Good morning and thank you for being here.
Now, if you remember I was talking about the
particle in a 1 dimensional box we solved
this equation for the particle. We found the
so called the stationary stare which I mentioned
which as I mentioned where just the standing
wave patterns that you would expect in the
one dimensional system.
And, let me just write the wave functions
and their loud energy levels; we found that
the wave functions were given by these were
the wave function if x was between 0 and L.
And the wave function would be equal to 0
if x is south side 
and these functions had an energy and the
n could take the value as 1, 2, 3 etc. And
if you remember n equal to 0 is not allowed
because then the wave functions would identically
vanish everywhere. And further this function
will give me a total wave function; when I
say total wave function it is a function of
x as well as t, which we have seen would be
given by the sine of (x) into a to the power
minus i E n t by h cross that is what it is.
So, this is the time dependent wave function
for the stationary stare. And this given by
such an expression each value of n will give
me a different possible stationary stare characterized
by a different function capital psi. And what
about this capital psi to this equation i,
h cross doe psi by doe t is equal to H psi
or maybe I will write this equation in a slightly
different fashion I will take this to the
left hand side. So that I shall get a negative
sign and the answer must be equal to 0 right.
So, any side that you have of this form maybe
I can put a subscription to n here because
I itself never depends upon n. And that psi
n will actually obey this equation correct
that is how it is this you does not matter
what the value of n is it will always obey
that equation. And maybe I can even write
it in as slightly better form by removing
this psi n outside. So, I will have i h cross
doe by doe t minus H operating psi n would
be equal to 0.
Let me remove that. So, these are the equations
that we have. I will also represent the allowed
energy levels of the system this is my box
extending from 0 to L. And along this axis
actually we are going to represent energy.
And if you are thinking of potential there
remember while we do it actually like this.
See in this region the potential of the infinity
in that reason also potential energy is infinity
well inside the box a potential is 0. And
then we look at the loud energy levels you
can put n equal to 1 L equal to 0 is not possible
as I said you can put n equal to 1 what is
going to happen is you we are going to get
energy; which may be somewhere pi square,
h cross square divided by 2 m L square can
represent that in this picture may be much
well; maybe I should write little bit lower.
Now, you notice that the energy this is actually
the stare that has the lowest energy but it
is energy is not 0. Now, if you are thinking
of about this system classically; suppose,
I have a classical particle which is trapped
in a in a box. And what can happen is the
particle can be just sitting in the box without
moving. I have a particle which is just inside
the box without moving. So, it will be sitting
somewhere where there is no absolutely no
kinetic energy because it is not moving right.
And the potential are inside the box you see
therefore, the total energy classical energy
can be 0.
So, the particle maybe just sitting here like
that if it was obeying classical mechanics.
And then it can have n as a 0 or in fact,
it may be moving. So, suppose it is moving
which is going in this direction then it will
hit the wall and be reflected. And I can give
it any energy I like. So, therefore classically
speaking the particle can have any energy
which is starting from 0 to any large value
right. But that is not possible in quantum
mechanics what is just happening in quantum
mechanics was first of all you find that energy
cannot be 0; not only that energy can have
only certain values. So, we say that only
certain discrete energy levels are possible
for the particle.
So, these stationary stares have only certain
discrete while release for their energies.
And what are their lowest possible one; we
have found is so much pi square, h cross square
by 2 m L square. And you mean you can actually
as why is it that the particle cannot be simply
be sitting at one location why is why is not
that this is possible in quantum mechanics?
The answer is actually quite simple; imagine
that you have a particle just sitting at a
particular location then I know where exactly
the particle is therefore, it is uncertainty
in procession will be 0. Because if it is
sitting somewhere precisely then uncertainty
momentum; so that an uncertain procession
will be 0.
And, if it is not moving it has no momentum.
So, you essentially means that momentum is
0. So, therefore there is no uncertainty in
procession and there is no uncertainty in
the momentum. And delta x is 0, delta p is
also 0 which means that I would have a situation;
where the uncertainty principle would be wile
average right. That is the reason why energy
equal to 0 is not possible for this particle
which is confined within a box; because it
would simply violate the uncertainty principle.
So, the lowest way possible energy level for
the particle is this one.
