the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu okay guys we're actually slightly ahead of where I thought we'd be at this point so I'm only gonna spend about half of today's lecture finishing up some new material on mass parabolas and stability I also got a comment in through the anonymous box that said please leave a little bit of time after class for questions so you can get them out right away and because I'm usually running off the teach some other class at like the IDC or some other building so from now on I'll try and leave about five minutes at the end of class for questions on today's material and we'll make up with it with the second half hour or 25 minutes of this class being for all the questions on the material so far in the first two weeks but first I wanted to give a quick review of where we were Wednesday and launch back into math parabolas which are ways of looking at nuclear stability in relative numbers and even or oddness of nuclei so you saw last time we intuitively derived the semi empirical mass formula as a sum of volume surface Coulomb asymmetry and pairing or whether things are even even or odd even terms with the coefficients in MeV gleaned from data and the forms of the and the exponents right here gleaned from intuition here we assume that the nucleus can be thought of like a big drop of liquid with some charged particles in it and so the droplet should become more stable the more nuclei there are or the nor nucleons there are but then you have some more outside on the surface that aren't bonded to the others all of the protons are repelling each other over linear length scales because the radius of this liquid drop would scale like a to the one third there's an asymmetry term which means if the neutrons and protons are out of balance there's going to be some some less binding energy and then there's this extra part that tells you whether the nuclei are even even or odd odd and this works pretty well if you remember we looked at theory versus experiment where all the red points here are theoretical predictions and all the black points are experimental predictions and for the most part they look spot-on it generates the classic binding energy per nucleon curve that you see in the textbook and can predict from the semi empirical mass formula zooming in and correcting for let's say just getting absolute values of errors you can see that except for the very small nuclei and a few Peaks which we explained by looking even closer the formula well it predicts nuclear stability quite well on average again this line right here if there's a dot that lies on this blue line it means that theory and experiment agree and a deviation by ofyou mev here and they are not too bad but we started also looking at different nuclear stability trends and we noticed that for odd mass number nuclei there is usually only one or sometimes none stable isotopes per Z whereas for even once there's quite a few more and we're going to be now linking up the stability of nuclei versus what mode of decay they will take in order to find a more stable configuration we looked quickly at the number of stable nuclei with even and odd Z and noted that this these places right here where there are no stable nuclei correspond to technetium and promethium there's no periodic table on the back of this wall but behind my back on the other wall there's that periodic table where you can see the two elements that are fairly light with no stable isotopes that's what those correspond to and the peaks correspond to what we call magic numbers or numbers of protons or neutrons where all available States some energy are pretty much filled and this goes for both protons and neutrons so something with a magic number for both n number of neutrons and Z number of protons is going to be exceptionally stable and we'll see how that's used as a tool to synthesize the super-heavy elements that we believe should exist and finally we got into these mass parabolas I found this to be a particularly difficult concept to just get mathematically if you remember we wrote out the semi empirical mass formula and said if you take the derivative with respect to Z as we did it you would get the most stable z for a given a and we started graphing for a equals 93 where naya beum is stable that was just the one I had on the brain from some failures in lab earlier this week we started plotting where those nuclei what is at the the relative masses are for a fixed a so let's regenerate that one right now because we were a little fast at the end of last lecture then I want to generate one for a equals 40 and you'll see something kind of curious I'm going to leave this up here for a sec if you notice for odd a nuclei there's only one parabola whereas for even a there are two why that is we're gonna see when we look at the table of nuclides but notice this nucleus right here can decay by either positron emission or beta emission to get to a more stable form and there are many real examples and I'm going to show you how to find them so let's start off by going back to the table of nuclides finding niobium 93 just go up one more chunk and there we are Niobe M 93 is a stable isotope and if you want to see where it came from you can scroll down a little bit and see it's possible parent nuclides right here so let's say that niobium will draw right there is stable we'll put it at the bottom of this parabola and