 We are going to learn about the principal quantum number finally. It is funny, is not it? When we had learned Bohr theory principal quantum number came first, then came L and then came S and that is how this picture was developed. What we have got is in exactly opposite sequence. So, if you remember we are separated the Schrodinger equation for hydrogen atom, well electron and hydrogen atom into 3 different parts, r dependent part, theta dependent part and phi dependent part. We solve the phi part, told you what the solution of theta dependent part is. Today we are going to tell you not solve what the solution of r dependent part is and just to recapitulate this is the important thing that we have learned so far. Capital phi is a e to the power i m phi, one thing I forgot and you should be able to do yourself. Can you normalize this? Can you try to find out what the value of a will be? That is something that I completely forgot to do sorry about that. Just remember something when you try to normalize you have to integrate between the limits of phi and the limits of phi are 0 and 2 pi, maybe I just do it here. It is a little bit of digression but let us do it anyway. I want to know what is the value of a if I normalize it. What do I do? I integrate capital phi star, capital phi d phi between the limits 0 and 2 pi and that probability should be equal to 1. So, what is phi? Let us say phi equal to capital phi equal to e to the power i m phi. e to the power i m phi. What is our phi capital phi star then? It is going to be e to the power minus i m phi d phi equal to 1 simple multiply this you get 1. So, you get 0 to 2 pi d phi equal to 1. But did I miss something? Yes I missed a remember this one has a this one also has a. So, I should actually write a star a mod square of a integral 0 to 2 pi d phi is equal to 1. So, what is a then? a has to be 1 by 2 pi well square root of that and it can be plus it can be minus we usually do not take the minus sign plus is good enough it does not matter whether you say that this direction is plus that direction is minus for wave function. So, here we get this is going to be 1 by root over 2 pi e to the power i m phi all right just a little bit of digression because I had forgotten to do it last time better erase this because I do not know where things are going to pop up. So, this is what it is the phi dependent part and from theta dependent part we get an idea of the total angular momentum and we have discussed why m has an upper limit of L. Now, this year all of a sudden bang on your face is the R dependent part we are not about to solve this equation like the theta dependent part the solution of this was already known when you had this when this Schrodinger equation was being formulated. So, just use the solution let me try to write this in a more compact less scary form. So, what am I trying to write here I am trying to write what is R nl capital R nl which is a function of small r what will it be first thing that I have here it is a constant and it is a constant it depends on the quantum number n as well as quantum number l. So, let me write nl m now I will change the color what do I have next I have this r to the power l I will write it as such that is not scary what do I have next I have this a polynomial which depends on n plus l as well as 2l plus 1 and this polynomial is in 2 z r divided by n a and this polynomials are called associated Laguer functions. So, this is the functional form that we need to know I am saying you need to know this because there is something important r to the power l and a polynomial we are going to see what kind of polynomials we have but it is important because that is what gives shape to the radial part of the wave function. No need to remember this nl all these things if you can remember this is good enough you cannot remember that also is fine. One thing that we need to see is that if you go back to this three equations you would see that energy appears only in the r dependent part. So, it tells us essentially that for a hydrogen atom energy does not depend on the subsidiary quantum number l or the magnetic quantum number m is that right is that wrong well it is right it is fine why is it that you get different why do you get that fine structure actually fine structure is obtained for other atoms and for other atoms this degeneracy is lifted we are not going to talk about that in this course. But what I am saying is the expression of e that comes out is this as the eigenvalue and it is exactly the same expression as that in Bohr theory. So, the strength of Bohr theory is retained here the strength of Bohr theory is was that it could give you the right value of energy remember it could determine it could give you the theoretical value of Rydberg constant correct to many many places of decimals and it turns out to be minus 13.6 electron volt divided by n square minus 13.6 electron volt another number that we should know it is like ABCD as far as quantum mechanics is concerned there is ionization potential of hydrogen atom. So, as you go higher up the energy ladder as you go higher up in n what happens to this number e n becomes well n becomes larger. So, the magnitude becomes smaller and e n becomes therefore, larger why does it become larger because the negative quantity it goes more