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[ Silence ]
>> Good morning.
Let's get started on time, we have a busy day today.
OK. First of all, those who turned in your sheets
from the seminar, thanks.
If anybody else has them you can turn them in to me
as you're leaving or the TAs, whoever you happened to see.
So the rules on when you have to turn those in is you can turn it
in to me or the TAs anytime you can find us,
office hours, discussion is fine.
But, it turns into a pumpkin one week
after the seminar you attended.
So you can't save them up for the rest of the quarter.
You have to turn it in promptly.
Usually, I'll all be at the seminar.
Yesterday, I had to skip it because I was working
on a Nanosep [phonetic] proposal that was submitted today.
But, you know, usually I'll be there
and your TAs will probably be there.
OK. So, I think last time I said that the material
from today's lecture was going to be
on the exam, I changed my mind.
When I started making up the exam,
we've covered a lot of stuff.
So-- I'm glad you're happy about it, too.
So the stuff from today's lecture,
I think it will actually help you understand some
of the material that's going to be on the exam.
But it, itself, is not going to be there.
Does anybody have any questions about the exam or extra credit
or that kind of stuff?
All right, good.
Let's continue talking
about diatomic molecules and selection rules.
OK. So, last time I briefly showed an anharmonic potential
and we're kind of running out of time, so I didn't talk
about it very much and I got a few questions about it.
All right.
So, there's a problem on how to deal with anharmonic potentials
and looking at the correction for the fact
that your diatomic molecule is not a perfect
harmonic oscillator.
That's in the homework, it's good to do, it's good practice.
What I want you to know about it right now is really conceptual.
OK. So, if we have the harmonic oscillator potential,
that's just a parabola.
And so that means as you go up to higher and higher energy,
you know, there's this potential it extends forever.
And so what that would mean in a physical sense,
is that your diatomic molecule is vibrating.
And of course the vibrational states are quantized,
but when you put in more energy and kick it up to next state,
it vibrates faster, and then on and on.
And if it were a perfect harmonic oscillator,
no matter how much energy you put into it, it would just do
that forever, it would vibrate faster and faster
but nothing would ever happen to it.
Of course, in reality we know that's not true.
If you put in enough energy, there's some threshold
which is the dissociation energy
above which the molecule just falls apart,
it has so much vibrational energy
that the nuclei just fly apart and go off into space.
And so that's what the anharmonic potential is
telling you.
So, first of all, you know,
we have this funny shape to the potential.
And that energy where you go
over the little threshold is the dissociation energy.
And another thing that you notice is
that the levels get closer and closer together as we approach
that dissociation energy.
So, that's all I want you know about it right,
it's just at a conceptual level what's going on with it.
And again, there is a homework problem about it where you have
to look at it more quantitatively.
It's good practice, but right now
if you understand this, you're in good shape.
OK, so as promised, now what we're going to do is we're going
to go through the gross and specific selection rules
for vibrational spectroscopy, IR spectroscopy
and for Rotational Raman.
And this is going
to tie together a bunch of things hopefully.
So it's going to tie together the threads between looking
at spectra of diatomic molecules,
looking there vibrations and rotations, and also talking
about symmetry, and in particular,
the things that we've talked about with the even and odd rule
and integrals going to zero because of symmetry.
And the plan is for this to give you a little bit better idea
of how you use this.
OK. So we know how to state the gross selection rule
for IR spectroscopy in words.
The molecule has to change its dipole moment
when it's vibrating in order for that vibration to be IR active
or for us to observe this spectrum.
So let's translate that into math.
So what that means is that it's--
we have this value of the transition dipole that's--
and that's need to be not zero in order for there
to be an IR active mode.
All right.
So, remember your direct notation, here we have
in the Ket, nu I, that's the initial state.
And then the Bra is nu F that's the final state.
And, you know, we're looking
at a transition probability between them.
But here, we're saying that those states are connected
by the dipole moment operator which mu.
And of course we have to remember our terminology
in context, mu is sometimes a reduced mass, here it's not,
it's a dipole moment operator.
OK. So, if we think about what our dipole moment looks
like in a little bit more formal way,
it comes from having two partial charges and I'm calling one plus
or minus delta Q on opposite sides of the molecule,
we're just going to assume it's diatomic and polar for the sake
of making the problem easy.
And, it's separated by some distance R.
So R is your equilibrium radius between the two molecules.
