 Hello, welcome to the lecture number 23 of my course, Quantum Mechanics and Molecular Spectroscopy. In the last class, we were looking at the transition moment integral and its connection with the absorption spectrum. So, let us quickly review that. So, if you have an absorption spectrum, which you usually record as a function of lambda and you get absorbance A, something like that, then this can be transformed as a function of epsilon as a function of nu. So, you get some other spectrum, different shape and this when as epsilon of nu by nu as a function of nu if you get another spectrum, okay. So, it is the same spectrum that is plotted in a different way, then integral epsilon of nu by nu d nu, which is nothing but area under this curve, k is proportional to the Einstein's B coefficients comma A coefficients comma transition moment integral, okay. So, I will write down the exact equation. So, B equals to 2303 divided by Na H eta integral epsilon of nu by nu d nu and A will be equal to 8 pi H nu cube by C cube 2303 by Na H eta epsilon of nu by nu d nu. And finally, your TMI that is nothing but your transition between the final state mu dot epsilon z axis i equals to modulus of the square or rather TMI square is equal to 2303 by 12 Na H bar eta integral epsilon of nu by nu, okay. Now, in this case of course, your Na is Avogadro constant and eta is refractive index of this solution, okay. One thing that you must remember the refractive index of the solution is not same as refractive index of the solvent. Of course, if you use very low concentrations, then you can approximate refractive index of the solvent equal to refractive index of the solution, okay. So, this is what we looked at in the last lecture. Now, in this lectures, I am going to take a look at what is known as line shapes. This is more of a descriptive, we will not derive much anything but I want to describe the line shapes. But before we go into line shapes, I want you to understand one thing. So, for example, if you have a function, okay, G of omega and this is given by integral minus infinity to plus infinity 1 over root 2 pi f of t e to the power of i omega t dt. If you have an integral and another integral where f of t is equal to 1 over root 2 pi minus infinity to plus infinity G of omega e to the power of minus i omega t d omega. Now, if you have two such integrals, these are called Fourier transforms, Fourier transform integrals. Now, you can see the two variables that I am using omega and t, these are the two variables and these two variables you can see are inverse with respect to it or conjugate it to each other, two variables are conjugate to each other, okay. That means whatever is there in time and its inverse is frequency, so time and frequency are conjugate with respect to each other, okay. Now, one can also write this in slightly in different way. So, G of nu is equal to 1 over 2 pi f of t e to the power of i. Now, you know omega is equal to 2 pi nu, so 2 pi nu t. So, all I am saying is that I am trying to you know instead of using the angular frequency I could use a linear frequency, okay. Now, it turns out that if you have an exponential function, okay. Now, let us just think of it this way, okay. If you have an exponential function e to the power of minus k t, if you have function e to the power of minus k t and this e to the power of k t, okay. If you take a Fourier transform of it Fourier transform of this function, okay. Then it turns out that this function will be nothing but, okay in k will be equal to 1 over 2 pi in variable k, 1 over 2 pi k naught k naught divided by k square plus k naught square, okay. Now, why am I talking about this? I will come to that. Now, let us suppose you have a excited state and a ground state. So, I can call it as 1 or i and this as 2 and this as f, okay. Now, excite go here and the population. So, what you had is initially you had some population, okay. Let us say call it as i naught that is the population excited state after light is switched off, okay. Once you will say what will happen? Once you will switch off the light what will happen? It will decay. So, your i of the fact 2 of the state 2 will be equal to i naught of state 2, okay e to the power of minus k t, okay. Now, k is here an exponential decay, okay. So, this is nothing but your first order kinetics or one could really write it as i of t is equal to i naught or i of 0, i of 0 into e to the power of minus t by tau and where we call tau as the lifetime. In such scenario what we will see that i Lorentzian of omega will be equal to now you can we have to understand one thing is that when you have a exponentially decaying function you can get a Lorentzian function as a Fourier transform, Fourier transform, okay. Now, if you get Fourier transform then what you get is i max of t by 4 tau square divided by omega minus omega naught square plus 1 over 4 tau square modulus, okay. Now, this is not very difficult you can use this formula and plug it in here and then you will be able to get this. Now, let us look at this little more carefully. When I say that my i Lorentzian of omega equals to i max of modulus 1 into modulus 4 tau square i max into multiplied by 1 by 4 tau square divided by omega minus omega naught square plus 1 by 4 tau square modulus, okay. Now, this when I plot this function it will look something like this. So, this will be my omega naught, okay. So, this will be like this and this width is called delta omega, okay, half. So, this is called half because if this one is the total height is i max and this height is i max by 2. So, delta omega half is nothing but full width at half max also known as F, W, H, M, delta omega half, okay. And it turns out this value will be nothing but delta omega half will be equal to 2 by delta nu half this is nothing but tau inverse this is nothing but A. We know that for a spontaneous decay process a lifetime is nothing but inverse of the first spontaneous lifetime. You can go back and check in the one of the earlier lectures, okay. So, by just by estimating this we can get the value of A. So, the line width function encodes what is A your Einstein coefficient A but we know once we know A from A we know how to get B and from B we have to we also know how to get TMI. So, just by measuring the line width, okay if this by measuring the line width one can get the lifetime and the Einstein's coefficient A, okay. So, it is rather easier to