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As is shown in e.g. Quantum Mechanics: A New Introduction by Paffuti and Konishi, the wavefunction $\psi$ of an electron (in the Hartree-Fock approximation) in a general atom with atomic number $Z$ scales such that $|\psi(a_Z)|^2 a_Z^3\sim 1/Z^2$ where $a_Z=a_0/Z$ with $a_0$ the Bohr radius.

I was wondering if this could be explained in a classical picture. From Kepler's third law, the time spent by the electron around $r=a_Z$ should be proportional to $a_Z^{3/2}$, while that spent by the electron far from the nucleus should be proportional to just $a_0^{3/2}$. Hence, the probability should be $|\psi(a_Z)|^2 a_Z^3\sim a_Z^{3/2}/a_0^{3/2}=Z^{-3/2}$ which is close but not quite right. What is wrong with this logic?

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The problem was that $a_Z$ is not the only thing dependent on $Z$ in Kepler's third law, and the proportionality constant of the force is needed as well. In particular, for a central force $F=k/r^2$, we find $T^2\propto R^3/k$ for period $T$ and some average radius $R$ (here $a_Z$). Close to the nucleus, $k$ is also proportional to $Z$, so we instead find that the time spent by the electron around $r=a_Z$ is proportional to $Z^{-2}$, which gives the expected result.

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