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[ Silence ]
>> So today we're going to finish up our discussion
of examples of vibrations in polyatomic molecules
and how we can figure out which ones are IR
and Roman active by symmetry.
And then we're going to move on to some cases
of more simplified molecules that we're going to talk
about linear molecules where we can calculate using some pretty
easy calculations.
What would be the bond length and the force constant
and things like that are for the bonds involved.
Of course we can do that for polyatomic molecules
but we need a computer.
So in the discussion of which vibrations are IR
and Roman active.
One of the things that came up last time is the fact
that we know anything quantitative
about these vibrations from that analysis.
We know were, you know, whether we can see them in the spectrum.
But that doesn't tell us what energy they show up at
and it doesn't tell us anything about how intense they are.
For that, we need to know some additional information we can't
get it just from group theory.
So we will get into that today for the cases
of diatomic molecules.
Then next time on Wednesday,
we're going to over selection rules
in a little bit more detail.
We've already talked about this in a qualitative way.
We're going to look at how you actually calculate it.
And then that's going to be it for the first midterm.
So let's see if we can get all that done.
Does anybody have any questions before we start?
Yes?
[ Inaudible Question ]
The material for the midterm does not cut off,
everything through, you know,
what we do on Wednesday it's going to be on it.
[ Pause ]
OK, so let's continue talking about bond vibrations.
All right so we've been doing this example of methane
which as I said is going to be a little bit harder
than probably what you're actually going
to have to do on the exam.
So if you get this, you're in good shape.
OK, so last time we talked about the bonding.
Now were going to talk about its vibrations.
And so the question is identify the vibrational modes
and determine whether they are IR or Roman active.
And so of course we're going to use this during--
we're going to do this using group theory.
And so as always when we talk
about molecular motion our basis is now a little coordinate
system on each atom.
And so we need to look at how these things transform
when we move the molecule around.
OK, so methane has four atoms and we have to take them all
into account because we're talking about these vibrations.
So remember our basis that we set up always has something
to do with the symmetry of what we're actually looking at.
And so here what we're interested
in is the displacement of the atoms relative to each other.
So we need our little coordinate system on each atom.
And so, there are 15 little unit vectors in our basis here.
And so if you want to set up the actual matrixes for this,
you need 15 by 15 matrixes.
So let's not do that, let's use the shortcut
to get the character.
So again I hope, you know, we look at the water example,
you get something that's 9 by 9.
It's useful to do that once, you know, for one operation,
just prove to yourself how it works
and then you're done once you've seen it that's good.
OK, so, for the identity operation,
nothing changes of course.
And so we get 15 for the character of that.
So of course identity matrix it just looks like a 15
by 15 matrix but ones on the diagonals
and zeros everywhere else.
Now let's look at what happens
when we look at the C3 rotations.
So this is where it starts
to get a little bit more complicated because when we look
at a C3 rotation of methane,
our basis is not just the individual bonds anymore
like it was when we were looking at the bonding.
You know, there we could use the shortcut
and it was simple enough because we were talking
about just the bond swapping places.
So here we have this little X, Y and Z axis on each atom.
And when you do 120 degree rotation,
X doesn't map on to Y directly.
You end up with things that are in between.
And so let's consider this, you know, I'm not going to set
up the whole giant matrix.
But let's consider this sort of, you know, piece by piece.
So we're thinking about this, you know,
around our as our C3 rotation, we're holding the molecule
by one of the Hs and rotating it around.
So let's look at the carbon and the H that's on top.
So for that, you know, these two are going to behave the same.
And we have to fill in our whole rotation matrix
as we learned before in order to get that.
And so you plug in the actual angle that we're rotating
through and you get that, the trace of that is zero.
[ Pause ]
And you can do this for the other hydrogens as well,
you know, everything changes places and space
and those contribute zero to the trace also.
And this is something that if you're not sure
about the rotation matrixes, that's a good thing to practice.
Also I would recommend that the general form
of how you make rotation matrixes is
if you don't remember.
That might view good thing to put on your cheat sheet
that you have to write down.
It's a useful thing to know how to do.
OK, so again the reason that we can't use the standard shortcut
in this case is that for the particular operation
that we're doing, the elements
in our bases don't map on to each other.
You know in a one to one fashion.
So we have to break it down like this.
And what we find is that the character of C3 is zero.
OK, so now let's look C2.
So again remember for the methane molecule,
the C2 axis are going between the hydrogen carbon bond, so,
you know, before we are holding the molecule by one hydrogen.