let's work down in Z so we'll move to zirconium zirconium 93 ostensively has a very similar atomic mass but if you remember that 93 amu is a rather poor approximation for the actual mass of all nuclei with a equals 93 in fact if you look very closely at the atomic masses zirconium 93 is 90 2.90 niobium 93 well it looks like we have to go all the way to another digit there 92.9 o6 you have to go to like even more digits ninety two point nine oh six three seventy five ninety two point nine oh six four seventy five so we go down in Z and we've actually gone up in mass by looks like the sixth or seventh digit in AMU if we go up in mass we go down in binding energy that tells us that there's something that's less stable and if you notice we went down by a very smaller or Wien up by a very small amount of mass notice also that it's beta decay energy is really really small ninety one kill electron volts so why don't we put zirconium just above and we note that that decay will happen by data where the beta let's say if we have an isotope with mass number a protons Z and let's just call it symbol question mark in beta decay we have the same a will have a different Z I'm gonna have to give these symbols let's call this parent and we'll call that daughter plus a beta plus an electron antineutrino and what has to happen to that Z in order for everything to be conserved it's the same reaction that we've gone here for I'm sorry for their code iam so you'd have to have one fewer proton to release one electron and so that becomes same a but different Z and this is the beta decay reaction let's go back a little farther we'll look at the possible parent nuclide for anyone have a question let's see which direction are we going in Z here that goes to yet trium actually looks like it's going down oh yeah for beta decay I'm thinking I have the reaction backwards sorry need one more proton to account for the extra negative charge you're right okay yep I was thinking backwards because we're now climbing up the decay chain in reverse order so this could have come from yttrium 93 with a much higher energy of three MeV so let's put it reom right here that gives a beta decay and we'll just go one more back to strontium 93 has an even higher beta decay energy so let's put that up here and let's take a look at its mass real quick the mass of strontium 93 92 point nine one four AMU if we go back to niobium 93 now it's noticeably different different to like four significant digits instead of six ninety two point nine one four versus ninety two point nine oh six and so that shows you that a tiny bit of mass and AMU corresponds to a pretty significant change in binding energy by that same conversion factor that we've been using everywhere nine thirty one point four nine AMU per big MeV yeah per C squared let's see yeah okay so let's go now in the other direction in the positron direction niobium can also be made by electron capture from molybdenum so let's put molybdenum right here let's say that around half an Emmy or what do we have here who was like half an MeV like that and let's see molybdenum 93 could have been made by electron capture from technetium 93 with an energy of three point two oh one MeV even more extreme and we'll go back one more because there's a trend that I want you guys to be able to see and this could have come from electron capture from ruthenium I think I may have said rubidium last time but are you is ruthenium 93 and that 6.3 MeV something like that and this is where we got to yesterday now I'd like us to take a closer look at the decay diagrams which tells us what possible decay reactions can happen in each of these reactions so let's since we're right here on the chart let's take a look at ruthenium turning into technetium by what it says electron capture so note that on the table you can click on electron capture and if it's highlighted then the decay diagrams are known it's not known for every isotope but for a lot of the ones you'll be dealing with it is and you get something I have to zoom out for a lot a lot of different decays what I want you to look at is this one here on the bottom that I'll zoom into that should be a little more visible so notice that if you want to go down the entire 6.4 something MeV it usually proceeds by B+ or positron decay by either method and as you go up the chain as these energy differences get smaller look what happens to the probability of getting positron decay it shrinks lower and lower and lower so they a trend that the larger the decay energy for this type of reaction the more likely you're going to get positron decay and in fact where we left off last time is in order to get positron decay the Q value of the reaction has to be at least 1.0 to 2 MeV better known as at least two times the rest mass of the electron because in this case to conserve charge and energy you shoot out a positron and you also have to inject an electron in order to conserve all the charge going on here so there you have it now let's look at the lower energy decay of technetium to molybdenum which had something like 3 MeV associated with it so we'll click on technetium and its energy is 3.2 MeV let's take a look at its electron capture significantly simpler already what do you notice about these positron to electron capture ratios anyone call it out indeed when the energy of the decay goes down notice that only these decays are allowed the electron