## Subtitles section Play video

• The following content is provided under a Creative

• Your support will help MIT OpenCourseWare

• 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