And at some point in time it's, it also has some
of displacement X that's through to the vibration.
So Re is the equilibrium distance between the nuclei
and then X is your displacement as it's moving around.
OK. So, the dipole that we have in any point is the radius
or the distance times the charge.
And so that's Re times delta Q plus X times delta Q,
so we have a term that's going to stay with us that it has
to do with the equilibrium distance.
And then we also have this little displacement.
And the term that's always going to be there, you know,
at equilibrium, is called mu not.
So that's the time independent sort of standard dipole moment
of the molecule that you've seen before,
and that's going to be a constant.
So it's just a value of that operator evaluated
at that distance.
OK. So now let's look at what happens
when these states are not equal.
So we're making a transition--
nu initial is not equal to nu final.
[ Pause ]
And here we can plug in what
that dipole moment operator represents.
Remember we're just talking
about a linear diatomic molecule,
all diatomic molecules are linear.
So we only have a one dimensional coordinate,
all it can do is vibrate along this axis that we're calling X.
If we had a more complicated system, we would have
to break this out and do it in all three directions
and we're going to see that in the case
of rotational spectroscopy where we look at the polarizability.
But for this case, we can simplify it a lot
because we're constrained to be in a one dimensional system just
because of the setup of the problem,
it's just a vibration in one direction.
OK. So my dipole moment operator just gets replaced by X,
that's the only component of the dipole that we're looking at.
And so I can evaluate that.
[ Pause ]
And so we get that Q is just D mu Dx.
And if we just look at that, we can see that that's going
to be zero unless the dipole moment is varying
with displacement.
Yes?
>> You said Q equals that.
If you point that in, where does your delta go?
Is it delta Q equals that?
>> Yes, sorry.
Hopefully that's the only one, there's a lot of math today.
OK. So, we can work this through.
And, you know, it takes a little bit
of getting used to the notation.
But hopefully you can see relatively straightforwardly
where this argument comes from.
So we're translating this statement
that we can already see how it works into the math.
OK. So that's the gross selection rule that tells us
that the molecule has to have a dipole moment and that--
sorry, I should be more careful with that.
The dipole moment should change as a result of that vibration
and that's what it gives us the IR spectrum.
So now let's talk about the specific selection rule.
So the specific selection rule, remember, is the statement
that we can only change vibrational states
by increments of plus or minus 1.
All right.
So to start to think about how this works,
remember what the harmonic oscillator way functions are.
They're the Hermite polynomials which have these kind
of functional forms, they're in your textbook
in the early chapters, I think maybe it's Chapter 3.
If you need to review what those look like,
it might be good to go check that out.
But here's what they look like sort of plotted out.
And one thing that that we need to remember
when we start talking about the selection rules
and how we know whether we can get a transition between one
or the other is the even-odd rule.
So, if we look at these just, you know,
without even seeing them mathematically,
we can see that the symmetry changes every other state.
So this one is even, the next one is odd,
the next one is even, one after that is odd, et cetera.
And remember, if we integrate and odd function
over a symmetric integral, that integral is going to go to 0.
And we also know that if you multiply two even functions
or two odd functions together, you get an even function
and that doesn't go to 0.
But if you multiply even and odd, you get odd and it does.
OK. So, what does this have to do with our selection rules?
We have to look at what happens when we plug
in our dipole moment operator in between those two states.
So we're going to be integrating over a pair of states connected
by that dipole moment operator.
And we're going to look at whether the resultant is even
or odd and that's going to tell us whether there's a
signal there.
OK. So before we do that, I just want to point out,
here's what that looks like for an anharmonic potential.
The picture I found only the even states are drawn in,
I think just for clarity but you get the idea.
So here's, you know, just what that looks
like in a more realistic potential
but the argument is the same.
OK. So, we're back to having our transition dipole.
And, again we know what that is, we've got an initial state
and a final state and they're connected
by the dipole moment operator.
And so now in order to figure out what we're doing,
we have to actually write down the wavefunctions.
And so we go look them up in our book or look on the Wolfram site
and find out what hermit polynomials are.
And they're written in terms of this term alpha,
which is not the polarizability.
Again, we're - we have to pay attention to context.
So here are Hermite polynomial functions.
And again we replace our transition dipole with just X
because we're still talking
about the harmonic oscillator where--
we just have displacement in one dimension,
that's all there is going on.
[ Pause ]
OK. So, just to remind you how this works, I wrote this
out in the long version, so not direct notation.
