Question

Pertaining to Fluorescence. What are the excitation and emission transition dipole axis and why are they important?

Answer 1

Ans- absorption of light by a molecule occurs very quickly, on a time scale of roughly 10-15 sec. This is too fast to permit any significant motions of the atoms (more specifically, of the nuclei) in the molecule itself or in the solvent molecules surrounding it. This is known as the Franck-Condon principle, and the excited state that the molecule is ‘promoted’ to immediately upon absorption of the photon is called the Franck-Condon excited state. As we will see, in order for a molecule (fluorescent or not) to return from the FranckCondon excited state to the equilibrium ground state, the molecule must undergo three distinct processes sequentially:

Step 1 - On a time scale in the picosecond range, the newly excited molecule adjusts to adopt a minimum free energy while remaining in the excited electronic state. After these adjustments are complete, the molecule will exist in a new, lower-energy excited state known as the equilibrium excited state. Step 2- The electron in the higher-energy orbital returns to the lower-energy orbital. This step may entail the emission of a photon as fluorescence, but there are also other, competing mechanisms to return the electron to the lower-energy orbital without photon emission. The return of the electron to the ground-state orbital is very rapid (time scale of the order of 10-15 sec). As a result, the molecule initially returns to a Franck-Condon ground state, in which it may have excess vibrational energy and a nonoptimal interaction with the surrounding solvent molecules. Step 3 - The molecule in the Franck-Condon ground state will undergo relaxation to the equilibrium ground state, in much the same way that the Franck-Condon excited state earlier relaxed to the equilibrium excited state. In the two following sections we will discuss in more detail these three processes and their consequences for the overall phenomenon of fluorescence. ‘Relaxation’ of the Franck-Condon Excited State The energy of the Franck-Condon excited state is typically higher than that of the equilibrium excited state. This is true for two reasons. First, in the Franck-Condon excited state the molecule may exist in a vibrational substate other than the lowest-energy vibrational level. One part of ‘relaxation’ of the Franck-Condon excited state is thus the conversion of the molecule to the lowest-energy vibrational level of the excited state:

Second, the electronic distribution, and hence various other properties such as the dipole moment, are different for a molecule in the excited state compared to the ground state. As a result, immediately after absorbing a photon the excited-state molecule will typically not exhibit an optimal interaction with the solvent molecules around it. The second aspect of ‘relaxation’ of the Franck-Condon excited state is thus that an excited-state molecule and the solvent molecules around it will adjust their positions and orientations to achieve a new arrangement of solvent molecules, which is optimally adjusted to the properties of the excited-state molecule. This is frequently a different, and lower-energy, arrangement of solvent and solute molecules than the one that characterized the Franck-Condon excited state. For these two reasons the equilibrium excited state attained after excited-state relaxation is of significantly lower energy than the Franck-Condon excited state

‘Deexcitation’ of the Excited State, and Relaxation Back to the Equilibrium Ground State In the equilibrium excited state just described, the molecule still has one electron in the excited-state orbital. The average delay between absorption of a photon and emission of fluorescence is of the order of nanoseconds (nsec) or tens of nsec for a typical fluorescent species. This means that after photon absorption a molecule typically will have time to reach the equilibrium excited state before emitting a photon. However, once a molecule ‘decides’ to return to the electronic ground state, the actual transition itself happens extremely rapidly (on a time scale of the order of 10-15 sec.), too fast to permit any motions of the nuclei within the molecule or in the surrounding solvent molecules. We thus say that the newly ‘deexcited’ molecule returns initially to a Franck-Condon ground state. In two important respects the transition just noted is similar to the earlier process of ‘promotion’ of the molecule to the Franck-Condon excited state. First, a given excited-state molecule may return to any of the vibrational (and rotational) substates of the electronic ground state, giving the possible transitions shown below:

‘The’ transition to the ground state is thus again a family of possible transitions to individual sublevels of the Franck-Condon ground state. Second, in the Franck-Condon ground state the newly ‘deexcited’ molecule may find itself initially surrounded by solvent molecules in an arrangement which was optimal to solvate the excited-state molecule but which is not optimal to solvate the ground-state molecule. As a result, the Franck-Condon ground state will typically dissipate some of its energy (both by returning to the lowestenergy vibrational level and by ‘reoptimizing’ its interactions with the solvent) in order to return to the equilibrium ground state (in which the molecule existed before photon absorption). These processes (and the reasons for them) are similar to those which drive the relaxation of the Franck-Condon excited state to the equilibrium excited state

