Branching, Chains and Dating

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Some nuclei can decay in two different ways. By the first way alone, the half-life would be two hours, and by the second way alone, six hours. So what is the real half-life? The average, four hours, right? Branching, chains and dating do come up in JEE, and by the end of this lesson, this puzzle will take you ten seconds. So, let's begin.

0:29Two ways to decay

So, take a nucleus that can decay in two different ways, by two modes. Let me ask you this: with two ways to decay, does it decay faster or slower than by either mode alone? Yes, exactly, it decays faster, because it now has two chances to decay. So the decay constants simply add: lambda equals lambda one plus lambda two. Lambda, the Greek letter, is the decay constant, and lambda one and lambda two, with a small one and a small two below, belong to the two modes. Okay? Now, for each mode, lambda equals l n two over its half-life T, where l n two is the natural log of two. The natural log of two cancels everywhere, and one over T equals one over T one, plus one over T two. Right? So the effective half-life T equals T one times T two, over T one plus T two: product over sum. T one and T two are called the partial half-lives of the two modes. Clear?

1:30Chains: A decays to B, B decays to C

Now, let us look at decay chains. The left graph is simple decay, from an earlier lesson; look at the right one. Both graphs show the number of nuclei, N, against time, t. Nucleus A decays to B, and B decays to C, which is stable; the arrows mean, decays to. Curve A falls, curve B first rises and then falls, and curve C keeps growing. Let me ask you this: why does B first rise, and then fall? Because two things happen to B at once. It gains nuclei from A at the rate lambda A times N A, and it loses its own at the rate lambda B times N B, where N A and N B, the N with a small A or B below, are the numbers of nuclei. So d N B over d t, the rate of change of N B, is the gain minus the loss. At first the gain wins, so B rises; later the loss wins. Okay? So B is greatest when d N B over d t is zero, that is, when lambda A times N A equals lambda B times N B, as the line under the graph also says. Now careful, this is a common trap: the rates are equal, not the numbers. In this graph, where A decays faster than B, the curves of A and B cross first, and B keeps rising after that. With me? And with no B at t equals zero, this happens at the time t equals the natural log of the ratio lambda A over lambda B, divided by lambda A minus lambda B, in that order. Keep this formula ready for the second example. Clear?

3:09Long-lived parent, and steady production

Now, a special case: the parent A lives very, very long. Then lambda A times N A equals lambda B times N B, and this is called secular equilibrium. Remember, lambda times N is the activity, the decays per second, so here the two activities are equal. Got it? Now put lambda equals l n two over the half-life, for each nuclide. The natural log of two cancels, and N B over N A equals T B over T A, where T A and T B are their half-lives. So a short-lived daughter is present only in a tiny amount. Okay? Next, suppose a reactor makes a nuclide at a constant rate q, the number of nuclei made per second, while it keeps decaying. Then d N over d t equals q, minus lambda times N: gain minus loss. Right? Let me ask you this: if the reactor keeps running, does the number N keep growing for ever? It does not, and here is why. Starting from zero, N equals q over lambda, times one minus e to the power minus lambda times t. The arrow means, tends to: as time goes on, N tends to q over lambda, its greatest value. Then the decays per second match the nuclei made per second. Clear?

4:30Radioactive dating

Finally, let us look at radioactive dating. Suppose a rock contains a parent nuclide P, which decays, as the arrow shows, to a stable daughter D, and all the D nuclei in the rock came from the parent. Let me ask you this: how can such a rock tell us its age? It tells us through the numbers of both kinds of nuclei. The parent left is N P equals N naught, times e to the power minus lambda times t, where N naught is the number of parent nuclei at the start. Since D is stable, every parent that decayed is now a D nucleus, so N D equals N naught minus N P, the number that has decayed. With me? Divide the two, and N D over N P equals e to the power lambda times t, minus one. So measure this ratio today, and you get t, the age of the rock.

5:23Example 1: effective half-life from 2 h and 6 h modes

Okay? Now, remember our puzzle from the start? Here it is, from the course. There is a nuclide that can decay by two processes, with half-lives two hours and six hours; the h on screen means hours. Find its effective half-life. Pause the video and try it first. Done? Okay, let's check. Product over sum: T equals two times six, over two plus six, and the cross is the times sign. Right? So twelve over eight gives one point five hours. Okay? So the answer is not four hours, the average. With two ways to decay, the nucleus decays faster, so the half-life is shorter than both. And that solves our puzzle. Got it?

6:09Example 2: when is B greatest in the chain A to B to C

Now a harder one, on a chain. Here is a nuclide A, of half-life one hour, which decays to B, of half-life two hours, and B decays to C, which is stable. Starting with pure A, after what time is the number of B nuclei greatest? Pause the video and try it first. Alright, let's check. Many students say one hour, the half-life of A, because by then half of A is gone. Is that right? It is not, and that is the trap. B is greatest when the rates are equal, so let us use our formula. Since lambda is l n two over T, and T B is twice T A, lambda A equals two lambda B, because a shorter half-life means a bigger lambda. Okay? Put this in: t equals the natural log of lambda A over lambda B, divided by lambda A minus lambda B, as before. The ratio inside the log is two, and the difference is two lambda B minus lambda B, that is, lambda B itself. Clear? So t equals l n two over lambda B, exactly the half-life of B, which is two hours. That is twice the popular guess. With me?

7:25Quick revision: effective half-life, chains, dating

So, quick revision. Two modes: the lambdas add, so T equals T one times T two, over T one plus T two. In a chain, B is greatest when lambda A times N A equals lambda B times N B, and with a very long-lived parent, N B over N A equals T B over T A, the ratio of half-lives. With constant production, N tends to q over lambda, and for dating, N D over N P equals e to the power lambda times t, minus one. Now go and try the practice problems on this page. You can do them. All the best!