Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
Since the electromagnetic and the weak force are closely related I wonder…. If charge accumulation causes electromagnetic fields to build charge screening strong enough to completely lower the coulomb barrier, why is it safe to assume that the weak force will still be strong enough to operate the standard radioactive decay function. I wonder if the strong force will rearrange the nucleus during the fusion process in a way that does not consider any interaction or interference with the weak force therefore leading to the exclusive production of stable elements. If the electroweak force goes away, does the strong force change its nature? Do protons attract each other inside the nucleus the way electrons do in cooper pair formation as happens in superconductivity? The bottom line is, can we assume that our understanding of the way isotopes form and behave in cold fusion is the same as happens in nuclear physics? The lack of unstable isotopes in cold fusion might mean that cold fusion is not caused by neutrons but by charge screening. Cheers: Axil On Tue, Jun 12, 2012 at 11:18 PM, Eric Walker eric.wal...@gmail.com wrote: On Tue, Jun 12, 2012 at 2:12 AM, Alain Sepeda alain.sep...@gmail.comwrote: one hypothesis about normal isotopic ratio in transmuted copper, is that the result is the same as nature, because the process is the same... Larsen talk about R and S nucleosynthetis process, not so different from WL (or similar neutron or hydrino absorption) I think the rate of flux is an important variable. With a high flux, the isotope ratios that result after everything has settled will be different than with those after an anemic flux. It is possible that you would need a similar flux to what occurs during r-process nucleosynthesis in supernovae to get similar ratios. Such a flux is generally very high. But another variable here is the speed of the neutrons. I suppose those emanating from a supernova will be traveling very fast, and if you had much slower ones, the flux might not need to be high to get comparable ratios. Ed Storms brings up an excellent point about neutron-based explanations. Here is my elaboration: it is true that the neutron-capture cross section goes way down when the neutrons are very slow. But that's a relative change of what is normally measured at higher energies, and even with a hypothesized momentum near or at zero, the cross section will not be infinite. So there will be some elastic collisions with atoms in the environment, and some of the neutrons can be expected to thermalize and exit the system. You would then expect to see a substantial number of these be picked up in a detector, but this is not seen. Eric
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
Eric, Your statement below - it is true that the neutron-capture cross section goes way down when the neutrons are very slow. - is not strictly correct. Check the 'Atlas of Neutron Capture Cross Sections' web page at: http://nucleus.iaea.org/sso/NUCLEUS.html?exturl=http://www-nds.iaea.org/ngatlas2/ Click on, say, 'Ni' in the periodic table shown, then check the cross sections for various nickel isotopes. They can be thousands of 'barns' (i.e., the cross section of a uranium nucleus). Most Ni cross sections go up quickly as neutron kinetic energy declines. The statement that escaping neutrons should be found may be correct. However, the interaction of low energy neutrons with nanoparticles, or surfaces with nano-topography seems complicated, so I'm unsure. -- Lou Pagnucco Eric Walker wrote: Ed Storms brings up an excellent point about neutron-based explanations. Here is my elaboration: it is true that the neutron-capture cross section goes way down when the neutrons are very slow. But that's a relative change of what is normally measured at higher energies, and even with a hypothesized momentum near or at zero, the cross section will not be infinite. So there will be some elastic collisions with atoms in the environment, and some of the neutrons can be expected to thermalize and exit the system. You would then expect to see a substantial number of these be picked up in a detector, but this is not seen. Eric
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
On Wed, Jun 13, 2012 at 12:11 AM, pagnu...@htdconnect.com wrote: it is true that the neutron-capture cross section goes way down when the neutrons are very slow. Good catch -- a typo. I meant to say that the cross section increases, i.e., neutron capture becomes more likely. I've been flipping concepts lately; I did that with wavelength/frequency in an earlier post. It's nice to see some graphs of the change in cross section. The statement that escaping neutrons should be found may be correct. However, the interaction of low energy neutrons with nanoparticles, or surfaces with nano-topography seems complicated, so I'm unsure. Makes sense. Eric
