Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Petrochemicals on 10/11/2023 13:49:22
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Why for example is U235 fissionable yet the more unbalanced u238 not? U238 has a greater number of neutrons, so I would assume it would be quicker to decay and be more unstable?
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U235 is fissile and u238 is fissionable. Fissile refers to an element than can undergo fission and release sufficient neutrons to enable a chain reaction. Fissionable refers to an element than can undergo fission but cannot sustain a chain reaction, due to insufficient neutrons being released. Why some elements are one or the other I can't tell you but I would suspect it is down to the neutron requirements of the favourable daughter nuclei. Most fissile elements have odd mass numbers but this is not absolute. Most of the energy in the big multi-megaton bombs derives from u238.
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Hi.
That's a good answer from @paul cotter. I can add a bit more detail and did start writing last night but it got long and the computer crashed.
To try and keep it short, I'll just take the comments you made in the OP and reply to each.
Why for example is U235 fissionable yet the more unbalanced u238 not?
Good question. There is an answer but it's quite long, see later.
U238 has a greater number of neutrons, so I would assume it would be quicker to decay and be more unstable?
"Stability" can be considered in two main ways.
1. Energetic stability: When some nucleons (protons and neutrons) come together to form a dense nucleus there is some binding energy released. A nucleus is more stable when the binding energy is higher.
Let's say this another way - some protons and neutrons have a particular mass when they are well separated from each other. When they come very close together (nuclear separation distances), the strong nuclear force really kicks in and takes over, the potential of each nucleon drops as they approach, so that energy is released. As such the mass of a nucleus is less than the mass of the individual component nucleons (by Einsteins famous E = mc^2).
In general, any two nucleons that approach each other to nuclear separation distances will result in some binding energy being released. So we expect the toal binding energy of a nucleus to be proportional to the number of nucleons in that nucleus. As such we usually consider the binding energy PER NUCLEON. If a large nucleus (say U-235) splits into two smaller nucleii then the binding energy per nucleon increases slightly and as a result the total energy (or mass) that remains in the daughter products is less than it was initially.
A nucleus can be considered to be more energetically stable when the binding energy per nucleon is increased. As you may know, the highest binding energy per nucleon occurrs when a nucleus has around 60 nucleons in total (with about half of those protons and half of them neutrons - so an element like Iron on the periodic table). U-235 of course has 235 total nucleons, so that splitting in half makes energetic sense. Indeed it would be reasonable to split in half and then split in half again.
2. Activation Energy: A nucleus can't just reach a new final state from an initial state. U-235 doesn't just spontaneously split because there is some activation energy required, to paraphrase this the nucleus must get over a small hill before it can roll down the big slope on the other side to an energetically stable final state. This is quite different from another type of radioactive decay (alpha decay).
We can consider a simple model, where the two daughter nucleii (well, packages of nucleons that will go on to become the individual daughter nucleii once they separate from each other) pre-form inside a region of space that we are still considering as the nucleus of the original parent. The strong nuclear force more or less switches off entirely once these two packages of nucleons separate from each other enough that we can consider them to be just touching rather than overlapping. When this happens the Coulomb force (the electrostatic force between the positively charged protons in each package of nucleons) becomes the important dominating force. There are textbooks and other sources of information that explain this in more detail. For now we'll just accept that the Coulomb potential at a critical distance = r1 + r2 where r1 and r2 are the radii of these daughter packages (the daughter nucleii that will form), is representive of the height of a potential barrier that would confine these two daughter packages to the parent nucleus region. While the daughter packages are confined, it all stays together as one parent nucleus instead of splitting into the two daughter nucleii.
