It turns out that the solution of the Schrödinger equation for the Morse potential can be expressed as
$$
G(v)=\tilde{\omega}_{\mathrm{e}}\left(v+\frac{1}{2}\right)-\tilde{\omega}_{\mathrm{e}} \tilde{x}_{\mathrm{e}}\left(v+\frac{1}{2}\right)^2
$$
Chapter 5 / The Harmonic Oscillator and Vibrational Spectroscopy
where
$$
\tilde{x}_{\mathrm{e}}=\frac{h c \tilde{\omega}_{\mathrm{e}}}{4 D}
$$
Given that $\tilde{\omega}_{\mathrm{e}}=2886 \mathrm{~cm}^{-1}$ and $D=440.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$ for $\mathrm{H}^{35} \mathrm{Cl}$, calculate $\tilde{x}_{\mathrm{e}}$ and $\tilde{\omega}_{\mathrm{e}} \tilde{x}_{\mathrm{e}}$.
 0.01961