 Solution

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.