I have done a lot of research on phosphorescence and luminescence, and I believe that UV markers rely on the effect of phosphorescence i.e. they take in UV radiation and give back radiation in the visible light spectrum (purple), which has a lower frequency/energy.

This is what I saw on Wikipedia

Fluorescence is the emission of light by a substance ... In most cases, the emitted light has a longer wavelength, and therefore lower energy, than the absorbed radiation

Where does the energy go? It surely cannot vanish!

Tell me if this effect applies for phosphorescence as well

BTW: Try to keep the language of the answer simple (I'm 14)


1 Answer 1


For simplicity let's consider a diatomic molecule, although the same ideas apply to larger ones. For reference, there's a nice diagram covering both fluorescence and phosphorescence on Wikipedia.

Typically, at room temperature, the molecule will be in its electronic ground state. It will be vibrating as well: for a diatomic molecule this vibration will stretch and compress the bond between the atoms. One can plot the potential energy as a function of atomic separation: the molecule will oscillate about the minimum potential energy (which defines the bond length). This is properly described by quantum mechanics: there will be a discrete set of energy levels, and most molecules will lie in the lowest one: the vibrational ground state. The gap between vibrational energy levels is typically much smaller than the electronic energy difference between the two electronic states. To a first approximation (the Born-Oppenheimer approximation) the electronic and vibrational energies can be treated separately, and simply added together.

When the molecule absorbs a photon of UV or visible light, it goes up to an excited electronic state. This has a different potential energy curve, with a different bond length (typically, longer) and a different set of vibrational levels. However, the electronic excitation happens rapidly, compared with the motion of the nuclei. It is represented as a vertical line on the diagram. So the molecule finds itself in a vibrationally excited level of the excited electronic state. What happens next is the dissipation of the vibrational energy, usually by collisions with other molecules in the liquid (assuming this is in a liquid). The molecule cascades down the vibrational energy levels, until it reaches the vibrational ground state (of the excited electronic state). This is where (some of) the energy goes: dissipated as heat into the surrounding liquid. No photons are involved in this process.

Fluorescence occurs by emission of a photon, and the molecule returns to the ground electronic state. However, again, because of the mismatch between the equilibrium bond lengths in the two electronic states, and the fact that the transition happens "vertically" on the diagram, the molecule returns to an excited vibrational ground state. Once again, the extra energy is dissipated as heat into the surrounding liquid, as the system returns to its vibrational ground state.

So, the photon emitted in fluorescence has a lower energy than the absorbed photon, because some of the energy is converted before, and after, the transition, into heat.

Something similar happens in phosphorescence, except that, before emission, there is an extra electronic transition happening in the excited molecule (which does not involve any photons). This is called intersystem crossing. But otherwise, the explanation (of where the energy went) is the same.

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    $\begingroup$ I do not understand, how come the energy from the electrons' energy levels is lost without the electron's energy level changing, because the energy can only be discrete, not variable $\endgroup$ Aug 20, 2018 at 10:52
  • 2
    $\begingroup$ The molecule contains both electronic energy and vibrational energy and, to a first approximation, you can think of these as being separately added together. The vibrational energy level spacings are much smaller than the gap in electronic energy between the ground state and the excited state. The collisions with the surroundings just involve transfer of vibrational energy. This typically goes only one way, from molecule to surroundings, because the vibrational energy levels are still larger than the typical thermal energies of the surrounding molecules. $\endgroup$
    – user197851
    Aug 20, 2018 at 11:00

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