 to the lecture number 21 of the course Quantum Mechanics and Molecular Spectroscopy. We will have a quick recap of the previous lecture and proceed with the present lecture. In the last lecture I was talking about rates of emission and absorption and their coefficients ok. So, when you have state 1 and 2 ok there is an absorption and absorption happens in the presence of h nu and there are two kinds of emission ok. One that happens in the presence of h nu is called stimulated emission ok and one this happens in presence of h nu and there is a spontaneous emission which happens in the absence of h nu ok. Now, we have the coefficients or rate constants and this is called B this is also called B and this is A and using our transient moment integral or transient dipole moment we figured out the relationship between A B and also A and the transient moment integral. So, the final equations that we came in the last lecture was A B equals to pi by 6 epsilon naught h bar square transition moment integral f mu mu z dot electric field along the z axis i or this could square or equivalently one could write pi 6 epsilon naught h bar square 2 mu z along the electric field 1 whole square and A was nothing, but 8 pi h nu cube C cube into P. So, this will turn out to be pi by 6 epsilon naught h bar square into 8 pi h nu cube pi C cube into 2 mu z along some electric vector 1 whole square. Now, this is nothing, but 2 pi mu cube divided by 3 epsilon naught h bar C cube mod plus of 2 mu z electric field vector 1 whole square. So, this is A. Now, we also said that both A and B are proportional to this transient dipole. Now, one of the most interesting part of spectroscopy using this methodology is that you can get both A and B theoretically as well as experimentally. One can measure A and B experimentally and calculate theoretically and one can compare the experiment and theory and that is the real test of quantum mechanics. For example, let us suppose there is a there are two states 1 and 2. Initially, there is a population of n 1 at time t is equal to 0 ok. Some thing I do not know some value and there is n 2 at time t is equal to 0, but if you start with a separation of very large separation of 1 and 2 then you can see that n 2 0 will be equal to 0 and let us call this as 100 percent ok and 0 percent something like that ok. And then what you do is you shane light at some point of time you shane light h mu corresponding to delta E that is equal to E 1 sorry E 2 minus E 1. Then what will happen? You will transfer some population let us say some population x percent has been moved to the x axis. So, what you get is at time t this will be 100 minus x and n 2 of t will be equal to x. Now, after that you switch off the switch off light. Now of course, the molecules will not be able to stay there in n 2 for all the while because thermodynamically n 1 is the most stable state. So, the molecules must return to n 1 state. So, the kinetics of that will be minus d n 2 by d t is going to be proportional to n 2 and this is happening in the absence of any light. So, the proportionality constant must be equal to the a coefficient. So, that is a. So, I can rewrite this equation and it turns out that n 2 of t is equal to it is a simple first order differential equation which you can solve and you get that n 2 of t is equal to 0 is equal to n 2. Now, I am going to use n 2 at t prime exponential minus a times t. Now, what is n 2 at t prime? n 2 at t prime is actually equal to x that is when the light interaction is complete. So, if you switch on the light and wait for a while and t prime is the state when you switch off the light or interactions complete or when you switch off the light, t prime is the time when light is switched off ok. So, what you will get is this equation ok. Now, this is an exponential you can see this. So, what it does is the following. So, you can think of it like this time n n 2 of t. So, it will have some value and then it exponentially decays ok. Now, that is this value will be nothing, but n 2 of t prime ok. It starts at t prime because that is when and it goes to infinity and this is this is asymptotically going to 0 ok. Now, if you have that this is a first order equation. So, your rate constant will be a. So, a is the rate constant. Now, if a is rate constant 1 by a is called lifetime ok. We will just come to that. So, let us just look at the first order kinetics ok. So, if you have a population that decays exponentially ok. So, this is let us say this is n 2 of t versus t and this is n 2 of t prime and I am starting time at t prime ok. Now, if it is exponentially decay I told you that the rate constant is equal to a and the equation is this n 2 of t is equal to n 2 of t prime exponentials minus a. Now, let us assume one thing. So, when t equals to a when t equals to say 1 by a ok t is equal to 1 by a and this I will call