 We have learnt about microwave spectroscopy and that has got to do with rotation of molecules. Now, we will try to see what information vibration of molecules can give us and vibrations if you might remember they are in infrared regime energy of vibrations of molecules is typically in the infrared regime. So, vibrational spectroscopy is also called IR spectroscopy and again this is this is a spoiler IR spectroscopy gives us an idea of bond strength. Rotational spectroscopy, microwave spectroscopy gave us an idea of bond length. Here we are going to get an idea of bond strength. So, again let us work with the same molecule say HCl. HCl is a diatomic molecule with a permanent dipole moment. So, what happens when it vibrates? When it vibrates essentially the dipole moment keeps changing. So, now you can think if you now plot the dipole moment along this direction what will happen? It will go up then go down and then go up again. So, again you get an oscillation kind of behaviour, oscillatory behaviour, but the graph will look a little different from what it did earlier because there is not going to be any minus sign because dipole moment will always point in one direction. It will be large dipole moment well something like this keep on increasing become large then keep on decreasing to that value keep on increasing to that. So, that kind of a variation. So, maybe sin square theta kind of sin square x kind of variation will be there. All right. So, the basic assumption in formulating the quantum mechanical treatment of HCl to start with at least is that the bond between two atoms behaves like a spring and not just any spring a spring that is well behaved a spring that follows Hooke's law. Hooke's law you must have studied is such where the force restoring force is proportional to the displacement. And it is always towards the centre that is why you get this minus sign and potential energy that comes here now for the first time we are encountering potential energy after hydrogen atom. For hydrogen atom molecules we did have potential energy, but here in rotations there was no potential energy for vibrations potential energy is there and that potential energy is a parabolic potential right half into k into R minus R Q is the equilibrium distance or equilibrium bond length. So, deviation from that and square of that that gives you the parabolic potential associated with harmonic oscillators. So, harmonic oscillator is the approximation we are going to use at least to start with. The question is this is a valid approximation to answer that we have plotted this I hope you are familiar with this shape now this is what we obtained when we talked about say dihydrogen molecule when two atoms very far away from each other energy of interaction is negligible then they come close, close, close energy keeps on increasing and decreasing stability keeps on increasing goes to a minimum value at equilibrium bond length and then for further approach there is a sharp increase in potential energy. So, this really is the potential energy surface which is approximately can be modeled by something called the Morse potential and this is definitely not parabolic, but if I draw a parabola on it you can see that in this region for small displacement the parabola more or less holds up to here then there is a deviation here also then there is a deviation. So, if you are content working with small vibrational energies, small displacements then the parabolic potential might work. We are starting with that kind of a hope. So, we are starting with the hope that good it is a good approximation simple harmonic oscillator for small displacements. So, this is our parabolic potential and we can plot it like this. Now, we will write Schrodinger equation as usual what is it? It is minus h cross square by 2 mu del square plus half k x square remember once again we are using mu because it is a 2 body problem h cl right this is h this is cl, but we do not like 2 body problems more difficult to handle. So, we reduce it to a 1 body problem and when we do that the mass that we need to consider is neither the mass of h nor the mass of cl, but rather the mass the reduced mass 1 by mu equal to 1 by m 1 plus 1 by m 2. So, minus h cross square by 2 mu del square this gives you the potential energy half k x square as we have discussed already is the potential energy minus h cross square by 2 mu del square gives you the kinetic energy term of the Hamiltonian. So, the Hamiltonian is there we have written the equation one can solve it. So, the boundary condition that we have to use here remember it is boundary condition that produces your quantization. So, the boundary condition that we have to use here is psi equal to 0 at x equal to plus minus infinity not at this surface right not at this surface right what is this surface we have just plotted this parabolic potential for this value of x what is the value of v there is no reason to think that at this point the wave function will become 0 it actually becomes 0 at x equal to plus minus infinity. So, using that boundary condition what you get is first of all you get quantization you get different energy levels and you get this kind of wave function. Do they look like particle in a box wave functions partially yes they do look like sine wave site and even particle wave function particle in a box wave function you had no node for the lowest energy one node in the next one two in the next and so on and so forth. But these are not really sine waves. So, these are something like this let me just write it here psi v is equal to normalization constant n v multiplied by e to the power minus for now I just write k and here I will just write k x square since we are just trying to give you an idea this is a Gaussian function and this is multiplied by what is called a Hermite polynomial in x. What is the Hermite polynomial please refer to our lecture series we are not going to delve further here unfortunately in this course you have to believe me when I say that this is the form and when you plot this this kind of a function is obtained it becomes 0 only at plus minus infinity. So, you can see some wave function outside the area that where we have drawn the potential energy surface also. This is the quantum mechanical description of simple harmonic oscillator quantization is there wave functions are there the only thing that remains to be told at this point is what are what is the expression for the energies this is the expression for energies e v equal to v plus half h cross omega h cross you know h by 2 pi omega is the angular frequency of vibration v is the vibrational quantum number goes from 0 1 2 3 so on and so forth. Now we have an interesting situation if you put v equal to 0 what is e v going to be it will not be 0 it is going to be half h cross omega. So, this energy is called 0 point energy what it essentially says is if it was a classical simple harmonic oscillator it could have gone down here to the bottom of the well and it could have had no energy at all which means it would not vibrate however for a quantum harmonic oscillator it is not possible to have 0 energy the oscillator is going to oscillate even at 0 Kelvin that is why it is called 0 point energy why is that so that is so because otherwise you know uncertainty principle will be violated. So, we have a quantum harmonic oscillator fine so it is vibrating let us say I keep on decreasing temperature and it stops it drops to the lowest possible level vibration energy goes down down down and