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 so you guys asked to do some numerical examples of the stuff that we've been learning today and I've got a fun one for you a very real one because it happens all the time in reactors all over the world so to set the stage for this they suppose I had a you know just suppose I had a radioactive cobalt-60 source that was calibrated in March 2011 to be approximately 1 micro curie and it's now October 2016 so let's start off with the easy part let's say I posed the question how active was this source actually when it was made because it just says what does it say 1 micro Curie and supposedly it was one micro Curie on March 2011 what does that mean the actual activity could be there's not a lot of confidence in that number who remembers significant figures from high school chemistry and physics this is when they're important because you got to know what you're buying so what is the actual activity of this source when it was calibrated what is our uncertainty on this anyone remember this from sig figs I hope so not I'll refresh your memory it's plus or minus the next decimal point so really this was 1 plus or minus 0.5 micro Kure so you might have 1/2 micro Curie source you might have a 1 and 1/2 micro Curie source they specifically decided to leave out the second decimal point so they're not liable for a source that's out of that calibration level so we supposedly have a 1 micro Curie source and I've measured it on let's say it's now October 2016 suppose I went and made a measurement and it was point 5 2 micro Curie's and we want to know how active was this source actually when we got it where would I begin okay yeah that's why I've got the table of nuclides right here so let's look up the half-life of cobalt-60 and there it is right there 1925 days so we know that the half-life for cobalt 60 is nineteen twenty five point four days which equals and I pre did the math out that I wouldn't spend lots of time on the calculator activity how many seconds is that one point six six times ten to the eighth seconds and we also remember that activity equation where the activity as a function of time equals the original activity times e to the minus lambda T the only thing missing here is the initial activity in lambda so how do we find lambda and remember remember that expression yep and we know that lambda equals log of two over the half-life which in this case is about 0.693 nine over one point six six times ten to the eighth which equals four point one seven times ten to the minus nine per second and so now it's pretty easy we know what lambda is we know what our current a is so we can say that our original activity is simply our current activity divided by e to the minus lambda T what's T well I made an approximation here it's been about five years and seven months I assumed that a month has 30 days on average which comes out to T I remember calculating this already one point seven four times ten to the eighth a second so we plug in that T right here plug in that lambda right there and we get our initial activity what did I get was 1.07 micro Curie's now hopefully this result is fairly intuitive because the half-life of cobalt 57 is 1925 days which is about five and a quarter years and it's just over five and a quarter years since this source was calibrated and we have just under half of the original activity so hopefully that's an intuitive numerical example now let's say how many atoms of cobalt 60 did we have or better yet what was the mass of cobalt 60 where would I go for that give you a hint it's on the screen yeah so for this plastic check source right here what's the actual mass of cobalt-60 that we put in in the beginning again supposing I had one of these right in front of you you could look at binding energy which is a form of mass or luckily we've got the atomic mass right up there in AMU so again this is a quick review of high school chemistry we need to find out how many atoms we have in this pellet and once we know the number of atoms and we have the molar mass up there and either amu per atom or same thing as moles per gram pretty much we'll know what the mass of cobalt-60 was so one important conversion factor to note is that one curie of radiation is three point seven times ten to the ten Beca rel and remember that one becker l is one disintegration per second so if we know the initial activity of our material using our decay constant we should know the number of atoms that we had right there so we know our initial activity a naught is 1.07 micro curie and anyone remember what's the relation between the activity and the number of atoms present just yell it out hope you'd know that by now this look familiar to anyone where the activity is directly proportional to the number of atoms there times its decay constant so since we now know a naught we can find n-not because now we know lambda as well so the number of atoms we had at the beginning is just our activity over lambda and our activity is 1.07 micro Curie's let's convert up to Curie's so we know