AIEEE and Earlier JEE Main Questions: Nuclear Energy

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. When a heavy uranium nucleus splits, a lot of energy comes out, right? Now split a neon twenty nucleus into two helium nuclei and one carbon nucleus. Does energy come out again? Most students say yes, of course. Hold on to your answer, because we will check it. In this video we solve older AIEEE and JEE Main questions on nuclear energy, a topic that comes up often, and this puzzle is one of them. Let's begin.

0:33The Q value: energy released in a reaction

First, in every nuclear reaction, two totals stay the same on both sides: the charge number, capital Z, the number of protons, and the nucleon number, capital A, the protons and neutrons together. Okay? Now, the Q value is the energy released in a reaction. Q equals, in brackets, the mass of the reactants minus the mass of the products, times c squared, where c is the speed of light. So Q is the mass lost, turned into energy. Right? One u, the atomic mass unit, corresponds to nine hundred and thirty one point five mega electron volts, written M e V on screen. The double-headed arrow means corresponds to. Clear? Many questions give binding energies instead. Now tell me, if the products are more tightly bound than the reactants, is Q positive or negative? Yes, exactly, Q is positive, because the extra binding energy comes out. So Q equals the total binding energy of the products, minus that of the reactants. Watch the order: the mass form is reactants minus products, but this one is products minus reactants. With me? So Q greater than zero, with the greater-than sign, means energy is released, and Q less than zero, with the less-than sign, means energy must be supplied. Got it?

1:53Fission and fusion on one curve

Now look at the graph. It plots E b n, the binding energy per nucleon, the E with a small b n below it, against capital A, the mass number. So the total binding energy is A times E b n, the factor many students forget. Okay? The curve rises steeply, peaks at iron fifty six, and falls slowly to uranium two thirty eight. So tell me, to release energy, should a nucleus split, or should nuclei join? It depends on where the nucleus sits. Energy comes out when the products are more tightly bound, nearer the peak. So light nuclei, on the steep left side, release energy by joining, which is fusion, the fusion arrow. Right? And heavy nuclei, on the far right, release energy by splitting, which is fission, the other arrow. Clear? Each fission of uranium two thirty five, with its mass number at the top left, releases about two hundred mega electron volts. With me? But fusion has a price. Both nuclei are positive and repel each other, so fusion needs temperatures from ten to the power seven up to ten to the power nine kelvin, the capital K on screen. Got it?

3:04From one reaction to a whole reactor

Now, for a whole lump of fuel, first count the nuclei. N equals the mass over the molar mass, times N A, Avogadro's number, the N with a small capital A below it. Okay? Then the total energy, E, is N times the energy released in one reaction. Right? One M e V equals one point six times ten to the power minus thirteen joules, the capital J on screen. Clear? If the fuel lasts a time t, the power, P, is E over t, with the time in seconds, so that P comes out in watts. With me?

3:37Example 1: energy to split neon-20 into helium and carbon

Now example one, our puzzle from the start, from JEE Main two thousand nineteen. Neon twenty gives two helium four nuclei plus carbon twelve, with each mass number at the top left of its symbol. The binding energies per nucleon are eight point zero three, seven point zero seven and seven point eight six M e V, in that order. Is energy released, or must it be supplied, and how much? Pause the video and try it first. Done? Okay, let's check. Never guess from the word split. So what exactly must we compare? We compare the total binding energies, not the values per nucleon. Before, neon has twenty nucleons, so twenty times eight point zero three gives one hundred and sixty point six mega electron volts. Right? After, two helium nuclei of four nucleons each, so two times four times seven point zero seven gives fifty six point five six. For carbon, twelve times seven point eight six gives ninety four point three two. Together, that is one hundred and fifty point eight eight mega electron volts. Okay? So the products have less binding energy than the neon we started with. Is energy released, or must it be supplied? Energy must be supplied, because Q comes out negative. Q is products minus reactants, so one hundred and fifty point eight eight minus one hundred and sixty point six, which gives minus nine point seven two mega electron volts. Clear? Since Q is less than zero, nine point seven two mega electron volts must be supplied, so the answer to our puzzle is no. If you said supplied, very good! Got it? That is because neon is a light nucleus, so splitting it moves away from the peak, not towards it. Splitting releases energy for heavy nuclei, not light ones. Right?

5:31Example 2: power of a reactor burning 2 kg of U-235

Now example two, the harder one, from JEE Main two thousand twenty. In a reactor, two kilograms, the k g on screen, of uranium two thirty five, with ninety two below, is fully used up in thirty days, at two hundred mega electron volts per fission. Find the power output. N A is six point zero two three times ten to the power twenty six per kilomole. Pause the video and try it first. Done? Okay, let's check. Many students write two over two hundred and thirty five, times six point zero two three times ten to the power twenty three. Is that right? That would be a thousand times too small, so be careful here. Here N A is per kilomole, so two kilograms is two over two hundred and thirty five kilomole of uranium. Units must match. Okay? So the number of fissions, N, is two over two hundred and thirty five, times six point zero two three times ten to the power twenty six, which gives five point one three times ten to the power twenty four. Right? Each fission gives two hundred M e V, and each M e V is one point six times ten to the power minus thirteen joules. So E is five point one three times ten to the power twenty four, times two hundred, times one point six times ten to the power minus thirteen, which gives one point six four times ten to the power fourteen joules. With me? Now the power. Can we just divide this energy by thirty? No, because thirty is in days, and watts need seconds. Thirty days is thirty times twenty four times three thousand six hundred seconds, the small s on screen, which gives two point five nine times ten to the power six seconds. Clear? So P equals one point six four times ten to the power fourteen, over two point five nine times ten to the power six. That gives about six point three times ten to the power seven watts, the capital W, where the wavy equals sign means approximately. So the power is about sixty megawatts, the M W on screen. Not bad for two kilograms of fuel!

7:35Quick revision: Q value, fission vs fusion, reactor power

So, quick revision. Q is the binding energy after, minus the binding energy before, and if Q is negative, energy must be supplied. Light nuclei release energy by fusing, and heavy nuclei by splitting. And for fuel, count the nuclei with N A, then P equals E over t, with matched units. Now try the other AIEEE and JEE Main questions on this page. All the best!