Energy Released in a Decay

Transcript

0:00Introduction

Hello everyone! Let me start with a puzzle. Uranium two hundred and thirty eight is radioactive, and it decays by throwing out an alpha particle, a whole helium nucleus with four nucleons in it. So why does it never throw out just a single proton, which is so much lighter? The answer is energy bookkeeping, and today we learn it: how much energy a decay releases, and who carries it away. Questions on this come up often in JEE, so let's begin.

0:31The Q-value: is the decay allowed?

So, every decay starts with one question: how much energy does it release? We call it capital Q, the Q value. Q equals the mass of the parent, minus the total mass of the products, times c squared, where c is the speed of light. The mass that goes missing comes out as energy. Okay? Now, in practice we never multiply by c squared ourselves. Take the masses in atomic mass units, written as a small u, and find the mass difference first. Then multiply it by nine hundred and thirty one point five, and Q comes out in mega electron volts, M e V for short. Clear? Now let me ask you this. Suppose Q comes out negative, so the products are heavier than the parent. Can the nucleus still decay on its own? The answer is no, because then the decay would need energy, and a nucleus sitting alone has none to give. So a decay is possible only if Q is greater than zero, that is, positive. Got it? Now, back to our uranium puzzle. The small number at the top left of each symbol is its mass number, and the arrow means changes into. Uranium would change into protactinium two hundred and thirty seven, plus hydrogen one, the proton. With the atomic masses, Q comes out minus seven point seven mega electron volts, less than zero. So uranium can never throw out a proton by itself. Puzzle solved, with me?

2:00Alpha decay: who carries the energy?

Alright, now alpha decay, with the Greek letter alpha. In square brackets, capital M with a small X below is the mass of the parent atom, M Y is the daughter atom, and M H e is the helium atom. Q is the parent minus the two products, times c squared, using atomic masses. Okay? Now tell me, does the alpha particle fly away with all of Q as its kinetic energy? Not all of it, and this is a common trap. The parent was at rest, so the alpha particle and the daughter fly apart with equal and opposite momenta, and the daughter recoils. With the same momentum, the lighter one gets more of the kinetic energy, which we call capital K, okay? And here is the result. K alpha, the kinetic energy of the alpha particle, equals A minus four, over A, times Q, where capital A is the mass number of the parent. And K daughter equals four over A, times Q, which is small. For a heavy nucleus, A is large, so the alpha particle takes almost all of Q, but not quite. Clear? And if a question gives you K alpha and asks for Q, just turn it around. Q equals A over A minus four, times K alpha, always a little more than K alpha. Got it?

3:21Beta and gamma decay

Now beta minus decay, the Greek letter beta with a small minus sign. With atomic masses, Q is simply M X minus M Y, times c squared, because the daughter atom needs one more electron than the parent, and that is exactly the electron which came out. This Q is the maximum kinetic energy of the electron, e minus, since it is shared with nu bar, the Greek letter nu with a bar on top, the antineutrino. Okay? Now beta plus decay, beta with a small plus sign, where a positron comes out. With atomic masses again, is Q just M X minus M Y once more? The answer is no, and many students lose marks right here. We also subtract two times m e, where small m with a small e below is the mass of one electron. It is two electron masses because the daughter atom has one electron fewer than the parent, so one electron is left over, and the positron has the same mass. With me? And finally gamma decay, with the Greek letter gamma. The photon energy equals the difference of the nuclear levels, less a tiny recoil energy, E squared over two times capital M times c squared, where E is the photon energy and capital M is the mass of the nucleus. Right?

4:37Example 1: maximum KE of electrons in carbon-14 beta decay

Now let's try one. Find the maximum kinetic energy of the electrons in the beta minus decay of carbon fourteen. The atomic masses are on screen: carbon fourteen, fourteen point zero zero three two four two u, and nitrogen fourteen, fourteen point zero zero three zero seven four u. Pause the video and try it first. Done? Okay, let's check. This is beta minus with atomic masses, so there is no electron mass to subtract. Q is the carbon mass minus the nitrogen mass, times nine hundred and thirty one point five mega electron volts. Okay? The difference is zero point zero zero zero one six eight atomic mass units, so keep every decimal place, because the answer sits in the last few digits. Times nine hundred and thirty one point five, that gives about zero point one five six mega electron volts, where the wavy equals sign means approximately. Clear? Now tell me, does every electron from carbon fourteen come out with zero point one five six mega electron volts? That is not the case, because this is only the maximum. Each electron gets up to this much, and the antineutrino, nu bar, carries away the rest. Right?

5:57Example 2: Q value and alpha KE in radium-226 decay

Now a harder one. Find the Q value of the alpha decay of radium two hundred and twenty six, and the kinetic energy of the alpha particle. The atomic masses of radium, radon two hundred and twenty two, and helium four are on screen, in atomic mass units. Pause the video and try it first. Done? Okay, let's check. Radium throws out an alpha particle, so the daughter is radon two hundred and twenty two. Q is the radium mass, minus the radon mass, minus the helium mass, all times nine hundred and thirty one point five. Okay? The mass difference is zero point zero zero five two two nine atomic mass units. Multiply by nine hundred and thirty one point five, and that gives about four point eight seven mega electron volts. It is positive, so the decay can happen. Got it? So the alpha particle comes out with four point eight seven mega electron volts of kinetic energy, right? That is exactly the trap, so be careful, because the radon nucleus recoils too. So K alpha equals A minus four over A, times Q, which is two hundred and twenty two over two hundred and twenty six, times four point eight seven. That gives about four point seven eight mega electron volts. Clear? And the small rest goes to the recoiling radon nucleus. So four point seven eight against four point eight seven, close, but not equal. See how quick that was?

7:34Quick revision: Q value, alpha KE share, beta minus and plus

So, quick revision. Q is the mass of the parent minus the mass of the products, times c squared, and a decay happens only if Q is positive. In alpha decay, the alpha particle gets A minus four over A, times Q, and the daughter takes the rest. And with atomic masses, beta minus is just M X minus M Y, times c squared, while beta plus also subtracts two electron masses. Now go and try the practice problems on this page. All the best!