Classical physics suggests that the only way to get over that barrier is to have the daughter nucleon packages with enough energy to get over that barrier. Quantum mechanics presents another option, the daughter nucleon packages could "tunnel" through the barrier even if they didn't have sufficient energy to legitimately get to the top of that barrier. Tunnelling is extremely dependant on the width of that barrier, which in turn will be dependant on the height of that barrier at the critical separation distance r1 + r2 since outside that parent nucleus region that potential is falling off like 1/r (the usual Coulomb potential). For an alpha particle (which is just 2 protons and 2 neutrons, so "not much"), the other package of nucleons is Z - 4 nucleons and for large Z (mass number or nucleon number, like we do have in Uranium-235 where Z= 235) we have that Z-4 is approximately Z anyway. The height of the Coulomb potential is q1.q2 / (4πε r) where r = r1 + r2 = the critical separation distance and q1 , q2 are the charges of the two daughter nucleon packages. The details aren't too important, we can simplify and assume r = r1 + r2 = the critical separation between the centres of charge is much the same for the situation where we have two daughter packages of about 235/2 ≈ 118 nucleons each, or when we have one big daughter package with 231 nucleons and a little alpha particle of 4 nucleons. So that the height of the potential barrier is then only dependant on the product q1 . q2 that appears.
For the situation where an alpha particle and a big daughter nucleon package form, we have q1 . q2 = 2 . P where 2 is the charge on the alpha particle and P is the charge on the big daughter nucleii left. (So P ≈ 90 for Uranium and it's large daughter nucleus left after alpha decay). For simplicity we have barrier height ~ P. Tunneling of the alpha particle out of the parent nucleus does sometimes happen.
For the situation where the Uranium 235 splits into two nucleii of about the same size, we have that the height of the potential barrier = (P/2) . (P/2) ~ P2. For large nucleii like Uranimum 235 with P ≈ 90, P2 is a shocking lot more than P. The barrier width is MUCH larger and tunneling probability falls of exponentially with this barrier width rather than just linearly, so this probaility takes another big hit. As a result the barrier is just too wide and tunneling almost never happens.
What this means is that, although U-235 will sometimes spontaneously split into two roughly equal sized daughter nucleii (after millions of years - exactly as quantum tunneling would predict), the more prevalent decay is just to have an alpha particle emission.
Anyway, let's get back to the main point.... Quantum tunnelling is not going to allow fission into two roughly equal sized daughter nucleii. The rule of classical physics does almost hold - the nucleon packages must have enough energy to legitimately get over that barrier, or at least get sufficiently close to the top that the remaining barrier width is now so small that tunneling is again realistic. This is the origin of the "activation energy" that is required for fission to proceed. The daughter nucleon packages must have enough energy to get over the potential barrier that confines them to the parent nucleus.
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OK, so that's the preliminaries all done.
Now U-235 and U-238 are almost the same things as far as nuclear physics is concerned, they differ by just a few neutrons. Both can decay by nuclear fission, both can produce two daughter nucleii of about half the original size of the parent. The total change in binding energy when this fission happens is about 200 MeV for both U-235 and U-238. They are almost the same when you consider the relative energetic stability of the parent nucleus compared to the daughter nucleii. So, energetic stability isn't the issue here.
That means that the difference in stability or fission-ability (if fission-ability is a word) must be due to activation energy. This is the energy required to lift the daughter nucleon packages over the barrier that confines them to the parent nucleus.
Now it would be nice if we could say that U-238 just has a higher activation energy than U-235. That would end our discussion because we could argue that the activation energy for U-238 is too big and that's why U-238 is more stable. However, it's NOT quite that simple....
As previously mentioned, U-238 and U-235 are almost the same things as far as nuclear physics is concerned. They both have similar activation energies. Typically (say in a nuclear reactor) we provide that activation energy by supplying a neutron to the nucleus. We have some high energy neutrons produced by earlier reactions and they crash into another nuclues of U-235 or U-238. Now this actually makes U-236 (or respectively U-239) for a brief while so it is really the activation energy of U-236 (or U-239) that we are considering. It's the U-236 (or U-239) that we want to undergo fission.
