 Welcome to the lecture number 38 of the course quantum mechanics and molecular spectroscopy. In the last class we solved couple of problems based on the rotational spectroscopy. This here in this class I will look at a problem on the vibrational spectroscopy and some aspects of the electronic transitions. In your vibrational spectrum okay one uses harmonic oscillator as a so we have talked about when we use harmonic oscillator because at the bottom of the potential one can always have the harmonic oscillator and the wave functions look like this. I told you that you know the potential k is given by d square v by dr square or dr square evaluated r naught. So if the r naught is your equilibrium distance. So this is nothing but the second derivative of the potential it evaluated at the equilibrium geometry okay and the bottom of the well one can always consider as a harmonic potential. Now when you have a harmonic potential the frequency v naught is given by by the way harmonic oscillator has only one frequency the fundamental frequency v naught that is given by 1 over 2 pi square root of k by mu okay. Now if I can measure v naught okay I will be able to get the force constant. Now force constant is nothing but it is the stiffness of the bond. What is stiffness of the it is like a stiffness of a spring whether the spring is strong or whether spring is weak okay. So force constant tells you whether the bond is strong bond or a weak bond. Now let us apply this concept to HCL again 1 and 35 and DCL 2 and 35 okay. When I measure the IR spectrum of this what I get is a fundamental frequency which is given by for this value this is nu naught bar because I am measuring in the centimeter inverse that is 2885.1 centimeter and for this it will be this will be 2091.1 centimeter inverse. So I know the value of nu and I also know nothing but your nu or v naught nu naught. So let us me call this as a nu call this as nu naught frequency. So nu naught is nothing but equal to nu naught bar into C that is just the transformation between the velocity and but you see since we are measuring in centimeter this must be measured in centimeters per second. So that will be nothing but 2.997 into 10 per 10 centimeters per second this is per second so this will be centimeter inverse so that will be nothing but second inverse. So you have that. Now if I now want to calculate this so nu naught so when I have what I have given is nu naught is equal to 1 over 2 by square root of k by mu. So I can slightly rewrite this equation. So this will be nothing but nu naught is nothing but nu bar into C. So nu bar into C is equal to 1 over 2 by square root of k by mu. So this I can always write as square root of k by mu is equal to 2 by nu naught bar C. Now I can take a square that implies k equals to 2 pi nu bar C into mu square. So I need to get that. Now I already know nu bar. So for nu bar or nu naught bar for C L 35 is 2885.1 centimeter inverse and nu naught bar for T C L is 291.1 centimeter inverse. I think I made a mistake this should not be like this. This should be square of this because I am taking a square. Sorry for that mistake. So now k will be nothing but 2 pi C nu naught bar whole square into mu. Now this will be nothing but 2 pi square is nothing but 4 pi square C square nu naught bar square into mu. So this will be nothing but 4 pi square C square is nothing but 2.997 into 10 power 10. I told you we have to take centimeters per second per second as a speed of light whole square into nu bar square is 2885.1 square into mu. Mu is nothing but for H C L H 1 35 C L mu will be nothing but 35 divided by 36 into 1.66 into 10 power minus 27 kg. So if I solve this I will get value of 476.4 Newton per meter. Similarly, if I take k is equal to 2 pi C nu naught bar square into mu but now it is 2 H C L 35. If I plug it in 4 pi square C square will be same 2.997 into 10 power 10 square. Now this will be 2885 in the case of H C L but this will be 2091.1 square because the frequency has changed into even the reduced mass has changed. So this will be 70 divided by 37 into 1.66 into 10 power minus 27. Now when I do this I will get 486.4 Newton per meter. Okay. Now you see the force constant of H C L and D C L is actually not exactly equal even though one would expect that the force constant should be equal. Now of course there is an assumption that the frequency everything else are fundamental constants nothing has changed except this frequencies are one which are measured experimentally. Okay. So if these if one assumes that the frequencies that you have measured are accurate then I will see that the force current of H C L is marginally different than force current of D C L. It is about by 10 Newton per meter. Okay. The stiffness of bonds is when you do isotopic substitution may not be exactly the same as exactly the same. Okay. So when you measure the vibrational frequency the moral of the story that I am going to tell you is that when you measure the vibrational frequencies you can convert the vibrational frequencies into the stiffness of the bond. Okay. One can say the bond is stronger or bond is weaker or you know something like the stiffness of the bond. Okay. By this token you can tell that the D C L bond is more stronger or more stiffer than the H C L bond. So the vibrational spectroscopy tells you about the stiffness of the bond or the strength of the bond but the bond distances are measured by the rotational spectroscopy which I talked about in the last lecture. There is one more factor that I want to teach you is something called related to electronic transitions. Now if I go back few lectures you will see the electronic transitions TMI is given by psi electronic excited state mu E psi electronic ground state multiplied by chi nuclear excited state multiplied by chi nuclear ground state. So let me define what it is psi prime E is wave function of the excited state. Of course this has to be electronic wave function psi E double prime is electronic wave function of the ground state mu E is the electronic dipole moment and chi N prime is the nuclear wave function of the excited state chi double prime is the nuclear wave function. Okay. There is one more thing that I have to tell you is that to get to this TMI we assume that mu naught is nuclear is approximately equal to zero. Okay. This is condon approximation which says that the nuclear dipole moment is does not change related to the electronic coordinates or changes very little with respect to the nuclear electronic coordinates. So this is the condon approximation. So to get this TMI we have