 Welcome to the lecture number 36 of my course, Fontomechanics and Molecular Spectroscopy. Essentially, until the lecture number 35, I have covered the derivation of the transition moment integral, the relationship of the transition moment integral with various experimental quantities such as the life time, line weights and also the absorption spectrum and in turn related to Einstein's coefficients A and B. Later, we looked at the transition moment integral and derived the selection rules for the rotations, vibrations and electronic transitions. Towards the end, we also looked at the rotations of the polyatomic molecules. However, the vibrations and the electronic spectroscopy of polyatomic molecules will require a knowledge of group theory. Since I have assumed that you will not need any other prerequisites than basic quantum mechanics, the vibrations spectroscopy and the electronic spectroscopy of the polyatomic molecules will be skipped. You can look at some other courses on molecular spectroscopy which include group theoretical treatment for the vibrations and electronic spectroscopy of polyatomic molecules. That in a sense completes the course contents. Now, in the next few lectures, I would like to look at some of the tutorial problems. Since spectroscopy simply does not mean transition moment integral. It also looks at how one interprets the spectra to be able to arrive at some useful atomic and molecular parameters. So, over the next few classes, I am going to teach you or we are going to solve some very simple problems as a tutorials. Okay. To begin with, I will start with a particle in a box. A particle in a box is nothing but something that you know you all might have learned is there are levels n is equal to 1, we will have a wave function like this, then there will be n is equal to 2, we will have a wave function something like that and n is equal to 3 we will have a function something like that and the wave function psi of n is given by root 2 by L by the way this is 0 and this is L where capital L is the length of the box. Okay. One more thing is that in a particle in a box these walls extend to infinity because any particle that is trapped inside this particle in this well or in this box cannot get out sin n pi x by L. Okay. That is our wave function and the corresponding energy value will be E n is equal to h square n square by 8 m L square. Okay. Now one thing that is let us look at this dimension as x. So, the box is along the x axis. So, if the box is along the x axis, so the dipole moment direction the mu this moment will be along x. Therefore, for a transitions part in a box okay you can write T mi as equals to integral. Of course, particle in a box is limited from the between 0 and L. So, the integral will be 0 to L. Okay. Now, let us say you go from n is equal to there is a n i that is the initial quantum number and you go to n f that is the final quantum number. Okay. So, what you have is you have two wave functions square root of 2 by L sin n f pi x by L your dipole moment mu root 2 by L sin n i pi x by L dx. So, this is your transition moment integral simply put if I have to write this will be nothing but psi n f mu psi n i. So, this is the integral that we need to solve. Okay. So, I am going to rearrange this integral slightly there is a square root of 2 by L square root of 2 by L. So, that can come out because 2 and L are constants for a given particle in a box. So, what you will get is 2 by L integral 0 to L this will be sin of n f pi x by L okay or dipole moment is along the direction x. It could be it could have something some multiplier. So, if I if I take mu is equal to some constant a into x okay then this will be a into x sin n i pi x by L dx. Now, a is also a constant. So, that also I can bring it out. So, this will be nothing but 2 a by L 0 to L sin n f pi x by L x sin n i pi x by L dx. So, essentially your transition moment integral will look something like that okay which is what we need to solve to get the selection rules. So, let us look at them little more carefully the transition moment integral will be now will be equal to 2 a by L integral 0 to L. Now, because these are product function what you had is the sin sin n f pi x by L x sin n i pi x by L dx. So, this I can slightly rewrite as 2 a by L 0 to L x sin n f pi x by L sin n i pi x by L dx. Now, I am going to use a small transformation that is if we have sin a x multiplied by sin b x this will be nothing but half of cos a minus b x minus cos a plus b okay. So, I am going to use this trigonometry transformation. So, this will be equal to this will be half. So, 2 a by L into half integral 0 to L x into cos of n f minus n i pi x by L minus cos of n f plus n i pi x by L dx okay. So, this I can write this 2 and this 2 I can cancel. So, what we will get is a by L 0 to L x cos of now what I will call it as n f minus n i that is the difference between the quantum numbers of the initial state and final state. So, that I will call it as delta n. So, this will be delta n pi x by L minus cos of. So, other thing is n f plus n i. So, that is the sum of the quantum numbers of the initial and final state