0:45

And we can simplify it and say that it's given by the negative of R, where

Â r is called the Rydberg constant, times z

Â squared, z remember is the, is the atomic number.

Â And then that's all divided again by,

Â by n squared your, your principle quantum number.

Â So R here is known as the Rydberg constant

Â [BLANK_AUDIO].

Â And, it's easy to remember this because it

Â corresponds to exactly 13.6 electron volts.

Â What you should also note from up here

Â is the [INAUDIBLE] the energy only depends on n.

Â Remember we had the three quantum numbers.

Â We had the n.

Â The l and the n sub l.

Â But the energies, levels for hydrogenic atom system

Â depend only on the principal quantum number n.

Â 1:45

So let's now try to plot these energy levels out into a

Â qualitative energy level scheme and see what they, what they look like.

Â So if we move down here.

Â What I've plotted out already is some energy levels, and basically

Â energy is increasing in this direction, in this direction here.

Â 2:33

Times z squared, all divided my n squared.

Â And that's giving the energies in electron, electron volts.

Â So if we consider here this n equals 1 and if we looked at this equation here then

Â n is 1 so n squared is 1, and z squared of course is the, is the atomic number.

Â 2:55

So if we take simplest system, the hydrogen.

Â So it's going to be 1, so the energy of e 1, is going to be

Â minus 13.6 electron volts.

Â If it was helium from number 2, then of course you would have z squared,

Â so z is 2, so z square is 4, so it'll be 4 times that.

Â 3:19

Likewise we go into lithium as we go on up the periodic table.

Â If z is 9 so it will be 9 times 13.6.

Â So let's stay with the hydrogen atom here.

Â So what we have on this side therefor we

Â know that the energy corresponds to exactly minus 13.60.

Â Electron-volts.

Â Next level, we know that n is equal to 2.

Â [BLANK_AUDIO].

Â So, now, we're going to stay with the the hydrogen atom for the, for the moment.

Â So, now we have minus 13.6 z again is 1.

Â So, now n is 2.

Â So, n squared is going to be 4.

Â So, the energy at this level is going to be 13.6 divided by.

Â 4, so if you do that you are going to get minus 3.40 electron volts.

Â The next one as we know is n equals 3, so now

Â you are going to get N squared is going to be 9.

Â So, it's going to be minus 13.6, divided by, by 9.

Â And if you do that, you should get a value of minus 1.51 electron volts.

Â And on up, n equals 4.

Â 4:56

And so on up, we are getting a little bit

Â tighter here at the moment, I'll explain that in a minute.

Â And then it becomes minus for n equals 5,

Â minus 0, sorry that's 0.54 sorry.

Â 5:13

So it's minus 0.54 electron volts.

Â Okay, so I'm just going to do one more here, so let's do n equals 6.

Â And then if you worked that out you're going to get the energy

Â level is m, [INAUDIBLE] divided by 36 or 30 plus 6 plus 6.

Â 13.6 by 6.

Â 5:55

And smaller.

Â And of course you can we know from the solutions of the the Schrodinger

Â equations that n can vary from one up to integer values up to n infinity.

Â And what you can keep doing is, you can imagine you come to

Â eventually n equals to infinity, and you get levels now are really coming close.

Â Really coming close together.

Â Coming in to a kind of continuum.

Â And of course when n is equal to infinity,

Â the energy infinity is going to be 0.00 electron volts.

Â 6:32

So this really is the energy levels that the solutions

Â that Schrodinger equation predict for the hydrogen, hydrogen atom system.

Â And as we say, the energy levels depend only on the principal quantum number.

Â So when we're talking about n equals 1 here.

Â 6:49

We've already said that this, of course, corresponds to the wave function,

Â psi 1, 0, 0, where 1 is the principal quantum number and the l and the nl are 0.

Â So, there's only one energy level.

Â Or there's only one energy state, for that energy level.

Â 7:22

Let's see if we can fit them in here.

Â So it's psi 2 1 0.

Â Which we would refer to as 2 p z.

Â But we've also got the other one, psi 2 1 1.

Â And last you've got psi 2 1 minus 1.

Â So what you can see here is for each, for

Â the integer level here it's just it's just one state.

Â But for energy level 2.

Â That's four states.

Â Likewise I'm sure you can show that for n equals 3.

Â You're going to get 9.

Â And if it's 4 you're going to get 16.

Â So if you like to we call this.

Â There's Force four-fold degeneracy for n equals 2.

Â Nine-fold degeneracy for n equals 3.