capture suddenly becomes much more likely but notice that it does not let you go directly from 3.2 mu v to 0 there's no allowable decay here so this is probably a change of that's 3.2 that's 1.3 a little less than 2 MeV all of a sudden electron capture becomes much more likely but not at positron decay is not disallowed yet so we can say electron capture or positron decay right there everyone with me so far so let's go to the really low energy one will click on molybdenum 93 and see how it decays with an energy of 0.40 5 MeV to niobium anyone want to guess what's allowed electron capture only there's not enough energy for positron decay and indeed it draws funny because there's a metastable state but if you go scroll down here there are two pathways allowed both of which by electron capture decaying diagrams quite a bit simpler so we leave this one here by saying it can only decay by electron capture any questions on the odd a before we move on to the even which is a little more interesting cool okay let's move on to the even case so for here I'm going to go back to the overall picture of the table of nuclides click on around where I think potassium-40 is looks like I got there and I want to point out one of these features if you wanted to go under go decay change and maintain the same mass number that's diagonally from upper left to lower right see how all the isotopes here have a forty in front of them the really interesting part is as you cross this line you go from stable to unstable to stable to unstable again the colors here is dark blue represents stable and dark gray represents long lifetimes of over a hundred thousand years so this is one of the reasons you find potassium-40 in the environment in fact point oh one one percent of all potassium in you and everything is potassium-40 it's what's known as a primordial nuclide it's not stable but it's half-life is so long that there's still some left since the universe began or whatever supernova that formed earth got accumulated into the earth but notice it can come from it can decay by a couple of different methods so let's pick one of those stable isotopes calcium for T and put that as the bottom of the parabola on this diagram so we'll put calcium here and in a relative sense we'll put a calcium point right there for its total mass and it could have come from beta decay from potassium-40 or electron capture from scandium 4040 it can beta-decay for about a 1.3 MeV so potassium it's right here let's say it could beta decay with about 1.3 MeV and we'll trace potassium back a little bit figure out where would it have come from interesting doesn't tell us okay forget that let's trace calcium back and say there's scandium 40 and scandium 40 can decay with Wow an enormous 14 point 3 2 MeV let's put that like here anyone want to guess which mode electron capture or positrons much more likely probably positron let's take a look oh boy another complicated one but the whole way down positron positron positron for all the most likely decays you won't find a drawing too every single line I believe that they know that at some point drawing extra lines is futile and they just all overlap each other so I don't know exactly how the algorithm works but it does draw up to some number of possible decay chains if you want to see every single one they are tabulated in a very very long list down below I'm never going to ask you to do something with all of these because that would be insane unless it's a relatively simple decay like that has two or three possibilities and let's see this could have come from electron capture from titanium forty with eleven point six eight I'm easy Wow okay here and there's titanium and let's go in the other direction so I do know that potassium-40 can decay into argon-40 so let's jump there argon is a stable isotope too so potassium-40 can decay into argon-40 by electron capture okay good a more respectable 1.50 5 MeV is positron decay allowed yes why is that / 1.02 to mev yeah anyone had a question no okay so we've got kind of a kink in our math parabola yeah that's correct yep so once the Q value satisfied it is technically possible but if you had something with the decay energy of like 1.0 to 3 MeV it would be exceedingly unlikely so in fact we can take a look at this this is I would say is also going to be on the exceedingly unlikely level and we can take a look so if we look at the decay diagram we know it makes positrons they're not even really listed interesting because so that process would not be allowed but this one because that's about 1.5 MeV should be allowed but since that branch ratio or the probability of that happening is already so low I wonder if it even says yep beta ray with a max or average energy of 482 point 8 MeV we're gonna go over why that energy is so low when we talk about decay next week with a relative intensity of something with a lot of zeros before the decimal place so there you go like you said energies near 1.0 to 2 MeV slightly above it are extremely unlikely but possible and measurable cool and then let's see what could have made argon-40 could have been beta decay from chlorine 40 so chlorine maybe was here and I don't think I have to draw any more so we've got an funny-looking parabola with a kink in it because really you have two bass parabolas overlapping I'm gonna go back to the screen so that the diagram from the notes makes a little more sense what