Energy of photons emitted by de-excitation of electrons
When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.


That is one of several pathways for deexcitation. Often, internal conversion, the radiationless deeexcitation happens instead.


  How does the change in energy level compare to the energy of the emitted photon?


It depends on the relaxed geometry of the excited state, as compared to the ground state. 

Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$. 

Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state. 

Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.

This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as Stokes shift.