So where we go to yesterday was that we had reduced the horrible partial differential equation in six variables,
which is posed by the time independent Schrodinger equation for a hydrogen like ion two to this equation here.
Sorry. I'm being premature. To this.
So the internal motion had been separated from the translational motion of the atom as a whole,
and we had exploited we by using a change of variables, the central mass coordinates and the separation variable.
And we had used the work that we did with the angular momentum operator earlier on to show that we could express the,
the internal Hamiltonian in in in this format in this format here.
So this was called this was called. Ah, I think that's right for the internal measurement.
Right. And then I pointed out that this Hamiltonian computed with the total with the angular momentum operators
because the only mention of the Angles feature and Phi sits inside this total angular momentum operator.
And therefore we can seek therefore we can seek stationary status.
There's going to be a complete set of stationary states, which is simultaneous eigen functions of E and L, right?
So we will also have the statement that L squared on E and L is equal to l, l plus one, e and L.
And then when we're using these stationary states, the action of this this operator can be replaced by this eigenvalue.
And now we do end up with what I wrote originally, which is that we're going to have a Hamiltonian which has a subscript L,
meaning it's the one that's valid for four stationary states which have total incrementing quantum number L which is going to
be PR squared over two mu plus l l plus one h bar squared over two mu squared minus z squared over four pi epsilon nought.
Oh. And this is fundamentally a Hamiltonian analogous.
This is a Hamiltonian analogous to that for the harmonic oscillator in the
sense that we have now only one surviving coordinate of our six coordinates.
We're down to one. That's this here, the modulus of the separation vector and it's conjugate momentum.
PR So we've made an enormous simplification, made tremendous progress,
and we're going to knock this into submission using the same approach as we did with the harmonic oscillator.
We're going to define what will turn out to be a letter operator ale, which is going to be defined to be dimensionless.
It's going to be a zero of a root two.
I better write this down from notes rather than my memory, because these factors do matter.
I p r over each bar minus l plus one over a plus z of l plus 1a0 where a zero is the
following not very helpful item for pi epsilon nought h bar squared over mu e squared.
So this is an object. I'll make it convincing. Well, let's make it convincing now.
What is this object? This is the. This is the so-called ball radius.
So it was introduced by Niels Bohr before there was proper quantum mechanics using what turns out to be a fallacious model of hydrogen, the poor atom.
And the way to see what is dimensions are are actually to to write, to bring this up to this side and say,
look, this is going to be squared over four pi epsilon nought a nought on this side.
And what's on the top, not the bottom. So therefore, I have to say this is equal to h bar over a nought squared of amu.
So out of this equation by putting that E on the top and this for perhaps in the nose on the bottom,
dividing through by a nought squared to get this on the bottom so that this is something we understand to be this is obviously electrostatic energy.
And what's this that we see on the right side? Well, H Bach is momentum, right?
So this is more or less P squared over mass. This is the reduced mass recall.
This is the reduced mass, more or less the mass of an electron.
So what we have on the right hand side is p squared.
When it is P squared for k is equal to one over a nought.
So if you have if you have a wave which has a wavelength which is comparable to one over the this scale radius here sorry.
A wavelength comparable to this scale radius here. Not worrying about two PI's at the moment.
Then what you have on the right here is.
So this side is two times kinetic energy associated with a particle which has a wavelength which is on the order of zero.
And what we have on the left side here is the electric energy. So the scale that's being said this, this, this dimension of this dimensional quantity,
the ball radius is the natural scale at which the kinetic energy associated with the uncertainty principle,
the zero point in it,
what with the harmonic oscillator we would have called the zero point motion is is on the is on the order of the electrostatic energy.
That's dimensionally where this number comes from. That number comes from so with.
The thing to notice now is that ail is dimensionless.
Why is it dimensionless? Well, this cancels the dimensions of this.
Obviously this in. This cancels this. This is all dimensionless.
This factor is zero. And here we're looking at again, this is PR divided by with this put on the bottom h for over a zero.
