Re: [Vo]:My current views on the 'Rossi's process'
In reply to Bob Cook's message of Tue, 25 Mar 2014 10:49:12 -0700: Hi Bob, When a fast electron interacts with other electrons it does so through a repulsive force, and imparts energy to them through collisions, knocking them away from their host atoms, and leaving them with some kinetic energy. A positron should also tear electrons away from their host atoms and leave them with excess kinetic energy. The only difference being that the force will be a mixture of attractive (and repulsive?) forces. Attractive at a distance (and repulsive in a head on collision ?). A near miss would be attractive forces and a whip around (conservation of angular momentum). Perhaps a head on collision results in annihilation? AFAIK stands for As Far As I Know. Robin-- The positron leaves the Ni-59 nucleus after an electron capture with about 1 Mev of energy--the disintegration energy is a little more than 1 Mev. However, I have not seen a cross section for the reaction we are talking about. I would agree, if the positron acts like an electron in a population of electrons, that it would slow down, but being a positive charge I not sure how that effects the slowing down. (I think you suggest its positive charge does not change the slowing down process?) The fact that the resulting photons total energy equal 2 x the electron mass probably means there is no excess energy and momentum that needs to be handled in the reaction. I am not sure whether neutrinos in the annihilation reaction have been ruled out by experiment. Probably ruled out only by theory. By the way what does AFAIK stand for? Bob - Original Message - From: mix...@bigpond.com To: vortex-l@eskimo.com Sent: Monday, March 24, 2014 1:32 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' In reply to Bob Cook's message of Sun, 23 Mar 2014 15:19:14 -0700: Hi, [snip] Your description is exactly as I understand it. The random walk is not very long however, since it probably occurs at the first electron it attracts and that is pretty quick after the nucleus gives it up. AFAIK annihilation usually only happens after the positron has slowed down. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:My current views on the 'Rossi's process'
Bob, et.al.: AFAIK = As Far As I Know -mark On Tue, Mar 25, 2014 at 10:49 AM, Bob Cook wrote: By the way what does AFAIK stand for? Bob
Re: [Vo]:My current views on the 'Rossi's process'
I should have guessed. Thanks Mark and Robin. You can tell I am an old timer. Bob - Original Message - From: MarkI-Zeropoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Tuesday, March 25, 2014 1:31 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' Bob, et.al.: AFAIK = As Far As I Know -mark On Tue, Mar 25, 2014 at 10:49 AM, Bob Cook wrote: By the way what does AFAIK stand for? Bob
Re: [Vo]:My current views on the 'Rossi's process'
Hello Robin, I am not good at magnetic forces and spin effect it is way above my paygrade. However, when reading your latest comment I saw your link and followed it. http://rvanspaa.freehostia.com/project.html I think this is a concept worthy as a starting point. I have no idea about where you are in this project but it would amaze me if you have been able to get funding with that approach. That is not because it is anything wrong with the synopsis. However, you lack all sorts of incentive to invest. 1. You need to quantify the possibilities and the obstacles. 2. You need time frames and 'waypoints' when you evaluate next step. 3. You need to show you have leadership / management involved that will measure with the $$ eyes of an investor. 'For the good of mankind' or for a possible Nobel prize is not an incentive for an investor. In addition you need to compare your idea and how it will come to fruition better, faster, cheaper etc. than the competition. The team must include all functions. A few years ago I had an operations manager who said that he was the the only person of any importance in the company as our revenue was directly proportional to his performance. At the first glance it is easy to agree with his findings. However, in a good organization that is true about all functions. The effort of the person sending out the invoices is also 100% proportional to the revenue.:) . . . Notice that I am positive to your thinking but if you want result you need to include the whole picture. I do think that you could organize this company very loosely and at low cost and then funding will come. Best Regards , Lennart Thornros www.StrategicLeadershipSac.com lenn...@thornros.com +1 916 436 1899 6140 Horseshoe Bar Road Suite G, Loomis CA 95650 Productivity is never an accident. It is always the result of a commitment to excellence, intelligent planning, and focused effort. PJM On Tue, Mar 25, 2014 at 1:25 PM, mix...@bigpond.com wrote: In reply to Bob Cook's message of Tue, 25 Mar 2014 10:49:12 -0700: Hi Bob, When a fast electron interacts with other electrons it does so through a repulsive force, and imparts energy to them through collisions, knocking them away from their host atoms, and leaving them with some kinetic energy. A positron should also tear electrons away from their host atoms and leave them with excess kinetic energy. The only difference being that the force will be a mixture of attractive (and repulsive?) forces. Attractive at a distance (and repulsive in a head on collision ?). A near miss would be attractive forces and a whip around (conservation of angular momentum). Perhaps a head on collision results in annihilation? AFAIK stands for As Far As I Know. Robin-- The positron leaves the Ni-59 nucleus after an electron capture with about 1 Mev of energy--the disintegration energy is a little more than 1 Mev. However, I have not seen a cross section for the reaction we are talking about. I would agree, if the positron acts like an electron in a population of electrons, that it would slow down, but being a positive charge I not sure how that effects the slowing down. (I think you suggest its positive charge does not change the slowing down process?) The fact that the resulting photons total energy equal 2 x the electron mass probably means there is no excess energy and momentum that needs to be handled in the reaction. I am not sure whether neutrinos in the annihilation reaction have been ruled out by experiment. Probably ruled out only by theory. By the way what does AFAIK stand for? Bob - Original Message - From: mix...