 We are now ready to discuss rotational spectroscopy and as we have said earlier rotational spectroscopy is performed in the microwave regime. So, you want to use microwave radiations if you want to study the rotation of molecules. But before I begin maybe it is time when I told you once again something about textbooks. So, most of the discussion that we have performed on quantum chemistry so far are from Atkins physical chemistry. There are portions that we have taken from McQuarrie and Simon forgive me if this spelling is wrong I keep forgetting whether this A is there between M and C or not and also whether there are 2 Rs or 1 because there is something some place called McQuarrie University their spelling and this McQuarrie author spelling are not the same. Simon is easier you can also study from Levine's book and another book that is very nice for quantum chemistry is Engel and Reed at this level. If you want to learn quantum chemistry in a little further detail then you should study the elementary quantum chemistry book by Pilar and there is another book by Just McQuarrie on quantum chemistry which is also excellent. And all these discussions that we are now doing on rotational vibrational spectroscopy you find them in all these books Atkins, McQuarrie and Simon, Levine, Engel and Reed maybe not Pilar and also you could study spectroscopy from the book by Banwell and McCash. This is the introductory book on spectroscopy that everybody studies at the beginning. And then if you are interested in spectroscopy there are many more texts available but we can talk about that later. That being said let us talk about rotation of a simple diatomic molecule HCl. Note we are not talking about H2, why we will see why. So, what we need really is a diatomic molecule with permanent dipole moment and this is a figure that is taken from Banwell and McCash's book to be very honest I do not like this too much because it is always better to do the more rigorous treatment unfortunately in this course there is no scope for it. So, this should be good enough. So, you can understand why it is that rotation couples with radiation by thinking like this. Think of this HCl molecule which has a permanent dipole moment rotating. What happens? The direction of dipole keeps changing constantly and if you plot the vertical component of dipole then what happens? Here it is negative, here it is 0, here it is positive and maximum, then again it is 0, negative and maximum 0 and so on and so forth it goes on like a negative cosine wave or something like that. What is light? Light is an electromagnetic radiation. So, it has electric field and magnetic field with the same frequency as a frequency of light. So, now you can think that this oscillating electric field produced by the rotating molecule can have constructive destructive interference with the oscillating electric field of light and that is the origin of the emission of spectroscopy interaction of radiation with matter. So, this is a very rudimentary hand waving kind of understanding that we may develop at the moment. The important point that come out from here is that it is this molecule better have a permanent dipole moment in order to be microwave active. This rotation is not going to produce any oscillating electric field unless the dipole moment is there in the first place and that is why we did not say hydrogen. H2 is a non-polar molecule. So, no matter how much H2 might rotate it is not going to interact with microwave. Does that mean that we cannot study rotational levels of H2? No, it does not mean that we can find it using something called rotational Raman spectroscopy. Unfortunately, in this course we are not going to get into Raman spectroscopy. Now, so that is the first condition it requires a permanent dipole moment and that immediately opens up several applications at least one of which is perhaps very well known to each other happens in our house out. So, this is the essential condition and HCl is microwave active. Water is also microwave active because water can also be approximated as a dipole right it is a polar molecule. So, even rotation of water can give you microwave activity and that is an important point remember in our discussion of application that is going to come up soon. To build a quantum mechanical description what we use commonly is a rigid rotor model. The assumption here is that during rotation the bond length does not change. Once again it is an application of Born-Oppenheimer approximation. So, let us say this is the molecule we are talking about HCl. This is the bond length separation between the nuclei and we are considering the molecule to be stationary. I mean we are considering the bond length to be same that means the molecule is not vibrating or anything during rotation. So, now see this fine print that you see here are the C here is the center of mass and of course, Cl is much bulkier atom than hydrogen. So, the center of mass will be displaced towards Cl. R0 is the bond length this R1 R2 are there and this is a little problematic model to use to start with because there are two bodies. If you try to write the equation you will have too many terms. So, common approach of something like this again coming from classical mechanics is that you reduce this two body problem to a bond body problem. This is what the one body problem looks like. So, the reduced mass I hope you know what reduced mass is we have discussed it earlier 1 by mu equal to 1 by H 1 by M H plus 1 by M Cl where M H and M Cl are the masses of hydrogen atom and chlorine atom respectively. So, this problem is reduced with a little bit of manipulation which I believe you have studied in rotational dynamics in class 11 physics. This problem is reduced to