Is it possible for a molecule to absorb and emit at the same wavelength? What is the reason behind it?

I’m working on charged tin porphyrins and got the excitation and emission (fluorescence) wavelength around 400 nm.

  • 2
    $\begingroup$ Yes, it is possible but it is not a single wavelength of overlap. Molecular spectra are quite wide. You have to see the excitation and emission spectrum and make sure you are not seeing Rayleigh scattering. $\endgroup$
    – AChem
    Oct 14, 2020 at 13:04
  • $\begingroup$ Thank you Sir. I have attached the graphs. Can you help me $\endgroup$
    – Pavithra J
    Oct 18, 2020 at 3:30
  • $\begingroup$ Since this is a simulation, I have no idea how to verify its correctness. You need to verify it experimentally. $\endgroup$
    – AChem
    Oct 18, 2020 at 12:47
  • $\begingroup$ Well, normally one has to explain why is not. $\endgroup$
    – Alchimista
    Nov 27, 2021 at 12:08

1 Answer 1


Yes it is very common particularly in the more rigid type of molecule. The best example is chlorophyll and this overlap of absorption and emission leads to energy transfer in photosynthesis. The reason for the overlap is that the excited state potential energy profile is very similar in shape but slightly shifted (as displacement) to that of the ground state and this results in a 'mirror image' spectrum. The spectra below are from anthracene.

anthracene mirror image

mirror image

You can see even in this rather crude figure that there can be absorption and emission at the same wavelength.

  • $\begingroup$ You should explain that in molecules the electronic states and vibrational states couple to yield microstates and that is why the molecular peaks are broad. $\endgroup$
    – MaxW
    Oct 14, 2020 at 19:20
  • 2
    $\begingroup$ In the vapour phase lines are v narrow and rotational levels can be resolved, limited by lifetime, collisional and doppler broadening, e.g. in benzene, naphthalene etc. (Density of states will ultimately limit this due to overlap/ crowding of levels). In solution the many collisions with solvent (which also contains electrons, of course, and so cause dipole and induced dipole (dispersion) effects) and these cause energy levels to broaden and is the dominant effect. This gives the appearance normally observed. $\endgroup$
    – porphyrin
    Oct 15, 2020 at 8:14
  • $\begingroup$ @porphyrin Say if we measure absorption and emission at one particular λ. I think the spectrum should show net absorption/ emission? So how come we get both spectra showing non-zero reading at that λ? I expect one spectrum to be non-zero, the other to be zero. $\endgroup$
    – TheLearner
    Oct 18, 2020 at 9:25
  • 1
    $\begingroup$ Usually absorption is measured through the sample in one experiment, but fluorescence is measured at right angles to this (in a separate experiment) so that the exciting light can be removed. In a laser the balance between absorption and emission determines if lasing is going to occur. $\endgroup$
    – porphyrin
    Oct 18, 2020 at 12:23

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