RE: [Vo]: Nuclear Stability and Proton or Neutron Addition
From: Eric Walker it is true that the neutron-capture cross section goes way down when the neutrons are very slow. Good catch -- a typo. I meant to say that the cross section increases, i.e., neutron capture becomes more likely. I've been flipping concepts lately. You got it right the first time – at least with regard to many elements and especially the most important one. The study of neutrons at cryogenic temperature is well-known. And in fact nickel, in particular, is known to reflect cold neutrons, when it would otherwise absorb them if they were warmer. Are ULM neutrons somehow warm? LOL. In fact, nickel is often used in experimental cryogenic neutron devices because it does not absorb ultracold neutrons. See the table here: http://en.wikipedia.org/wiki/Ultracold_neutrons If you want to defend W-L, which is impossible, start by trying to imagine how “ultra low momentum” differs from “ultra cold” and then ask yourself how the authors and supporters of such a brain-dead theory can continue to overlook the 800 pound gorilla in the closet? attachment: winmail.dat
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
On Wed, Jun 13, 2012 at 5:57 AM, Jones Beene jone...@pacbell.net wrote: The study of neutrons at cryogenic temperature is well-known. And in fact nickel, in particular, is known to reflect cold neutrons, when it would otherwise absorb them if they were warmer. Are ULM neutrons somehow warm? LOL. According to the Wikipedia article, 58Ni (the most abundant natural isotope) has the highest neutron optical potential (likelihood of reflecting a neutron). That's really funny. Thanks for the pointer. The page that Lou mentioned indicates that certain isotopes of nickel have a neutron capture cross section in the range of 10^3 - 10^4 barns for very low energies, and the graphs increase as you go from right to left. But the ones for 58Ni and 60Ni have a sharp drop at around 10,000 neV -- is this due to the neutron optical potential? http://nucleus.iaea.org/sso/NUCLEUS.html?exturl=http://www-nds.iaea.org/ngatlas2/ 58Ni (68 percent) and 60Ni (26.2 percent) are the most abundant isotopes. The remaining 5.8 percent will be the others, all with relatively high cross sections. But what this seems to mean is that in order to have neutron-generated power and avoid seeing thermalized neutrons that would be picked up by a detector, you need next to no neutrons below the neutron optical potentials for 58Ni and 60Ni or you need a change to occur in the optical potential. Assuming the latter does not happen, there appears to be an energy threshold below which you will see neutrons in a detector -- 335 neV in the case of 58Ni. That seems to eliminate a convenient asymptotic trend to arbitrarily high cross sections that would be capable of making every neutron that is generated disappear before it leaves the system. Without a decent simulation or some number crunching, it is hard to know exactly how the numbers would turn out for different starting energies. But needless to say there's a prima facie case against slow neutrons. If you want to defend W-L, which is impossible, start by trying to imagine how “ultra low momentum” differs from “ultra cold” No desire on my part to defend W-L, in particular, although I don't have the scorn for them that some do. It's a luxury of my having no training in physics to not have the gag reflex that many more knowledgeable people do when they hear about W-L. I took physics 101 in college and did fine, but not physics 201, so most of this stuff is new to me. Can you elaborate on what you have in mind in contrasting ultra low momentum with ultra cold? and then ask yourself how the authors and supporters of such a brain-dead theory can continue to overlook the 800 pound gorilla in the closet? The plausibility of neutron-based explanations aside, I think there are many 800 pound gorillas. I'm distressed that my nano-Polywell reactor will probably have to deal with the usual branching ratio gorilla and with a prompt gamma to soft x-ray/extreme ultra violet conversion gorilla. I guess it partly comes down to which fantastic set of modifications to physics that one is willing to defend. Eric
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
