Following the absorption of light by an arbitrary small molecule, the molecule is found to be excited both vibrationally and electronically. The excess vibrational energy of the molecule will usually be dissipated into the solvent in less than ~10 ps, and fluorescence emission will occur in about ~10 ns.

Since there are many transitions available to accept the excess energy form the vibrations in the surrounding solvent it makes sense that the relaxation occurs very rapidly. On the other hand, electrons can move much faster than nuclei can. What must happen for an electronically excited molecule to relax through the emission of a photon and why does it take so much longer than the vibrational relaxation?


2 Answers 2


As you point out, there are many places for this relatively small amount of vibrational energy to go. There are both radiative and non-radiative transitions which can take place to vibrationally quench the system in the electronic excited state. The explanation for both parts of your question (the vibrational and electronic parts) can be understood by Fermi's Golden Rule. I won't go into a lot of detail of deriving this or anything else because you can find anything you wanna know about it by searching online.

Fermi's golden rule says that the transition rate, $\Gamma_{i\to f}$, from some initial state, $|i\rangle$, to a set of final states, $|f\rangle$, is given by, $$ \Gamma_{i\to f}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho $$ where $\rho$ is the density of final states of the system and $H'$ is a perturbing Hamiltonian of some kind. For instance, the electric field of a photon. This is what it would usually be for absorption. In the case of decay, it may also be the field of a photon, but I am not as familiar with using Fermi's golden rule in this case.

Notice that this rule has dimensions of inverse time. It can be shown that the inverse of this rate is equal to the mean lifetime of the excited state (see this page).

Basically all of the physics you are asking about is contained within the fact that this transition rate is proportional to the density of possible states to which the system will decay. To put it simply, there are many many more intramolecular and intermolecular transitions which can take place to redistribute this energy. The vibrational density of states for this problem could probably be approximated quite well as being continuous because there will be many vibrational levels which are slightly perturbed due to distortions from equilibrium geometries. Think of the width of vibrational bands in IR.

On the other hand, a very large amount of energy must be dissipated for electronic relaxation to take place, and the density of these high-energy states is very low. Especially if one excites to the lowest electronic state. If, however, excitation occurs to a higher electronic energy level, decay should take place faster because the density of electronic states becomes much larger.

These processes have almost nothing to do with the speed of nuclear motion compared to electronic motion, as the condition which must be satisfied is really conservation of energy, and this is not always easy to do when the energy levels are quantized. From this perspective, the result is not surprising because vibrational energy levels are certainly closer to being continuous than electronic energy levels. Also, the spacing of vibrational state is generally smaller on an excited state than the ground state, so this adds to the quickness of the decay to the ground vibrational state.

As an aside, this paper is of some interest as it shows that vibrational relaxation times in the gas phase (for this one molecule) are slower than vibrational relaxation in solution. This can also be understood by the density of states argument discussed above.

  • $\begingroup$ Thank you very much for the clear explanation and the referenced paper! I had also not considered the case of higher electronically excited states decaying faster, which helps explain Kasha's rule, and that is definitely something interesting to think about. $\endgroup$
    – Max
    Dec 13, 2017 at 12:55

The processes available to the excited molecule are vibrational relaxation/redistribution, emission (fluorescence , phosphorescence), inter-system crossing and internal conversion. We assume no chemistry occurs. These processes occur in competition with one another. Often it is observed in solution that vibrational relaxation is far faster than other processes so that emission occurs only from the lowest vibrational level of the excited state.

When a photon is absorbed forming an excited state only one type of vibrational mode is excited, and this is because only certain vibrations have a large Franck-Condon factor and so increase the transition dipole which couples the two electronic states. Because of the relative disposition of the electronic states the excited state is usually formed with more vibrational energy than has the ground state, i.e. the excited state is vibrationally hot. This proceeds to cool by transferring (redistributing) energy to other vibrational modes in the molecule. (The fact that electrons move far faster than nuclei do is why we can draw potential energy as a parabola with vibrational levels. This is often called Born-Oppenheimer approximation. The nuclei are effectively stationary while the electron 'moves' from ground to excited state. )

Now normal modes are supposed to be orthogonal but in reality all vibrations are anharmonic and this allows coupling between vibrational modes and so the energy flows out of the initial excited mode into others. The larger the number of vibrational quanta excited the more this occurs and this is also aided by the increasing density of states in the accepting modes: these can be many millions /cm$^{-1}$. Sometimes this is called the Franck-Condon weighted density of states ($\rho$) and the product of this with the operator $H$ coupling the levels forms the Fermi Golden rule expression ($k\approx |H|^2\rho)$ for the rate constant of the process.

All this happens in the isolated molecule (as in the gas phase) but the effect is small compared to that in solution where the continuous collisions with solvent cause far larger perturbations and larger coupling together of levels and so energy is lost faster out of the initial vibrational levels, generally of the order of a few picoseconds. (In passing we note that there are exceptions that have been observed in some rigid aromatic molecules, such as Nile-Blue, where excitation of the Franck-Condon mode in the singlet excited state lasts for hundreds of picoseconds. This is observed as fast periodic oscillations in the transient absorption).

Once in the vibrationless singlet state inter-system crossing to the triplet can occur, where just as in the singlet, vibrational energy can be lost before phosphorescence occurs. Also from the excited singlet, internal conversion to the ground state occurs. Internal conversion directly populates the ground state and so produces a vibrationally very hot ground state (as all the electronic energy is dumped into the ground state) and this looses energy rapidly to the solvent. You can observe this in transient absorption, but only for short lived excited states where this is the dominant pathway and lasts about 10 ps before the energy diffuses away in the solvent.

Internal conversion does not have to be fast, if it were always of picosecond duration then fluorescence would be almost unknown as the fluorescence yield is $\displaystyle \varphi=\frac{k_f}{(k_f+k_{ic}+k_{isc})}$ where subscripts mean fluorescence, internal conversion and intersystem crossing. The rate constant fro fluorescence is typically $10^8$ s$^{-1}$ and so if $k_{ic}=10^{12}$ the yield would be approx $10^{-4}$ whereas very many types of molecules have yields > 10%.

In competition to these processes fluorescence can occur, the rate constant for this is governed by the transition dipole moment between the ground and excited state $\bar \mu=\int \psi_i\mu\psi_fdRdr$ where $i$ and $f$ are label the total wavefunctions (electronic + vibrational) for the initial and final states and $\mu$ is the sum of electronic and nuclear dipoles.

  • $\begingroup$ Whoops, I missed the notification and did not see your answer until now. The mention of anharmonicity allowing different 'orthogonal' modes to couple has given me some more valuable insight, and I will look further into this and into the FC weighted density of states. Thanks a lot! $\endgroup$
    – Max
    Jan 23, 2018 at 13:13
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    $\begingroup$ There is more detail in my answer here chemistry.stackexchange.com/questions/28883/…. The theory papers were done in the late 1960's and early 70's. There are numerous papers. The names to look for are papers by W. Siebrand, by S.H Lin, by J. Jortner and by K. Freed and their collaborators. There are also many paper on experiments done mainly on isolated molecules in the vapour phase. $\endgroup$
    – porphyrin
    Jan 23, 2018 at 14:23

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