JEE Advanced and IIT-JEE Questions: Nucleus

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Nitrogen fifteen and oxygen fifteen both have fifteen nucleons and, in a problem we solve today, the same radius. Yet oxygen fifteen is held together less tightly. Why should that be? Hold on to that question, because this small difference will give us the radius of a nucleus. Today we solve JEE Advanced and IITJEE questions on the nucleus, a topic that comes up often. So, let's begin.

0:30Size of a nucleus, and its density

First, the size. The radius is R equals R naught, times A to the power one third, with capital A the mass number. R naught, the R with a small zero below, is about one point two femtometres, written f m, that is one point two times ten to the power minus fifteen metre; the wavy sign means approximately. Okay? Now the volume, four thirds pi R cubed, becomes four thirds pi R naught cubed, times A, as on screen. So it is proportional to the mass number; the curly sign means, is proportional to. Right? Now, a question from IITJEE two thousand three. How does the mass, small m, of a uranium nucleus vary with its volume, capital V? The mass is about A times the mass of one nucleon, so it grows with A too, and m is proportional to V; the double arrow means, it follows. So mass over volume is constant: every nucleus, light or heavy, has the same density. Clear?

1:32Mass defect and binding energy

Now a question from IITJEE nineteen ninety four. Weigh a stable nucleus, then all its protons and neutrons separately. Which is greater? The nucleus is the lighter one, which surprises many students. Its rest mass, capital M, is less than the total for its separate nucleons; the less-than sign says so. With me? This missing mass is the mass defect, delta m, with delta, the Greek letter like a little triangle. It is Z protons of mass m p, with a small p below, plus A minus Z neutrons of mass m n, minus M, the nuclear mass. Times c squared, the speed of light squared, it gives the binding energy, E b, with a small b below. Okay? Most questions give atomic masses. Then use m H, the hydrogen atom's mass, in place of m p, and M atom in place of capital M; the electron masses cancel. And one atomic mass unit, u, is worth nine hundred and thirty one point five mega electron volts, M e V, as the arrow shows. Right?

2:36Binding energy per nucleon

Now the graph. It plots E b n, the binding energy per nucleon, equal to E b over A, against the mass number. The higher a nucleus sits on this curve, the more tightly it is bound. Okay? It rises steeply for light nuclei and peaks at about eight point seven five mega electron volts for iron fifty six; the fifty six at the top left is its mass number. For A between thirty and one hundred and seventy, it stays near eight. Right? Then it falls slowly, to seven point six at uranium two thirty eight. Splitting a heavy nucleus releases energy, and so does joining two light nuclei. How can both be true? Yes, exactly, both move up the curve. Fission splits a very heavy nucleus into two middle-mass nuclei, fusion joins two very light ones, as the arrows show, and either way the products have a larger E b n, so energy comes out. Got it?

3:34Example 1: order of density of a uranium nucleus

Example one is from IITJEE nineteen ninety nine. Find the order of magnitude of the density of a uranium nucleus. R naught is one point one times ten to the power minus fifteen metre, and m p, the proton mass, is one point six seven times ten to the power minus twenty seven kilogram. Pause the video and try it first. Done? Okay, let's check. The question never tells us the mass number of uranium. Do we need it? No, and we just saw why. The mass is A times m p, the volume four thirds pi R naught cubed times A, so the mass number cancels. So rho, the Greek letter like a p, is three m p over four pi R naught cubed. Right? On top, three times one point six seven gives five point zero one, times ten to the power minus twenty seven. Below, one point one cubed is one point three three one, and four pi times that is about one point six seven times ten to the power minus forty four. Okay? Five point zero one over one point six seven is about three, and the powers give ten to the power seventeen. So rho is about three times ten to the power seventeen kilograms per cubic metre, of order ten to the power seventeen. So the IITJEE nineteen eighty nine statement that it is of order ten to the power four is false. Clear?

5:00Example 2: nuclear radius from N-15 and O-15 binding energy

Now the harder one, example two, from JEE Advanced two thousand sixteen. The electrostatic energy of Z protons in a nucleus of radius R is three fifths, times Z times Z minus one, e squared, over four pi epsilon naught R; e is the proton's charge, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. The four masses are on screen, with each mass number on top and atomic number below. The radii are equal, and the binding energies differ only through electrostatics. Find the radius. Pause the video and try it first. Done? Okay, let's check. How many protons and how many neutrons does each nucleus have? Nitrogen has seven protons and eight neutrons, and oxygen has eight protons and seven neutrons, simply swapped. Oxygen's extra proton means more electrostatic repulsion, which is why it is bound less tightly. With me? Write each binding energy with atomic masses. Nitrogen has one more neutron and oxygen one more hydrogen atom, so on subtracting, only m n minus m H, plus M O minus M N, survive inside the square brackets, times c squared; M O and M N, with small letters below, are the atomic masses. Right? Neutron minus hydrogen gives zero point zero zero zero eight four, and oxygen minus nitrogen gives zero point zero zero two nine five six. Together, zero point zero zero three seven nine six u, times nine hundred and thirty one point five, is about three point five four mega electron volts. Okay? Now the electrostatic side. Oxygen has Z equal to eight, so we use eight squared, which is sixty four, right? Careful, because that is exactly the trap here. The formula has Z times Z minus one: oxygen gives eight times seven, which is fifty six, and nitrogen gives seven times six, which is forty two, a difference of fourteen. With e squared over four pi epsilon naught as one point four four, the electrostatic difference is three fifths times one point four four over R, times fourteen, which is twelve point one over R, with R in femtometres. Got it? Set twelve point one over R equal to three point five four, and R is about three point four two femtometres. So a tiny mass difference measured the radius of a nucleus, and our puzzle is solved. Well done!

7:28Quick revision: radius, density, binding energy per nucleon

So, quick revision. The radius is R naught times A to the power one third, and every nucleus has a density of order ten to the power seventeen kilograms per cubic metre. Binding energy is the mass defect times c squared, with one u worth nine hundred and thirty one point five mega electron volts. And E b n peaks at iron fifty six, so fission and fusion both release energy. Now try the other questions on this page. All the best!