understand that one can measure all the quantities or estimate the quantities like transition moment integral Einstein's coefficients A Einstein coefficient B just by measuring the spectra either in the time domain or in the frequency domain. Of course, I can in the time domain I can measure tau from there I can get A from that I can get the from there I can get the line width or the full vector half max, okay. All these quantities are interrelated. Of course, there is one problem in measuring these quantities directly. One thing that these are called what is known as natural line width that means this function we should be able to fit it to a Lorentzian. If it is a not a Lorentzian then we would not be able to extract these parameters, okay. So, this will only happen if it is a natural line width that means the spectrum is not getting influenced by any other external factors but is a pure spectrum of the molecule itself. So, this is what I will call as a intrinsic behavior and when do we get if you can somehow isolate an atom or a molecule from the external influence of other molecules of its kind or from the solvent only then you will be able to understand this or under the approximation that the solvent has very little role to play, okay. However, in the presence of solvents or any other molecules or the molecules of the same kind this approximation can break down quite easily. So, in such scenario when you do not get Lorentzian line shapes all these quantities cannot be extracted. Now, in general one finds that there are other effects like temperature and because of the temperature there are molecular speeds, okay which are given by Maxwell Boltzmann distribution in the gas phase and the solution they are given by the how the path length or mean free path. So, in such scenario the line shape is given by a Gaussian function i g of omega is equal to i max into exponential minus omega minus omega naught whole square divided by 2 sigma square, okay. Now, in such scenario your omega naught is a central frequency and sigma is the standard deviation of the distribution. So, for example, you could have so if this is your omega naught you could have a Gaussian distribution which looks like this and this is your omega naught, okay and you have value of sigma that you can calculate in such scenario delta omega half is equal to nothing but 2 pi delta omega mu this is given by 2 to 2 ln sigma 2 sorry that is your so Fwh in this case and you will see that all of these will always be more than. So, let us say delta omega half of a Gaussian distribution will always be greater than delta omega half of the Lorentzian distribution because that is a natural line width. So, Lorentzian is a natural line width and Gaussian distribution comes because of external influences and when external influences come in then the peak width increases, okay. So, always okay natural line width is the most narrowest line width you cannot that is why it is called natural line width and you cannot go below that in fact it is actually controlled by the uncertainty principle, okay. You cannot go below that but generally you never reach that value you always have line widths which are wider than the natural line width that is because the atom or a molecule in question is always getting influenced by the external parameters could be temperature could be influence of the next molecule intermolecular interactions or the solvent effect any of this such external parameter. So, this it could be greater or greater than or the worst case scenario should be greater than or equal to. So, the natural line width is the narrowest line width any transition will have. Now, sometimes in the gas phase what you have a Doppler bonding because you know atoms and molecules are moving around when you have Doppler bonding you are then you are then you have delta nu half is given by 2 nu into 2 k B T l and 2 divided by m c square to the power of half and of course you can clearly realize it will descend on the temperature it will depend on the mass and it will depend on the speed of light okay and it will depend on the. So, this is a temperature dependent because you know temperature will affect the molecular speeds. So, the Doppler broadening will depend on the temperature okay. Now, there is something called Void plofer Void plofer okay profile Voi G T. Now, this Void profile is basically a convolution of natural rodent line it turns out this natural lemurs is also called homogeneous homogeneous and this is heterogeneous homogeneous because it is intrinsic intrinsic to the molecule or heterogeneous because it is getting influenced by the external parameters. When you have a combination I Void of omega is given by I integral of I G omega prime I L omega minus omega prime d omega prime okay. So, it is a convolution okay. So, essentially the entire line shape. So, if you have some shape something like that then it is a linear it is a convolution that means the product function of the Lorentzian line shape and the Gaussian line shape and that is called a Void profile. Unfortunately, when you do not have Lorentzian line profile okay you will not be able to get information about the transient moment integral or A or B. So, the only when you have homogeneous line weights can the experimentally measurable quantities be equated to or experiment measurable quantities can be connected to the theoretically evaluate quantities such as transient dipole, Einstein A coefficient and Einstein B coefficients. Therefore, when I write such equations by Na H eta integral epsilon of nu by nu d nu or A is equal to 8 pi h nu cube by C cube 2303 divided by Na H eta integral epsilon of nu by nu nu side epsilon by modulus square equals to 2303 divided by 12 Na H bar eta epsilon nu by nu d nu. When you write all these quantities okay and you have a band this is nu versus intensity and this band okay this band must be Lorentzian. So, that means I L is equal to I max 1 by 40 square divided by omega minus omega naught whole square plus 1 by 4 square. So, only if we have this shape all this will be right otherwise not. So, we can only evaluate or only connect the theoretically evaluated quantities such as A Einstein coefficient A Einstein coefficient B transient moment integral etc to the experimentally absorbed spectra only if you have a Lorentzian line shape otherwise we cannot okay. We will stop here and take it up in the next lecture. Thank you.