The next one is how much you put n equal to
into 2. So, you are going to get energy which
may be somewhere here; that will be 4 pi square
h cross by 2 m L square. So, this has n equal
to let me, write the value of n here this
is n equal to 1, this is n equal to 2 and
then you will have n equal to 3 somewhere
here; notice the way I draw this picture.
This is the next aloud energy level it is
value will be 9 pi square h cross square upon
2 m L square. And notice you see the this
gap is only pi square h cross square by 2
m L square; while that gap 
it will be 4 minus 1; here, you have a 4 there
you can say there is a 1.
So, this gap is actually 3 times that gap;
that is important to notice this while you
think of this gap how much it will be it will
9 minus 4; which is actually going to be 5
times pi square h cross square by 2 m L square.
And this will go on happening if you go. If
you looked at the next level actually may
be I can just draw it here. But it may not
be accurate actually the spacing I mean it
is difficult for me to show it now. But you
know that spacing in going to be how much?
This is n equal to 4 it would have an energy
of 16 pi square, h cross square by 2 m L square.
And this gap will be how much it is going
to be 7 times pi square, h cross square divided
by 2 m L square.
So, therefore this is another interesting
thing this spacing between successive energy
levels actually increase as you go up. And
further what I will do is I will also represent
the wave function in this same picture, so
that whatever wave length is there in this
picture. So, the wave functions the way I
am going to draw them is a percentage here;
as well as, this function is concerned that
will be the wave function it is 0 there, it
is 0 there and the maximum value is right
in the middle.
The next stage I will represent with the different
color the wave function will be looking like
this it has a node in the middle so it would
like that. Now, while the lowest one the wave
function is positive everywhere so that there
is no node. The next one has actually 1 node
right in the middle of the box. So, there
the function is 0 to this side function is
positive, while to the other side right the
function is negative. What about the next
one it would have 2 nodes and it would look
like this. Well, the way I have drawn this
figure you see it is not easy; you see it
is not possible for me to draw the fourth
wave functions.
So, I will just draw it here the important
thing is that you see here there is 3 nodes.
So, let me put the nodes 1, 2 this nodes you
see these are the nodes. So, the function
would be like this. So, these are my wave
functions. But then let us imagine that it
is sitting in the particle is sitting in the
ground stare; then the wave function would
be square root of 2 by L, sine pi x by L this
is the lowest of possible stage. The lowest
possible stage is actually referred to as
the ground stage of the system. The next possible
one is referred to as first excited stage,
second excited stage, third excited stage
right.
And, imagine that you have a prepared system,
imagine that it is in contacts I mean, this
my system is a particle. But imagine that
this interacting with the surroundings. And
the surroundings is at a temperature of t;
that may be a room temperature. Then what
will happen is that I mean I suppose, you
know a little bit about these you will have
a Boltzmann distribution of the particle right.
When, the particle may be in this stare or
may be in that stare or may be in that stare
with what is refer as a Boltzmann distribution;
which will depend exponentially upon the aloud
the energy of the particle. But suppose your
surroundings is a 0 Kelvin which is an idealized
situation. And if the surroundings is at 0
Kelvin what will happen is that your particle
will be in the lowest possible stage.
But even in that lowest possible stage the
particle has an energy. So, even if the system
is at 0, even if the surroundings are at 0
Kelvin, even when it is in the lowest possible
stage the system has an energy. And so therefore,
this energy that is processed by the particle
even at 0 Kelvin is referred to as 0 point
energy; where, I can did may be here 0 point
energy why do you say that this is the pseudo
code energy. Because even in the lowest possible
stage the system has a non 0 energy; that
is why you say it is pseudo code energy. I
mean, I have considered 0 leave this system
of ((Refer Time: 16:25)).
Actually, classical mechanics what will happen
is that it will not do anything it will not
move it will be just sitting there and the
energy would be actually 0; but quantum mechanics
does not allow that right. And that even at
absolute 0 this system would have that much
energy. And we have seen the reason for this
the reason is simply your uncertainty principle.
If the particle just sat there without moving
that will violate the uncertainty principle
would. So, it has been moving.
Now, let us look at this particular function;
this is the function that we refer to us as
psi 1 of(x) remember this is only the time
in independent part; that means, it is only
the position dependent part of the wave function
and the function has this form. In fact, yesterday
I had drawn the value of psi 1 square; how
it would look like. And the maximum or the
most probable position I showed that these
are the middle of the box. But now suppose,
I can do some experiment where I will measure
the position of the particle right. Then what
do you find? Most of the time you will find
that the particle is right in the middle but
then of course, other procession are also
probable.
