 Welcome to the lecture number 29 of the course, Quantum Mechanics and Molecular Spectroscopy. In the last class, we were looking at the rotational transitions. One of the features of the rotational transitions is the spacing between the rotational lines. So, I showed you that the rotational lines are evenly spaced by 2 h b e, 2 h b e, 2 h b e. So, this is from, so let us call it as 2 h b e, 4 h b e, 6 h b e, 8 h b e. This is from 0 to 1, this is from 1 to 2, this is from 2 to 3 and this is from 3 to 4. This is the values of j, j is equal to 0 to j is equal to 1, j is equal to 1 to j is equal to 2, etc. We know that h b e is nothing but h bar square by Re square. When I multiply 2 h b e, this will come out to be h bar square by mu Re square. So, if 2 h b e can be experimentally measured, then based on this one can estimate Re square, okay. So, 2 h b e will give us Re square in time, we can get Re. Now, what was Re? Re is nothing but if you had 2, if you had a molecule ab, Re was the equilibrium distance. So, by measuring the rotational spectrum, one can measure the equilibrium bond distance of molecule ab. So, rotational spectrum gives rise to equilibrium, which is nothing but your R e. So, rotational spectrum or rotational spectroscopy is the only spectroscopic method by which one can estimate the bond distances, that is geometrical parameters, okay. No other spectroscopic technique can measure bond distances as the rotation spectra can do. So, that is the power of the rotational spectroscopy, okay. However, one must always remember that this is only possible if mu naught is not equal to 0. That means, you can only measure the bond distances of hetrodiatomic molecules or molecules that have permanent dipole moment, okay. Now, one of the problems of rotational spectroscopy is something called centrifugal distortion, okay. Now, what is centrifugal distortion, okay? Now, let us suppose if you have a molecule ab with a distance Re and it is rotational energy, E rot is given by H BE into J into J plus 1. And we see that as E rotation is H BE into J plus J plus 1 and it turns out that as the BE decreases and we see BE is equal to H BE is equal to H bar square by 2 mu Re square. So, BE is inversely proportional to 1 over Re square or that is what it is. Or Re is inversely proportional to 1 over square root of BE. That means, as the equilibrium bond length increases, the rotational constant goes down. So, longer bonds have smaller rotational constants. Longer bonds have rotational constants. So, this is something that you must remember, okay. As if I take AB, okay, the same molecule and keep increasing its bond length, the rotational constant keeps going down, okay. Now, fortunately or unfortunately molecules are not rigid rotors. That means that molecules rigid. That means you cannot hold molecule AB such that the position of A and B are at the, are locked up, okay. They are going to move, okay. So, one of the things, of course, they vibrate, but the rotational motion and vibrational motion have different time periods, okay. So, we will not able to, we will not consider this. What is that, what does it mean? What does that mean? Vibrational time period is far smaller than the rotational time period. So, since there is a mismatch in time periods, the effect of vibrations on rotations is minimal and one can ignore that. So, molecules are not rigid, but rotational and vibrational time periods much different. How different? For example, rotational time periods are picoseconds and vibrational time periods are femtoseconds. So, there are three orders of magnitude different, okay. Therefore, the coupling between the rotational or influence of vibrations on rotations can be neglected, okay. This is a very general statement, okay. There will be cases when you cannot neglect it, okay, but that is kind of advanced level spectroscopy, but at a level that this course is intended, one can always assume that the rotational motion and the vibrational motion are independent of each other, okay. Now, if that is the case, if vibration is not influence the rotation, what else can that? Now, think of it like this, if you have an object AB or a molecule AB such that this is RE and you want to rotate it, okay. So, you rotate. Then what happens, there is something called centrifugal force, which kind of throws away A and B away from each other. So, what happens is that A and B start moving away from each other and this is called centrifugal distortion. One starts getting centrifugal distortion and you will see the faster you rotate, the more will be centrifugal distortion. Of course, you will realize that for J is equal to 0 state, J is equal to 0 state is essentially no vibrational energy is 0. That means for J is equal to 0, so there will be no centrifugal distortion, okay. So, as the J states keep increasing, the centrifugal distortion