And you can start to understand why people
who do quantum mechanics for a living
like to use direct notation, it's really compact,
it's nice to not have write
down all these integrals all the time.
Of course the trade off is you have to make sure
that everybody knows exactly what you're talking
about because there's a lot of ambiguity that could be hidden
in the notation otherwise.
But-- so here, we're talking about the well defined case
of a harmonic oscillator, we all know the wavefunctions,
we can look them up and so that's OK.
All right.
So, if we work out this integral we can, you know,
we can pull out the constants.
And we need to know an identity relating to Hermite polynomials.
And again, we don't have time to go through this
but you can check it out on the Wolfram Mathematica website or,
you know, resources like that.
But-- so here's a nice identity involving Hermite polynomials
that's going to help us evaluate this integral
or at least simplify it in a form
where we can see what it does.
OK. So, having rewritten the sacker [phonetic] in the form
that this identity gives us, I swear that's going
to do something useful for me in a second.
We need to use another identity in order to get there.
[ Pause ]
This integral equal 0, if nu prime doesn't equals nu.
Again, where did I get this, I looked it up.
And one of the things about, you know,
going through this kinds of arguments.
When you're first learning how to do it, you really wonder,
you know, how does anybody think of this stuff or when, you know,
when do you know when to do what.
It takes a lot of trial and error.
I mean so I'm running through right here in lecture.
You know, we don't have very much time,
it's all nice and neat.
If you actually want to work out stuff like this
for yourself starting from scratch
and if you don't know the answer,
it takes a really long time
and there's a lot of trial and error.
So if it doesn't, you know,
immediately seem trivial that's completely fine,
that's because it's not.
OK. So, this identity equal 0, if nu prime doesn't equal nu.
And it equals this function of nu if they are equal.
And so that means that the first term
in this whole thing equals 0 unless the final state equals
the initial state minus 1.
The second one equals 0 unless you jumped by a plus 1.
[ Pause ]
And so we get the fact that the transition dipole equals 0
unless our change in nu is plus or minus 1.
So it's called the specific selection rule for a reason,
it's very, very specific, you have to plug
in the actual functions for the wavefunctions here.
And look at these integrals in great detail
to see what's going on.
So, anyway now you know where that comes from
and you see how our symmetry argument and being able
to use the even-odd rule and decide which integrals go to 0
by a symmetry can help us out in looking
at fairly complicated things.
So here it's nice because we happen to know this identity
so we can actually get the value.
But even without having the value, you could say, you know,
from symmetry arguments, "This is either zero or it's not."
All right.
There's one last thing that I'm not going to spend a lot of time
on right this minute, we'll come back to it later.
I want to point out that there are some spectra
for which you do see a peak in the middle of the spectrum.
So I just spent all this time proving to you
that you don't see a peak in the middle
of the spectrum ever for a reason.
But it turns out, if your molecule has orbital angular
momentum, so that means not all your electrons are paired.
In other words, it's a radical.
Then you can have a transition delta J equals 0.
And so remember, we have, you know, on this side
of the spectrum we're going from zero to one
in the vibrational state.
And, you know, J to J minus 1 in the rotational state,
the opposite on the other side.
And we just proved that for a normal diatomic molecule,
it's illegal to go from zero to one in the vibrational state
and stay the same in the rotational state.
However, if you have orbital angular momentum,
this is allowed.
And we're going to talk about why later on when we talk more
about electronic structure and electronic spectroscopy.
But I just want to point out that this is there.
One of the things that that a lot of people find hard
or frustrating about chemistry is that there are lots
of rules that are given.
And then later you find out all the exceptions to the rules.
Well, we told you this but it's not entirely true.
You know, I want to point out those things upfront that we're,
you know, we're operating under a particular set of assumptions
and we're using some rules that apply to specific cases,
they're not universal, there are other cases.
We'll talk about that one later.
OK. So, now we have gone over our selection rules
for IR spectroscopy and we're going to talk about Raman.
Question first.
>> Yes, you know, just put everything in perspective.
All the selection [inaudible] things going
to like designing the IR in which, you know, --
what is it, what do you use this for.
>> What do you use this for, that's a good question.
So, designing the IR instrument is another whole area
of research that actually we have some really advanced
researchers at UCI doing this.
Professor Ge designs new IR spectrometers.
Hopefully, we'll get a chance to talk
about some applications at the end.
