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welcome back to the lectures on chemistry
in todays lecture we will examine microwave
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spectrum
some of the reasons for why we have to study
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microwaves spectra let me give you in detail
if you want to know the geometry of the molecule
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or the equilibrium structure of the molecule
and if the molecule processes in dipole moment
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then in the gas phase microwave spectra gives
you spectral give you the most accurate geometry
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data that you can obtain rotation which is
the phenomenon that is associated with microwave
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spectra is one of the most fascinating phenomenon
phenomenon in quantum mechanics angular momentum
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associated with that number of rotations of
molecules the locations of ah molecular species
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are molecular complexes all of them give rise
to very rich spectra in the microwave region
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whose analysis tells you more about the electric
charge distribution that is present in the
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molecule and therefore also gives you a feel
for how the atoms are bonded to each other
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and so on
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so the precise dominantly and the shape of
the molecule is something that we always worry
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about in the gas from the gas phase specter
of many of the compounds but what is important
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is of course for the rotational spectrum our
microwave spectrum to be obtained the molecule
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must have here permanent dipole moment are
they charge a symmetry the plus and minus
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charge centers must be separated from each
other the other the most important reason
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as a chemist or as a physis that one is worried
about microwave is that this also led to the
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first and most important discovery in spectroscopy
called the majors or the microwave amplification
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by stimulated emission of radiation and microwave
spectra or microwaves were very important
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in the second world war the off shore of the
second world war led to a lot of this research
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in the spectroscopy of these molecules and
eventually this let this was a precursor to
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the most important discovery i would think
that the after the second world war namely
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the lasers the microwave amplification respectfully
stimulated emission of radiation was a precursor
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to laser
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so therefore in in both historical sense as
well as in the real sense of studying the
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molecular geometry it is one of the most important
spectroscopic tools even as a
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taritation or as someone who is interested
in chemical physics rotations molecule rotations
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and their study is a fascinating subject so
let us look at the microwave spectrum for
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a typical diatomic molecule which has a dipole
moment
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let us assume that the diatomic molecule is
a widget molecule this is an important assumption
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because no molecule is legit even at zero
kelvin in harmonic oscillator model you have
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studied that molecules have zero point kinetic
energy zero-point energy and therefore molecules
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vibrate even at zero kelvin therefore the
assumption of rigid molecule is something
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that we will do for convenience and if necessary
this can be relaxed depending on the molecule
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energies the rigid diatomic molecule essentially
means the following that the molecule geometries
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dont change if the molecular geometries dont
change then its easy to calculate the moment
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of inertia its easy to calculate using the
bond angles and are tentative model of band
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angels and bonds lens calculate the moment
of inertia calculate the spectral parameters
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verify them with the experiments and then
go back and redo it again
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lets to a simple example of a rigid diatomic
molecule and let us assume a classical picture
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to begin with supposing the night diatomic
molecule as two different masses m one and
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m two connected by a your bond length which
is connected to their centers of masses and
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the distance is r then the rotational kinetic
energy of this molecule about the axis of
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rotation there are three axis of rotation
there is an axis which passes through the
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bond coaxial to the bond then there is a bond
axis which passes through the center of mass
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perpendicular to that mean remove this r for
the time being and then there is an axis which
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is perpendicular to both this bond axis and
the bond here basically perpendicular to the
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plane of the screen that you are watching
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so there are three axis now when we calculate
moments of inertia or this molecule in order
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to calculate the rotational kinetic energy
for the system for its rotation about one
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of the axis
one of the three axis what do we do classically
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we calculate the distance of the atoms the
perpendicular distance of the atoms from the
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axis multiplied by their masses then we do
what is called them i r i square i classically
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about any axis a is essentially some over
all the masses multiple by their perpendicular
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distances from the axis and summed over all
the all the atoms here there are two atoms
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and therefore what you will have is the distances
from the axis as m one r one squared plus
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m two r square a simple to body kinetics kinematics
tells you that the moment of inertia i can
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also be expressed by this formula mu r squared
where mu is given as the reduced mass m one
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m two by m one plus m two and r is the inter
atomic distance
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now the question is about what axis i mentioned
that there are three access associated with
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the simple linear molecule a diatomic molecule
there are three mutually perpendicular axis
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and if we assume the atoms to be point mass
then the perpendicular distance of the atoms
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m one and m two about the bond axis is zero
therefore there is no moment of inertia associated
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with rotation about this axis there is no
kinetic energy that is no rotational kinetic
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energy associated with that axis now what
about the rotational kinetic energy associated
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with this axis or with the axis perpendicular
to this line as elastic bond axis the perpendicular
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distances are the same for both of them and
the mass being m one m two this formula tells
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you for both of those access the moments of
inertia i is given by the simple formula mu
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r squared its a classical it's a very elementary
classical mechanical formula you can derive
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that and in fact that be one of the exercises
for you to drive this the mu being reduced
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mass r being the inter atomic distance leads
you to this formula that the moment of inertia
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i is that given this as the moment of inertia
