Placeholder Image

字幕表 動画を再生する

  • 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.

  • PROFESSOR: OK, guys.

  • Welcome back.

  • As you can see, we're not using the screen today.

  • This is going to be one of those fill-the-board lectures.

  • But I am going to work you through every single step.

  • We're going to go through the Q equation

  • and derive its most general form together,

  • which, for the rest of this class,

  • we'll be using simplified or reduced

  • forms to explain a lot of the ion

  • or electron-nuclear interactions as well

  • as things like neutron scattering

  • and all sorts of other stuff.

  • We'll do one example.

  • For any of you that have looked at neutrons slowing down

  • before, how much energy can a neutron

  • lose when it hits something?

  • We'll be answering that question today in a generally

  • mathematical form.

  • And then a few lectures later, we'll

  • be going over some of the more intuitive aspects

  • to help explain it for everybody.

  • So I'm going to show you the same situation that we've

  • been describing sort of intuitively so far,

  • but we're going to hit it mathematically today.

  • Let's say there's a small nucleus,

  • 1, that's firing at a large nucleus, 2, and afterwards,

  • a different small nucleus, 3, and a different large nucleus,

  • 4, come flying out.

  • And so we're going to keep this as general as possible.

  • So let's say if we draw angles from their original paths,

  • particle 3 went off at angle theta

  • and particle 4 went off at angle phi.

  • So hopefully those are differentiable enough.

  • And if we were to write the overall Q equation showing

  • the balance between mass and energy here,

  • we would simply have the mass 1 c squared

  • plus kinetic energy of 1.

  • So in this case, we're just saying

  • that the mass and the kinetic energy

  • of all particles on the left side and the right side

  • has to be conserved.

  • So let's add mass 2 c squared plus T2 has to equal mass 3 c

  • squared plus T3 plus mass 4 c squared plus T4, where,

  • just for symbols, M refers to a mass,

  • T refers to a kinetic energy.

  • And so this conservation of total mass or total energy

  • has got to be conserved.

  • And we'll use it again.

  • Because, again, we can describe the Q, or the energy consumed

  • or released by the reaction, as either the change in masses

  • or the change in energies.

  • So in this case, we can write that Q--

  • let's just group all of the c squareds together

  • for easier writing.

  • If we take the initial masses minus the final masses,

  • then we get a picture of how much mass was converted

  • to energy, therefore, how much energy

  • is available for the reaction, or Q,

  • to turn it into kinetic energy.

  • So in this case, we can put the kinetic energy

  • of the final products minus the kinetic energies--

  • I'm going to keep with 1--

  • of the initial products.

  • And so we'll use this a little later on.

  • One simplification that we'll make now

  • is we'll assume that if we're firing particles at anything,

  • that anything starts off at rest.

  • So we can start by saying there's no T2.

  • That's just a simplification that we'll make right now.

  • And so then the question is, what

  • quantities of this situation are we likely to know,

  • which ones are we not likely to know,

  • and which ones are left to relate together?

  • So let's just go through one by one.

  • Would we typically know the mass of the initial particle coming

  • in?

  • We probably know what we're shooting at stuff, right?

  • So we'd know M1.

  • What about T1, the initial kinetic energy?

  • Sure.

  • Let's say we have a reactor whose energy we know,

  • or an accelerator, or something that we're

  • controlling the energy, like in problem set one.

  • We'd probably know that.

  • We'd probably know what things we're firing at.

  • And we would probably know what the masses

  • of the final products are, because you guys have been

  • doing nuclear reaction analysis and calculating

  • binding energies and everything for the last couple of weeks.

  • But we might not know the kinetic energies

  • of what's coming out.

  • Let's say we didn't actually even know the masses yet.

  • We'd have to figure out a way to get both the kinetic energies.

  • And what about these angles here?

  • This is the new variable that we're introducing,

  • is the kinetic energy of particles 3 and 4

  • is going to depend on what angles they fire off at.

  • Let me give you a limiting case.

  • Let's say theta was 0.

  • What would that mean, physically?

  • What would be happening to particles 1, 2, 3,

  • and 4 if theta and phi were 0, if they kept on moving

  • in the exact same path?

  • Yeah?

  • AUDIENCE: Is it a fusion event, or [INAUDIBLE]

  • PROFESSOR: We don't know.

  • Well, let's see.

  • Yeah.

  • If it was a fusion event-- let's say there was one here and one

  • standing still--

  • then the whole center of mass of the system

  • would have to move that way.

  • So one example could be a fusion event.

  • A second example could be absolutely nothing.

  • It's perfectly valid to say if, let's say, particle 1 scatters

  • off particle 2 at an angle of 0 degrees,

  • that's what's known as forward scattering, which

  • is to say that theta equals 0.

  • So this is another quantity that we might not know.

  • We might not know what theta and phi are.

  • And the problem here is we've got, like, three or four

  • unknowns and only one equation to relate them.

  • So what other-- yeah?

  • Question?

