Question

The so-called hot bands correspond to transitions that originate from vibrationally excited levels of the electronic ground state. They are called hot bands because they require population in the excited vibrational level.

Assuming that both ground and excited electronic potential-energy surfaces are approximately
harmonic, with the same frequency in the ground and excited states, show that there is a hot-band peak
arising from the v" = 1 (first excited) vibrational level of the ground electronic state and that this peak is of
lower frequency than the adiabatic transition.

Answer 1

Hot bands result from transitions originating from already excited vibrational levels of the ground state to the excited electronic state, and thus, require less energy than the adiabatic transition from the ground vibrational level, which results in a lower frequency peak.

Step-by-step explanation:

The hot bands in the context of molecular spectroscopy refer to absorptions due to transitions from vibrationally excited levels of the ground electronic state to various vibrational levels of an excited electronic state. According to the question, if we assume harmonic potential-energy surfaces for both the ground and excited states with the same frequency, then we indeed expect hot-band transitions.

Considering the ground (v" = 1) and excited electronic states with the same vibrational frequency, a hot band arises when a molecule transitions from the first excited vibrational level of the ground state (v" = 1) to a vibrational level in the excited electronic state. Given that the energy of the v" = 1 level is already elevated due to its vibrational excitation, the additional energy required to transition to the excited electronic state is less than the energy required for the transition from the v" = 0 level (adiabatic transition). Thus, the frequency of the hot band peak, which corresponds to the energy difference between these levels, is lower than that of the adiabatic transition.

In summary, the hot band peak associated with the v" = 1 level will indeed have a lower frequency than the peak associated with the adiabatic transition from the ground vibrational level (v" = 0) due to the initial energy present in the excited vibrational level.