## Related questions with answers

(a) For ${ }^{16} \mathrm{O}_2, \Delta G$ values for the transitions $v=1 \leftarrow 0,2 \leftarrow 0$, and $3 \leftarrow 0$ are, respectively, 1556.22, 3088.28, and $4596.21 \mathrm{~cm}^{-1}$. Calculate $\tilde{v}$ and $x_{e^*}$. Assume $y_e$ to be zero.

(b) For ${ }^{14} \mathrm{~N}_2, \Delta G$ values for the transitions $v=1 \leftarrow 0,2 \leftarrow 0$, and $3 \leftarrow 0$ are, respectively, 2345.15, 4661.40, and $6983.73 \mathrm{~cm}^{-1}$. Calculate $\tilde{v}$ and $x_{e^*}$. Assume $y_e$ to be zero.

Solution

VerifiedThe **vibrational energy levels of a diatomic molecule** are given by:

$\begin{aligned} \tag{12.33}E_v = \left(v+\dfrac{1}{2}\right)\hbar\omega \end{aligned}$

where $\omega$ is defined as:

$\begin{aligned} \omega = \sqrt{\dfrac{k_\text{f}}{m_{\text{eff}}}} \end{aligned}$

where $k_\text{f}$ represents the **force constant** and $m_{\text{eff}}$ is **effective mass** which is defined as:

$\begin{aligned} \tag{12.32}m_{\text{eff}}=\dfrac{m_1m_2}{m_1+m_2} \end{aligned}$

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