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welcome back to the lectures on chemistry
and introduction to molecular spectroscopy
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we shall continue the lecture from the harmonic
vibrational spectroscopy of a diatomic molecule
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to look at one model for the anharmonic vibration
and this model is due to professor philip
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m morse from m i t around nineteen twenty
nine ah he came up with the molecular ah motion
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being an harmonic and the vibrational motion
eventually leading to for very large frequencies
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of vibration or very large the energies of
vibrational quantum with very large quantum
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number the molecule eventually dissociate
in the harmonic model dissociation doesnt
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exists because no matter how high the energy
is the parabolic nature of the harmonic ah
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potential energy curve tells you that the
molecule eventually reaches back to its equilibrium
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state and therefore there is nothing called
dissociation or a break way of a diatomic
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molecule accounted for in the harmonic model
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therefore it is very important for vibrationally
induced dissociation of chemical structures
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that the vibrational motion be anharmonic
and the model that proposed by philip morse
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has the following form for the potential energy
as a function of the ah distance from the
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equilibrium the v r is given by the specific
functional form a constant d e multiplied
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by this particular mathematical quantity alpha
exponential of minus alpha r minus r e whole
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square so the potential energy has a very
specific form due to exponential and alpha
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is a constant which we shall see in a minute
how its identified with the ah the equilibrium
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or which is called the harmonic oscillated
frequency
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ok
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now its important to visualize this potential
energy first to understand why it was ah why
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it is meaningful now if you plot v of r as
a function of r it is d e times one minus
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exponential of minus alpha r minus r e whole
square therefore at r is equal to r e the
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equilibrium bond distance
v of r is v of r is v of r is zero because
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this becomes one because the exponent is zero
therefore one minus one zero therefore v of
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r is the minimum at r is equal to r e and
ok for very large values r v of r as r goes
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to very very large value or say the element
being the infinity you see that the exponential
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of minus alpha times r approximately r e is
too small and as r goes to infinity this goes
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to zero therefore the potential energy v e
becomes d e which is a positive constant therefore
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for very large values of r if we go back to
the graph and as r increases from the equilibrium
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value you see there is exponential of minus
value for e therefore the potential energy
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and this is a square
this is a square therefore as r is slightly
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different from r e this whole graph sort of
grows up and eventually it reaches ah a plato
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and the value which doesnt change for very
large values of r equals to r e is the asymptotic
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value ok which you can call this graph reaches
the asymptotic value and that value is d e
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ok for r less than or r e this is negative
and therefore the exponent of the exponential
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becomes positive and if this becomes ah more
negative then this increases forever and therefore
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what you see here is that ok
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so this is the form of the potential energy
for a given value for alpha if alpha is very
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large then you see that this graph is narrower
if alpha is very small this graph is ah more
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elongated therefore the alpha gives the spread
roughly between what is called the harmonic
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area this area looks more like a parabola
so therefore you can see that for small values
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of r minus r e this function will actually
become parabolic in the in the limit of r
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minus r e being very very small and at r equals
r e this is the minimum therefore this is
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the the parabolic potential which you have
with the half k x square half k r square that
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you use for the harmonic model it has that
as the limit of small amplitude oscillations
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and for very large amplitude of oscillations
you see that the molecules is such that the
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atoms go apart from each other they never
come back and this is the dissociation limit
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ok
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for this potential energy with is form if
you write down the as minus h bar square by
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two mu d square by d r square plus the dissociation
energy d e times one minus e to the minus
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alpha r minus r e square the way functions
h psi n is equal to e n psi n as actually
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close form solutions that is analytic solutions
given by philip morse and later corrected
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by professor ter haar that e n is h omega
a times n plus half and there is another term
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which contains a small constant called anharmonic
constant x e but with n plus a half whole
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square so you can see that the energy level
e zero for example is half h let me not right
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now we i think we are using it for angular
frequencies so let me write this as nu e and
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nu e ok then e n for e zero where n is zero
is half h nu e minus number this is one by
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four h nu e x e what is the next energy level
for this problem is n is equal to one e one
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is three of h nu e minus this is three half
therefore it is nine by four h nu e x e
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please remember for motion very near equilibrium
the x e is the small constant and its called
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the anharmonic constant and therefore the
energy is not precisely a half h nu but its
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slightly lower than h half h nu x e is positive
ok therefore you see that the energy levels
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as you go from zero one two three etcetera
or more and more away from what is called
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the harmonic oscillator energy levels so if
you look at the harmonic oscillator energy
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levels if you write that its suppose you call
this is e zero and then this is e one and
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this is as e two and compare that with the
potential graph which goes something like
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that what you see is the lower level is the
lowest level is slightly lower than the original
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e naught this is e naught because it contains
this minus half h x e nu e minus one by four
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sorry and then the e one is even lower than
the harmonic value of e one e two is even
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closer and e three is closer and so on and
so finally you see that the energy level becomes
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very dense and so on now you can see that
as you go further and further up if the hormone
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if the oscillation is sufficiently large if
the amplitude of the oscillation is fairly
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large and quantum energy levels are very high
you see that the molecules eventually breaks
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down and dissociation takes place
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the therefore the energy differences which
