Expectation of Radius ⟨r⟩

How big is a hydrogen atom in a given state? The sharpest single answer is the expectation value of the radius, $\langle r\rangle$ — the mean distance of the electron from the nucleus, averaged over the probability density.

Definition

For a state $\psi_{n\ell m}$, the expectation value of any function of $r$ is

where the angular integral over $|Y_\ell^m|^2$ gives 1. Setting $f(r) = r$ gives $\langle r\rangle$. The crucial $r^2$ from the volume element means we are really averaging $r$ against the radial distribution $P(r) = r^2|R_{n\ell}|^2$.

The closed-form result

Carrying out the integral with the hydrogen radial functions yields a clean formula that depends on both $n$ and $\ell$:

A few consequences:

• $\langle r\rangle$ grows roughly as $n^2$, so excited states are much larger — the atom literally swells as the electron climbs the ladder.
• At fixed $n$, larger $\ell$ gives a slightly smaller $\langle r\rangle$. Higher-$\ell$ orbits are more circular and stay nearer the most probable radius, while low-$\ell$ orbits are elongated and probe both small and large $r$.

Worked examples

For the ground state $1s$ ($n=1, \ell=0$):

The mean radius is $1.5\,a_0$ even though the most probable radius is $a_0$ — the long exponential tail pulls the average outward.

For the $2p$ state ($n=2, \ell=1$):

The takeaway

The mean radius of a hydrogen state is $\langle r\rangle = \tfrac{a_0}{2}[3n^2 - \ell(\ell+1)]$, computed from the radial distribution $r^2|R_{n\ell}|^2$. It scales like $n^2$ and decreases mildly with $\ell$ at fixed $n$.

Try it

This is a numerical exercise — return a number. Working in units of the Bohr radius ($a_0 = 1$), use $\langle r\rangle = \tfrac{a_0}{2}[3n^2 - \ell(\ell+1)]$ to compute $\langle r\rangle$ for the $2p$ state ($n=2$, $\ell=1$).

Sign in on the full site to ask questions and join the discussion.