Your question of whether Rossi or W-L are correct (if either is) made me want to check whether a cascade of neutron captures suggested by W-L were consistent with Rossi's reports. Using the data from the wiki-page on masses and half-lives for Ni-isotopes http://en.wikipedia.org/wiki/Isotopes_of_nickel -- it appears that the sequence of neutron captures are all exothermic - 58Ni | (8.2 Mev) 59Ni | (10.5 Mev) 60Ni | (7.0 Mev) 61Ni (energy released) | (9.8 Mev) 62Ni | (6.1 Mev) 63Ni | (8.9 Mev) 64Ni | (5.3 Mev) 65Ni -- 65Cu (stable) | (8.1 Mev) 66Ni -- 66Cu -- 66Zn (stable) (The ~780 Kev cost of electron capture is subracted from energy figures.) From 58Ni through 64Ni, half-lives are very long. After 66Ni, half-lives become too short to provide much transmutation ash. If we start with off-the-shelf Ni (normal isotopic mix), it looks like very little Cu63 would result, but there could be significant amounts of 65Cu and some 66Zn. The distribution of 58Ni-64Ni should show enhanced concentrations of the heavier Ni-isotopes. Rossi claims he use Ni enriched in 62Ni and 64Ni in the e-cat. Unless they have significantly larger cross-sections to capture low energy neutrons, the enrichment would probably not help if W-L neutrons are responsible. (The Lattice Energy website on Nickel-Seed LENR Networks that may have a more complete analysis than mine.) Rossi claims the e-cat LENR results from Ni-proton capture. -- Lou Pagnucco Dave Roberson wrote: I have been reviewing a table of nuclides in an attempt to make sense of the process suggested by WL proponents and those of Rossi. In the WL theory a neutron is formed by the combination of an electron and a proton with the .78 MeV of energy being supplied by their process. This neutron then finds its way into a nucleus of nickel in this version of devices and energy is released. The final result is the next heavier isotope of nickel plus a significant amount of energy. The Rossi process involves the insertion of a proton into the nucleus of the subject nickel atom forming a new copper atom along with release of energy. Some of the copper isotopes formed by addition of a proton into their parent nickel isotopes decay by beta plus action into the next heavier nickel isotope along with a release of additional energy. The above two paragraphs offer an extremely brief description of the two theories. They are not intended to get into details which can be located within many documents. My purpose for writing this document is to reveal an interesting observation that I have made concerning the two processes. This may be well known to many of the people on the list, but it is new to me and I offer it as a refresher. If you take any stable isotope of an element, for example nickel 60 and either add a neutron as with the WL process or overcome the Coulomb barrier by forcing a proton into the nucleus you find an interesting result. In virtually every case only one of these processes leads to a stable isotope in a single reaction. There are only a couple of exceptions to this observation and that appears to be when neither process results in a single step stable new atom. Of course the newly created atoms will all eventually decay in steps until a stable result is obtained. I further notice that the end result of the two processes is the same nuclide. An example is as follows: Start with Ni60 and add a proton to it by forcing the particle against the Coulomb barrier and you obtain Cu61. Some immediate energy is released by the new element and at a half life later a Beta Plus decay process occurs which releases more energy. The Beta Plus decay leaves us with Ni61. The energy release is composed of two parts as we progress from Ni60 to Ni61. Now, instead of adding a proton, letâs allow a neutron to encounter the Ni60 nucleus. In this case a stable isotope of nickel Ni61 is directly formed and a significant amount of energy is released. I followed both of these processes through several different elements and can state that the same total energy is released regardless of the path taken when I start with an isotope of an element and end at the same final product. I consider this an important and useful observation. A second issue I would like to discuss is also interesting and leads to some neat results. The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be. Or, since Ni61 is stable, then Cu61 is unstable. This appears to apply throughout the entire list of elements and I would appreciate it for others to verify this conclusion. I have a couple of additional concepts that I plan to present at a later time, so for now review what I have observed and please make relevant comments. Dave
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