And, therefore, what will happen is that when
I do the experiments remember our process
number 3; each time you may get the different
answer. And suppose I make an average of all
this measurements what will be the answer?
As you remember the average of large number
measurements would be given by the expectation
value of x which I can calculate from my knowledge
of the wave function.
How would I calculate that? Well, actually
the way I would have to calculate this I would
have to have the operate associated with procession
which is just x itself. I will have to allow
that to operate upon in principle I should
say psi 1(x, t) and psi 1 star (x, t). And
then multiply it by d x integrate from minus
x infinity to plus infinity. And divide this
by integral minus infinity to plus infinity
d x psi 1 (x, t) star, psi 1(x, t) d x this
is our more very detail expression. But you
would realize from old discussion yesterday
that there is a time dependency here, there
is a time dependency there. But stationary
stages are such that the time dependence from
here will become exactly cancelled by the
time dependence from there. And what happens
is that this psi 1; this is not really necessary
to you use that capital psi 1. But it is enough
if you just use this psi 1 right.
So, let us do that let me remove make that
simplification remove this with psi 1 of (x)
and that with psi 1 star of (x). Because there
is time dependency but that time dependency
gets exactly cancelled. Time dependence of
one is cancelled by the time dependence of
the other that is what we saw yesterday. And
not only that it is necessary to integrate
from minus infinity to plus infinity but it
is enough if it integrates from 0 to L. Because
it is only in that region that psi 1 of (x)
is non 0. And you can do this same kind of
thing below down here; you can say well you
would have psi 1 star f(x) and psi 1 of (x).
And, again you do not need to integrate from
0 to sorry, minus infinity to plus infinity.
But it is enough if you integrate from 0 to
L and to do this. But then of course the wave
we have taken this function remember we did
calculate this. And how did we calculate that
we arrived at this expression; this square
root of 2 by L by usineg the normalization
condition by imposineg the condition that
this integral is when evaluated the answer
is 1; that is how this is square root of 2
by L was done. Therefore, if I am usineg that
function actually this integral I do not have
to perform because I already know what the
answer is, it is 1.
So, this is the object that we have to evaluate
and even here you can make simplification.
Because if you look at psi 1 there is no effect
of star operation taking the complex of integrate
does not effect it; because it does not contain
square root of minus 1. So, even that can
be removed. And therefore effectively what
is it that I have after all these simplifications;
what I have is integral d x of psi 1 of (x)
is actually referring 2 times. So, psi 1 square
x into x is the integral that I have; I mean,
of course I can calculate this I will do that.
But let us just think about the problem physically.
I have a particle which I know is somewhere
between 0, x equal to 0 and x is equal to
L. And further if I look at the most improbable
position; which is this point L by 2. If you
look at the wave function you will find that
this wave function is actually symmetric about
that point. And if you look at the square
of the wave function you will find that the
even this square of the wave function is symmetric
about that point; which means, that when I
make a measurement it may be to the particle
may be found to the left of this point or
to the right of the point with equal chance.
And, therefore what would you expect the average
to be? Obviously, the average has to be equal
to L by 2. Because the particle may be to
the left of that point or may be to the right
of that point with equal likelihood. And therefore
even without doing this calculation I can
make a prediction that it has to be equal
to L by 2. And this is exactly what happens?
I will not do the complete integral because
the integrals are not that difficult. But
I will tell you how it comes ((Refer Time:
23:41)). If you integrate from 0 I mean, the
integrals that you have to perform is integral
0 to L, d x, sine square (pi x divided by
L) into x.
So, here you see you have sine square. But
there is a trigonometry identity that you
can use sine square theta is actually equal
to 1 minus cos 2 theta divided by. So, you
can use this trigonometry identity. So, instead
of that sine square by x, by L you can actually
write sine square pi x by L; may be written
as 1 minus cos (2 pi x by L) divided by 2.
Now, what you can do is; we can substitute
that in here then you will see that there
are 2 times correct there are 2 times in the
integral; what happens is that the first time
is non 0 value; where at the second time you
will find that the second time is identically
equal to 0.