that there is one Curie in 10 to the 6 micro Curie's and we'll use that conversion factor right there times three point seven times ten to the ten becquerels per one Curie divided by our decay constant four point one seven times ten to the minus nine for a second let's check our units to make sure everything comes out we have a micro curious on the that's right I brought a canceling color we have micro Curie's on the top micro Curie's on the bottom Curie's on the top Curie's on the bottom and we have a becker ELLs which is a disintegrations per second and we have a per second down on the bottom so the per second goes away and the becquerels just becomes atoms and so we actually get and not is I think we say something times 10 to the 12th yeah nine point five times ten to the twelfth Adams that's the way it was so the final step how do we go for a number of atoms to mass anyone tell me yep so the last thing we'll do is we'll say we have nine point five times ten to the twelfth atoms times Avogadro's number which is one mole in every six times 10 to the 23rd atoms and then we go to the table of nuclides to get its atomic mass this is one of those situations where you don't need to take the eighth decimal place because you're not calculating you're not converting from mass to energy you're just getting mass so if you might wonder where did all the decimal points go think of what type of calculation we're doing if we started off with a one significant digit number do we really care about keeping the eighth decimal place and the rest of everything no definitely not we're not turning a mu into MeV in which case the sixth decimal point could put you off by half an MeV or like the rest mass of the electron we're just getting masses here so let's just call that fifty nine point nine that's enough for me fifty nine point nine grams in one mole of cobalt 60 and just to confirm we have atoms here atoms there moles there holes there and we should get a mass in grams and this came out to 0.95 nano grams of cobalt 60 not a lot of cobalt 60 can pack quite a wallop in terms of activity so even though this has a pretty long half-life as far as isotopes go it takes very little of it to have quite a bit of activity certainly enough for our fairly inefficient handmade Geiger counter to what's going on pretty neat huh cool now let's say I ask you another question a simpler question how many disintegrations per second are coming out of that cobalt-60 source let's say we wanted to find the efficiency of our Geiger counter you'd have to know how many counts you measure you'd have to know how far away your source is and you'd have to know how many disintegrations per second are happening so how do I get to the number of disintegrations per second that's right well just take our current activity which is 0.5 to micro Curie's and so we'll say point 0.5 2 times 10 to the minus 6 Curie's times three point seven times ten to the ten Becquerel in 1 Curie is well we have curious here we have Curie's there and this comes out to what was our current activity was in the tens of thousands but 19,000 barrels so we know that right now this source is giving off about 19,000 disintegrations per second is that how many gamma rays it's giving off do we know that yet why not I heard a lot of knows what other information do we need to know sure luckily we've got that right here so if you look here the mode of decay is beta decay to nickel 60 so cobalt 60 is actually primarily a beta source however it's used for its characteristic gamma rays so let's take a quick look if we look at its decay diagram it's pretty simple and let's just say somewhere near a hundred percent of the time it beta decays upto this energy level and can undergo any number of transitions like this the only two we really tend to see is like that one and that one are the most likely ones so on average each disintegration of a cobalt-60 atom is going to produce two highly energetic gamma rays so actually what you'd have to know for this source is despite it being nineteen thousand Becker ELLs of cobalt-60 it's giving off 38,000 gamma rays per second so that way your source calculations wouldn't be off by a factor of two then if you know the distance between your source and your detector and you know how many of those gamma rays are going through the detector itself which we can count calculate with a solid angle formula which I'll give you a little later you'll know how many of them should interact in here once you learn photon nuclear interactions and you'll know how many of them actually get captured and that's how you can get the efficiency of the detector so we've kind of filled in half the puzzle you now know for a fixed source how many atoms there are how many disintegrations there are and how many gamma rays you expect it to give off and then later in the course we'll tell you how to figure out how many of them make it into the detector and how many of those should interact in the detector does everyone clear on what we talked about here yep [Music] double the number of