I will call it as a capital N okay. So, that will be nothing but capital N pi x by L dx okay. Now, we will see this will be nothing but a by L whole thing multiplied by 0 to L x cos delta n pi x by L minus 0 to L x cos capital N pi x by L dx dx. Now, you can see this as a these 2 integrals they are x cos A x type. So, integral x cos A x dx has to be integrated by part. So, this will come out to be 1 over A square cos A x plus x by A sin A x. So, this integral is given by this. So, we just have to plug it in. So, this will come out to be this is equal to A by L okay. Now, the first integral is 1 over A square 1 over A square will be 1 over delta n pi by L whole square cos delta n pi x by L okay evaluated from 0 to L plus 1 by 1 by delta n pi by L into x sin delta n pi x by L evaluated from 0 to L. Similarly, the next integral will be negative of 1 over n pi by L whole square cos n pi x by 0 to L minus again minus of minus minus x pi n pi by L into sin n pi x by L evaluated from 0 to L. So, you need to get these 4 times 1, 2, 3, 4 okay. So, now your tantrum moment integral will be equal to A by L thing multiplied by the first term will be was 1 over delta n pi by L whole square cos delta n pi by L evaluated from 0 to L that was my first term I am going to go 1 term at a time. So, this will be equal to A by L integral. So, this will be L square divided by delta n square pi square cos of delta n pi by L. So, there is an x here x 0 to L okay. So, when I evaluate this, this will be nothing but 0 to L okay. Now, if I take L square by delta n square pi square as a common then what I will get is cos cos L is when I plug it in L here L will go so we will get cos delta n pi okay then cos 0 is 1. So, this is minus 1 okay. Now, the second term was plus x by delta n pi by L sin delta n pi x by L evaluated from 0 to L okay. So, this will be nothing but plus okay. Now, if I plug L value then x will be L and this will go to so this will be L square by delta n pi sin of sin of L. So, this will be sin of delta n pi delta n pi because when x is equal to L this will be delta sin delta n pi okay. When x is equal to 0 of course this term will go to 0 so there will be nothing left okay. This is your second term. Similarly, you have the other two terms. So, when I can simply write this as L square by n square pi square cos n pi minus 1 minus L square by n pi sin capital n pi this whole thing. So, this is the equation that we get okay. It is a bit complicated you can see there is a L square everywhere. So, one can but I do not have to really worry about L square and all. So, there is a formula. So, sorry this is not 0 this is 8. Now, you can see that there are two factors one is delta n and there is that is nothing but f minus ni and the other one is capital n that is nothing but nf plus ni. Now, I am there are if you take any two quantum numbers by the way in the case of particle in a box n can only take values of 1, 2, 3, etc. So, they are you know integers real I mean positive integers. So, nf minus ni would only be equal can either be odd or even. So, if ni. So, nf minus ni can be odd even okay. Now, if nf minus ni is even when that will happen when both ni and nf are odd or but both nf and ni are even. So, that will give even. So, when if nf minus ni is even it turns out that even n which is nothing but nf plus ni is also equal to even okay. So, if both of them if the difference between the ni or delta n nf that is nothing but your delta n if delta n is odd sorry if a delta n is even then your n also is even. So, when you have even okay then you can see that sin n pi delta n pi okay. So, that will be sin 2 pi 4 pi etc will go to 0. So, sin delta n pi okay will go to 0 if delta n is even. Similarly, sin n pi will also go to 0 if delta n is even. Similarly, you can see that cos delta n pi okay cos delta n pi will go to 1 if delta n is and cos capital N pi will also go to 1 when n is even. So, if this goes to 1 1. So, 1 minus 1 is 0 this is 0 1 minus 1 is 0 and this is 0. So, for this will go to 0 this will go to 0 this will go to 0 this will go to 0 all the term will go to 0 when nf minus ni or delta ni is equal to even and and capital N is even. So, your tmi will go to 0 if delta n is equal to even and n is also even. So, you can go back and work out that tmi will not be equal to 0 if delta n is odd okay. If delta n is equal to odd delta n is equal to odd implies delta n must be equal to plus minus 1 plus minus 3 plus minus 5 etc. So, for a particle in a particle in a box box the change in the quantum number must be odd for the transition for the transition to be allowed okay. So, now which means if you have a particle in a box. So, n is equal to 1 n is equal to 2 n is equal to 3 n is equal to 4 n is equal to 5. So, this will be allowed this will not be allowed this will be allowed this not be allowed. So, every alternate okay will not be allowed. So, every alternate so similarly this will be allowed this will not be allowed and this will be allowed similarly this will be allowed and this will be allowed this will not be allowed okay. So, I will stop here and we will continue in the next lecture. Thank you.