 Solution

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.