# Fission continued ## Метаданные - **Канал:** DrPhysicsA - **YouTube:** https://www.youtube.com/watch?v=1c7OBwYnMkY ## Содержание ### [0:00](https://www.youtube.com/watch?v=1c7OBwYnMkY) Segment 1 (00:00 - 05:00) hello today we're continuing in our series on nuclear physics looking in more detail at Nuclear fishion So last time in the last video we explained why a large nucleus does not split as you might expect that it would into two smaller nuclei whose binding energy is greater and therefore whose mass energy is less and the Reon that it doesn't spontaneously do that is because it first according to the liquid drop model has to deform into an emerging uh two separate nuclei and when it does that the surface area of this increases and so the surface term of The Binding energy increases which reduces The Binding energy which increases the equivalent mass energy and consequently if you look at the chart of this is essentially deformation this is when the nucleus is one big nucleus and this is the when the nuclei have split this is a mass energy chart and we said the mass energy when you've got one big nucleus is larger than when you've got two small nuclei total mass energy and therefore you expect that nature will want fall down this uh this slope but in fact what happens is that as you go via the deformed shape in fact the mass energy increases and you get a hump or a barrier and that barrier the height of that barrier is called the activation energy and ordinarily the nucleus Nature has no energy spare energy around this can be of the order of 6 me nature does not have the energy to get over that barrier in order to fall down the hill to where it wants to be which is the reason we said why your gold and your silver jewelry do not spontaneously Decay into smaller nuclei ultimately ending up at i56 which has the highest binding energy and yet some elements do for example if you have a neutron in the presence of 235 uranium has 92 protons that will produce two smaller nuclei X and Y we'll consider what they are a bit later on plus some neutrons plus of course energy in other words the uranium is here and X and Y are these two smaller nuclei we get some additional neutrons and we get some energy now looking at that I can see that's a jolly good thing I could conceive of a nuclear power reactor this energy which is going to be kinetic energy is going to for example heat water that could produce steam steeve steam can drive a turbine which drives a generator and produces electricity brilliant this is an excellent thing but how does it happen given that we showed in the last video that you've got this activation energy barrier to get over and I can hear those of you who are cynical saying I know what you're going to do you're going to give this Neutron sufficient energy that it will enable the UR to get over the barrier that's cheating well no I'm not I'm actually going to use what's called a thermal Neutron the point about a thermal Neutron is it is a neutron whose energy is entirely dictated by its temperature so if it's at room temperature it will have the energy associated with room temperature and in thermodynamics and we've done this in the thermodynamics videos thermal energy is simply defined by K boltzman's constant time T the temperature and in the units that I want to work K is 9 * 10- 5 EV per Kelvin you can give it in various units and that's the units I want to use that's not SI units but it's the units I want to use so that's K multiplied by T what is room temperature in Kelvin that is 300 and if you multiply that you've got 9 * 3 is 27 10 - 5 * this is 10^ 2 is 10 - 3 EV and that's I think 0. 27 electron volts so a thermal neutron has an energy of 0. 027 electron volts pitifully small that is not going to assist anybody to get over an activation energy of 6 me so I haven't Che ### [5:00](https://www.youtube.com/watch?v=1c7OBwYnMkY&t=300s) Segment 2 (05:00 - 10:00) that Neutron does not supply the energy required to get over the activation energy so what's going on and how does it do it well I've actually missed out a stage when I say that this goes to this because actually we first got to ask what happens here this is a slow moving Neutron it's got hardly any energy at all for all intents and purposes it's more or less stationary and so it's easily captured by the uranium nucle remember neutron has no positive charge so it's not going to be repelled by the coolon force so it can be absorbed by the uranium 235 to make uranium 236 so what actually happens is the neutron the thermal Neutron becomes absorbed by the 235 uranium to produce uranium 236 still with 92 protons now if you do the get the semiempirical mass formula out and you calculate the total mass energy of this and then of course you're going to find there's a difference otherwise n wouldn't bother to do it and the difference is that there is a change in energy equal to 6. 6 meev so you've got 6. 6 meev of energy being released as a consequence of this Neutron being absorbed simply because the total mass energy of this is greater than the total mass energy of this by 6. 