And, the first time so let me just look at
the first time what is the first time? Integral
into L, d x. The first time is simply well
I did make a mistake did not I? At this point
I did make a mistake. Because psi 1 of (x)
actually condemns where is it? Square root
of 2 by L as I did not write this; square
root of L I just wrote the sine square there.
So, therefore I have to correct for that mistake;
there will be a 2 by L sitting here correct.
So, therefore I should have written that I
would have 2 by L. And then I would of course
have this sine square which is actually going
to give me 1 minus cos 2 pi x by L divided
by 2 into x.
So, that is what integral lies. But then obviously
as I told you there are 2 times the first
time will come from here, the second time
will come from there. But that second time
I am not actually going to evaluate other
than ((Refer Time: 26:47)). If you did evaluate
the contribution from that time it is going
to be 0; maybe you can do this as an exercise.
And then what will happen to the other time
I will get 2 by L, integral 0 to L, d x into
x is what you are going to get from the first
term.
And, there this is a half. And this is pretty
straight forward integral what will happen
is that you will get 2 by L into L square
by 2 as there is a result of doing the integral
there is a half already there. So, therefore,
this is what will happen. And if you cancelled
all these things properly what is the answer.
The answer is going to be L by 2 this is which
I would have expected physically any way.
Now, interestingly you see I have done this
calculations for the first the lowest possible
stage. But I can do the calculation for the
any stage.
But do I need to do this calculation you can
use your physical argument; whatever, be the
stage you take sine coir effect you see you
will find that it is symmetric around the
middle point. And therefore it does not really
matter what the stage is; what I would expect
is that the average value of the procession
should be L by 2. And that you can easily
calculate and show but it is not necessary,
because the physical intuition tells me that
has to be equal to L by 2, because simply
because it does not matter what the value
of n is psi n square is going to be symmetric
about the matrix of the box. So, having completed
that calculation let me look at the expectation
value of momentum the particle is moving only
in one direction. So, any momentum that I
would like to think off would be denoted as
p x.
Say, if you wanted to calculate the expectation
value of p x actually to be very precise.
I should say the operate associate with momentum
that is the expectation value what is it that?
I am trying to calculate the particle is within
the box. I want to calculate the average value
of momentum; if I can make it large number
of measurements of momentum. And take it is
average what would be the answer. The answer
is going to be this with of course p x of
((Refer Time: 29:40)).
So, what is actually p x? Well, actually I
should start from the beginning where I would
start with the time dependent wave functions
and did exactly what I had done earlier, but
because we already done it in the case of
procession I not going to do that. The experiment
is going to simply simple because you will
have for this the same kind of simplifications
as earlier. So, be the operate will be minus
x cross doe by doe x; that is the momentum
operator and it is going to operate 1 psi
of x.
So, that is the object that I have to evaluate.
But then again I mean, we can think about
physically you see the box is here let me
say the particle is inside; what is going
to happen basically, the particle will be
maybe it is moving in this direction at one
instant. But then of course you have to heat
the wall; let us, the physical way to think
of about it than deflect. So, therefore the
particle may be moving in this direction or
may be going in the opposite direction with
equal latitude right. If it is moving in this
direction you would expect momentum is 0;
while, if it is moving in the opposite direction
you would expect that sorry, made a another
mistake point out my mistakes it may.
If the particle is moving in this direction
its momentum will be positive; while, if it
is moving in opposite direction the momentum
will be negative So, what do you expect on
an average to happen the answer is there has
to be pseudo; without doing any calculation
I could have said that the answer had to be
0. But here this integral is actually fairly
easy see this minus h cross I can take out.
I would have integrals 0 to L d x psi 1 of
x and I have doe by doe x operating upon psi
1 of (x). But that is nothing but the derivative
psi 1 of x with respect to x correct that
it what it is.
And, what would that be; well, it is actually
minus x cross well you have sine 1 effect
it is multiplied by the derivatives f 1 of
(x) very easy to do in this integral. It is
nothing but integral ray is nothing but psi
1 square divided by 2. Because the derivatives
of this is nothing but psi 1 of x into d psi
1 by d x; in this integral is nothing but
1 by 2 psi 1 by x square. But then you have
to substitute the limits. The limits are from
0 to l. But you have the nice thing when x
is equal to 0 psi 1 I know it must be 0. And
when x is equal to L psi 1 again must be 0.
