Lecture 15 :Microwave spectra of Diatomic molecules


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

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