Discussion
In electronic spectra of molecules, the vibrational quantum number leading to convergence of vibrational spectral lines is given by :
Answer: C
The convergence occurs at v = 1/(2xe) − 1.
For an anharmonic oscillator the vibrational levels are
\[ \begin{aligned} \varepsilon_{v} &= \bar{\omega}_{e}(v+\tfrac{1}{2}) \\ &- \bar{\omega}_{e}x_{e}(v+\tfrac{1}{2})^{2} \end{aligned} \]
The spacing between neighbouring levels shrinks as v rises. Setting the spacing to zero gives the level where the progression converges:
\[ \begin{aligned} \Delta\varepsilon &= \bar{\omega}_{e}[1 - 2x_{e}(v+1)] = 0 \\ v_{max} &= \frac{1}{2x_{e}} - 1 \end{aligned} \]
This is the form given in Banwell. The other options drop the factor 2, or put the wavenumber \(\bar{\omega}_{e}\) in the denominator, which makes v carry units of cm instead of being a pure number.
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