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The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrodinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.