 We are broken off at this point in the last module now we will complete the discussion we are in the middle of the question about probability density and actual probability and we said that it is important to remember that psi psi star is really probability density in this case R R star. So, for S orbitals maximum probability density of finding the electron at the nucleus is well probability density is maximum at the nucleus for an S electron as we will see probability is actually 0. So, let us go ahead with that and now I do not remember whether we have done it in a previous module but we will just do it again if required. Let us work out an expression for the volume element in spherical polar coordinates. I think we have done it earlier once but we will still do it. See how do I get a volume element if it is if you are talking about x, y, z then it is simple increase x by a small amount dx increase y by a small amount dy increase z by a small amount dz. So, the volume element would be dx, dy, dz when we talk about spherical polar coordinates what we do is this is one coordinate R up to here is R we increase that coordinate by an amount dr. So, that is one side of our volume element. Next what we do is next coordinate is theta. So, we increase theta this is your theta by a small amount d theta. So, the side that we generate of the volume element there is this essentially this one and what is the length of this arc well R multiplied by d theta. d theta is a small angle remember. So, since it is a small angle is going to be this R multiplied by d theta. This is again something that we learned in very simple three-dimensional coordinate geometry in 11, 12. The expression for the length of an arc R d theta. So, two sides of the volume element are defined now the third side is this one. To get this volume element what we should do is we should go down to the x, y plane because that is where phi is defined this here this angle is phi the angle between the projection of the position operator in the x, y plane with the x axis. So, increase that angle by a small amount d phi remember this length what is this length this is R sin theta. So, what will the length of this arc be R sin theta d phi and essentially what we do is this is the length of the arc we sort of take it back. The length of this arc is equal to the length of this arc it is the same value R sin theta d phi. What then is the volume element you have one R from here another R from here. So, even though it is written later on I will just write it we get R square then let us write everything in theta we get a sin theta from here. So, sin theta might as well write d theta and anything else in phi no I should write dr maybe R square dr sin theta d theta d phi this is the volume of the volume element and this is what is going to lead to our answer what is this volume element for R equal to 0 for R equal to 0 see R square multiplied by we do not care what else is there R is equal to 0 means R square equal to 0. So, volume element for R equal to 0 is equal to 0. So, no matter how high psi square might be since the volume is 0 the probability which is a product of probability density and the volume is equal to 0. We take the heaviest thing in the universe if it is volume is 0 it is not even there. So, mass is 0 similarly probability is a product of volume and probability density if the volume itself is 0 then even for a high probability for a high probability density probability will be equal to 0. So, this is the reason volume element is R square dr sin theta d theta d phi. Now, let us actually do a little more rigorous calculation of the expression for probability p of course is equal to integral psi psi star d tau within the required limits. So, let us write this out. So, we get what is psi psi is capital R into capital theta into capital phi and you might remember that capital R is your capital R is real capital phi is also real sorry capital theta is also real capital phi is an imaginary quantity. So, when we write mod psi square that is psi star psi that boils down to R square capital theta square capital phi star multiplied by capital phi of course and what I have written here is essentially what you have here is R square R square it is capital theta square here capital phi star capital phi is here and you might remember what capital phi is capital phi is 1 by root over 2 pi e to the power i m phi. So, you can figure out what capital phi star is going to be. Now, let us simplify this a little bit. So, this here capital R square into R square into dr is called the radial probability distribution function. I will come back to this dr quantity a little later we will see why we should consider that also. The second one is capital theta square sin theta d theta for an S orbital we know that this these do not make any contribution. So, capital theta square you can take as 1. So, that is the case then integral 0 to pi sin theta d theta what will it be minus cos theta d theta cos theta is equal to 1 for theta equal to 0 minus 1 for theta equal to pi. So, we are going to get 1 minus minus 1 that is equal to 2 and what will I get here if you integrate this we will get from the normalization constant we will get 2 pi. So, what we get is 4 pi square. So, for an S orbital the radial probability distribution function that we work with is 4 pi square multiplied by capital R square into R square dr. For others we just work with capital R square into R square dr this part you can study from some basic book like Atkins physical chemistry. So, now what we want to do is we want to find the probability of finding the electron in a shell of thickness dr at radius R this is