Therefore H Bach. Therefore something with the dimensions of P. So this thing here, this operator here is dimensionless.
Same as the letter operations in the harmonic oscillator with dimensionless.
Okay. So what do we do with that? What we do is we calculate what a.
And this thing, of course, carries this subscript little L because little L the orbital quantum number is appearing in in its definition.
And what we do is we now calculate l dagger a l.
Right. So what does this give us?
We have all this in a zero squared of two out front because we're going to have two zeros and root twos and then we're going to have PR.
We showed was a commission what we we engineered that it was a commission operator.
So when I take the commission now joint of what I have up there, I get a minus IPR.
The minus sign is from the I over H bar and then the rest of course is the same because it's it's commission, it's just boring numbers plus z.
Well, actually, this is an operator, strictly speaking, but we're in the position representation,
so it looks like a number and that has to be multiplied onto IPR over each bar.
So no minus sign because this is Al I'm writing down now minus L plus one overall plus Z overall plus 1a0.
So we have to multiply this stuff out and the way we do it is we regard this in the back here as one factor and that in the front is another factor.
So this is looking like the product of a number minus minus some number, a number plus some number.
Right? So the usual pattern, just so this is mirroring very closely what we did with the harmonic oscillator.
So we get a zero squared over two.
These two obviously multiply together and we get PR squared over each bar squared and then these I have to multiply together.
So what do we get? We get the square basically of this number, this of this number here.
So we have plus l plus one overall squared plus Z over L plus 1a0 squared minus the cross term here which is going to be 12.
There'll be two minus twice twos and those are going to cancel and we will have a zero or so.
That's, that's the, the two easy parts. Right, because.
It's the front things squared, plus the back thing squared.
And now we have to think about the cross terms which would vanish because this is a plus B into A minus B spiritually,
these cross terms would vanish if we weren't dealing with operators,
and they fail to vanish only because we are dealing with operators so that there's a failure of computation.
Otherwise the PR well, okay, so PR commutes with this.
So for the cross terms, we do we don't get anything from this thing on this,
but we do get something from this thing on this, namely we get the relevant commentator.
So the extra term that arises because we're working with operators is going to be.
I. Overreach bar.
That's that. So. So this minus sign and this minus sign cancel.
We have that L plus one and then I have a p r comma one overall plays big bracket.
That's what I'm going to get from this term on this term.
Not cancelling on this term, this term on this term.
Okay. So that's going.
So what? So what is this going to come to? Well, let's rearrange things.
He's very squared of two PR squared over each bar squared plus L plus one squared of R squared minus.
I'm going to put this one down next to Z of a zero.
Then this term plus z squared over L plus one squared.
A zero squared. When I have to do this.
What I remember is something we handled very last term that when I had to do it arose when I was doing that,
when we were doing the comitato of P and V, the potential function of X that turned out to be the right DV by the X Times.
The commentator of P and X. This is exactly the situation we have here because this is the operator canonically conjugate to r.
So what we have here is the derivative of one overall, which is so minus i l plus one l plus one over h bar two times minus one over
all squared because that's the derivative of one over all times PR commercial.
But this is a piece of a canonical commutation relation.
This is equal to minus I. All right. We show that oh come a PR is equal to H bar.
So in this order it's minus H bar. So we have rather a load of minus signs.
Let me see. I think we have one, two, three minus signs and another minus sign coming from here.
So I think in total we have a plus sign. So this stuff here, I believe comes to L plus one over each bar R squared.
Sorry, the bar get the bar cancels this edge bar on top.
That's on the bottom. So we get an L plus one over squared and this can now be combined with this.
Except I've got the wrong blinking sign. Let me just double check that.
Yep. I'm looking for minus. Is it mine as well?
I want it to be mine. So let's declare it to be minus. And I'm sure it is minus.
I'm sure it is minus. Let's not spend time chasing down some wretched sign because now what we're going to do is combine this.
So let's this is the side calculation here. We're going to have L plus one overall squared brackets, L plus one from up here.
