Nuclear Reactions and Q-values

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Suppose a nuclear reaction absorbs one point one nine mega electron volts of energy, so its Q value is negative. You fire an alpha particle carrying exactly that much kinetic energy at a nitrogen nucleus sitting at rest. Will the reaction happen? Many students would say yes. Hold on to your answer; we will check it with real numbers. Questions on Q values do appear in JEE, and they are quick marks once you know the method. So, let's begin.

0:35Writing a nuclear reaction

First, how we write a nuclear reaction. On screen, the projectile, small a, hits a target nucleus, capital X, and out come the products, capital Y and small b. The arrow means gives. Okay? Each nucleus is written like this nitrogen. The small raised number at the top left, fourteen, is the mass number A, the number of nucleons. The small number at the bottom left, seven, is the charge number Z, the number of protons. With me? Now, in every nuclear reaction, four things are conserved. The total charge number, the total nucleon number, energy, and momentum. The last two will matter a lot today. Clear? So let us use the first two right away. Nitrogen fourteen plus helium four gives a question mark, plus hydrogen one. Tell me, what is the missing nucleus? Balance the bottom numbers first. Seven plus two is nine, so the missing nucleus has Z equal to eight, because eight plus one is nine. Now the top numbers: fourteen plus four is eighteen, so its mass number is seventeen. Charge number eight is oxygen, so it is oxygen seventeen. Easy, right?

1:51The Q-value: energy released in a reaction

Now the main idea, the Q value, the energy released in a reaction. Q equals, in brackets, m sub a plus m sub X, minus m sub Y, minus m sub b, all times c squared. Here m sub a, the m with a small a below it, is the mass of the projectile, and the same way for the others. And c is the speed of light; the small raised two means squared. Okay? So Q is the mass lost, times c squared. Now tell me, if the products are lighter, where did that missing mass go? Yes, exactly, it turned into kinetic energy! So Q also equals the kinetic energy of the products, minus the kinetic energy of the reactants. Got it? For energy units, use this. 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. And the other way round, an energy E corresponds to a mass, E over c squared. Clear? One more form uses binding energies. Q equals the total binding energy of the products, minus that of the reactants. Watch the order here. The mass form is reactants minus products, but the other two forms are products minus reactants. With me?

3:15Exothermic, endothermic and the threshold

Next, let us look at the sign of the Q value. Q greater than zero, with the greater-than sign on screen, means the reaction is exothermic, and energy is released. Okay? And Q less than zero, with the less-than sign, means the reaction is endothermic. Energy must be supplied, as the kinetic energy of the projectile. Right? Now, this is our puzzle. Mod Q, with the two vertical bars around Q, means the size of Q without its sign. If Q is negative, is a kinetic energy of just mod Q enough for the projectile? It is not enough, and here is why. Momentum is also conserved, so the products cannot come out at rest, and part of the kinetic energy stays as motion. So, for a target at rest, the threshold kinetic energy, K sub t h, equals mod Q, times, one plus m sub a over m sub X, the projectile's mass over the target's mass. Got it? One last result. Suppose a reaction starts from rest and gives two products, one light and one heavy. Which one carries away more energy? Yes, the lighter one gets more. From rest, the two products have equal and opposite momenta, so they share the energy in the inverse ratio of their masses. K one over K two equals m two over m one, where the small one and two below mark the two products. Clear?

4:40Example 1: Q value of N-14 plus alpha giving O-17

Now example one, the same reaction we just balanced. Find the Q value of nitrogen fourteen plus helium four, giving oxygen seventeen plus hydrogen one, and state whether it is exothermic. The atomic masses, in atomic mass units, are in the same order. Pause the video and try it first. Done? Okay, let's check. First, add the reactants. Fourteen point zero zero three zero seven four, plus four point zero zero two six zero three, gives eighteen point zero zero five six seven seven atomic mass units. Okay? Now the products. Sixteen point nine nine nine one three two, plus one point zero zero seven eight two five, gives eighteen point zero zero six nine five seven atomic mass units. Right? Look at the two totals. Which side is heavier, and so, is Q positive or negative? Yes, the products are heavier, so the total rest mass goes up instead of down. Delta m, the little triangle delta meaning the change, is reactants minus products, and that is minus zero point zero zero one two eight zero atomic mass units. With me? Now change it to energy. Minus zero point zero zero one two eight zero, times nine hundred and thirty one point five, gives about minus one point one nine mega electron volts. The wavy equals sign means approximately. Clear? Q is less than zero, so the reaction is endothermic, not exothermic. And this is exactly the reaction from our puzzle, the one that absorbs one point one nine mega electron volts. Next, let us see whether an alpha particle with that much energy can make it happen.

6:24Example 2: threshold energy of the alpha particle

Now example two, the harder one, and it settles our puzzle. For the reaction of example one, find the least kinetic energy of the alpha particle, alpha being the Greek letter on screen, that can make it happen with nitrogen at rest. Pause the video and try it first. Done? Alright. From example one, mod Q is one point one nine mega electron volts. So is the answer simply one point one nine mega electron volts? No, no, careful, that is exactly the trap, because momentum is conserved too, so we need the threshold formula. The projectile is the alpha particle and the target is nitrogen fourteen, so m sub a over m sub X is about four over fourteen, the ratio of their mass numbers. Okay? So K sub t h equals one point one nine, times, one plus four over fourteen. One plus four over fourteen is eighteen over fourteen, which simplifies to nine over seven. Right? Now, one point one nine times nine over seven gives about one point five three mega electron volts. So the answer to our puzzle is no. An alpha particle with only one point one nine mega electron volts cannot make this reaction happen; it needs about one point five three, which is more than one point one nine.

7:41Quick revision: balance Z and A, Q from masses, threshold

Got it? So, quick revision. First, balance Z and A on both sides of the reaction. Second, Q equals the mass of the reactants minus the mass of the products, times nine hundred and thirty one point five M e V per u, where the slash means per. And third, when Q is negative, the projectile needs the threshold energy, mod Q times, one plus m sub a over m sub X, and not just mod Q on its own. Now go and try the practice problems on this page. All the best!