After some research, I figured out that all EM waves/photons are absorbed by atoms by exciting an electron from an orbital to an other. However, atoms emit only certain EM waves with specific wavelengths that we then see as color in our eye. What determines which wavelength are "completely" absorbed and how will the energy eventually get released?
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1$\begingroup$ Photons do not need to change the orbitals that are occupied at all. Exciting molecular rotations and vibrations uses photons and has nothing to do with the orbital structure. $\endgroup$– Matt HansonCommented Jan 1 at 0:54
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2 Answers
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Exciting an electron from one orbital to another is one of the processes of photon absorption. Emission of EM waves by the inverse process is similarly only one of the possible processes.
There are many such processes. Antennas, molecular rotation, molecular vibration, bound-free interactions, free-free interactions, cyclotron radiation, synchrotron radiation, ...
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The factor that affects the probability of transition the most is the transition dipole moment integral, $\langle \psi_f | \hat{\mu} | \psi_i \rangle$, in accord with Fermi’s Golden Rule. This rule is a result of time dependent perturbation theory in quantum mechanics. For each different type of state or interaction (electronic, vibrational, rotational states), the important terms in this integral will differ. This leads to the concept of selection rules for state transitions. This is intimately related to the Franck-Condon principle via the Born-Oppenheimer approximation. But to have any more detail it would be necessary to have more context of what sort of problems you are considering.