cobalt disintegrations so yep so I'm I'm ignoring the point O two and point O six percent decays because they're extremely unlikely and having looked this up ahead of time I know that this transition and that transition are by far the most likely ones and if you don't know that sure do those right there the 99.9 something intensity ones it's all there so I had some questions on Piazza what happens when we can't read the pixilated decay diagram that's just there for fun the actual table that will tell you all the information you need to know is right below and I just decided all right forget the ones that are to e to the minus six likely or 0.007 whatever too many zeros so you can say that on air - probably two or three significant digits that gamma ray and that gamma ray are the only ones you tend to see is everyone clear on how I made that determination cool yes and each one of those that's right the number of disintegrations is the number of atoms that leave this position so let's take a crazier example but I've already had you look at americium-241 sort of bring home the message that the number of americium disintegrations does not necessarily equal the number of gamma rays that you will expect to see because any one of these is possible there's an 84% likely one that looks like it goes to the third level from the bottom because it's just a third number I'm seeing here if you don't believe that then check how much you have to zoom out to see what's going on here right let's see intensity yep the third most energetic alpha ray so the third from the bottom is indeed the 84% likely one then there's another one that's 13% likely so we can't discount that either so you're gonna hit anything from like the second to the fourth energy level and any number of those gamma cascades that come off there so the intensity of your source in Becker ELLs or Curie's does not immediately tell you how many gamma rays or other disintegration products you should expect there are some pretty simple ones like the dysprosium one that we saw was it like 150 one I've been here before it's not purple no that's a complicated one all right don't remember which dysprosium isotope just had a single decay but the only time that you would get one particle per disintegration is if your decay diagram looked like this that's your parent and that's your daughter that's the only time you can expect one thing if you have a more complex two K diagram than that you'll be looking at more than one quantum of radiation of some form per disintegration so is that unclear to anybody cool now let's get into a more fun problem and I'll leave this up here since we're gonna refer to it a fair bit so this is a good example of activity half-life mass number of atom calculations we're gonna use this information to answer a more interesting question how is this made anyone have any idea what sort of intentional nuclear reaction could have produced cobalt-60 and I'll go back to cobalt 60s we can get its proton number all right you say a little louder sure you can have a neutron bombardment so you can have cobalt 59 absorbs a neutron becomes cobalt 60 with some half-life and then will decay with some decay constant lambda and become in this case it undergoes beta decay to nickel 60 let me make sure I got the proton number right going from memory here hooray okay how do we set up the series radioactive decay and production equations to describe this phenomenon so we have a physical picture of what's going on here how do we construct the differential equation model the same way we've been doing it Friday and Tuesday so someone kicked me off what do we start with [Music] okay destruction times flux [Music] okay let's make a couple of designations let's call cobalt 59 I'll use another color for this just so it's clear we'll call cobalt 59 and one we'll call it cobalt 60 and to coal nickel 60 and three just to keep our notation straight and then continuing our notation pile here you mentioned that there's going to be some Sigma some microscopic cross section times some flux the number of neutrons whizzing about the reactor times the amount of cobalt 60 so let's stick our DN 1 DT equals no production minus some destruction does this look eerily familiar to any of you forget the fact that it's a sigma and a flux just treat those as constants what does this look exactly like exactly it looks just like DN 1 DT equals minus lambda and 1 right yeah that's exactly it this in effect is like your artificially induced decay so the probability that one atom of N 1 absorbs a neutron times the number of neutrons are there gives you some rate at which these atoms are destroyed just like a lambda gives you the rate at which those atoms naturally self-destruct so you can think of this Sigma times flux like a lambda mathematically they're treated identically the only difference is we're imposing this Neutron flux so it's like an artificial lambda which means solving the equations and setting them up is exactly the same so how about n 2 what's the production and destruction rate of 12 