 Hello, welcome to the lecture number 35 of the course quantum mechanics and molecular spectroscopy. In the last class we talked about the rotations of the polyatomic molecules and discussed the spherical rotors. In this lecture we will talk about the symmetric rotor and a symmetric rotor. Now, in the case of symmetric rotor two of the rotational constants let us call it as I b and I c they are equal and this is not equal to I a. Further we will also consider I a not equal to 0 ok. If I a is equal to 0 then it will of course it will become a linear molecule. Now, in this case there are major two major axis of rotation ok I a and I b of course since they are equal you will not be able to distinguish, but since I b and I c are equal you will not be able to distinguish rotations around I b and I c or axis b and c, but I a will be different. So, in such case you will have total angular momentum j which will be along the axis you know b and c. So, that will have the values 1 2 etcetera. And to represent the rotation along a axis you will have additional count number k which will take the values of j, j minus 1 etcetera, etcetera 0, minus 1, minus 2, etcetera up to j ok. Now, this is something that is already familiar to you. What is that? This is like your orbital angular momentum, rotational angular momentum ML in the case of hydrogen atom ok. It turns out that energy for this will be of course you will see that you know k will have 2 j plus 1 possibilities. In such scenario your I square a by omega a square plus I square b omega b square plus I square c omega c square will be equal to j into j plus 1. In addition this is the total angular momentum apart from that you have I a square omega a square will be equal to k square h bar square. And so your energy total energy will be e j k is equal to half sum over i. And from this what you will get is the following. So, your energy of j k will be equal to b into j j plus 1 minus sorry plus a minus b into k square. So, when you have such an equation you will see it will depend on the value of absolute value of k and plus a square minus k will not. So, similar to the ML values ok. So, the ML value does not even poly atom poly electron atoms ML value does not change the energy only the L value changes the energy. Similarly, in this case the value of k will not change the energy ok. So, now if you had I b equals to I c and not equal to I a and I a is not equal to 0 then you have delta j transitions delta j is equal to j minus j prime. So, going from or j prime or j prime minus j double from going from one value of j to other value of j. Now if you now subtract this delta e will be equal to b j prime j prime plus 1 plus a minus b k square minus b j double prime j double prime plus 1 plus a minus b k square. So, if you take this this will be nothing but 2 b j into j plus 1 when j prime is equal to that is because you know this is the selection rule for j ok j double prime is equal to j plus 1. Now we will see delta e does not have any k component. Within the rigid rotor approximation you will see that the symmetric rotor for symmetric rotors the k has almost no almost not k for the symmetric rotor the k has no contribution to the rotational spectrum. But that is not always true because that is under rigid rotor approximation ok. However if it is not rigid so if there is in the presence of centrifugal distortion if you have centrifugal distortion and let us call the centrifugal distortion as d j d k and of course you know there is always a mixing term that is the d j k. Now in that case your energy is given by of j and k is given by b j into j plus 1 plus a minus b k square minus d j j square plus j plus 1 whole square minus d j k into j into j plus 1 k square minus d k into k to the power of 4 ok. Now if you take delta e between j and j prime j prime and j double prime with delta j is equal to 1 then what we will get is that 2 j plus 1 into b minus 2 d j into j plus 1 square minus d j k ok. Now we will starting see that this will depend on the value of k and the centrifugal distortion constant d j k. So, only when you have the centrifugal distortion or when the molecule is no longer rigid rotor then you will start seeing the effects of the k the rotational constant k. So, what will happen then the values of the values will move up and down depending on the value of k sorry value of d j k. You generally we generally know that when you have rotational we generally know when you have centrifugal distortion because of this the lines come closer and now you have one more quantity. So, learn becomes even more closer because the value of k and that goes as a value of k square. So, which means as the k keeps increasing ok the rotational constants are sorry rotational levels are more and more closely spaced. Now in the case of asymmetric rotor then you will see that I a is not equal to I b is not equal to I c. So, in principle one could have 3 rotational constants. So, we will have j and then we will have k and one will have k prime. So, the analysis of spectrum becomes even more difficult. So, one way to look at is that when you have j and what you do is that you think that k is approximately equal to k prime when k is approximately equal. So, any of the two rotational constants are close to each other than the third one sorry this has to be not equal ok. So, when any two of them are very close to each other then what you can think of is the same treatment in that this approximation one can do the same treatment as a symmetric rotor ok. It is a possibility ok, but this cannot be guaranteed then it becomes even more complicated ok, but one can think of ok then your energy value will depend on you know a minus b and b minus c and c minus. So, this will be much more complicated. However, one can do an approximation when k is almost equal to k prime that means two of the rotational constants are very close to each other and third one is much larger than one can do a treatment which is very similar to the symmetric rotors. But having all this one of the things that you must remember it is the rotation transitions generally divided by j to j prime or j prime to j double prime with delta j is equal to plus minus 1 and delta k is equal to 0. So, this is the selection rule selection rule for the polytomol molecules. We will stop here and continue in the next lecture. Thank you.