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A molecule exposed to a photon with some energy is put into an excited state, and emits a photon of some energy when it returns to ground state. The photon that provided the energy for excitation corresponds to a particular wavelength, at which we see a peak in an absorption spectrum. The photon that was released when the molecule returned to its ground state also corresponds to a particular wavelength, at which we see a peak in an emission spectrum. Fluorescence is when a molecule emits a photon of lower energy when it returns to ground state than what it absorbed initially to be put in an excited state. That photon corresponds to a particular wavelength, at which we see a peak in the fluorescence spectrum. But isn’t the emission of a lower energy photon the case for all molecules, since all molecules lose some energy in form of vibration and rotation before they reach ground state? How is fluorescence distinguished from emission?

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  • $\begingroup$ Emission is the generic term for fluorescence (from excited singlet state) and/or phosphorescence (from triplet) to singlet ground state. Look up Jablonski diagrams for molecules in gas phase and in solution to see the effect of solvent and vibrational/rotational effects $\endgroup$ – porphyrin Jul 1 '18 at 21:10
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You can google all of this for more detail...

Basically an "emission spectrum" is the most general term for detecting photons as a function of energy or wavelength. All the term implies is that your measuring photons.

A "fluorescence spectrum" is a little more specific. It means that a photon was used to excite the atom/molecule and that the atom/molecule decayed very rapidly with the emission of a photon of longer wavelength (lower energy).

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