The problem is that you're thinking of the electron as a particle. Questions like "what orbit does it follow" only make sense if the electron is a particle that we can follow.
But the electron isn't a particle, and it isn't a wave either. Our current best description is that it's an excitation in a quantum field (philosophers may argue about what this really means; the rest of us have to get on with life). An electron can interact with its environment in ways that make it look like a particle (e.g., a spot on a photographic plate) or in ways that make it look like a wave (e.g., the double slits experiment) but it's the interaction that is particle-like or wave-like, not the electron.
If we stick to the Schrödinger equation, which gives a good description of the hydrogen atom, then this gives us a wavefunction that describes the electron. The ground state has momentum zero, so the electron doesn't move at all in any classical sense. Excited states have a non-zero angular momentum, but you shouldn't think of this as a point like object spinning around the atom. The angular momentum is a property of the wavefunction as a whole and isn't concentrated at any particular spot.
The following is not well-known, but (modified) Maxwell equations can indeed describe both electromagnetic field and electrons.
@Quantumwhisp commented: "Maxwell's equations don't describe charged particles at all", and then asked: "Can you derive the Lorentz-Force from maxwell's equations?"
I am not saying these comments are unreasonable, but, surprisingly, Dirac did derive the Lorentz force from Maxwell equations (Proc. Roy. Soc. London A 209, 291 (1951)).
I summarized Dirac's derivation elsewhere as follows.
Dirac considers the following conditions of stationary action for the free electromagnetic field Lagrangian subject to the constraint $A_\mu A^\mu=k^2$:
\begin{equation}\label{eq:pr1}
\Box A_\mu-A^\nu_{,\nu\mu}=\lambda A_\mu,
\end{equation}
where $A^\mu$ is the potential of the electromagnetic field, and $\lambda$ is a Lagrange multiplier. The constraint represents a nonlinear gauge condition. One can assume that the conserved current in the right-hand side of the equation is created by particles of mass $m$, charge $e$, and momentum (not generalized momentum!) $p^\mu=\zeta A^\mu$, where $\zeta$ is a constant. If these particles move in accordance with the Lorentz equations
\begin{equation}\label{eq:pr2}
\frac{dp^\mu}{d\tau}=\frac{e}{m}F^{\mu\nu}p_\nu,
\end{equation}
where $F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}$ is the electromagnetic field, and $\tau$ is the proper time of the particle ($(d\tau)^2=dx^\mu dx_\mu$), then
\begin{equation}\label{eq:pr3}
\frac{dp^\mu}{d\tau}=p^{\mu,\nu}\frac{dx_\nu}{d\tau}=\frac{1}{m}p_\nu p^{\mu,\nu}=\frac{\zeta^2}{m}A_\nu A^{\mu,\nu}.
\end{equation}
Due to the constraint, $A_\nu A^{\nu,\mu}=0$, so
\begin{equation}\label{eq:pr4}
A_\nu A^{\mu,\nu}=-A_\nu F^{\mu\nu}=-\frac{1}{\zeta}F^{\mu\nu}p_\nu.
\end{equation}
Therefore, the last three equations are consistent if $\zeta=-e$, and then $p_\mu p^\mu=m^2$ implies $k^2=\frac{m^2}{e^2}$ (so far the discussion is limited to the case $-e A^0=p^0>0$).
Thus, the first equation with the gauge condition
\begin{equation}\label{eq:pr5}
A_\mu A^\mu=\frac{m^2}{e^2}
\end{equation}
describes both independent dynamics of electromagnetic field and consistent motion of charged particles in accordance with the Lorentz equations. The words "independent dynamics" mean the following: if values of the spatial components $A^i$ of the potential ($i=1,2,3$) and their first derivatives with respect to $x^0$, $\dot{A}^i$, are known in the entire space at some moment in time ($x^0=const$), then $A^0$, $\dot{A}^0$ may be eliminated using the gauge condition, $\lambda$ may be eliminated using the first equation for $\mu=0$ (the equation does not contain second derivatives with respect to $x^0$ for $\mu=0$), and the second derivatives with respect to $x^0$, $\ddot{A}^i$, may be determined from the first equation for $\mu=1,2,3$.
However, the above is about classical electrodynamics. What about quantum theory? It turns out that modified Maxwell equations can be equivalent to the Klein-Gordon-Maxwell electrodynamics or (with some caveats) to the Dirac-Maxwell electrodynamics (see my article Eur. Phys. J. C (2013) 73:2371 at https://link.springer.com/content/pdf/10.1140/epjc/s10052-013-2371-4 ).
Best Answer
The only atoms for which the Schrodinger equation has an analytic solution are the one electron atoms i.e. H, He$^+$, Li$^{2+}$ and so on. That's because with more than one electron the forces between electrons make the equation too hard to solve analytically. However, over the 90 or so years since Schrodinger proposed his equation a vast array of numerical methods for solving it have been developed, and of course modern computers are so powerful they can calculate the (electronic) structure of any atom with ease. This applies even to heavy atoms where relativistic effects need to be taken into account.
The Rydberg equation is an approximation because it does not take the electronic fine structure into account. However it's a pretty good approximation. It works because for a one electron atom the energy of the orbitals (ignoring fine structure) is proportional to 1/$n^2$, where $n = 1$ is the lowest energy orbital, $n = 2$ is the second lowest and so on.