That's the old plus one squared minus one times this stuff.
So you can see we're going to fit this and this is going to give me an L, l plus one, which is exactly what I want.
So this is going to be a nought squared over to p r squared over each bar squared plus l l plus one.
Overall squared minus two Z over 800.
Close brackets, and then I'm going to take this and join in on that.
So we get plus garbage in the garbage term is z squared over two L plus one squared.
This should all be dimensionless. I think it probably is dimensionless on the grounds that it's the product of two dimensionless operators.
Now what is the point of this ridiculous exercise? The point is that we should see the original hamiltonians peaking out here.
We should basically have our original Hamiltonian plus garbage.
So the. In order to get our original Hamiltonian, we need to have a new under here and a mew under here.
So why don't we multiply by a meal on the top? On the bottom?
So this is a nought squared mew and we want to take this bar outside and then that won't be under there as we want.
We can allow that to to wander inside and then this bracket becomes hl this stuff here becomes HL.
And we still got unwanted garbage in the back. But that's exactly how it worked with the harmonic oscillator.
Remember, a dagger A was equal to H.
The Hamiltonian overreached by Omega minus a half.
There was garbage in the back, which in that case was a half.
So this is obviously some constant with the dimensions of energy, and I better make sure that I've done that right.
It's the instruction B squared. Yes, of course it should, because it was the bar I took out from there.
Otherwise it is correct. And that's that's just just the business.
So we've expressed h let's write down what we let's write this down in the other way.
We've said that H is equal to H bar squared over a nought squared mu which must provide the dimensions of energy,
a dagger l a l minus z squared over two l plus one squared.
Another way of writing. We've expressed h basically as dimensionless constant times as of a dagger.
A dagger. So this is the harmonic oscillator trick and it's all just looking a little bit, little bit messier.
But this is only a boring number, right? It's just I mean, what's the difference in a half?
And this it's just a number. They're both. They're both. They're both just numbers.
Rational numbers. Okay. So what do we do next? What do we do?
In the harmonic oscillator, we calculated a a dagger comma, a, we found the commentator.
That was the next thing that we did. And that's what we do just right now.
Make sure I'm don't know it's more convenient to work it out the other way around later use.
All right. So what is that? We have to write this horrible thing down again so we can have an in or squared of a to open over in a square bracket
because we're talking comitatus now write down I pr on each bar because I'm writing down a now which is just fair.
I need to write down minus L plus one over R.
I don't need to write down the boring constant in the back,
the z of real plus 1a0 that will commute with everything in sight and that will make no contribution to the commentator.
Then I have to write down the corresponding parts of a dagger, which is minus EPR over bar, minus L plus one overall.
And now I can rest easy. So this is what the commentator that I have to do obviously.
So PR commutes with itself nothing doing their PR on the other hand does not compute with this.
So we are going to get a zero squared over two i.
And I l plus one that's that l plus one over h bar that's that age bar p r comma one overall commentator I get, I get that from that list.
Live in hope that I do. What did I do with the minus sign? I put it in the bin.
I shouldn't have done right. There should be a leading minus sign. Then I have the same term actually,
because here I am going to get plus but everything plus plus this except I'm going to have a one overall comma PR, which of course is minus this.
So I'm going to get this all over again. So why don't you just rub out the two and then it's right.
What's this? Commentator? We've already discussed that problem.
It's going to be the rate of change. It's going to be D by the R of this times.
The commentator. So this is equal to minus a zero squared i l plus one over h bar.
And then I'm going to have a minus one over all squared for the derivative of this times.
PR Commodore. Commentator Because that's how these things work.
But this once again is minus h bar.
So the two minus is here.
Cancel the two i's, make another minus sign which kills this minus sign all being well, this is equal to a zero squared l plus one.
The poles cancel all squared checks sine.
Yeah. Uh, should have a.
No, that's correct. That's correct. Good. All right. So we want to express this.
Remember the commentator? So in the harmonic oscillator case, what was this commentator?