I won't be surprised if he use raw nickel, and pretend it enriched just as he say COP is limited to 6, that gamma heat the lead... to fool the pursuers... by the way the idea that only the beginning and end of the reaction account for energy release is normal. We just have to remind about the neutrino kinetic energy, but it is small (linked to conservation of momentum, and thus initial and final momentum of other particles, that we should notice if huge). one hypothesis about normal isotopic ratio in transmuted copper, is that the result is the same as nature, because the process is the same... Larsen talk about R and S nucleosynthetis process, not so different from WL (or similar neutron or hydrino absorption) another hypothesis is manipulation or mistake. With Rossi this is not to exclude. about guessing the theory, the results of Iwamura give strong data. in fact not the results (which just show nucleon+nucleus reaction), but the initial condition which are very controlled and simpler compared to SPAWAR, FP, Celani, Piantelli... Some should analyse if they match ES crack theory, WL surface theory, Brillouin bulk Q-wave theory, Takahashi bulk TSC theory, Kim zubarev theory. if we can divide by 2 the number of theories it should be good. This experiment was very controlled, so it should eliminate some hypothesis. 2012/6/12 pagnu...@htdconnect.com Your question of whether Rossi or W-L are correct (if either is) made me want to check whether a cascade of neutron captures suggested by W-L were consistent with Rossi's reports. Using the data from the wiki-page on masses and half-lives for Ni-isotopes http://en.wikipedia.org/wiki/Isotopes_of_nickel -- it appears that the sequence of neutron captures are all exothermic - 58Ni | (8.2 Mev) 59Ni | (10.5 Mev) 60Ni | (7.0 Mev) 61Ni (energy released) | (9.8 Mev) 62Ni | (6.1 Mev) 63Ni | (8.9 Mev) 64Ni | (5.3 Mev) 65Ni -- 65Cu (stable) | (8.1 Mev) 66Ni -- 66Cu -- 66Zn (stable) (The ~780 Kev cost of electron capture is subracted from energy figures.) From 58Ni through 64Ni, half-lives are very long. After 66Ni, half-lives become too short to provide much transmutation ash. If we start with off-the-shelf Ni (normal isotopic mix), it looks like very little Cu63 would result, but there could be significant amounts of 65Cu and some 66Zn. The distribution of 58Ni-64Ni should show enhanced concentrations of the heavier Ni-isotopes. Rossi claims he use Ni enriched in 62Ni and 64Ni in the e-cat. Unless they have significantly larger cross-sections to capture low energy neutrons, the enrichment would probably not help if W-L neutrons are responsible. (The Lattice Energy website on Nickel-Seed LENR Networks that may have a more complete analysis than mine.) Rossi claims the e-cat LENR results from Ni-proton capture. -- Lou Pagnucco Dave Roberson wrote: I have been reviewing a table of nuclides in an attempt to make sense of the process suggested by WL proponents and those of Rossi. In the WL theory a neutron is formed by the combination of an electron and a proton with the .78 MeV of energy being supplied by their process. This neutron then finds its way into a nucleus of nickel in this version of devices and energy is released. The final result is the next heavier isotope of nickel plus a significant amount of energy. The Rossi process involves the insertion of a proton into the nucleus of the subject nickel atom forming a new copper atom along with release of energy. Some of the copper isotopes formed by addition of a proton into their parent nickel isotopes decay by beta plus action into the next heavier nickel isotope along with a release of additional energy. The above two paragraphs offer an extremely brief description of the two theories. They are not intended to get into details which can be located within many documents. My purpose for writing this document is to reveal an interesting observation that I have made concerning the two processes. This may be well known to many of the people on the list, but it is new to me and I offer it as a refresher. If you take any stable isotope of an element, for example nickel 60 and either add a neutron as with the WL process or overcome the Coulomb barrier by forcing a proton into the nucleus you find an interesting result. In virtually every case only one of these processes leads to a stable isotope in a single reaction. There are only a couple of exceptions to this observation and that appears to be when neither process results in a single step stable new atom. Of course the newly created atoms will all eventually decay in steps until a stable result is obtained. I further notice that the end result of the two processes is the same nuclide. An example is as follows: Start with Ni60 and add a proton to it by forcing the particle against the Coulomb
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