then it stops now we know the separation r 0 that means you know position we know the momentum exactly it is 0. So, position and momentum are both determined accurately that violates uncertainty principle right. So, in order to ensure that it does not violate uncertainty principle poor quantum harmonic oscillator has to oscillate even at 0 Kelvin with 0 point energy. In fact, at room temperature only this level is populated because the separation is something like 1000 centimeter inverse if you use Boltzmann distribution we will find that population of this even your v equal to 1 state is going to be 0.008 or something only 8 will be in v equal to 1 state and it will fall off rapidly. So, if the ratio is 1000 to 8 what will the ratio 8 is to something p it will be practically 0. So, only the v equal to 0 level is populated at round state for all practical purposes. So, all transitions actually originate from there that is very important to understand. Alright that being said there is more interesting stuff coming up we know that angular frequency for a simple harmonic oscillator is root over k by mu k is force constant and mu is reduced mass. So, if you can somehow determine omega then we should be able to determine force constant provided I know which atoms I am working with if I know which atoms I am working with I know the masses. How do I find this let us see what is the energy gap between two successive levels e v plus 1 minus e v do the math very simply write v plus 1 here v and v will cancel 1 minus half is half so you left with h cross omega. So, half will also cancel since I made a mistake I will write it e v plus 1 is equal to v plus 1 plus half multiplied by h cross omega. So, e v plus 1 minus e v is equal to well v and v cancel half and half cancel you are left with h cross omega what is this omega remember it is the angular frequency of vibration and if you think why I am saying this again you will see very soon why angular frequency of vibration. So, instead of writing h cross omega I can write h omega by 2 pi fair enough what is omega by 2 pi omega by 2 pi is nu. So, this becomes e v plus 1 minus e v is equal to h nu what is nu nu is the frequency of vibration I might as well write like this, but now we are in an interesting situation what is there on the left hand side e v plus 1 minus e v that is delta e do you remember what delta e is we have written that it is called Bohr resonance condition that is equal to h nu light. So, what it turns out is that this h nu light is equal to h nu vibration h and h cancel you are left with nu light is equal to nu vibration. So, a simple harmonic oscillator would absorb only that light well for delta v equal to one kind of transition which is the only transition that is allowed it is going to absorb that light which whose frequency is exactly equal to the vibrational frequency that is something that I find to be very amusing. So, you can write like this vibrational frequency is 1 by 2 pi root over k by mu we know that. So, that is equal to frequency of light as well and it is conventional in vibrational spectroscopy like in rotational spectroscopy to work with not frequency, but wave number. So, nu bar becomes 1 by 2 pi c multiplied by root over k by mu. So, this here is an expression of light that would be absorbed for a v to v plus 1 transition and that has k in the numerator mu in the denominator. If you know mu you can easily find the force constant k. What would the force constant k tell us about? It tells us about how strong the bond is. So, it can determine bond strength. What would the spectrum look like? Actually it would look like this one single line of course, there would be some width and all, but for delta v equal to 1 even if the transitions originated earlier it does not matter it will always be just h nu wave. So, a single line is expected in the spectrum of simple harmonic oscillator more so, because in any case all transitions begin from v equal to 0 at room temperature. The selection rule is delta v equal to 1. How do we get it? Please refer to our spectroscopy lectures. As usual we will not close the discussion without showing you an actual spectrum. This is the expectation one single line. This is the reality if you use a good enough IR spectrometer a high resolution IR spectrometer instead of one band you get well two branches of lines. Where do these branches arise from? These branches arise from the fact that every vibrational level is associated with rotational levels. Now you might remember here that we said that rotational levels are such that the higher levels are populated. Why are they populated? Because the energy gap is so small. Higher vibrational levels are not populated. Here we have written n this is not populated much because the energy gap is too large. But for n equal to 0 all these levels will be populated in the same ratio as is predicted by Boltzmann distribution. So, now upward transitions can actually begin from any of these levels. So, you get arrows of different length and without going into further detail let us just believe this again that the selection rules are delta v equal to 1 and delta j equal to plus minus 1 meaning if you start from a j value let us say I cannot make out which j it is let us say it is j equal to something 3. So, from here you will be able to go to only particular j values associated with the v equal to 1 level. Now I recommend that you work out the expression of energies associated with this kind of transition from Banwell and McCash's book. I am not doing it here because it is very very simple algebra for those who are going to take this as an NPTEL course if you want we can do it in the interactive session others if there is any problem please write me an email or something and we will do it. So, we get two lines these are called the P branch and the R branch as you might notice that there is nothing at this at the center where the line was expected to be that is because you have delta v equal to 1 delta j equal to plus minus 1 delta j equal to 0 is not there that is why you get this kind of an expression and this kind of a line. So, from one spectrum if your spectrometer is good enough you can get an idea of IR which gives you an idea of bond strength as well as you can get an idea of separation of rotational levels the separation is going to be 2B once again. So, you can get an idea of bond length and bond strength from the same experiment if your IR spectrometer has high enough resolution. So, one more thing if you zoom into one of these lines unfortunately it is flipped here for whatever reason you see that each line is actually a doublet there are 2 lines associated with it why go back to class 8 level chemistry remember chlorine occurs in nature as 2 isotopes chlorine 35 and chlorine 37 atomic weight of 35.5 is just an average. So, when you do the experiment you actually sample individual molecules. So, they are either HCl with chlorine 35 or HCl with chlorine 37 and here you get 2 different lines from for chlorine 37 and chlorine 35 the isotope effect shows up very very nicely in this high resolution what is called row vibration on spectrum as well. So, we get an idea of isotope abundance isotope abundance relative abundance of isotope we get an idea of relative abundance of isotopes as well from high resolution IR spectra which are often called row vibrational spectrum rotation vibration right. So, row vibrational spectrum so much for vibration and rotation next we are going to go into electronic spectroscopy.