  • AUDIENCE: For forward scattering,

  • when you say theta equals 0, do you mean they just sort of move

  • together forward, kind of like an inelastic collision,

  • and they just keep moving in the same direction?

  • PROFESSOR: An inelastic collision would be one.

  • And since we haven't gone through what inelastic means,

  • that would mean some sort of collision where--

  • let's see.

  • How would I explain this?

  • I'd say an inelastic collision would be like

  • if particles 1 and 2 were to fuse, like a capture event,

  • for example, or a capture and then a re-emission, let's say,

  • of a neutron.

  • Yeah.

  • If it was re-emitted in the forward direction,

  • then that could be an inelastic scattering event--

  • AUDIENCE: Oh, OK.

  • PROFESSOR: --but still in the same direction.

  • Or an elastic scatter at an angle of theta

  • equals 0 could be like there wasn't any scattering at all.

  • Because really in the end, can matter--

  • let's say if you have a neutron firing at a nucleus,

  • depends on what angle it bounces off of,

  • in the billiard ball sense.

  • If it bounces off at an angle of 0, that means it missed.

  • We would consider that theta equals 0.

  • But the point here is that we now

  • have more quantities unknown than we

  • have equations to define them.

  • So how else can we start relating

  • some of these quantities?

  • What else can we conserve, since we've

  • already got mass and energy?

  • What's that third quantity I always yell out?

  • AUDIENCE: Momentum.

  • PROFESSOR: Momentum.

  • Right.

  • So let's start writing some of the momentum conservation

  • equations so we can try and nail these things down.

  • So I'm going to write each step one at a time.

  • We'll start by conserving momentum.

  • That's what we'll do right here.

  • And we can write the x and the y equations separately.

  • So what's the momentum of particle 1?

  • How do we express that?

  • AUDIENCE: Mass times velocity.

  • PROFESSOR: Yep.

  • So it would be like M1 V1.

  • So we'll have a little box right here for momentum.

  • We could say mass times velocity--

  • or, how do we express that in terms of the variables

  • that we have here, like we did last week?

  • What about in terms of kinetic energies?

  • Well, another way of writing mass times velocity

  • would be root 2MT.

  • Because in this case, we would have root 2 times

  • M1 times 1/2 M1 V1 squared.

  • The 2s here cancel.

  • Let's see.

  • You have M1 squared.

  • You have a V squared.

  • And the square root of M squared V squared is just MV.

  • So this is an equivalent way of writing the momentum

  • in the variables that we're working in already.

  • And so since that doesn't introduce another variable

  • like velocity-- which we do know,

  • but it's kind of confusing to add more symbols-- let's

  • keep as few as possible.

  • So what's the x momentum of particle 1?

  • Just what I've got up there.

  • 2 M1 T1.

  • What's the x momentum in this frame of particle 2?

  • 0.

  • We're assuming that it's at rest.

  • And now, what's the x momentum of particle 3?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: I heard a couple of things.

  • Can you say them louder?

  • AUDIENCE: Square root of 2 M3 T3?

  • PROFESSOR: Yep.

  • Root 2 M3 T3.

  • But in this case, if we're defining,

  • let's say, our x-axis here, it also

  • matters what this angle is.

  • So you've got to multiply by cosine theta in this case.

  • And that's the x momentum of particle 3.

  • And we've also got to account for particle 4.

  • So we'll say add 2 cosine phi.

  • Now let's do the same thing for the y momentum.

  • What's the y momentum of particles 1 and 2?

  • 0.

  • They're not moving in the y direction to start.

  • And how about particle 3?

  • I hear whispers, but nothing vocalized.

  • AUDIENCE: Sine?

  • PROFESSOR: Yep.

  • Same thing, but root 2 M3 T3 sine theta.

  • And M4-- I almost wrote the wrong sign

  • there-- has got to be a minus.

  • If the momentum of the initial particle system in the y

  • direction is 0, so must the final momentum

  • in the y direction.

  • So these two momenta have to be equal and opposite.

  • And that's times sine phi.

  • So now we actually have sets of equations that relate

  • all of our unknown quantities.

  • We have the mass conservation equation,

  • we have the Q equation, we have the x momentum,

  • and we have the y momentum.

  • And from this point on, it's a matter of algebra

  • to express some of these quantities

  • in terms of some of the others.

  • So let's get started with that.

  • Because angles are kind of messy,

  • and theta should uniquely define phi,

  • let's try and get things in terms of just one angle.

  • So I'm going to start by separating

  • the thetas and the phis on either side of the equals sign,

  • so that hopefully later on we can eliminate one

  • in a system of equations.

  • So all I'm going to do is I'm going to subtract or add

  • the theta terms to the other side of the equation.

  • So let's say we'll separate angles.

  • So we'll have root 2 M1 T1 and minus root 2 M3 T3 cosine

  • theta.

  • I'll be depending on you guys to check for sign errors

  • here because those will be messy.

  • I do have notes in case, but I'm hoping

  • I won't have to look at them.

  • And all we have left on this side

  • is root 2 M4 T4 cosine phi.