6 me in what way will that 6. 6 Me Be manifested Well normally when we looked at Alpha and beta particles we said that the energy is exhibited in the form of kinetic energy in the emerging particles but that cannot be here why because this uranium is just in a block of uranium it's to all intense and purpose is stationary its momentum is zero the neutron has such a small kinetic energy that you might as well regard its momentum as essentially zero so the total momentum beforehand must equal the momentum afterwards because momentum is conserved so the momentum of this uranium 236 is zero and kinetic energy I remind you kinetic energy is p^2 2 m and if P the momentum is zero then p^2 is z and the kinetic energy is zero so that 6. 6 me cannot be in the form of kinetic energy what is it then well what it does is it simply takes uranium 236 in its ground state and it whacks it up to a higher level excited state of 6. 6 me so in other words it is manifested not as kinetic energy but as potential energy now of course as we've already seen one of the obvious things can then happen is that the nucleus simply de excites and in doing so emits a gamma photon with an energy of 6. 6 me that's one option but it isn't what it always does and that's the key what it can also do is to use this 6. 6 me to get over the activate energy which for these purposes for Uranium 235 to what's called fish break up into smaller nuclei the activation energy is actually 6. 2 me so if you've got 6. 6 me of potential energy in this excited state let's put a star there to show that it's in an excited state then one option is to use that energy to get over the activation energy barrier and that is what is happening and that's why uranium 235 can as it were spontaneously fishing into two smaller nuclei as I said we'll work out what those are later plus some neutrons plus the energy you need for running your fishing reactor now uranium when it is mined from the ground natural uranium consists of basically two isotopes there are others but the two major ones are uranium 238 which is 99. 3% of the natural uranium and uranium 235 which is about 0. 7% of natural uranium there are other Isotopes but they are very small percentages so overwhelmingly what you dig up from the ground is uranium 238 and what I've shown you is what happens to uranium 235 so it' be really rather nice wouldn't it if uranium 238 did the ### [10:00](https://www.youtube.com/watch?v=1c7OBwYnMkY&t=600s) Segment 3 (10:00 - 15:00) same thing because we got an awful lot of that so let's ask the question if I change uranium 235 to uranium 238 will I achieve the same outcome well let's work it through if this is uranium 238 then if it absorbs a neutron it will become uranium 239 which will be in an excited state because there will be a mass energy difference between the two but unlike in the case of 235 the mass energy difference is not 6. 6 me it's 4. 8 me so the uranium 239 goes to an excited state of 4. 8 me from which of course it can de exite and release a photon at 4. 8 me but can it also get over this activation energy barrier well the problem is that for uranium 238 the activation energy barrier is broadly the same it's still 6. 2 me and you've only got 4. 8 me in the potential energy of this excited state so uranium 238 sadly will not spontaneously fish so although that's what you get most of you cannot get it to get over the activation energy because you don't get enough energy from the merger of the uh Neutron and the uranium 238 to form 239 so that does not fish into smaller nuclei so uranium 235 is known as file material that means it will spontaneously break up into smaller nuclei and give you energy but uranium 238 sadly won't what a shame so let's have a look at the experimental results for fishing of uranium 235 and we'll also look at uranium 238 now these scales are what are called logarithmic scales that is to say that if this is 10us 3 every Notch up increases not by 1 but by tfold so this is 10- 2 10-1 1 10 2 10 cubed all right so we're going up by a factor of 10 in each case and similarly and this is what's called the cross-section if you can see that I'll rewrite it there cross-section is for these purposes a measure of probability How likely will this reaction How likely will the fishing reaction happen and along the bottom we've got the energy of the neutron in electron volts and again I'm going to put that as a logarithmic scale so we start at 10- 3 10 - 2 10 -1 1 10 SAR 10 cubed 10 4th 10 5th 10 the 6 and I don't think I can get any further but that'll do us for the time being so that's the energy of the neutron room temperature is about here and the cross-section for that is approximately 500 and what you find is that the uh graph will drop this is for Uranium 235 rather like that and then it sort of flattens out and what this shows you is that as the neutron energy increases the cross-section falls off dramatically remember this is a log scale this goes from 500 down to about one so it's 500 times more likely to get the fishing products when you have