becomes more and more effective, okay. Now, what is that it is doing? It is increasing the bond length. So, your AB is no longer RE, but something else slightly more than RE and we know as the as the distance increases, the rotational constant goes down. So, effectively when the rotational constant goes down, the delta E goes down. So, therefore, EJ rotation will now be equal to HPE J into J plus 1 minus HDE J square into J plus 1 square. So, essentially what you are doing is you are writing a series. This is the first term that is J into J plus 1 that is a linear in of course, this is linear in J and this is quadratic in J. So, you are trying to add up more terms, okay. So, essentially what you have done is nothing but J HBE J into J plus 1 minus HDE J into J plus 1 whole square. So, this is for J into J plus 1, this is a linear in J into J plus 1 and this is quadratic in J into J plus 1, okay. So, this is just the first correction. Then you can have the third order correction, etc. But it turns out that the second order term or this DE itself is so small, do not have to worry about the terms later on. Now, this is what you get. And you can see that this is subtraction that means as you keep increasing the DE, okay by the way DE is called centrifugal distortion. Now, as you keep increasing DE or as you keep increasing J value, the DE becomes more and more effective. So, now let us look at and all having done this the way nature of the wave function has not really changed. You are still using spherical harmonics. So, delta J is equal to plus minus 1 and delta M is equal to 0. These selection rules are still valid. However, what you are changing is the energy spacing. So, what you are doing is you are changing the energy spacing. Now, when you did not have the rotation energy E rot is given by H BE J into J plus 1. If we have centrifugal distortion in the presence of centrifugal distortion. So, your E rotation will now be given by H BE J into J plus 1 minus H DE J square into J plus 1 square. So, in general, if you look at the transitions of the rotations, then one can think of 0 to 1, 1 to 2, 2 to 3. This will be 2 H BE, this will be 4 H BE, this will be 6 H BE and separation will be 2 H BE where this is your energy, okay. Now, if you have centrifugal distortion, so in the sense of centrifugal distortion, what you will have your 0 to 1 delta E will be equal to H BE 1 into 2 minus H DE 1 square into 2 square minus H BE 0 1 minus H DE 0 square into 1 square. So, this term essentially is going to be 0. So, your delta E 0 to 1 will be equal to 2 H BE minus 4 H DE. Now, for 1 to 2 delta E will be equal to H BE 2 into 3 minus H DE 2 square into 3 square minus H BE 1 into 2 minus H DE 1 square into 2 square. So, this will be nothing but 6 H BE minus 2 square is 4, 3 square is 9, 4 into 9 is 36, 36 H DE minus 2 H BE minus 4 H DE. So, this will turn out to be 4 H BE minus 32 H DE. Similarly, when you have 2 to 3 transition delta E that will be equal to is equal to H BE 3 into 4 minus H DE 3 square into 4 square minus H BE 2 into 3 minus H DE 2 square into 3 square. So, this will be equal to 12 H BE minus 3, 3 square is 9, 4 square is 16. So, 144 144 H DE minus H BE 6 minus 36 H DE. So, this will be 12 minus 66 6 H BE minus 144 36 that is equal to 108. So, this will be 108 H DE. Now, if you start plotting the spectrum earlier what we had 0 to 1, 1 to 2, 2 to 3. So, this for 2 H BE, 4 H BE, 6 H BE. Now, it will transform to 2 H BE 0 to 1 minus 4 H DE. Second one will be 1 to 2 that will be 4 H BE minus 32 H DE. Third one will be 2 to 3 this will be 6 H BE minus 10 to 8 H DE. So, now the lines are becoming narrower. So, this was just 2 H BE separation was 2 H BE. Now, in this case this will be 2 H BE minus 4 minus 32 is 28 H DE and this will be 2 H BE minus 108 minus 32. So, 76 minus 76 H DE. So, the lines are getting closer. So, if you keep moving to the higher values of J, the lines will get closer and closer. Generally, the BE and DE BE is about 10 power 4 times larger than DE. So, DE is approximately equal to BE into 10 power minus 4. So, 10,000 more. So, that means, when you start multiplying this number at least to a value of that order of BE it will start getting affected. That means, J into J plus 1 square should be of the order of 10,000 or at least of the order let us say even 10 percent of it should be of the order of 100. Only then it will start affecting the the rotational line spacing. That means, we have to go to very high values of J. What high values of J? The values of J that are at least in double digits. So, double digit J values will start affecting the rotational spacing or rotational line spacing. Okay. So, for large values of J, the lines become closer and closer and that is the indication of saying the presence of centrifugal distortion. We will stop it here and continue in the next class. Thank you.