Schrödinger's Cormorant

(6-10) ① At any distance r from the nucleus, there is a spherical surface with an area given by 4πr2. Immediately surrounding this surface is a thin spherical shell with a thickness Δ, where Δ is very small compared to r. Every point within this shell is, more-or-less, at a distance r from the nucleus, so to answer our question, we need to find the volume of the shell and multiply it by the value of the radial wavefunction at this r. This volume is 4πr2Δ, so we need to multiply the value of ψradial by r2 to find a value that is proportional to the probability of finding the electron at a distance close to r.

② Just as we saw in Chapter 5, the electron can't be confined too close to the nucleus because the necessary curvature of the wavefunction would make the kinetic energy prohibitively high. However the electrostatic attraction to the nucleus opposes this and makes it highly unfavourable for the electron to stray too far away. Thus, ψ2 becomes very small and tends towards zero.

③ We know from the angular distribution functions, reflected in the 90% contour plots, that 2p has one angular node, while 2s has none:

Looking, however, at the radial distribution distribution functions, we can see that the situation is reversed: 2s has a radial node but 2p has none:

So the answer is that they each have one node.

In fact, the number 2, in the designations 2s and 2p, is called the principal quantum number and this defines the the total number of nodes in the wavefunction (this is always n-1). So the 4p wavefunction, for example has a radial function that looks a bit like this:

2 radial nodes, plus one angular node gives total 3 nodes, consistent with n=4.

④ The greater mean distance from the nucleus for the electron in a 2s orbital means that it will, on average, have a higher potential energy than a 2p electron electron.

However the 2p function has a greater curvature - linked to the angular node that 2s lacks - and hence an electron in 2p has, on average, a higher kinetic energy.

Add these energies up and the total energy of a 2s and a 2p electron can be the same.

⑤ The way for an electron to beat the shielding is to have a radial distribution function that burrows through the density of electrons that are, on average, closer to the nucleus. It's the part of the function that's closest to the nucleus that matters because if the electron is that close in, the repulsion from other electrons that opposes the nuclear attraction will be minimised.

Looking at the radial functions, you can see that, although the most probable distance from the nucleus is greater for a 2s electron, the 2s function has a greater average ψ2 in that key region close to the nucleus:

And that's the key to the lower energy of 2s electrons compared to 2p in atoms with multiple electrons.

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