 Hello everyone, so we go to the second lecture of this course on nuclear and radiochemistry. So far we have discussed the radioactive decay, we learnt that radioactive decay follows the first order rate law and we also discussed the different types of conditions like you know the two independently decay radioisotopes, how does the activity change with time and also branching a radioisotope decaying with by two modes like beta plus and beta minus, so how does the half life depend upon the partial half lives. Today we will discuss the decay of an isotope to another isotope which is also radioactive, so there is a chain, the decay happens in a chain of radioisotopes and so there will be different aspects of that and lot of applications also which I will be discussing in this lecture. So, when we mean by radioactive decay chain, I mean that radioisotope to isotope A undergoes some decay to B and B is also radioactive to which also decay in further to C. So, we have already discussed the radioactive decay of A which follows the exponential decay, so N1 is the number of atoms of A at any time and which we discussed in the previous lecture that it follows an exponential decay with N10 the initial atoms of A and the decay follows an exponential path. Now, let us discuss how the activity of B will change with time if it is growing from a freshly purified isotope A. So, the lambda values for A is lambda 1 and for B is lambda 2 and the number of atoms of A and B at any time t are N1 and N2 respectively. Let us assume the initial conditions that when time is 0 then there is no daughter product. We have only the N10 the initial number of atoms of A and N2 is 0. So, we set up a differential equation for the number of atoms of B how it is changing with time and solve that equation to find out at any point of time t how does the number of atoms and hence the activity of B change with time. So, let us see here the rate of change of number of atoms of B will be given by d and 2 by dt. B is growing from A so that this N1 lambda 1 that is the rate of decay of A is the rate of growth of B and the B is changing decaying by the rate d and B by dt equal to N2 lambda 2. So, there is a growth part and there is a decay part now this is a differential equation and we have to solve it for N2. So, N2 is the number of atoms of B at any time t. So, let us see how do we solve first we separate the variables. So, we say d and 2 by dt plus N2 lambda 2 equal to N1 lambda 1 and N1 lambda 1 N1 we know so we can write N10 e raised to minus lambda 1 t lambda 1 I have replaced N1 by N10 e raised to minus lambda 1 t. So, such a differential equation I want to solve then we can we can have there is something called as the integration factor this is e raised to lambda 2 t. If I multiply both such this equation by this factor it becomes easy to integrate. So, I say d N2 by dt e raised to lambda 2 t plus N2 lambda 2 e raised to lambda 2 t equal to N10 lambda 1 e raised to it will become lambda 2 minus lambda 1 t. Now, let us integrate this equation. So, you see here this this term becomes differential of N2 e raised to lambda 2 t differential of e raised to the differential of N2 e raised to lambda 2 t equal to N10 lambda 1 e raised to lambda 2 minus lambda 1 t. So, if I integrate then we can see the solution of the integration of this will become N2 e raised to lambda 2 t equal to N10 lambda 1 upon lambda 2 minus lambda 1 e raised to lambda 2 minus lambda 1 t plus c. So, now you can now find out the value of the integration constant c that is when t equal to 0 N2 equal to 0. So, if you put N2 equal to 0 here this term becomes 0 and so it becomes c equal to minus of this term. So, c becomes minus N10 lambda 1 upon lambda 2 minus lambda 1. So, if you substitute in this equation you get N2 e raised to lambda 2 t equal to N10 lambda 1 upon lambda 2 minus lambda 1 e raised to lambda 2 minus lambda 1 t minus 1. So, now you can see here the the final equation becomes N2. Now, if I take this e raised to lambda 2 on the other side then N2 equal to N10 lambda 1 upon lambda 2 minus lambda 1 e raised to lambda 1 t minus e raised to minus lambda 2 t. So, this is the equation for the number of atoms of B at any time t. So, it contains the exponential terms due to the decay of A and B. Now, if you want to find out the activity then activity of dotted product B activity A2, A2 will be N2 lambda 2 equal to N10 lambda 1 lambda 2 upon lambda 2 minus lambda 1 e raised to minus lambda 1 t minus e raised to minus lambda 2 t and so you can see here. So, this N10 and lambda 