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# Exponential decay formula proof (can skip, involves calculus)

## Video transcript

the notion of a half-life is useful if we're dealing with increments of time that are multiples of a half-life for example after when at time equals zero we have a hundred percent of our substance then after time equal one half-life we'd have 50% of our substance at time is equal to two half-lives we'd have 25% of our substance and so on and so forth so if I say that three half-lives have gone by and in the case of carbon that would be what roughly fifteen thousand years I can tell you roughly or almost exactly how much what percentage of my original element I still have in the case of carbon-14 I'll tell you what percentage of my original carbon-14 has not decayed into nitrogen as yet nitrogen-14 and that's useful but what if I care about how much carbon I have after half a year or after half a half-life or after three billion years or after ten minutes what if I want a general function a general function as a function of time that tells me the number or the amount of my decaying substance I have so that's what we're going to do in this video and it's going to be a little mathy but I think the math is pretty straightforward especially if you've if you've taken a first year course in calculus and this is actually a pretty neat application of it so let's just think a little bit about the rate of change or the probability of or the number of particles that are changing at any given time so if we say then the difference our change in our number of particles or the amount of particles and in any very small period of time in any very small period of time what's this going to be dependent on this is the number of particles we have in a given period of time this is our rate of change so we one thing we know that our rate of change is going down we know it's a negative number right we know that in the case of radioactive decay I could do the same exercise with with with compounding growth where I would say oh no it's it's not a negative number that our growth is dependent on how much we have in this case the amount we're decaying is proportional but it's going to be the negative of how much of the actual compound we already have let me explain that so what I'm saying is look our amount of decay is proportional to the amount of the substance that we already are dealing with and just a to maybe make that a little bit more intuitive imagine a situation here where you have 1 times 10 to the ninth you have a billion carbon atoms and let's say over here you have 1 times 10 to the 6 carbon atoms and if you look at it at over some small period of time let's say if you look at it over one second let's say our DT DT is an infinitesimally small time but let's say it's a change in time it's a delta T and let's say over one second you observe that this sample had I don't know let's say you had you saw one thousand carbon particles you really wouldn't see that with carbon-14 but this is just for the sake of our intuition let's say over one second you saw one thousand carbon particles per second here well here you have one thousandth of the number of particles in this sample as this one so for every thousand particles you saw decaying here you would really expect to see one carbon particle per second here just because you have a smaller amount now I don't know what the actual constant is but we know that no matter what substance we're talking about this this constant is dependent on the substance carbon is going to be different from uranium is going to be different from you know we looked at radon it's quote they're all going to have different quantities right here and we can see that we'll actually do in the next video you can actually calculate this from the half-life but the rate of change is always going to be dependent on the number of particles you have right that's the I mean we saw that here with half-life when you when you when you have half the number of particles you'll you you lose half as much right here if we start with one hundred particles here we went to fifty particles and we went to twenty-five when used twenty-five when you start with 50 in a period of time you lose 25 when you start with 100 you lose 50 so clearly the amount you lose is dependent on the amount you started with right over any fraction of time and here it's this very small fraction so what I've set up here is really a fairly simple but it doesn't sound so simple a lot of people if you say it's a differential equation we can actually solve this using pretty straightforward techniques we just have to this actually a separation of variables problem and so what can we do let's divide both sides by n so if we get we want to get all the ends on this side and all the T stuff on the other side so if we have 1 over N DN over DT is equal to minus lambda I just divided both sides of this by N and then I can multiply both sides of this by DT and I get 1 over n D n is equal to minus lambda DT now I can take the integral of both sides of this equation take the integral and what do I get what's the antiderivative I'm taking the indefinite integral of the antiderivative of of 1 over