This commentator was actually one. So the bad news is, it's not looking very promising at this point.
You think, oh, no, it's not good, because I'll come to some damn function of Oh,
but we've seen that damn function of while somewhere before, up in the Hamiltonian.
Basically, I've lost that. I've lost Hamiltonian. There it is.
So we've got l l plus one plus squared of A to B to mu r squared is appearing in the Hamiltonian.
So supposing I would take h l plus one and from it I would take h l.
Then everything in the hamiltonians would cancel except that middle term which has the right form, in namely it contains a 1 to 4 squared.
And what would we have. We have h squared over two mu I think R squared brackets L plus one l plus two minus l l plus one to do that.
Right. Given hope. Okay. So obviously there's going to be a factor of L plus one common.
And then we're looking at the difference between L plus two and L, in other words,
to the two is going to cancel this and this is going to equal l plus one h squared over over mu r squared.
So we can express with this little side calculation,
we can go back up the board and express this as an appropriate multiple of the differences in the Hamiltonian.
So it's going to be a nought squared. That's the say nought squared.
Then we will want to multiply by mu divide by h bar squared and then we'll be able to say This is h l plus one minus eight gel.
Check that we haven't got any horrible factors in this. Okay.
All right. What was the next thing we did in the harmonic oscillator?
Having got the commentator of the A and the Dagger and having expressed H is a product of A and a dagger.
The next thing we did was calculate the commentator. Use these results to calculate the commentator of A with H.
So that's what we do now. I will do it here so that we can see our results.
So I want to calculate the commentator a l comma h l and I can do that by expressing this.
HL HL is basically a product of the A's.
Now I've only got to locate the wretched product. It's at the top there, isn't it?
Uh oh. So hard from this position to see what you need to see this product, etcetera, etcetera, etcetera.
There's a statement. This is the statement I'm looking for. I want that statement. All right, so this h l can be traded in for that product.
So we have h bar squared over a0 squared mu times.
The commentator of a l and a l i l dagger round and I can rest easy.
There are no need to put this stuff inside a commentator because it be with everything inside so it can't contribute to the commentator.
So this is the commentator I have to evaluate and this is easy peasy because this is a commentator, a product, some A with B and C,
which in principle is the commentator of A with B, C standing idly by, and then the commentator of A with C, B standing idly by.
But A, of course, computes with itself. So there's only one term which is an L.
Al mutated with Al Dagger so this is equal to h bar squared over a zero squared mu al comma al dagger.
And we just worked that one out. And the answer was and the answer was, it was here.
Sorry, I did something wrong. And. Yes, it's very important.
I need an extra factor, Al, and thank you very much. In the back that stands out.
This is standing idly by while Al works on his companion. All right, so.
So I now need to plug this in for this commentator. And you can see that all these all these factors are going to cancel.
This factor is going to cancel on this factor.
And so we're simply going to get h l plus one minus h l times the al I've been very helpfully told to include.
So let me just. Right. So we're now we're now in wonderful shape.
We so we've completed all three steps of the simple oscillator calculation.
And now we just need to go for the point of the exercise which was which is so we're given that we always were given that HL on L is equal to e of L,
right. What we want to do is make ourselves a new.
So we've got one stationary state. We want to make ourselves a new stationary state by multiplying by al obviously both sides of the equation.
So let me write down the right side of the equation first.
This implies that E which is a boring number times h l being l is equal to a l l of e an l usual business.
Swap these over h l a l which I'm not entitled to.
Plus the commentator for that restores order.
On E!
Now, we just worked that out and found that it was the difference of two ages times ail so this becomes h l plus one minus h l with an al in back.
Guess what? We have an el al here with a minus sign and hll here with a plus sign.
So the whole thing is equal to h l plus one ail e and l.
So we have achieved what we wanted to achieve. That is to say, we have shown that this state.
Is all is an iron state, not of HL, but of l plus one and for the same energy.
So the map is looking a little different now from harmonic oscillator, but nonetheless we have a very powerful result.