First of all, the energy is not really supplied by the kinetic energy (or speed) of the neutron when it hits the parent nucleus. It's natural to think that to smash apart a nucleus you must use a neutron with some speed but it's not really like that. Most of the energy is supplied to the parent nucleus just because you have supplied it with a neutron. We have already mentioned binding energy much earlier in this post. Initially we had a neutron that was not in a nucleus, so it had mass (energy) ~ the usual mass of an individual neutron. Once that neutron gets into a nucleus then we have some binding energy that can be released, the mass of the new nuclues is less than the sum of the initial nucleus plus the individual neutron. This binding energy that can be released is usually much more important than the kinetic energy of the original inbound neutron. This is important so we'll say it again another way: Most of the energy you supply to a Uranium nucleus is just beause you have supplied it another neutron, the amount of kinetic energy that the neutron also carries in is just a bit of "fine tuning". Indeed, for U-235 we have a better cross-section (a probability of causing a fission event) when we present it with a low speed neutron. Much of a nuclear reactor is concerned with slowing down (moderating) the high speed neutrons that are produced and presenting a nice slow neutron to the U-235 nucleii. Ideally, if you could present U-235 with a neutron that had 0 kinetic energy then that would be just great, that's almost exactly what it wants - but the neutron obviously won't move into the nucleus unless it does have some velocity relative to the U-235 nucleus. Give the neutron too much speed and it tends to go past or "through" the U-235 nucleus and not result in a fission event at all.
So, the activation energy for U-235 to undergo fission after acquiring a neutron (so that's really the activation energy of U-236) is 6.2 MeV (that one I know). The activation for U-238 for fission after acquiring a neutron is also about 6 MeV. the exact figures for U-238 you can find in .... books and stuff, they may differ by about 1 MeV. To keep it simple we'll just say the activation energy is approximately 6 MeV for either the U-238 and U-235 that you use as the fuel. That is to say, they are almost exactly the same. As previously mentioned U-235 and U-238 just are much the same from the point of view of nuclear physics, they differ by just a few neutrons. So it's NOT that the activation energy for U-238 is much higher than U-235, that isn't the case and it does NOT explain why one is readily fissile after being supplied with a neutron and the other is not.
The difference in fissionability is mainly explained by the availability of that 6 MeV just from absorbing a slow neutron. The simple model for fission following neutron absorption is that for a brief while we can imagine that 235U will form a particle we label as 236U* , the asterisk * that appears after the symbol is important. 236U* is 236U in an excited state, while 236U is just the ground state (normal state) of U-236. You can find information and data about 236U from...... books and stuff. They got their information by doing experiments with U-236. For example, you can measure the mass of one mole of U-236 and thus obtain the mass of one atom of U-236. A little subtraction for the electrons in that atom and you have the mass of the (ground state) U-236 nucleus. Meanwhile the mass of 236U* is not something you can measure in practice, even if a particle like 236U* does exist, it doesn't exist for long. It will undergo fission or revert back to 235U plus a neutron quickly. The particle 236U* is really just a model or idealisation that is used. The idea then is that 236U* has a mass which is equal to the mass of 235U plus the mass of one individual neutron, i.e. it has a mass which has not been adjusted by the release of binding energy that would result when the 235U picks up a neutron and adds it to its nucleus.
Now the activation energy available when 236U is formed from the addition of a (slow) neutron to the 235U nucleus is equal to the difference in energy (mass) of 236U* and 236U (the difference in energy between the excited state that is briefly formed and the rest state that would be possible after the binding energy from incorporating the neutron into the nucleus is released). A similar thing applies for U-238 (where the activation energy is the difference between 239U* and 239U ). It turns out that this difference in energy between the excited states and the ground state is just over 6 MeV for the U-235 starting nucleus and just under 6 MeV for the U-238 starting nucleus. That is what matters, when U-235 absorbs a slow neutron this is just enough to lift it over the 6 MeV activation energy barrier, while for the U-238 absorbing a slow neutron it isn't.
We actually have a good explantion for this small difference in excitation energy and it is precisely due to some of the stuff that @paul cotter has alluded to. We have studied the binding energy of a nucleus and we have a fair model of how it depends on the number of protons and neutrons. There is a "pairing term" that is important in this binding energy formula. It turns out that the binding energy gets a boost when neutrons come in pairs and also when protons come in pairs, so that there is an overall boost when both the neutron number and proton number is even, there is no net adjustment when only one is even and the other is odd, there is negative adjustement when both are odd numbers. So the key thing to note is that with our U-238 and U-235, they are both Uranium and so both have proton number = 92 which is even. However one has an odd number of neutrons while the other has an even number, this makes a small adjustment to the binding energy. Indeed U-233 is just like U-235, it is readily fissile with thermal neutrons. Meanwhile, U-236 is just like U-238, it is not readily fissile with thermal neutrons. It's just whether the number of neutrons they have is odd or even which makes a small, but as it turns out very important difference.
