Why does $2p$ have highest RDF at $4a_{0}$?

Question

I was reading notes from my first class in Quantum Physics that I received and left confused at the following statement:

For each principal quantum number $n$, the orbital set with the highest $\ell$ (orbital angular momentum quantum number) has its maximum electron density at the corresponding Bohr radius $n^2a_{0}$.

I just couldn't understand how the highest $\ell$ number corresponded to the Bohr radius i.e. maximum electron density of the $2$p orbital was calculated to be at $4a_{0}$.

Answer 1

Short of actually doing the calculation here is a crude estimate (which happens to be exact):

1. Orbitals with maximal angular momentum $\ell=n-1$ (for given $n$) is expected to have a relatively well-defined notion of radius, cf. e.g. my Phys.SE answer here.

2. Due to the virial theorem we have $\langle \frac{1}{r} \rangle=\frac{1}{n^2a_0}.$ It is then natural to expect OP's sought-for equation $\langle r \rangle=n^2 a_0.$

3. For a more refined argument, see e.g. my Phys.SE answer here.