we've kind of traced out here is that there's two overlapping parabolas here there's the one with the what is it the odd Z and the even Z so there you go just like the one on here which i think is for a different mass number yep 102 we get the same kind of behavior where things will mostly follow the lower-mass parabola but sometimes if something gets stuck here it can go either way to get more stable so I want to stop here for new stuff because this is precisely where I thought we'd be at the end of the week and in the next half an hour I'd like to open it up to questions or working things out together on the board or anything else you might have had yeah sure it's going to go in whatever when whatever way makes it more stable so you're never going to have a nucleus is going to spontaneously gain mass in order to get to a different path you can only go down on the mass axis but let's say you happen to be starting here at potassium-40 you can go down via either mechanism to the next mass parabola down that's quite if you're at argon you're stuck and in fact if you want to take a look what do I mean scientifically by stuck I mean stable argon-40 is a stable nucleus that comprises ninety nine point six percent of the argon so that's what I mean by stock is stable and if we look at the rest of the table of nuclides for similar looking places so let's hunt near potassium-40 so notice potassium-40 right here has got stable isotopes to the upper left and the lower right if we look back over here manganese 54 same deal it's got a stable isotope to the upper left and a stable one to the lower right how much you want to bet that when we click on manganese 54 it's got two possible parent new clients I'm sorry two possible decay math methods so let's take a look manganese 54 can either electron capture and positron decay to chromium 54 or beta decay to iron 54 let's take a look at one more to hammer the point home and I think that'll probably be enough cobalt 58 right near nickel 58 and iron 58 interesting that one's not allowed unless there's more down here so you can electron capture to iron 58 but there's no allowed decay to what was it nickel 58 okay let's look for more chlorine 36 has argon and sulfur on either side there it is beta-decay and electron capture and how much you want to bet there's basically never a positron here but basically not actually never so you get a positron 0.01 percent of the time and electron capture one point eight nine percent of the time where's the other 98 and change percent right here in the beta decay so in this case chlorine 36 will preferentially beta decay if you also notice it's a let's see you don't know if that actually matters so but I'm gonna say it's more likely to beta decay so when you sum these up you get a hundred percent of the possible to case let's see how many energy levels there are there too hopefully not too many that qualifies as not too many yeah John they'll be due to those as well as some other processes which we're going to cover on decay but if you notice I've been giving you a lot of flash-forwards in this class we've introduced cross-sections as a thing that a proportionality constant between interaction probabilities we're going to hit them hard later I've also been kind of introducing or flash forwarding different methods of decay so there's also alpha decay there's also isomeric transition or gamma emission there's also spontaneous fission this is the whole basis behind how fission can get working without some sort of kick-starting elements so maybe now's a good time to show you let's go to uranium-235 and see how it decays a de ganar D undergoes alpha decay de thorium 231 most of the time and if you look how it's no it's not terrible we can make sense of this it also undergoes SF which stands for a spontaneous fission so one out of every seven Lords it you know seven out of every billion times it will just spontaneously fizz into two fission products and this is why if you put enough uranium-235 together in one place you can make a critical reactor in reality you don't tend to want to put enough u-235 together to just spontaneously go critical we use other isotopes as Kickstarter's for example californium I think it's two fifty-two let's take a quick look there we go californium 252 undergo spontaneous fission three percent of the time it's even heavier even more unstable so there is a reactor called high fer or the high flux isotope reactor at Oak Ridge National Lab one of its main outputs is californium Kickstarter's for reactors so to get things going you put a little bit of californium in as a gigantic neutron source and then you don't really need it anymore once it gets going so it's one of the safer ways of starting up a reactor is put in a crazy neutron source and then once it gets going take it out or leave it in and burn it I'm not actually sure which one they do yep it is when we get to now is a good time to introduce the super-heavy element since you asked so a lot of these older elements were named after all actually this is kind of a hobby of mine so I don't know if you guys saw the periodic table outside I collect elements because if you're gonna collect something you might as well collect everything that everything else is made of it's the same reason I went into nuclear energy I started off course 6 or 6/1 specifically electrical and I was like well I could be designing like the next