But, why do you need to know the selection rules?
So imagine that you made a new molecule
and you don't know its properties
and you get its IR spectrum.
We need to be able to interpret in terms
of which levels we'll be looking at.
So everywhere we see a peak that represents a transition
between one level and another.
And if we don't know like, can we only go up by one,
can we go up by two or three or four.
Then we wouldn't know what that spacing represents,
we wouldn't be able to interpret the spectra.
And so that's why we need to know these things in order
to look at the spectra.
And so if you're physical chem is working in this area,
there's a really tight connection
between theory and experiments.
So, you go to the experiments in the lab and then you have
to work out the theory
to explain why your results look like they do.
And sometimes that's done by the same groups.
Sometimes that's done in collaboration.
So, I'm an experimentalist but I have collaborations with people
like Professor Tobias who does theory, you know,
and then some other things we work out ourselves.
But it's always really important to understand the physical basis
for the stuff that you measure.
OK. So that was IR.
And again, I want to emphasize, remember our whole lecture
about the differences between absorption and scattering.
Don't get confused about the difference between IR
and vibrational Raman spectroscopy.
They have a different physical basis.
In IR, we're talking about absorbing a photon and going
up in an energy level.
In Raman, we're talking about a photon being scattered
and either giving up or getting a quantum
of energy from the molecule.
OK, so right now, we're going to talk about rotational Raman.
So we just talked about IR spectroscopy
which is direct vibrational spectroscopy.
Now we're talking about Raman which is scattering
and we're also talking about rotations.
[ Pause ]
OK. So, let's think about a molecule in an electric field.
And again, remember in words our gross selection rule
for a rotational Raman.
When the molecule rotates around the polarizability,
it has to change with time, otherwise we can't see anything.
And so again, mu is our induced dipole, that's induced
by the electric field.
But now we have a more complicated situation,
it's not a one dimensional problem,
we still have a diatomic molecule
but has this three dimensional electron cloud that's shaped
like a football.
And it has components in all three dimensions.
It's a second rank tensor.
So, we have to breakdown our mu into X, Y, and Z components.
And so here's what these are.
And again if you're not quite comfortable with where that came
from going review, you know, conversions between Cartesian
and spherical coordinate systems again.
So-- and here, I'm being a little bit careful to try
to call it mu induced because this molecule may
or not have a dipole on its own.
It's got one induced by the electric field
and that's what we're interested in rotational Raman.
OK. So this thing has an X, Y, and Z component.
The electric field also has an X, Y, and Z component.
So we're applying our electric field in whatever direction
and it's got components in each dimension.
OK. So then what it makes sense to do is to split
up my induced dipole in terms of the polarizability contributions
from the parallel
and perpendicular orientations of the molecule.
So remember we said, you know, to a first approximation,
it doesn't matter whether the molecule is pointing like this
or like this, it just matters whether it's parallel
or perpendicular.
And so we can rewrite that in terms of alpha parallel
and alpha perpendicular.
So remember, alpha here is our polarizability
and it's time-dependent.
It's going from parallel
to perpendicular twice every rotation.
[ Pause ]
And so we can simplify this expression
for the induced dipole.
[ Pause ]
All right, so again, remember the point of why we're working
through this is that we want to see, you know, what happens
to the transition dipole.
So that is-- the transition is allowed
or the molecule shows us something
in the rotational Raman spectrum only
if the transition dipole is non-zero.
So, here's the transition dipole,
we have a quantum numbers for our rotational states.
And we have the X, Y, and Z component
of the induced dipole sandwiched in between.
So again, we've got our final state and our initial state
and it's connected by the operator.
And the operator in this case is this transition dipole
that is an interplay between the polarizability of the molecule
and the applied electric field that's coming
from our spectrometer.
Yes?
>> The J and the MJs, that's the thing--
those are the quantum numbers the angular--
>> That's right.
J and MJ are the quantum numbers for the angular momentum.
And I put both of them in here to,
you know, to make it general.
Yes.
>> Is the bottom [inaudible] suppose to be F, I, Z?
>> Yes. Thank you, sir.
>> Can be any direction that just has to be--
if only like say only X has a value and the rest are 0,
is that still forbidden?
>> That's a really good question.
So, you know, the question is, do you have to have a component
in X, Y, and Z for it to be allowed.
No, if any of them is allowed, then it's allowed.
You know, we'll-- you'll see a spectrum.