and more and more towards 0 that is all. When energy is equal to 0 that means the electron has left the atom it is no longer experiences the potential energy due to the nucleus. So, that is what it is. So, the energy depends only on the radial part which is in agreement with Bohr theory there is a very important take home message. So, we have got 3 of the 4 quantum numbers that we know what we have not got is S spin angular momentum quantum number spin angular momentum quantum number well spin quantum number comes from Dirac statement which is very different we are not going to talk about spin at all in this course I think maybe a little bit when we talk about bonding maybe not. So, n, l and m are obtained and they make sense when we get them using this Schrodinger treatment of the hydrogen atom problem. Now, that being said let us have a look at the radial functions. So, this is what it is remember some constant multiplied by r to the power l e to the power minus r in it did I write that in the previous part perhaps not. So, sorry about that e to the power minus r by n n a 0 is required because the function has to vanish a vanishing infinity. If you just write r to the power l multiplied by this function will not be right. So, there was a typo in the previous slide e to the power minus r by n a 0 is very much there. So, these are the wave functions please do not try to find out the numbers here from this n minus l minus 1 and all that because this to do that you will need to know what this Laguier polynomials are because they also have some constants which get multiplied with this to get give you the constants that you see here. But let us have a look at the wave functions n equal to 1 l equal to 0 1 s that is a decay in terms of r e to the power minus r by a 0 n equal to 2 l equal to 0 has that same decay multiplied by 2 minus r by a 0 n equal to 2 l equal to 1 again there is no polynomial here it is just r by a 0 multiplied by e to the power minus r by a 0 where is that r to the power l term here well l equal to 0 so r to the power 0 is 1 what about here l equal to 1 so see r by a 0 is there n equal to 3 l equal to 0 now you have a polynomial of second order n equal to 3 l equal to 1 this has a polynomial of first order 4 minus 2 r by 3 a 0 so on and so forth and we will see how that determines the shapes of this wave functions what are nodes I think you are familiar with radial nodes and angular nodes and all that radial nodes are points where the radial part of the wave function goes through 0 and changes sign please remember and changes sign part of it so how do you get the number of radial nodes well it is n minus l minus 1 as you know where does it come from it comes from what is the degree of this polynomial as you see when l equal to 1 l equal to 0 what is the degree of that polynomial 1 minus 1 minus 1 1 minus 0 minus 1 that is 0 so that polynomial is not there when n equal to 2 l equal to 1 2 minus 0 minus 1 1 so this is this polynomial is of order 1 what about n equal to 3 l equal to 0 3 minus 0 minus 1 there is 2 a polynomial of order 2 so on and so forth so number of radial nodes comes from equating these radial parts of the wave function to 0 so higher the degree of polynomial more nodes you get okay so if I want to know what the where the where are the nodes you just equate this to 0 and you get a quadratic equation in case of 3 n equal to 3 l equal to 0 case you get a linear equation for n equal to 2 l equal to 0 case and solutions of r give you the position of nodes one question that students often ask us here is that how do you know that when this part is equated to 0 you are going to get real roots well we know because these are not just any polynomials the Laguerre functions Laguerre functions are such that when they are equated to 0 you do get real roots only you do not get imaginary roots that is the saving grace okay now let us take a closer look at the orbitals now before that let me formally say what an orbital is what is an orbital many of you believe that an orbital is a region of space in which probability of finding the electron is maximum that that definition is wrong an orbital is an acceptable solution of Schrodinger equation for a one electron atom that is what an orbital is so orbital is a wave function not just any wave function a wave function that you get by solving Schrodinger equation for a hydrogenic atom hydrogenic means only one electron is there z can be whatever one electron is there so helium plus would be a hydrogenic atom but you get from there is an orbital lithium 2 plus same okay please remember an orbital is an acceptable solution of Schrodinger equation for a one electron atom okay so now let us look at this 1s and 2s orbitals exponential decay in r this is what it looks like psi 2 0 0 where is this going to become 0 r by a 0 minus well 2 minus r by a 0 equal to 0 so r equal to 2 a 0 so at r equal to 2 a 0 we have a node what is a 0 the Bohr radius all right now here we show you the 3 radial parts 1s 2s and 2p okay as you see this 2p goes through a maximum we will come back to this later also but if you just go back to the equations I think you can see it very clearly where is 2p n equal to 2 l equal to 0 is 2s n equal to 2 l equal to 1 this is your 2p what do I have in 2p I have r by a 0 multiplied by e