Now, we have to hold it with the two of them sticking
out then one is coming out at you and one is pointing at me.
So if we look at our C2 axis there, then we have
to treat the carbon and the hydrogen separately.
So all of our hydrogen switch position when we do
that so those contribute zero to the character.
And on the carbon atom, Z is going to stay the same and X
and Y contribute minus one.
And so if we add all those up, we get minus one for C2.
OK, how about S4?
So remember this one is hard to visualize in methane.
We hold the molecule by the top hydrogen again
so that we have the other three hydrogens sticking
out like this.
We rotate 90 degrees and then reflect
through a plain perpendicular to that rotation axis.
And when you do that, you see that the Z axis
on the carbon changes sign and everything else gives you zero.
So we get minus one for S4.
And again if you have trouble visualizing that, you know,
go check out the molecular models and make sure
that you can prove it to yourself.
OK, so the last thing that we have to deal
with is our dihedral plains.
So on these vertical plains, we have to look
at what each atom is doing.
So for the carbon X and Z stay the same, Y changes sign.
And the same thing happens for the hydrogens that are in plain.
And then the other ones get reflected.
And so we have to add
up all these contributions to get the character.
And when we do that, we end up with 3.
So now we have a reducible representation
for the motions of methane.
So remember we're not just looking at the vibrations yet.
We have these bases tells us something
about the displacement of all the atoms.
And so we have to reduce our reducible representation.
And we use the reduction formula to do that.
And here's what we end up with.
So we get A1 plus E plus T1 plus 3T2.
And we had 15 vectors in our basis so we should end
up with 15 elements here.
And it might look like we don't have enough until you remember
that E is doubly degenerate.
So that counts as two.
And T is triply degenerate.
So every T counts us three symmetry species
in that representation.
OK, so after we reduce our representation, then we have
to take out the symmetry species that correspond to translations
and rotations because those don't show
up in our vibrational spectra.
And so if we look at the character table,
we see that for translations that's going to take care of one
of the set-- the degenerate sets of T2.
So there's and X, Y and Z in there.
So translation of this molecule, you'll need the axis
that corresponds to that.
So that removes one of them from consideration.
OK, so how about rotations, so RX,
RY and RZ are going to be all degenerate.
And those fall into the T1 category.
And again we're just reading this right
up the character table.
So the vibrations are everything that's left over,
so we have A1, E and 2T2.
And so you can look at the character table and decide
which ones are IR and Roman active.
OK, so let's look at what the actual vibrational modes are
for methane.
So they're a little bit hard to visualize
but this is a good picture of what's going on.
So we have this wag where, you know,
some of the bonds are moving with respect
to the other ones, there's a twist.
The symmetric stretch is easy to see that's the one where all
of the bonds are flexing in and out.
And then there's the scissors motion.
So if we try to put this in terms of our answer that we got,
we see that the one that's A1 is inactive.
There's no component of the dipole moment and no component
of the polarizability that's corresponding to that.
And that's fine, it just means we can't see it.
That vibrational mode doesn't give us anything
in our spectroscopic methods.
We find that the ones that are--
that have symmetry E are Roman active.
And the T2 vibrations are IR and Roman active.
And so let's take a look at what this looks like in practice.
So if we look at the IR spectrum of methane, we expect it
to have a couple of a few vibrational modes,
so T2 is degenerate.
So here's our IR spectrum of methane.
And so we can see that we get these peaks, you know,
again the group theory analysis just enables us
to say that they're there.
It doesn't tell us anything about where they are,
what the intensities are.
And if we look at all this fine structure,
those are the rotational transitions.
So when we excite this molecule to an excited vibrational state,
there's enough energy in there to excite all
of the rotational transitions too.
And so we see all of these rotational lines.
And we're going to talk about that in more detail in a minute
for diatomic molecules.
Here's the Roman spectrum of methane.
And so, you know, notice that the intensities here are higher
than the ones on this side.
This are the strokes lines versus the anti strokes line.
So this is the one where the electromagnetic radiation is
giving up a quantum of energy of to the molecule rather
than taking it from the molecule.
So we can rationalize the intensities that way.
But otherwise these things look kind of complicated
and the rotational fine spectrum
of different modes are overlapping.
And to really predict what this is going
to do, you need a computer.
You can do it very accurately especially for molecules
that are relatively small like this.
But if you really want to understand the details
of what's going on, it's best to do it computationally.
For smaller things like diatomic molecules,
we really can do these calculations just
with a pencil and paper.