@bigpond.com To: vortex-l@eskimo.com Sent: Monday, March 24, 2014 1:32 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' In reply to Bob Cook's message of Sun, 23 Mar 2014 15:19:14 -0700: Hi, [snip] Your description is exactly as I understand it. The random walk is not very long however, since it probably occurs at the first electron it attracts and that is pretty quick after the nucleus gives it up. AFAIK annihilation usually only happens after the positron has slowed down. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:My current views on the 'Rossi's process'
In reply to Bob Cook's message of Sun, 23 Mar 2014 15:19:14 -0700: Hi, [snip] Your description is exactly as I understand it. The random walk is not very long however, since it probably occurs at the first electron it attracts and that is pretty quick after the nucleus gives it up. AFAIK annihilation usually only happens after the positron has slowed down. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:My current views on the 'Rossi's process'
I lost that data in my memory indeed. Thanks for refreshing it. Do you have a public reference on fusion of odd count nuleons not fusing? I respect your knowledge but like to understand this a bit better. It looks like there are many variaties possible. Looks like much is depending on what type/size of NAE occurs locally. On Fri, Mar 21, 2014 at 4:53 PM, Axil Axil janap...@gmail.com wrote: You should take a look at the table 2 and table 3 element list from the DGT ICCF-17 document. http://cdn.coldfusionnow.org/wp-content/uploads/2013/04/2012-08-13-ICCF-17__Paper_DGTGx.pdf The is a large increase in very light elements and not much nickel to copper transmutation. This means that Cluster fusion of many nuclei including many protons and a heavy metal nucleus is occurring per fusion event. In the Rossi ash, iron was 10% of the element assay. *1H+1H+62Ni = 4He + 4He + 56Fe + 3.495 MeV this one produces iron.* Fusion cannot happen if the nucleon count is odd, e.g. Ni61. This indicates photofusion. Gamma Radiation is converted to huge magnetic fields and will result in EUV radiation from the eventual destruction of the EMF soliton that will be thermalized by election capture.
Re: [Vo]:My current views on the 'Rossi's process'
Eric, on the little info I could find in public domain, I understand that ß+ decay happens within the nucleus. Are you saying that there are quite some exceptions? On Sat, Mar 22, 2014 at 7:10 AM, Eric Walker eric.wal...@gmail.com wrote: On Fri, Mar 21, 2014 at 7:54 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: 1. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. I think the electron-positron annihilation photons from the radioactive decay of certain isotopes of nickel would escape the system. Since the mean free path of these photons is long, they would be unlikely to thermalize, unless some sort of 100 percent efficiency gamma thermalization mechanism is at play. (Only handfuls of gammas are typically seen.) Eric
Re: [Vo]:My current views on the 'Rossi's process'
I think a little mass lead, for example. will pretty well shield the 511 ev x-rays coming from Beta+/electron annihilation. Rossi had such shielding in his earlier E-Cat design. Bob - Original Message - From: Teslaalset To: vortex-l@eskimo.com Sent: Sunday, March 23, 2014 7:12 AM Subject: Re: [Vo]:My current views on the 'Rossi's process' Eric, on the little info I could find in public domain, I understand that ß+ decay happens within the nucleus. Are you saying that there are quite some exceptions? On Sat, Mar 22, 2014 at 7:10 AM, Eric Walker eric.wal...@gmail.com wrote: On Fri, Mar 21, 2014 at 7:54 AM, Teslaalset robbiehobbiesh...@gmail.com wrote: 1.. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. I think the electron-positron annihilation photons from the radioactive decay of certain isotopes of nickel would escape the system. Since the mean free path of these photons is long, they would be unlikely to thermalize, unless some sort of 100 percent efficiency gamma thermalization mechanism is at play. (Only handfuls of gammas are typically seen.) Eric
Re: [Vo]:My current views on the 'Rossi's process'
Eric-- I think the key to looking for beta+/beta- annihilation is to do co-incident counting in detectors looking for photons coming in opposite directions (from a single point) and to look for radiation of any wave length and above--say 50 ev. The annihilation photons come off back-to-back. I did this experiment on a radioactic sodium isotope that decays by beta+ emission to determine the rest mass (energy) of an electron in 1959. It was quite an accurate experiment. The photons will loose energy as they pass by atoms with electrons in any shield. The denser the material the quicker the loss of energy. Until they get to real low energies they mostly move in a straight line. Thus even shielding will not destroy the coincident events that detectors will record. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Friday, March 21, 2014 11:10 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' On Fri, Mar 21, 2014 at 7:54 AM, Teslaalset robbiehobbiesh...@gmail.com wrote: 1.. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. I think the electron-positron annihilation photons from the radioactive decay of certain isotopes of nickel would escape the system. Since the mean free path of these photons is long, they would be unlikely to thermalize, unless some sort of 100 percent efficiency gamma thermalization mechanism is at play. (Only handfuls of gammas are typically seen.) Eric
