The vibrational wavenumber of the oxygen molecule in its electronic ground state is 1580 cm−1,...

Question

The vibrational wavenumber of the oxygen molecule in its electronic ground state is 1580 cm−1, whereas that in the first excited state (B 3Σu−), to which there is an allowed electronic transition, is 700 cm−1. Given that the separation in energy between the minima in their respective potential energy curves of these two electronic states is 6.175 eV, what is the wavenumber of the lowest energy transition in the band of transitions originating from the v = 0 vibrational state of the electronic ground state to this excited state? Ignore any rotational structure or anharmonicity.

Robert · Accepted Answer

PE vs R curve for dioxygen molecule
From the figure above , d= 6.175 eV= 6.175 x 8065.54 cm-1  = 49804.7 cm-1  . 
a = 1580 cm-1  , b = 700 cm-1  . 
Therefore wavenumber for the said transition is : d-a+b = (49804.7- 1580 +700) cm-1  =  
48924.7 cm-1  = 6.066 eV.

The vibrational wavenumber of the oxygen molecule in its electronic ground state is 1580 cm −1 , whereas that in the first excited state (B 3 Σ u − ), to which there is an allowed electronic...

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