And so it's more useful to zoom in
and take a closer look at that.
OK, so let's go back to thinking about what this looks
like in a theoretical sense.
So we're talking about motions of the molecule.
For a diatomic, this is a lot easier to visualize
because we just-- all we have is just this molecule vibrating
in and out.
There's only one motion that's going on.
So the spectra are a lot easier to interpret.
And we can do things a lot more quantitatively
without using computational methods.
And again I don't want to imply
that the computational methods aren't very precise
that you can't get good answers of it.
You really can, you can't just--
you can't just do it easily in class.
So we're going to focus on this for purposes
of calculating things that are quantitative.
OK, so things to notice
about the harmonic oscillator formalism.
So these are all things that you learned last quarter.
So we have this system
where there's a harmonic potential wall.
Of course that's an approximation where treating it
as harmonic oscillator
that doesn't mean it's always really like that.
But for small displacements, it works pretty well.
Notice that there is a zero point energy,
so the lowest state is not exactly
at the bottom of the potential.
That's important for the harmonic oscillator treatment.
And again remember it's different
from the rigid rotor approximation.
There we were allowed to have a zero rotational state,
here were not.
Remember that these things are quantized
and we have energy increments of each mu, as we go up in energy.
And of course we're looking at, you know,
this is potential energy so we can write it as KX squared.
And it's only dependent on, you know,
it's a one dimensional system.
It's only dependent on that internuclear separation.
And X equals zero isn't zero,
that doesn't mean the nuclear attaching each other that's its
equilibrium internuclear separation.
And so when we measure bond lengths of molecules,
of course what we're measuring is the equilibrium distance
or the average distance in reality,
these things are always moving round
and vibrating back and forth.
OK, so we can use this to get quantitative information
about things like the bond length and the force constant
which tells us something about how stiff the bonds are.
And we can calculate it quantitatively using some pretty
simple approximations.
And if you don't remember some of this from last quarter,
it's useful to go back
and review what the harmonic oscillator way
of functions look like.
So they're from the polynomials,
you should have seen this last quarter.
It's useful to go look at them particularly
since next lecture we're going to talk about selection rules
and you might have to look at different symmetries
of different vibrational states.
OK, so here's what that corresponds
to when we actually look at the spectra.
So here are the lowest energy two states
of the harmonic oscillator.
So remember, our quantum number here is called mu typically
for the harmonic oscillator way of functions.
So we have mu equal zero and mu equals one.
And then all of these little transitions in between
that are represented by the red
and blue arrows are the rotational transitions.
So remember, vibrational transitions take a lot more
energy than rotational ones.
So when we excite the vibrational transitions,
all the rotational ones come along for the ride
and we see them in our spectra.
And it turns out that's kind to useful because we can use them
to get some valuable information.
And so we're representing these states in direct rotation
as the state that has quantum number mu for the rotational--
sorry for the vibrational state and quantum number J
for the rotational state.
And if we look at the IR spectrum, in this case it's HCl,
we have two sides of the spectrum.
So on this side, we're going from zero J to 1J minus 1.
So we're starting in the lowest energy vibrational state taking
it up to the first excited state.
And while we're doing that,
we're going down in rotational transitions.
On the other side, we're going the other way.
So we're going from zero to one in the vibrational transition.
And we are going up in the vibrational,
oh sorry in the rotational state.
So that needs to get fixed.
So let's look at-- I have that correct on the next one
so let's look at that.
Another thing that I want to point out before we move
on in IR spectra is
that sometimes you'll see them plotted
with the peaks pointing down.
Sometimes you'll see them with the peaks pointing up.
It's just different conventions it doesn't really matter.
And you know, that's something
that you'll see in the literature.
It's also important to pay attention
to whether your spectrum is plotted in frequency units
or in wave numbers because of course the energy--
the direction of increasing energy is going
to go the opposite way depending on-- oh sorry frequency units
or wave length-- the energy will go the opposite way.
OK, so these things also have historical names.
So if we're going down in rotational states that's called
the P-Branch.
And if we're going up in the rotational states that's called
the R-Branch.
These are just historical names it's useful to know the exist
for purposes of reading the literature.
But, you know, for the most part, I'm mostly concerned
that you understand the physical basis
for this and what's going on.
Again here are some examples of spectra.
Here's one in wave numbers, here's one in frequency units.
You will see both and you should definitely know how
to deal with both.
Like being able to fluently convert between frequency
and wave numbers and energy units is definitely something
that you should be able to do.
