Questions: The Franck-Condon Principle and Vibronic Transitions

Question 1

A diatomic molecule has an excited electronic state with a significantly longer equilibrium bond length than the ground state. Starting from v''=0 in the ground state, which vibronic transition do you expect to be the most intense in the absorption spectrum?

• A. The 0–0 transition, because it requires the minimum photon energy and is therefore most probable
• B. The 0–0 transition, because ground-state vibrational wavefunctions always overlap most with v'=0 of the excited state
• C. A transition to a high vibrational level v' > 0 of the excited state, because the vertical jump from v''=0 lands near the turning point of that level
• D. The transition to the highest accessible vibrational level, because greater energy transfer maximizes photon absorption

Answer: (C)

The Franck-Condon principle says transitions are vertical: the nuclear geometry is frozen during the femtosecond-timescale electronic jump. When the excited state has a longer equilibrium bond length, its potential curve is displaced to the right. The v''=0 wavefunction (peaked near the ground-state equilibrium geometry) now overlaps most with a high vibrational level of the excited state, whose wavefunction has a turning-point maximum near that same geometry. Option A and B represent the common misconception that the 0–0 transition dominates regardless of geometry — this is only true when the two states have nearly identical equilibrium structures.

Question 2

A molecule absorbs UV light and its fluorescence (emission) spectrum is measured. According to the Franck-Condon principle and Kasha's rule, the emission spectrum compared to the absorption spectrum should be:

• A. Identical to the absorption spectrum, because the same FC overlap integrals govern both processes
• B. A mirror image of the absorption spectrum, displaced to lower energy (redshifted)
• C. A broad featureless band, because rapid vibrational relaxation destroys the vibrational structure before emission
• D. Dominated by a single sharp 0–0 line, because emission always terminates at the lowest vibrational level of the ground state

Answer: (B)

After absorbing light, the molecule rapidly relaxes to v'=0 of the excited state (Kasha's rule) before emitting. Emission then proceeds as a vertical downward transition, and the same geometric displacement that shaped absorption also shapes emission — but now traversed in reverse. The FC factors are symmetric, producing the same progression envelope but shifted to lower energy (the 0–0 gap must be crossed before vibrational levels of the ground state are accessed). This mirror-image rule is diagnostic in spectroscopy: a clear mirror-image relationship between absorption and emission spectra confirms vibrational progressions governed by the same potential energy displacement.

Question 3

When two electronic states have nearly identical equilibrium geometries and force constants, the 0–0 transition dominates the absorption spectrum and transitions to higher vibrational levels of the excited state are weak.

• T. True
• F. False

Answer: True

If the two potential energy curves sit nearly directly above each other (no horizontal displacement), then a vertical transition from v''=0 lands near the bottom of the excited-state curve — precisely where v'=0 has its largest amplitude. The FC overlap integral |⟨χ_v'=0|χ_v''=0⟩|² is large, while overlaps with higher v' levels are small because their wavefunctions have low amplitude near the equilibrium geometry. The misconception to avoid is assuming this is always the case; strong vibrational progressions (e.g., in I₂) reveal significant geometry change between states.

Question 4

A molecule initially in v''=0 of the ground electronic state can be excited to multiple different vibrational levels of the excited electronic state in a single absorption experiment, each with a different probability.

• T. True
• F. False

Answer: True

This is the direct meaning of the vibrational progression in an absorption spectrum: each band corresponds to a distinct v''=0 → v' transition, and the intensity of each band reflects its Franck-Condon factor |⟨χ_v'|χ_v''=0⟩|². Absorption is not restricted to a single final state — it populates a distribution of vibrational levels, with the envelope shaped by the FC factors. This is why UV-Vis spectra of diatomics often show a series of regularly spaced bands rather than a single line.

Question 5

Why does the 'vertical' nature of an electronic transition mean the 0–0 band is not always the most intense, even though it involves the smallest energy change?

Model answer: Transition probability is governed by the Franck-Condon factor — the square of the overlap integral between the vibrational wavefunctions of the initial and final levels — not by the energy gap. A vertical transition preserves the nuclear geometry, so the molecule arrives at whatever point on the excited-state potential energy surface sits directly above its starting geometry. If the excited state has a different equilibrium bond length, this vertical landing point does not correspond to v'=0 (which sits at the new equilibrium) but to a higher vibrational level whose turning point is near the old equilibrium geometry. The 0–0 FC overlap is therefore small, and the most intense band shifts to higher v'.

This insight — that band intensities encode geometry changes between electronic states — makes FC analysis a powerful structural tool: the position and shape of the vibrational progression in an absorption spectrum directly reveals how much the equilibrium geometry changes upon electronic excitation. Long progressions (like I₂) signal large geometry changes; short progressions signal minimal changes. The principle connects quantum mechanical wavefunctions to observable spectral shapes.