Now, I mentioned that the kinetic energy of the neutron is some fine tuning. U-238 will undergo fission after absorbing a neutron but that neutron must be carrying in the extra needed in the form of kinetic energy. It must be fast. However, for reasons that we won't explain (and I'm not sure if we have a good explanation), fast neutrons tend to have a lower cross-section for interaction with a nucleus, so fast neutrons just do not get absorbed by a nucleus as often as you would like. In the case of U-238 and U-235 the cross-section for interaction is important for nuclear power generation and it has been studied in detail. The cross-section for interaction with U-238 is greatest when the neutron has a kinetic energy of about 1 MeV, below this and there just isn't enough energy to get over that activation barrier and fission to occur, above that and we have the usual effect where fast neutrons just tend to go through or past the nucleus and not interact. 1 MeV is the sweet point where there is enough energy but the reduction in chance of interaction due to speed hasn't become too important. Meanwhile, U-235 has a preference for the slowest neutrons it can get because just having been supplied with the neutron is enough to get it over the activation energy. Over-all U-235 has a much higher cross-section for interaction at all neutron energies ranging between 0 (stationary) and 107 eV = 10 MeV. I have attached a diagram.
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I hope that helps a bit. The fissionability of U-235 and U-238 is so different just because one will get just over the activation energy for fission by absorbing a neutron, while the other doesn't quite get there.
Best Wishes.
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WOW ES, that sure is a detailed answer, thank you. I don't want to derail this thread but I have a quick question since you seem to be up to speed on these matters. u235 when it captures a neutron becomes u236 with a subsequent existence of ~10 to the power of minus twelve seconds before scission occurs OR such is what I had previously believed. In my copy of the Merck index an isotope of uranium, namely u236 is listed with a t1/2 of 2.342 x10 to the power of 7 years! This apparent contradiction confuses me. I am aware of metastable isotopes but this long lived u236 is not listed as metastable.
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I hope that helps a bit. The fissionability of U-235 and U-238 is so different just because one will get just over the activation energy for fission by absorbing a neutron, while the other doesn't quite get there.
Best Wishes.
Yes it does, thank you very much. I had heard before about the even number nucleons but it had slipped my memory. As for the CSA the simple explanation I received was the simple slow moving target explanation, but as a whole near physics seems like a dark art still that is being slowly revealed.
A follow on question similar to Paul's is that the ground state of U236 has a shorter half life than ground U235. It has the paired nucleons, what I am assuming is the need for thermal neutrons and is not fissile?
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Hi.
I have a quick question since you seem to be up to speed on these matters. u235 when it captures a neutron becomes u236 with a subsequent existence of ~10 to the power of minus twelve seconds before scission occurs OR such is what I had previously believed. In my copy of the Merck index an isotope of uranium, namely u236 is listed with a t1/2 of 2.342 x10 to the power of 7 years! This apparent contradiction confuses me.
I haven't read the original descriptions you saw. However, it seems that when "they" say U235 captures a neutron and becomes U236 with a half-life of 10^-12 s they were really taking about it becoming the excited state 236U* and how short lived that would be. It was just slightly sloppy notation on their part, the 235U didn't really become 236U (which is the ground state of U236) by capturing a neutron. In all fairness, our understanding and choice of models has changed over the past few years and the notation 236U* may not have been in common use at the time that text was written.
The usual (ground state) 236U is reported as having a half-life of 23.42 million years exactly as you have stated. The excited state 236U* is a completely different kettle of fish - that will be incredibly short lived.
Just for your information (and to the best of my knowledge) the existence of a particle like 236U* isn't even a certainty or required. It is just a convenient model, it's helpful to imagine a particle like that existed for a short while after 235U absorbed a neutron. The incredibly short half life they are reporting is probably based on theoretical prediction rather than actual measurement and observation. I mean, you have 10^-12 seconds in which you must (i) identify that you really are looking at a particle like 236U - which I suppose means confirming that 92 protons and 144 neutrons can be recovered by dismantling the nucleus and then also (ii) confirming that it isn't just your regular ground state 236U you were looking at - which I suppose you would do by confirming that it has mass (rest energy) which is ever so slightly greater than regular 236U. This is a lot of precision and not a lot of time to do it in.
