AIEEE and Earlier JEE Main Questions: Radioactivity

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. The half-life of a radioactive substance is twenty minutes. At one instant, one third of it has decayed, and at a later instant, two thirds of it has decayed. How much time passes in between? It looks like a long calculation with logs, right? In fact it takes one line, and we will see why. Today we solve radioactivity questions from AIEEE and the earlier JEE Main papers, and this puzzle, from AIEEE two thousand eleven, is one of them. So, let's begin.

0:37The decay law on one screen

So, first the whole decay law on one screen. N equals N naught, times e to the power minus lambda times t. Here 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 falling curve; the chain on the right belongs to another lesson. Okay? Next, the half-life, T half, written as 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. On the graph you also see tau, the Greek letter where the red dashed line meets the time axis: tau is the mean life, one over lambda, so the half-life is also zero point six nine three times tau. Right? Now let me ask you this: if you start with N naught nuclei, how many are left after three half-lives? Yes, exactly, N naught over eight. Each half-life halves what is left, so after n half-lives, N equals N naught over two to the power n. You can see it on the graph: N naught over two, then over four, then over eight. With me? Now a line that many students forget. Questions often ask how much has decayed, not how much is left. The fraction decayed is one minus N over N naught. Clear? And the activity, the number of decays per second, equals lambda times N, as you see on screen. Lambda is fixed, so the activity also halves in every half-life, just like N does. Got it?

2:24Our puzzle, from AIEEE 2011

Now, back to our puzzle. One third decayed, then two thirds decayed. Many students stare at one third and two thirds and reach for logs. So tell me, which quantity should you look at, the part decayed or the part left? The part left, always, because the decay law is about what is left. At the first instant, one minus one third, so two thirds is left. At the second, one minus two thirds, so one third is left. Okay? Now look, two thirds to one third is exactly halving, as the arrow shows. And halving takes one half-life. So the answer is just twenty minutes, with no logs at all, and our puzzle is solved in one line. Right?

3:10What each decay does to A and Z

Now the second kind of question: decay chains. Here capital A is the mass number, the number of nucleons, and Z is the atomic number, the number of protons. In alpha decay, alpha being this Greek letter, the nucleus throws out a helium nucleus, so the mass number falls by four and Z falls by two. Okay? Next, beta minus decay, written beta with a small minus sign at the top: an electron comes out. So the nucleus loses a negative charge. Tell me, does Z go up or down? It goes up by one, and this surprises many students. Inside the nucleus, a neutron becomes a proton, and the electron is thrown out. One more proton, so Z rises by one, while the mass number stays the same. Right? Beta plus decay, with a small plus sign, is the opposite: a proton becomes a neutron. So Z falls by one, and the mass number again stays the same. Clear? And in gamma decay, gamma being this Greek letter, only a photon comes out. So neither the mass number nor Z changes; an AIEEE question from two thousand seven asks exactly this. With me? Behind all these rules is one idea: charge and nucleon number are conserved in every decay. Got it?

4:32Example 1: Th-232 chain with 6 alpha and 4 beta decays

Now, example one, from JEE Main two thousand nineteen. Thorium, written T h, with the mass number two hundred and thirty two on top and the atomic number ninety below, starts a decay chain. Six alpha particles and four beta particles are emitted. Find the mass number and the atomic number of the final nucleus. Pause the video and try it first. Done? Okay, let's check. Only the alphas change the mass number. Six times four gives twenty four, and two hundred and thirty two minus twenty four gives two hundred and eight. Okay? Each alpha also takes away two protons. Six times two gives twelve, and ninety minus twelve gives seventy eight. With me? Now the four betas. Many students subtract four again and write seventy four, but is that right? It is not, and that is the trap. Each beta minus turns a neutron into a proton, so Z goes up by one each time. Seventy eight plus four gives eighty two. Clear? So the final nucleus has mass number two hundred and eight and atomic number eighty two. And the betas did not touch the mass number at all. Very good!

5:50Example 2: ratio decayed after 80 min, half-lives 20 and 40

Now a harder one, example two, from JEE Main two thousand sixteen. Element A has a half-life of twenty minutes and element B, forty minutes, and at the start the two samples have equal numbers of nuclei. Find the ratio of the numbers of nuclei that have decayed after eighty minutes. Pause the video and try it first. Done? Okay, let's check. For element A, eighty over twenty gives four half-lives. So the fraction left is one half to the power four, which is one sixteenth. Okay? For element B, eighty over forty gives two half-lives. So the fraction left is one half squared, which is one quarter. Clear? So the ratio is one sixteenth to one quarter, which is one is to four, and we are done, right? That is exactly the trap, so read the question again. It asks for the nuclei that have decayed, not the ones left. So use one minus the fraction left: one minus one sixteenth gives fifteen sixteenths, and one minus one quarter gives three quarters. With me? Now divide. Fifteen sixteenths times four thirds gives five over four. So the ratio of the decayed nuclei is five is to four.

7:06Quick revision: half-life, fraction decayed, A and Z rules

Got it? So, quick revision. After n half-lives, N naught over two to the power n is left, and the fraction decayed is one minus N over N naught. Alpha lowers the mass number by four and Z by two; beta minus raises Z by one, beta plus lowers it by one, and gamma changes nothing. And always read the question: does it ask what has decayed, or what is left? Now try the other AIEEE and JEE Main questions on this page. All the best!