got a thermal Neutron at room temperature as it were then if you have a neutron at 10 the 5 that's 100,000 electron volts or 100K EV you're a thousand times more likely to get fishing to happen and there's a kind of a hand waving argument to explain that if you've got a neutron of 100 K that's a fast moving Neutron and the uranium won't be able to capture it because it'll just Whiz On by so it doesn't get absorbed and consequently you don't get the fishing taking place now I can also draw the graph so this is all for 235 uranium 235 I can draw the graph for uranium 238 which looks a bit like this is 238 there is no probability of uranium 238 fishing at low Neutron energies ### [15:00](https://www.youtube.com/watch?v=1c7OBwYnMkY&t=900s) Segment 4 (15:00 - 20:00) because the excitation state of uranium 239 isn't big enough to get over the activation energy so it just doesn't happen but when the neutrons have got sufficient energy it's possible that energy com combined with the 4. 8 me of excitation energy will be enough to provoke some kind of fishing reaction to happen but look it's a very small it's nothing like the size of the uranium 235 remember again this is a log scale this is now uh 10 to 100,000 times less likely than the fishing you get with uranium 235 at room temperature even up here when you've got uh neutrons with a megga volt me one me of energy so that gives you an idea of why you need slow moving neutrons that can be easily absorbed to cause fishing to happen and even when you give neutrons very high energy the probability of fishing 238 is very much smaller so now let's ask the question what are the X and Y what do the uh what does uranium 236 split up into so once again I'm going to have probability on this axis and a number on this axis and this is uranium 236 that's the excited state of uranium caused by the absorption of the neutron and here is halfway house 118 so the question is do you find that a uranium 236 nucleus will split into two nuclei each having 118 nucleons and what you actually find experimentally is that if this is 95 and this is 140 the probabilities look something like this that there's a very high probability that you'll get a nucleus with 95 uh nucleons and there's a high probability that you'll get a nucleus with about 140 nucleons but actually there's a very low probability that you will get a nucleus with 118 nucleons suggesting that nature does not want to split exactly half and the liquid drop model doesn't explain that and let me just explain why indeed the new the uh liquid drop model um does precisely the opposite so let's have a look at what the liquid drop model would say this is The Binding energy per nucleon against nucleon number and here's the famous shape again building up to iron 56 after which there's a gentle Decay that's probably too great a um a gradient but it'll make my point so this is Ion 56 so let's say we have a nucleus which has a nucleons and then here is a over2 and we'll go either side so this is now a over 2 minus Epsilon and this is exactly the same on the other side so that's a over2 plus Epsilon what are these binding energies per nucleus well let's say that the starter is e0 let's call this point E1 which means that this will be E1 minus Delta and plus Delta all right this is pretty much a straight line here so it's perfectly legitimate to say that this will be as much below E1 as this one is above now we ask ourselves what was The Binding energy at the start well The Binding energy is e per nucleon time a nucleon so The Binding energy to start with so is equal to e not which is The Binding energy per nucleon times the number of nucleons which is a what is The Binding energy to finish with is going to be equal to well we've got a/ 2us Epsilon nucleons each having E1 plus Delta energy and we've got a over2 plus Epsilon nucleons each having E1 minus Delta energy so the total energy is going to be a/ 2us Epsilon * E1 + Delta plus a/ 2 + Epsilon * E1 minus Delta a 2 - Epsilon * E1 1 plus Delta a ### [20:00](https://www.youtube.com/watch?v=1c7OBwYnMkY&t=1200s) Segment 5 (20:00 - 25:00) 2+ Epsilon * E1 minus Delta I'm afraid we just have nothing more to do than to actually work that lot out so that's going to be a over 2 or rather a E1 / 2 plus a Delta / 2 then this one minus Epsilon E1 minus Epsilon Delta that's this here what about this one here you're going to get plus here ae1 / 2 - a Delta / 2 and then here plus E1 minus Epsilon Delta and you can see some of those are going to cancel I've got a plus a Delta over 2 and a minus a Delta over two so they go I've got a minus Epsilon E1 and a plus Epsilon E1 so they go and I've now got a E1 /2 + a E1 / 2 which is just a E1 and minus EP epon Delta minus Epsilon Delta is - 2 Epsilon Delta so that is The Binding energy at end and this was the beginning and what energy do you get out well the energy you get out is the larger amount minus the smaller amount so that's going to be so Q the energy you would expect to get out is going to be what you end up with a E1 - 2 Epsilon Delta minus what you started with which