what it boils down to. So, what we are saying is that let us consider a shell of uniform thickness dr at different values of R 4 pi R square into capital R square into dr dr is a constant. So, this is for S orbital for others this 4 pi will not come R square is of course a parabolic increasing function in R and what about this capital R capital R for S orbital is maximum at R equal to 0 then it becomes 0 asymptotically. So, now see since R square is small R square equal to 0 at R equal to 0 this product 4 pi R square capital R square dr becomes 0 as 4 pi R square dr becomes 0 at R equal to 0. So, this is what we get. So, actually the maximum is somewhere in the middle maximum probability would be somewhere in the middle for one S orbital for 2 S orbital we are going to get a little more interesting plot. So, for S orbitals maximum probability density of finding the electron is on the nucleus no doubt, but probability of finding the electron on the nucleus is actually 0 why because the volume element volume of that element is actually equal to 0. So, it does not matter how high the value of R R square or psi star S for R equal to 0. This is what gives an answer to the apparently problematic question of why is it that the electron S electron why is it that it does not reside at the nucleus most of the time. So, here we are these are the different wave functions what we have said. Now, if you work out this radial distribution functions radial probability functions what will we get for 1 S orbital we already know what we are going to get it will go through a maximum for 2 S orbital it has to go through 2 maximum. Now, remember that okay maybe I show you the answers first this is the answers let us go through the pictures one by one. For 1 S orbital we have discussed it already somewhere it is going to be maximum the probability density is R square R square. Now, for 2 S orbital remember if you just look at psi or psi square the inner lobe is much bigger, but when you look at the probability distribution along radius you see the outer lobe becomes the major lobe why because you are multiplying by small R square as you go out R increases R square increases as its second power. So, even though the value of the wave function itself or its square is very small at say 5 or 10 units of R the value of R square multiplied by capital R square small R square multiplied by capital R square blows up because this R increases as second order of small R that is why as far as probability is concerned the outer lobe is the major lobe. Similarly, for 3 S orbital you have only one maximum for 3 P orbital this is what you get okay no sorry I think this is 2 P orbital not 3 S sorry you have carried away little bit for 2 P orbital there is only one maximum no node for this one for 3 S orbital there are 2 nodes remember and if you just look at psi square it will be something like this will decay. But since you are multiplying it by R square the innermost band is smallest followed by the next one the biggest one is on the outside. So, when we draw this what we are really drawing is the plot of finding the electron along plot of probability of finding the electron along the radius as R increases how does the probability of finding the nucleus change okay this is what we are generating using the orbital in this case capital R and right now we are only working with R we have not worked with theta phi and this is how we can generate those regions of maximum regions of maximum probability of occurrence of the electron by using the orbitals okay and then if you just overlay all of them this is the kind of figures that we get this is where maximum probability of finding 1 S is 2 S interestingly 2 P has a maximum at a smaller value of R compared to 2 S then again 3 S of course has a maximum that is at a much larger R compared to 2 S but 3 P has a maximum for a smaller value than 3 S 3 D has a maximum at smaller value than 3 P these become very important when we talk about multi electron atoms and when we consider phenomena like shielding okay but then how do we find where these are how do we know what is the max what is the probability where the probability is maximum these are 2 very important things we can find out the average value of radius in the usual way by calculating the expecting expectation value okay of course here the expression means that we are working with the normalized wave functions and also we can find the most probable value by differentiating the probability with respect to R and equating it to 0 that will give us the maximum okay so this is these are things that we can do using the radial part of the wave function and this is what we will do in our assignments as well. So, we close this discussion what comes up next is very interesting we have talked about phi theta R dependent parts separately now we are going to learn what happens when we put them together and we put them together we generate beautiful pictures like this what is this picture this axis is actually the wave function these are the other axis so of course the problem here is that we have R theta phi and we have psi so if I have to draw a picture of psi as a function of R theta phi all together then I need 4 dimensional space but I only have access to 3 in fact when I draw on a paper I only have access to 2 so how do we handle this problem how do we represent the orbitals which seem to be a 4 dimensional we seem to require 4 dimensional space to depict how do we depict them in 3 dimensions or 2 dimensions do we take sections do we use some other trick how do we generate figures like this that is what we are going to discuss in the next couple of modules.