# 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