On Tue, Jun 12, 2012 at 2:12 AM, Alain Sepeda alain.sep...@gmail.comwrote: one hypothesis about normal isotopic ratio in transmuted copper, is that the result is the same as nature, because the process is the same... Larsen talk about R and S nucleosynthetis process, not so different from WL (or similar neutron or hydrino absorption) I think the rate of flux is an important variable. With a high flux, the isotope ratios that result after everything has settled will be different than with those after an anemic flux. It is possible that you would need a similar flux to what occurs during r-process nucleosynthesis in supernovae to get similar ratios. Such a flux is generally very high. But another variable here is the speed of the neutrons. I suppose those emanating from a supernova will be traveling very fast, and if you had much slower ones, the flux might not need to be high to get comparable ratios. Ed Storms brings up an excellent point about neutron-based explanations. Here is my elaboration: it is true that the neutron-capture cross section goes way down when the neutrons are very slow. But that's a relative change of what is normally measured at higher energies, and even with a hypothesized momentum near or at zero, the cross section will not be infinite. So there will be some elastic collisions with atoms in the environment, and some of the neutrons can be expected to thermalize and exit the system. You would then expect to see a substantial number of these be picked up in a detector, but this is not seen. Eric
[Vo]: Nuclear Stability and Proton or Neutron Addition
I have been reviewing a table of nuclides in an attempt to make sense of the process suggested by WL proponents and those of Rossi. In the WL theory a neutron is formed by the combination of an electron and a proton with the .78 MeV of energy being supplied by their process. This neutron then finds its way into a nucleus of nickel in this version of devices and energy is released. The final result is the next heavier isotope of nickel plus a significant amount of energy. The Rossi process involves the insertion of a proton into the nucleus of the subject nickel atom forming a new copper atom along with release of energy. Some of the copper isotopes formed by addition of a proton into their parent nickel isotopes decay by beta plus action into the next heavier nickel isotope along with a release of additional energy. The above two paragraphs offer an extremely brief description of the two theories. They are not intended to get into details which can be located within many documents. My purpose for writing this document is to reveal an interesting observation that I have made concerning the two processes. This may be well known to many of the people on the list, but it is new to me and I offer it as a refresher. If you take any stable isotope of an element, for example nickel 60 and either add a neutron as with the WL process or overcome the Coulomb barrier by forcing a proton into the nucleus you find an interesting result. In virtually every case only one of these processes leads to a stable isotope in a single reaction. There are only a couple of exceptions to this observation and that appears to be when neither process results in a single step stable new atom. Of course the newly created atoms will all eventually decay in steps until a stable result is obtained. I further notice that the end result of the two processes is the same nuclide. An example is as follows: Start with Ni60 and add a proton to it by forcing the particle against the Coulomb barrier and you obtain Cu61. Some immediate energy is released by the new element and at a half life later a Beta Plus decay process occurs which releases more energy. The Beta Plus decay leaves us with Ni61. The energy release is composed of two parts as we progress from Ni60 to Ni61. Now, instead of adding a proton, let’s allow a neutron to encounter the Ni60 nucleus. In this case a stable isotope of nickel Ni61 is directly formed and a significant amount of energy is released. I followed both of these processes through several different elements and can state that the same total energy is released regardless of the path taken when I start with an isotope of an element and end at the same final product. I consider this an important and useful observation. A second issue I would like to discuss is also interesting and leads to some neat results. The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be. Or, since Ni61 is stable, then Cu61 is unstable. This appears to apply throughout the entire list of elements and I would appreciate it for others to verify this conclusion. I have a couple of additional concepts that I plan to present at a later time, so for now review what I have observed and please make relevant comments. Dave
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