screen for a cell phone or we could solve the energy problem which is the problem all others are based off of so my whole life theme has been go to the source that's why I came here in high school and never left that's why I declared course 22 that's why I collect elements and probably is the reason for many other things which only a psychiatrist could diagnose but let's look at some of the other elements for example yttrium I think it has an isotope 40 anyone know no it doesn't have a 40 what about a 50 60 hundred whatever at least it knew that why was you a trio anyone know where this is coming from 89 seems high oh my god you're right okay gotcha you work with it awesome anyone know what this is all about get trium there's a town called Iturbi in sweden where large deposits of yttrium and ytterbium or YB tend to be found or DB name for dubnium so let's say the really basic elements tend to come from Latin F F stands for iron which actually stands for theorem led is plumbum gold as Orem silver AG is argenteum I don't know if I'm saying that right I never took Latin and I've never heard it spoken of course and then a lot of the heavier and heavier elements as we go our being named for more and more famous scientists or places where they tend to be made like dB I'm gonna guess 264 a mass there oh nice for a doob nah in Russia that has got one of the few gigantic for heavy element colliders where they're constantly synthesizing and characterizing these super-heavy elements so finally they said you know what they've made enough of these in Dubna let's name one of the elements after them or SG seaborgium for glenn seaborg or n o nobelium for Alfred Nobel yep ah do they intersect I've never seen a case where they intersect that would make for a crazy situation indeed however part of what the homework assignments about is to derive an analytical form for a mass parabola and then check the data to see how well it works so for any cases where you have an even mass number and you have either odd odd or even even nuclei you can check those equations analytically to see if they'll intersect so the top parabola for potassium-40 let's take a quick look at how many protons and neutrons it has potassium has a proton number of 19 which means it has a neutron number of 21 so the top parabola is odd n and odd Z where the bottom one is even N and even C whereas for odd mass number nuclei it has to be either odd even or even odd else it would be even which is a funny sentence when you say it all out loud yeah so that's the idea here is that notice that the even even parabola tends to be further down all those nuclear magic numbers - 8 20 28 I'm not going to quote the rest those little ones I know all even numbers so any other questions on these math problems before we launch into super heavy elements yep uh-huh analytically or experimentally which question so analytically analytically there should be some isotope of well not potassium that wouldn't be allowed so in this case the the stable element positions have got to kind of switch off shouldn't they so if that's potassium 40 that would still have to be potassium you don't really have another choice there isn't really a position there is there analytically that's the interesting thing is that you can either be odd odd or even even for an even mass number but you can't just take off one Neutron from potassium-40 and then you've got potassium 39 then you're on a different mass number or if you exchange a proton and a neutron which you pretty much do in either of these directions there's no way to get straight down here yep fraud what proof sorry yeah let's go back to that slide just to make sure you mean the pairing term in the semi empirical mass formula yeah yeah so for aw dog nuclei and D Delta is negative which means lower binding energy which means higher mass and that's why we see it bump up on the mass right here yep and you have a second part of the question Oh ah we can actually relate so we can relate the binding energy to the mass and the mass parabola analytically because the binding energy is equal to Z times protons plus n times mass of neutron minus the actual mass of that same nucleus a comma Z so they're actually directly related just negatively so something with a higher mass is going to have a low binding energy which means it's less bound and less stable and indeed the further up the mass scale we go the higher those beta or electron capture or positron energies are and there's another thing you can check to which is the half-life half-life is what we'll be talking about on Tuesday it's how long before an average amount of a substance has undergone radioactive decay so let's look at some of these isotopes and start looking at half-life trends as another measure of stability so potassium-40 has an exceptionally long half-life so it's relatively stable let's take a look not at either the stable isotopes but let's go up the mass parabola chain in one direction calcium for T scandium-46 can diem for T scandium 45 less than a second and it's got quite a high decay energy by whatever method you want to use let's go up to titanium for T anyone want to guess is eating the half-life is gonna go up or down let's see if the half-life goes down we know the decay energy goes up indeed half-life goes down from 182 milliseconds to 50 milliseconds and let's say titanium forty could have come from Wow two proton decay from