was e0 * a and the question is if you want the maximum amount of energy out that is to say that the nucleus goes into the lowest state of binding energy which is what nature wants to do if you want Q to be as large as possible then this term 2 Epsilon Delta has to be zero because that is subtracting from the total energy out and if 2 Epsilon Delta is zero that means Epsilon is zero which means according to the liquid drop model the two nuclei that should be formed should both be of a over2 because that is what will give you the maximum energy out but actually experimentally you find that is not what happens so the liquid drop model in this respect is wrong and maybe in a later video we'll explain why that is okay so we now know that when the uranium 235 forms uranium 236 in an excited state and then splits it's likely to split it's most likely to split into two nuclei one having about 140 nucleons and one having about 95 and there are a variety of um variations as it were in what you can get there is no single outcome so let me write one example so the neutron is absorbed by uranium 235 to produce uranium 236 in an excited state and that can produce Zenon with a 140 nucleons and of course it has 54 protons plus strontium which has 94 nucleons 38 protons plus two neutrons plus of course the energy that you want to drive your nuclear reactor if that's what you're going to use you get energy out but these two elements and you'll notice the 140 and 94 are very close to what I said would be the preferred numbers 140 and 95 we showed were the preferred but this is close enough 140 + 94 comes to 234 there were 20 36 to start which is why you get two extra neutrons given off but these two nuclei are unstable they are going to Decay and they're going to Decay by Beta Decay they are Neutron Rich so the beta Decay is going to be beta minus Decay so that means that the Zenon is going to Decay by Beta minus Decay to 140 and now um a neutron will convert to a proton so you've got an extra proton and that means the element changes so this is now cesium but that to is unstable so it decays by Beta minus Decay to 140 56 barium and that's unstable so that decays by Beta minus we' still got too many neutrons so we're converting neutrons into protons that's going to be 140 57 ### [25:00](https://www.youtube.com/watch?v=1c7OBwYnMkY&t=1500s) Segment 6 (25:00 - 29:00) lannum and that's unstable so that decays by Beta Decay to 140 58 serium and that's stable so that's where the line comes to an end but that's a lot of decays all of those decays have got Associated half lives so this process is producing an awful lot of radioactivity as all the products gradually decay similarly the strontium is unstable that will Decay by Beta minus Decay that will go to itum which is uh 9439 and that will then Decay to by Beta minus Decay to zirconium which is 9440 and that is stable so you can see that you've got a lot of Decay products being produced each of which is itself um Radioactive by Beta Decay until you eventually get down to a stable state but that isn't the only method of Decay as I said there are lots of options so when you've got to your uranium 236 excited state another option is to produce 137 iodine which is 53 plus 97 itum which is 39 plus two neutrons of course plus energy again and what I particularly want to focus on here once again these two are unstable they will Decay but I particularly want to look at the iodine Decay because iodine Decay does what you think it might do bet minus Decay it's got too many neutrons so it changes a neutron into a proton so we now get one 37 54 Zenon but then an interesting thing happens instead of beta Decay you now get Neutron emission you actually shed a neutron and that produces 136 54 Zenon the if you shed a neutron you don't change the element and that is stable but you've got a new Neutron coming out I just want you to observe this formula here the initial fishing process produced two neutrons as a consequence of the fishing that fishing is very fast that will take place in about 10us 12 seconds or so these are called prompt neutrons they come out immediately you have the fishing but this Neutron is called a delayed Neutron why because you don't get it immediately because you first have to wait for the iodine to Decay into Zenon through beta minus emission and only when you've got this Zenon 137 can you then Decay by Neutron emission 2136 Zenon so you've got to wait essentially for the half life associated with this Decay before you get this Neutron so in radioactive decay of this kind and fish of this kind you get prompt neutrons which are the direct result of this uh fishing action and you get delayed neutrons which arise after a certain amount of Decay has already taken place and the time scale for this Neutron to be emitted will of course be dependent upon the halflife associated with the Decay that you have to have first before you get Neutron emission and how those delayed neutrons can be used in a fishen reactor we will see in the next video --- *Источник: https://ekstraktznaniy.ru/video/40015*