steam when they try to fuse Ni56 with an alpha particle to form Zn60. My calculations suggest the same occurrence if I assume that the activation barrier energy is lost into the mass of the Zn60 nucleus. I guess I must have a mental barrier that is difficult to overcome! Dave -Original Message- From: Harry Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, May 24, 2012 4:22 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? I guess this is also Frank Znidarsic contention: If the range of the strong nuclear force increased beyond the lectrostatic potential barrier a nucleon would feel the nuclear force efore it was repelled by the electrostatic force. Under this ituation nucleons would pass under the electrostatic barrier without roducing any radiation. Could this author's original idea that lectron condensations increase the range of the nuclear foces be orrect? http://www.angelfire.com/scifi2/zpt/chapter4.html harry On Thu, May 24, 2012 at 11:38 AM, Harry Veeder hveeder...@gmail.com wrote: As another way to over come the coloumb barrier, I vaguely recall a paper proposing that the range of the strong force may reach further under some circumstances. Harry
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
*It isn't clear to me why a cooper pair of protons would be of nuclear dimensions, nor why they would be able to surmount the Coulomb barrier.* Essentially, there exists no Coulomb barrier at the point of charge concentration if that concentration is dense enough. These days, I am interested in concentration of electron charge in a small volume. This is how the Chin reaction works. Rossi’s reaction is inferior in my opinion as hard to control. In the Chin reaction, this negative electric charge concentration on a nano tube will induce a large number of positive charge holes of equal by opposite charge. Now See http://en.wikipedia.org/wiki/Electric-field_screening *Electric-field screening* The main point here is that as long as there are many positive ions between two positive charges; say a proton and a nucleus, their interaction is * screened* strongly, simply because these many positive charge carriers can terminate electric field lines. So a free ion attracts ions of opposite sign, making a little `counter ion cloud' which neutralizes its charge, and therefore by Gauss's law, basically eliminates the electric field. The size of this `cloud' is roughly the screening length yD, the parameter that determines when the exponential `cuts off' the Coulomb interaction in U(r). A useful formula for yD is due to Debye, which comes from a certain relatively-easy-to-solve limiting case of interaction of charges with free ions present where the sum over j is over *all* the ions, and where j counts the number of ions. As you can see, as you add more and more positve charges, because the induced charges enter squared, the screening length goes down, down, down. See the function for the Debye-Hückel length where Zj = Qj/C is the integer charge numberhttp://en.wikipedia.org/wiki/Charge_numberthat relates the charge on the j-th ionic species to the elementary chargehttp://en.wikipedia.org/wiki/Elementary_charge . http://en.wikipedia.org/wiki/Debye_length *Debye length*** ** This formula is often called the *Debye screening length*, and provides a good first estimate of the distance beyond which Coulomb interactions can be essentially ignored, as well as the size of the region near a point charge where opposite-charge counter ions can be found. On Wed, May 23, 2012 at 10:08 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 22 May 2012 21:44:13 -0400: Hi, [snip] The cooper pair of protons speculation It isn't clear to me why a cooper pair of protons would be of nuclear dimensions, nor why they would be able to surmount the Coulomb barrier. (They only have a reasonable chance of tunneling through it if they get close enough). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
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
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
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
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
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson 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.
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
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson 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.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson 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.
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
In the case of regular fusion the kinetic energy of the proton would be conserved in the reaction products (i.e. the total energy of the system would be 5.6 MeV + 3.41MeV) these reaction products might be a Cu59 nucleus with 9.01MeV of kinetic energy or more likely some combination of gamma rays, neutrinos, nuclear excitation, and other nuclear reaction products. I am not familiar with the specifics of this reaction. The fusion of Iron inside a star is a different matter. Iron can fuse with hydrogen in an exothermic reaction because hydrogen has zero binding energy due to the fact that it has a single proton nucleus. Iron cannot fuse with itself inside a star because the resultant reaction would be endothermic, this is why stars burn out, not because of H + Fe fusion. To: vortex-l@eskimo.com Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 17:53:30 -0400 That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson 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.
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Awkshully, the most stable isotope in the periodic table is not one of iron’s major isotopes. 62Ni has the highest binding energy per nucleon of all isotopes, including iron, even though the average binding energy per nucleon for iron is slightly higher than nickel. Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. From: David Roberson I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.com wrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly * borrowed* and then a short time later, the debt is repaid – before the net gain is obvious. *From:* David Roberson 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.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
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 WL 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 WL 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
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I also suspect that the reaction is a bit more complex than a single hydrogen fusion. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 9:44 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.com wrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson 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.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
The process suggested by Rossi has major issues as you point out. I do think that one could force a reaction like this with very high energy protons impacting nickel, but it seems quite unlikely to me that this will happen at the low temperatures encountered within Rossi's reactors. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 6:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Awkshully, the most stable isotope in the periodic table is not one of iron’s major isotopes. 62Ni has the highest binding energy per nucleon of all isotopes, including iron, even though the average binding energy per nucleon for iron is slightly higher than nickel. Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. From: David Roberson I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
On Tue, May 22, 2012 at 3:50 PM, Jones Beene jone...@pacbell.net wrote: ** Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. Or a gambit intended to divert attention from what is really going on. Eric
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
The hard part is making the coulomb barrier go away; after that the fusion is easy. On Tue, May 22, 2012 at 11:24 PM, David Roberson dlrober...@aol.com wrote: I also suspect that the reaction is a bit more complex than a single hydrogen fusion. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 9:44 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.comwrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly * borrowed* and then a short time later, the debt is repaid – before the net gain is obvious. *From:* David Roberson 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.
