 Hello, welcome to lecture number 22 of the course quantum mechanics and molecular spectroscopy. In the last lecture we looked at the relationship between the transient moment integral or the transient dipole with the Einstein's coefficients a and b. And we ended up with a equals to 2 pi mu cube divided by 3 epsilon naught h bar c cube into 2 mu z epsilon 1 whole square. This could also be written as 2 pi mu cube by 3 epsilon naught h bar c cube f mu z along epsilon direction 1 or i whole square ok. Here I have assumed that the state 1 is identically equal to state i and state f is identically equal to state 2. So, which is nothing but looking at transition between state 1 or a i to state 2 or f ok. Similarly, we have the absorption coefficient or the stimulate emission coefficient to be pi by 6 epsilon naught h bar square 2 mu z 1 whole square ok this along the dot product of. Now, it turns out that these two coefficients are related to the transient moment integral all this is called transition dipole and square of the transient dipole ok. Now, there are experimental quantities the one you can easily measure and one should be equal to a and b. So, one such experimental quantity is the absorption coefficient. So, we will try to derive the relationship between the a and b the Einstein coefficients a and b and the absorption coefficient. Now, when we think of absorption one thing that comes to our mind is the Lambert-Bier law ok. What does Lambert-Bier law says? Lambert-Bier law simply says that the i equals to or i by i naught log of this is equal to epsilon c and log of i by i naught is also called absorption a this is equal to epsilon c absorbance. So, this is absorbance and this is the extinction coefficient ok. So, extinction coefficient epsilon will have ok the units of mole inverse meter inverse ok because absorb bands by itself does not have any units slightly differently if I can write this relationship. So, if I write log of i by i naught epsilon cl. So, if I take you know exponential on both sides ok. So, this will be equal to i by i naught will be equal to tenth power of epsilon c we can always write i equals to i naught tenth power of epsilon cl ok. Now, this is the form of the. So, one can write a differential form. So, d i equals to 2.303 epsilon which is a function of nu ok c times i times dx ok. What is your length? Length is nothing, but l is the length. So, if you have going from say x 1 to x 2 this is going to be l, l is equal to modulus of x 1 minus x 2. So, one if you look at the differential form ok. So, this is nothing, but differential form ok. So, essentially comes from this equation or in fact, if you integrate you will get back this equation ok. So, the differential form of differential form of Lambert Bayer law ok will be nothing, but minus d i equals to 2.303 into epsilon of nu c is the concentration i is the intensity dx ok. Now, let us assume that there are n number of molecules, n number of molecules volume of centimeter cube ok and that is going from initial state i to initial state f final state f under the light condition ok. So, your minus d i will be equal to n times h nu h nu is nothing, but your energy w fi that is the rate coefficient into dx ok. So, the absorption the incremental absorption that will happen will depend on number of molecules the energy at which and it is the rate of absorption multiplied by dx ok. Now, minus d i according to Lambert Bayer law will also be equal to 2.303 epsilon of nu c times i c times i dx ok. Now, you can see these two are equal. So, this is tried in terms of molecule and this is terms of Lambert Bayer law. So, we can equate them. So, n h nu w that is the rate constant between f and i rate of f ok dx is redundant now because it is an incremental change eraser you can remove that. Now, this will be equal to 2.303 epsilon of nu at this nu and this nu are the same c times i. So, I can write rate of absorption is equal to 2.303 epsilon nu c dot i divided by n h nu ok. Now, you can see that concentration is nothing, but 1000 times n divided by n a n by n a is number of moles is not it n by n a is number n a is Avogadro constant and concentration is 1000 times because it is written in less than meter cube ok. So, when I plug in that. So, c is 1. So, what you get is w f i is equal to 2.000 when you 2.303 epsilon of nu this is 1000 times n ok into i divided by n into n a h nu and this n and this n a will cancel. So, what you will get 1000 into 2.303 will give you 2.303 epsilon of nu i divided by n a h nu ok that is your rate of absorption. So, now the rate of absorption is given by w f i going from state i to state f. So, this will be equal to 2303 epsilon of nu times intensity of light divided by Avogadro