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The normalized spherical harmonics separate the angular variables. The Radial Schrodinger equation for the hydrogen atom isWrite $R=A{r}^l{e}^{-\kappa r}$. ThenSubstitution separates a constant term and a $1/r$ term. The latter vanishes precisely when ${\hslash }^2\kappa (l+1)/m=e^2/(4\pi {ϵ}_0)$. With $\kappa =1/[a(l+1)]$ this determines the Bohr radiusThe constant term determinesThese are the maximal-angular-momentum hydrogen states, with principal quantum number $n=l+1$.

Because the angular integral is one, normalization requiresThereforeThe phase of $A$ remains arbitrary.

In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the Gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.