volt 60 first of all someone yell it out what's the creation rate which lambda so what is lambda 1 this one yeah okay so let's keep with these these notations will say I'm gonna say this is like a lamb to artificial biome to keep with Sigma flux and 1 and just so we keep our notations straight I want to be able to cleanly separate the natural and the artificial production and destruction and what's the destruction rate they said Sigma Phi n 2 is that the physical picture we have up here not quite because like we can see here cobalt-60 self-destructs on its own right yeah so however that actually is a correct term to put in the other one that we're missing would be minus lambda n - okay so you know what let's let's escalate this a bit into reality and say we're going to we're gonna do this it's the actual equation is not gonna be that much harder so thanks Sean but let's do this yeah yes good question that's what I was getting to so there is an absorption cross-section for every reaction for every isotope and now I'd like to show you guys where to find them on the 2201 site there is a link to what's called the Janus database which is tabulated and plottable cross-sections of all kinds so every single database that we know of it's updated continuously by the OECD so you can trust that the data is updated I won't say accurate because some of these cross-sections are not very well known so I've already started it up in this case we don't have Neutron capture data for cobalt-60 let's keep it in the symbols for now and then later on we're just going to say it's probably zero but let's keep it together so for cobalt 59 let's double click on that and you are presented with an enormous host of possible reactions like right here you have n comma total which is the total cross-section for all interactions of neutrons with cobalt 59 and you can see that this varies with the same sort of shape that I was haphazardly drawing it's low there are some resonances and then it continuously increases at lower energies is this the right cross-section to use if it accounts for every possible interaction of a neutron with cobalt 59 no so what other reactions besides the one that we want which is capture could this account for yep scattering vision end-to-end production sometimes one Neutron goes in and to come out like for beryllium like we talked about the beginning of class so you've got to know to choose the right cross-section and in nuclear reaction parlance that shorthand right there that's n comma total that accounts for elastic scattering inelastic scattering to any energy level capture fission end-to-end reactions sometimes proton release sometimes exploding whatever nuclei do so let's look at our other choices shrink that we have the elastic cross-section compare that with a total and you get a general idea of how much the total and elastic cross-sections actually matter so a lot of those resonances are elastic cross-section residences but there are other reactions that are responsible for a lot of the other craziness going on so let's unselect that oh I'm sorry I'm an empty one let's compare mt1 and mt2 so mt2 elastic ah there we go okay that's what I was hoping for so right now the red one is the total cross-section and the green one is the elastic cross section and you can see that at high energies the total cross section is mostly the elastic cross section but at low energies especially right around here at the thermal energy of neutrons and a reactor there's something else responsible that's probably what we're going after so let's keep looking through this janice database hey there's the end-to-end reaction if you guys want to see how likely this is it's another one of those reactions that look at that it's zero until you get to eleven MeV so what do you guys think the Q value for end-to-end production of cobalt 59 is very how negative it's on the graph yeah negative looks like 10 or maybe ten and a half MeV oh hey awesome so there you go yeah indeed it's very energetically unlikely to produce to fire in one neutrons and get to but if you have a ten point four five four MeV Neutron you can make it happen pretty cool huh that's not the reaction we're going for what we want let's see if I can find it proton plus Neutron Neutron plus deuteron all the in elastic energy levels there it is capture the gamma reaction here is what's referred to as capture and there we go a nice normal-looking cross-section so for cobalt 59 if we go down to about 0.025 evie here read it off it's about 20 barns because you guys asked for a numerical example so let's say that our capture cross section for cobalt 59 is about 20 barns which is to say 20 times 10 to the minus 24 centimeters squared let's also put up our capture one for cobalt 60 so let's go back to our table take a look at cobalt 60 I think I know what the answer is going to be which is we don't know not in this database unfortunately so for symbolism let's keep it there but we're gonna say well we don't know yeah so let's designate these different cross sections let's call