IIT-JEE Questions (1981–1992): Radioactivity

Transcript

0:00Introduction

Hello everyone! Here is a strange fact. Even a neutron is radioactive when it is free. Suppose slow neutrons, with a half-life of seven hundred seconds in this question, fly along a tube ten metres long. What fraction decays on the way, half, or only a tiny bit? Keep your guess in mind. Today we solve old IITJEE questions on radioactivity, from nineteen eighty one to nineteen ninety two, and this puzzle is one of them. So, let's begin.

0:33What each decay does to A and Z

First, what each decay does to a nucleus. Capital X is the parent and capital Y the daughter; capital A at the top left is the mass number, and capital Z at the bottom left, the atomic number. In alpha decay, alpha being the Greek letter, a helium four nucleus flies out, so the arrow leads to A minus four on top and Z minus two below. Okay? Now beta minus decay, the Greek beta with a small minus up top. The beta particle is an electron from the nucleus, but a nucleus holds no electrons. So, from nineteen eighty seven: what happens inside the nucleus in negative beta decay? Yes, exactly, a neutron in the nucleus decays, emitting an electron. Small n, the neutron, becomes small p, the proton, plus e minus, the electron, plus nu bar, the Greek nu with a bar on top, the antineutrino. The mass number stays the same, and the atomic number rises by one. Right? Beta plus decay is the reverse. Here a proton becomes a neutron, giving out e plus, the positron, and nu, a neutrino, with no bar this time. So the atomic number falls by one, and the mass number stays the same. Clear? Now a topper's shortcut for long chains, from a parent with A and Z to a final nucleus with A prime and Z prime; the little dash is the prime. Only alpha decay changes the mass number, so n alpha, the n with a small alpha below, is A minus A prime, over four. And n beta, for beta minus, is two n alpha, minus bracket, Z minus Z prime. Got it?

2:16The decay law, half-life and mean life

Next, the decay law: N equals N naught, times e to the power minus lambda times t. N is the number 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 right one is for another lesson. Okay? Now the half-life, T half, the T with one over two below it, is l n two over lambda, where l n is the natural log, that is, zero point six nine three over lambda. And the mean life, tau, the Greek letter like a small t, is one over lambda; the coloured dashed line on the graph, the tangent at the start, meets the time axis at tau. With me? Next, from nineteen eighty three: a sample decays at ten to the power three disintegrations per second, and its half-life is one second. What is the rate after one second, and after three seconds? After one second, the rate is five hundred per second. But some students take away five hundred every second, and go below zero by three seconds! That is the trap: each half-life halves what is left. So after n half-lives, N is N naught over two to the power n, and the rate halves the same way. Follow the arrows: one thousand, then five hundred, then two hundred and fifty, then one hundred and twenty five per second. Right? Now one more shortcut: when t is much less than the half-life, shown by the double less-than sign, e to the power minus lambda times t is about one minus lambda times t; the wavy sign means approximately. So the fraction that decays is about lambda times t, which our neutron puzzle needs.

4:05Example 1: alpha and beta count, U-238 to Pb-206 (1985)

Okay? Now, example one, from IITJEE nineteen eighty five. The uranium series starts with uranium two thirty eight, atomic number ninety two, and ends with lead two oh six, atomic number eighty two; P b is lead. How many alpha and beta particles are emitted? Pause the video and try it first. Done? Okay, let's check. The atomic number falls from ninety two to eighty two, that is, by ten. Each alpha takes away two, so ten over two gives five alpha particles, right? No, that is the trap, because beta decays change the atomic number too. Only alpha decay changes the mass number, so count the alphas from it: two hundred and thirty eight minus two hundred and six is thirty two, and thirty two over four gives eight. Okay? Now the atomic number. Eight alphas, two each, would lower it by sixteen, but it falls only by ninety two minus eighty two, which is ten. So what pushes the atomic number back up? Yes, exactly, the beta minus decays, each one raising the atomic number by one. That is our formula: n beta equals two n alpha, minus bracket, Z minus Z prime, so sixteen minus ten gives six. Clear? So the answer is eight alpha and six beta minus particles. Quick check: ninety two, minus sixteen, plus six, gives eighty two, which is lead. With me?

5:43Example 2: fraction of slow neutrons decaying in 10 m (1986)

Now example two, the harder one: our puzzle, from IITJEE nineteen eighty six. Neutrons with kinetic energy zero point zero three two seven electron volts, the e V on screen, have a half-life of seven hundred seconds. What fraction decays before travelling ten metres? Pause the video and try it first. Done? Okay, let's check. The decay law works with time, but the question gives us a distance. So what must we find first? Yes, the time of flight, so we need the speed, v, which is the square root, the tick-shaped sign, of two K over m, with K the kinetic energy and m the mass. But K must be in joules, the capital J: zero point zero three two seven, times one point six times ten to the power minus nineteen. Okay? The neutron's mass, m n, the m with a small n below, is one point six seven times ten to the power minus twenty seven kilogram. So the speed is two point five times ten to the power three metres per second, written m s to the power minus one. Right? So the time of flight is ten over two point five times ten to the power three, which is four times ten to the power minus three seconds. That is much less than seven hundred seconds, so our shortcut works. Clear? So the fraction is just t over T half, that is, four times ten to the power minus three, over seven hundred, right? That is the trap, because t over T half drops the zero point six nine three. The fraction is about lambda times t, and lambda is zero point six nine three over T half. So it is zero point six nine three times four times ten to the power minus three, over seven hundred, which is three point nine six times ten to the power minus six. Our puzzle is solved: only about four in a million decay on the way. Got it?

7:48Quick revision: alpha-beta count, half-life, lambda t

So, quick revision. Count the alphas from the mass number alone, then the beta minus particles from the atomic number: n beta equals two n alpha, minus bracket, Z minus Z prime. T half is l n two over lambda, tau is one over lambda, and each half-life halves N and the rate. And for a time much less than the half-life, the fraction decaying is about lambda times t. Now try the other old IITJEE questions on this page. All the best!