JEE Advanced Questions (2013–2018): Radioactivity

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. How much blood is there in your body? Doctors can find out without taking it out, using a tiny radioactive tracer. They inject it, wait a few hours, draw a little blood, and measure how fast it decays. So how can that little blood tell us the total? JEE Advanced asked exactly this in two thousand seventeen, and today we solve it, with one more JEE Advanced question. Questions like these come up often, so let's begin.

0:32The toolkit: decay law and half-life

So, first, our toolkit. The decay law says N equals N naught, times e to the power minus lambda times t. N is the number of nuclei left at time t, and N naught, the N with a small zero below, is the number at the start. Lambda, the Greek letter, is the decay constant, the probability that a nucleus decays, per unit time. The left graph shows N falling with time; the one beside it, a chain of decays, is from an earlier lesson. Okay? Now the half-life, written T half, the T with one over two below it. It is 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. Right? So after one half-life, half the sample is left. Let me ask you this: after two half-lives, is the whole sample gone? It is not, and this is the classic trap. In each half-life, only half of what is left decays. Look at the left graph: N naught over two at T half, N naught over four at two T half, and N naught over eight at three T half. So after n half-lives, N equals N naught over two to the power n. Clear? And the mean life, tau, the Greek letter like a small t, equals one over lambda. On the graph, the red dashed line, the tangent at the start, meets the time axis at tau. And since T half is l n two over lambda, T half is zero point six nine three times tau, just as the graph writes it. Got it?

2:13Activity, and a shortcut for short times

Now, the activity, capital A, is the rate of decay, the number of decays per second. It is d N over d t inside two vertical bars, which mean we take its size without the minus sign. That equals lambda times N, and so it equals A naught, times e to the power minus lambda times t, where A naught is the activity at the start. With me? Its unit is the becquerel, written B q, and one becquerel means one decay per second, okay? Now tell me: the activity is lambda times N, so as N halves, what happens to it? Yes, exactly, it halves too, because lambda is a constant. So the activity falls with the same half-life as the number of nuclei. Right? Now a topper's shortcut. When the time t is much less than the half-life, and the double less-than sign means much less than, e to the power minus lambda times t is about one minus lambda times t. The wavy equals sign means approximately. Clear? So the fraction of nuclei that decays in that short time is just about lambda times t. No logs needed, and example two uses exactly this. Got it?

3:26Example 1: 64 times the safe level, half-life 18 days (2016)

Now, example one, from JEE Advanced two thousand sixteen. Some radioactive material of half-life eighteen days was spilt in a laboratory, and the radiation was found to be sixty four times the permissible level, the safe level. After how many days can the laboratory be used safely? Pause the video and try it first. Done? Okay, let's check. The radiation is sixty four times too high, and each half-life cuts it in half. So sixty four over two gives thirty two half-lives, right? No, careful, that is wrong, and it is the same trap we saw on the graph. The radiation comes from the decays, so it follows the activity and halves in every half-life of eighteen days. Halving means dividing by two, again and again. With me? So tell me, starting from sixty four, how many halvings bring the radiation down to the safe level? Very good if you said six, because sixty four is two to the power six, so we halve it six times, and that is what the double arrow on screen says: six half-lives. Okay? So the time is six times eighteen, which is one hundred and eight days, and then the laboratory is safe. Powers of two make these questions quick, right?

4:47Example 2: total blood volume from an I-131 tracer (2017)

Now example two, a harder one: our puzzle from the start, from JEE Advanced two thousand seventeen. Iodine one thirty one, written with the symbol for iodine and its mass number at the top left, decays with a half-life of eight days. Some serum labelled with it, of activity two point four times ten to the power five becquerel, is injected into a person's blood, and it spreads uniformly in less than half an hour. After eleven point five hours, the h on screen, two point five millilitres of blood, the m L, is drawn, and it gives an activity of one hundred and fifteen becquerel. Find the total volume of blood. You may use e to the power x is about one plus x, and l n two is about zero point seven. Pause the video and try it first. Done? Okay, let's check. The plan is to find the activity left in the body, then compare it with the sample. But look at the units: the half-life is in days, and the time is in hours. What must we do first? Yes, exactly, put both in the same unit. Eight days is eight times twenty four hours, which is one hundred and ninety two hours. Skip this, and lambda times t comes out far too big. Okay? Now, lambda is l n two over T half, so with l n two as zero point seven, lambda times t is zero point seven times eleven point five, all over eight times twenty four, which gives zero point zero four two. That is tiny, since eleven point five hours is much less than the half-life, so the shortcut works. Right? So e to the power minus lambda times t is about one minus zero point zero four two, which is zero point nine five eight. That is the fraction of the iodine still there. Clear? So the activity left in the whole body is two point four times ten to the power five, times zero point nine five eight, which is about two point three times ten to the power five becquerel. Got it? Now, two point five millilitres of blood gives one hundred and fifteen becquerel. So how much blood holds all of the two point three times ten to the power five? Since the serum is spread uniformly, the activity is proportional to the volume of blood. So capital V, the total volume, over two point five millilitres, equals two point three times ten to the power five over one hundred and fifteen, which gives two thousand. With me? So V equals two thousand times two point five millilitres, which is five thousand millilitres, that is five litres; the capital L on screen means litres. And our puzzle is solved: five litres of blood, from just two point five millilitres. Right?

7:43Quick revision: N0 over 2 to the n, A halves, lambda t rule

So, quick revision. First, N equals N naught times e to the power minus lambda times t, and after n half-lives, N naught over two to the power n is left, never zero. Next, the activity, lambda times N, halves in every half-life, just like N does. And for a time much less than the half-life, the fraction decaying is about lambda times t, with both in the same time units. Now try the other JEE Advanced questions on this page, and all the best!