On the German Wikipedia page, we can read: "The higher the azimuthal quantum number $\ell$ for a fixed principal quantum number $n$, the more the average distance of the electron from the nucleus increases." Can someone explain to me why this is correct?
I thought that the average distance should decrease for increasing $\ell$ and fixed $n$, as \begin{equation} \langle r\rangle _{n\ell} = \frac{a_\mathrm{B}}{2}(3n^2-\ell(\ell+1)). \end{equation}
In the classical limit (the standard two-body $1/r$ potential problem), the energy of an orbit does not depend on the angular momentum:
$$ \epsilon \propto -\frac{\mu}{2a} $$
where $a$ is semi-major axis of the ellipse, and $mu$ depends on the interaction strength.
The state with highest angular momentum at fixed $a$ is the circle, and that spends all its time $a$ from the middle; meanwhile, as you approach zero angular momentum, part of the orbit approaches $r=0$.
