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

The electronic ground state refers to the most stable and low energy state that an electron can occupy in a molecule. In this state, the molecule has the least amount of energy and is at the lowest point on the potential energy curve. Understanding the electronic ground state is critical because it represents the starting point for any excitation process.

In the context of the oxygen molecule discussed in the exercise, we start with electrons in their electronic ground state, which has a vibrational wavenumber of 1580 cm^-1. The electronic ground state is key to determining how much energy will be required to move an electron from this baseline up to an excited state. When studying molecular spectra, transitions are often considered from the v = 0 vibrational level in the ground state because it is the most populated level at typical temperatures.

Excited state transitions are processes by which electrons in a molecule absorb energy and move from the ground state to a higher energy level, known as an excited state. These transitions are not random; specific amounts of energy are required to promote the electrons to different energy levels, and these are often quantified in terms of their corresponding wavenumbers.

When an oxygen molecule goes from the v = 0 vibrational level of its electronic ground state to the v' = 0 level of its first excited state, a certain amount of energy, expressed in terms of a wavenumber or electron volts, is absorbed. In the exercise, the wavenumber for the vibrational level of the excited state is 700 cm^-1. This means that the excited state has less energy associated with its vibrational movement compared to the ground state. The total energy absorbed to reach this state encompasses not just the increase in electronic energy level but also includes the change in vibrational energy characterized by the difference in wavenumbers of these two states.

Potential energy curves graphically represent the energy of a molecule as a function of the position of its nuclei, essentially describing how the energy changes with the vibrational state. These curves are critical in understanding vibrational and electronic transitions because they illustrate the relative energies of different states and the transitions between them.

In our case, the exercise mentioned potential energy curves with minima that are separated by 6.175 eV. The difference between the minima of the electronic ground state and the excited state provides insights into the electronic transition energy. By converting this energy separation to wavenumbers using the formula (1 eV = 8065.54429 cm^-1), and then accounting for the change in vibrational energy (difference in vibrational wavenumbers), we can determine the wavenumber for the lowest energy transition in the band of transitions, which ultimately is the energy required for the electron to transition between these states.