Question

(a) For 16O2,ΔG{ }^{16} \mathrm{O}_2, \Delta G values for the transitions v=10,20v=1 \leftarrow 0,2 \leftarrow 0, and 303 \leftarrow 0 are, respectively, 1556.22, 3088.28, and 4596.21 cm14596.21 \mathrm{~cm}^{-1}. Calculate v~\tilde{v} and xex_{e^*}. Assume yey_e to be zero.

(b) For 14 N2,ΔG{ }^{14} \mathrm{~N}_2, \Delta G values for the transitions v=10,20v=1 \leftarrow 0,2 \leftarrow 0, and 303 \leftarrow 0 are, respectively, 2345.15, 4661.40, and 6983.73 cm16983.73 \mathrm{~cm}^{-1}. Calculate v~\tilde{v} and xex_{e^*}. Assume yey_e to be zero.

Solution

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The vibrational energy levels of a diatomic molecule are given by:

Ev=(v+12)ω(12.33)\begin{aligned} \tag{12.33}E_v = \left(v+\dfrac{1}{2}\right)\hbar\omega \end{aligned}

where ω\omega is defined as:

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

where kfk_\text{f} represents the force constant and meffm_{\text{eff}} is effective mass which is defined as:

meff=m1m2m1+m2(12.32)\begin{aligned} \tag{12.32}m_{\text{eff}}=\dfrac{m_1m_2}{m_1+m_2} \end{aligned}

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