numbers h nu h nu is where the energy is happening. Now, this is a very simple equation except that we need to now slightly rearrange this equation. Now, before I go I want to look at something. So, let us suppose you have a you have a solution of some molecules and then you shine light. So, you get i naught and you get back i. Now, what happens is that if you generally if you record the solution what you get you get a band. So, what you plot is absorbance a versus lambda. Now, we know that absorption is nothing but epsilon cl. So, if you know absorption if you know the concentration if you know the length. So, epsilon will be equal to a divided by cl ok. So, and we know lambda is equal to nu lambda is equal to c. So, c nu is equal to c by lambda. So, we know these two equations. So, I can convert this plot as plot of epsilon of nu versus nu ok. So, what I am trying to tell you here is that if I have an absorption band some molecule getting absorbance solution and which is generally positive plot at absorbance versus lambda. One can transform this using appropriate conversion factors is that one can plot the same thing as nu versus epsilon of nu. This is just a different representation ok. So, what you can get is that you can get the absorption coefficient or the extinction coefficient as a function of the wavelength or the frequency. Of course, wavelength and frequency are related with respect to the both multiplication of that is seen here. So, one can always plot generally we plot it like this ok, but one could also plot like this. Now, why I want to use it I will come to this little, but just remember that one can also plot an absorption spectrum as a function of nu the epsilon nu ok. Now, let us look at the equation that we had we had the equation of rate equals to 2.303 epsilon of nu times i divided by n a h. Now, the intensity of light i in any medium is given by u dot c by eta ok, where u is the total energy density and c is speed of light eta is refractive in this ok. Now, if I plug in that then I will get w fi equals to 2.303 epsilon of nu u dot c divided by n a h nu eta, but the energy density u is given by integral over nu ok, rho radiation nu d nu that is a total. So, if you want the total energy density ok, what you need is the energy density at each frequency and you integrate over all the density ok. So, when I use this is you have w fi is equal to 2.3 sorry this is not 2.3, but it is I am sorry about this. So, it is it should be 23.03. So, that decimal does not exist ok, this is 23.03 epsilon of nu u instead of u I write I will write it later into c divided by n a h nu into eta integral over nu nu radiation of nu d nu. Of course, this is not the right way to because you see there is a nu here as well ok. So, in principle one should write it as 23.03 c because this also depends on nu. So, this is actually wrong way of writing ok. Now, because this is integral over nu any quantity that has nu must be inside the integral. So, 2.303 c divided by n a h eta integral over nu rho radiation nu epsilon nu by nu d nu. So, that is the right way to write ok even because you are integrating over all value nu. So, any variable that has nu dependence must be integrated ok. So, the rate of absorption is equal to 23.03 c divided by n a h eta integral over nu rho radiation of nu epsilon of nu by nu d nu ok. Now, what I am going to do is the following. Now, think of it like this you have an absorption spectrum absorption as a function of lambda you will get some peak ok. Now, this I told you I can transform in a way that I can plot epsilon of nu versus nu thing like that ok. So, that is what I have done. Now, when I record from say nu 1 to nu 2 which is the 2 ends of the spectrum ok. Then what I will assume is that the rho radiation or the radiation density is constant. So, radiation density is constant over the absorption band ok. If I that is a constant then which means this one I can bring it outside because when you integrate constants just get multiplied. So, what you get is W fi is equal to 2303 c rho radiation of nu divided by n a h eta integral over nu epsilon of nu by nu. Now, what you are doing is the you get this you get the rate constant ok. Now, we have 2 equations ok. One is the radiation density. So, sorry one is the rate constant rho f i is equal to 2303 c divided by n a h eta rho radiation of nu integral nu epsilon nu by nu d nu ok. But we know rate of absorption W fi per molecule is just given by b times rho radiation nu. Now, when I equate these 2 constants I will get b equals to 2303 c divided by n a h eta integral nu epsilon nu by nu d nu ok. So, I need to just get this integral. Now, what is this integral? Now, I told you that you could also have an absorption