IIT-JEE Questions (2001–2012): Radioactivity

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Picture a detector near a radioactive sample: it counts some particles in two seconds, and only three quarters as many in the next two. We never count the nuclei in the sample. So can we still find how long a nucleus lives, on average? It looks impossible, right? Today we solve some IITJEE questions from two thousand one to two thousand twelve on radioactivity, and this puzzle is one of them. So, let's begin.

0:33The decay law in four lines

So, first, let us put the chapter on one screen. The decay law: N equals N naught, times e to the power minus lambda times t. N is the number of nuclei left at time t, N naught, the N with a small zero below, is the number at the start, and lambda, the Greek letter, is the decay constant. The left graph shows this curve; the decay chain on the right is for another lesson. Okay? Next, the half-life, T half, the T with one over two below it: the time in which half the nuclei decay. It equals the natural log of two over lambda, written l n two, which is zero point six nine three over lambda. Right? Now the mean life, tau, the Greek letter like a small t. Tau equals one over lambda, which is one point four four times T half. So the mean life is longer; in the graph, the red dashed line, drawn along the starting slope, reaches the time axis at tau, a little after T half. Today's last question tests this. Clear? And after n half-lives, N equals N naught over two to the power n. On the graph: N naught over two after one half-life, N naught over four after two, and N naught over eight after three. Got it?

1:52A trap from IIT-JEE 2006

Now, a short question from IITJEE two thousand six. The half-life is four days. What is the probability that a nucleus decays within two half-lives, that is, eight days? Half decay in the first four days, the other half in the next four, so the probability is one, right? Oh no, careful, that is wrong, and it is the trap here. In each half-life, only half of what is left decays. So after two half-lives, one half in brackets, squared, gives one quarter still there. With me? So the probability of decay is one minus one quarter, which is three quarters. For one nucleus, the fraction left is its chance of surviving. Okay?

2:35Activity: decays per second

Now let us look at activity. The activity, capital A, is the number of decays per second: d N over d t, the rate of change of N, with two straight bars that mean we take its size. It equals lambda times N, which is A naught, times e to the power minus lambda times t, with A naught the activity at the start. Right? Now let me ask you this: does the activity fall with the same half-life as the number of nuclei, or a different one? Yes, exactly, it falls with the same half-life, because lambda is fixed, so the activity halves whenever N halves. So every half-life trick works for activity too. Clear? The unit of activity is the becquerel, written capital B, small q: one becquerel is one decay per second. Got it? Next, the number of nuclei that decay within a time t is N naught, times bracket one minus e to the power minus lambda times t. Each nucleus survives with probability e to the power minus lambda times t, so it decays with probability one minus that. With me?

3:43Example 1: initial activity from two later readings

Now, example one, from IITJEE two thousand four. After two hundred and eighty days, the activity of a sample is six thousand disintegrations per second, and after another one hundred and forty days, it is three thousand. Find the initial activity. Pause the video and try it first. Done? Okay, let's check. In the extra one hundred and forty days, the activity falls from six thousand to three thousand, as the arrow shows. That is exactly half, so the half-life is one hundred and forty days. Okay? Now, two hundred and eighty over one hundred and forty gives two. So the first reading came two half-lives after the start. Right? So, going back two half-lives, many students double six thousand and write twelve thousand per second. Is that right? Oh no, careful, it is not, and that is the trap. Going back, we double once for every half-life, so we double twice. So A naught, the initial activity, is six thousand times two squared, which is twenty four thousand per second, where s to the power minus one means per second. Clear? Now check it forwards, the topper's way. Twenty four thousand at the start, twelve thousand at one hundred and forty days, six thousand at two hundred and eighty, and three thousand at four hundred and twenty. Every number matches. Well done!

5:12Example 2: mean life from counts in back-to-back 2 s

Now, example two, from IITJEE two thousand three, is our puzzle. An element decays by beta emission, beta being the Greek letter. The detector records a number n of beta particles in two seconds, the s on screen, and zero point seven five times n in the next two seconds. Find the mean life, to the nearest whole number, using the given natural logs. Pause the video and try it first. Done? Okay, let's check. We do not know how many nuclei there are, so tell me, how many of them decay in the first two seconds? Let us use the formula from our activity screen: with N naught nuclei at the start, N naught times bracket one minus e to the power minus two lambda decay in the first two seconds. Right? In the next two seconds, we start with only the survivors, N naught times e to the power minus two lambda. And the same fraction of them, one minus e to the power minus two lambda, decays. Clear? Now divide the second count by the first. N naught cancels, the bracket cancels, and e to the power minus two lambda equals zero point seven five times n over n, which is three quarters. And that solves our puzzle, because the unknown number of nuclei cancels. Got it? Taking natural logs, two lambda equals the natural log of four thirds, that is, two times the natural log of two, minus the natural log of three. Two times zero point six nine three one is one point three eight six two, and subtracting one point zero nine eight six gives zero point two eight seven six. With me? So lambda is zero point two eight seven six over two, which gives zero point one four three eight, about zero point one four four per second; the wavy equals sign means approximately. Okay? Now many students write zero point six nine three over lambda and stop. Is that what the question wants? It is not, and that is the trap: zero point six nine three over lambda is the half-life, about four point eight seconds. The question asks for the mean life, tau, one over lambda, which is about six point nine five seconds, so seven seconds to the nearest whole number. Very good!

7:35Quick revision: decay law, half-life, mean life, activity

So, quick revision. N equals N naught, times e to the power minus lambda times t; T half is zero point six nine three over lambda, and tau, one over lambda, is one point four four times T half. Activity, lambda times N, halves in every half-life too. And for counts in two back-to-back intervals of length t, the second count over the first is e to the power minus lambda times t. Now try the other IITJEE questions on this page. All the best!