cr.42 with a half-life of 350 nano seconds so as we go up the mass ladder and down the stability ladder the half-life decreases which kind of follows intuitively something that's exceptional Stables should have a half-life of infinity and something that's exceptionally unstable should just blow apart instantly like remember the first week of class we talked about helium for grabbing a neutron becoming helium 5 and instantaneously going back to helium 4 if you look at helium 5 it's half-life is measured in MeV or 7 times 10 to the minus seventh femtoseconds so if helium 4 absorbs a neutron it simply doesn't want it and gets rid of it in ten to the minus seven femtoseconds which would tell us that it's exceptionally unstable so I hope that's a long-winded answer to that question but what does it mean to go to be going up in the mass levels any other questions on mass parabolas or the liquid drop model or stability in general yes so if you're changing one neutron to a proton in each case you're switching back and forth from the odd odd to the even even mass parabolas so if I were to redraw these dots more to scale this would have to be on the odd odd and oh let me draw them a little better yep that's on the odd odd and that's on the even even yeah so excuse my poor drawing skills but if you're switching one proton to a neutron or vice versa by definition you're jumping back and forth between these parabolas that's a good question for clarification you had a question too [Music] the semi-empirical I'm sorry the semi empirical mass formula is a good way to get an analytical guess at most of them if you want an exact answer always use the actual binding energy I would not say it's used much now except well that's gonna be one of your homework questions is this formula predicts that as you get heavier and heavier and heavier nuclei should just continuously get less stable and that was as of when this was derived let's say decades ago we now know something different is happening so if you look at the table of nuclides you can sort of see some swells in the number of black pixels until it cuts off and this region actually where we think super-heavy elements happen I want to jump to the actual table of nuclides which I'll say is our snapshot of knowledge today and go all the way to the top and our knowledge kind of cuts off at these elements which are for now temporarily named in a very uncreated way we don't even know anything about them uun probably going to have a proton number of what 110 I don't know what the prefixes are but you you you would be on one one one probably has 111 protons beyond here off the screen or probably up into the next room it's predicted that once you approach the next magic number in nuclei there should be an island of stability where and it may not necessarily be totally stable but the half-life should go up again and we should be able to synthesize super-heavy matter and if you actually graph Neutron number versus half-life so notice how we were looking at half-life as a measure of stability it starts to go up then comes down and then to the extent of our knowledge it is going back up again to the next predicted magic number so what we think should be happening is half-life should be continuously going up and yeah you had a question well whatever you want it's gonna be it should be dense as all heck because nuclear matters quite a bit denser than ordinary matter and quite a bit as quite an understatement so what would you do with super heavy matter a lot of it could be used to probe the structure of matter there's a lot about how the nucleus is constructed that we don't know and beyond the scope of this course would also be an understatement there's folks that are making their careers now and figuring out what are the forces between nucleons why do things spontaneously fizz at the rates that they do you'll even hit a little bit of this in 22:02 when you can calculate the rough half-life for alpha decay using quantum tunneling through the potential barrier in a nucleus and so the more nuclei we have to mess around with the more data and real examples we have to study but practical applications well I could imagine we might find something denser than osmium osmium right now is a density of about 22 grams per cubic centimeter this stuff zirconium it's about 6.9 or so Steel's like eight LEDs like eleven Mercury's like nineteen via any of you ever played with liquid mercury before yeah this is a don't try this at home kids kind of moment my grandfather happened to be a dentist so we happen to have a lot of mercury to mess around with and it's like unintuitive ly heavy it's unbelievable a one-pound jar is about that big I think would be cool if we could time something even denser and then really really dense matter happens to make really really good photon shields gamma seen on the Star Trek thing I mean this in the actual nuclear physics sense is the best way to stop gamma rays for gamma shielding is just put more matter in front of it and if we find a denser state of matter that's earth stable you then have a smaller gamma shield so there are practical applications to in radiation shielding they also might make awesome nuclear fuel because you better believe they're gonna fizz like crazy so who knows maybe we can I don't think that would be cost-effective but it would probably work so the way they're doing this is actually slamming calcium 48 