Excited level, fluorescence spectrum

Figure 9.49 Single vibronic level fluorescence spectrum of styrene, in a supersonic jet, with excitation in the Ojj band...

Radiative transitions may be considered as vertical transitions and may therefore be explained in terms of the Franck-Condon principle. The intensity of any vibrational fine structure associated with such transitions will, therefore, be related to the overlap between the square of the wavefunctions of the vibronic levels of the excited state and ground state. This overlap is maximised for the most probable electronic transition (the most intense band in the fluorescence spectrum). Figure... 

However, it should be noted that most fluorescent molecules exhibit broad and structureless absorption and emission bands, which means that each electronic state consists of an almost continuous manifold of vibrational levels. If the energy difference between the 0 and 1 vibrational levels of So (and Si) is, for instance, only about 500 cm4, the ratio Ah/No becomes about 0.09. Consequently, excitation can then occur from a vibrationally excited level of the S0 state. This explains why the absorption spectrum can partially overlap the fluorescence spectrum (see Section 3.1.2). 

In general, the differences between the vibrational levels are similar in the ground and excited states, so that the fluorescence spectrum often resembles the first absorption band ( mirror image rule). The gap (expressed in wavenumbers) between the maximum of the first absorption band and the maximum of fluorescence is called the Stokes shift. 

The excited molecules normally release their energy by spontaneous emission of fluorescence, terminating not only in the initial ground state but on all vibronic levels of lower electronic states to which transitions are allowed. This causes a fluorescence spectrum which consists, for instance, in the case of an excited singlet state in a diatomic molecule, of a progression of either single lines (A/ = 0 named Q-lines) or of doublets (A7 = 1 P- and i -lines) ... 

Fig. 5. Fluorescence spectrum of Naj, from the 5 IIu state (u = 10, J= 39), excited by the argon laser line = 4726 A. The fluorescence lines represent transitions with A/ = 1, terminating at different u" levels (u" is labelled at each line) of the electronic ground state. (From ref. )...

During reactions (b) and (d) SO2 can be formed in the excited ACBi) state, as indicated by a peak at X = 3640 A in the fluorescence spectrum, observed during the reactionLower vibrational levels (u" = 10, 11) of CO are excited by the reaction... 

The energy difference between the ground and the first excited levels Sj can be obtained from the 0-0 band in absorption and/or fluorescence. The 0-0 frequency in fluorescence is a better measure. For a Tj state, the 0-0 band or the blue edge of the phosphorescence spectrum is considered. Other parameters are ... 

The fluorescence spectrum is found to be markedly non-Boltz-mann and sharply peaked at the directly excited level throughout the laser pulse. This is due to two effects the competition between electronic quenching and rotational relaxation processes (4) and the short length of the laser pulse. Because the pulse is so short, steady state is not established throughout the upper rotational levels. The peaks of the fluorescence pulses from levels which are not directly excited by the laser lag the laser pulse peaks by one to four nanoseconds, depending on the energy gap between the given level and the directly excited level. 

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