As you sign ChemEguy I will suppose physics is not your stronger suit.
F=ma does not hold at atomic dimensions because it is a continuous function.
At atomic dimensions quantum mechanics reigns, it does not hold as such because it is an emergent classical formula from the underlying quantum mechanical interactions. At this level fundamental forces are defined as exchanges of momentum using the formalism of feynman diagrams. The link expands on this.
But since we know from the Einstein-Plank relation E=nhν, energy is quantized and not continuous.
As answered in the comments, energy may or may not be quantized, depending on the solutions of the boundary value problems using quantum mechanical equations. Bound states are quantized.
Therefore, F=ma does not apply at very low quantum numbers.
It has a different manifestation at dimensions commensurate with h, the Planck constant. Quantum numbers are the numbers distinguishing elementary particles from each other, and are not immediately connected with exchange forces; various interactions conserve or not quantum numbers, and it will depend on the exchange forces whether they are conserved or not.
$E = n h \nu
between single dollar-signs would result in $E = nh\nu$. $\endgroup$ – dmckee --- ex-moderator kitten Feb 25 '18 at 16:46