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The equation provided is not a good one, mostly because it ends up weighing more than the starting mass, while the remainder of the topic speaks of it as if mass was "lost" in the process of the reaction. Just reverse the equation to simplify things.
Hello Patrick, I think that should be that one particle is impossible for an exothermic reaction, and has very low probability for an endothermic reaction.
Consider the reference frame in which, before the collision, all the incoming particles' momenta sum to zero (i.e., the "zero momentum frame"). By conservation of momentum, they still must do so after the collision. But if we have zero momentum and only one particle, that therefore means it must be stationary in this frame. But then what about conservation of energy? Since the system initially had kinetic energy but now has zero kinetic energy, we can only balance energy if the reaction is endothermic, with the kinetic energy of the incoming particles in this frame exactly equal to the energy absorbance - which in turn means that given the position and velocity of one particle, for such a "sticking" reaction to occur the position and velocity of the other particle must exactly equal a calculated value. If the energy absorbance of the reaction was precisely defined, this would have zero probability, but if it has some width there will be a small probability of such a reaction. (Of course, on the macroscopic scale our absorbance line can be extremely broad, so such a collision is much more likely to occur with macroscopic particles.) However as soon as we have at least two particles, we can balance momentum, kinetic energy and reaction energy, of any amounts, in any frame. Hence, an exothermic reaction must have at least two products, an endothermic reaction might in principle have only one but in practice rarely if ever does. Or something like that; it's late here. Securiger 16:30, 22 Nov 2004 (UTC)
Why is that Nuclear reaction is one of the high-importance articles while Nuclear fusion, which is just one reaction type, is a top one? -- Crwx ( talk) 16:16, 21 April 2012 (UTC)
Nuclear reactions should transmute the nuclei into different species. The article in general is rather clear on this point, but under Notable Types it has inelastic scattering. This is decidedly not a nuclear reaction. Unfortunately, the problem is that the main article does not even mention nuclear physics at all! So, I'm not sure what to do about this. Most likely the best case is to move all the scattering material to the scattering page. I do think, however, that some kind of note somewhere in the article that elastic and inelastic scattering are not nuclear reactions, with links to the respective articles, is a good idea. The elastic scattering page also does not mention nuclear reactions. At least I'll go fix that now. DAID ( talk) 10:45, 27 September 2012 (UTC)
See inelastic scattering. A transfer of simple kinetic energy from one particle to another is kinetically elastic (K-E) since the sum of KE is conserved in COM frame and not transformed to other types of energy-- but in nuclear terms it is IN-elastic scattering (I-S), because the energy of the incoming particle is not conserved. Momentum-transfer results in I-S, but does not imply anything else, as some I-S is K-E, and some is not. The article referenced uses the example of Compton scattering, but that's for electron targets, and we can easily generalize to nuclear interactions such as Bremsstrahlung and pair production. Neither of these are "elastic scattering" events (and they aren't kinetically elastic either, since some KE gets turned into radiation or new particles). Both involve interactions between an electron (classically) and a nucleus. But because the nucleus and the electron emerge unchanged from the event, neither of these are nuclear reactions, either. So nuclear reactions are a subset of inelastic scattering off nuclei, and inelastic scattering interactions in turn may be kinetically elastic or not. The Venn diagram is three overlapping circles.
You might consider the simple straight-on collision of a neutron hitting a deuteron, after which the neutron is greatly slowed and the deuterium gets (let me see) 8/9ths of the neutron's initial kinetic energy. This is kinetically elastic, but certainly an inelastic scattering event for the neutron, since it loses all but 1/8th of its energy. But it's not a nuclear reaction. If it produces a lepton-pair now it becomes not only inelastic scattering but it's kinetically inelastic now as well, since new particles are created and even the total kinetic energy of the reactants is not conserved. Still not a nuclear reaction, though! But if the deuteron is broken up, now you have a kinetically inelastic event, an inelastic scattering, AND now (finally) also a nuclear reaction. S B H arris 01:22, 28 September 2012 (UTC)
This article is internally inconsistent. The organisation is poor (at best). I have a tentative plan to do it fresh. It's more time to list the issues than to fix them, but I want to hear comments. If I do re-create the article, I will be sure to archive the previous version here. No promises for sitting down to actually do the work, though. From this talk page, I feel like this is something people are requesting from an expert for some years. I am happy for the work everyone has done, but fixing the article piece-wise seems crazy to me, because the layout has no flow, nor does it give a good concept of the topic. But I'm only here to help, so let's play nice. DAID ( talk) 17:00, 27 September 2012 (UTC)
Kiana20 ( talk) 19:15, 8 October 2012 (UTC)Kiana20
This is the
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Nuclear reaction article. This is not a forum for general discussion of the article's subject. |
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The equation provided is not a good one, mostly because it ends up weighing more than the starting mass, while the remainder of the topic speaks of it as if mass was "lost" in the process of the reaction. Just reverse the equation to simplify things.
