Mass, Energy and Binding Energy

Transcript

0:00Introduction

Hello everyone! Here is a strange fact. Weigh one proton and one neutron separately. Now join them into a deuteron, the nucleus of hydrogen two, and weigh it again. The deuteron is lighter than its two parts put together! So where did the missing mass go? Hold on to that question, we will answer it today. Questions on binding energy come up often in JEE, so let's begin.

0:28The atomic mass unit and E=mc²

So, first, a unit for such tiny masses. One atomic mass unit, the small letter u, is one twelfth of the mass of a carbon twelve atom; the small twelve at the top left is the mass number. That is one point six six zero five times ten to the power minus twenty seven kilogram. Tiny, right? Now, Einstein says mass is energy: E equals m times c squared, where c is the speed of light. So tell me, how much energy is one kilogram of mass worth? One kilogram, times c squared, that is three times ten to the power eight, the whole squared, gives nine times ten to the power sixteen joules. That is a huge amount of energy, okay? In the same way, one atomic mass unit is worth nine hundred and thirty one point five mega electron volts, M e V for short. We use this number in almost every problem. Got it? And here are the masses, with a small letter below each m: m p for the proton, m n for the neutron, m e for the electron, and m H for a whole hydrogen atom. Notice, m H is just m p plus m e, to these six decimal places, okay?

1:37Mass defect and binding energy

Now, back to our puzzle. Every nucleus has less mass than its separate protons and neutrons added up. So how would you write down the missing mass? Yes, exactly, add up all the separate nucleons, then subtract the nucleus. We call this missing mass the mass defect, delta m; the little triangle is the Greek capital delta. Delta m equals Z times m p, plus bracket A minus Z, times m n, minus capital M, the mass of the nucleus. Capital Z is the atomic number and capital A the mass number, so A minus Z counts the neutrons. With me? And where did the missing mass go? It became energy, the binding energy. E b, with a small b below, equals delta m times c squared; or simply, delta m in atomic mass units, times nine hundred and thirty one point five, in M e V, okay? But questions usually give atomic masses, which include electrons. So can we still use m p here? No, we cannot, and this is a classic trap. With atomic masses, use m H, the whole hydrogen atom, so the electrons cancel against those in M atom, the capital M with the small word atom below. Use m p instead, and you are off by Z electron masses. Right? The binding energy is released when the nucleus forms, and the same energy is needed to pull it apart. Clear? Finally, E b n, the binding energy per nucleon, is E b over A, okay? This, not the total, measures the stability of a nucleus.

3:15Reading the binding energy curve

Now look at this graph: binding energy per nucleon against the mass number, A, okay? For light nuclei it rises steeply, with peaks at helium four, carbon twelve and oxygen sixteen. The maximum, about eight point seven five M e V, comes at iron fifty six; the wavy equals sign means approximately. Right? Between A equal to thirty and one hundred and seventy, shown by the two less-than signs, it stays near eight M e V, nearly flat. Now, why so flat? Because the nuclear force is short-ranged, and it saturates. Each nucleon binds only its few nearest neighbours, so extra nucleons do not bind the others any tighter. With me? For very heavy nuclei it falls slowly, to seven point six M e V at uranium two thirty eight. So careful: uranium's total binding energy is huge, but per nucleon it is less tightly bound than iron. Okay? Now suppose a heavy nucleus splits into two middle-mass nuclei. Does it release energy, or does it need energy? It releases energy, yes, exactly! The pieces sit higher on the curve, and the extra binding energy comes out. That is fission. And two very light nuclei joining in fusion also move towards larger E b n, and release energy. Clear?

4:40Example 1: binding energy of the deuteron from atomic masses

Now, example one, and it finishes our puzzle. Find the binding energy of the deuteron, using the three atomic masses on screen: hydrogen one, the neutron, and hydrogen two. Pause the video and try it first. Done? Okay, let's check. The deuteron has one proton and one neutron, and these are atomic masses. So do we use m p, or m H? We use m H, correct, so that the electrons cancel. Delta m equals m H plus m n, minus capital M of hydrogen two; the bracket names the atom, okay? Put in the numbers: one point zero zero seven eight two five, plus one point zero zero eight six six five, minus two point zero one four, one zero two. That gives zero point zero zero two three eight eight atomic mass units. There it is, our missing mass! Multiply by nine hundred and thirty one point five: about two point two two M e V, okay? That energy came out when the deuteron formed, and the same energy breaks it apart. Right? Per nucleon, divide by two, and we get one point one one M e V, far below the eight M e V in the middle of the curve, so the deuteron is only loosely bound. Very good, that is example one done.

6:02Example 2: mass of Li-7 and energy of p + Li-7 to 2 alphas

Now a harder one. The binding energies per nucleon of lithium seven and helium four are five point six zero and seven point zero six mega electron volts. Find the mass of a lithium seven atom from these, and the energy released when a proton converts it into two alpha particles, with the Greek letter alpha. Pause the video and try it first. Done? Okay, let's check. Lithium seven has Z equal to three, so three protons and four neutrons. Total binding energy: seven times five point six zero, which is thirty nine point two M e V, okay? Now turn the mass defect formula around. Capital M equals three m H, plus four m n, minus the mass defect, thirty nine point two over nine hundred and thirty one point five, in atomic mass units. Clear? Three m H plus four m n is seven point zero five eight one three five, and the mass defect is zero point zero four two zero eight three. Subtract: seven point zero one six zero five two, so the mass is about seven point zero one six atomic mass units, and that finishes part one. Now part two. The letter p is the proton: a proton plus lithium seven gives two helium four nuclei, the alpha particles; the arrow shows what it turns into. Okay? Let me ask you this: how much binding energy does the lone proton bring in? It brings zero, exactly, because a single proton has nothing to be bound to. So energy released is binding energy after, minus binding energy before: eight times seven point zero six, minus seven times five point six zero. Right? Fifty six point four eight, minus thirty nine point two zero, gives about seventeen point three M e V released. See, the nucleons moved up the curve, and energy came out. Perfect, that is the whole problem.

7:57Quick revision: mass defect, 931.5 MeV, Eb per nucleon

So, quick revision. Mass defect: Z times m H, plus bracket A minus Z, times m n, minus the atomic mass. Binding energy: delta m in atomic mass units, times nine hundred and thirty one point five, in mega electron volts. And compare stability by E b over A, which peaks near iron fifty six. Now go and try the practice problems on this page. All the best!