  • So that's the x momentum equation.

  • Let's do the same thing with the y momentum equation.

  • So all we'll do is take the theta term

  • and stick it to the left of the equals sign.

  • So that would give us minus root 2 M3 T3 sine theta equals

  • minus root 2 M4 T4 sine phi.

  • Right away, we can see that the minus signs can cancel out,

  • just for simplicity.

  • And what else is common to these that we can get rid of?

  • Yep?

  • AUDIENCE: Square root of 2.

  • PROFESSOR: Everything here has a square root of 2.

  • So we'll just get rid of all of the square root of 2s

  • to simplify as much as possible.

  • And now we look a little stuck.

  • But now is the time to remember those trigonometric identities

  • back from high school that I don't think--

  • has anyone used these since?

  • In 1801 or 1802, anyone used a trig identity?

  • A little bit?

  • OK.

  • I would hope so.

  • But I don't know what other people are teaching nowadays.

  • At least this way I'll make sure you remember the high school

  • stuff.

  • We're going to rely on the fact that we already

  • have got a cosine and a sine.

  • We have a set of simultaneous equations.

  • If we can add them together and destroy the angles somehow,

  • that will make things a lot easier.

  • So for the thetas, we have a cosine, a sine,

  • and an unangled term that looks kind of messy.

  • Here we have a cosine and a sine.

  • Anyone have any idea where we could

  • go next to destroy one of these angles?

  • Anyone remember any handy cosine or sine trig identities?

  • AUDIENCE: If you squared both terms,

  • you could get square root of cosine squared, square root

  • of-- sorry.

  • You get cosine squared and sine squared

  • and then you factor out the square root of M4T4

  • and then cosine squared plus sine squared equals 1.

  • PROFESSOR: Exactly.

  • So we can rely on the fact that if we

  • can square both sides of both equations and add them up,

  • we would have a cosine squared of phi

  • plus a sine squared of phi, which also equals 1.

  • So we can destroy this phi angle and make things a lot simpler.

  • So we'll start by squaring both sides.

  • Let's start with the x momentum equation.

  • So if we have--

  • let's see-- root M1T1.

  • So we're going to take that stuff squared.

  • And that squared is not too hard.

  • Neither are those.

  • So we'll have root M1 T1 squared, which just gives us

  • M1 T1, minus root M1 T1 times root M3 T3 cosine theta.

  • Let's just lump those terms together as root M1 M3 T1 T3

  • cosine theta.

  • Also, anyone, raise your hand or let

  • me know if I'm going too fast.

  • I'm trying to hit every single step.

  • But in case I skip one, please slow me down.

  • That's what class is for.

  • OK.

  • And then we've got another one.

  • Let's just stick a 2 in front of there

  • and plus that term squared.

  • So we'll have M3 T3.

  • Let's see.

  • Yeah.

  • Looks like cosine squared of theta.

  • Yep.

  • Equals-- this one's easier--

  • M4 T4 cosine squared of phi.

  • OK.

  • Now we'll do the same thing for the y momentum equation.

  • Much easier because there's no addition anywhere.

  • And we have M3 sine squared theta--

  • over here-- equals M4 T4 sine squared phi.

  • So this is quite nice.

  • Now if we add these equations together,

  • we get rid of all of the cosine and sine squared terms.

  • So let's add them up.

  • Let's see.

  • We'll add the two equations.

  • Add equations.

  • And let's try and group all the terms together.

  • So we have M1 T1 minus 2 root M1 M3--

  • it's getting hard to write over the lip of the chalkboard

  • here--

  • cosine theta.

  • And we have M3 T3 cosine squared theta plus M3 T3

  • sine squared theta.

  • Equals M4 T4 cosine squared theta plus sine squared theta--

  • or phi.

  • I'm sorry.

  • Cosine squared phi plus sine squared phi.

  • OK.

  • Hopefully that's as low as I'll have to write.

  • And like we saw before, cosine squared

  • plus sine squared equals 1.

  • So that goes away.

  • That goes away.

  • And let's keep going over on this side of the board.

  • I told you this would be a fill-the-board day.

  • Let's see if we actually get all six

  • instead of just the four visible.

  • But I think we'll finish this derivation in four boards.

  • So let's write what we've got left.

  • Let's see.

  • Remaining.

  • So we have M1 T1 minus 2 root M1 M3--

  • so much easier to write standing up--

  • cosine theta equals M4 T4.

  • Quite a bit simpler.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Did I miss a term?

  • AUDIENCE: The M3 T3.

  • PROFESSOR: Ah.

  • Thank you.

  • You're right.

  • You're right.

  • And we had a plus M3 T3.

  • Yeah, that would be important.

  • Thank you.

  • Equals M4 T4.

  • So we now have a relation between the masses,

  • the energies, and one angle, which is getting a lot better.

  • We still have one more variable than we can deal with.

  • So let's say if we're--

  • let's see.

  • Which of these variables do you think

  • we can eliminate using any of the equations you see,

  • let's go with, on that top board over there?