it Sigma 59 Sigma 59 and Sigma 60 so those will be our two cross sections we'll just call this one Sigma 59 we call this one Sigma 60 and we already know the lambda for cobalt-60 so let's say the lambda for a 60 cobalt from this stuff up here four point one seven times ten to the minus nine per second let's just refer to that as lambda for ease of writing things down so we've got a complete set of reactions for a DN 2 DT what about DN 3 what's the production rate of n 3 yep lambda n 2 anything else what about the destruction rate is what it's a stable isotope but you're on to it Shawn so what would you add based on what you added to end to yeah there's gonna be some new Sigma let's call it Sigma nickel 65 and 3 find out let's go to our tables there there's data for nickel 60 let's look up its absorption cross-section so we'll scroll down to our Z gamma our capture cross-section plot it take a look at around 1 minus 8 and it's like two barns not negligible so our capture cross-section for nickel 60 keep overriding myself and then I remember it's a blackboard two barns which is tell you 2 times 10 to the minus 24 cm squared so we can just refer to that as Sigma nickel for the purposes of this problem we don't particularly care how much nickel 60 we're making nickel 60 is a stable isotope of nickel yeah forget it so for the purposes of this problem forget the N 3 equation we don't care how much stable nickel 60 they were making because it's a lot cheaper to get it out of the ground probably something like 10 orders of magnitude cheaper yep even one okay yep so the incident energy I picked because one thing we had gone over is that the thermal energy of the neutron is around 0.025 evie which is equal to 2.5 times 10 to the minus 8 MeV so I took that value about 2.5 times 10 to the minus 8 MeV so around here and just went up and on a log scale it looks to be closest to two barns you can always like get the actual value so if you want to zoom right in I'm gonna keep zooming in to the 10 to the minus 8 3 giin maybe not now you can set your bounds accordingly so o 0 or negative values yeah okay whatever let's just read off the graph for now you can use this tool to get the actual value but for problems like this I think estimating it from the graph is gonna be fine when you get into 2205 and you're like what's the actual flux in the reactor to within 1% or something estimating from the graph is no longer allowable there are tabulated values of these things oh yeah so you can actually set the plot settings get the tables cool Oh tabulated there we go so you can read off values of the cross-sections from a table like that but for now since we want to make sure to get this problem done let's just stick with the graph so we don't care about the end 3 equation we're also going to say well let's not ignore Sigma 60 yet so how do we solve this set of equations let's make a little separation here first of all the easy one what's N 1 is a function of time 59 5 and what else and there's an n1 naught and there's a T because as you're burning it out it matters how much time you have doesn't this look eerily similar to an N equals n naught e to the minus lambda T again it's like exponential artificial decay because we're burning those things out for a fixed amount of neutrons going in the amount that we burn is proportional to the amount that is there so that's the burn rate that's the amount that's there this is our artificial lambda so that equations easy what we really want to know is what is n2 as a function of time that's the 60 million dollar question and I'm not exaggerating there because cobalt-60 is expensive so the question I'm posing to you guys on the homework so for those of you who've started problems at 4 I'm gonna be swapping out the noodle scratcher problem to this problem that we're going to begin together in class and you guys are going to finish on the homework so I'm kind of giving you a help on the homework what is n 2 the amount of cobalt-60 in your reactor as a function of time and what is your profit for running the reactor as a function of time assuming a few things so let's set up some parameters I'm gonna say that the Neutron flux is the same as that in the MIT reactor which is about 10 to the 14th neutrons per centimeter squared per second we already have all of our lambdas we have all of our Sigma's I'm going to say that the cost of running the reactor is and I have a quote on this $1,000 per day which is the same as 1 cent per second not a bad rate to stick something in the reactor right it's not bad at all if you had to build your own reactor your daily operating costs would actually be a million dollars a day who are a commercial power plant so every time a plant goes down you lose a million dollars a day plus the lost electricity or whatever that you have to buy so let's put that in there and from the cost of this hypothetical cobalt-60 source we know that cobalt-60 runs about $100 per micro Kure because these sources run for about $100 and so the eventual problem that we're going to set up here