Re: [Vo]:My current views on the 'Rossi's process'
On Sun, Mar 23, 2014 at 7:12 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: Eric, on the little info I could find in public domain, I understand that ß+ decay happens within the nucleus. Are you saying that there are quite some exceptions? Perhaps Robin or Bob can correct me if I'm wrong, but I think the way beta-plus decay works is that the unstable nucleus emits a positron during the transition to the daughter. The positron does something of a random walk around the (extra-nuclear) environment until it encounters an electron, at which point you get the annihilation and resulting 511 keV photon pair (each going off in opposite directions). Eric
Re: [Vo]:My current views on the 'Rossi's process'
Your description is exactly as I understand it. The random walk is not very long however, since it probably occurs at the first electron it attracts and that is pretty quick after the nucleus gives it up. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Sunday, March 23, 2014 3:14 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' On Sun, Mar 23, 2014 at 7:12 AM, Teslaalset robbiehobbiesh...@gmail.com wrote: Eric, on the little info I could find in public domain, I understand that ß+ decay happens within the nucleus. Are you saying that there are quite some exceptions? Perhaps Robin or Bob can correct me if I'm wrong, but I think the way beta-plus decay works is that the unstable nucleus emits a positron during the transition to the daughter. The positron does something of a random walk around the (extra-nuclear) environment until it encounters an electron, at which point you get the annihilation and resulting 511 keV photon pair (each going off in opposite directions). Eric
Re: [Vo]:My current views on the 'Rossi's process'
On Sun, Mar 23, 2014 at 2:31 PM, Bob Cook frobertc...@hotmail.com wrote: Until they get to real low energies they mostly move in a straight line. Thus even shielding will not destroy the coincident events that detectors will record. Interesting; I didn't realize that. Somewhere I got the impression that the remission of a photon after a scattering with an electron would be in a random direction. Eric
Re: [Vo]:My current views on the 'Rossi's process'
To conserve momentum and spin they go off in opposite directions and polarized in opposite directions so that the net spin is 0 as well as the linear momentum of the two photons.. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Sunday, March 23, 2014 3:29 PM Subject: Re: [Vo]:My current views on the 'Rossi's process' On Sun, Mar 23, 2014 at 2:31 PM, Bob Cook frobertc...@hotmail.com wrote: Until they get to real low energies they mostly move in a straight line. Thus even shielding will not destroy the coincident events that detectors will record. Interesting; I didn't realize that. Somewhere I got the impression that the remission of a photon after a scattering with an electron would be in a random direction. Eric
Re: [Vo]:My current views on the 'Rossi's process'
On Fri, Mar 21, 2014 at 7:54 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: 1. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. I think the electron-positron annihilation photons from the radioactive decay of certain isotopes of nickel would escape the system. Since the mean free path of these photons is long, they would be unlikely to thermalize, unless some sort of 100 percent efficiency gamma thermalization mechanism is at play. (Only handfuls of gammas are typically seen.) Eric
[Vo]:My current views on the 'Rossi's process'
I've been reading quite some theories and views on what exactly Rossi's / Defkalion's processes might be. Here's my current view focussing on the main effects only. Comments and (dis)agreements are welcome: The main chain of fusions/transmutations is in my view: Ni58+p Cu59 + e- Ni59 +p Cu60 + e- Ni59 + p Cu60 + e- - - - - - Cu63 + e-. All Cu isotopes in the range of Cu59 - Cu62 have relative short half-life. The longest half-life is that of Cu61 (3.3 hours). This is why Rossi's process needs quite some time to shut down. The fusion/transmutation chain stops at Cu63 because Cu63 is stable with an extreem long half-life. Protons (p) are provided by (absorbed) Hydrogen ions. Electrons (e-) are released due to Vibrationally Promoted Electron Emission (VPEE). The released energy is caused by two sources: 1. The emitted electrons e- (with very high kinetic energy, 5 - 8 MeV); the electrons are absorbed by the reactor wall causing eddy currents that are converted into heat due to resistance of that wall material. Those eddy currents also may be the cause of the extreemly high magnetic fields that have been observed (Defkalion). 2. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. Some ballpark figures on the total energy generated and the amount of fuel involved: Assuming all the Nickel in the reactor in the form Ni58 and finally all transmutted into Cu63: Ni58 mass is calculated to be 57.95380± 15 amu. The actual mass of a copper-Cu63 nucleus is 62.91367 amu. Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu. Delta mass is 62.95380-62.91367=0.04013 amu. 1 amu = 931 MeV is used as a standard conversion 0.04013×931 MeV=37.36 MeV. So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. So, without further energy losses it requires 2 - 3 grams of Ni and approx. 0.2 grams of H2 to produce 10KW of heat over a 6 months period continuously.