We have generated ourselves from a state which identity and angular momentum quantum number.
L We've made ourselves a state. Which can only be characterised as e l plus one.
We've made up of a state of the same energy but more angular momentum.
So what have we done? This operator. This is. This is.
This is some normalising constant. Right? We had just this kind of thing in the case of the harmonic oscillator.
So. Physically.
What have we done? Well, classically.
What have we done? We've taken an orbit that might look like this, and we've turned it into an orbit.
I'm trying to make an orbit that has about the same Semi-major axis and is rounder.
You know about Kepler orbits. So we weave with the same supply of energy.
We have increased the angular momentum. So we've made the orbit less eccentric.
So in the civil law case, what did we do? We made a service in orbit with less energy.
And then we argued that the. The energy, we were able to show that the energy could never be negative.
So we said to ourselves, So there must be some.
So so given given this state now we could apply a L plus one,
a sub l plus one to this and make ourselves e comma l plus two with even more angular momentum.
So like an isolated case, we said we can make ourselves an orbit with even less energy.
Is this possible with a given supply of energy, with a bound orbit two to have more and more more angling momentum?
No, it's not. At some point, you've got the maximum angular momentum you can have for that given energy,
which in classical physics is what we would call a circular orbit. We've completely destroyed the the radial motion.
So what we've been doing here is we've been shifting kinetic energy.
We've shifted k e from.
PR squared over two mu 2ll squared h bar squared over two mu squared.
Right. This was the tangential kinetic energy. This was the regular kinetic energy.
We've shifted energy from here to here when we've got no energy left in the arrow as little as the theory, you know,
as quantum mechanics allows, which won't be zero but will be some amount, then we won't be able to shift anymore.
So there must be a maximum angular momentum for a given energy.
We'll call this. Curly L All right.
This is the maximum angular momentum and it is a function of energy.
But we won't write. We won't write that it's a function, as you will. But we're going to find out what function of energy it is.
So what does that mean? That means if we take the if we take the what are we going to call this?
The the secularisation operator a l belonging to this maximum angular momentum.
And we use it on the states which has the maximum angular momentum for the given energy.
What do we get? Nothing. That's the only way we can be prevented from getting a state which has even more
angular momentum for the same energy is if this operator simply kills this state.
So we've used this argument twice before, once in the harmonic isolated case, and also in the case of the angular momentum operators.
What do we mean by nothing? What we mean is the mod square of this is nothing.
What does that map to? That maps to e maximum angular momentum, a l dagger, a l e curly thingy is nought.
Where have we seen a dagger a before? I think we must have seen it in the Hamiltonian.
We need to replace that by the Hamiltonian times.
Some horrible factors. Yeah.
All right. So. Well, that we already have it here. So a dagger a comes right down to this line here.
So this line here can replace the dagger in here.
So we get to have that e curly l onto hainaut squared mu over h bar squared uh h h curly l plus
z squared over to curly l plus one squared close a bracket e curly l ain't much but eight.
But this thing, this is an eigen function of this operator with eigenvalue e this is a boring number.
So it stands by was this bangs into that and makes a one.
This gives me e times this and this is left over.
It bangs into this and makes a one. So this implies that h0 squared mu over h bar squared e plus z squared over two l plus one squared is nothing.
Or perhaps I should write this as equals minus. So what have we done?
We've got a relationship between the energy and the maximum allowed angular momentum.
More than that,
we know that these angular momentum quantum numbers because this is orbital increment and we're talking about we prove that those had to be integers.
So in being defined to be is an integer.
Integer? What integer?
We know that Curly L is allowed to be nothing. One, two, three, four.
So n is equal to the numbers it's allowed to be is 1 to 3, four, blah, blah.
Nothing not included in the list because of that plus one.
So we have shown that e the energy has to be of the form minus z squared h bar squared over a nought move.
I done that right. Not squared mu one over.
Now we need to hear one over n squared. So we have found the possible energies of a hydrogen atom.
Well, in fact, for a hydrogen like Ion, because Z remembers this integer which controls the number of charge units on the nucleus.