a single mass a single atom a single particle with mass equal to reduce mass mu rotating around a note this now massless charge less center separation between the center and this mass is R0 the bond length. Now, couple of comments here in HCl what is mu going to be like is it going to be like the mass of hydrogen or is it going to be like the mass of chlorine the answer is is going to be like the mass of the hydrogen. So, it is almost like hydrogen is going around chlorine and chlorine is stationary it is just that since we have used reduced mass we do not need to consider the mass of chlorine anymore. So, the center can be considered to be massless completely. Have we encountered this situation earlier actually we have is it not? When we set up Schrodinger equation for hydrogen we had written the Hamiltonian theta phi this was actually the inherent assumption. So, what we will do now is that we will try to see what the solutions are. In fact, since the Schrodinger equation is similar to that of hydrogen atom angular part the wave functions are also the same as in hydrogen atom spherical harmonics. Since we know that the wave functions are spherical harmonics same as hydrogen atom we can try to work with them what do we know about the spherical harmonics of hydrogen atom. We know that the spherical harmonics we have written them as y which are functions of theta and phi they are eigen functions of the l square operator with eigen value of l into l plus 1 multiplied by h by 2 pi h cross not really h square divided by 4 pi square. Here the difference with hydrogen atom is that this R is constant. So, we can try to find out what is the rotation energy here how remember the relationship between kinetic energy and angular momentum it is l square by 2i is not it. So, if I take this l square operator and divide by 2i what is i i equal to mu R0 square then what do I get I get h square by 4 pi square multiplied by 2i multiplied by l into l plus 1. We will make only one change here we will not write l when you talk about rigid rotor it is conventional to write j as the quantum number instead of l what is j j is the rotational quantum number very very similar or well for all practical purposes same as the azimuthal quantum number in hydrogen atom. So, let us write it a little better this turns out to be h square divided by 8 pi square i into j into j plus 1 I forgot to write this y theta phi here multiplied by y theta phi. What does this mean we said earlier that this here is the kinetic energy operator. So, I can write it as maybe t theta phi operating on y of theta and phi. So, this is what we get an eigenvalue equation this here is the eigenvalue. So, this is kinetic energy it is as simple as that is there a potential energy for rigid rotor no because we have defined it as a one body problem as a mass going around a massless charge less center. So, in that case if the center does not have any mass or charge then there is no question of potential energy that can arise the energy of a rigid rotor is purely kinetic energy. So, it is now so this is what the expression is E j equal to h square by 8 pi square i j into j plus 1 in joule. Now, generally it is conventional to not work in joules when we talk about microwave spectroscopy we like to work in wave numbers. So, how do I write the energy in wave numbers like this this is the energy h by 8 pi square i c how did we get it how did we get this in the first place how do we get energy in centimeter inverse well this thing has to be divided by h c. So, just bring in an h c here what happens this h and this h goes and you have a c left in the denominator that is what gives us the expression actually it is conventional to write epsilon j here you have written E E j equal to h by 8 pi square i c multiplied by j into j plus 1 centimeter inverse where j is 0 1 2 exactly like your subsidiary quantum number of hydrogen atom. So, in short it is written E j equal to b into j into j plus 1 centimeter inverse where b h by 8 pi square i c this here is the rotational constant in centimeter inverse. There is a selection rule I told you earlier that not all transitions take place. So, selection rule without going into the derivation here is delta j equal to plus minus 1. So, what would be delta E for delta j equal to plus 1 you just substitute the values and work out b what is the energy for the jth level b j j plus 1 to know the energy of the j plus 1th level I just have to write j plus 1 in place of j. So, it becomes b into j plus 1 into j plus 2 subtract this from here you get 2 b into j plus 1. So, what does it mean? It means that it means 2 things first of all the energy levels are separated by 2 b into j plus 1 we are going to show you the diagram. Secondly, your rotational spectral lines will also occur at 2 b into j plus 1 of course we get line spectra. So, what do the spectral look like? First let us draw this energy level ladder that we talked about here it is it ok b j j plus 1 right. So, if it is b into j into j plus 1 when j equal to 0 it is 0 j equal to 1 b into j into j plus 1 is 2 b when b equal to 2 then what happens j is 2 j plus 1 is 3 right 2 into 3 6 b so on and so forth. So, as you see the energy gaps keep on increasing as you go higher up the ladder does that ring a bell does that remind you of particle in a box ok. Now, what about the spectrum? Where will the lines be? The first line will be for transition between 0 to 1 that will be at 2 b. Second line will be for transition between 1 and 2 that is 6 b minus 2 b 4 b. Third line might be between 12 b for transition from 6 b to 12 b n equal to j equal to 2 to j equal to 3 level that is 6 b and so on and so forth. So, this is what the spectra look like you get an equi-spaced spectra where