2 will become A2, N10 lambda 1 will become A10 and so the final equation will be A2 equal to A10 lambda 2 upon lambda 2 minus lambda 1 e raised to minus lambda 1 minus e raised to minus lambda 2 t. So, this is a important relationship. I hope you remember the derivation of this which will be used many times in subsequent lectures also. So, the net result of that the radioactive decay chain is that if you want to find out the activity of B at any time t, we determine the number of atoms of B and 2 equal to N10 and lambda 1 upon lambda 2 minus lambda 1 e raised to minus lambda 1 t minus e raised to lambda 2 t and activity equal to N lambda. So, activity of dotted becomes A2 equal to A10 lambda 2 upon lambda 2 minus lambda 1 e raised to minus lambda 1 t minus e raised to minus lambda 2. So, this is a important relationship and there are several corollaries of this equation which we will discuss in details subsequently. Now, there will be situations where not only the dotted product is radioactive, but the grand dotted is also radioactive. So, like grand dotted means when I say C. So, when C is also radioactive one can set up similar equation like we discussed for B, the equation for the activity of C as a function of time. So, to set up that equation you can write this equation dN3 by dt that is the growth of the C atoms as a function of time equal to the growth of C is from B that is N2 lambda 2 and the decay of C is to dN3 lambda 3. Again you can put the conditions that when the initial conditions when t equal to 0 N1 equal to N10 and there is no B and C. So, you only purely freshly purified parent isotope is there. So, under those conditions how the activity of C will change with time that is the problem we have to solve here. So, already you know the number of atoms of A how they will change as a function of time N1 equal to N10 e raised to minus lambda 1 t. Similarly, for B we have just now solved the equation N2 equal to N10 e raised lambda 2 by lambda 2 minus lambda 1 e raised to minus lambda 1 t minus e raised to minus lambda 2 t. And so, if you set up this equation set up substitute the value of N1 and N2 in this equation rather N2 you have to put. So, you that N2 will contain N10 and then you solve the differential equations similar way like we solved the one for N2 then the solution I am not going into the solution of the differential equation because the same way you can do for this also. So, the solution of that equation will be N10 now we can generalize this the lambda factors in terms of k1 k2 k3. So, k1 e raised to minus lambda 1 t plus k2 e raised to minus lambda 2 t plus k3 e raised to minus lambda 3 t. So, k1 you can see here the k1 is related to lambda 1 lambda 2 upon lambda 3 minus lambda 1 into lambda 2 minus lambda 1 k2 lambda 1 lambda 2 upon lambda 1 minus lambda 2 into lambda 3 minus these two are in the denominator. Similarly, k3 will be lambda. So, the numerator is constant you can see same in all the cases that denominator is very. So, when you have k1 lambda 1 is this minus lambda 1 when you have k2 is minus lambda 2 when you have k3 this minus lambda 3. So, that is how you can see the order in the values of formally for k1 k2 k3. And then you multiply this number of atoms by lambda 3 to get the activity. So, the activity of C as a function of time will be a10 initial activity of parent into lambda 2 lambda 3. So, lambda 3 term has come because you are multiplying n 3 by lambda 3 and the lambda 1 term has gone into a10. So, a3 equal to lambda 2 lambda 3 upon lambda 3 minus lambda 1 into lambda 2 minus lambda 1 e raised to minus lambda 1 t plus lambda 2 lambda 3 upon lambda 1 minus lambda 2 into lambda 3 minus lambda 2 e raised to minus lambda 2 t plus lambda 2 lambda 3 upon lambda 1 minus lambda 3 into lambda 2 minus lambda 3 e raised to minus lambda 3 t. So, due to this equation you can calculate even the activity of the grand order product. And there are several cases you will see that we in fact, we do produce the radio isotopes by two decay processes. So, that will come in the discussion subsequently. Now, there can be situations where the grand order instead of being radioactive is a stable isotope. In fact, this is very common just we discussed when you discuss A going to B going to C that is the same situation and the grand order is stable which is not undergoing further radioactive decay. But you can use the formula for the number of atoms of C which we derived just now to calculate the number of atoms of C that is formed at the end of the time some particular time. So, what