We have found this with the possible values, right?
It's given by this constant, which we know what it is. We'll simplify it in the moment.
We know what it is. Times one over n squared where n is one, two, three, four.
So this gives the energy levels. We write this as minus z squared.
Times are over and squared where.
Oh. Curly.
Ah is whatever you see it to be which is page bar squared over two a nought squared MMU which is not very intuitive.
The way to make this intuitive is to take those zeros.
There are two of them and turn one of them back into its h bars and things.
Now where did we define a zero? For heaven's sake, it was right over here somewhere.
Right? There it is. So take. So one of those two I'm going to replace by that garbage there.
All right. So this is going to become on the bottom.
We're going to have an eight pi epsilon nought a zero.
All right. That's the four pi epsilon nought. The edge bar squared will cancel top and bottom.
So that goes away. The mu in the E squared. Well the mu will go away with this and the E squared will sit on the top.
So the red book is is what is squared over for pi ips and not a zero would be the potential energy of two charges of charge,
you know, two electric charges that was separated by a zero.
So this is half of the potential energy. A separation of a zero.
And. So this is the fundamental energy scale of atoms.
And what does it equal to 13.6 electron volts plus, you know, 13.6 to 3 significant figures.
So the energy range of which we work, you know,
the the battery that you stick into your you stick your camera or something has 1.5 volts basically because of that 13.6 EVs,
it's all of all of condensed matter. Physics is a mere reflection of that number.
We're all you know, that's why we live at one TV, not a one movie or 1 million every year or whatever.
So what do we need to do next? Yeah.
Jargon. This is called the principle.
Principle. Al Quantum number.
So in these hydrogen like ions,
we've discovered that there are a whole series of different distinct states which have the same energy and different tangle of momenta.
So let's talk a bit about degeneracy. Okay.
So if a principal quantum number and equals one, we have that L which is equal to n minus one is equal to zero.
And what is the largest angular momentum you can have is nothing. And on the ground, state of hydrogen.
There's one electron. It sits in the state with the lowest energy, which is going to be associated with any equals one.
And it has no angular momentum. It's on a totally radial orbit in classical physics.
Right. Not going round and round at all. It just goes in and out.
In and out. I mean, in classical physics.
Quantum mechanics. But it doesn't have any angular momentum.
So that's a surprising result for any equals.
Two. L is the maximum angular momentum is equal to one.
That means that l can be nought if you like, and l can be one.
Right? This is the maximum angular momentum. So there's a slightly funny thing going on here.
And was introduced as one plus the maximum angular momentum.
But now I'm saying it's better to what we one standardly thinks about it.
One thinks about what's the value of N from it.
One one takes as N minus one the maximum angular momentum.
So that's right. And in this sense, we have one state.
It'll be two states.
Well, this one here, basically we have this is for spin this party where it'll turn out to be two states when we include the spin of the electron.
But remember, we were doing the growth strategy, which means we said we were going to forget about the spin of the electron here.
We would have one state and here we would have three states, right?
Because four equals one where we got total an element of one, which means we've got three possible orientations of it.
M can be one nothing or minus one. So we have three quantum states here, one here.
So we've got four states all with the same energy for an equal to one, four equals one.
And so it goes down the line. So the number of states is is increasing rapidly.
Because there'll be five states for any equals three. The maximum incremental will be two for two units of Anglo mentum.
You've got five possible orientations and then you've still got three of these and one of those.
So that's nine states, etc.
So, so the structure that we've derived is extremely degenerate.
What does this have to say about experiments?
So stick some hydrogen atoms in a in a vessel and pass a electric current through and get the electrons knocked out of their out of their comfortable,
out of their comfort zone. And you will get photons coming out at discrete frequencies.
New is going to be the difference in the energies over H, which is going to be z squared.
The rip constant over H of regional h bar one over NW prime squared minus one over and squared.
This is for. And goes to end primes.
So if you were in one of these higher states, for example, and equals to you will have less.
