IIT-JEE Questions (1983–1999): Nuclear Energy

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Suppose a fusion reactor burns deuterium, the heavy form of hydrogen, and sends out two hundred megawatts of electricity. How much deuterium does it need in one full day? Is it a tonne, or a whole truckload? Keep that guess; you will work it out yourself by the end. Today we solve IITJEE questions on nuclear energy, from nineteen eighty three to nineteen ninety nine. So, let's begin.

0:29Why fission and fusion both release energy

So, today everything comes from one curve. E with a small b below it is the binding energy, the energy needed to pull a nucleus apart into its nucleons. Divide it by capital A, the mass number, and you get E with small b n below it, the binding energy per nucleon. Okay? The curve is highest near iron fifty six, F e with a small fifty six on the top left, the mass number. There it is about eight point seven five mega electron volts, M e V for short. Right? Now let me ask you this. Higher on the curve means more tightly bound. So, to release energy, must the nuclei move up, towards iron, or down? Yes, exactly, they must move up. Deuterium, H with a small two on top, sits very low, at the left. When light nuclei join, the new nucleus sits higher, as the fusion arrow shows, and energy is released. Clear? And uranium two thirty eight sits at seven point six, on the right. When one heavy nucleus splits into two middle-sized ones, the value goes from seven point six to about eight point five, as the arrow shows. That is fission. So both climb towards iron and release energy. Got it?

1:43Two ways to find the energy released

How much energy comes out? There are two ways. First, the energy released equals the total binding energy after, minus the total before. And each total is A times E b n, the mass number times the binding energy per nucleon. Total, not per nucleon; remember this. Okay? The second way uses masses. Q, the energy released, is the mass before minus the mass after, times nine hundred and thirty one point five mega electron volts, because that is what one atomic mass unit, the small u, is worth.

2:15Fusion, fission, or not possible at all?

Right? Next, from IITJEE nineteen eighty four. In each symbol, the small number on top is the mass number and the one below is the charge number; n is a neutron, and gamma, the Greek letter, is a photon. All four release energy. So, are all four possible fusion reactions? Not all of them, and here is the trap: releasing energy is not enough, because Z, the charge number, and A, the nucleon number, must balance. In the first, Z on the left is six plus one, which gives seven, but carbon fourteen has Z equal to six, so it is impossible. Got it? In the second, six plus one gives seven, and twelve plus one gives thirteen. In the third, seven plus one gives eight, and fourteen plus one gives fifteen. Light nuclei join into a heavier one, so both are fusion. With me? In the last one, uranium two thirty five splits into xenon, X e, and strontium, S r. That is fission, not fusion. So the answer is only b and c. Clear?

3:20What fusion and fission need

Next, what does each reaction need? For fusion, two positive nuclei must come within about one femtometre, f m, while the Coulomb repulsion pushes them apart. Okay? That needs temperatures from ten to the power seven, up to ten to the power nine kelvin, which is why it is called thermonuclear fusion. Right? Now fission, where fast neutrons must be slowed down. This IITJEE question from nineteen ninety four asks which of these slows them easily: lead shielding, heavy water, collisions with heavy nuclei, or a strong electric field? The answer is heavy water, the moderator. An electric field does nothing, because a neutron has no charge. Got it? And why not heavy nuclei? In a head-on elastic collision, the neutron gives all its energy to hydrogen, zero point eight nine of it to deuterium, but only zero point two eight to carbon. The heavier the nucleus, the less energy it takes. Clear?

4:20Example 1: energy released when two deuterons form He-4

Example one is from IITJEE nineteen eighty eight. The binding energy per nucleon is one point one mega electron volts for the deuteron, the deuterium nucleus, and seven point zero for helium four, H e with a small four on top. Find the energy released when two deuterons fuse to form a helium nucleus. Pause the video and try it first. Done? Okay, let's check. Many students just do seven point zero minus one point one, and get five point nine mega electron volts. Is that right? Oh no, careful, those are per nucleon values, and we need totals. Before, two deuterons with two nucleons each: two times two times one point one gives four point four mega electron volts. Okay? After, one helium nucleus with four nucleons: four times seven point zero gives twenty eight mega electron volts. Right? So the energy released is twenty eight minus four point four, which is twenty three point six mega electron volts. That is exactly four times the trap answer!

5:28Example 2: grams of deuterium per day for a 200 MW reactor

Clear? Now, example two, from IITJEE nineteen ninety: our puzzle from the start. Deuterium fuses into helium in a reactor of two hundred megawatts, M W, of electrical power, at twenty five percent efficiency. How many grams of deuterium are needed per day? The masses are on screen, in atomic mass units. Pause the video and try it first. Done? Okay, let's check. Delta m, with the Greek letter delta, a little triangle, is the mass lost per reaction. Two times two point zero one four one, minus four point zero zero two six, gives zero point zero two five six atomic mass units. Okay? Times nine hundred and thirty one point five, that gives about twenty three point eight mega electron volts, or about three point eight two times ten to the power minus twelve joules per reaction. Right? Now the power. Must the fusion itself produce two hundred megawatts, or more? It must produce more, because only twenty five percent becomes electricity. So it produces two hundred megawatts over zero point two five, which is eight hundred megawatts, or eight times ten to the power eight watts. Use two hundred, and your answer is four times too small. Got it? One day has eighty six thousand four hundred seconds, so the energy per day is eight times ten to the power eight, times eighty six thousand four hundred, which is about six point nine one times ten to the power thirteen joules. With me? Divide by the energy of one reaction to get about one point eight one times ten to the power twenty five reactions. But careful, each reaction uses two deuterons, so we need about three point six two times ten to the power twenty five deuterons. Clear? Divide by Avogadro's number, six point zero two times ten to the power twenty three, to get about sixty moles, and each mole is about two grams. So the answer is one hundred and twenty grams. That is all it needs for a whole day, not a truckload. That is the power of fusion!

7:33Quick revision: total BE, Q from mass, Z and A, efficiency

So, quick revision. Energy released is the total binding energy after minus before, each total being the mass number times the binding energy per nucleon. From masses, Q is delta m, in atomic mass units, times nine hundred and thirty one point five mega electron volts. Check Z and A first, and divide the electrical output by the efficiency. Now try the practice problems on this page. All the best!