0:30

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.

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1:00

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.

1:57

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.

2:52

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.

3:07

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.

3:34

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.

4:04

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.

4:57

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.

5:23

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.

5:58

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.

6:30

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.

8:02

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.

8:26

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

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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.

9:42

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.

10:01

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.

10:27

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.

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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.

11:25

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.

12:27

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.

12:53

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.

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