So, therefore at the boundary at these are
the boundaries; at the boundary the function
is 0. So, that the function at boundary there
is psi 1 of (x) is 0. So, therefore you get
that answer to be 0. And you would realize
that it does not matter whether it is the
lowest possible stage or the first x stage;
you can have any value for the quantum number
n same derivation can be done. And you will
find that for all the stage the average value
of the momentum has to be 0.
But what about I mean that is all fine what
about; suppose, I am going to think of this
square of the momentum; while, momentum may
be positive or negative. But this square is
guaranteed to be positive. So, if I was going
to calculate the average of the square of
the momentum. Let me, just modify what is
here d x is square let me remove the other
parts. So, I want to calculate the x square
so how can I do that? You will have to put
d x square here.
And, what is going to happen is integral 0
to L, d x, psi 1 of x, square of p x I know
it is just minus h cross square, doe square
upon doe x square, psi 1 of x; remember discussion
of the in which we did calculate the square
the operator corresponding to the square momentum.
And it is just h cross square minus h cross
by 2 sorry, minus h cross square doe square
up on doe x square that is what it is. But
then what is going to happen is I am going
to substitute for psi 1 as I will get this
2 by L square, integral 0 to L, d x, sine
pi x by L, doe square upon doe x square, sine
pi x by L.
So, you have to take this second derivatives
of this sine function. If you remember how
to calculate derivatives the first time you
differentiate psi you will get different sine.
And the next time you differentiate you are
going to you are going to get the sine pack.
So, doe square upon doe x square operating
upon sine what will it give me? Let me continue
from here on to here or may be somewhere up
it would be equal to 2 by L, integral 0 to
L, d x sine pi x by L. And carry 2 differentiations
the answer is going to be minus pi by L the
whole square. Because the each time you differentiate
the factor of pi by L is going to come out.
So, you are going to get minus x pi by L square
sine pi x by L and this object is just a constant.
So, again another mistake I will you please
point out when I make mistakes this negative
sign should have a forgotten to write, h cross
also I have forgotten to write this keeps
on happening to me. So, I that means I shall
have to modify my equations this minus h cross
I will have to put it outside. So, I should
not forget that which means that I shall have
to put a minus h cross square here also.
So, therefore what is going happen this factor
and that factor I can combine what would I
get? h cross square there is a negative sign
there, another negative sign there; with this
I can forget the 2 negatives sign h cross
square, pi square divided by L square this
is what you are going to get. And then you
will have of course this 2 by L. And you are
going to get integral 0 t L, d x sine pi by
x, another sine pi by x. So, you will get
sine square pi x by L this is what you are
going to get.
And, if you look at this integral is nothing
but pi 1 square of (x), d x integral 0 t L
this whole this you can say psi 1 of (x) square
d x. And what is that your function remembers
is normalized. And therefore that integral
you do not have to actually perform. Because
you know what it is; it is equal to 1. And
hence this is equal to 1. And therefore this
object which actually is equal to this integral
that we were trying to evaluate; this object
is nothing but h cross square pi square by
L square. So, we have evaluated the expectation
value of d x square the average value of this
square of the momentum.
Now, I have done it for the ground stage n
equal to 1. But suppose, I had done it for
a general stage with an arbitrary what would
you expect? The thing that is going to happen
is that you will have an n pi here and an
n pi is going to occur everywhere. But then
when you carry out the differentiation with
respect to x; you see in this case it is a
pi by x that is coming here out. But if I
had sine n by x by L; then what will happen
an n pi will come out each differentiations
1 n pi will come out. So, therefore if the
final answer in such a situation would be
not just pi square, h cross square by L square.
But n square, pi square, h cross square by
L square. So, for a general stage what would
the value of p x square average? It would
be simply n square, pi square, h cross, pi
square by L square.
Now, it is very simply straight forward to
calculate average value of kinetic energy
what is kinetic energy? Operator it is going
to be p x square divided by 2 m; this is the
kinetic energy operator shall we calculate
the average value of this what will happen
while you see 1 by 2 m is the step constant.
So, you will have the average value of p x
square by 2 m. And the we have already calculated
the average value of p x square it is going
to be; n square, pi square, h cross square
divided by 2 m L square. If you want if you
calculated the average value of kinetic energy
in the n stage; this is going to be the answer.
