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Orthogonality of Vibrational Eigenfunctions in Different Electronic States and Fluorescence Behavior

This text discusses the orthogonality of vibrational eigenfunctions in different electronic states of a harmonic oscillator. It also explores the fluorescence behavior of an excited dye molecule after excitation with a photon energy higher than the S0 to S1 energy difference.

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Orthogonality of Vibrational Eigenfunctions in Different Electronic States and Fluorescence Behavior

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  1. Recall that the vibrational eigenfunctions of a harmonic oscillator are orthogonal: If yj and yk belong to different electronic states as shown here, is this still true? (A) Yes, because two vibrational eigenfunctions are always orthogonal. (B) No, because the potential curves are anharmonic oscillators, and two eigenfunctions of anharmonic oscillators are no longer orthogonal. (C) No, because they belong to different oscillators (different el. curves)

  2. Recall that the vibrational eigenfunctions of a harmonic oscillator are orthogonal: If yj and yk belong to different electronic states as shown here, is this still true? (A) Yes, because two vibrational eigenfunctions are always orthogonal. No, see (C) (B) No, because the potential curves are anharmonic oscillators, and two eigenfunctions of anharmonic oscillators are no longer orthogonal. Not true! Any two non-degenerate eigenstates of the same system are orthogonal. (C) No, because they belong to different oscillators (different el. curves). Orthogonality is only given within one and the same curve.

  3. S4 S3 S2 S1 S0 (ground state) Consider the experimental evidence you just saw, and look at the level diagram shown here. What happens after excitation with a photon energy higher than the S0 S1 energy difference if the excited dye molecule can fluoresce? (A) The molecule does not absorb. No fluorescence is detected. (B) The molecule absorbs on a vibronic transition of the S1 state with higher vibrational quanta can fluoresce back into S0 from that higher-up state. (C) The molecule absorbs into S2 or higher singlet states. From there, it quickly relaxes by other processes than photon emission into the S1(v=0) state, from where it fluoresces. (D) The molecule absorbs into S2 or higher singlet states. From there, it fluoresces.

  4. S4 S3 S2 S1 S0 (ground state) Consider the experimental evidence you just saw, and look at the level diagram shown here. What happens after excitation with a photon energy higher than the S0 S1 energy difference if the excited dye molecule can fluoresce? (A) The molecule does not absorb. No fluorescence is detected.Wrong! The experiment shows that green and all higher photon energies are absorbed. (B) The molecule absorbs on a vibronic transition of the S1 state with higher vibrational quanta can fluoresce back into S0 from that higher-up state.Wrong! The experiment shows that the wavelength (i.e. color) of the fluorescence light is the same for all excitation wavelengths. (C) The molecule absorbs into S2 or higher singlet states. From there, it quickly relaxes by other processes than photon emission into the S1(v=0) state, from where it fluoresces. → CORRECT! (Kasha’s Rule) (D) The molecule absorbs into S2 or higher singlet states. From there, it fluoresces. Wrong! The experiment shows that the wavelength (i.e. color) of the fluorescence light is the same for all excitation wavelengths.

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