You don't really need the existence of 236U* to make sense of how or why fission may occur following neutron absorption by 235U. All you need to know is that you had rest mass (energy) equal to that of
[ the mass of regular 235U ] + [ the mass of a separate neutron ]
to start with, so you must have that much energy after the absorption. This is slightly more than the rest mass of (ground state) 236U because we know that some binding energy would have been released when the neutron was incorporated into that nucleus. This surplus energy is what you have available to overcome the activation barrier of the nucleus which would be much like 236U. The ineterim stages are less important and may or may not involve there ever having been a 236U* particle.
The latest models that I'm aware of are still based on what was called "the liquid drop model" of a nucleus - which does model the thing as a drop of incompressible fluid. The process of fission caused by an incoming neutron is thought to be explained by a deformation in the shape of the nucleus as the neutron comes in. This deformation can either increase or decrease the binding energy of the nucleus at that moment. There are two competing terms in the binding enery formula for a nucleus, "the surface term" and the "the Coulomb term". When you deform a nucleus (e.g. make it a bit prolate instead of spherical) then you increase the average distance between charged particles (the protons), this can increase the binding energy or equivalently lower the mass you would observe when you go and measure the mass of that deformed nucleus. Meanwhile, "the surface term" represents the reduction in benefit of being surrounded by other nucleons when you are only at the surface of the nucleus instead of in the middle of it. This "surface term" comes from considering the nuclear force which is effectively only a nearest neighbour force (it's so short range that you only feel the effect from nearest neighbours and nucleons on the surface have less nearest neighbours then those in the middle of the nucleus). When the spherical nucleus is deformed in any way then the amount of surface it has inceases. So this increase in the surface reduces the benefit from the attractive nuclear force. Since that was an attractive force, it was increasing the binding energy. So reducing that will lead to the mass (energy) of the deformed nucleus being more than it would be for the spherical un-deformed nucleus. Hopefully you see what I mean about there being competing terms. Which one wins out can depend on exactly how the nucleus is deformed and what shape it has.
So to reconcile the model where we consider 236U* as being something that really exists with the deformation and liquid drop model we would be saying that 236U* is what you have when the nucleus is "this much deformed" and in "this shape", it has a rest mass which is slightly more than the rest mass of an undeformed 236U nucleus. However, the main point is that the deformation in the liquid drop model considers the process as being something that is continuous, as the nucleus deforms the rest mass changes smoothly and continuously, all values of the rest mass between that for 235U and that for 236U* are obtained (and quite possibly some interim shapes that had rest mass below or above that). Meanwhile the kinetic energy of the incoming neutron was being changed (usually reduced) as it starts impacting on the surface of the nucleus, so that there was (in general) a smooth and continuous exchange of energy from k.e. in the neutron to the rest energy of the deforming nucleus. A bit later the neutron is mostly inside the nucleus and it's a bit of grey area how much of its mass is now the mass of the nucleus. You just don't have a clearly identifiable separate neutron and a separate nuclues on which it is impacting any longer.
By comparison, the model involving the existance of 236U* just breaks it down into a series of discrete steps, you have this, then this, then this : you start with 235U plus a separate neutron, then you have 236U*, which is like 236U with enough energy to overcome the activation barrier and so then you will quickly have the two separate daughter nucleii in existance.
Anyway, I hope you can see what I mean about the existence of a particle like 236U* being just a model we can use. Whether the real process is a smooth and continuous one like deformation in the liquid drop model OR a series of discrete steps remains uncertain. Indeed it doesn't even have to be one or the other, it could be both. Quantum mechanics can model the situation as a wave function where the discrete states or eigenfunctions are the discrete stages suggested by the discrete model. The wavefunction is just a superposition of those eigenfunctions where the coefficients of the eigenstates evolve over time. What this means is that at time t = 0 units = initially the wavefunction is 100% that of there being a separate neutron and a separate undeformed nucleus, as time increases the blend of eigenstates starts to shift so that at time t=1 there is a 50% chance of finding the thing in the initial state if you observed it and a 50% chance of findining it in the 236U* state. As time evolves the proportion of the eigenstate for 236U* increases and the amount of the initial state decreases etc. etc. In this way you still have only the existance of discrete states when you go and observe the system BUT the transition from initial state to final state does progress in a smooth and continuous way, so that the average or expectation value for the rest mass of the nucleus would be a continuous function of time.