to the power minus r by 2 a 0 so r by a 0 increases with r naturally e to the power minus r by 2 a 0 decreases with r take a product you are going to get a function that goes through a maximum and we will have more to talk about that what are these these are square of capital R this is what your square of 1s orbital is again a fall which is a little steeper and for 2s orbital what do you get it falls off becomes 0 then increases again and tapers to 0 later on okay for 3s orbital sorry rather for 2p orbital what is it take square again it is a curve that looks like this what are these figures that are shown in in set we will take a rain check on that we will talk about it later but let me ask you a question what I see is that capital R square is maximum at r equal to 0 that is the nucleus for s orbital does that mean that probability of finding the 1s electron or 2s electron at the nucleus is maximum because if it is then we are back to square 1 we are back to the other 4 model what have we gained so here it is important to remember what bond interpretation is what is psi psi star or r r star in this case well no r star is there r square r is a real function anyway please remember that r square or psi psi star is probability density it is not probability so it is absolutely okay to say that for s orbital is the maximum probability density and after all these years I notice the density spelling is wrong here maximum probability density is at the nucleus fine probability density not probability what am I talking about you will see what I am talking about so remember what probability is probability density is defined at a point you cannot talk about probability of a point you have to define a volume element about that point and you can talk about probability in that small volume element and the volume element as we had established earlier is r square dr sin theta d theta d phi so if you want probability you need to integrate psi star psi with respect to this d tau so this is what it is going to be suppose I want to know what is the probability of finding the electron between some r 1 and r 2 even if r 1 is 0 I have to write some value of r 2 there that is going to be integral capital R square small r square dr between r and r 2 capital theta square sin theta d theta between 0 and pi capital phi star capital phi d phi between 0 and 2 for s orbitals what happens is in fact I do not even need to know for the discussion that we are going to do later if you understand that these two integrals would be some constant that is enough because for an s orbital s for s orbital only for s orbital what happens is that there is no theta phi dependence so these integrals naturally will be some constant so this capital R square multiplied by small r square is called the radial probability distribution function so it tells you how what is the probability of finding the electron at some distance small r well it means what is the probability of finding the electron between small r and small r plus dr for s orbital there is no theta phi dependence so this becomes 2 and this integral becomes 2 pi I recommend that you read this part from Atkins physical chemistry book it is discussed very very nicely there so if it is not an s orbital you only work with capital R square small r square dr if it is an s orbital and if you want to find out some probability you need to work with 4 pi multiplied by capital R square small r square dr but it does not matter for the discussion we are performing if you remember capital R square small r square that is enough what does it now mean for an s orbital especially what it means is that if I want to find out the probability of finding the electron s electron in a shell of thickness dr at a radius r so this is small r this is let us say when I write like this it means the separation between these two arrows is dr so what is the local locus of this point a circle what is the locus of this point actually a sphere not even a circle so what I get is a spherical shell right so probability of finding the electron in a shell of thickness dr is actually this small r square multiplied by capital R square is density and volume of the sphere is 4 pi r square well sorry what am I saying capital R square is the density probability density and 4 pi r square multiplied by dr that is your volume of the shell that is what gives us the probability now what happens for s orbital probability density is very high at small r equal to 0 but remember that it is being multiplied by small r square so the for the product the value at small r equal to 0 is actually 0 so you multiplying r square which is an increasing function of r by capital R square which is a decreasing function in r I am talking about s orbitals what will the product be the product will be something that goes through a maximum so maximum probability of finding the s electron is somewhere here wherever this curve becomes maximum it is not at r equal to 0 so this discussion brings about a very important understanding that probability density and probability are not one and the same probability density is one thing but if the volume element is very small then probability will be 0 so probability density is like your intelligence or competence or smartness and this r square dr this is like the amount of work you put