Re: [Vo]:swellings protons and fusion
On Mon, Jun 17, 2013 at 6:48 PM, mix...@bigpond.com wrote: I calculate them. If you have access to a windows machine, you can download the program from http://rvanspaa.freehostia.com/Isotopes.zip (if it doesn't work, please let me know ;) (Elements beyond Uranium are excluded). Sorry, It didn't work. First there is an Unhandled exception error, then I choose continue and it doesn't work. harry
Re: [Vo]:swellings protons and fusion
Ni62 + H -- Cu63 63Cu is a stable copper isotope Pd106 + H -- Ag107 Pd108 + H -- Ag109 107Ag and 109Ag are stable silver isotopes. Standard binding theory says these should be endothermic reactions but if the charge radius of the proton is not fixed then binding theory may need to be revised too. Harry On Mon, Jun 17, 2013 at 1:42 AM, Axil Axil janap...@gmail.com wrote: This is a global causation theory that applies to all types of nuclei equally. This idea does not address LENR reactions in only nuclei with an even number of nucleons: Ni58, Ni60, Ni62, Ni64. It also does not explain how no radioactive isotopes are produced. On Mon, Jun 17, 2013 at 1:31 AM, Harry Veeder hveeder...@gmail.comwrote: Recent evidence indicates the charge radius of proton is smaller in the presence of a muon than in the presence of an electron. http://en.wikipedia.org/wiki/Proton#Charge_radius Since a muon is a massive cousin of the electron, a muon's orbit is much smaller than an electron and therefore its orbital speed is much greater. This could mean that the charge radius of proton might depend inversely on the speed of negatively charged particles in the proton's neighborhood. If so, then protons (hydrogen ions) will tend to swell even more inside a lattice because they are bathed by even slower moving electrons. If the swelling also extends the reach of the strong force, this might enable protons to fuse with the lattice nuclei. Harry
Re: [Vo]:swellings protons and fusion
I am interested in what goes on inside those nuclei. How do those expanding protons effect the weak force? On Mon, Jun 17, 2013 at 2:12 AM, Harry Veeder hveeder...@gmail.com wrote: Ni62 + H -- Cu63 63Cu is a stable copper isotope Pd106 + H -- Ag107 Pd108 + H -- Ag109 107Ag and 109Ag are stable silver isotopes. Standard binding theory says these should be endothermic reactions but if the charge radius of the proton is not fixed then binding theory may need to be revised too. Harry On Mon, Jun 17, 2013 at 1:42 AM, Axil Axil janap...@gmail.com wrote: This is a global causation theory that applies to all types of nuclei equally. This idea does not address LENR reactions in only nuclei with an even number of nucleons: Ni58, Ni60, Ni62, Ni64. It also does not explain how no radioactive isotopes are produced. On Mon, Jun 17, 2013 at 1:31 AM, Harry Veeder hveeder...@gmail.comwrote: Recent evidence indicates the charge radius of proton is smaller in the presence of a muon than in the presence of an electron. http://en.wikipedia.org/wiki/Proton#Charge_radius Since a muon is a massive cousin of the electron, a muon's orbit is much smaller than an electron and therefore its orbital speed is much greater. This could mean that the charge radius of proton might depend inversely on the speed of negatively charged particles in the proton's neighborhood. If so, then protons (hydrogen ions) will tend to swell even more inside a lattice because they are bathed by even slower moving electrons. If the swelling also extends the reach of the strong force, this might enable protons to fuse with the lattice nuclei. Harry
Re: [Vo]:swellings protons and fusion
In reply to Harry Veeder's message of Mon, 17 Jun 2013 02:12:21 -0400: Hi, [snip] Ni62 + H -- Cu63 63Cu is a stable copper isotope Pd106 + H -- Ag107 1H+106Pd = 107Ag + 5.788 MeV (exothermic) Pd108 + H -- Ag109 1H+108Pd = 109Ag + 6.488 MeV (exothermic) 107Ag and 109Ag are stable silver isotopes. Standard binding theory says these should be endothermic reactions but if the charge radius of the proton is not fixed then binding theory may need to be revised too. Harry [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:swellings protons and fusion
On Mon, Jun 17, 2013 at 3:16 AM, mix...@bigpond.com wrote: In reply to Harry Veeder's message of Mon, 17 Jun 2013 02:12:21 -0400: Hi, [snip] Ni62 + H -- Cu63 63Cu is a stable copper isotope Pd106 + H -- Ag107 1H+106Pd = 107Ag + 5.788 MeV (exothermic) Pd108 + H -- Ag109 1H+108Pd = 109Ag + 6.488 MeV (exothermic) 107Ag and 109Ag are stable silver isotopes. Standard binding theory says these should be endothermic reactions but if the charge radius of the proton is not fixed then binding theory may need to be revised too. Harry [snip] Regards, Robin van Spaandonk Robin, Do you look these values up in a table or do you calculate them? I thought the addition of protons to any element above Ni62 will not release heat, or does it depend on the isotopes involved? On this Italian video about Rossi http://www.classmeteo.it/web/portale/video/prometeo-fusione-fredda-e-cat-e-lenergia-del-futuro/ @5:27 it says: 62Ni + H -- 63Cu + 6.12 MeV I thought this was incorrect according to theory but is it correct? Harry
Re: [Vo]:swellings protons and fusion