we saw just now that the number of atoms of C is given by initial atoms of A into this term k 1 in terms of k 1, k 2, k 3 where k 1, k 2, k 3 are functions of lambda 1, lambda 2 and lambda 3 and these subsequent the exponential terms for the three isotopes. Now, this is a hypothetical case now here because lambda 3 the grand order is not radioactive. So, in such a situation lambda 3 becomes 0. If it is not decaying means the half-life is infinite. So, lambda 3 is 0. So, k 3 the k 1, k 2, k 3 these terms are same as we discussed the previous slide and I have highlighted this lambda 3 in those terms this lambda 3 will become 0 and you can see the implication of this in this equation. So, when C is stable lambda 3 is 0 and therefore, k 3 becomes equal to 1. You can see here if lambda 3 is put equal to 0 then it become lambda 1, lambda 2 upon lambda 1, lambda 2 that is equal to 1 that term is simplified. Similarly and now the the terms now you can see lambda 3 you put equal to 0 then it becomes n 1 0 minus lambda 2 upon lambda 2 minus lambda 1. So, this term lambda 1 will get cancelled out. He raised to lambda 1 t minus lambda 1 upon lambda 2 lambda 1 minus lambda 2. So, the second term k 2 term. So, that minus term will come because of this lambda minus lambda 2 here and lambda 2 will get cancelled with this lambda 2 we have and then he raised to lambda 2 t plus 1 k 3 becomes equal to 1 and you can now in fact rearrange this equation in such a way that these terms represent n 1 and n 2. What is n 1? n 1 is n 1 0 1 minus he raised to minus lambda 1 t and n 1 is n 1 0 raised to minus lambda 1 t. So, this is n 1 0 minus n 1 that is at any time t and minus n 2. So, n 2 is n 1 upon lambda 1 upon lambda 1 minus lambda 2 raised to minus lambda 1 t minus is 1 lambda 2 t. So, the number of atoms of c at any time will be the initial atoms of a minus the number of atoms of a and b at that particular time that is n 1 and n 2. So, this is how you can calculate like for example, if you have a priestly purified parent and after a long time you want to know how many atoms of band otter will be formed which is stable you can use this equation provided you know the number of atoms of n 1 and n 2 at that point of time. In fact, if you know initial activity you can calculate n 1 and n 2. So, that is how we can try to find out the activity or the number of atoms of the grand otter. To extend this further into a large series like natural radioactive series in fact, this this condition was there in the early 20th century when the scientists were trying to separate the otter products from the natural radioactive series and there are n number of decays like as I mentioned 238 uranium undergoes 8 alpha and 6 beta decay 2 206 lead. So, it suppose at any point of time you want to find out what will be the activity of thorium 230 which is the 1 2 3 4th product otter product or radium 226 which is the fifth otter product. So, in such situation whatever we discussed just now may not be useful. So, you can go for a generalized equation and that is called as the Bateman equation. So, generalized equation for a decay chain where there are the parent isotope a decays by subsequent decay to b c so on to m n and so on then you can write a generalized equation for n th otter product like d n n upon d t where this is the number of atoms and this is the n th isotope n m lambda m that is the the growth part minus n n lambda n that is the decay part. So, this is the similar equation which we just now solved. Now, the solution of this equation will be similar to that we found for the grand otter, but in a generalized way now. And so, the number of atoms of n this atom this particular isotope at any time t will be given by initial number of atoms of a into same terms k a k b instead of k 1 k 2 I am putting here k a k b. So, k a k b k m k n so, and the corresponding exponential terms it is minus lambda a it is my lambda b t's and so on. Now, here again you can see the order in terms of the lambdas that are being used k a will be lambda a upon lambda n minus lambda a lambda b upon lambda b minus lambda a. So, you can see lambda a is in the subtraction mode minus lambda a in all terms. The k b is again lambda b minus lambda b everywhere and k m and k n. So, you can set up the equation a generalized if you can make a computer program to calculate the number of atoms of any dotter product at any point of time using this generalized Bateman equation. And this becomes very useful when you have to now calculate the activity of a dot product. Like for