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Discussion

The variation of intensities of spectral lines shown in the figure corresponds to :

Intensity of spectral lines plotted against wavenumber in cm-1, falling steadily from the first line

  • A.Ground and excited state potential curves with the excited minimum almost directly above the ground minimum and a short vertical arrow
  • B.Ground and excited state potential curves with the excited minimum slightly displaced and a vertical arrow reaching the excited curve's left wall
  • C.Ground and excited state potential curves with the excited minimum displaced further, the vertical arrow reaching the excited curve on its right side
  • D.Ground and excited state potential curves with the excited minimum far displaced and a long vertical arrow reaching its left wall

Answer: A

The pattern corresponds to (A), upper and lower potential curves with the same equilibrium distance.

By the Franck-Condon principle, an electronic transition is vertical: the nuclei do not move during it, so the most intense vibrational band is the one whose upper-state level lies directly above the ground-state equilibrium geometry.

In (A) the excited-state minimum sits directly above the ground-state minimum, so the vertical transition lands on v′ = 0. The 0-0 band is the strongest and the higher bands (0-1, 0-2 and so on) get steadily weaker, just as the figure shows: intensity highest at the first line and falling away.

When the upper curve is displaced to a longer or shorter distance, as in (B), (C) and (D), the vertical transition reaches a higher v′, so the intensity first rises to a maximum in the middle of the progression; in (D) it reaches above the dissociation limit and gives a continuum.

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