and you guys are gonna solve on the homework and we can keep going a little bit on Friday if you want is at what point at what tea do you shut off your reactor and extract your cobalt 60 to maximize your profit and this is an actual value judgment that folks that make cobalt 60 have to make how long do you keep your nickel target in there and not hit diminishing returns because you're always going to be lets say increasing the amount of cobalt 60 yeah increasing the amount of cobalt 60 that you make if your source is basically undeletable until you reach some certain half-life criterion but it might not make financial sense to do so so let's start getting the solution to end to let's see so I'm going to rewrite our end to equation and we can start solving it I'm sorry that's adn to DT equals Sigma 59 flux and 1 minus lambda n 2 minus Sigma 60 flux and 2 so how do we go about solving this differential equation yep the old integrating factor first we want to get rid of the n1 because that's another variable and we've already decided right here that N 1 is N 1 naught times e to the minus Sigma 59 flux t and we haven't specified what's N 1 naught let's do that now last number we'll put in is we started with a 100 gram source of cobalt 59 that when we write it in isotope parlance it sounds exotic but that's actually the only stable isotope of cobalt so that's just a lump of cobalt from the ground that we stick in so we know what N 1 naught is so we can now rewrite this equation as let's just go with n2 prime for shorthand and we'll put everything on one side of the equation so we'll have plus lambda plus Sigma 65 and 2 minus Sigma 59 Phi times e to the minus Sigma 59 Phi T equals zero so what is our integrating factor here e to the lambda yep so it is e to the integral of whatever's in front of our n to lambda plus Sigma 60 Phi DT which equals e to the lambda plus Sigma 60 Phi T so we now multiply every term in this equation by RM you are integrating factor so let's say we have an to prime times I'm just gonna use mu for shorthand so it's gonna take a long time to write plus let's see that right there is like mu prime because we have if we take the integral of mu times and let's see yeah oh yeah this is right so we have lambda plus Sigma 60 and 2 times mu minus mu times Sigma 59 Phi e to the minus Sigma 59 by T equals 0 this stuff right here is like the expanded product rule so we can write it more simply so we can say n 2 times mu prime let's see equals mu sigma 59 phi e to the minus sigma 59 Phi times T so next we integrate both sides and we get n 2 times mu equals let's see let's expand everything out now so that stuff would be Sigma 59 phi e to the lambda plus sigma 60 Phi minus Sigma 59 Phi times T so if we integrate all of that we're going to get Sigma 59 Phi I think there's an end not missing here isn't there let's see there should be an n1 not missing here yep there's an n1 not that I dropped for some reason let's stick that back in and one nut and and one not and one nut over that stuff lambda plus Sigma 60 Phi minus Sigma 59 Phi times whatever is left e to the lambda plus 6 Sigma 60 Phi minus Sigma 59 Phi T plus C so now we can say what's our integration constant C the last thing we haven't specified is our initial condition so let's assume that when we started our reactor there was no cobalt-60 that makes for the simplest initial condition so we can substitute that in here so at T equals 0 n equals 0 so we would get 0 equals if T is 0 then that just becomes Sigma 59 Phi and 1 naught over lambda plus Sigma 60 Phi minus Sigma 59 Phi plus C and so that makes things pretty easy because we now see equals minus Sigma 59 Phi and 1 not over that stuff that I keep saying over and over again and then we've pretty much solved the equation the last thing we have to do is divide by mu and we'll end up with the same solution that we got on Friday and the same solution that we got on Tuesday so I realized the second after I said it last time that oh we can't just absorb some eetu the something T into our integration constant C because there's a variable T in it so I would say look at this derivation to know the whole solution and so finally we end up with anyone mind if I erase a little bit of this stuff up top okay all erase the decay diagram because we're not using that anymore and hopefully everybody knows our conversion factor so the end solution and two of t would look like let's see Sigma 59 Phi and one not over lambda plus Sigma 65 minus Sigma 59 Phi e let's see times e to the - what's lambda 2 in this case lambda plus Sigma 65 t minus e to the minus sigma 59 5t and that right there is our full equation for n2 now you guys said let's make this numerical okay we have every numerical value already chosen i've already plugged these into the desmos thing so you can see generally how this goes so we've solved it theoretically so now let's make this numerical and make some sort of a value judgment right we know Sigma 59 because we just looked that up we know Phi we impose that as 10 to the 14th neutrons per