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Mike_Fontenot
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Anyone on this forum know what the highest-probability fission chain reaction of plutonium-239 is?
Mike Fontenot
Mike Fontenot
Mike_Fontenot said:Anyone on this forum know what the highest-probability fission chain reaction of plutonium-239 is?
Mike Fontenot
QuantumPion said:Are you looking for fission product yield? If so you can see a plot http://www.nndc.bnl.gov/sigma/getFissionYieldsPlot.jsp?submit=Update+Plot&ymin=3.737E-19&ymax=6.084E-2&yscale=lin&ylimits=auto&energy=0&zplot=24&zplot=25&zplot=26&zplot=27&zplot=28&zplot=29&zplot=30&zplot=31&zplot=32&zplot=33&zplot=34&zplot=35&zplot=36&zplot=37&zplot=38&zplot=39&zplot=40&zplot=41&zplot=42&zplot=43&zplot=44&zplot=45&zplot=46&zplot=47&zplot=48&zplot=49&zplot=50&zplot=51&zplot=52&zplot=53&zplot=54&zplot=55&zplot=56&zplot=57&zplot=58&zplot=59&zplot=60&zplot=61&zplot=62&zplot=63&zplot=64&zplot=65&zplot=66&zplot=67&zplot=68&zplot=69&zplot=70&evalid=4615&mf=8&mt=454" .
Mike_Fontenot said:No. I want to know what the most likely IMMEDIATE fission products are. I.e., what are the two large nuclei that are immediately produced by the fission of a Pu-239 nucleus most likely to be? Surely someone somewhere knows the answer ... I'm just amazed that the answer isn't easy to find on the web.
Usually, those two initial large nuclei are very unstable, with half-lives in minutes, seconds, or even less. They will then decay into longer-lived nuclei, as time passes after the explosion.
I didn't find any of those "yield charts" on the web to be useful, because they seem to be mostly concerned with the eventual longer-lived products. That's probably because the issues that usually are motivating those charts are issues of the health effects of nuclear reactions, or issues of nuclear reactor design, or of issues of spent-fuel-rod disposal. The extremely short-lived initial products don't exist long enough to be important in those issues.
Mike Fontenot
QuantumPion said:The link I posted is direct fission product yields, not cumulative yields. For example, the highest yield isotope indicated is Zr-100 which has a half-life of 7 seconds.
QuantumPion said:The link I posted is direct fission product yields, not cumulative yields. For example, the highest yield isotope indicated is Zr-100 which has a half-life of 7 seconds.
If one goes to the chart of nuclides (as indicated by QuantumPion), and selects 239Pu FY and uses Zoom 1, then one will see the yields. The highest yield is 40Zr100 (at 0.048) and its complement 52Te134 (at 0.044). Two fission (prompt) neutrons would also be emitted.Mike_Fontenot said:I looked at your chart again, but I couldn't tell what the highest-probability REACTION was. The chart does show that the most likely smaller nucleus is 40_Zr_100, but what is the larger nucleus in that reaction? The chart says that the most likely larger nucleus is 52_Te_134, but I doubt that the most likely reaction produces those two nuclei (because I THINK that the initial reaction won't involve the conversion of any protons ... so Z will be conserved).
Is there a way to tell from that chart which initial REACTION is the most probable? That's what I'm looking for.
Mike Fontenot
Astronuc said:The proton number Z is preserved 40 + 52 = 92 and 100 + 134 + 2n = 236 = 235 + 1n
Astronuc said:If one goes to the chart of nuclides (as indicated by QuantumPion), and selects 239Pu FY and uses Zoom 1, then one will see the yields.
QuantumPion said:[...] which does happen to have a positive Q value although I don't know if that reaction actually can occur).
According to the http://www.nndc.bnl.gov/qcalc/" the complimenting fragment of Zr-100 is Xe-140
Mike_Fontenot said:That "Q calculator" IS helpful. I assume that "Q" is the energy released (when Q > 0) for the indicated reaction. But I'm confused about your comment about whether a reaction "can actually occur".
It STILL seems to me that the question "What is the most probable immediate reaction when Pu-239 is hit with a low-energy neutron?" is a well defined question, that should have a simple and straightforward answer. I'm surprised that the answer doesn't seem to be widely known.
Mike_Fontenot said:Before asking my question on this forum, I tried to come up with what I thought might be a very likely reaction when Pu-239 absorbs a neutron, and breaks up into two large nuclei, plus a few neutrons that are capable of causing further Pu-239 fissions.
I started with a reaction that I had found somewhere on the web, for U-235 fission:
92_U_235 + 1n -> 56_Ba_141 + 36_Kr_92 + 3n.
I looked for a similar reaction involving Pu-239, and picked the one that gave the largest amount of released energy. Here's what I got:
94_Pu_239 + 1n -> 58_Ce_148 + 36_Kr_89 + 3n.
But in arriving at the above, I didn't know how to determine the energies of the emitted neutrons. From my limited understanding, in order for the emitted neurons to cause further fission of Pu-239, they must have a relatively low energy. So my above reaction may or may not be capable of producing a chain reaction. How do I determine the energies of the emitted neutrons in any given proposed reaction, and what is the range of acceptable neutron energies that will produce a chain reaction? I think that is the missing link for me.
Mike Fontenot
QuantumPion said:Prompt fission neutrons have an energy distribution ranging from 1 to 10 MeV with a peak around 2 MeV (plot http://www.nndc.bnl.gov/sigma/getMF...max=0.011453942787843738&yscale=lin&endf-6=0"). Fast neutrons may cause fission however thermal neutrons have a much higher probability.
QuantumPion said:It's not widely known because the question is meaningless qualitatively. Fission has a wide range of possible products. This fission product yield distribution IS well known as you can see from the linked plot. While one particular isotope might technically have the highest yield probability, it is still less than 5% so it is kind of pointless to characterize it as "the most likely reaction".
Pu-239, or Plutonium-239, is a radioactive isotope of plutonium that is commonly used as a fuel in nuclear reactors. It is important in fission chain reactions because it is fissile, meaning it can be split by absorbing a neutron, which releases energy in the form of heat.
In a fission chain reaction, a neutron is absorbed by a Pu-239 nucleus, causing it to split into two smaller nuclei and releasing more neutrons. These neutrons can then be absorbed by other Pu-239 nuclei, creating a chain reaction that releases a large amount of energy.
The most common materials used to induce a fission chain reaction with Pu-239 are other fissile isotopes, such as uranium-235 or plutonium-241. These materials can provide the necessary neutrons to sustain the chain reaction.
The energy from a fission chain reaction with Pu-239 is harnessed by controlling the rate of the chain reaction and using heat exchangers to transfer the released energy to a working fluid, such as water or gas. This energy can then be used to generate electricity.
The main risks associated with a fission chain reaction using Pu-239 are the potential for nuclear accidents and the production of radioactive waste. Proper safety measures must be in place to prevent accidents, and the disposal of radioactive waste must be carefully managed to avoid harm to human health and the environment.