used the condon approximation. Now this will tell you whether the transition is going to be allowed or not. But one of the important factors is this okay is the overlap so chi N prime chi N double prime. This is nothing but overlap of the nuclear wave functions of the ground and the excited states. But probability of transition P is equal to modulus of TMI square. So when I use that the probability must be proportional to modulus of chi N prime chi N double prime square. And this value is called Franck-Condon factor. Franck-Condon factor is nothing but the overlap of the nuclear wave functions of the ground and the excited states square of it. Okay. Now there is one small theorem that I want to prove is that. Now let us suppose chi N prime is a wave functions okay. Let me call it some quantum or let me give some quantum number F because final now excited state okay. So wave functions of the excited state nuclear Schrodinger equation and chi double prime N i is given by or given by wave functions of the ground state nuclear Schrodinger equation. Okay. Now because these are the solutions of Schrodinger equation they should form complete set. So you have a complete set chi prime N with quantum number F and forms complete set. Similarly F chi double prime N i also forms a complete set. Now there is something that you must always remember is that whenever I have a complete set okay. Any other wave function can always be written as a linear combination of those complete set wave functions. So which means I can always write chi double prime N F should be equal to sigma over let us say k C k chi N and I can always write sorry this is i F is equal to sigma over k let us say j C j chi N j double prime. So one can always express the nuclear wave function of the excited states as a linear combination of the ground state wave functions and one can always express the nuclear wave function of the ground state as a linear combination of the excited state wave functions. Now if I have any if I take chi ground state nuclear wave function i this wave function to be equal to sum over j chi C j N j. Now if I assume that all functions are orthonormal functions are then it terms out that C j square modulus of j should be equal to 1 because total coefficients should add up to 1 okay. So this is the let us look at chi prime N let us say somewhere function F integral chi double prime. So I want to look at this. So this can I can always write it sum over chi N F but this wave function I can always write it as this. So this is equal to sum over j chi prime N j C j okay. Now whenever this j will become equal to F whenever so this value of the integral will be will be equal to C j delta F j. So whenever F and j are not equal then this overlap integral will go to 0 otherwise it will survive and become C j. Now this is only for one wave function now if I want to take over all the wave all the final wave function if I f chi prime N comma F sigma over j chi prime N j C j this will be nothing but sigma over F each of the time will get j okay but j is equal to F okay. So if I get F so what I will only survive when F is equal to say what we will get is C F. Now I do not want this what I want is the modulus of chi N F F chi double prime N i square sigma over F this will be nothing but sigma over F C F actually not like this because you have to individually square them not the sum of the square so square of this. So this is nothing but sum over F C F square so we know that when they are orthogonal and F or j or anything they are all dummy indices should be equal to 1. So the moral of the story is when you have an electronic transient sum of the Franck-Condon factors must be equal to 1 which simply means that when I have a probability P that is equal to and then I have probability P for electronic transition going from a ground state to excited state this must be equal to square of the tension moment integral square. So this will be nothing but psi E excited state mu E psi E ground state square multiplied by chi nuclear prime chi nuclear double prime square but this is only for one wave function so if you have to look at all the nuclear wave function so then it should be sum over all the nuclear wave functions F then but this sum over all nuclear wave function is equal to 1 so that is nothing but psi E because this is equal to 1. So if you take the electronic transients over various nuclear states vibrations and rotations and if you add all of them it should be equal to get equal to 1 so the the contribution of the nuclear wave function is only partial every nuclear wave function is only partial it can go from 1 to 0 if only if it is 1 then there are other do not others do not contribute it is 0 that would not contribute everything else it is contribute. So the sum of all the nuclear wave functions overlap must be equal to 1 okay so that is the factor. So what is the sum of Franck-Condon factors for an electronic transition should be equal to 1. Now see the equilibrium distance r0 in the ground this is ground state this is excited state they are equal so what happens the transition from v is equal to 0 to v is equal to v prime is equal to 0 this is going to be most overlapping but instead of that if you have something like that where r0 is the same so if you make a vertical transition then so v the lowest energy transition will not have the overlap overlap will be the next vibrational level not the v is equal to 0 but v is equal to 1 and sorry not v is equal to 1 sorry 0 but v is equal to 1 and v is equal to 2. So the transition will shift so based on whether the transition will be shifted or not shifted one can tell whether the geometry has changed or not the diameter changes is largest larger than the the more and more the geometry changes the more and more the shift in the transition will happen okay. So by the way this is called a vertical transition and the principle that governs is called Franck-Condon principle we will stop it here and this brings to the end of my lecture series in my discourse we have covered in this course how the transition moment integral comes by and use the transition integral to get to various various spectroscopic techniques rotational vibrational and the electronic transitions. You can always give me a feedback about you about this course and I am always reachable on my email address that are even when this course is not running you can always write to me on nourishchem.itb.ac.in my email address at the department of chemistry IIT Bombay to reach out on any questions that are related to this course. Thank you very much.