**Pre-script** (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

**Original post**:

This post is, in essence, a continuation of my series on electron orbitals. I’ll just further tie up some loose ends and then – hopefully – have some time to show how we get the electron orbitals for other atoms than hydrogen. So we’ll sort of build up the periodic table. Sort of. 🙂

We should first review a bit. The illustration below copies the energy level diagram from Feynman’s *Lecture* on the hydrogen wave function. Note he uses √E for the energy scale because… Well… I’ve copied the E_{n} values for *n* = 1, 2, 3,… 7 next to it: the value for E_{1 }(-13.6 eV) is *four *times the value of E_{2 }(-3.4 eV).

How do we know those values? We discussed that before – long time back: we have the so-called *gross* structure of the hydrogen *spectrum *here. The table below gives the energy values for the first seven levels, and you can calculate an example for yourself: the *difference *between E_{2} (-3.4 eV) and E_{4 }(-0.85 eV) is 2.55 eV, so that’s 4.08555×10^{−19 }J, which corresponds to a *frequency *equal to *f *= E/*h* = (4.08555×10^{−19 }J)/(6.626×10^{−34} J·s) ≈ 0.6165872×10^{15} Hz. Now that frequency corresponds to a wavelength that’s equal to *λ = c*/*f *= (299,792,458 m/s)/0.6165872×10^{15}/s) ≈ 486×10^{−9} m. So that’s the 486 *n**ano*-meter line the so-called Balmer series, as shown in the illustration next to the table with the energy values.

So far, so good. An interesting point to note is that we only have *one *solution for *n *= 1. To be precise, we have one *spherical *solution only: the 1*s* solution. Now, for *n* = 2, we have one 2*s *solution but also *three *2*p *solutions (remember the *p *stands for * principal *lines). In the simplified model we’re using (we’re

*not*discussing the

*fine*or

*hyperfine*structure here), these

*three*solutions are referred to as ‘degenerate states’: they are

*different*states

*with the same energy*. Now, we know that any

*linear combination*of the solutions for a differential equation must also be a solution. Therefore,

**any linear combination of the 2**. In fact, a superposition of the 2

*p*solutions will also be a stationary state of the same energy*s*and one or more of the 2

*p*states should also be a solution. There is an interesting

*app*which visualizes how such superimposed states look like. I copy three illustrations below, but I recommend you

But we’ve written enough about the orbital of *one *electron now. What if there are two electrons, or three, or more. In other word, how does it work for *helium*, *lithium*, and so on? Feynman gives us a bit of an intuitive explanation here – nothing analytical, really. First, he notes Schrödinger’s equation for *two *electrons would look as follows:

Second, the ψ(* x*) function in the ψ(

*,*

**x***t*) =

*e*

^{−i·(E/ħ)·t}·ψ(

*) function now becomes a function in*

**x***six*variables, which he – curiously enough – now no longer writes as ψ but as

*f*:The rest of the text speaks for itself, although you might be disappointed by what he writes (the

**bold-face**and/or

*italics*are mine):

“The geometrical dependence is contained in *f*, which is a function of six variables—the simultaneous positions of the two electrons. * No one has found an analytic solution*, although solutions for the lowest energy states have been obtained by numerical methods. With 3, 4, or 5 electrons it is hopeless to try to obtain exact solutions, and

**it is going too far to say that quantum mechanics has given a precise understanding of the periodic table**.

*It is possible, however, even with a sloppy approximation—and some fixing—to understand, at least qualitatively, many chemical properties which show up in the periodic table*.

The chemical properties of atoms are determined primarily by their lowest energy states. We can use the following approximate theory to find these states and their energies. First, **we neglect the electron spin, except that we adopt the exclusion principle and say that any particular electronic state can be occupied by only one electron**. This means that any particular orbital configuration can have up to *two* electrons—one with spin up, the other with spin down.

**Next we disregard the details of the interactions between the electrons in our first approximation**, and say that each electron moves in a

*central field*which is the combined field of the nucleus and all the other electrons. For neon, which has 10 electrons, we say that one electron sees an average potential due to the nucleus plus the other nine electrons. We imagine then that in the Schrödinger equation for each electron we put a V(

*r*) which is a 1/

*r*field modified by a spherically symmetric charge density coming from the other electrons.

I**n this model each electron acts like an independent particle**. The angular dependence of its wave function will be just the same as the ones we had for the hydrogen atom. There will be *s*-states, *p*-states, and so on; and they will have the various possible *m*-values. Since V(*r*) no longer goes as 1/*r*, the radial part of the wave functions will be somewhat different, but it will be qualitatively the same, so we will have the same radial quantum numbers, *n*. The energies of the states will also be somewhat different.”

So that’s rather disappointing, isn’t it? We can only get some *approximate *– or qualitative – understanding of the periodic table from quantum mechanics – because the *math *is too complex: only *numerical *methods can give us those orbitals! ** Wow! **Let me list some of the salient points in Feynman’s treatment of the matter:

- For
*helium*(He), we have two electrons in the lowest state (i.e. the 1*s*state): one has its spin ‘up’ and the other is ‘down’. Because the shell is filled, the ionization energy (to remove*one*electron) has an even larger value than the ionization energy for hydrogen: 24.6 eV! That’s why there is “practically no tendency” for the electron to be attracted by some other atom: helium is chemically inert – which explains it being part of the group of*noble*or*inert*gases. - For
*lithium*(Li), two electrons will occupy the 1*s*orbital, and the third should go to an*n*= 2 state. But which one? With*l*= 0, or*l*= 1? A 2*s*state or a 2*p*state? In hydrogen, these two*n*= 2 states have the same energy, but in other atoms they don’t. Why not? That’s a complicated story, but the gist of the argument is as follows: a 2*s*state has some amplitude to be near the nucleus, while the 2*p*state does not. That means that a 2*s*electron will feel some of the triple electric charge of the Li nucleus, and this*extra*attraction*lowers*the energy of the 2*s*state relative to the 2*p*state.

To make a long story short, the energy levels will be roughly as shown in the table below. For example, the energy that’s needed to remove the 2*s* electron of the lithium – i.e. the *ionization *energy of lithium – is only 5.4 eV because… Well… As you can see, it has a higher energy (*less *negative, that is) than the 1*s* state (−13.6 eV for hydrogen and, as mentioned above, −24.6 eV for helium). So lithium is chemically active – as opposed to helium.

You should compare the table below with the table above. If you do, you’ll understand how electrons ‘fill up’ those electron shells. Note, for example, that the energy of the 4*s* state is slightly *lower *than the energy of the 3*d* state, so it fills up *before *the 3*d *shell does. [I know the table is hard to read – just check out the original text if you want to see it better.]

This, then, is what you learnt in high school and, of course, there are 94 naturally occurring elements – and another 24 heavier elements that have been produced in labs, so we’d need to go all the way to no. 118. Now, Feynman doesn’t do that, and so I won’t do that either. 🙂

Well… That’s it, folks. We’re done with Feynman. It’s time to move to a physics *grad* course now! Talk stuff like quantum field theory, for example. Or string theory. 🙂 Stay tuned!

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