second thing there it is we know our lambda we know our we don't know our Sigma 60 so we're just going to forget that for now but the point is for everything except time and n 2 we have numerical constants for this so once we plug it all in I modified the decimals example from last time to have the actual unit so you can see that our fake l1 our lambda I'll call it lambda 1 equals Sigma 59 times Phi which is 20 Barnes 20 times 10 to the minus 24 centimeters squared times 10 to the 14th neutrons per centimeter squared per second and the centimeter squares cancel that becomes to the minus 10 and we get that our lambda 1 is like 2 times 10 to the minus 9 per second okay our lambda 2 well we already have that 4.1 seven times 10 to the minus 9 per second so this is one of those cases where lambda 1 approximately equals lambda 2 so just like you see in the book when you plug in all the numbers you get very similar equation which is to say there's going to be some maximum of cobalt-60 produced and in this case the x-axis I have in seconds because that's the units we're using for everything the y-axis is number of atoms so right there 6 times 10 to the 23rd that's 1 mole so at most you can make about a third of a mole of cobalt-60 out of a hundred moles of cobalt 59 which means you're never actually going to have 1 mole of cobalt 60 because of the way that our natural and our artificial decay constants work out because they're fairly equal you're never going to be able to convert and harvest at all that's the numerical output of this now I have another question for you is the top of this curve necessarily the profit point for this reactor no good answer why do you say no [Music] exactly that's what you guys are gonna do on the homework I think I've done we've done the hard part together here and so now I want you guys to decide given those profit parameters and I will write them down on the pset how long do you run your reactor to maximize your cobalt-60 profit so this is one of those examples where we did the whole theoretical derivation we decided yes let's escalate the situation for reality everything works out just fine the final answer well you just tack on this extra like artificial bit of decay from the cobalt-60 being in the reactor but the form of the equation is exactly the same there's just a couple other constants in it for reality then you plug in all the numbers for the constants so you pretend like that stuff is lambda2 and that stuff is lambda 1 there's lambda 1 again there's lambda 2 there's lambda 1 and it's exactly the same equation will form as the original solution that we had when you plug in all the numbers you get something remarkably similar to what's in the book just scaled for actual units of atoms and seconds is there a question yeah sure if you can find it that's great but I couldn't find it that easily you can hunt through the different databases and Janus to try to find it but I'm not going to penalize it if you can't find it if I couldn't find it so yeah a lot of the homework is gonna be redoing this derivation for yourself so I want to make sure that you can go from a set of equations that models 8th actual physical solution and I guarantee you you'll have another physical solution on the exam solve them using your knowledge of 1803 get some sort of a solution which looks crazy but it comes from straightforward math then plug in some realistic numbers an answer an actual question how long should you run your reactor to maximize your profit so it's kind of neat we've had we've been here one month together you can already start answering these value judgment questions about running a reactor and so again I don't know I don't know who did but I'm glad you asked is this filled mostly simulation and the answer is no you actually have to use math to make value judgments if you want to go and make isotopes and then you go and make the isotopes and you sell them to people like me so I can bring them into class and scare unwitting members of the public theoretically yeah hypothetical source indeed so any questions on what we did here from start to finish yep the equation right in the middle there yeah yeah yeah was like oh yeah that was a mistake thank you yep it probably means I was talking and thinking at the same time and forgot to write that but indeed it's back everywhere else thank you yep this lambda right here is the actual lambda for cobalt-60 yep this right here is just an analogy I'm drawing to say that it's almost like that stuff is lambda 1 that's our original artificial decay constant and this stuff here is like our lambda 2 because there's natural decay and then there's reactor induced destruction yep that comes from this solution right here there's another interesting bit too did we necessarily save let's see did we necessarily say the no never mind that's fine that comes from let's trace it through so mu contains no yeah it comes from C that's right so we had it over here because that's part of our solution for let's see where we trace it back to it starts off here in the