Translating Molecular Behavior into Observable Fluorescence Properties We will now consider two questions of critical importance in understanding fluorescence spectroscopy: first, at what wavelengths will the fluorescence be emitted (i.e., what will be the shape and the position of the fluorescence spectrum), and second, what will be the intensity of the emitted fluorescence? To answer the first question, recall that ‘the’ transition of a molecule from the equilibrium excited state to the Franck-Condon ground state is not a unique transition but rather a family of possible transitions to various vibrational (and rotational) substates of the electronic ground state. This means that the shape of a fluorescence spectrum will exhibit the same general features as does the absorbance spectrum; it will in generally be fairly broad, may be asymmetric and may reveal ‘vibrational fine structure’ (shoulders, partially resolved component peaks, etc.). What about the position of a fluorescence spectrum, and more precisely, what will be the relationship between the fluorescence and the absorption spectrum? The general answer to this question can be understood by considering the energy diagram below:

the energy of the equilibrium excited state must be less than or equal to that of the Franck-Condon excited state. Likewise, the energy of the equilibrium ground state must be less than or equal to that of the Franck-Condon ground state. Recall as well that the energy of the photon absorbed equals the energy difference between the equilibrium excited state and the Franck-Condon excited state, and that the energy of the photon emitted equals the energy difference between the equilibrium ground state and the Franck-Condon ground state (arrow lengths in the above diagram). From the above four statements we can conclude that the energy of the photon emitted as fluorescence must be less than or equal to (typically less than) the energy of the photon initially absorbed. This means in turn that the wavelength of light emitted as fluorescence will be greater than or equal to (and typically greater than) the wavelength of the light absorbed, since λ = hc/Ephoton. If we extrapolate this last statement from the behavior of a single molecule to that of a whole population of fluorescent molecules, we conclude that the emission (fluorescence) spectrum for a fluorescent molecule will be shifted to longer wavelengths with respect to the excitation (absorption) spectrum. We now turn to the question of the intensity of fluorescence emitted by a sample, which can also be a useful source of information about the behavior and environment of a molecule. For this discussion we will have to introduce some additional information about how molecules can return from an excited state to the ground state. Franck-Condon Excited State Equilibrium Excited State Equilibrium Ground State Franck-Condon Ground State Photon Emitted Photon Absorbed Energy 7 Determinants of Fluorescence Intensity: Competition among Energy-Dissipating Pathways You may have wondered whether every molecule that absorbs a photon must emit fluorescence to return to the ground state. The answer is ‘no,’ because emission of a photon is not the only way that a molecule in an excited electronic state may return to the ground state. There are typically other, competing pathways to achieve this end as well, and for many molecules these alternative, radiationless (or nonradiative) processes are so efficient that the molecules emit little if any fluorescence at all. There are a number of detailed pathways by which a molecule that has absorbed a photon can return to the ground state without emitting a photon; these are known collectively as pathways of radiationless (or noradiative) decay of the excited state. For our purposes it is useful to group these processes into three classes: (1) The molecule may be able to return the electron to the ground-state orbital while converting its excess electronic energy to excess vibrational or rotational energy, which can then be transferred to the surrounding medium. The transformation of energy within the molecule in this manner is known as internal conversion. For many molecules this is the major or exclusive means of dissipation of the energy provided by photon absorption. (2) The molecule may transfer its excess electronic energy to a special type of molecule called a quencher and thereby return to the electronic ground state without emitting a photon. Quenchers such as acrylamide, Br- and I- ions, etc. act by transiently colliding or complexing with excited-state molecules. One special class of quenchers, however, can 'accept energy at a distance' by a phenomenon known as resonance energy transfer, which we will discuss in a later section. (3) The molecule may undergo intersystem crossing from the excited state discussed above, which is technically known as a singlet state (one in which all electron spins are paired), to a triplet state (which has two unpaired spins). It is much 'harder,' in quantummechanical terms, for a triplet-state molecule to convert to the ground state than for a singlet-state molecule to do so. As a result, molecules that convert to the triplet state return to the ground state either by eventual internal conversion or by a distinct, relatively slow process of photon emission called phosphorescence.

The above diagram summarizes the alternative possible 'fates' of a molecule in the excited state; here kf , kic, kq and kix are respectively the first-order rate constants for emission of fluorescence, internal conversion, quenching and intersystem crossing (the latter three processes, all of which lead to outcomes other than fluorescence, are those described in (1)- (3) above). If (as is usually the case) the rate of relaxation of the Franck-Condon excited state to the equilibrium excited state is much faster than is the overall rate of decay of the Fluorescence (kf ) Internal Conversion (ki ) Quenching (kq) Intersystem Crossing (kx) Equilibrium Excited State Triplet State (Phosphorescence) (Relaxation) Franck-Condon Excited State Absorption Time (log scale) 8 equilibrium excited state, we can relate the above rate constants to two important characteristics of a molecule’s fluorescence. The first is the overall rate of decay of the excited state, which is more often reported as the excited state lifetime, τf :