AIEEE and Earlier JEE Main Questions: Nucleus

Transcript

0:00Introduction

Hello everyone! First, a strange fact. Weigh fifty protons and seventy neutrons one by one, then join them into a tin nucleus and weigh it again. The tin nucleus is lighter, by more than one whole atomic mass unit! So where did that mass go? Today we solve older JEE Main questions on the nucleus, and this puzzle is one of them. Let's begin.

0:25How big is a nucleus?

First, the size. The radius R equals R naught, times A to the power one third, the cube root of the mass number. R naught, the R with a small zero below, is about one point two f m, the femtometre, ten to the power minus fifteen metre. The wavy equals sign means approximately. Okay? Now let me ask you this. One nucleus has eight times as many nucleons as another. Is its radius also eight times bigger? It is not, and that is the trap, because the radius grows only as the cube root. The cube root of eight is two, so mass numbers in the ratio one is to eight give radii in the ratio one is to two; the double arrow means, it follows. Right? So for two nuclei, R two over R one, with small numbers below, equals A two over A one, to the power one third. R naught cancels in the ratio. Clear?

1:18Mass defect and binding energy

Now, the mass of a nucleus. X is the element's symbol, with the mass number A at the top left and the atomic number Z at the bottom left. So the nucleus holds Z protons and A minus Z neutrons. With me? Let me ask you something. Is the nucleus heavier than all its protons and neutrons weighed separately, or lighter, or exactly equal? If you said lighter, yes, exactly, the nucleus is lighter than its separate parts. The missing mass is the mass defect, delta m, where capital delta, the Greek triangle, means a difference. Delta m equals Z times m p, plus, bracket A minus Z, times m n, minus capital M, the mass of the nucleus. Here m p and m n, with a small p and a small n below, are the masses of a proton and a neutron. Okay? The binding energy, E b, with a small b below, equals delta m times c squared, where c is the speed of light. It is the energy released when the nucleus forms, and the energy needed to break it completely into its nucleons. Got it? Now, one atomic mass unit, written u, is about nine hundred and thirty one point five mega electron volts, the M e V on screen. So just multiply delta m in u by nine hundred and thirty one point five. Clear?

2:39Binding energy per nucleon

Now, to compare nuclei, we share the binding energy out per nucleon. Tell me, do we divide by Z, the protons, or by A, all the nucleons? If you said A, that is correct, we divide by all the nucleons. So E b n, with a small b n below, the binding energy per nucleon, equals E b over A, and it measures how stable a nucleus is. The figure plots it against the mass number. Right? Look at the curve. It starts low at hydrogen two on the far left, rises steeply for light nuclei, and reaches its peak, about eight point seven five mega electron volts, at iron fifty six. The fifty six at the top left of F e is its mass number. With me? Notice the long flat part. For A between thirty and one hundred and seventy, shown with the less-than signs, it stays near eight mega electron volts. That is because the nuclear force is short-ranged and saturates: each nucleon binds only its few nearest neighbours. Okay? At the far right, uranium two thirty eight drops to about seven point six mega electron volts. So a heavy nucleus splitting into two middle ones, fission, and two very light nuclei joining, fusion, both move towards larger E b n, the two arrows in the figure, and so both release energy. Got it?

4:02Example 1: radius of Te-125 from Al-27 (AIEEE 2005)

Now, example one, from the AIEEE two thousand five paper, the older form of JEE Main. If the radius of the aluminium twenty seven nucleus is estimated as three point six femtometres, find the radius of the tellurium one twenty five nucleus. Pause the video and try it first. Done? Okay, let's check. Many students write three point six times one hundred and twenty five over twenty seven, which is about sixteen point seven femtometres. Is that right? That would be wrong, and it is the trap we just saw, because the radius goes as the cube root. So R of tellurium over R of aluminium, with their symbols below, is one hundred and twenty five over twenty seven, to the power one third. One hundred and twenty five is five cubed and twenty seven is three cubed, so the ratio is five over three. Right? So the radius of tellurium is three point six times five over three, which gives exactly six femtometres. Clear? And a quick check: three point six over three, the cube root of twenty seven, gives R naught, the usual one point two femtometres, and one point two times five is again six. Well done!

5:10Example 2: binding energy per nucleon of Sn-120

Now, example two, from JEE Main two thousand twenty, and this is our puzzle from the start. Find the binding energy per nucleon of tin, S n, with one hundred and twenty on top and fifty below. Then m p is one point zero zero seven eight three u, m n is one point zero zero eight six seven u, and the tin mass, m with a small S n below, is one hundred and nineteen point nine zero two one nine nine atomic mass units. And here one u is nine hundred and thirty one mega electron volts. Pause the video and try it first. Done? Okay, let's go. Tin has Z equal to fifty, so fifty protons, and N, the number of neutrons, is one hundred and twenty minus fifty, which gives seventy. With me? Now weigh the parts. Fifty times one point zero zero seven eight three, plus seventy times one point zero zero eight six seven, gives one hundred and twenty point nine nine eight four atomic mass units. Okay? Subtract the tin's mass, one hundred and nineteen point nine zero two one nine nine, and delta m is about one point zero nine six two atomic mass units. There is our puzzle: more than one whole atomic mass unit is missing! It became the binding energy that holds the nucleus together. Right? So E b equals one point zero nine six two times nine hundred and thirty one, which gives about one thousand and twenty point six mega electron volts. Notice, we used nine hundred and thirty one, the value this question gives. Clear? Now the last step, per nucleon. So do we divide by fifty or by one hundred and twenty? We divide by one hundred and twenty, all the nucleons. One thousand and twenty point six over one hundred and twenty gives about eight point five mega electron volts per nucleon. Got it? The figure gives a quick check. Dividing by fifty, the division sign on screen, would give about twenty point four, far above the peak, so it is off the curve. But eight point five sits right on the flat part. Well done!

7:11Quick revision: nuclear radius, mass defect, binding energy

So, quick revision. The radius is R naught times A to the power one third, so radii go as the cube root of the mass number, not as the mass number itself. The mass defect, delta m, is the mass of the separate nucleons minus the mass of the nucleus, and the binding energy is delta m times c squared, with one u about nine hundred and thirty one point five mega electron volts. And per nucleon, divide by A, all the nucleons. Now try the other JEE Main questions on this page. All the best!