And, what is this actually if you look at
this expression; this is nothing but the expression
for the energy level well. So, if you just
calculate the average value of left kinetic
energy you are getting the total energy of
the system is that surprising? No, it is not
because the particle has 0 potential energy.
So, there is no contribution for the potential
energy the energy is purely kinetic that is
why this happens. Now, I want to illustrate
a simple a chemical application of this model.
Let us think of beta dine, 1- 3 beta dine
into a precise. This is the molecule it has
a 2 conjugated carbon carbon double bonds.
And if you do experiments you can measure
this distance using x ray techniques these
1 line these 1.38 and that this is 1 point
roughly I mean 1 .48. So that the distance
between the 2 end carbon items how much would
be; you can somehow these things. So, 2 .76
plus 1 .48 will find there it is 4.24 Armstrong
well, I wonder whether I mentioned even x
these are in Armstrong. So, how much is an
Armstrong is again let me remind your memory;
it is 10 to the power minus 8 centimeter or
it is equal to 10 to the power of minus10
meters.
Now, beta dine has a 4carbon items remember
my discussion; yesterday I said that each
carbon is being to this will leave the orbit
perpendicular the pane of the molecule. And
they show there are 4 p orbital which are
un hybridize one sitting on each carbon and
these 4 p orbital can overlap right. And if
you put an electron to this pi system; this
system is referred to as the pi system of
the molecule. If you put an electron into
the pi system electron can be thought as moving
from one end of the molecule to the other
end. And this model you can try to apply to
the system.
So, what is it you think of any particle electron
in the pi system; it is moving from 1 end
of the molecule to the other end of molecule.
And what would you expect? You would expect
that the length of the box is perhaps something
like 4.24 roughly; I mean, it up the order
of 4 or may be 5. So, therefore I say, I have
a box which is of a length I mean; let me
not worry about the this number that is given.
Let me try to estimate the length using experiment.
So, I say that I have a box which is of length
L. And if you think of these electrons which
are confined to that length L; their loud
energy levels are given here this is the first
energy level, this is second, that is a third
and so on.
But remember each carbonate will contribute
only 1 electron to the pi system; the other
electrons of each carbon is used to form sigma
bonds with the hydrogen and with other carbon
.So, therefore how many pi electrons should
I worry about. Because there are 4 carbon
items, there are 4 pi electrons. And how are
they going to set; well, these are the loud
energy levels. So, the out these 4 pi electrons
2 of them will see here in this energy level;
well, I am anticipating your previous knowledge;
you would have something previous knowledge.
So, therefore, only 2 electrons can be sitting
in anyone particular stage. So, these are
2 electrons there and you have 2 electrons
which are sitting here. And that is all there
are 4 electrons all of them are accountable;
for the first you are sitting in the lowest
way possible stage. And the second 2 electrons
second stage is sitting in the next stage.
So, that I what would what you would expect
to happen in the molecule. And the pi electrons
are accommodated in the first and the second
loud energy levels; just the remember we are
thinking only of the pi electrons; we are
not worried about the sigma electrons which
form the framework of the molecule.
And, now suppose I take this molecule and
allow it to interact with electromagnetic
radiation; as I was mentioning though it is
possible for electromagnetic radiation to
make the potential energy time dependent;
and as the result of that possible for the
wave function to get modified. And effectively
what would happen is that quantum may be absorbed
as a result of each 1 electron from here may
be promoted from that orbit from that stage
to this stage ok. So, 1 electron could be
promoted from here to there. And you can see
how much energy will be required to cause
this promotion energy of the initial stage
as far as that electron is concerned it was
having an energy which is equal to so much.
And as the result of the absorption of the
radiation it has gone on to that stage.
So, therefore the change in energy which I
will denote as delta E, in this case will
be actually the energy of these stages with
the n equal to 3 minus the energy of the stage
with n equal to 2; that is the change in energy.
And that will be equal to 5 pi square h cross
square divided by 2 m L square. And that should
be equal to the change in energy. And if this
is brought about by photon of frequency nu;
then h nu has to be equal to change in energy
and that must be equal to so much, and so
it is possible you to calculate nu. And it
is possible for you to estimate nu or a or
a not estimate nu not can determine nu by
studying the absorption of the molecule; you
can find out at what frequency the radiation
is absorbed; what is the frequency of the
radiation that is absorbed by the molecule
experimentally you can determine that. And
then you use that I think I will stop here.
Thank you for listening.