Axil, I don't know. Perhaps it affects a neutron's susceptibility to decay. Harry On Mon, Jun 17, 2013 at 2:25 AM, Axil Axil janap...@gmail.com wrote: I am interested in what goes on inside those nuclei. How do those expanding protons effect the weak force? On Mon, Jun 17, 2013 at 2:12 AM, Harry Veeder hveeder...@gmail.comwrote: Ni62 + H -- Cu63 63Cu is a stable copper isotope Pd106 + H -- Ag107 Pd108 + H -- Ag109 107Ag and 109Ag are stable silver isotopes. Standard binding theory says these should be endothermic reactions but if the charge radius of the proton is not fixed then binding theory may need to be revised too. Harry On Mon, Jun 17, 2013 at 1:42 AM, Axil Axil janap...@gmail.com wrote: This is a global causation theory that applies to all types of nuclei equally. This idea does not address LENR reactions in only nuclei with an even number of nucleons: Ni58, Ni60, Ni62, Ni64. It also does not explain how no radioactive isotopes are produced. On Mon, Jun 17, 2013 at 1:31 AM, Harry Veeder hveeder...@gmail.comwrote: Recent evidence indicates the charge radius of proton is smaller in the presence of a muon than in the presence of an electron. http://en.wikipedia.org/wiki/Proton#Charge_radius Since a muon is a massive cousin of the electron, a muon's orbit is much smaller than an electron and therefore its orbital speed is much greater. This could mean that the charge radius of proton might depend inversely on the speed of negatively charged particles in the proton's neighborhood. If so, then protons (hydrogen ions) will tend to swell even more inside a lattice because they are bathed by even slower moving electrons. If the swelling also extends the reach of the strong force, this might enable protons to fuse with the lattice nuclei. Harry
Re: [Vo]:swellings protons and fusion
In reply to Harry Veeder's message of Mon, 17 Jun 2013 16:54:06 -0400: Hi Harry, [snip] Robin, Do you look these values up in a table or do you calculate them? I thought the addition of protons to any element above Ni62 will not release heat, or does it depend on the isotopes involved? I calculate them. If you have access to a windows machine, you can download the program from http://rvanspaa.freehostia.com/Isotopes.zip (if it doesn't work, please let me know ;) (Elements beyond Uranium are excluded). You will get heat from adding a proton to almost any element in the periodic table. Though obviously more from some than from others. While the per nucleon binding energy is at a maximum for Ni62, the mass excess of a free proton is so large compared to that of a bound proton that it more than makes up for the slight decrease in binding energy that gradually builds above Ni62. Take a look at the graph at the bottom of this page:- http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html A lone proton has a binding energy of zero (i.e. it's on the X-axis of the graph). So there is always lots of energy released when it moves to another point on the line (by binding with the nucleus that is just to the left of the target position). The energy release is equal to the vertical difference on the graph as measured in MeV on the left hand scale corrected by the total change in mass of the other nucleons that were already present as they change from the original isotope to the new one. (Since a free proton is zero, you can just read the value of the target off the scale on the left, i.e. target - 0 = target.) The correction factor is usually positive below 62Ni, and usually negative above 62Ni, however it still works, even for some of the heaviest nuclei e.g. H + 238U = 239Np + 5.29 MeV On this Italian video about Rossi http://www.classmeteo.it/web/portale/video/prometeo-fusione-fredda-e-cat-e-lenergia-del-futuro/ @5:27 it says: 62Ni + H -- 63Cu + 6.12 MeV I thought this was incorrect according to theory but is it correct? The calculation is correct. However most physicists today will tell you that the reaction is impossible because 62Ni has a charge of 28 and strongly repels a lone proton. They don't take into account the possibility that the proton may be accompanied by an electron. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:swellings protons and fusion
This is a global causation theory that applies to all types of nuclei equally. This idea does not address LENR reactions in only nuclei with an even number of nucleons: Ni58, Ni60, Ni62, Ni64. It also does not explain how no radioactive isotopes are produced. On Mon, Jun 17, 2013 at 1:31 AM, Harry Veeder hveeder...@gmail.com wrote: Recent evidence indicates the charge radius of proton is smaller in the presence of a muon than in the presence of an electron. http://en.wikipedia.org/wiki/Proton#Charge_radius Since a muon is a massive cousin of the electron, a muon's orbit is much smaller than an electron and therefore its orbital speed is much greater. This could mean that the charge radius of proton might depend inversely on the speed of negatively charged particles in the proton's neighborhood. If so, then protons (hydrogen ions) will tend to swell even more inside a lattice because they are bathed by even slower moving electrons. If the swelling also extends the reach of the strong force, this might enable protons to fuse with the lattice nuclei. Harry