OK. So here's our induced dipole again.
And, you know, see we again hid our wavefunctions
with the bra-ket notation which looks nice and clean.
Now let's go back to what the wavefunctions
for our rigid rotor are.
They are spherical harmonics.
And, they're kind of big and ugly
but you can see their symmetry easily enough if you plot them.
So again that this is something to look up if you need to do it,
go check it out in the early chapters of the book
or on the Wolfram website.
OK. So, here's some spherical harmonics.
And we can use a trig identity to simplify our expression
for the induced dipole in this case.
[ Pause ]
And so, if we're looking at the transition
between two rotational states, it has two components.
One has to do with the polarizability
and the other it has to do with the electric field.
[ Pause ]
And so, this part goes to zero unless delta J equals 0.
So, this is the one that contributes the Rayleigh line.
So this is the in no transition and rotational energy.
That photon hits it and it just bounces off elastically
and it doesn't change frequency.
And so if you're doing this experiment to find out something
about the molecule, this is really boring.
So, you shine your laser at it
and the same color laser light comes back off
and that's all you get.
The other one goes to zero unless you have a change
in rotational state of plus or minus 2.
And you have to have an isotropic polarizability,
so that delta alpha has to not equal 0.
So between the parallel and perpendicular,
you have to have a difference in alpha,
otherwise this whole thing is going to go zero.
Yes?
>> You have JF equals JI there and JF equal JI,
would that be JF equals JI plus 2?
>> JF equals JI plus or minus 2.
And delta alpha does not equal zero.
>> Does not?
>> Right. Because if the change in polarizability equals 0,
then that term is going to go to zero right away.
Just making sure everybody is awake.
Thank very much for catching up.
OK. So we can put this in terms
of an integral involving spherical harmonics.
But you get most of the take home message just
by looking at this.
OK. So, just for completeness, here's what you get if you write
out the X and Y components.
This is not-- the details are not the most important things
to know right now.
But I just wanted to write it out for completeness.
OK. So, let's-- now that we got through the selection rules
and we understand where they come from,
let's talk about how this actually gets used
in a practical sense.
It's-- it can be hard to make the connection
between the examples that are simple enough
that we can work them out in class,
and to what you would actually do in a research context.
OK. So, one of the things that IR and Raman spectra are used
for most frequently is identifying functional groups
in organic molecules.
And, as you can see here-- so here's a spectra--
here's a pair of spectra of styrene and butadiene rubber.
So this is a real material that somebody made
and they're interested in knowing what's, you know,
what's going on in its structure.
And I want you to notice a couple of things about it.
So, one is there are these peaks in the IR and Raman.
They're plotted on the same scale.
So here we have this in wave numbers,
it's given as Raman shift, but, you know,
the IR one is on the same scale.
And so a couple of things to notice, some of the IR
and Raman bonds coincide, so you'll see some vibrations
that have a transition in both the IR and the Raman spectrum.
Just like, you know, when we do our symmetry analysis
and we come up with, you know, some of the vibrations
of the molecule are IR active and Raman active,
and some of them or one or the other.
So, you know, we see for instance this set
of transitions is really strong in both.
But if we look at this one, that has a big peak
in the Raman spectrum and only a teeny tiny one in the IR.
So, what that means is, you know,
we have some vibrational mode
that really changes the electric polarizability a lot.
It means there's a large overlap between those two states
when they're connected by that polarizability operator.
But in the vibrational--
in the transition between vibrational states,
it doesn't change the dipole moment very much.
So again, these symmetry arguments alone don't tell us
anything about the magnitude,
it just tells you whether there's a peak there.
But you have to actually plug in what the values are
and calculate the intensities if you want that information.
So you can have something like this where there's a huge value
in the Raman and a tiny little one in the IR.
And symmetry will just tell you that they both exist, that's it.
Other things that are useful to know here,
for really complex molecules, a lot of times if you're like--
so for instance if you're a synthetic chemist,
it's not really worth the effort to fit every single bond
in the spectrum and really understand everything
about all the details.
Physical chemist like to do that and that's, you know,
that's an important part of p-chem research
when you're trying to understand why things are the way they are.
For synthetic chemists though,
this is still incredibly useful even if you don't go through
and fit everything because people have really well
documented tables of where bonds belonging
to certain functional groups show up in the spectra.
And so you've probably used this in lab or--
you know, some of them, some
of you were doing synthetic research.