it as tau. So, when t is equal to tau its value is 1 by a and what happens then you will have 1 2 into t is equal to n 2 t prime ok. This will become a into 1 by a. So, that will become that that will become minus 1. So, this will be e to the power of minus 1. So, this is equal to 1 by e. So, this is called lifetime. Lifetime is when lifetime is a time tau is called lifetime and lifetime is a time when initial population n 2 of t prime ok decreases to 1 over e of n 2 of t prime ok. So, if you look at that. So, for a given exponential the lifetime is fixed that is 1 over e ok and the rate constant in this case the rate constant is equal to a. Therefore, lifetime will be equal to 1 over a. So, it is also called radiative lifetime tau equals to. So, tau now equals to 1 by a this is equal to now what is a equal to a was equal to some constants multiplied by the transient moment integral or transient dipole square ok. So, some constants multiplied by 1 over 2 views are 1. So, one can think of the radiative lifetime tau is proportional to because these are all constants 1 over mu z. So, what is that? So, lifetime the excited state in this case that is 2 is inversely proportional to the transition dipole ok. So, what does it really mean? It means that if you measure the lifetime tau that is inversely proportional to 1 over that means, if this number is large this number will be small. So, lifetime is inversely proportional transient dipole moment. So, if there is a transition strong transitions strong transitions have smaller lifetime ok. So, something absorbs very strongly ok will have a shorter lifetime ok. This is something that now if you quickly remember in the chemistry there is something called potassium dichromate it is very strongly absorbing ok because the extinction coefficient is much stronger. So, if you excite potassium dichromate it must decay much faster that is what it means ok. Now, this kind of tells you that once you by the way the tau this value can be independently measured can be measured experiment this is measured experimentally which means now these are this tau of course, is multiplied by lot of constants ok, but there is another factor that is also important is that A has this nu cube. So, tau is also proportional to 1 over nu cube ok 1 over nu cube into mu z epsilon 1 square. Now, if this value is equal let us say there are 2 there are 2 transitions on molecule A and molecule B such that ok. So, that is that I will call it as 1 A 2 A 1 B 2 B and you see this the way I have drawn is this h nu A and this is h nu A such that h nu A is greater than h nu B and it so happens that this integral 2 mu z 1 square of A is equal to modulus of 2 mu z 1 square of B. If the transition dipole square in both cases in A and B if they are equal ok this assume a scenario when these 2 are equal then what happens is that then your lifetime will depend on 1 over nu cube that means larger separation the decay will be the faster ok. So, the tau is 1 over nu cube because nu is larger in the case of A with respect to B 1 over nu cube will be smaller. Therefore, tau so based on this one can write tau A will be smaller than tau B and the ratio of tau A to tau B will be equal to nu P cube by nu A cube ok. So, this is the ratio of tau A to tau B so that is only when these 2 transition dipoles are equal of course, if they are not equal then it will be complex mixture of the ratio of the transition dipole squares and the. So, this is kind of a way one can estimate. So, if T tau can be experimentally measure then depending on where you have excited what is the nu then one can actually experimentally calculate or estimate transition dipole ok. Now, what I am trying to say if tau is experimentally measured if tau is expansion measured then from this one can estimate transition dipole or other way around if an calculate using theory transition dipole then from here one can estimate and of course, this tie is a lifetime. So, one can go back and forth ok if you can measure one then you can estimate the other one or if you can calculate the transition dipole you can estimate the tau and one can go back and forth and see whether you have done it in a in a very consistent manner. So, both experimentally and theoretically the transition dipole and the lifetimes can be measured or estimated. So, that is the connection between the transition dipole and the lifetime and this comes via the Einstein coefficient A, but once you know the Einstein coefficient here you also know the Einstein coefficient B because A and B are connected. So, it just takes one experiment to estimate all the values ok. So, if I can measure the lifetime then I can if I can measure tau then I can measure then I can estimate 1 transition dipole 2 A coefficient and 3 P coefficient ok. So, I am going to stop it here and we will continue in the next lecture. Thank you.