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## Homework Statement

The first vibrational band in the absorption spectrum of H

_{2}CO is at 28871 cm

^{-1}. The first two bands in the emission spectrum (from v'=0 level) lie at 27021 cm

^{-1}and 24687 cm

^{-1}.

Explain these observations. Determine the band origin of the electronic transition and the wavenumber of the vibrational mode in the excited electronic state that is responsible for the vibronically induced intensity.

The vibrational numbers of H

_{2}CO in its ground electronic state are:

v

_{1}=2766 cm

^{-1}(A

_{1})

v

_{2}=1746 cm

^{-1}(A

_{1})

v

_{3}=1501 cm

^{-1}(A

_{1})

v

_{4}=1167 cm

^{-1}(B

_{1})

v

_{5}=2843 cm

^{-1}(B

_{2})

v

_{6}=1251 cm

^{-1}(B

_{2})

## Homework Equations

G(v)=w

_{e}(v +1/2) - w

_{e}x

_{e}(v + 1/2)

^{2}

## The Attempt at a Solution

From ground state A

_{1}the molecule is excited to A

_{2}(absorption) it then rapidly relaxes to v=0 of A

_{2}. The emission spectrum is from this level.

I do not know where to begin with the band origin or how to find the vibrational mode wavenumber.

Any help greatly appreciated.