IIT-JEE Questions (1994–2000): Radioactivity

Transcript

0:00Introduction

Hello everyone! Let me start with a puzzle. Iodine one thirty one has a half-life of eight days. Take a fresh sample, and watch one particular nucleus. When will that nucleus decay? Will it be exactly on day eight? Hold on to your answer. This very question came in IITJEE nineteen ninety eight, and the idea behind it is worth knowing well. Today we sharpen three tools, then solve two questions from those years. Let's begin.

0:31Tool 1: the decay law, and one single nucleus

So, tool number one, the decay law. d N over d t, the rate of change of N, equals minus lambda times N, where lambda, the Greek letter, is the decay constant. The double arrow means, which gives: N equals N naught, times e to the power minus lambda times t, where N naught, the N with a small zero below, is the number at t equals zero. Okay? Now, back to our puzzle. This law tells us how many nuclei are left. But does it tell us when one particular nucleus will decay? It does not, because decay is a random process. So the law is statistical, true only for large numbers of nuclei. With me? Here is iodine, with its mass number, one thirty one, at the top left. One particular nucleus may decay at any time after t equals zero, long before eight days, or long after. The eight days only tell us when half of a large sample is gone. Puzzle solved, right?

1:32Tool 2: half-life and mean life

Tool number two is the half-life, written T half, the T with one over two below it, the time in which half the nuclei decay. T half equals l n two over lambda, where l n means the natural log, and that is zero point six nine three over lambda. The left graph shows N falling with time t; the right graph belongs to another lesson. Okay? Now, in each half-life, half of what is left decays. On the left graph, N naught falls to N naught over two, then N naught over four, then N naught over eight. So after n half-lives, N equals N naught over two to the power n. Got it? Next, the mean life, the average time a nucleus lives. Tell me, which is longer, the half-life, or the mean life? Yes, the mean life is the longer one. The mean life, tau, the Greek letter, equals one over lambda, which is one point four four times T half. On the left graph, the red dashed line, the tangent at the start, meets the time axis at tau, a little after the first half-life. Clear?

2:43Tool 3: nuclei made at a constant rate

Now tool number three. Suppose nuclei are made at a constant rate q, while they also decay. Then d N over d t equals q minus lambda times N: q nuclei come in, and lambda times N decay away, per unit time. Okay? Starting from N equals zero, the solution is N equals q over lambda, times, in brackets, one minus e to the power minus lambda times t. With me? Now tell me this. Does N keep on growing for ever? No, it does not, it levels off. The arrow means tends to, and the sideways eight is infinity. As t tends to infinity, e to the power minus lambda times t tends to zero, so N tends to q over lambda. Got it?

3:30Example 1: mean life, and 6.25% of the reduced number (1996)

Now example one, from IITJEE nineteen ninety six. At a certain instant, twenty five percent of the nuclei of a sample are undecayed, and ten seconds later, twelve point five percent are; the small s means seconds. Find the mean life, and the further time in which the number falls to six point two five percent of this reduced number. Pause the video and try it first. Done? Okay, let's check. The arrow here means falls to: twenty five percent falls to twelve point five percent in ten seconds. That is exactly half, so the half-life, T half, is ten seconds. Okay? For the first part, the mean life, tau, is T half over zero point six nine three, so ten over zero point six nine three seconds. The wavy equals sign means approximately: about fourteen point four seconds. Clear? Now the second part. Twelve point five percent down to six point two five percent is just one more halving. So the answer is ten seconds, right? That is not right, and this is the trap. Read it again: six point two five percent of this reduced number, not of the original sample. Six point two five percent is one sixteenth, which is one half to the power four. With me? So the reduced number must be halved four times, which takes four half-lives. Four times ten seconds gives forty seconds, not ten. Got it?

5:02Example 2: nuclei made at a constant rate, 3N0/2 (1998)

Now example two, a harder one, from IITJEE nineteen ninety eight. Nuclei of a radioactive element, called capital A here, are produced at a constant rate alpha, with decay constant lambda, and N naught nuclei at t equals zero. First, find N at time t, that is, as a function of time. Then, if alpha equals two N naught lambda, find N after one half-life, and as t tends to infinity. Pause the video and try it first. Done? Okay, let's check. Careful, this alpha, the Greek letter, is not an alpha particle here; it is the rate of production, like q in tool three. So, which equation do we start from? Yes, exactly, the same one as tool three: d N over d t equals alpha minus lambda times N, okay? Now separate the variables. The tall S is the integral sign: the integral from N naught to N, of d N over alpha minus lambda N, equals the integral from zero to t, of d t. Clear? Integrate, and multiply by minus lambda: l n of alpha minus lambda N, over alpha minus lambda N naught, equals minus lambda times t. Take e to the power of both sides, and alpha minus lambda N equals alpha minus lambda N naught, times e to the power minus lambda times t. With me? Solve for N, and the first part is done. N equals one over lambda, times the square bracket: alpha, minus the quantity alpha minus lambda N naught, times e to the power minus lambda times t. Right? For the second part, put alpha equals two N naught lambda. Then alpha minus lambda N naught is just lambda N naught, and dividing by lambda, N equals two N naught, minus N naught times e to the power minus lambda times t. Okay? Now, after one half-life, many students write N naught over two. Is that correct? It is not, because new nuclei keep coming in. After one half-life, lambda times t is l n two, and e to the power minus l n two is one half. So N equals two N naught, minus N naught over two, which is three N naught over two, more than we started with! Clear? And as t tends to infinity, e to the power minus lambda times t tends to zero, so N tends to two N naught. That is alpha over lambda, just like q over lambda in tool three. Got it?

7:46Quick revision: decay law, mean life, N tends to q/lambda

So, quick revision. N equals N naught, times e to the power minus lambda t; T half is zero point six nine three over lambda; and tau is one over lambda, one point four four times T half. One nucleus is random: it may decay at any time. For a percent of the reduced number, count the halvings from there. And for nuclei made at a rate q, d N over d t equals q minus lambda N, and N tends to q over lambda. Now go and try the practice problems on this page. All the best!