Gavroy, The energy correction of an atom due to gravity would be incredibly small. Before we start it must be clear what we're not talking about. An atom falls in a gravitational field and this adds to its total energy, but does not affect the atomic energy levels. When you fall into a black hole, for example, it's not the fall that hurts you, it's the tidal gravitational forces. That is, you're stretched in one direction and squashed in the other. Tidal gravitational forces are quadrupole.
This means as far as atomic levels go, the effect of gravity will be quadrupole. It would resemble Stark splitting but with ΔJ = 2. One could write down the selection rules and line patterns, but more important first is to realize what a small effect we're talking about.
Take an extreme example - let the atom be near a one solar mass black hole. The Schwarzschild radius of the sun is 3 x 105 cm. The gravitational potential is GMm/r, the usual gravitational force is GMm/r2, but the tidal gravitational force is GMm/r3. And for its effect on an object of diameter d this means ΔE = (GMm/r3) d2.
Okay, how do we work out the value. Easiest way is to realize that the gravitational potential energy GMm/r of an object of mass m near the surface of a black hole is roughly mc2, and that's most of the factors in our expression. What we're left with is ΔE = mc2 (d/r)2. Numerically the diameter of an atom is d = 10-8 cm, the rest mass of an electron is mc2 = 0.5 MeV, and as we said for a solar mass black hole, r = 3 x 105 cm.
So ΔE = 0.5 MeV (10-8 cm/3 x 105 cm)2 = about 10-21 eV for the level shifts. Putting that in terms of frequency, it means a Δν of about 1 Hz.