in the harmonic oscillator model where identical
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between nearby levels or not so in the case
of morse oscillator this energy level is slightly
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more than this difference between the two
energy levels e naught and e one is definitely
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more you call it as delta e one then this
is more than delta e two therefore the frequency
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at which the molecules absorb if it is anharmonic
molecule and if satisfies thus anharmonic
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molecule model this frequency of absorption
is slightly more than this frequency of absorption
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and therefore what you see here is of course
a spectra line corresponding to delta e one
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ok if this is the increasing e the next is
the delta e two if you see it it is slightly
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lower lesser than the delta e one and of course
the intensities well also decrease because
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the higher the energy level is the fewer the
molecules are or at any given temperature
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ssubject to thermal equilibrium conditions
and that's about that's about the distribution
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law
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therefore you see that delta e two if this
is called delta e two and then delta e three
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is even smaller and so on but something else
also happens in the harmonic oscillator model
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its not possible for us to actually undergo
a transition actually force molecule to undergo
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a transition from e zero to e two this doesnt
exist it cannot be same the dipole moment
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operator does not connect to that however
in the anharmonic oscillator model in the
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morse oscillator model it is possible for
you to see dis transition it is also possible
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for you to see let me put it on the point
of medium point with some other color it is
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also possible for you to see dis transition
it is possible for you to see the dis transition
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and even this and so on and these are vibrational
overtones there is no overtone in harmonic
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oscillation model there is only one line what
you see is only one line corresponding to
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this delta e and that's the same for every
other transition as well therefore the morse
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oscillator provides you slightly more realistic
what is called vibrational spectroscopy and
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the vibrational spectrum that you see in the
case of diatomic molecules but please remember
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we have kept the rotational motion completely
out of this picture we assume that the molecule
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is purely vibrating and we do not worry about
the rotational energy is associated with that
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but after we do the microwave spectroscopy
we will see how to look at the vibrational
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rotational spectrum together
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but for the time being the simple picture
of the harmonic oscillator model gives you
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no transition other than one line the anharmonic
model due to morse oscillated gives you several
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energy levels which are different from each
other and therefore the gap between them is
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also different so lets calculate the gap for
a simple example say e naught was written
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as h nu e minus one by four h nu e x e e one
was written as h nu e three by two sorry this
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is half h nu in this is three by two h nu
e minus nine by four h nu e x e and e two
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is written as ah three by two so then its
five by two h nu e minus twenty five by four
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n plus the half whole square therefore its
five by two whole square h nu e x e and so
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on therefore if you calculate e one minus
e naught the answer is h nu e minus nine by
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four plus one by four so you get two h nu
e x e x e is a very small number therefore
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this is an extremely small number compared
to h nu e ok essentially you can write this
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an h nu e times one minus two x e x e been
very small this is close to x e but what about
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e two minus e one if you look at that that
again h nu e but the difference is twenty
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five by four minus nine by four therefore
you get one minus four x e ok the difference
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is sixteen by four
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and the next one if you want to write is minus
seven by two h nu e and it is forty nine by
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four h nu e x e and therefore you see e three
minus e two is h nu e times forty nine minus
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twenty five is twenty four therefore you get
one minus six x e see how the successive energy
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differences are becoming smaller and smaller
due to the larger contribution of the x e
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this is minus two x e here its minus four
x e and here its minus six x e therefore its
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possible for us to actually obtain values
for nu e and x e if we get two experimental
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spectral lines if we get a transition due
to this and if we get a transition due to
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this then the two equations involving the
nu e and the nu e x e can be solved and is
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possible for us to obtain numerical values
for the unharmonicity constant and therefore
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use it for fitting experimental spectra of
diatomic molecules where the motion is slightly
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unharmonic
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there are molecular problems where the motion
is very very high unharmonic and in the case
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of polyatomic molecule we will come ah to
ah look at at least for a brief moment what
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are called the non rigid molecular motions
and so on therefore it is easy to understand
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that vibrational spectroscopy ah starts with
the elementary model of a harmonic oscillator
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but then the corrections to the harmonic oscillated
and the real molecular spectrum or usually
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taken into account by correcting the potential
energy in such a way that unharmonic corrections
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can be done
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the previous lecture in the previous lecture
i mentioned the unharmonic corrections can
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be x cube like terms the potential energy
terms you have here that is r minus r e cube
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terms r minus r e to the power four terms
and so on they are called cubic and unharmonisities
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and corpic and unharmonisities and the r minus
r e square is called the quadratic harmonic
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term so keep this in your mind in solving
some of the problem related to unharmonic
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vibrational motions of diatomic molecule
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in the next lecture we will continue this
and look at polyatomic molecular motions and
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then in a similar wave we will extend the
harmonic oscillator model to molecules with
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many vibrational degrees of freedom what are
called the normal modes of the vibration they
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will also some picture pictorial representation
of some of the normal modes of vibration ah
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through a calculation tool that is quite well
known today called the gaussian zero nine
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the gaussian program and the gaussian program
is a computational chemistry program which
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allows you to calculate molecular properties
quite accurately we will see the harmonic
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oscillator model for a polyatomic molecule
in the next lecture following this until then
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thank you very much