Your energy would be a smaller negative number. Right? You'll have one of a one of a two squared of a quarter here.
And this will be if you could then fall down to the state and primed equals one, in which case this will be one.
So this will be this bracket will be, say, three quarters and you will get three quarters of this number coming out.
So that gives you that gives you some frequency. And what we have is a series.
Of lines of the way we think about this is that we have a series of lines each.
Four fixed. And primed, i.e. bottom level.
So if we fix and primed at one, we can have transitions from MN is 2 to 1 or and is 3 to 1 or and is 4 to 1.
And these are the successive lines of the lineman series.
So here we have the. So here, here, here is the energy of an equals one.
His n equals two has an equal three. And Lyman Alpha is the name used for the for the spectral line associated with an electron tumbling from equals
two down to when equals one and Lyman beta is associated with from an equals three down to and equals one,
which is further to fall. So it emits more energy. So the line appears at higher frequency.
Longer shorter wavelength. So so the Lyman series this.
Is fat and primed equals one, if any equals two.
We're looking at Lyman Alpha. That's what it's conventionally called, if any equals three.
It's Lyman Beta. And this has, I think is 112 nanometres.
Is that right? 121. Sorry. And as you go down to end equals infinity.
In other words, if you fall all the way from not being bound into the bottom of the atom,
then this is the Lyman at the beginning of the Lyman continuum. And that's what.
What is it? 92. Nanometres.
Roughly speaking, I've got a more accurate number here. 91.2.
So these lines are all in the. These are all vacuum ultraviolet lines.
They all carry. So this one is carrying 13.6 EV of energy and these are carrying this is carrying three quarters of 13.6 KV of energy.
So they're carrying enough energy to kick electrons out of the air molecules.
So, so the so these photons are heavily are absorbed by all kinds of things.
They're very these. These are very easily absorbed photons because they carry enough energy to lift electrons out of most atoms.
And then we have the next is the bottom of Ceres, which is where the whole story started.
Which is so end prime just two and equals three.
If you go from 3 to 2, that's called Boma Alpha, but it's written as H Alpha because that stands for Hydrogen Alpha.
So Palmer was a was a Swiss schoolteacher who empirically fitted the formula we've we've derived.
So I've lost it. There it is.
He fitted that formula empirically to to measured frequencies of of lines that he identified as being the Balmer series lines.
Well, a series of lines equal to a hydrogen series. So this this is called H Alpha.
And it's a it's a pink photon. It's 600 and something nanometres six, five, six.
So it's pink light. So many astronomical objects are pink because they are shining in alignment.
Sorry. In Alpha. In Palma, Alpha Beta.
If you. Then you go to passion. That's four in prime, just three and obviously and can be four or five, etc., etc., etc.
So these start off as pink and they get bluer.
So as you go down this list, the wavelengths get shorter as you go to infinity with Series Limited.
I did write it down here. 364.
Six nanometres. So they. They go from pink light right through the optical spec.
The rest of the optical spectrum to the to the ultraviolet region.
And the Passion series starts at 1875, I think.
So any calls for you were looking at 18?
Yep. So that's already these are sort of more or less optical.
By now we're in the near-infrared, etc., etc., etc.
So. That's pretty much the right place to stop.
I think what we should do. Uh, there's just one other thing I would point out is that so you can apply these
formulae to the inner electrons of to the innermost electrons of other atoms,
like atoms that have more than one electron. You can't apply them to the outer electrons with any useful way because we've done all this right.
Remember, with with no other electrons present, we got one nucleus and one electron.
But there's one very important thing to take home, which is that this energy scale goes like Z squared.
So the energies, the characteristic energies of the innermost electrons are going up like Z squared.
And by the time you get uranium, which has 92 units of charge, the Z is 92.
You're almost a factor of of ten to the four.
You're almost a factor of of 10,000 higher in energy, which means that these electrons are moving essentially relativistic.
LI So that's just the thing to bear in mind.
Okay. And we'll look at the wave functions that go with this lot tomorrow.