in so if you are super smart but you do not work you are not going to do well so that total probability of doing well is 0 this is something similar to that let us say this is just an analogy do not take it too far but that is what it means it is not sufficient to have high probability density your volume element must also play a very very important role. So now let us see what this 4 pi r square capital R square is for different orbitals this is what they look like for 1 s orbital it goes through a maximum for 2 s orbital well if you just take square of it what will you get? Again if you search Anindya Datta Quantum in YouTube you will get a lecture series and in that series we have actually plotted in front of your eyes and shown how these functions arise I will not do it here I encourage you to do it use any graph plotting software of your preference and please work out it is a lot of fun. So what is capital R square for 2 s starts from a high value goes down to 0 is not a straight line increases and then tapers off to 0 at a long long r so see this volume is so much more than this volume is not it? So capital if you just look at capital R square that is what it is it is much higher in the vicinity of the nucleus but do not forget you have to multiply it by 4 pi r square in order to get this radial distribution function and 4 pi r square what will it look like it will be something like this that is why when you multiply this small value of capital R square by a large value of capital R square sorry when you multiply a small value of capital R square by a large value of 4 pi small r square this part actually blows up and in the vicinity of the nucleus since r is equal to 0 or very small value this becomes very small that is how you get this kind of a distribution two hums the one close to the nucleus is small the outer one is much much larger. Similarly these are the shapes that you get for 3 s, 3 p and 3 d radial parts and if I just plot them all together you see we will talk about two quantities here quickly average value of radius and most probable value of radius what is the most probable value of radius it is essentially the radius at which this essentially the radius at which these curves become maximum how do I find it just take dpr and differentiate with respect to dr please remember what dpr is please remember you have to take small r square multiplied by capital R square does the 4 pi matter not in this discussion because you are equating to 0 it does not matter whether it is 4 4 pi or whether it is 400 pi you are equating it to 0 anyway so when you equate when you differentiate r square multiplied by capital R square equate to 0 for whichever value this becomes 0 that gives you the r value at which you get a maximum in the curve that is called the most probable radius or most probable value of radius and see this if you can compare 3 p well 3 s 3 p and 3 d the most probable radius of 3 s is actually larger than that of 3 p which is much larger than that of 3 d this is something that I guess we have learned in the high school and it has been said that d orbitals are attracted more towards the nucleus because they are more diffuse we do not even have to go that into that in that course in this course please remember that well do not even have to remember this I am just showing you here that the most probable radius of 3 s is more than that of 3 p more than that of 3 d that is one quantity and what is the average value of radius that you know very well since I am working with normalized wave function this is sufficient expression for average value of radius when will they be same when will they not be same well if you have a distribution that looks like this a symmetric distribution then the average value and the most probable value will be one and the same most probable value is just the modal value however when the distribution is cubed something like this then what will happen this will be average value but look at the areas the area on the left hand side of average value is so much less than the area on the right hand side of the average value so the average value not average value sorry most probable value so the average value will actually be larger than the most probable value here you can just work this out to satisfy yourself taking some numbers you will see that average and most probable values need not be same unless you have a perfectly symmetric distribution which one is important both are important for different things. So we have performed a discussion of r dependent, theta dependent, phi dependent parts separately now to get those beautiful pictures that I had shown you earlier we need to put them all together that is what we will do in the next lecture and we will see what kind of pictures that we get and that is what gives us what we call shapes of orbitals and using these orbitals we will be actually able to generate the we will be able to work out the regions of space where probability of finding the electron is 90 percent or 95 percent 98 percent or something so what you see here is a depiction of the wave function the z axis vertical axis is actually wave function the other two axis can be x and y or whatever and you will see how we arrive at this kind of shapes for the orbitals that we are going to study in the next lecture.