Re: [Vo]:My current views on the 'Rossi's process'
You should take a look at the table 2 and table 3 element list from the DGT ICCF-17 document. http://cdn.coldfusionnow.org/wp-content/uploads/2013/04/2012-08-13-ICCF-17__Paper_DGTGx.pdf The is a large increase in very light elements and not much nickel to copper transmutation. This means that Cluster fusion of many nuclei including many protons and a heavy metal nucleus is occurring per fusion event. In the Rossi ash, iron was 10% of the element assay. *1H+1H+62Ni = 4He + 4He + 56Fe + 3.495 MeV this one produces iron.* Fusion cannot happen if the nucleon count is odd, e.g. Ni61. This indicates photofusion. Gamma Radiation is converted to huge magnetic fields and will result in EUV radiation from the eventual destruction of the EMF soliton that will be thermalized by election capture. On Fri, Mar 21, 2014 at 10:54 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: I've been reading quite some theories and views on what exactly Rossi's / Defkalion's processes might be. Here's my current view focussing on the main effects only. Comments and (dis)agreements are welcome: The main chain of fusions/transmutations is in my view: Ni58+p Cu59 + e- Ni59 +p Cu60 + e- Ni59 + p Cu60 + e- - - - - - Cu63 + e-. All Cu isotopes in the range of Cu59 - Cu62 have relative short half-life. The longest half-life is that of Cu61 (3.3 hours). This is why Rossi's process needs quite some time to shut down. The fusion/transmutation chain stops at Cu63 because Cu63 is stable with an extreem long half-life. Protons (p) are provided by (absorbed) Hydrogen ions. Electrons (e-) are released due to Vibrationally Promoted Electron Emission (VPEE). The released energy is caused by two sources: 1. The emitted electrons e- (with very high kinetic energy, 5 - 8 MeV); the electrons are absorbed by the reactor wall causing eddy currents that are converted into heat due to resistance of that wall material. Those eddy currents also may be the cause of the extreemly high magnetic fields that have been observed (Defkalion). 2. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. Some ballpark figures on the total energy generated and the amount of fuel involved: Assuming all the Nickel in the reactor in the form Ni58 and finally all transmutted into Cu63: Ni58 mass is calculated to be 57.95380± 15 amu. The actual mass of a copper-Cu63 nucleus is 62.91367 amu. Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu. Delta mass is 62.95380-62.91367=0.04013 amu. 1 amu = 931 MeV is used as a standard conversion 0.04013×931 MeV=37.36 MeV. So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. So, without further energy losses it requires 2 - 3 grams of Ni and approx. 0.2 grams of H2 to produce 10KW of heat over a 6 months period continuously.
