Questions: Electronic Spectroscopy and the Franck-Condon Principle

Question 1

In a UV absorption spectrum, which transition has the greatest intensity according to the Franck-Condon principle?

• A. Always the v=0 → v'=0 transition
• B. The transition to the v' level whose wavefunction has the greatest overlap with the ground-state v=0 wavefunction
• C. The transition to the highest vibrational level of the excited state
• D. The transition with the largest energy gap

Answer: (B)

The Franck-Condon principle says intensity is proportional to |⟨v'|v⟩|². The most intense band corresponds to the excited-state vibrational level whose wavefunction overlaps most with the ground-state wavefunction. When excited and ground electronic states have the same equilibrium geometry, this IS the 0→0 transition — but when geometries differ (the common case), the maximum shifts to a higher v'.

Question 2

Phosphorescence is slower than fluorescence because it involves a spin-forbidden transition from a triplet excited state back to the singlet ground state.

• T. True
• F. False

Answer: True

Fluorescence is a singlet→singlet (S₁→S₀) radiative transition, which is spin-allowed and occurs on nanosecond timescales. Phosphorescence involves intersystem crossing from S₁ to the triplet state T₁, followed by a T₁→S₀ emission. Because this requires a spin flip, it violates the spin selection rule and is slow — occurring on microsecond to second timescales. This is why phosphorescent materials continue glowing after the excitation source is removed.

Question 3

Why is the Franck-Condon transition described as 'vertical' on a potential energy diagram?

Model answer: Because electronic transitions occur so rapidly (~10⁻¹⁵ s) that nuclei cannot move during the event — both nuclear positions and momenta are essentially frozen. On a potential energy vs. internuclear distance diagram, the transition appears as a vertical arrow from the ground-state vibrational wavefunction up to the excited-state potential curve at the same nuclear coordinate.

This follows from the Born-Oppenheimer approximation: electrons move on a timescale many orders of magnitude faster than nuclei. The electronic transition is complete before the nuclei have time to respond, so the nuclear geometry at the moment of absorption is the same as in the initial state. This verticality determines which vibrational levels of the excited state are populated.