OK, so let's look a little bit closer
at why these spectra look the way they do.
So we have an R-Branch and a P-Branch
and there's all this rotational fine structure in there.
And it's symmetric because on one side,
we're going down a quantum in rotational transitions
and then we step through all the different transitions
that there are.
Remember a line doesn't represent the state.
It represents the transition between two states.
And if we're looking at, you know,
we're going down a rotational transition and stepping
through that on one side and going up on the other side.
But there's no peak in the middle.
So the fundamental frequency that we're looking
at here that's telling us about the energy of going
from the zero to one vibrational state doesn't have a peek there.
And the reason for that is the selection rules.
So we have to have a difference
in vibrational state of plus or minus one.
So zero is not allowed.
So we don't see a transition here.
So when we going to get that fundamental frequency,
we have to pick the point
in between these two sets of vibrational lines.
OK, what else is there to say about that,
yeah so the Q-Branch is the name
for the central line which is missing.
As we'll see next lecture there are some times
when you do get a peak there.
And well talk about what situations that arises in
and when it's there it's called the Q-Branch.
The other thing I want to mention again is look
at the intensities here.
So the lowest energy transition is not the one
that has the highest intensity.
There are some maximum up here and again that's
because the degeneracy.
When we get to a little bit higher energy states,
there are more ways for the system to be in that state.
And so it's more highly populated.
If we were to raise the temperature
so that more energy is available, we would see
that those curves would flatten out.
We would get the higher energy states more populated
and also just the whole thing would spread
out because there's more degeneracy.
There's more ways to occupy those states.
OK, so we've talked about what the spectra look like and how
to interpret them in a really qualitative way.
Let's get into actually calculating some things
from them.
OK, so here's our spectrum of HCl
and this particular one is plotted in frequency units.
And we have this pair of lines in the center that's
on either side of the fundamental transition
from mu equals zero to mu equals one
without a change in rotational state.
And so that frequency is what's going to tell us
about the energy
of the vibrational transition that we want.
Another thing I didn't bring up before is if you look
at this now that it's blown up like that.
You see that the lines are split.
Does anybody know why that is?
So we're, you know, we're talking
about these bond vibrations,
there's really only one vibration that it can undergo.
So why would it have two different lines?
It's like there's very slightly different energy there
in the bond vibrations, yes?
[ Inaudible Response ]
That's a good guess but let's see.
[ Inaudible Response ]
Yeah it's an isotope effect.
So if we have say deuterium instead of hydrogen,
there's going to be some, you know,
little natural abundance population
of different isotopes in the sample.
And, you know, deuterium is going
to be a lot heavier than hydrogen.
And so we'll see, you know, the difference
in the vibrational frequency.
And a lot of times when we're interested in looking
at vibrational states of molecules,
one way that you can do that is isotopically label specific
things in order to make the frequency a little
bit different.
If we have time to get into some applications next time,
I'll show you some examples of that
where people have been very clever
about using isotope labeling to break the degeneracy.
And look at different vibrations in a complicated molecule.
OK, so for now let's stick to a not very complicated molecule
because we can calculate a bunch of stuff about quite easily.
OK, so I like this picture because it's showing us
on the energy level diagram what we're seeing in the spectrum.
OK, so we're looking at this transition
from mu equals zero to mu equals one.
And then we have the corresponding rotational
transitions showing up here.
So we have to take the average position of these two lines
and get the center frequency.
And that's going to tell us the fundamental frequency
of that vibrational transition which is what we want to know.
And so we know that in general, the spacing between the lines
for rotational spectra, same thing here is 2B,
it's two times the rotational constant.
But so that's for the ones that are spaced apart evenly.
Here there's a center one missing.
So that center pair of lines is spaced apart by 4B.
And so our change in frequency, here it's called delta F is 4B
over H. And that's what is it in hertz.
So let's put this in terms
of what we're actually interested in finding out.
So our kinetic energy is, you know,
what we're describing with the potential.
And that's one half Iomega squared and we can write this
down in terms of the reduced mass.
So here our moment of inertia can be written pretty easily
in terms the reduced mass
because it's a diatomic molecule.
And, you know, again as we've seen for, you know,
a diatomic molecule like this where one
of the atoms is really heavy and the other one is light.
You've hopefully seen this in some of the homework problems.
You end up with the vibration, you know,
looking like the chlorine is just staying still
and the little proton is bouncing in and out.