nuclei into other super-heavy elements that have exceptionally long half-lives so if you can't read what the screen says this here is berkelium for berkeley with proton number 97 mass number 249 let's take a look at EK 249 which happens to be way beyond uranium so it's definitely not a stable isotope but it has a half-life of 320 days that means you can make a bunch of it chemically separated make it into a target and fire calcium 48 nuclei into it anyone want to guess why do we use calcium 48 and I'll give you a hint and write the proton number for calcium the isotope that we use it's calcium 28 Coliseum 48 anyone want to take a guess why start here why not just smash to berkelium Xin to each other calcium 48 happens to be exceptionally stable because it's got two magic numbers it's proton number is twenty one of those peaks of stability and it's Neutron numbers 28 so start with something super stable something with a lot more binding energy to begin with and you maximize your chance of making something with more binding energy that won't just spontaneously disappear so there are reasons calcium 48 was chosen and not something heavier or lighter if we go back to that article you can see what happens here is you make some element 117 which has yet to be made and it undergoes alpha decay until it reaches some rather stable you know 17 seconds that's pretty exceptional and if you notice the trends here as you decay the alpha energy steadily goes down and the half-life steadily goes up and so what you do is you make an super super-heavy element hoping that it will decay and rest in one of these islands of stability beyond the magic numbers that we know right now which I thought this was super cool because this is actually how opening now like new elements are made I think we've been seeing one a year or so for the past few years on average there might have been a year when there was more than one announced recently there's only a few places in the world doing them but you can start to already with two weeks of 2201 you can start to get a handle for why do they use the nuclei they do and then what sort of things are you looking for decays with lower and lower energy mean you're already starting to gets less steep on whatever imaginary mass parabola don't quite know how to draw this one because it's beyond anything we know and as the half-lives keep going up you can tell that it's reaching a measure of stability however to get you started on the homework for the open ended problem I think I'll bring it up right now so you guys can take a look so let's go to the stellar site hopefully it doesn't call me good at the problem set 2 this is the way I know that everyone's seen the pset 7 days before it's due cuz I'm gonna put it up on the screen so you can see it all the way at the end predicting the island of stability does the semi empirical mass formula predict the island of stability well let's start you off with the easier part of the question which is yes or no and I'm gonna leave you to the why and the how if we graph binding energy per nucleon versus mass number the semi empirical mass formula predicts something like this what happens as we go beyond the realm of known mass numbers anyone how should I extend this curve would you say Alex just keep going yeah does this predict an island of stability I don't think so so that's one of the few questions I'm asking you in this homework and it's up to you guys use your creativity again this is an open-ended problem I'm not looking for a specific answer I want to see how you think and how you would change this formula to account and actually predict the island of stability while still satisfying the mostly correct predictions from the elements we know so sorry go ahead well yaar you're on the right track if you wanna make if you want to show stability you'd want it to maybe have a higher value right here higher binding energy per nucleon would correspond to a lower mass which would correspond to higher stability so how would you predict this island of stability and then more specifically how would you reconcile the misty inaccuracies in the semi empirical mass formula because we know it doesn't work very well for all cases there are some cases like right around here where it works great and there's some like right here and right here where it really doesn't and get things wrong by like ten MeV which is pretty significant you know that's like four digits on the mass scale like the fourth decimal place that's huge to a nuclear engineer so that's something to get thinking about and remember I did tell you that there will be some open-ended problems I'm gonna mark them as open-ended so you actually know we're not looking for a right or wrong answer this is one of those kinds of things where we want to see how you think and what do you think is missing there's other hard problems where we give you the answer because I'm not interested in you deriving some explainin expression and getting it right I'm interested in the derivation process what are the steps you choose what sort of assumptions do you make what sort of terms can you neglect and say that's in the ninth decimal place I'm going to forget it so in this case we give you the answer because we're going to grade you on the process and you can use the answer to check your process and see if you're on the right track or not for the skill building questions we actually do