spacing is 2 b in each case spacing between 2 successive lines is 2 b of what use is that? Well since they are equally spaced and from the spacing you can find out what the b is you can actually determine the bond length can you? They do go through a maximum why they go through a maximum will come to that right. But even before that let me at least mention that is not it clear that from the spacing of the spectral lines we should be able to find out what the if you know the reduced mass which there are easier ways of finding you can find out what the bond length is we are going to come back to that. Before that let us address this question why? Why is it that the spectrum goes through a maximum? This is why see each rotational level actually is 2 j plus 1 fold degenerate remember these remember those l levels all of them where 2 l plus 1 fold degenerate and as we said we just substitute small l by capital J here. So, you get this angular momentum pointing in specific different orientations and how many such directions are allowed depends very much on which j you are working with. So, g j equal to 2 j plus 1 here we have given you an example of j equal to 2. So, degeneracy is there what happens to degeneracy then as you go higher of the level degeneracy increases and what is Boltzmann distribution? It is g j multiplied by e to the power minus epsilon j by k t. So, epsilon goes up with j. So, e to the power minus epsilon j epsilon here is just that e that we wrote earlier e to the power minus epsilon j by k t will go down with increase in j. So, that is a decreasing function degeneracy is basically a straight line with j. So, multiply a straight line this positive slope by a decreasing function what do you get? You get something that goes to a maximum. This is why rotational spectra goes to a maximum. Homework for you is find out an expression for j in which population is maximum. I will just discuss how to do it this is how you do it essentially you take d n j d j equal to 0. At maximum the first derivative will be equal to 0. So, take a derivative of this what is this 2 j plus 1 multiplied by e to the power minus here I will write j minus e j divided by k t this is equal to 0. So, what are the solutions? Is this right what I have written not really. So, this d d j of this is equal to 0. Now you can do this differentiation find out what you will get and what you will get always is you will get an expression with 2 terms. I will not tell you what the 2 terms are but you substitute you will get j equal to may be 10.23. So, what does that mean? It means that see here we are using calculus which inherently requires continuous variation we do not have continuous variation here we actually have discretization. So, that is why this point 23 or whatever it is it comes you have to neglect this point just work with the whole number that comes out. You can take the nearest integer that is going to be the value of j where the population is going to be maximum. Please do it yourself if you have any difficulty it is worked out in Atkins book as well as Manuel and Maccash's book. So, what is the information that we find I have played a spoilsport and we have already told you that when you look at this spectrum from the spacing you know what to be is that is going to give you bond length because I after all is mu R0 square if you know mu from here you find B is a very elegant easy method of finding bond length what kind of molecules only those molecules that are dipolar H2 cannot be done this way you need Raman spectroscopy. Now, in the last lecture we talked about what Planck had said about experiment. So, you cannot close it without showing you an actual spectrum this here is a spectrum of carbon monoxide rotational spectrum you can see more or less equi-space and here they are plotted transformation that is why it is negative going but you can understand that if you work out the spacing here between any two lines you can knowing the masses of carbon and oxygen should be able to work out the bond length. So, this is one of the applications but there are others. Second application is that commonplace thing that we are talking about microwave oven why is it that food gets hot if you place it in a microwave oven why is it that food gets hot when you place it inside a microwave oven because unless you are somebody who eats nails and stones for dinner most of the food that we eat is 70 percent water. So, when you place it in microwave oven microwave of the frequency that is broadly absorbed by water is incident on your food and it actually goes through other glass and all container. So, all this water inside your food starts rotating it gets energized but they cannot keep rotating always they have to stop when they come to stop that is because of dielectric friction and that is when this rotational energy they are acquired is given to the surroundings that is the food in the form of heat and since there is 70 percent water it is sort of like fire within that is why microwave is a very efficient way of heating up food or cooking it. Mind this you cannot fry anything in a microwave because to fry you need to go beyond 100 degree centigrade you need to get your oil heated oil is non-polar it is never going to heat up you can cook but you cannot fry in microwave. Astrophysics is another area where microwave spectroscopy has a lot of application look at microwave radiation and from there you can tell which molecule is there in that the environment of that particular star or even planet I think and it is used for things like fault analysis in analysis of materials. So, microwave spectroscopy is very useful not only for chemists or physicists but also for engineers and people working in many other fields.