 Welcome to the lecture number 18 of the course Quantum Mechanics and Molecular Spectroscopy. In the last class we looked at the transient probability to a state f from state t and this was given by E naught pi square pi 4 h bar square omega fi pi omega square sin square del omega by 2 into t divided by delta omega by 2 whole square modulus of f epsilon dot mu i whole square and we know this is the transition moment integral, okay and this for a continuum of states around f. We showed that this P f of t was equal to pi E naught square by 2 h bar omega fi by omega square modulus of f epsilon dot mu i square rho E fi into t and from here we defined rate, rate of absorption is equal to P f of t divided by t, so that was nothing but pi epsilon naught pi 2 h bar omega fi pi omega square modulus of f epsilon dot mu by i whole square rho E f, so rho f small mistakes, so we called it as rho f of E and rho f of E, okay. This we finally wrote as this is nothing but W f, okay, so this was equal to from state i, so rate from state i to state f that is buried in a continuum is equal to 2 pi h bar mod mu square rho f of E, okay and where mod mu square was given by E naught square pi 4 h bar square modulus of f epsilon dot mu i square and this we call it as transition dipole, okay r square of the transition integral because mu is the transition dipole. So now we know two things that the rate of absorption from a state i to f is given by mu square and some constants that published multiplied by density of states at f, okay. Now these constants turn out to be 2 pi h bar and that mu square I just wrote is equal to E naught square by 4 h bar square and your transition moment integral square, okay. So this is called transition modulus of this is transition dipole and this transition dipole is proportional to transition moment integral. Now there is one thing that you must remember that is that this transition moment integral you know dictates what selection rules are going to be. Of course in this course up till now we have not come across the selection rules but I will come to the selection rules towards the end of this course, okay. So the last few lectures will be based on the selection rules of rotation, vibration and electronic transitions. Now this is something that is rate of absorption, okay. In Chemica's picture we had two things, okay remember when you had this delta functions we had two of the delta functions delta omega fi plus omega plus delta omega fi minus omega modulus maybe couple of lectures below you can go and look at it and this I told you corresponds to stimulated emission and this will correspond to absorption. So the stimulated emission and absorption are a similar process. So in the presence of the classical light, okay a molecule or an atom or a quantum mechanical particle can absorb energy and go from a ground state to the excited state or can come back from the excited state to the ground state. So if you have a state i and state f, okay you can either go up that is your absorption and or come down that is your stimulated emission and this is in the presence of, however when you excite a molecule, okay it does not stay there forever it has to come back even if you switch off the light and that is called spontaneous emission. So emission can happen by two different pathways one is the stimulated emission and other is a spontaneous emission. Unfortunately in semi-classical picture spontaneous emission cannot be dealt with directly. So it has to be introduced in an ad hoc manner and that ad hoc manner was developed by Einstein it is called Einstein's coefficient. So we will look into that now. Now for example if you have a light impinging on a particle, okay so of course it is never going to be a single photon generally when you do spectroscopic measurements you will have bunch of photons some intensity of light, okay and you have to that intensity of light you have some radiation density. So I will define a quantity called radiation density that is new radiation density rho or rho radiation density at some frequency new this is given by du by dv d new du by d new where okay u is the energy per unit volume unit frequency. Now why want I want to do this so the reason why I want to do it is that I want to in introduce a concept of spontaneous okay because I told you that in presence of light okay in a semi-classical picture there is no spontaneous emission there is only stimulated emission but we all know that the spontaneous emission does happen. So we have to introduce the concept of spontaneous emission. So there are totally three processes that happen when the light is absorbed by when light is impinged on a particle. So one is the absorption second one is the spawn stimulated emission and third one is spontaneous emission and both of these can be related to the transition all transition dipole but I still do not know how to look at this okay and that is what we are looking at. Now let us consider two levels. So let us start with the simplest of the problem there are two levels okay let us call it as initial level i with energy E i and final level f with energy E f okay and let N i be the population of the lower level and N f be the population of the so N i so N i, N f are the populations of the initial final levels and we have radiation density rho radiation at a new frequency is du by du okay. Now if you have that then what can happen there are three processes that can happen so the particle can go from top to bottom that is the absorption it can come down from top to bottom by stimulated emission and it can come down from top to bottom by spontaneous. Now let us look at the rate of functions now if I want to go from N i to from top to bottom so that means rate of absorption. So the rate of absorption W going from the initial state to final state okay for the sake of convenience and the way it is written in the textbooks well I will also call i as 1 and f as 2 okay so initial state is 1 and final state is 2 so we can also i to 1 and 2 and corresponding energies are E 2 and E 1 okay. Now if you have omega f i this is nothing but sorry rho w f i i f this is nothing but w w 1 2 that is the rate constant from going from level 1 to level 2 that is given by this rate will of course depend on number of particles there are or now what is the population in the ground state okay and the population in the ground state is N 1 and N excel state is N 2 so this is proportional to number of molecules or the population in the ground state so this is proportional