  • Well, what other quantities are we

  • likely to know about this nuclear reaction?

  • Let's bring this back down.

  • Are we likely to know the Q value?

  • AUDIENCE: Yeah.

  • PROFESSOR: Probably.

  • Because like you guys have been doing

  • on problem sets one and two, if you know, let's say,

  • the binding energies, or the masses, or the excess masses,

  • or the kinetic energies of all your products,

  • any combination of those can get you

  • the Q value of that reaction.

  • And if you just look up those reactions like, let's say,

  • radioactive decay reactions, on the table of nuclides,

  • it just gives you the Q value.

  • So chances are we can express some of these kinetic energies

  • in terms of Q.

  • And all we've got left is T1, T3, and T4.

  • So which of these are we most likely to be

  • able to know or measure?

  • T1, we probably fixed it by cranking up our particle

  • accelerator to a certain energy.

  • T3 or T4, what do you guys think?

  • Let's say we had a very small nucleus

  • firing at a very big one.

  • Which one do you think would be more

  • likely to escape this system and get

  • detected by us standing a couple feet away with a detector?

  • Yep?

  • AUDIENCE: T3.

  • PROFESSOR: Probably T3, the smaller particle.

  • We've just arbitrarily chosen that.

  • But for intuitive sake, let's say, yeah.

  • Why don't we try and get T4 in terms of Q T1 and T3?

  • That's not too hard, since it's addition.

  • So our next step will be substitute.

  • And we'll say that Q equals--

  • I'm just going to copy it up from there--

  • T3 plus T4 minus T1.

  • So we can isolate T4 and say T4 equals Q plus T1 minus T3.

  • And continue substituting.

  • I usually don't like to have my back to the class this much.

  • But when you're writing this much, it can be a little hard.

  • So let's stick this T4 in right here and rewrite the equation

  • as we've got it.

  • M1 T1 minus 2 root M1 M3 T1 T3 cosine theta plus M3 T3 equals

  • M4 times Q plus T1 minus T3.

  • I anticipate us needing to see this side of the board soon.

  • I also apologize for the amount of time

  • it takes to write these things.

  • There's another strategy one can use

  • at the board which is defining intermediate symbols.

  • And here's why I'm not doing that.

  • When I was a freshman, back in--

  • whoa-- 2001.

  • Who here was born after 2001?

  • Nobody.

  • OK.

  • Thank god.

  • I don't feel so old.

  • I was in 18023, which was math with applications,

  • which was better known as math with extra theory.

  • And in one class, not only did we fill nine boards,

  • but we ran out of English letters--

  • symbols-- and we ran out of Greek letter symbols,

  • and we moved on to Hebrew.

  • Because they were distinct enough from English and Greek.

  • And being, I think, the only Hebrew speaker in the class,

  • I was the only one that could follow the symbols,

  • but I couldn't follow the math anymore.

  • So I am not going to define intermediate symbols for this

  • and just keep it understandable, even

  • if it takes longer to write.

  • OK.

  • So let's start off by dividing by M4.

  • Our goal now is to try to isolate Q.

  • Because this is something that we would know or measure.

  • And it will relate all of the other quantities, only one

  • of which we won't really know yet.

  • So let's divide everything by M4.

  • So we have T1 times M1 over M4 minus 2 over M4 times root

  • of all that stuff plus T3 times M3 over M4

  • equals Q plus T1 minus T3.

  • And we've almost isolated Q. I'll call this step just

  • add and subtract.

  • And I'm going to group the terms together.

  • So let's, for example, group all the T1s together

  • and group all the T3s together.

  • So if I subtract T1, I get T1 times M1 over M4 minus 1,

  • minus 2 over M4 root M1 M3 T1 T3 cosine theta, plus--

  • and if I add T3, then I would get M3 over M4 plus 1 equals Q.

  • So this is a good place to stop, turn around, and see you guys,

  • and now ask you, which of the remaining quantities

  • do we probably not know?

  • So let's just go through them one by one,

  • just to remind ourselves.

  • Are we likely to know what T1 is?

  • Probably.

  • How about the masses M1 and M4?

  • If we know what particles are reacting,

  • we can just look those up, or measure them, or whatever.

  • We know M4.

  • We know our masses.

  • We know T1.

  • What about T3?

  • We don't necessarily know yet.

  • So T3 is a question mark.

  • How about cosine theta or theta?

  • We haven't said yet.

  • And T3 we don't know.

  • And the masses we know.

  • And the Q we know.

  • So finally, to solve for--

  • well, we only have two variables left, T3 and theta.

  • So this here-- this is actually called the Q equation in its

  • most complete form--

  • describes the relationship between the kinetic energy

  • of the outgoing particle and the angle at which it comes off.

  • How do we solve this?

  • How do we get one in terms of the other?

  • Anyone recognize what kind of equation we have here?

  • It's a little obscure.

  • Well, it's not obscure.

  • But it's a little bit hiding.

  • But it should be a very familiar one.

  • Think back to high school again.

  • Yes.