differential equation starts off here as well so that's part of what's inside mu okay that's where it came from so mu contains this stuff yep then once you it doesn't actually become land it becomes e to the lambda plus 65 times G yeah but when we when we do the integration of that we don't get any factors around oh yeah we do actually there's a Mew stuck in right here and so I just wanted to say that expanding this term comes out to lambda plus Sigma 65 minus lambda of Sigma 59 Phi times T so when you integrate this whole term and again there's an N naught that should be there you do bring this whole pile in front of the T down on the bottom of the equation so that's where it comes from yeah cool I don't think there's any more missing terms okay maybe time for a last question because it is 10 o'clock yep [Music] I so the crop the cross-section is almost like if you fire a neutron at an atom the bigger the atom appears to the neutron the more likely it's going to hit it so it's kind of a theoretical construct to say if something has an enormous cross-section it's like shooting a bullet at a gigantic target with a high probability of impact or interaction something with a small cross-section there's still only one atom in the way but it's like you're shooting a bullet at a tiny target and have less of a chance of hitting it does that make sense cool they can can't relate it to a physical cross-sectional area as far as I know it's not like a certain nucleus has a larger cross-sectional area otherwise things like gadolinium which has a cross-section of a hundred thousand barns would just be a larger atom and it's not in yeah Sean good question they can be theoretically calculated in some cases in the Yip Book nuclear radiation interactions he does go over how to calculate those from quantum stuff and so you'll get a little bit of that in 22:02 in terms of predicting the cross section for hydrogen and the cross section for water and molecular water is not just the sum of its parts that's the kind of crazy part cross sections do change when you put atoms and molecules together just tend to tend to be at lower energies around thermal energies and such some can be let's say all of them probably can be theoretically calculated just not that easily but the really simple ones you can predict titties theoretically predicting the resonances in those cross sections that's tough let's look at a simple cross section like hydrogen yes if you like if you know let's say the full wavefunction for a given atom or for all the electrons on an atom you should be able to so let's do n total for hydrogen much simpler this is the kind of thing that can be predicted from Theory quite easily in fact you will be doing this in 22:02 the other one where was it the other one no I don't expect you to be able to predict this but you will learn why the resonances are there and why they take the shape that they do so if we want last thing for a juco like three minutes late but everyone's still here you can go if you have to by the way I can't keep you here if you want to know if you want to make this equation more realistic and account for every possible energy in the reactor you can make these cross-sections a function of energy and integrate over the full energy range and this is actually how it's done and you will do this in 2205 where you'll be able to take the energy dependent cross section in tabulated or theoretical form and then integrate this whole equation and also account for the fact that the flux as an energy component usually it looks something like analyte water reactor if this is energy and this is flux there'll be a bit of a thermal spike there won't be much going on in the middle I'm sorry a fast spike and there'll be a thermal spike and knowing how many neutrons are at every energy level what's the probability of every Neutron at energy level interacting and what are the cross-sections at every energy level integrated over the full energy range is what gets you the accurate correct solution what we've done here is called the one group approximation where we've assumed that all the neutrons have the same energy thermal energy which is a okay assumption for thermal light water reactors and it'll get you a good estimate the more neutrons you have at different energies the less good that estimate becomes yeah that's that's the energy of a neutron let's say you have a neutron at about 298 Kelvin from the maxwell-boltzmann temperature distribution you can turn that temperature into an average kinetic energy and that average kinetic energy will give you a velocity and that velocity is around 2200 meters per second and that average kinetic energy happens to be about 0.025 evie so thermalized neutrons like the ones flying about in the reactor are moving quite slowly at just 2200 meters a second compared to the fast neutrons which can be moving closer to the speed of light not that close but much much much closer cool I'll take it as a good sign that you all voluntarily stayed a little late so did you guys find this example useful you