Re: [Vo]:My current views on the 'Rossi's process'
What would be the testable predictions of your theory? What should we be looking for when someone tests a device and publishes data about it? On Fri, Mar 21, 2014 at 7:54 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: I've been reading quite some theories and views on what exactly Rossi's / Defkalion's processes might be. Here's my current view focussing on the main effects only. Comments and (dis)agreements are welcome: The main chain of fusions/transmutations is in my view: Ni58+p Cu59 + e- Ni59 +p Cu60 + e- Ni59 + p Cu60 + e- - - - - - Cu63 + e-. All Cu isotopes in the range of Cu59 - Cu62 have relative short half-life. The longest half-life is that of Cu61 (3.3 hours). This is why Rossi's process needs quite some time to shut down. The fusion/transmutation chain stops at Cu63 because Cu63 is stable with an extreem long half-life. Protons (p) are provided by (absorbed) Hydrogen ions. Electrons (e-) are released due to Vibrationally Promoted Electron Emission (VPEE). The released energy is caused by two sources: 1. The emitted electrons e- (with very high kinetic energy, 5 - 8 MeV); the electrons are absorbed by the reactor wall causing eddy currents that are converted into heat due to resistance of that wall material. Those eddy currents also may be the cause of the extreemly high magnetic fields that have been observed (Defkalion). 2. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. Some ballpark figures on the total energy generated and the amount of fuel involved: Assuming all the Nickel in the reactor in the form Ni58 and finally all transmutted into Cu63: Ni58 mass is calculated to be 57.95380± 15 amu. The actual mass of a copper-Cu63 nucleus is 62.91367 amu. Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu. Delta mass is 62.95380-62.91367=0.04013 amu. 1 amu = 931 MeV is used as a standard conversion 0.04013×931 MeV=37.36 MeV. So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. So, without further energy losses it requires 2 - 3 grams of Ni and approx. 0.2 grams of H2 to produce 10KW of heat over a 6 months period continuously.
Re: [Vo]:My current views on the 'Rossi's process'
That sorta goes to my point about looking for experimental results that either lend support or reduce support for a particular theory. I'm noticing that a lot of the experiments are veering towards testing nuclear products, which is going to be expensive. It won't matter much if Rossi is selling reactors, it soon becomes someone else's problem to properly theorize how it's happening. But it will matter a bunch if Rossi stalls and we need to know what's going on in order to get to production. On Fri, Mar 21, 2014 at 8:53 AM, Axil Axil janap...@gmail.com wrote: You should take a look at the table 2 and table 3 element list from the DGT ICCF-17 document. http://cdn.coldfusionnow.org/wp-content/uploads/2013/04/2012-08-13-ICCF-17__Paper_DGTGx.pdf The is a large increase in very light elements and not much nickel to copper transmutation. This means that Cluster fusion of many nuclei including many protons and a heavy metal nucleus is occurring per fusion event. In the Rossi ash, iron was 10% of the element assay. *1H+1H+62Ni = 4He + 4He + 56Fe + 3.495 MeV this one produces iron.* Fusion cannot happen if the nucleon count is odd, e.g. Ni61. This indicates photofusion. Gamma Radiation is converted to huge magnetic fields and will result in EUV radiation from the eventual destruction of the EMF soliton that will be thermalized by election capture. On Fri, Mar 21, 2014 at 10:54 AM, Teslaalset robbiehobbiesh...@gmail.comwrote: I've been reading quite some theories and views on what exactly Rossi's / Defkalion's processes might be. Here's my current view focussing on the main effects only. Comments and (dis)agreements are welcome: The main chain of fusions/transmutations is in my view: Ni58+p Cu59 + e- Ni59 +p Cu60 + e- Ni59 + p Cu60 + e- - - - - - Cu63 + e-. All Cu isotopes in the range of Cu59 - Cu62 have relative short half-life. The longest half-life is that of Cu61 (3.3 hours). This is why Rossi's process needs quite some time to shut down. The fusion/transmutation chain stops at Cu63 because Cu63 is stable with an extreem long half-life. Protons (p) are provided by (absorbed) Hydrogen ions. Electrons (e-) are released due to Vibrationally Promoted Electron Emission (VPEE). The released energy is caused by two sources: 1. The emitted electrons e- (with very high kinetic energy, 5 - 8 MeV); the electrons are absorbed by the reactor wall causing eddy currents that are converted into heat due to resistance of that wall material. Those eddy currents also may be the cause of the extreemly high magnetic fields that have been observed (Defkalion). 2. The ß+ decay energy of Cu(x) Ni(x) + e+ + ve (2 -4 MeV) of each decay step in the chain, causing the Ni/Cu powder to heat up. Some ballpark figures on the total energy generated and the amount of fuel involved: Assuming all the Nickel in the reactor in the form Ni58 and finally all transmutted into Cu63: Ni58 mass is calculated to be 57.95380± 15 amu. The actual mass of a copper-Cu63 nucleus is 62.91367 amu. Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu. Delta mass is 62.95380-62.91367=0.04013 amu. 1 amu = 931 MeV is used as a standard conversion 0.04013×931 MeV=37.36 MeV. So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. So, without further energy losses it requires 2 - 3 grams of Ni and approx. 0.2 grams of H2 to produce 10KW of heat over a 6 months period continuously.