to N 1 okay and it is also proportional to let me write this way so w 1 2 is proportional to N 1. Now w 1 2 is also proportional to amount of radiation if the more radiation is there if you have more intensity more number of intensity of light more number of transition so it is proportional to radiation density so that means it is proportional to at the appropriate frequency nu so it is proportional to two quantities N 1 that is the population of the ground state and the radiation density nu. So instead of this so I need to remove the proportionality constant so what I will get is w 1 2 equals to proportional constant I will call it as b 1 2 N 1 rho rad so this is rate of absorption so that is when process are when the process of going from the state 1 to state 2. Now under same condition if I want to come back from state 2 to state 1 so rate of absorption for coming down w 2 1 or rate of coming down w 2 1 that is means you have state 2 and state 1 and you have to come down so this coming down is by two processes one is the spontaneous emission other is the stimulated emission okay in the case of stimulated emission it will be proportional to the population N 2 and it will be proportional to radiation density and the proportionality constant I will call it as b 2 1 while the spontaneous process will only depend on the population it does not depend on the radiation density because spontaneous emission does not need radiation if you excite the molecule and switch of the light it will come down by itself okay and that is the spontaneous emission. So you will have N 2 but you do not need radiation density but it should be proportional constant and that proportional constant I will call as a 2 1 okay so that is the a 2 1 that is nothing but your so this will this rate will be for the stimulated emission and this rate for a building spot you have two rates now you have two rates one is w 1 2 that go takes molecules from top to bottom sorry bottom to top and then you have w 2 1 which brings molecule now if let us assume there is a thermal equilibrium states this is my one this now if you take thermal equilibrium then you will have to follow Boltzmann law and what this Boltzmann population distribution law says that N 1 by N 2 equals to exponential delta E by k t okay by the way it is slightly written usually written as other way around N 2 by N 1 is equal to exponential minus delta E by k t so I am just writing the inverse of this okay now this delta E is equal to E 2 minus okay so if you there is an equilibrium okay and what you have is 1 by N 2 is equal to exponential by E by k t okay by the way this k t is nothing but your Boltzmann constant okay equilibrium also means rate of transition going from bottom to top is equal to rate of transition going from top to bottom so that is nothing but w 1 2 must be equal to w 2 1 equilibrium line you have let us suppose you have an equilibrium between A and B so that means so there is a k forward and k backward multiplied by rate constant so rate of forward reaction must be equal to rate of backward reaction so simply means k f into A should be equal to k B into P so that is what I am doing so under equilibrium rate of absorption should be equal to rate of spontaneous and stimulated emission put together so when I put do this this will be equal to B 1 2 N 1 rho rad mu must be equal to P 2 1 N 2 rho rad plus A 2 1 N 2 okay so this is the equilibrium so we have now have 2 equations for the equilibrium okay so that is the equilibrium according to Boltzmann law and this is the equilibrium because of the radiation that is present and that is making things go up and down. Now according to blackbody radiation this is given by Planck okay rho radiation mu is given by 8 pi h mu cube pi c cube into 1 over e to the power of h mu by k t minus 1 okay so this derivation we have to look it up blackbody radiation theory okay by the way there was something called you know ultraviolet catastrophe you know this equation was proposed by Planck where the energy is given in terms of h mu okay for avoiding the black ultraviolet catastrophe in the blackbody radiation. Now let us look at this little bit more carefully radiation at mu a blackbody radiation is given by 8 pi h mu cube pi c cube mu is the frequency h mu cube 1 over e to the power of h mu by k t minus 1 okay now we know that n 1 by n 2 is equal to e to the power of delta e by k t so this is nothing but e to the power of delta e is nothing but h mu h mu by k t so these are the 2 equations that we have and that we need to know. Now let us go back to our equation n 1 p 1 2 n 1 rho rad of mu is equal to b 2 1 n 2 mu rad rho rad mu plus a 2 1 n 2 so I am going to slightly rearrange okay I am going to bring this equation to this side so that will be nothing but then I can take rho rad into mu as common is equal to sorry into b 1 2 n 1 minus p 2 1 n 2 equals to a 2 1 n 2 so your rho rad mu is given by a 2 1 n 2 divided by b 1 2 n 1 minus p 2 1 n 2. Now what I am going to do is the following is that now I am going to slightly rearrange this equation in a such a way that I will be able to use this one okay now what I will do is I will divide by n 2 all through the denominator and the numerator so this will be nothing but rho this implies rho rad of mu okay let me do it in the next page is equal to a 2 1 by n 2 divided by b 1 2 n 1 minus b 2 1 n 2 now I am going to divide by n 2 fourth numerator denominator okay so when I do that this is equal to a 2 1 divided by b 1 2 n 1 by n 2 minus b 2. Now that is the equation that we get of mu and my n 1 by n 2 is equal to exponential h mu by k t right okay now and I also know further that rho rad of mu is equal to heat by h mu cube by C cube 1 over e to the power of h mu by k t minus 1 okay now I am going to slightly rewrite this equation rho rad of mu is equal to a 2 1. Now what I will do is I will take b 1 2 as common if I take b 1 2 as common sorry b 2 1 as common then I will get b 1 2 by b 2 1 n 1 n 2 minus 1 okay now you can see n 1 by n 2 is this exponential h mu by k t so this will give me a 2 1 divided by b 2 1 now n 1 by n 2 is divided by b 1 2 by b 2 1 to e to the power of h mu by k t okay now you can see quickly that there is some semblance between this and this okay so we can see that these two equations are looking similar but they are not really similar yet so we need to do little bit of more of mathematical manipulation to be able to look at that which I will continue in the next lecture okay I will stop it here and thank you very much.