  • AUDIENCE: Is it the cosine angle for the triangle [INAUDIBLE]

  • PROFESSOR: Let's see.

  • Certainly, there's probably some trig involved in here,

  • in terms of, yeah, if you know the cosine,

  • then you know, let's say, the x or the y

  • component of the momentum.

  • But there's something simpler, something that

  • doesn't require trigonometry.

  • Yep.

  • AUDIENCE: Is it quadratic?

  • PROFESSOR: It is.

  • It's a quadratic-- so who saw that?

  • It's actually a quadratic equation, where the variable

  • is the square root of T3.

  • That's the trick here, is you have something without T3,

  • you have something with square root T3,

  • and you have something with T3, better

  • known as root T3 squared.

  • And there.

  • So this is actually a quadratic equation.

  • Despite the fact that it may not have

  • looked that way in the first place, there we go.

  • So now, someone who remembers from high school,

  • tell me, what are the roots of a quadratic equation?

  • Let's say if we have the form y equals ax squared plus bx

  • plus c, what does x equal?

  • Just call it out.

  • AUDIENCE: Negative b--

  • PROFESSOR: Yeah.

  • AUDIENCE: [INAUDIBLE] square root--

  • PROFESSOR: Yep.

  • AUDIENCE: --b squared minus 4ac--

  • PROFESSOR: Over--

  • AUDIENCE: 2a.

  • PROFESSOR: 2a.

  • And in this case, a is that stuff.

  • b is that stuff without the T3.

  • And c is that stuff.

  • Because we have, like, 15 minutes

  • before I want to open it up to questions

  • and I don't think we have to repeat the quadratic formula

  • stuff, I will skip ahead.

  • Skip ahead.

  • This is when I'd normally say it's an exercise to the reader.

  • But no.

  • It's not the phrase I like to use.

  • It's boring.

  • And I can just tell you guys what it ends up as.

  • It ends up with root T3 equals-- and this is the one time I am

  • going to define new symbols because it's just easier

  • to parse--

  • ends up being, we'll call it s, plus or minus root

  • s squared plus t, where s--

  • let's see if I can remember this without looking it up.

  • No.

  • I have to look at my notes.

  • I don't want to get it wrong and have you all write it down

  • incorrectly because of me.

  • There we go.

  • The remaining stuff in the square root, E1

  • cosine theta over M3 plus M4.

  • And t equals M4 Q--

  • is it a minus?

  • It is a plus.

  • Over M3 plus M4.

  • So these are the roots of this equation.

  • This is how you can actually relate

  • the kinetic energy of the outgoing particle

  • directly to the angle.

  • So I want to let that sink in just for a minute,

  • stop here, and check to see if there's

  • any questions on the derivation before we

  • start to use it to do something a little more concrete.

  • Yep.

  • AUDIENCE: Where did the E come from?

  • PROFESSOR: The E. Oh, I'm sorry.

  • That's a T. Thank you.

  • Kinetic energy.

  • Again, we should be consistent with symbols.

  • And I think--

  • I don't see any other hanging Es.

  • Good.

  • Thank you.

  • So any other questions on the derivation as we've done it?

  • We managed to do it in less than four boards.

  • There we go.

  • OK.

  • Since I don't see any questions, let's

  • get into a couple of the implications of this.

  • So let's now look at what defines an exothermic reaction

  • where we say if Q is greater than 0--

  • which is to say that some of the mass becomes kinetic energy--

  • if an exothermic reaction is energetically possible, then

  • what is the minimum T1?

  • Ah.

  • That's why I brought it.

  • What's the minimum T1 up here to make that exothermic reaction

  • happen?

  • We'll put a condition on T1.

  • So if the reaction's exothermic, which

  • means it will happen spontaneously,

  • how much extra kinetic energy do you have to give to the system

  • to make the reaction happen?

  • Let's think of it in the chemical sense.

  • If you have an exothermic chemical reaction,

  • is it spontaneous or is it not?

  • It is spontaneous.

  • Same thing in the nuclear world.

  • If you have an exothermic nuclear reaction,

  • do you need any kinetic energy to start with

  • to make it happen?

  • No.

  • OK.

  • There we go.

  • So that's kind of the analogy.

  • So T1 has to be greater than or equal to 0.

  • It's pretty much not a condition, right?

  • It happens all the time.

  • So if we were to say T1 were to equal 0--

  • let me get my crossing out color again.

  • If T1 were to equal 0, then s could equals 0.

  • And T1 is 0 here.

  • And then you just get--

  • that's an s-- t equals M4 Q over M3 plus M4.

  • And this just kind of gives you a relation

  • between the relative kinetic energies of the two particles.

  • Another way of writing this relation

  • would just be that E3 plus E4 has to be

  • greater than or equal to E1.

  • AUDIENCE: T?

  • PROFESSOR: All this-- hmm?

  • AUDIENCE: T?

  • PROFESSOR: Ah.

  • Thank you.

  • Because Es will be used in a different point of this class.

  • So we'll stick with T for kinetic energy.

