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yeah welcome back to the lectures in chemistry
in this lecture we shall continue from the
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previous screen if you recall on microwave
spectroscopy of the diatomic molecules we
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talked about the degeneracy of the wave functions
the wave functions associated with each of
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these energy levels have the same form as
the spherical harmonics
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now lets look at the energy level diagram
or the microwave spectrum before i do this
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let me introduce the convention that spectroscopy
is use when we write the rotational kinetic
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energy rotational energy on mechanical rotational
energy associated with the system as h bar
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squared by two i into j into j plus one thats
right this out explicitly it is h square by
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four phi squared into two i j into j plus
one which is h square by h square by eight
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phi square i into j into j plus one
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now remember that there is another way of
writing the energy in terms of wave numbers
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wave numbers new bar such that eis given by
this formula h c newborn therefore if we write
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the e rotational energy as h c new bar corresponding
to the quantum number j h is a constant c
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is a constant h is planks constant c is the
speed of light therefore there is no association
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the new bar is associated with the quantum
number j and that is given by hate square
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by eight phi square i j into j plus one and
therefore if we write the wave numbers new
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j new bar j as h by eight phi squared i see
j into j plus one spectroscopy have a notation
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for this constant and you see this is a molecular
constant h is a planks constant the c is the
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speed of light and i is of course the moment
of inertia associated with the molecule
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so this is a constant associated with each
molecule and this is given the symbol b and
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its called the rotational constant for diatomic
molecule therefore the value new bar j is
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be j into j plus one is what everybody uses
now what is the dimension of b its very clear
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from the way it is written h by eight phi
square i c and the fact that j takes only
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j is equal two zero one two three ask quantum
numbers it is very clear that j doesnt have
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any dimension and therefore the dimension
of b has to be the same as the dimension of
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the new bar j which is a wave number unit
and wave number unit is one by length that
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is the number of waves in a given unit length
if you remember the definition of wave numbers
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therefore this is per unit length or per centimeter
in our centimeter inverse lets see if b has
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the same dimension h by eight phi squared
i c is nothing other than h by eight phi square
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mu r square c recalling the definition of
phi and h has the units juil second which
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is kilogram meter square per second divided
by mu in kilograms its a reduced mass remember
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mu is m one m two by m one m two and therefore
it has the dimension of the mass kilogram
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r square is meter squared and c speed of light
is nothing but meter per second therefore
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canceling out to the appropriate quantity
what you end up with is one by meter plus
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b has the dimension of the wave number given
that j is a quantum number therefore dimensionally
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we are saying right things that is correct
and the quantity bis a characteristic of every
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diatomic molecule how it is it is
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dependent on two parameters corresponding
to the molecule one is the reduced mass of
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the molecule and the other is the internal
comic distance between the two atoms in the
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molecule now for on both counts it depends
on the given molecule given by atomic molecules
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and therefore be is very specific to the given
molecule it is a property of the molecule
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under the rigid wrote our assumption
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now let us look at the energy levels e as
a function of new j bar thats what we want
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right since the formula for nu j bar is b
j into j plus one that is right a few values
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j equal to zero corresponds to e zero which
is seen on j equal to one corresponds to e
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one which is two b j into j plus one j is
equal to two corresponds to e two the second
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energy level which is two into three it is
six b and j equal to three for example e three
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is twelve band so on ok lets me write just
the last next one more quantity j equal to
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four and write e four also as twenty b
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therefore what you see is the energy levels
increasing as a function j square j into j
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plus one for very large values of j functions
like j square before we see that the energy
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levels increase and the differences between
the energy levels which is what you see as
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a spectroscopic transition now will be determined
by the differences that you have between these
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levels so lets draw the energy level diagram
now e zero is zero that is j equal to zero
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e one if we draw this it is two b e two is
six b therefore on a scale of the appropriate
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energy this is the increasing energy scale
in
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the system value e two is six b the difference
between e not and e one is two b the difference
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between e one and e two is four b and write
the next one it is eight eight three is twelve
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b and the difference between this two is six
b and of course e four goes out of the screen
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here is twenty b somewhere let me write of
the top and the difference here is now eight
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b
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so successive energy levels corresponding
to the value j equal to zero j equal to one
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j equal to two j is equal to three and j is
equal to four as you see it the successive
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energy levels defer by two b four b six b
and eight b there is a selection rule in quantum
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mechanics for spectroscopic transitions ah
that can take place in a rigid diatomic molecule
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the selection role is that what transitions
are allowed what transitions are allowed or
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can be seen or can be seen for rigid diatomic
molecule the transitions that are allowed
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correspond only to this value delta j is equal
to plus or minus one which means that if the
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molecule is in the state j equal to one it
can undergo a transition if the microwave
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radiation is shown on the molecule it can
undergo a transition to the next level j equal
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to two by the process of absorption or it
can undergo a transition from j equal to one
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to j equal to zero by the process of emission
which is either spontaneous emission are stimulated
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emission either one of these processes but
it cannot jump from j equal to one to be equal
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to three under this assumption or within this
model of setting up the rigid hamiltonian
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as a classical hamiltonian converting it into
the quantum and following through this rigid
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approximation this model does not permit a
transition from j from j plus two or a j minus
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two or a j two j plus three or j minus three
delta j has to be plus minus one
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now with that you see that the first new zero
two one new bar corresponds to two b because
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that is nothing but the energy difference
between e one and e zero the new bar between
