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A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

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.

Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is thenThereforeEquivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=({\gamma }^3/\pi {)}^{1/2}$.

The expected radius isFor the ground state, $\gamma =m{q}^2/{\hslash }^2$. The Bohr radius is $a_0={\hslash }^2/(m{q}^2)=1/\gamma$, soThe mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the radial normalization of the hydrogen ground state.

For fixed $l$, the radial equation isSubstituteCancellation of the coefficient of $1/r$ requiresthe reduced-mass Bohr radius. The remaining constant term givesEach such no-radial-node state is a circular Coulomb bound state with principal quantum number $n=l+1$.