spectrum as a function of nu epsilon nu. So, for each value of nu there is a different epsilon. So, what you do is you get you divide epsilon nu by nu. So, what you do is you transform this plot as epsilon nu by nu into nu. So, if you plot that plot and this area of that this integral ok. So, this area is this integral. So, your rate b is 2303 times c that is speed of light n a Avogadro constant h eta and this integral whatever is that you measure epsilon nu by nu d nu and this integral can be experimentally measured ok. So, by measuring the extinction coefficient epsilon nu by nu or other way around. So, what we get is the following ok. You are Einstein's coefficient b Einstein's coefficient b is given by 2303 c by n a h eta integral over nu epsilon nu by nu d nu ok. Now, you see b that is Einstein's coefficient of absorption is now proportional to the extinction coefficient. So, there is a relationship between the extinction coefficient and the absorption coefficient of the extinction. Similarly, one can write a is equal to now a was nothing, but 8 by h nu cube by c cube ok b. So, 2303 c by n a h eta integral nu epsilon nu by nu d nu. So, this c and becomes c square and this and this h it can set. So, what you get is 8 pi into 2303 divided by n a eta c square nu cube integral nu epsilon of nu by nu d nu that is a. So, you can by measuring the absorption spectrum and plotting. So, when you measure the absorption spectrum a versus lambda can transform it as epsilon of nu versus nu which can again be transformed as epsilon of nu by nu by nu. So, one can do this transformation and when you have this spectrum and this is nothing, but this ok. Then you can actually you can actually calculate the Einstein's a coefficient and b coefficient. So, Einstein's a coefficient and b coefficient can be related to the absorption spectrum ok. Now, it turns out we also know that Einstein's a coefficient b coefficient are also related to transition moment integral. So, we know b is is not equal to pi by 6 epsilon naught h bar square f mu z i whole square, but this is same as 2303 c by n a h eta integral over nu epsilon over nu by nu d nu. Now, when you equate these two, so let us do that. So, pi by 6 epsilon naught h bar square ok f mu z i whole square is equal to 2303 c by n a h eta integral over nu epsilon nu by nu d nu. So, which means your transition dipole f mu z epsilon i whole square equals to 6 epsilon naught h bar square by pi into 2303 or 2303 c n a h eta integral over nu epsilon nu by nu d nu ok. Now, I can going to do. So, this is equal to 6 epsilon naught h bar square is h square by 4 pi square there is already pi. So, 4 pi cube 2303 c by n a h eta. So, this h and this square gets cancelled this 4 2 times and 3 times 2 pi epsilon naught 2303. So, this can one can write it as 3 pi 2 epsilon naught pi square 2303 c by n a eta there is h on the bar into h integral over nu epsilon nu by nu d nu. So, one can think of the transition dipole momenta square is proportional to integral over nu epsilon over nu by nu d nu ok. This is the relationship that I wanted to bring. So, epsilon nu is nothing but your extinction coefficient. So, the transition dipole of a transition between state i and state f or state 1 and state 2 is proportional to the this integral which is made of extinction coefficient. So, once we know extinction coefficient or rather we know. So, if you have a band ok and we know the integral underneath it. So, the entire band integral the way you plotted is epsilon of nu by nu versus nu if you know this integral then this integral is proportional to your transition dipole. So, if you have a by the way all these are constants ok except you know if you change the solvent the refractive index will change other than that everything else is a constant ok. So, once you change the once you change a transition from say molecule A to molecule B if the area under the curve changes then you know the transition moment transition dipole is changing or not. So, this is a theory theoretical quantity and this is an experimental quantity. So, this is theory and this is experiment ok. So, one can always relate the transition dipole to the extinction coefficient not in a straight way but you know in the kind of integral. So, there is an integral here but you know there is it is basically the entire transition or extinction coefficient of a band not exactly at one position not at the peak or lambda max it is the extinction coefficient over the entire band ok. So, which can be experimentally measured using an absorption spectrum and theoretically one can also get the transition dipole. So, these two are related to each other. So, we will stop here and continue in the next lecture. Thank you.