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  • 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