example, you want to separate 226th radium from a uranium sample with old sample, but you want to know a priori what will be the number of atoms or what will be the activity of 226th radium. So, there you can do this calculation and arrive at whether it is worth going for suppression at this point of time or not. So, these Bateman equations are very useful in finding out the activity of dotter products in a natural radioactivity series. Now, we will come to another aspect of this radioactive decay chain. Then many a times we will come across situations where depending upon the half lives of parent and daughter, you may have their growth and decay of the dotter product in a different fashion. So, we will discuss that in subsequently. So, again I give this typical case of a going to b going to c with the number of atoms of a and b being n 1 and 2 and their decay constant being lambda 1 lambda 2 respectively. And we derived the expression for the activity of b as a function of time in terms of the initial activity of parent into lambda 2 upon lambda 2 minus lambda 1 e raised to minus lambda 1 t minus e raised to minus lambda 2 t. So, now there are two situations a where the parent is short lived than daughter. So, when we say parent is short lived than daughter means that t half the half life of parent is short less than that of daughter. And in terms of the decay constants lambda 1 the decay constant of parent is more than that of the daughter. So, you take note of this one this is important because there will be many different situations depending upon the magnitude of lambda 1 and lambda 2. And the second situation is when the parent isotope is long lived than daughter isotope. So, that means t 1 is more than t 2 or conversely lambda 1 is less than lambda 2. So, depending upon the magnitude of lambda 1 and lambda 2 or t 1 and t 2 we will discuss different situations. Now here the first case where the half life of the parent is short lived than that of daughter. And I give you a typical example where you have 131 tellurium having the half life of 30 hours really a day 1.25 day or so. It indicates to 131 iodine having half life 8 days and which is decaying to 131 genome which is stable. So, assume that we have initially a freshly purified 131 tellurium. So, at t equal to 0 the activity of daughter that is 131 iodine is 0. In such situations you if you see the if you see that activity profile of a daughter product this is the daughter product activity profile what will happen is at t when. So, the general equation what was the general equation a 2 equal to a 1 0 e raise to minus a 1 0 lambda 2 upon lambda 2 minus lambda 1 e raise to minus lambda 1 t minus e raise to minus lambda 2 t. This is the generalized equation for the activity of iodine 131. Now here tellurium 131 is short lived than iodine 131. So, the activity of tellurium 131 will decay faster than the growth and decay of tellurium iodine 130. So, what happens let us put this condition here when t is very high compared to half life of parent isotope then this e raise to minus lambda 1 t tends to 0. So, that means this term becomes 0. When this terms become 0 then the activity of daughter will good will change as a 1 0 lambda 2 upon now that negative sign you can bring here 9 1 minus lambda 2 e raise to minus lambda 2 t. So, you can see here that the daughter activity after some time after in fact the significant number of half lives of the parent will decay with its own half life. Well there is a constant term due to the ratio of lambdas. So, this I have tried to illustrate in this graph here this is in the linear scale. So, the activity of a 1 the parent of tellurium 131 will decay exponentially and from this activity of parent the daughter activity is growing and then it will reach a maximum and then it will start decaying with its own half life. So, there is no case of a equilibrium we will discuss the equilibrium case in the second part of this lecture. So, essentially what is happening that the total activity if you try to add this to a total activity will change and then it is difficult to resolve the parent and daughter activity from this data this is a total activity. So, in this is a case of no equilibrium that means the after the decay of the parent isotope the daughter isotope will decay with its own half life. There are several cases like these cases, but there are there are more useful cases when the parent is long live than daughter and then we have a case of an equilibrium. So, that I will discuss in the next part of this lecture. Thank you.