Do you think that a double proton reaction is possible as per Kim http://www.freerepublic.com/focus/f-chat/2746057/posts I offer this possibility because Piantelli has seen protons at an energy of 6 MeV emanate from the nickel bars that he uses in his reactor when they are placed in a cloud chamber after they are removed from his reactor immediately after use. Cheers: Axil On Mon, Jun 11, 2012 at 2:49 PM, David Roberson dlrober...@aol.com wrote: I have been reviewing a table of nuclides in an attempt to make sense of the process suggested by WL proponents and those of Rossi. In the WL theory a neutron is formed by the combination of an electron and a proton with the .78 MeV of energy being supplied by their process. This neutron then finds its way into a nucleus of nickel in this version of devices and energy is released. The final result is the next heavier isotope of nickel plus a significant amount of energy. The Rossi process involves the insertion of a proton into the nucleus of the subject nickel atom forming a new copper atom along with release of energy. Some of the copper isotopes formed by addition of a proton into their parent nickel isotopes decay by beta plus action into the next heavier nickel isotope along with a release of additional energy. The above two paragraphs offer an extremely brief description of the two theories. They are not intended to get into details which can be located within many documents. My purpose for writing this document is to reveal an interesting observation that I have made concerning the two processes. This may be well known to many of the people on the list, but it is new to me and I offer it as a refresher. If you take any stable isotope of an element, for example nickel 60 and either add a neutron as with the WL process or overcome the Coulomb barrier by forcing a proton into the nucleus you find an interesting result. In virtually every case only one of these processes leads to a stable isotope in a single reaction. There are only a couple of exceptions to this observation and that appears to be when neither process results in a single step stable new atom. Of course the newly created atoms will all eventually decay in steps until a stable result is obtained. I further notice that the end result of the two processes is the same nuclide. An example is as follows: Start with Ni60 and add a proton to it by forcing the particle against the Coulomb barrier and you obtain Cu61. Some immediate energy is released by the new element and at a half life later a Beta Plus decay process occurs which releases more energy. The Beta Plus decay leaves us with Ni61. The energy release is composed of two parts as we progress from Ni60 to Ni61. Now, instead of adding a proton, let’s allow a neutron to encounter the Ni60 nucleus. In this case a stable isotope of nickel Ni61 is directly formed and a significant amount of energy is released. I followed both of these processes through several different elements and can state that the same total energy is released regardless of the path taken when I start with an isotope of an element and end at the same final product. I consider this an important and useful observation. A second issue I would like to discuss is also interesting and leads to some neat results. The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be. Or, since Ni61 is stable, then Cu61 is unstable. This appears to apply throughout the entire list of elements and I would appreciate it for others to verify this conclusion. I have a couple of additional concepts that I plan to present at a later time, so for now review what I have observed and please make relevant comments. Dave
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
In reply to David Roberson's message of Mon, 11 Jun 2012 14:49:06 -0400 (EDT): Hi, [snip] Now, instead of adding a proton, lets allow a neutron to encounter the Ni60 nucleus. In this case a stable isotope of nickel Ni61 is directly formed and a significant amount of energy is released. I followed both of these processes through several different elements and can state that the same total energy is released regardless of the path taken when I start with an isotope of an element and end at the same final product. I consider this an important and useful observation. This is only true if you include the 0.782 MeV required to convert a Hydrogen atom to a neutron (and is simply a restatement of the conservation of mass-energy; hardly surprising). Otherwise you end up with a 0.782 MeV discrepancy due to starting with H in one case and a neutron in the other case. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