You can use it to just see what kind
of functional groups you have in your molecule
and it's very powerful.
Here's another one that's even more complicated
where somebody has assigned, you know,
what types of bonds are likely to vibrate in this regions.
And so, you know, again you see things
like the CO stretch shows up in both.
But it's a lot more intense in the IR than the Raman
which kind of makes sense.
If you think about a carbon yield that's stretching out,
that is going to change the dipole moment a lot more
than it changes the polarizability.
So some of them are easy to visualize
and rationalize, some of them less so.
If you actually want to get into all the details
and see how this works, you have to do it computationally.
And people do have pretty good success predicting these things
and understanding all the modes computationally.
A good example of somebody who does this
for research is Filip Furche at UCI,
so that's something that his group does.
There are many others in different places.
OK. So what else can we do with vibrational spectroscopy
in a more advanced context where we're really interested
in the physical chemistry?
This is an example from the website of Charles Harris
and his group at Berkeley.
And, they are illustrating how we can use vibrational
spectroscopy in a time-resolved way.
So, instead of just taking a spectrum at a particular point
in time and measuring the bond vibration, they're interested
in looking at some chemical process that's happening
and using fast laser pulses to capture snapshots of the system
as it's undergoing some kind of a change.
And so here's their schematic picture of what that looks like.
There's a chemical reaction and it's going
between different states here throughout the course
of the reaction.
And, they're able to probe this thing with an IR pulse and look
at it at different time points
and actually watch bonds being broken and formed.
So they have their IR spectrum.
And as they watch it as a function of time,
they can see vibrational modes corresponding to one set
of bonds disappearing and another one growing in.
So you can actually use these kinds of things
in a time-resolved way to learn something about processes
that are happening in real time.
So that was an example from Berkeley, here's an example
of another IR application from UCI.
So this is from Professor Ge's lab.
One of the things that she does is two-dimensional
IR spectroscopy.
So in this case, we have our IR spectrum that is-- so--
basically each of these two dimensions is a correlated
IR spectrum.
And what that gives us is something that looks
like a topological map when we're looking down on it.
And we're going to talk about this in a lot more detail as far
as how would it actually works when we get
in to two-dimensional NMR.
But for now just treat it us it's two IR spectra
that are correlated to each other, and if you took a slice
through vertically or horizontally,
you would see the normal one-dimensional
infrared spectrum.
And what you get by doing this in a correlated fashion
and looking at it this way is you see connections
between different vibrational states.
And so what that is telling us is
that for a large complicated molecule like this peptide
in solution, if you have one bond vibration somewhere
in the molecule, it's not completely independent
from everything else.
If you wiggle this bond over here, that's going
to affect how one on a different part of the molecule vibrates.
That's what these peaks are telling you,
they're correlations between pairs of vibrations.
And so by looking at that, you can learn interesting things
about the molecular structure, you know, how it interacts
with the solvent or its environment
and also how it changes over time.
OK. So here is our grand summary for vibrational
and rotational spectroscopy
where for the most part done talking about it.
This enables us to learn things
about molecular geometry for simple molecules.
We can identify functional groups and figure
out what bonds are being destroyed or formed
from more complicated ones.
And of course we can look at things that are as complex
as all the way up to proteins.
But in that case, we need a lot more sophisticated computational
methods to be able to fit the data
and understand what we're seeing.
And for those of you who are interested in doing research.
These are the groups that, you know,
are doing vibrational spectroscopy research at UCI.
And, I'm sure that I forgot at least one group.
But, here's some examples.
Yeah, I forgot, Professor Nizkorodov, that's one.
And as I mentioned Professor Furche looks
at theoretical modeling of different kinds
of spectroscopy including but not limited
to vibrational spectroscopy.
So, yeah, of course I couldn't do
that without forgetting someone.
All right, so, we are done a little bit early today that's
because I'm going to stick around and answer questions.
Good luck on the exam.
Wait-- the-- there-- one thing I want to announce.
For those who didn't see this John Mark [assumed spelling] is
having a review session this evening.
It's 8:00 to 9:20 PM and it's in Rowland Hall 104,
is that right, John Mark?
>> Yeah.
>> OK. Cool.
TAs, can you guys stick around for a minute after class.
All right, we're done.
Don't forget to check the seeding chart tomorrow night
and make sure that you're all ready to go
and setup for the exam. ------------------------------7b17aea0cdac--