0:30

So, the classic case you all know about is beta hydrogen atom,

Â but this could also apply to the helium plus atom, and so

Â on, up the periodic table, so long as you make sure you

Â take all the electrons off, and you're just left with one one electron.

Â So, another example said would be carbon 5

Â plus and right on, say uranium, uranium 91 plus.

Â [BLANK_AUDIO].

Â 1:00

Okay.

Â So the good thing about the systems, if you like, is

Â that you can solve the Schrà¸£à¸–dinger equation exactly for these systems.

Â When you get more than one electron, like you would have say

Â in the helium atom, then you cancel all the sure equations exactly.

Â There are approximations you can make and this really is the basis for

Â what we call electronic structure calculations that

Â are carried out now with large computers.

Â But when you just got the one electron interacting with

Â the nuclear charge then you can solve the Schrà¸£à¸–dinger equation exactly.

Â And of course you may not, may not already

Â know this, but the solutions that come out of

Â Schrodinger equation for the hydrogenic atoms are the s,

Â and the p, and the d and the f orbitals.

Â 1:57

So how would we go about solving

Â the Schrà¸£à¸–dinger equation for a system like this?

Â Well it's a little bit more difficult than the, just simple systems

Â we've been looking at so far, like the parked killer in the box.

Â So first of all you have to look at the energy interactions that are going on.

Â And the the model that you will pick for the

Â for the atom is the model proposed first by Rutherford.

Â And Rutherford purposed that you have an atom

Â as a, you have a central nuclear charge.

Â 2:52

Charge electron, and that's circulating around

Â the positively charged nucleus, and you described

Â the distance between the electron and the

Â centrally charged nucleus as corresponding to r.

Â 3:07

Now from our preliminary knowledge of quantum mechanics already, you

Â should, this is probably a simplistic model, because we now

Â know that for a microscope particle like an electron, it's movement

Â cannot be described in a trajectory, like we're implicating here.

Â 3:34

So how would you describe that that system in terms of the Schrodinger equation.

Â Well, first of all let us write

Â out the Schrodinger equation that we saw previously.

Â And let's write it out in, in, in three dimensions here.

Â So what we has was, we have minus h bar squared all over

Â 2m, del squared, this is out laplacian operator.

Â 4:04

And then that was our potential energy term and then you have the potential

Â energy which we designated as V that operates on a, y function and

Â we are talking about three dimensions here, so that's x, y, z.

Â And that's equal to E psi of x, y, z.

Â And then we said that this is our Hamiltonian operator, so we couldn't

Â write that also shorthand notation x, y, z is equal to E psi of x, y, z.

Â Now, in our particle in a box model, what we did was, we pretty much

Â said that, we said that the potential energy term here was 0.

Â 4:57

Now in this case, of course, we can't do

Â that because we've said here in the Rutherford model,

Â that you have a nuclear charge, positive nuclear charge

Â interacting with a, the negative electron at distance r.

Â So we know that the potential energy is given by, by

Â Coulomb's law between, the interaction between two, between two, two charges.

Â 5:23

We'll write the potential energy V, and let's say

Â it's the function of r the distance between it.

Â And we know, we should know, that that's the product

Â of two charges, so its minus z times plus z e.

Â So it's going to get minus z e squared.

Â And the coulomb losses is proportional to 1 over, the distance

Â between the two charges r and just because we like to

Â work in SI units, you also need to include 4 pi

Â epsilon 0, where epsilon 0 here is the permittivity of a vacuum.

Â 5:58

Now we have written this in terms of r the distance, but you can also write r, as the

Â square root of x squared plus y squared plus

Â z squared if you are using a cartesian coordinate system.

Â This is just an extension of the Pytharogus relationship.

Â 6:30

And this would be, if we move down here a

Â little bit, we would have the the kinetic energy term.

Â So we're going to have minus h bar squared all over 2m times del

Â square and that's the kinetic energy term for the electron.

Â And now we have to write in our potential energy

Â term and we said that above there that's z e

Â squared, minus z e squared, all over 4 pi epsilon

Â 0 and then we have, we write it using cartesian coordinates.

Â So, we have the square root of x squared plus y squared plus z squared or if

Â you like, or the distance between the electron and the nucleus.

Â And then we're operating on our wave function

Â psi x, which is a function of x, y, and z, and

Â then you said that's equal to E psi of x, y, z.

Â And if we use our shorthand notation, we can say

Â H psi of x, y, z is equal to E psi of x, y, z.

Â So now we've written, we can write out our Schroedinger equation.

Â So now we should be able to, we should be able to solve that.

Â The problem, however, is that we can't really solve

Â the Schroedinger equation for this system in cartesian coordinates.

Â 8:02

Because to do to solve the Schroedinger equation for this system, we

Â have to be able to do what is called separation of the variables.

Â And for separation of the variables, you can't do that in the x, y, z system.

Â For the system as, for the, for the hydrogenic atom system as defined.

Â So what we must do, we move down here.

Â 8:26

Is we must transform our system into what we call spherical polar coordinates.

Â [BLANK_AUDIO]

Â And the, spherical polar coordinates, is the

Â easiest way to, to understand these is if you

Â write them on a, on a globe type fixture here.

Â So very similar to the idea of locating a point.

Â So let's say we have our point here So we have a point here

Â and, and this is our x, y, z coordinates in our cartesian coordinate system.

Â 9:42

And then to define the exact point we, on that line

Â of latitude, we would have to define here an angle phi with

Â respect to the x-axis or if you like in the, in

Â the global scale we would be defining a line of, of longitude.

Â 10:01

So in that fashion we can transform our core system into an r theta and a 5.

Â It's not that difficult to transform between these two coordinate

Â systems and you may have done this in some of your

Â school mathematics, but let's just show that you can say

Â that z is a coordinate, z equal to r cosine theta.

Â 10:27

Your y coordinate is equal to r sine

Â theta, sine of phi and your x coordinates

Â r sine of theta cosine of phi.

Â So these are easily, quite easy to work out using simple trigonometric factors.

Â So the key point here now, as we've transformed our Schrodinger equation from

Â our cartesian coordinate system x, y, z and we wrote it up there as, like that.

Â [BLANK_AUDIO].

Â And what we've done is we've transformed

Â that now and we write it as H psi of r theta phi

Â and that's equal to E psi of r theta phi.

Â 11:25

So the key thing that this transformation has

Â allowed us to do now is we can, as I mentioned

Â above, we can separate, the variables.

Â We couldn't do this in the x, y, z system, but we

Â can't when we define as in terms of r theta and phi.

Â So what it means really is that if we have our psi of r

Â theta and psi then we can define that as a function.

Â 12:27

And we call this the angular part of the wave function.

Â And indeed, we might go into this little later, this angular

Â part, y of theta phi can itself be separated it into two variables.

Â One just a function of theta and one just a function of, of phi.

Â 12:53

And just to point out here that an angular

Â part, we'll show this later on, this angular part.

Â Describes the shapes of the orbitals for the halogenic type atoms.

Â So the familiar spherical shape and the dumbbell shaped

Â that you see for the s and p orbitals respectively.

Â That is defined by the angular wave

Â functions, that come out after

Â solution of the Schrà¸£à¸–dinger equation

Â for the hydrogenic type atoms.

Â [BLANK_AUDIO]

Â