The Law of Radioactive Decay

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You have a radioactive sample, and in ten days, half of it decays. So in the next ten days, the other half decays, and after twenty days, nothing is left, right? It sounds so logical, doesn't it? Keep your answer in mind, because by the end of this lesson you will know for sure. Half-life questions come up often in JEE, and they all rest on one law, the law of radioactive decay. So, let's begin.

0:32The decay law

So, imagine a sample of radioactive nuclei, decaying all the time. Let me ask you this: if you double the number of nuclei, what happens to the number of decays every second? Yes, exactly, it doubles. The rate of decay is proportional to the number of nuclei present, and the sign after the word decay means, is proportional to. So, with N the number of nuclei, d N over d t, the rate of change of N, equals minus lambda times N, where lambda, the Greek letter, is a constant. The minus sign tells us that N keeps falling. Okay? Solve this, and you get the decay law: N equals N naught, times e to the power minus lambda times t, where e is the base of natural logs. N naught, the N with a small zero below, is the number of nuclei at time t equals zero. So N falls exponentially with time. Clear? And lambda has a name, the decay constant. It is the probability that a nucleus decays, per unit time. Got it? Now here is a question that traps many students. If you heat the sample strongly, or put it under huge pressure, does it decay faster? No, it does not, and many students get this wrong. Lambda does not depend on temperature, pressure, or chemical state, whether the atom is free or part of a compound. So for a given nuclide, lambda is a fixed number. With me?

2:05Half-life

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. Put N equals N naught over two in the law, and you get T half equals the natural log of two, over lambda; the l n on screen means the natural log. That is zero point six nine three over lambda. In the figure, the left graph shows N against time t; the graph beside it, a chain of decays, is for a later lesson. Okay? Now back to our puzzle. Half the sample decays in ten days, so the other half decays in the next ten days, right? No, that is wrong, and it is the classic trap. In each half-life, only half of what is left decays. Look at the left graph: N naught at the start, N naught over two after one half-life, N naught over four after two, and N naught over eight after three. So after n half-lives, N equals N naught over two to the power n. After twenty days, one quarter of the sample is still there! Right? Now, the number of half-lives in a time t is t over T half. So we can also write N equals N naught, times one half to the power t over T half; that is the half in brackets, raised to that power. This form works for any time, even when t is not a whole number of half-lives.

3:30Mean life

Clear? Next, the mean life, the average time a nucleus lives before it decays. Tell me, which is longer, the half-life or the mean life? The answer is the mean life; it is longer. The mean life, tau, the Greek letter, equals one over lambda, which is one point four four times T half. In the left graph, the orange dashed line meets the time axis at tau, a little after the first half-life. Okay? Now see what is left after one mean life. Put t equals tau in the law, and N becomes N naught over e, that is thirty six point eight percent of N naught. Remember this number. Got it?

4:14One nucleus: a game of chance

Now, think about one single nucleus. The probability that it survives a time t is e to the power minus lambda times t. Okay? So the probability that it decays within that time is one minus e to the power minus lambda times t. Right? And for a short time, t much less than T half, shown by the double less-than sign, the fraction that decays is about lambda times t. The wavy equals sign means approximately. Clear? And here is the last point. Decay is a random process: no one can say when a particular nucleus will decay. So the law is statistical, and it holds for large numbers of nuclei. With me?

4:55Example 1: fraction left after 30 days and 25 days

Now, let us try one from the course. The half-life of a nuclide is ten days. What fraction of a sample remains after thirty days, and after twenty five days? Pause the video and try it first. Done? Okay, let's check. Here is a topper's trick: first count the half-lives. Thirty days over ten days gives three half-lives. So the fraction left is one half to the power three, which is one eighth, just like N naught over eight on the graph. Okay? Now twenty five days, which is between two and three half-lives. So the fraction is halfway between one quarter and one eighth, right? No, it is not halfway, because the curve is not a straight line. Twenty five days over ten days gives two point five half-lives, so the fraction is one half to the power two point five. With me? That is one quarter, times one over root two, so one over four root two; the tick-shaped sign is the square root. So the fraction is about zero point one seven seven. See, the formula works for any time, a whole number of half-lives or not. Got it?

6:05Example 2: time for 90% of a sample to decay, half-life 12 h

Now a harder one. The half-life of a nuclide is twelve hours, the h on screen. How long does it take for ninety percent of a sample to decay? Pause the video and try it first. Alright, let's check. Many students write one half to the power t over twelve, equals zero point nine. Is that right? It is not, and that is the trap. The law gives the fraction left, not the fraction decayed. Ninety percent decays, so ten percent is left, and one half to the power t over twelve equals zero point one. Clear? Now flip both sides: two to the power t over twelve equals ten. The double arrow means it follows, so t over twelve equals the log of ten to the base two, the log with a small two below. That is about three point three two two. Okay? So t equals twelve times the log of ten to the base two, about thirty nine point nine hours. Right? Now let us do a quick check. Thirty six hours is three half-lives, which leave one eighth, and forty eight hours is four half-lives, which leave one sixteenth. With me? One sixteenth is less than one tenth, and one tenth is less than one eighth. So t must lie between thirty six and forty eight hours, and thirty nine point nine fits perfectly. Got it?

7:29Quick revision: N = N0 e^-lambda t, half-life, mean life

So, quick revision. N equals N naught, times e to the power minus lambda times t, and lambda does not depend on temperature or pressure. T half equals zero point six nine three over lambda, and after n half-lives, N naught over two to the power n is left. Tau equals one over lambda, one point four four times T half, and in one mean life, N falls to N naught over e. Now go and try the practice problems on this page. You can do them. All the best!