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

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 $$4a_{0}$$.

• Are you looking for an explanation other than just look at the wavefunctions and calculate where maximum is? Commented Dec 28, 2018 at 22:12
• Consider to spell out acronyms. Commented Dec 28, 2018 at 22:55

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 $$\langle r \rangle=n^2 a_0.$$