@Petrochemicals , I think your question will be shorter to answer. I'll be writing that now, after a cup of tea.
Best Wishes.
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Hi.
I had heard before about the even number nucleons but it had slipped my memory. .....
A follow on question similar to Paul's is that the ground state of U236 has a shorter half life than ground U235. It has the paired nucleons, what I am assuming is the need for thermal neutrons and is not fissile?
What you are missing is that both U236 and U235 tend to decay naturally by alpha emission. This was briefly mentioned in one of the earlier posts (tunneling of an alpha particle is realistic but tunneling of a larger daughter nucleus is not). So spontaneous emission of an alpha particle does happen, although it's rare and the half-life is long (23 million years for U236 and 700 million years for U235). By comparison the spontaneous emission of a larger daughter nucleus is so unlikely we can completely ignore it, it almost never happens. We must have the classical physics rule obeyed where the daughter nucleon packages were given enough energy to legitimately get over the potential barrier instead of just tunneling through. To get fission into two larger daughter nucleii we must actively do something to the isotopes - for example put them in a nuclear reactor where we can keep a steady stream of neutrons hitting them.
Recall that an alpha particle has 2 protons and 2 neutrons. So when an alpha particle leaves the parent nucleus the odd or even nature of the number of protons and neutrons is completely unchanged. As such it really doesn't matter whether you started with a nucleus with odd neutrons or even, you will end up with the same nature in the daughter nucleus. So however the binding energy was adjusted (a positive, negative or no adjustment according to odd/even nature of the protons and neutrons), you have almost exactly the same adjustment in the daughter nucleus anyway. The jump or change in binding energy per nucleon is therefore the same as if there just was no adjustment at all from the pairing of nucleons. It doesn't matter.
This is very different from what happens when there is a neutron capture (in let's say, a nuclear reactor). Then you do change the number of neutrons from being odd to even OR from even to odd. This does change the gap (the available excitation energy) between the excited state Z+1 U* and the ground state Z+1 U where Z = the mass number of the original isotope (235 or 236 for your examples). Recall that Z+1 U* is by definition the particle with an amount of energy that is the rest mass of Z U plus the rest mass of a separate neutron. So, just to be clear about it, the excited state has an adustment as if there were an even/odd number of neutrons where the ground state has an adjustment as if there is an odd/even (the opposite way round) number of neutrons.
Looking at the isotopes U235 and U236, they are almost the same in their half-life, there's one order of magnitude difference and that's all. When you're talking about millions of years anyway, this is all the accuracy that we would claim from our models. However, we can do a bit better here. The process of alpha decay is different than neutron capture and subsequent fission and we do get surprisingly good results modelling the phenomena entirely as quantum tunneling through a barrier. The difference in half life is a consequence of Q (the amount of energy released from the decay process) being slightly higher for U236 then for U235. Recall that the overall energtic stability of a nucleus is NOT just a consequence of having an odd / even number of protons and neutrons, there are many terms in the binding energy formula. I've mentioned "the surface term", "the Coulomb term" and "the pairing term" - but there is also "the volume term" and "the asymmetry term". I'm not going to explain all that, there are books and few people like the mathematics anyway. Instead we'll just state what it means in English.
A nucleus is "more energetically stable" when a whole raft of things are met: Overall, the stable region is shown on a Segre chart. The number of neutrons in U235 and U236 is just different enough that alpha emission puts U236 just a bit more into the bingo, perfect balance of protons and neutrons area then alpha emission from the U235. So the Q value is slightly better. We have a relationship called the Geiger-Nuttall rule and which is easily explained with our model based on quantum tunelling, where the half life t?/sub] varies with Q (the energy released from the nuclear decay). The half-life of a substance that decays by alpha emission is actually exponetially dependant on the energy released by the decay, so even a small change in this Q value can make a big change in the half life.
