Hydrogen emission and absorption spectral lines are typically depicted as the same:

enter image description here(source)

However, in more complex systems, the emission and absorption spectra are significantly different. For example:

enter image description here

Absorption and photoluminescence spectra of DCJTB doped in a PS film. (source)

What causes this shift of $\lambda_{\text{max}}$ etc. between absorption and emission?


3 Answers 3


As @MikaelKuisma mentioned, the reason for this difference is the contribution of nuclear vibrational overlap to the transition moment. When an electron is excited from a ground state $S_0$ to the first excited state $S_1$, the bond is stretched and the internuclear separation increases:

enter image description here (source: Martin Pope and Charles E. Swenberg, Electronic Processes in Organic Crystals and Polymers, 1999)

Assuming harmonic oscillations, the overlap between each two vibrational wavefunctions (different energy level for different frequencies/number of nodes) is defined as the nuclear vibrational overlap, which in turn contributes to the total transition moment.

The intensity of absorption/emission is proportional to the transition moment squared, $d_{nm}^2$ (following Beer-Lambert's law):

$$I = I_0 10^{-\varepsilon c l}$$ $$A = \log \frac{I_0}{I} = \varepsilon c l \propto d_{nm}^2$$

which means that factors contributing to the transition moment are expressed in the absorption/emission spectra.

For more information see Franck-Condon principle.

  • 1
    $\begingroup$ To summarize this a bit, when absorbing from the ground electronic + nuclear state the absorption can be blueshifted since higher nuclear states can be populated; when emitting from the ground nuclear state of an excited electronic state the emission can be redshifted since not all the electronic energy needs to be given back and some can be retained as vibrational energy. $\endgroup$ Commented Feb 18, 2016 at 22:52

In general, absorption and emission spectrum of a medium containing some active molecules are not the same, because when molecules absorb radiation, they are in low-energy state where their environment (other molecules, solvent or solid matrix) is in corresponding low-energy state, so the medium as a whole has certain corresponding low-energy spectrum; while when they radiate accumulated energy, they are in a higher-energy state that leads to different interaction with the environment and this generally leads to different spectrum with different positions of peaks. Typically one observes shifts and broadenings. Search Stokes shift, for example.

For ionized gas such hydrogen or sodium gas this effect is very weak, as the medium is very rarified so the interactions of the atoms with other things are probably much weaker and even when energy of an atom changes, this has almost no consequence on its geometry and effective Hamiltonian. So it is possible that the spectrum remains almost the same as for emission as for absorption.

  • $\begingroup$ You are only speaking of solvent reorganization here. The key to longer emission wavelengths in molecules is however vibrational relaxation. The exited state electrons emit phonons (or heat) as the nuclei thermalise to excited state landscape minimum. $\endgroup$ Commented Feb 18, 2016 at 21:13
  • $\begingroup$ @MikaelKuisma you're right, I forgot to mention the effect of vibrating nuclei. You may add your thoughts on that to this answer or write a new answer if you wish. $\endgroup$ Commented Feb 18, 2016 at 21:26

The figure in Sparkler's answer described the energy levels but it needs some further explanation.

First this picture only applies in the condensed phase or if in the gas phase only at extremely high pressure.
When a photon is absorbed from the ground state, provided the electronic transition is allowed, transitions from v=0 in the ground state to v'=0,1,2 etc in the excited state can occur. The relative magnitude of each vibrational transition (also called vibronic) depends on the square of the overlap of the wavefunctions in the ground state and excited state vibrational levels. This is the Franck-Condon principle. This overlap depends also on the relative displacement of the excited state potential wrt to ground state. Usually bonds are longer in the excited state as an electron has been placed into an anti-bonding orbital.
Once the energy is in the excited state, this can fluoresce or undergo non-radiative transitions to other excited states such as a triplet state or back to the ground state. In competition with this is vibrational relaxation in the excited state. This is caused by frequent interactions with the many surrounding solvent molecules as each collision has a chance of removing some vibrational energy. This process is very rapid in the condensed phase with a rate constant typically >$10^{13} s^{-1}$ compared to the fluorescence rate constant usually not much more than $10^{10} s^{-1}$. Thus all the excited molecules are now in v'=0 in the excited state. Assuming that the ground and excited state have a similar shape the fluoresce and absorption spectra are mirror images of one another as shown in the figure. In the question the molecule also shows a mirror image but the picture is in wavelength not wavenumber (or energy) so is a bit distorted.
In the gas phase collisions are rare compared to those of solvent in the condensed phase and so vibrational relaxation can no longer compete with fluorescence and so the fluorescence spectrum can arise from a mixture of vibrational levels not just v'=0.


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