OK, so we know the spacing between the rotational lines
which is the same for the fine structure
in the vibrational spectrum as it was
for the pure rotational spectra that we looked at.
And it's 4B for the central transition.
And so we can just plug that in and solve for the change
in energy between these states.
[ Pause ]
And so, you know, we're interested in getting the change
of it, the change in energy which is just equal
to H times the frequency difference.
So as with any kind of spectra, can be probably
like this E equals H mu.
That energy difference for the transition is the same
as the energy of the proton that went in and was absorbed
to promote that molecule to the higher state.
The only thing that's a little bit tricky here is that we have
to remember what the states actually mean.
So we're not able to see a direct line for the transition
that we're interested in.
We have to infer it from the structure of the rest
of the rotational lines.
OK, so we have the energy there for--
I didn't work out the number
but we know what it is it's just H times that frequency.
So we have the microwave value that was calculated
for the bond length of HCl.
So that means like somebody went
and measured the direct rotational spectrum
of this molecule and got the bond length
which again we know how to do, we looked at that.
And that's given us 0.127 nanometers which I looked up.
So now we're going to calculate the bond length of HCl
from the vibrational spectrum and see how well at degrees.
OK, so we have our mu R squared equals 2H4 squared
over the energy here.
And if we plug in values, and solve for R
within the significant figures
that we have here given the numbers that we got,
we do get the same answer.
So hopefully this convinces you
that you can use this pretty simple analysis
to get information about the molecules at least in the case
of diatomic molecules.
And again, obviously this is a lot more useful
for larger molecules that we don't already know everything
the motions of.
And there we use the same procedure it's just
that you need computational methods to do it.
OK, so let's continue our discussion.
We need to look at just how to deal
with different units, just as a reminder.
So again here's the same spectrum, this time it's plotted
with the peaks going down and it's in wave numbers.
They're just different conventions
and you'll see both in the literature.
Spectroscopy is really the land of confusing notation
and things being given in different units
and draw in different ways.
That's just because of historical convention,
so one issue is that chemists and physicists are looking
at a lot of the same things.
And they have different conventions
in how they prevent-- how they present things.
If I were the dictator of the world that when you be that way
but we don't got o pick, you know,
how these things are represented.
And so that's part of it, it's juts learning the conventions
that you're going to see in different parts of the field.
And particularly, you know, we're talking, you know,
maybe going to some of the [inaudible] seminars
and seeing people's current research.
That's part of the language and part of being able
to understand what's going on there.
So let's look at this.
So if we want to calculate the bond length
and you have the spectrum and wave numbers,
you want to pick a pair of lines that's relatively close
to the center.
So why do we want to pick a pair
of lines that's close to the center?
We know that the rotational constant is the same
and these lines are spaced by 2B
in the rotational fine structure.
But remember we have centrifugal distortion.
So when we get the molecule excited to higher
and higher rotational states, then it doesn't behave
as a perfect rigid rotor anymore stretching more and more.
And that's the same thing here.
So if we pick lines corresponding
to really high energy rotational states, we're not going
to get the best value for our rotational constant there.
And there will be unnecessary error
in the calculation of the bond length.
So we want to pick something from a part of the spectrum
where our assumptions are more likely to be correct.
So pick a coupe of them that are close to the center.
And so that gives us B in wave numbers.
So here it has it has a tilde over it.
So it's in inverse centimeters rather than frequency units
with the appropriate conversion.
And then we can plug all our stuff in
and we get something for I.
And again it's good to keep track of your units.
So for our moment of inertia,
we get kilogram meter squared which we should.
[ Pause ]
And then we can solve for R given
that we know the expression from the moment of inertia
of a diatomic molecule.
[ Pause ]
And then the last step is check
that your answer is in the right units.
And that it's a reasonable order of magnitude
so we did get length units which is good
because we're talking about a bond length.
And, you know, we got something that's on the order of angstroms
which is a reasonable value.
So as far as calculating these things, paying attention
to units and whether you got a reasonable order
of magnitude is a large part of the battle.
That will really get you a long way in terms of figuring
out whether your answer makes sense.
OK, so this is one type of information that we can get
from vibrational spectroscopy.
So we can learn about the lengths of bonds.
We can also learn about the force constant of bonds.
So the first constant is telling us something about how floppy
or how stiff that bond is.
So we're going with the harmonic oscillator approximation we're
just assuming that we have little springs
in between our atoms and their bouncing back and forth.
And of course you can have a really soft spring
or you could have a very stiff spring.