Final note: If you want to be controversial, you could argue that "natural decay" of U235 isn't just limited to alpha decay as the most prevalent form. There were some caves found which contain large Uranium deposits and these caves would sometimes flood with water. When the water was there, this would moderate the neutrons released by the Uranium. All the other factors you would want, like the geometry of the Uranium deposits just happened to be there. You had a perfectly natural (not man-made) nuclear reactor. When the process became critical or super-critical, the heat generated would evaporate the water. Then the nucelar fission chain reactions would stop since the neutrons were too fast again. You had a genuinely natural, self-regulating nuclear reactor. The Uranium in those caves was not decaying predominantly by alpha decay, it was undergoing fission induced by thermal neutrons. It's half-life (which is "natural" in that no man had made it happen) was much less than 24 million years.
See https://www.scientificamerican.com/article/ancient-nuclear-reactor/ for one article that describes these natural reactors, if you're interested. I had a passing interest because it's another example of something that we would have eventually discovered by some means even if people like Einstein and Oppenheimer had not done their work. Eventually we would have found a natural nuclear reactor and we would have realised something was going on.
Best Wishes.
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I'm surprised that nobody has mentioned these in the context of nuclear stability.
https://en.wikipedia.org/wiki/Magic_number_(physics)
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Hi ES. In relation to your post #5, I am going on what I remember from physics lectures over 50 years ago, so I would imagine the understanding of the processes involved has advanced considerably since then. And believe it or not, I have another question for you: u235 on exposure to a thermal neutron what ratio of u236/ u236* would one expect and are there conditions that affect this ratio?
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t?/sub] varies with Q (the energy released from the nuclear decay). The half-life of a substance that decays by alpha emission is actually exponetially dependant on the energy released by the decay, so even a small change in this Q value can make a big change in the half life
That is quite a revelation.
Thank you once again Eternal Student for 2 detailed concise posts. They give me a start into the very complex world of nuclear physics.
I had read of the natural reactors in the past, they are extraordinary things.
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Hi.
I have another question for you: u235 on exposure to a thermal neutron what ratio of u236/ u236* would one expect and are there conditions that affect this ratio?
I'm not really an expert on Nuclear reactions. I've just been reading some lecture notes and a textbook on it recently (they weren't my lecture notes, one of the family just kindly made them available to me).
U236* should be so short lived, you won't find any significant amount of it. As fast as you may produce it, it will decay and usually by fission. So you'll just find the later decay products. You ( @paul cotter ) mentioned it should have a half life of 10^-12 seconds. I previously mentioned it may not even be a real particle that exists, it's a useful model and/or may be something like U236 with a deformed nucleus etc. There should be very little you can do to increase the proportion of U236*, the main problem would be keeping it alive (undecayed).
U236* is what you shoud be producing (as the first step) from U235 and a thermal neutron. You shouldn't produce U236 (the ordinary ground state). However, there probably are some routes where the excited state U236* can lose its surplus energy (say by emitting a gamma ray) and become U236 (the ground state). There are probably also some chains (of neutron absorption and/or decay events) that would also lead to the production of U235. For example, we can already use neutron bombardment to change one element into another, it's a bit random and the yield isn't great - you produce all sorts of stuff and only a small portion was what you wanted - but it happens.
Nothing goes absolutely as it should in the textbooks inside a nuclear reactor. Sometimes a contaminant will be present, that might absorb some neutrons and form something unexpected. There are various things that could happen, I'll avoid presenting examples but I'm sure you can imagine the situation. We know that you end up with traces of all sorts of stuff being left in the reaction vessel and spent fuel rods when reactors are finally decommissioned.
More significantly, you never really start with just 100% U235. Natural Uranium is a mixture of isotopes and it's difficult to enrich the proportion of the (desirable for reactors) U235. Getting to 5% U235 is very good and is considered a high quality fuel. So, although on paper we can write down what should happen to U235 with a thermal neutron, in practice you never really had pure U235 to start with, there was quite a lot of other Uranium isotopes in it already.
Best Wishes.
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Thank you ES, that was a badly worded question that I posed. What I should have said was that given a u235 nucleus impacted by a thermal neutron what ratio of possible reactions would take place, u235+n→u236*→bang, or u235+n→u236. Or in other words can a thermal neutron impact lead to u236. I think if u236 was a possible reaction product it would be a negative influence on fission processes. No need for further replies, I am just thinking aloud, and thanks again for your detailed responses.