Â So the degeneracy and energy levels is equal to, is equal to n squared.

Â 8:17

At this stage your, I suppose a valid question.

Â Free to ask is, how do we know that this is, is, is true?

Â Is there any experimental evidence to support

Â these energy levels that the Schrodinger equation predicts?

Â And the answer is that there,

Â there is indeed, and this evidence was

Â presented well before quantum theory was presented.

Â And what I show down here is both the, what

Â we call the absorption, sorry this is the absorption spectrum.

Â 9:05

Now this is in the visible,

Â just talking about the visible region of the electromagnetic spectrum.

Â And you should be know of the spectrum of colors going from, from red.

Â Up to violet, this is the well known visible region spectrum.

Â So what's happening here, we aren't going to go into any of the experimental

Â details of how this is done, but this is when you say you shine, visible light.

Â And you pass through hydrogen atoms.

Â 9:41

What you find is what comes out is you have your, your visible spectrum

Â again, but you notice these lines in

Â the spectrum, these dark lines in the spectrum.

Â 9:58

Now you can also perform this on what we call an emission spectrum, and this is

Â where you have an excited state of an atom, in this case, and it emits energy.

Â And what you you'll find is it's the opposite here to the absorption spectrum.

Â You get a line.

Â 10:30

So these as I say were known

Â well before people knew about quantum, quantum mechanics.

Â And they were an intriguing.

Â Intriguing spectrum and intriguing mystery if you like.

Â 10:44

Now what I've done here is, this is in the visible region.

Â So I've actually given the exact values that people measure for these lines.

Â So the red line is 656.

Â The first blue line here, blueish green line is 486.

Â 434 410, and so forth for, for the other lines.

Â So what's interesting, though, if you go back up to

Â the, spectrum or the energy levels for, for, for hydrogen.

Â As I said for the particle in a box, what quantum mechanics is

Â telling us is that these are the only energy levels allowed for this system.

Â You can't get intermediate levels, there a ladder

Â as I said before of, of energy levels.

Â You can't have intermediate, intermediate levels.

Â So when the, the hydrogen atom the will exist in it's.

Â Lowest energy level, which is n equals 1.

Â But to get up to these other energy levels that are also with

Â the quantum mechanics that is also available

Â for it, then it will require energy.

Â To get there.

Â Now what's intriguing is, if you, you can work out, so

Â you can work out the energy difference between these different levels.

Â 12:01

Now what's intriguing from what we've been talking about here is that

Â if you do measure the energy level, say between n equals 2.

Â And n equals 3, so that's simply

Â a difference between 3.4 and 1.51 electron volts.

Â What that comes out to be is exactly equal to the energy,

Â of this line here, the 656 nanometer line.

Â Because this stuff corresponds exactly.

Â So the difference between 3.40 and 1.51 electron volts.

Â So what we can do is we can, let's just draw a line in there.

Â 12:59

Likewise you find that for the light-blue, bluish-green line,

Â it corresponds exactly to the energy difference between the n equals

Â 4 and the n equals 2 level and the blue

Â line corresponds to the difference between the n equals 5.

Â And n equals 2 line, and the last line in the indigo

Â corresponds to the n equals 6 and the n equals 2, 2 level.

Â So, this is pretty impressive evidence

Â that the Schrodinger equation is predicting.

Â 13:43

These experimentals transitions exactly, because

Â the energies here that are measured

Â experimentally correspond exactly to the predicted

Â values given by the Schrodinger equation.

Â Now, over here, this series of lines, as I said,

Â this was originally known well before quantum theory was known.

Â And, this series of lines was called the, the,

Â the Balmer series after the person who first noticed them.

Â There's also other lines that people, line spectrum

Â people observe, and these are more into the,

Â in this region here, as you go into UV and also as you decrease in energy.

Â Long wavelength.

Â You go into the infra-red.

Â And these, actually.

Â Again, if you can show that these correspond to transitions between the one.

Â From the 1 level.

Â What from the 1 level absorption or to the 1 level in the emission spectra.

Â So you have 1 to 2, 1 to 3, 1 to 4, and so forth.

Â And then they're called [UNKNOWN] series

Â again after the scientist who discovered 'em.

Â And you'll also find other transitions, from the

Â n equals 3 levels [UNKNOWN] and these of

Â course are lower in energy so they're going

Â to fall in the infrared region of the spectrum.

Â And again, you can find the

Â exact correspondence between the values predicted.

Â From the solutions of the

Â Schrodinger equation and the experimental measurements.

Â [BLANK_AUDIO]

Â