I think that a double proton reaction is possible for a brief period of time until a beta plus decay occurs bringing stability. The energy released can be easily determined by looking at the energy released when a neutron is added to H1. This is a common energy level at 2.22402 MeV. My hypothesis is that there would be a two part energy release. The first would occur immediately after the protons joined and the second would be at the beta decay time. A positron and neutrino would escape during the second release. This concept is based upon the behavior of numerous different elements which hopefully continues in this unique condition. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jun 11, 2012 3:47 pm Subject: Re: [Vo]: Nuclear Stability and Proton or Neutron Addition Do you think that a double proton reaction is possible as per Kim http://www.freerepublic.com/focus/f-chat/2746057/posts I offer this possibility because Piantelli has seen protons at an energy of 6 MeV emanate from the nickel bars that he uses in his reactor when they are placed in a cloud chamber after they are removed from his reactor immediately after use. Cheers: Axil On Mon, Jun 11, 2012 at 2:49 PM, David Roberson dlrober...@aol.com wrote: I have been reviewing a table of nuclides in an attempt to make sense of the process suggested by WL proponents and those of Rossi. In the WL theory a neutron is formed by the combination of an electron and a proton with the .78 MeV of energy being supplied by their process. This neutron then finds its way into a nucleus of nickel in this version of devices and energy is released. The final result is the next heavier isotope of nickel plus a significant amount of energy. The Rossi process involves the insertion of a proton into the nucleus of the subject nickel atom forming a new copper atom along with release of energy. Some of the copper isotopes formed by addition of a proton into their parent nickel isotopes decay by beta plus action into the next heavier nickel isotope along with a release of additional energy. The above two paragraphs offer an extremely brief description of the two theories. They are not intended to get into details which can be located within many documents. My purpose for writing this document is to reveal an interesting observation that I have made concerning the two processes. This may be well known to many of the people on the list, but it is new to me and I offer it as a refresher. If you take any stable isotope of an element, for example nickel 60 and either add a neutron as with the WL process or overcome the Coulomb barrier by forcing a proton into the nucleus you find an interesting result. In virtually every case only one of these processes leads to a stable isotope in a single reaction. There are only a couple of exceptions to this observation and that appears to be when neither process results in a single step stable new atom. Of course the newly created atoms will all eventually decay in steps until a stable result is obtained. I further notice that the end result of the two processes is the same nuclide. An example is as follows: Start with Ni60 and add a proton to it by forcing the particle against the Coulomb barrier and you obtain Cu61. Some immediate energy is released by the new element and at a half life later a Beta Plus decay process occurs which releases more energy. The Beta Plus decay leaves us with Ni61. The energy release is composed of two parts as we progress from Ni60 to Ni61. Now, instead of adding a proton, let’s allow a neutron to encounter the Ni60 nucleus. In this case a stable isotope of nickel Ni61 is directly formed and a significant amount of energy is released. I followed both of these processes through several different elements and can state that the same total energy is released regardless of the path taken when I start with an isotope of an element and end at the same final product. I consider this an important and useful observation. A second issue I would like to discuss is also interesting and leads to some neat results. The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be. Or, since Ni61 is stable, then Cu61 is unstable. This appears to apply throughout the entire list of elements and I would appreciate it for others to verify this conclusion. I have a couple of additional concepts that I plan to present at a later time, so for now review what I have observed and please make relevant comments. Dave
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
Yes, that is true. I probably should have included that addition in my document to clarify the complete reaction. I was starting with the individual proton and an electron plus the isotope. The neutron is of course constructed by taking the proton and electron and adding that amount of energy. This neutron construction energy must be removed from the final reaction in order to step back to the initial equal conditions. Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jun 11, 2012 6:02 pm Subject: Re: [Vo]: Nuclear Stability and Proton or Neutron Addition In reply to David Roberson's message of Mon, 11 Jun 2012 14:49:06 -0400 (EDT): i, snip] Now, instead of adding a proton, lets allow a neutron to encounter the Ni60 ucleus. In this case a stable isotope of nickel Ni61 is directly formed and a ignificant amount of energy is released. I followed both of these processes through several different elements and can tate that the same total energy is released regardless of the path taken when I tart with an isotope of an element and end at the same final product. I onsider this an important and useful observation. This is only true if you include the 0.782 MeV required to convert a Hydrogen tom to a neutron (and is simply a restatement of the conservation of ass-energy; hardly surprising). Otherwise you end up with a 0.782 MeV iscrepancy due to starting with H in one case and a neutron in the other case. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]: Nuclear Stability and Proton or Neutron Addition