N well that's the natural log the natural log of n plus some constant plus some constant I'll just do that in blue plus some constant and then that equals what's the antiderivative of just some constant well it's just that constant times the derivative the variable we're taking the antiderivative with respect to so minus lambda times T plus some constant these are different constants but they're arbitrary so if we want we can just subtract that constant from that constant and put them all on one side and then we just get another constant so this boils down to our solution to our differential equation is the natural log of N is equal to minus lambda T plus some other constant out of it I'll call it C 3 doesn't matter and now if we want to just make this a function of of N in terms of T let's take let's take both let's take both of these or both take e to the power of both sides of this you can view that as kind of the inverse natural log so e to the power of Ln of n that's Ln of n is just saying what power do you raise e to to get to N so if you raise e to that power you get n so if you I'm just raising both sides of this equation I'm raising e to both sides of this equation so this e to the Ln of n is just n and that is equal to e to the minus lambda T plus C 3 and now this can be re-written as n is equal to e to the minus lambda T times e to the C 3 and now once again this is an arbitrary constant so we can just relate rename that is I don't know let me rename it as c4 c4 so our solution to our differential equation n as a function of T is equal to our c4 constant c4 e to the minus lambda T now let's say even better let's say at N equals 0 let's say that n at N equals 0 we have n sub 0 of our of our sample that's how much we're starting off with so let's see if we can substitute that into our equation to solve for c4 so we say at n sub 0 n sub 0 is equal to let's put 0 in here so let's see that's is equal to n sub-nought and that's equal to C 4 times e to the minus lambda times 0 well minus anything times 0 is 0 so it's e to the 0 so that's just 1 so C 4 is equal to n not our starting amount for the sample so we've actually got an expression we have the number of particles or the amount as a function of T is equal to the amount that we start off with at time is equal to 0 times e to the minus lambda times time and we just have to be careful that we're always using the time constant when we solve for the different coefficients so the seems all abstract you know how does this relate to half-life well let's let's try to figure out this equation for carbon this will be true for anything where we have radioactive decay if we actually had a plus sign here to be radioactive it would be a exponential growth as well we know that carbon C 14 has a as a 5,000 seven-year half-life half life so the way you could think about is this if at time equals zero you start off with T at time equals zero so time equals zero T equals let me write that down if at end of zero is equal to and we could write a hundred there if we want actually why don't we do that if n of 0 we start off with a hundred and then at n of 5,700 years so we're going to take T to be in years you just have to be consistent with your units how much will we have left we'll have 50 left we could have written x and x over 2 here and it would all worked out in the end so let's see let's apply that to this equation and try to solve this for a lambda so we know n of 0 is equal to 100 so we immediately know that we can write this equation as as n of T is equal to 100 e to the minus lambda T at least in the in this exact in this circumstance in this circumstance right and we also know that n of 5,700 so that means n of 5,700 that is equal to we just said we're after that's one half life away so we have half as much of our compound left that's equal to 50 which is equal to this to the 5,700 power times lambda so it's equal to 100 times e to the minus lambda times 5,700 and now we just solve for lambda then we'll have a general equation for how much carbon we have at any given moment in time so if you divide both sides of this by 100 what do we get we get 0.5 we have 1/2 1/2 is equal to e to the let me just write minus 5,700 lambda and we could take the natural log of both sides so then we get just scroll down a bit natural log of 1/2 is equal to the natural log of this is just minus 5,700 lambda to solve for lambda you get lambda is equal to the natural log of 1/2 / - 5,700 so let me see what that is let's see what that is so 0.5 natural log is that / - 5757 100 negative is equal to 1.2 times 10 to the negative 4 it's equal to 1.2 1 times 10 to the 9 is 4 so there you have it we figured out our lambda so the general equation for how much carbon-14 we can expect at any moment in time T where T is and years is n of T is equal to the amount of carbon we start off with times e to the minus lambda e - um is 1.2 1 times 10 to the minus 4 times T in years so now if you say after half a year you just plug it in and you have to tell me how much you started off with and then I can tell you how much you have after half a year or after a million years or after a gazillion years and we'll do a lot more of these problems in the next video