  • Thank you.

  • So all that this condition says is

  • that if mass has been converted to energy,

  • then that kinetic energy at the end

  • has to be greater than at the beginning.

  • And that's all it is.

  • So it makes this equation quite a lot easier

  • to solve for an exothermic reaction.

  • You can also start to look to say, well, what happens

  • as we vary this angle theta?

  • What does the kinetic energy do?

  • Let's take the case of an endothermic reaction.

  • Now we are running out of space.

  • For an endothermic reaction where Q is less than 0,

  • you would have to have T1 to be greater than 0.

  • Otherwise the reaction can't occur.

  • So you have to impart additional energy into the system

  • to get it going.

  • And it also means that not every angle of emission is possible.

  • You might wonder, why do we care about the angle,

  • because the reaction still happens anyway?

  • Well, it doesn't happen at every angle.

  • And reactions have different probabilities

  • of occurring depending on the angle at which the things come

  • out.

  • So you could see here that as you vary T1

  • and as you vary cosine theta, you still

  • have to make sure that this quantity on the inside here--

  • so, s squared plus t--

  • always has to be greater than or equal to zero or else

  • the roots of this are imaginary and you don't have a solution.

  • So it's kind of nice that this came out quadratic.

  • Because it lets you take some of the knowledge you already know

  • and now apply it to say, when or when are nuclear reactions not

  • or are they allowed?

  • Wait.

  • Let me rephrase that.

  • When are nuclear reactions allowed or not allowed?

  • You can now tell, depending on the angle of emission

  • and the incoming energy and the masses, which are all things

  • that you would tend to know.

  • So is everyone clear on the implications here?

  • If not, let me know.

  • Because that's what this class is for.

  • AUDIENCE: Yeah.

  • Can you just go over it one more time?

  • PROFESSOR: Yes.

  • So, for exothermic reactions where Q is greater than 0,

  • all that says from our initial part of the Q equation,

  • if Q is greater than 0, then we have this thing right here,

  • where the final kinetic energies have to be larger

  • than the initial one.

  • Which is to say that some mass has turned

  • into extra kinetic energy.

  • And the solution to these is pretty easy

  • because you don't need any kinetic energy to make

  • an exothermic reaction happen.

  • So you can just set T1 equal to 0, which makes s equal to 0,

  • because they're all multiplied here.

  • And then it simplifies lowercase t

  • as just a ratio of those masses times the Q equation,

  • which will tell you pretty much how much kinetic energy is

  • going to be sent off to particle 3 right here.

  • Up there.

  • Particle 3.

  • Because then we have this condition, if root T3 equals s

  • plus square root of s squared plus t,

  • and we've decided that s equals 0,

  • that just means that T3 equals lowercase t, which equals that.

  • So then you've uniquely defined the kinetic energy

  • for an exothermic reaction, as long

  • as you have no incoming kinetic energy.

  • For the case of an endothermic reaction, first of all,

  • we know that the incoming kinetic energy

  • has to be greater than 0.

  • It's like the excess energy that you

  • need to get a chemical reaction going.

  • Has anyone here ever played with--

  • what's the one, a striking one here?

  • Well, has anyone ever lit anything on--

  • no, that's-- yeah.

  • Of course you have.

  • And that's not a good explanation.

  • Hmm.

  • What's a good, striking endothermic chemical reaction?

  • Can anyone think of one?

  • Yeah?

  • AUDIENCE: When you put tin foil in Liquid-Plumr and it

  • releases--

  • PROFESSOR: And it's a hydrogen generator?

  • AUDIENCE: Let's see.

  • I guess that's an explosion.

  • PROFESSOR: I think that happen-- yeah.

  • That's more like an explosion.

  • That's, like, the intuitive definition of exothermic.

  • Yeah.

  • Actually, there's a fun one you can do, too.

  • This is great that it's on video.

  • You do that plus put manganese dioxide in hydrogen peroxide

  • and you have an oxygen generator.

  • And then you have the purest, beyond glacially pure,

  • spring water.

  • You just mix H and O directly.

  • Just don't get near it.

  • Because it tends to be pretty loud.

  • We do this for our RTC or reactor technology

  • course, where I've got to teach a bunch of CEOs enough

  • basic high school chemistry so they can understand reactor

  • water chemistry.

  • And the way I make sure that they're paying attention

  • is with a tremendous explosion.

  • So folks come here, pay about $25 grand apiece for me

  • to fire water-powered bottle rockets at them.

  • It's a pretty sweet job.

  • So if you guys are interested in academia,

  • you know, these things happen in life.

  • It's pretty cool.

  • Yeah.

  • All right.

  • Since I can't think of any endothermic chemical reactions

  • off the top of my head, I'll have

  • to keep it general and abstract and say,

  • if you have an endothermic reaction,

  • you have to add energy in the form of heat

  • to get the reaction going.

  • In an endothermic nuclear reaction,

  • heating up the material does not impart

  • very much kinetic energy.

