Subtitles section Play video Print subtitles The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: So I want to do a quick review of what we did last time, because I know I threw-- I think we threw the full six boards of math and physics at you guys. We started off trying to describe this general situation. If you have a small nucleus 1 firing at a large nucleus 2, something happens, and we didn't specify what that was. A potentially different nucleus 3 could come shooting off at angle theta, and a potentially different nucleus 4 goes off at a different angle phi. Just to warn you guys, before you start copying everything from the board, starting last week I've been taking pictures of the board at the end of class. So if you prefer to look and listen or just take a few notes rather than copy everything else down, I'll be taking pictures of the board at the end of class from now on and posting them to the Stellar site. So up to you how you want to do it. We started off with just three equations. We conserve mass, energy, and momentum. Mass and energy-- let's see-- come from the same equation. c squared plus T1 plus M2 c squared plus T2 has to equal M3 c squared plus T3 plus M4 c squared plus T4. We started off making one quick assumption, that the nucleus 2, whatever we're firing things at, has no kinetic energy. So we can just forget that. What we also said is that we have to conserve x and y momentum. So if we say the x momentum of particle 1 would be root 2 M1 T1 plus 0 for particle 2, because if particle 2 is not moving, it has no momentum. Has to equal root 2 M3 T3 cosine theta, because it's the x component of the momentum, plus root 2 M4 T4 cosine phi. And the last equation for y momentum-- we'll call this x momentum, call that mass and energy, call this y momentum-- was-- let's say there's no y momentum at the beginning of this equation. So I'll just say 0 plus 0. Equals the y component of particle 3's momentum, root 2 M3 T3 sine theta, minus-- almost did that wrong-- because it's going in the opposite direction, root 2 M4 T4 sine phi. We did something, and we arrived at the Q equation. I'm trying to make sure we get to something new today. So the Q equation went something like-- and I want to make sure that I don't miswrite it at all. So when we refer to the Q equation, we're referring to this highly generalized equation relating all of the quantities that we see here. So I'm not going to go through all of the steps from last time, because, again, you have a picture of the board from last time. But it went Q equals T1 times M1 over M4 minus 1 plus T3 times 1 plus M3 over M4 minus 2 root M1 M3 T1 T3 cosine theta. And last time we talked about which of these quantities are we likely to know ahead of time and which ones might we want to find out. Chances are we know all of the masses involved in these particles, because, well, you guys have been calculating that for the last 2 and 1/2 weeks or so. So those would be known quantities. We'd also know the Q value for the reaction from conservation of mass and energy up there. And we'd probably be controlling the energy of particle 1 as it comes in. Either we know-- if it's a neutron, we know what energy it's born at. Or if it's coming from an accelerator, we crank up the voltage on the accelerator and control that. And that leaves us with just three quant-- two quantities that we don't know-- the kinetic energy of particle 3 and the angle that it comes off at. So this was the highly, highly generalized form. Recognize also that this is a quadratic equation in root 3, or root T3. And we did something else, and we arrived at root T3 equals s plus or minus root s squared plus t, where s and t-- let's see. I believe s is root M1 M3 T1 cosine theta over M3 plus M4. And we'll make a little bit more room. t should be-- damn it, got to look. Let's see. I believe it's minus M4 Q plus-- oh, I'll just take a quick look. All right, I have it open right here. I don't want to give you a wrong minus sign or something. I did have a wrong minus sign. Good thing I looked. 1 times E1 over M3 M4. And so we started looking at, well, what are the implications of this solution right here? For exothermic reactions, where Q is greater than 0, any energy E1 gets this reaction to occur. And all that that says, well, it doesn't really say much. All that it really says is that E3-- I'm sorry-- T3-- and let me make sure that I don't use any sneaky E's in there-- plus T4 has to be greater than the incoming energy T1. That's the only real implication here, is that some of the mass from particles 1 and 2 turned into some kinetic energy in particles 3 and 4. So that one's kind of the simpler case. For the endothermic case, where Q is less than 0, there's going to be some threshold energy required to overcome in order to get this reaction to occur. So where did we say? So, first of all, where would we go about deciding what is the most favorable set of conditions that would allow one of these reactions to occur by manipulating parameters in s and t? What's the first one that you'd start to look at? Well, let's start by picking the angle. Let's say if there was a-- if we had what's called forward scattering, then this cosine of theta equals 1. And that probably gives us the highest likelihood of a reaction happening, or the most energy gone into, let's say, just moving the center of mass and not the particles going off in different directions. Let's see. Ah, so what it really comes down to is a balance in making sure that this term right here, well, it can't go negative. If it goes negative, then the solution is imaginary and you don't have anything going on. So what this implies is that s squared plus t has to be greater or equal than 0 in order for this to occur. Otherwise, you would have, like I said, a complex solution to an energy. And energy is not going to be complex. That means the reaction won't occur. So this is where we got to last time. Yes. AUDIENCE: When you say s squared plus t is greater than 0 or greater than or equal to 0, it's endothermic, wouldn't it also be greater than or equal to 0 for an exothermic? MICHAEL SHORT: It would. But there's a condition here that-- lets see. In this case, for exothermic, Q is greater than 0 and that condition is always satisfied. For an endothermic reaction, Q is negative. So that's a good point. So if endothermic, then Q is less than 0, and it's all about making sure that that sum, s squared plus t, is not negative. What that means is in order to balance out the fact that you've got a negative Q here, you have to increase T1 in order to make that sum greater than or equal to 0. Yes. AUDIENCE: So that condition, s squared plus t is greater than or equal to 0, is that basically a condition for the endothermic reaction to occur? MICHAEL SHORT: That's correct. If s squared plus t is smaller than 0, which is to say that this whole