Hello Patrick, I think that should be that one particle is impossible for an exothermic reaction, and has very low probability for an endothermic reaction.
Consider the reference frame in which, before the collision, all the incoming particles' momenta sum to zero (i.e., the "zero momentum frame"). By conservation of momentum, they still must do so after the collision. But if we have zero momentum and only one particle, that therefore means it must be stationary in this frame. But then what about conservation of energy? Since the system initially had kinetic energy but now has zero kinetic energy, we can only balance energy if the reaction is endothermic, with the kinetic energy of the incoming particles in this frame exactly equal to the energy absorbance - which in turn means that given the position and velocity of one particle, for such a "sticking" reaction to occur the position and velocity of the other particle must exactly equal a calculated value. If the energy absorbance of the reaction was precisely defined, this would have zero probability, but if it has some width there will be a small probability of such a reaction. (Of course, on the macroscopic scale our absorbance line can be extremely broad, so such a collision is much more likely to occur with macroscopic particles.) However as soon as we have at least two particles, we can balance momentum, kinetic energy and reaction energy, of any amounts, in any frame. Hence, an exothermic reaction must have at least two products, an endothermic reaction might in principle have only one but in practice rarely if ever does. Or something like that; it's late here. Securiger 16:30, 22 Nov 2004 (UTC)
Why is that Nuclear reaction is one of the high-importance articles while Nuclear fusion, which is just one reaction type, is a top one? -- Crwx ( talk) 16:16, 21 April 2012 (UTC)
Nuclear reactions should transmute the nuclei into different species. The article in general is rather clear on this point, but under Notable Types it has inelastic scattering. This is decidedly not a nuclear reaction. Unfortunately, the problem is that the main article does not even mention nuclear physics at all! So, I'm not sure what to do about this. Most likely the best case is to move all the scattering material to the scattering page. I do think, however, that some kind of note somewhere in the article that elastic and inelastic scattering are not nuclear reactions, with links to the respective articles, is a good idea. The elastic scattering page also does not mention nuclear reactions. At least I'll go fix that now. DAID ( talk) 10:45, 27 September 2012 (UTC)
See inelastic scattering. A transfer of simple kinetic energy from one particle to another is kinetically elastic (K-E) since the sum of KE is conserved in COM frame and not transformed to other types of energy-- but in nuclear terms it is IN-elastic scattering (I-S), because the energy of the incoming particle is not conserved. Momentum-transfer results in I-S, but does not imply anything else, as some I-S is K-E, and some is not. The article referenced uses the example of Compton scattering, but that's for electron targets, and we can easily generalize to nuclear interactions such as Bremsstrahlung and pair production. Neither of these are "elastic scattering" events (and they aren't kinetically elastic either, since some KE gets turned into radiation or new particles). Both involve interactions between an electron (classically) and a nucleus. But because the nucleus and the electron emerge unchanged from the event, neither of these are nuclear reactions, either. So nuclear reactions are a subset of inelastic scattering off nuclei, and inelastic scattering interactions in turn may be kinetically elastic or not. The Venn diagram is three overlapping circles.
You might consider the simple straight-on collision of a neutron hitting a deuteron, after which the neutron is greatly slowed and the deuterium gets (let me see) 8/9ths of the neutron's initial kinetic energy. This is kinetically elastic, but certainly an inelastic scattering event for the neutron, since it loses all but 1/8th of its energy. But it's not a nuclear reaction. If it produces a lepton-pair now it becomes not only inelastic scattering but it's kinetically inelastic now as well, since new particles are created and even the total kinetic energy of the reactants is not conserved. Still not a nuclear reaction, though! But if the deuteron is broken up, now you have a kinetically inelastic event, an inelastic scattering, AND now (finally) also a nuclear reaction. S B H arris 01:22, 28 September 2012 (UTC)
This article is internally inconsistent. The organisation is poor (at best). I have a tentative plan to do it fresh. It's more time to list the issues than to fix them, but I want to hear comments. If I do re-create the article, I will be sure to archive the previous version here. No promises for sitting down to actually do the work, though. From this talk page, I feel like this is something people are requesting from an expert for some years. I am happy for the work everyone has done, but fixing the article piece-wise seems crazy to me, because the layout has no flow, nor does it give a good concept of the topic. But I'm only here to help, so let's play nice. DAID ( talk) 17:00, 27 September 2012 (UTC)
Kiana20 ( talk) 19:15, 8 October 2012 (UTC)Kiana20