  • You might raise it from a fraction of an electron

  • volt to maybe a couple of electron volts

  • if things are so hot that they're

  • glowing in the ultraviolet.

  • That doesn't cut it for nuclear.

  • So you have to impart kinetic energy to the incoming particle

  • such that the kinetic energy plus the rest masses

  • is enough to create the rest masses of the final particles.

  • And that's the general explanation I'd give.

  • I forget who had asked the question.

  • But does that help explain it a bit?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Cool.

  • OK.

  • I'll take five minutes.

  • And let's do a severely reduced case

  • of this, the case of elastic neutron scattering.

  • It's kind of a flash forward to what we'll be

  • doing in the next month or so.

  • Does everyone have what's behind this board here?

  • I know that was, like, three boards ago.

  • So I hope so.

  • So let's take the case of elastic neutron scattering.

  • Remember I told you that after we developed this highly

  • general solution to the Q equation,

  • everything else that we're going to study

  • is just a reduction of that.

  • And this is about as reduced as it gets.

  • So in elastic neutron scattering,

  • we can say that M1--

  • well, what's the mass of a neutron in AMU?

  • And let's forgive our six decimal points' precision

  • for now.

  • What's it about?

  • AUDIENCE: 1.

  • PROFESSOR: 1.

  • So we can say that M1 equals 1.

  • And in the case of elastic scattering,

  • the particles bounce into each other

  • and leave with their original identities.

  • So that also equals M3.

  • If we're shooting neutrons at an arbitrary nucleus, what's M2?

  • Yep?

  • AUDIENCE: A?

  • PROFESSOR: Just A, the mass number.

  • Same as M4.

  • Now, we don't have M2 in this equation.

  • Whatever.

  • But the point is, yeah.

  • We're going to use these two.

  • We're going to use these two.

  • So let's substitute that in.

  • Oh, and one last other thing I mentioned.

  • What is the Q value for elastic scattering?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Right.

  • 0.

  • Because the Q value is the difference in the rest

  • masses of the ingoing and outgoing particles.

  • If the ingoing and outgoing particles are the same,

  • M1 equals M3, M2 equals M4, that sum equals 0.

  • Therefore, Q equals 0.

  • So let's use these three things right here

  • and rewrite the general Q equation in those terms.

  • Which board is it on?

  • Right there.

  • So let's copy that down.

  • So let's say we have T1 times M1 is 1 over M4 is A minus 1

  • minus 2 over M4 is A. This is where it gets nice and easy.

  • M1 and M3 are just 1.

  • So 1 times 1 times T1.

  • We don't know what that is yet.

  • So let's call it the Tn, T of the neutron coming in.

  • How about this?

  • We'll call it T in and T out for ease of understanding.

  • Cosine theta.

  • What do we have left?

  • Plus T out.

  • And let's make T1 into an in right there.

  • Times M3 over M4.

  • M3 was 1.

  • M4 is A. Plus 1 equals Q, equals 0.

  • This is quite a simpler equation to solve.

  • So let's group this all together.

  • There's a couple of tricks that I'm

  • going to apply right now to make sure

  • that everything has A in the denominator

  • to make stuff easier.

  • We can call 1 A over A here.

  • We can call 1 A over A there.

  • That lets us combine our denominators

  • and stick the sine right there.

  • That becomes an A. Same thing here.

  • I'll just connect the dashes and stick the minus sign there,

  • leaving an A right there.

  • Now we can just multiply everything

  • by A, both sides of the equation.

  • So the As go away there.

  • We have a much simpler equation.

  • 0 equals T in of--

  • let's see-- 1 minus A over 1.

  • OK.

  • We'll just call it 1 minus A. Minus 2 root T in T

  • out cosine theta plus T out A plus 1.

  • And, OK, it's 10 minutes of, or it's five minutes

  • of five minutes of.

  • So I'm going to stop this right here

  • at a fairly simple equation.

  • We'll pick it up on Thursday.

  • And I want to open the last five minutes to any questions

  • you guys may have.

  • Since that request came in on the anonymous rant

  • forum, which hopefully you all know now exists.

  • Yep.

  • AUDIENCE: So what exactly is forward scattering?

  • I didn't really get that before.

  • PROFESSOR: So let's look at elastic scattering

  • as an example.

  • So in elastic scattering, two particles

  • bounce off each other like billiard balls.

  • In forward elastic scattering, the neutron,

  • after interacting somehow with particle 2,

  • keeps moving forward unscathed.

  • So in the elastic scattering sense,

  • forward scattering is also known as missing.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: You can have other reactions,

  • let's say, where you have a particle at rest,

  • another particle slams into it, and the whole center of mass

  • moves together.

  • I don't know if you'd call that forward

  • scattering as much as, let's say,

  • capture or fusion or something.

  • But in this case, scattering means that two particles go in,

  • two particles leave.

  • Whether it's elastically, which means

  • with no transfer of energy into rest mass,

  • or inelastically, where, let's say,

  • a neutron is absorbed and then re-emitted

  • from a different energy level.

  • And that's something we'll get into in, like, a month.