want you to come up with some sort of an answer like explaining the terms in the semi empirical mass formula or modifying an equation to calculate something else we will be looking for a right answer there but those are questions to make sure that you get the basics of the material if you can answer all of the questions in the first half of these P sets fairly quickly let's say in three or four hours you're totally on the right track the hard ones is because this is MIT and we want you to think beyond just knowing what's in the Turner book or the Yip book like I said you guys are the leaders of this field so any other questions on instability in general yes you say correct so that would be even N and even Z are on N and odd Z like in the reading and like on these mass parabolas yep any other questions yes there's a few roots so the question was is the only is the only reason people think they'll be super heavy elements because the mass increases right so in this case the the mass will always are you talking about now the total mass or okay we have a few things to go on so all there are a number of different aspects of nuclear stability that are all pointing to the same conclusion one of them you can see on this graph here if you look at the alpha decay half-life as a function of Neutron number it doesn't just increase or decrease monotonically it swells up and down and it reaches a relative maximum near certain magic numbers we can confirm that with the lower mass nuclei it doesn't work for really low mass because tiny things don't tend to go under undergo alpha decay but there are patterns that were simply recognizing and saying well if this is the next magic number it should continue to increase and I should mention to this scale is logarithmic so the top right here is like 10 to the 4 seconds just so you know there are 86,400 seconds in what is it a day and 3 times 10 to the 7 seconds in a year so if this graph let's say for Z 111 were to continue on its track it should reach like 10 to the 9 or 10 to the 10 which could be like hundred year lifetimes or 100 year half-lives which means definitely you can chemically separate them and do things with them I don't know if they would be safe enough to deal with but we also don't really know what's going to happen you can see that there is some uncertainty and things don't always follow the trend even the error bars are outside the dashed lines but so we have this to go on we have the alpha decay half-life we also have the alpha decay energy as you approach an island of stability something that's more stable won't give off as much kinetic energy to its alpha particle there's also for the ones that you can actually measure that live long enough you can measure their mass-to-charge ratio and actually get a good picture of their actual mass so we would expect the mass defect to follow a certain trend as we go up the mass is always going to increase if you add more nucleons it's going to increase but the mass defect which is the real mass minus the atomic number mass if stability were to increase do you think the mass defect would increase or decrease with more stability let's take a quick look at this if a were to stay the same a shrinking real mass and remember lower mass means more stability would mean a higher or a lower mass defect I mean a lower mass defect or a lower excess mass as you'd call it so in this case you would expect the mass of the nucleus to be smaller than its a if it was going more stable and all of these trends work in the same direction which is saying okay so we have the Alpha energy we have the mass defect we have the half-life all pointing to the same thing that something should be more stable and we have some patterns to go on but our understandings kind of incomplete so yep whoo good question so the question is if super-heavy elements exist do they exist out there in space I think there would be a couple places they would exist the source of most of the elements beyond iron is supernovas where a regular old fusion doesn't cut it anymore when you hit the maximum of this binding energy per nucleon curve you're at about iron-56 that's why stars tend to form a core of iron before it goes really bad in whatever way it does for a star there are multiple ways when you get a supernova you have an insane explosion and the core gets compressed from the outside forcing fusion of heavy elements to happen that's because you're putting in extra kinetic energy so it's like you have an endothermic reaction where if Q is less than zero how do you make that reaction happen add kinetic energy which can come from a tremendous explosion outside of the outer regions of the star so who's to say that some of these super-heavy elements aren't formed in supernovas I think they would be but would they actually make it out to be part of earth and then let's say live the five billion years that Art's been around we don't know if their half-lives are long enough there very well may have been some five billion years ago or when the supernova was made but we haven't detected any here on earth so we know that they're not five billion years stable rather I wouldn't even need to say we know that but we have a pretty good idea that's a great question is like are they naturally made probably yeah any other questions I like these outside the material ones we can take things beyond our known universe start to explain them you