[BLANK_AUDIO] Hi there. So, in this presentation we're going to move on to discuss the application of the Schrodinger equation to what we call hydrogenic. Hydrogenic atoms. Now, hydrogenic atoms are one electron atoms. They just contain one electron, and that is interacting with the, the nuclei. So, the classic case you all know about is beta hydrogen atom, but this could also apply to the helium plus atom, and so on, up the periodic table, so long as you make sure you take all the electrons off, and you're just left with one one electron. So, another example said would be carbon 5 plus and right on, say uranium, uranium 91 plus. [BLANK_AUDIO]. Okay. So the good thing about the systems, if you like, is that you can solve the Schrรถdinger equation exactly for these systems. When you get more than one electron, like you would have say in the helium atom, then you cancel all the sure equations exactly. There are approximations you can make and this really is the basis for what we call electronic structure calculations that are carried out now with large computers. But when you just got the one electron interacting with the nuclear charge then you can solve the Schrรถdinger equation exactly. And of course you may not, may not already know this, but the solutions that come out of Schrodinger equation for the hydrogenic atoms are the s, and the p, and the d and the f orbitals. That you're probably already familiar with from your general chemistry course. So how would we go about solving the Schrรถdinger equation for a system like this? Well it's a little bit more difficult than the, just simple systems we've been looking at so far, like the parked killer in the box. So first of all you have to look at the energy interactions that are going on. And the the model that you will pick for the for the atom is the model proposed first by Rutherford. And Rutherford purposed that you have an atom as a, you have a central nuclear charge. And the charge on that is going to be positive, and it's going to be the atomic number z times the electronic charge. And then his model proposed that you have an electron here. Charge electron, and that's circulating around the positively charged nucleus, and you described the distance between the electron and the centrally charged nucleus as corresponding to r. Now from our preliminary knowledge of quantum mechanics already, you should, this is probably a simplistic model, because we now know that for a microscope particle like an electron, it's movement cannot be described in a trajectory, like we're implicating here. So, we'll stick with this model at the moment, because it's, it's, it's our purpose for now. So how would you describe that that system in terms of the Schrodinger equation. Well, first of all let us write out the Schrodinger equation that we saw previously. And let's write it out in, in, in three dimensions here. So what we has was, we have minus h bar squared all over 2m, del squared, this is out laplacian operator. And then that was our potential energy term and then you have the potential energy which we designated as V that operates on a, y function and we are talking about three dimensions here, so that's x, y, z. And that's equal to E psi of x, y, z. And then we said that this is our Hamiltonian operator, so we couldn't write that also shorthand notation x, y, z is equal to E psi of x, y, z. Now, in our particle in a box model, what we did was, we pretty much said that, we said that the potential energy term here was 0. Now in this case, of course, we can't do that because we've said here in the Rutherford model, that you have a nuclear charge, positive nuclear charge interacting with a, the negative electron at distance r. So we know that the potential energy is given by, by Coulomb's law between, the interaction between two, between two, two charges. So, for this system here, we could write the potential energy. We'll write the potential energy V, and let's say it's the function of r the distance between it. And we know, we should know, that that's the product of two charges, so its minus z times plus z e. So it's going to get minus z e squared. And the coulomb losses is proportional to 1 over, the distance between the two charges r and just because we like to work in SI units, you also need to include 4 pi epsilon 0, where epsilon 0 here is the permittivity of a vacuum. Now we have written this in terms of r the distance, but you can also write r, as the square root of x squared plus y squared plus z squared if you are using a cartesian coordinate system. This is just an extension of the Pytharogus relationship. So now, let's write out the Schrodinger equation for our, hydrogenic atom system. And this would be, if we move down here a little bit, we would have the the kinetic energy term. So we're going to have minus h bar squared all over 2m times del square and that's the kinetic energy term for the electron. And now we have to write in our potential energy term and we said that above there that's z e squared, minus z e squared, all over 4 pi epsilon 0 and then we have, we write it using cartesian coordinates. So, we have the square root of x squared plus y squared plus z squared or if you like, or the distance between the electron and the nucleus. And then we're operating on our wave function psi x, which is a function of x, y, and z, and then you said that's equal to E psi of x, y, z. And if we use our shorthand notation, we can say H psi of x, y, z is equal to E psi of x, y, z. So now we've written, we can write out our Schroedinger equation. So now we should be able to, we should be able to solve that. The problem, however, is that we can't really solve the Schroedinger equation for this system in cartesian coordinates. Because to do to solve the Schroedinger equation for this system, we have to be able to do what is called separation of the variables. And for separation of the variables, you can't do that in the x, y, z system. For the system as, for the, for the hydrogenic atom system as defined. So what we must do, we move down here. Is we must transform our system into what we call spherical polar coordinates. [BLANK_AUDIO] And the, spherical polar coordinates, is the easiest way to, to understand these is if you write them on a, on a globe type fixture here. So very similar to the idea of locating a point. So let's say we have our point here So we have a point here and, and this is our x, y, z coordinates in our cartesian coordinate system. What you do in spherical coordinate system, is you define this point here in terms of a distance here from say the origin zero. So it defines a distance r. And then you define it in terms of an angle theta that it makes with the z-axis. So if you like that we'll then, if we define the distance r and then the angle theta we would describe a cone along our globe here. And then our in that line of latitude. And then to define the exact point we, on that line of latitude, we would have to define here an angle phi with respect to the x-axis or if you like in the, in the global scale we would be defining a line of, of longitude. So in that fashion we can transform our core system into an r theta and a 5. It's not that difficult to transform between these two coordinate systems and you may have done this in some of your school mathematics, but let's just show that you can say that z is a coordinate, z equal to r cosine theta. Your y coordinate is equal to r sine theta, sine of phi and your x coordinates r sine of theta cosine of phi. So these are easily, quite easy to work out using simple trigonometric factors. So the key point here now, as we've transformed our Schrodinger equation from our cartesian coordinate system x, y, z and we wrote it up there as, like that. [BLANK_AUDIO]. And what we've done is we've transformed that now and we write it as H psi of r theta phi and that's equal to E psi of r theta phi. So the key thing that this transformation has allowed us to do now is we can, as I mentioned above, we can separate, the variables. We couldn't do this in the x, y, z system, but we can't when we define as in terms of r theta and phi. So what it means really is that if we have our psi of r theta and psi then we can define that as a function. Simply of R, which depends on the r, and we can multiply that by another function which depends solely on theta and phi. And what we call these things then, as what we call is we call this thing [COUGH] radial part. And we call this the angular part of the wave function. And indeed, we might go into this little later, this angular part, y of theta phi can itself be separated it into two variables. One just a function of theta and one just a function of, of phi. And just to point out here that an angular part, we'll show this later on, this angular part. Describes the shapes of the orbitals for the halogenic type atoms. So the familiar spherical shape and the dumbbell shaped that you see for the s and p orbitals respectively. That is defined by the angular wave functions, that come out after solution of the Schrรถdinger equation for the hydrogenic type atoms. [BLANK_AUDIO]