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one and two transition from energy level one
to two is for b which is the energy difference
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between e two and e one and the new bar two
to three is six b the energy differences between
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e three and e two and what you see is nothing
but if you were to obtain the spectrum of
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this molecule you will see if we plot this
as the wave number unit and we plot the absorption
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or the absorbance along the way axis what
you will see is a transition corresponding
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to the frequency two b which is a transition
from the ground state rotational state to
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the j equal to one state you will see one
other transition if there are enough molecules
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in the j equal to one state you will see a
transition from one to t two if you plot the
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absorbance we will see a transition corresponding
to four band if the molecule is in the j equal
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to two state the absorption spectrum from
j equal to j equal to three will give you
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align corresponding to six b so what you see
is a series of equidistant lines spectral
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lines remember that the energy levels are
not equidistant the energy levels separate
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are separated from each other by different
orders different values zero two b to two
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four six b six to twelve the energy levels
of not equidistant but the spectrum that you
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obtained which are due to the transition between
these energy levels the spectrum is equidistant
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therefore any two lines adjacent line the
gap between them gives you a value of two
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b if you recall that b is nothing other than
h by a phi square i c and if you know from
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the experimental spectrum the gap between
two successive lines as two b two b as the
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experimental value between two adjacent lines
then a measurement of this from the experimental
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spectrum immediately tells you how to get
the value for i by simple multiplication and
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given that you know what molecule you are
taking the spectrum whether it is hydrogen
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chloride our carbon monoxide for example molecules
which are permanent dipole moment which are
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the only ones that you can see using microwave
spectra you see that the reduced was is something
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that you know immediately and therefore knowing
i from the experimental spectrum allows you
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to calculate the inter atomic distance with
that molecule no more accurately you know
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that value of b the more accurately you can
calculate the value of the inter atomic distance
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and so on today after fifty sixty years of
research and microwave spectroscopy one can
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get get the bond distances in experimental
diatomic spectra up to about the third or
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the fourth decimal in angstroms which is a
very very high level of accuracy
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therefore experimental microwave spectroscopy
is the most important means for determining
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the intercom existences in a diatomic molecule
experimentally and informing the value through
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various theories you can predict the value
of the moment of d intercom distance and verify
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with the experiments let us see a typical
a few the atomic molecular spectra before
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we move on to the next topic in this subject
i i will show you two spectra here the spectrum
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that you see in this picture is the spectrum
of carbon monoxide and let me read the lines
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of the text here it is the the x axis is wave
number axis which corresponds to centimeter
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inverse ten centimeters twenty centimeters
etcetera in verses and then the y axis corresponds
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here to the absorption the extent of absorption
and you see a beautiful equidistant spectrum
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as you see between the nearby picks and the
scale here tells you that this is an overlapping
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spectrum of two molecules molecule carbon
monoxide with the carbon isotope the naturally
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amended isotope c twelve and o sixteen thats
a lower line
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corresponding to these tips that you see here
the small the the larger ones c twelve o sixteen
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and then you have the c thirteen o sixteen
its natural abundance of c thirteen is very
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low and you should know immediately why the
isotopic muscles will give rise to different
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spectrum but for both cases what you see is
between the different lines that you have
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here one two three four five or between the
lines here one two three four five they are
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equidistant that the two different isotopes
of carbon give two different spectra should
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be obvious from this formula which is given
by that the the to be is nothing but two h
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by eight pi squared i c and i is m one m two
by m one plus m two
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therefore if the mass of one of the atoms
is oxygen sixteen and the other atom is carbon
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twelve you get one value for the this times
r squared you get one value for the reduced
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mass and if the molecule is oxygen sixteen
and the carbon thirteen then you get another
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value for reduced bars and therefore you see
that you get two different spacings rotational
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spacings for the same molecule dependent on
the isotopic masses of the compound now where
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do you think that this difference will be
maximum the difference will be maximum you
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recall the reduced mass will differ by a maximum
value if for example one of the masses doubled
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you take the hydrogen spectrum take h c l
and if you compare the hydrogen chloride spectrum
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with deuterium chloride d c l you see the
reduced mass will differ by a large amount
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and therefore any replacement of hydrogen
by a deuterium or tritium will give you a
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very larger ship in the microwave spectrum
of the compound therefore isotopic masses
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do play a role in a significant role in the
microwave spectra of many of these compounds
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in fact we will use this to determine all
the different moments of inertia of a polyatomic
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molecule and thats how experimentally this
is done
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so the energy level is one part of the story
as you recall from the first lecture in a
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spectrum we are interested in at least two
or three different things and as far as this
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course is concerned we are interested in two
of the three things namely the line positions
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and the line intensity the line which are
very complicated so we will try avoid the
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description online bits is usually a subject
for the advanced course in molecular spectroscopy
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lets look up to the we have looked at the
line positions for a diatomic molecule as
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basically happening in a spectrum with respect
to plot with respect to the frequency or a
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wave number that it will happen at two b four
b six b eight b etcetera
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so there is the the discretization of the
energy of the molecule due to the fact that
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we solve the schrodinger equation and which
gives raise to quantum numbers here both the
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degrees of rotation degrees of freedom have
the same moments of inertia and therefore
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we have only one free running parameters namely
the moment of inertia and we get a quantum
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number dependence j into j plus one as the
energy level so in a sense the line positions
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are now very clearly understood as far as
the rigid microwave spectrum of molecules
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concerned