  • So you can have forward elastic or inelastic scattering.

  • In this case, I'm talking about elastic scattering, which

  • is the simple case of, like, the billiard balls miss each other.

  • Which is technically a case that can be treated by this.

  • Because all you have to do is plug in theta equals 0

  • and you have the case for how much energy

  • do you think the neutron would lose if it misses particle 2.

  • AUDIENCE: [INAUDIBLE].

  • PROFESSOR: Yeah.

  • It wouldn't lose any energy.

  • Right?

  • It would have the same energy.

  • So that's the case for forward scattering.

  • A neutron, when it interacts somehow with another particle,

  • can lose as little as none of its energy.

  • If it misses, no one said it had to lose any energy.

  • And by solving this equation here,

  • which we'll do on Thursday, we'll

  • see what the maximum amount of energy that neutron can lose

  • is, which is the basis for neutrons slowing down

  • or moderation in reactors.

  • Yeah.

  • AUDIENCE: Are T in and T out equal there, in which case

  • that equation is used to solve for theta?

  • PROFESSOR: T in and T out are not always equal.

  • But in the case of forward elastic scattering,

  • they would be.

  • Because the neutron comes in with energy T in

  • and it leaves with energy T in.

  • For any other case in which the neutron comes off

  • of particle 2 at a different angle,

  • it will have bounced off of particle 2, moving particle 2

  • at some other angle phi, and giving it

  • some of its energy elastically.

  • The total amount of that kinetic energy will be conserved.

  • So let's say-- what did we call it?

  • What is it?

  • Yeah.

  • So T1 would have to be the same as T3 and T4

  • together for this Q equation where

  • Q equals 0 to be satisfied.

  • So what you said can happen.

  • But it's only the case for forward scattering.

  • Any other questions?

  • Yep.

  • AUDIENCE: In the case of an exothermic reaction,

  • we assume that T1 equals 0.

  • Can you re-explain why we made that assumption?

  • PROFESSOR: So the question was, in an exothermic reaction,

  • why did we say T1 equals 0?

  • It's not always the case.

  • But it provides the simplest case for us to analyze.

  • So an exothermic reaction can happen when T1 equals 0.

  • It can also happen when T1 is greater than 0.

  • So we're not putting any restrictions on that.

  • But in the case that T1 equals 0, s is destroyed

  • and the harder part of T is destroyed,

  • making the solution to this equation very

  • simple and intuitive.

  • Which is to say that if you just have two particles that

  • are kind of at rest and they just merge and fire off

  • two different pieces in opposite directions,

  • their energies are proportional to the ratio

  • of their single mass to the total mass.

  • So that's like a center of mass problem.

  • You'll notice also I'm not using center of mass coordinates.

  • Center of mass coor-- who here has used those in 801 or 802?

  • And who here enjoyed the experience?

  • Oh.

  • Wow.

  • No hands whatsoever.

  • So center of mass coordinates and laboratory coordinates

  • are different ways of expressing the same thing.

  • Usually you can write simpler equations

  • in center of mass coordinates.

  • But for most people-- and I'm going to go with all of you,

  • since none of you raised your hand-- it's not that intuitive.

  • That's the same way for me.

  • So that's why I've made a decision

  • to show things in laboratory coordinates,

  • so you have a fixed frame of reference

  • and not a moving frame of reference of the center of mass

  • of the two particles.

  • But the center of mass idea does kind of make sense here.

  • If you have two particles that are almost touching

  • and then they touch and they break into pieces and fly off,

  • the total amount of momentum of that center of mass was 0.

  • And it has to remain 0.

  • And so each of these particles will take a differing ratio

  • of their masses away.

  • We already looked at this for the case of alpha decay,

  • where if you have one nucleus just sitting here-- let's

  • say there was no T1.

  • There was just some unstable T2 that was about to explode

  • and then it did.

  • Remember how we talked about how the Q

  • value of an alpha reaction is not the same energy that you

  • see the alpha decay at?

  • Same thing right here.

  • So this Q equation describes that same situation.

  • Notice there's no hint of M1.

  • There was really no M1 in the end.

  • We don't care what the initial mass of the particle that

  • made alpha decay is.

  • All we care about is what are the mass ratios and energy

  • ratios of the alpha particle and its recoil nucleus.

  • So it all does tie together.

  • That's the neat thing, is this universal Q

  • equation can be used to describe almost everything we're

  • going to talk about.

  • So this is as complex as it gets.

  • And from now on, we'll be looking at simpler reductions

  • and specific cases of each one.

  • So it's five of.

  • I want to actually make sure to get you to your next class

  • on time.

  • And I'll see you guys on Thursday.

The following content is provided under a Creative

字幕と単語

ワンタップで英和辞典検索 単語をクリックすると、意味が表示されます

B1 中級

6.Q方程式 - 最も一般的な核反応 (6. The Q-Equation — The Most General Nuclear Reaction)

  • 2 0
    林宜悉 に公開 2021 年 01 月 14 日
動画の中の単語