the rotational kinetic energy classically
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is given for such bodies as the rotational
velocity times rotational the angular velocity
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times the moment of inertia here multiplied
by half half i omega squared or there omega
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is the angular velocity
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which is nothing but the speed of rotation
about the given axis in terms of rotational
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angular momentum j which you know again from
classical mechanics j is given by i omega
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you know that the rotational kinetic energy
is given by
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j square over two i this is a classical formula
and the rotational kinetic energy is the same
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about both of the axis and under the point
mass approximation remember the third moment
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of inertia is zero before there are only two
degrees of freedom associated to rotational
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degrees of freedom associated with in linear
molecule under the point most approximation
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if the masses are not point muscles but the
atoms have a mass distribution the size and
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the charge and all those things you might
find out that the moment of inertia is so
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small that you still need not have to be concerned
with the rotational degree of freedom about
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the axis its almost a free rotation with more
energy associated with it
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therefore there are only for a linear molecule
there are only two rotational degrees of freedom
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and for the rigid molecule both these degrees
of freedom have the same moment of inertia
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and therefore there is only one moment of
inertia associated with the rotational motion
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of a linear molecule now this is classical
mechanical formula the spectroscopes studied
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by looking at the molecule energy levels which
are obtained by solving the molecule schrodinger
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the equation and the Schrodinger equation
as you know from the previous models and the
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lectures we write down the classical kinetic
energy and the potential energy we write down
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the classical hamiltonian and then we transform
that into the quantum mechanical formula and
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when we do that the angular momentum j in
classical mechanics which is given H by r
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cross p formula becomes the corresponding
the quantum-mechanical operators for j become
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the j x j y and the j components just like
you had the in the case of momentum and these
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can be written down and you can go through
a whole lot of algebra to write down the hamiltonian
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for the system again as the operator j square
by two i with the difference that this is
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quantum mechanical and the angular momentum
j is now a quantum-mechanical quantity and
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therefore it has very special properties which
are not the same as the classical angular
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momentum we will look into this more as and
when we required the details just let me take
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you aside to the rotational to the problem
of the hydrogen atom
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which was done several lectures ago in hydrogen
atom when we solved the hamiltonian we express
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to the hydrogen atom hamiltonian in terms
of spherical polar coordinates and you might
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recall that the angular part of the hydrogen
atom hamiltonian
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you recall remember that that was nothing
but minus h bar squared by h bar squared sin
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one by sin theta though by the theta sin theta
though by though theta plus one by sin squared
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theta though squared by though y squared this
was the angular part that you recall from
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the hydrogen atoms solution now the properties
of the angular momentum operators j x j y
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and j z said the components of the angular
momentum operator in an x y z coordinate system
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that is associated with the center mass of
the molecule if it is expressed in spherical
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polar coordinates you get exactly the same
form as you have in the case of hydrogen atom
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angular parts
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therefore the hamiltonian for the rotational
motion becomes exactly ditto of what you have
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here with the one difference that is it too
i in the denominator which corresponds to
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the j squared by two i part of the j squared
the angular momentum part is given by this
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formula therefore what is obvious immediately
is that if we have the schrodinger equation
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written as h psi is equal to e psi and if
the h is written as minus h-bar squared away
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two i times the formula that you have sin
theta sorry one way sin theta though by though
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theta sin theta though by though theta plus
one by sin square theta though square by though
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phi square sigh which is a function of theta
and phi is equal to e times sin theta and
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phi then the solution sin theta phi you know
exactly what it should be and this is nothing
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other than the spherical harmonics that you
derived earlier sin theta phi is the spherical
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harmonics why i use the symbol l m theta phi
earlier for denoting the orbital angular momentum
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but here now the quantum number is the rotational
angular momentum quantum number and this is
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now replaced by y j k theta phi
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where k has the same role as the m and j is
the quantum number associated the rotational
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motion of the molecule this also tells you
immediately what should be the value of the
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energy and the energy is you remember that
there is h square h bar squared by to i which
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is there on the left-hand side here h psi
equal e psi gives you earlier it gave you
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l into l plus one now you will get j into
j plus one which is the quantum number associated
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with the rotational motion and therefore what
you have is h bar squared by two i j into
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j plus one is the rotational kinetic energy
associated with the diatomic molecule e rotational
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there is no potential energy here we are only
worried about the rigid item rigid body motion
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the rotational kinetic energy is the total
energy after rotating diatomic molecule
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therefore e rotation is now given by h bar
squared by two i into j into j plus one this
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is a simple rigid by atomic model with the
values of j being zero one two three etcetera
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all the way up and the value of k if you remember
is the same as the value of m earlier k goes
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from minus j minus j plus one to j minus one
up to j so there are two j plus one values
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for each j which means that the rotational
energy levels associated with the diatomic
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molecule for each value of j have two j plus
one way functions associated with them associated
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with each j the all of them have the same
energy given by this formula h bar squared
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by two i j into j plus one therefore the energy
levels are degenerate or two j plus one fold
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degenerate