The periodic table

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. ūüôā

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 ‚ąöE for the energy scale because… Well… I’ve copied the En¬†values for n = 1, 2, 3,… 7 next to it: the value for¬†E1¬†(-13.6 eV) is four times the value of E2¬†(-3.4 eV).

exponential scale

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 E2¬†(-3.4 eV) and¬†E4¬†(-0.85 eV) is 2.55 eV, so that’s 4.08555√ó10‚ąí19¬†J, which corresponds to a¬†frequency¬†equal to¬†f¬†= E/h¬†= (4.08555√ó10‚ąí19¬†J)/(6.626√ó10‚ąí34¬†J¬∑s) ‚Čą 0.6165872√ó1015¬†Hz. Now that frequency corresponds to a wavelength that’s equal to¬†őĽ = c/f¬†= (299,792,458 m/s)/0.6165872√ó1015/s)¬†‚Čą 486√ó10‚ąí9¬†m. So that’s the 486 nano-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 1s solution. Now, for n = 2, we have one 2s solution¬†but also three¬†2p¬†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 2p¬†solutions will also be a stationary state of the same energy. In fact, a superposition of the 2s and one or more of the 2p 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 google¬†for stuff like this yourself: it’s really fascinating! You should, once again, pay attention to the symmetries planes and/or symmetry axes.

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 Schr√∂dinger’s equation for¬†two¬†electrons would look as follows:

two electronsSecond, the Ōą(x) function in the Ōą(x, t) = e‚ąíi¬∑(E/ńß)¬∑t¬∑Ōą(x) function now becomes a function in six¬†variables, which he – curiously enough – now no longer writes as Ōą but as f:formulaThe 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‚ÄĒthe 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‚ÄĒand some fixing‚ÄĒto 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‚ÄĒone 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.

In 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 1s 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 1s¬†orbital, and the third should go to an n¬†= 2 state. But which one? With¬†l¬†= 0, or¬†l¬†= 1? A 2s state or a 2p state?¬†In hydrogen, these two n¬†= 2¬†states have the same energy, but in other atoms they don‚Äôt. Why not? That’s a complicated story, but the gist of the argument is as follows: a¬†2s state has some amplitude to be near the nucleus, while the 2p state does not. That means that a 2s¬†electron will feel some of the triple electric charge of the Li nucleus, and this extra attraction lowers the energy of the 2s¬†state relative to the 2p¬†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 2s 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 1s¬†state (‚ąí13.6 eV for hydrogen and, as mentioned above, ‚ąí24.6 eV for helium). So lithium is chemically active – as opposed to helium.¬†energy values more electrons

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 4s state is slightly lower¬†than the energy of the 3d state, so it fills up before¬†the 3d¬†shell does. [I know the table is hard to read – just check out the original text if you want to see it better.]

periodic table

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. ūüôā

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. ūüôā Stay tuned!

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