And force constant
of the molecules what tells us about that.
And so that has to do with the vibrational frequency.
So notice that in order
to get the bond length we didn't actually use anything that had
to be gotten from the vibrational spectrum.
We're just using the fine structure
of the rotational transitions which we could have gotten
from microwave spectroscopy.
But microwave spectroscopy is not typically used very much
because we can get all of this information
from something like IR.
OK, so omega is our frequency in radiance
which we have our center of frequency
that we measured off the spectrum in hertz
and we have to convert it.
And that equals the square root of K over the reduced mass.
So what's K here?
That's the spring constant.
So just-- we're using Hooke's law from introductory physics
since we're making this assumption
that our molecule behaves like a simple spring.
And that's the force constant
that we're interested in getting.
So if we plug in our values, again remembering
to pay attention to units.
One of the things that could cause somebody to slip
up here is that you have your mass in atomic mass units.
You know, when you calculate to reduced mass.
And then when we have our force constant that needs to be
in kilograms because we'll end up with something in meters.
And so you need your conversion factor from kilograms to amu.
And so we get at force constant for HCl
of almost 500 Newton meters.
So just because I imagine people are going to be concerned
about this, on the exam I will give you things
like the conversion between kilograms to amu.
So unit conversions and physical constant that you might need
like that I'll give you.
Equations, I'm not going to give you
because you have a cheat sheet.
OK, so we learned something
about our force constant the bond
that tells us whether it's stiff or floppy.
But since we only talk about HCl so far,
we haven't really put it in the context.
We don't know what that means.
So let's juts look at this
for a few different diatomic molecules.
So if we look at HF, here's it's vibrational frequency in hertz
and its force constant is 970 Newtons per meter.
So we can see that HCl is a lot floppier than HF
which makes sense if we think about it you know,
again with our intuition from general chemistry.
So we know that, you know,
fluorine is a much smaller molecule,
it has a smaller electron cloud.
It's more electron negative,
those electrons are held more tightly to the nucleus.
And so we expect it to have a stiffer bond
when it's making a bond with hydrogen than HCl.
If we keep going down, bromine is a little bit less different,
but HBr is floppier still than HCl.
And for HI we have a much more dramatic effect.
So as we're going down the halogens,
we see that the force constant
in the bond is getting smaller corresponding
to those electrons being held less tightly
and making a longer bond.
So looking in the other direction, you know,
here we're comparing going down to halogens
where everything is a single bond.
If we look at carbon monoxide, so that's a triple bond
between two atoms that are both relatively small.
We get a very large force constant for that.
So that CO triple bond is very stiff.
And the nitrogen monoxide triple bond is a little bit less stiff.
So again knowing these exact numbers isn't all
that important except that it gives you some insight
into what we're talking about when we look
at that force constant or relates back to two things
that we know from general chemistry.
OK, so the last thing that I would like to do today is
to talk about what happens when we have anharmonic potentials.
So, so far we've just assumed that we're looking
at harmonic potentials everything is ideal.
We can treat everything as a perfect harmonic oscillator.
What happens if we can't?
So if we have an anharmonic potential that looks like this,
and our harmonic approximation isn't perfect,
that means that we have to have a correction term
to our vibrational state.
And so here's the equation for the Morse potential
which is a commonly use potential.
And we just end up having to add some correction terms to account
for the fact that our potential is not a perfect
harmonic oscillator.
As far as what you need to be able to do with this right now,
I just want you to know that's it there.
That in some cases, we are dealing with molecules
that aren't going to behave as perfect harmonic oscillators.
And in that case, there are things that we can do
and there are corrections that can be made
for the anharmonicity of that potential.
And in computational chemistry, one of the important things
in various problems is coming up with potentials
that accurately represent the physical reality.
So not just for vibrational states but for all kinds
of processes that happen in atoms
and molecules, spectroscopically.
If you can come up with the potential that describes,
you know, what this-- what the energy differences
between these states look like that's a large part
of being able to solve the problems.
OK, so we're going to quit there for today.
So you should be able to do a lot of the practice problems
that are online because, you know, now we've gone through how
to look at these things qualitatively.
All right, we still have five minutes,
please pay attention for a second.
So you should now have the information to do most
of the practice problems that are online.
Next time, we're going to talk
about how you calculate the selection rules
and figure whether particular transitions are going
to happen or not.
And hopefully if we have some time we'll talk
about applications.
Anybody have anymore questions right now?
All right, I will see you on Wednesday.
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