From: David Roberson The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be The possible exception is 40 nucleons. The “impossibility” depends on how precise you wording is. Is 10 billion years “stable”? If so, there is one exception. 40Ar is 99+ of all argon. Argon is element 18. Element 19 is potassium. 40K is radioactive, but with an extremely long half-life, over one billion years, so there is still primordial potassium on earth, in the natural mineral, and there will be a diminishing amount for billions of years in the future. In fact, there should be some primordial potassium here when out sun expires. That is relatively stable. So, to that extent 40K is both stable but radioactive. Of course, you can define “stable” to be “non-radioactive” but then you must take note that some grand unification theories (including the Sheldon Glashow original) predict that even the proton will decay eventually – making all mater radioactive in a long time enough time scale so that there are no stable and non-radioactive elements. Semantics is a bitch but 40 is magic. Which is to say that 40 is a magic number for nucleons – so much so that calcium, element 20 also has a stable 40 nucleon isotope. Bottom line there are three adjoining elements in the periodic table which all have isotopes of 40 a.m.u. and all retain at least an important percentage of that magic numbered isotope - from when our solar system formed 4-5 billion years ago.
Re: [Vo]: Nuclear Stability and Proton or Neutron Addition
I understand what you are saying about the 40 nucleon stability line and it follows my analysis as stated. The Potassium isotope does decay into Ar40 or Ca40 with a half life of 1.248 Billion years. Decay half life is relative and even though this represents a long period of time, it is still decaying as my hypothesis suggests. If you take Potassium 39 and add a neutron to it as with a WL reaction you obtain Potassium 40. This isotope then will decay into either Ar40 or Ca40 according to the chart I have. On the other hand, if I take Potassium 39 and add a proton and electron then I obtain Ca40 which is totally stable. The 40 nucleon stability region comes close to violating the rule but does not quite make it. Are you aware of any violations that meet the absolute criteria? Any form of radioactivity would not qualify an element as totally stable. I appreciate your finding this close call and there are others where the suspect isotope falls inside a parallel set of vertical lines of stability with proton addition. Cl36 is located in a similar position and has a half life of 300 k years. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Mon, Jun 11, 2012 10:27 pm Subject: RE: [Vo]: Nuclear Stability and Proton or Neutron Addition From: David Roberson The above rule that I found makes it impossible to have two stable isotopes of elements with the same number of nucleons that are one level apart. An example of this rule would be that since He3 is stable, then H3 cannot be The possible exception is 40 nucleons. The “impossibility” depends on how precise you wording is. Is 10 billion years “stable”? If so, there is one exception. 40Ar is 99+ of all argon. Argon is element 18. Element 19 is potassium. 40K is radioactive, but with an extremely long half-life, over one billion years, so there is still primordial potassium on earth, in the natural mineral, and there will be a diminishing amount for billions of years in the future. In fact, there should be some primordial potassium here when out sun expires. That is relatively stable. So, to that extent 40K is both stable but radioactive. Of course, you can define “stable” to be “non-radioactive” but then you must take note that some grand unification theories (including the Sheldon Glashow original) predict that even the proton will decay eventually – making all mater radioactive in a long time enough time scale so that there are no stable and non-radioactive elements. Semantics is a bitch but 40 is magic. Which is to say that 40 is a magic number for nucleons – so much so that calcium, element 20 also has a stable 40 nucleon isotope. Bottom line there are three adjoining elements in the periodic table which all have isotopes of 40 a.m.u. and all retain at least an important percentage of that magic numbered isotope - from when our solar system formed 4-5 billion years ago.
