For CO molecule the rotational spectral lines appear at 3.8424 cm^-1 for the transition J = 0 to J = 1. Calculate the moment of inertia and bond distance for the molecule.

Question

For CO molecule the rotational spectral lines appear at 3.8424 [latex]\mathrm{cm}^{–1}[/latex] for the transition J = 0 to J = 1. Calculate the moment of inertia and bond distance for the molecule.

Accepted Answer

We know that

[latex]\bar{v}_{J ightarrow(J+1)}=\bar{v}_{0 ightarrow 1}=2 B[/latex]

Therefore [latex]2 \mathrm{~B}=3.8424 \mathrm{~cm}^{-1}[/latex]

or [latex]\mathrm{B}=1.9212 \mathrm{~cm}^{-1}[/latex]

Moment of inertial I:

[latex]\begin{array}{l}I=\frac{h}{8 \pi^{2} B c}=\frac{6.624 imes 10^{-27}}{8 imes 3.14 imes 3.14 imes 1.9212 imes 3 imes 10^{10}}\ \I=1.4567 imes 10^{-39} \mathrm{~g} \mathrm{~cm}^{2} \ \=1.4567 imes 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}\ \ ext { reduced mass of } \mathrm{CO} ext {, } \ \\mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=\frac{\frac{12}{N_{A}} imes \frac{16}{N_{A}}}{\frac{12}{N_{A}}+\frac{16}{N_{A}}}=\frac{12 imes 16}{N_{A}(12+16)}\ \ ext { where } N_{A}= ext { Avogadros number } \ \\mu=\frac{12 imes 16}{6.023 imes 10^{23} imes 28}=1.14 imes 10^{-23} \mathrm{~g} \mathrm{~mol}^{-1}=1.14 imes 10^{-26} \mathrm{~kg} \mathrm{~mol}^{-1}\ \\mathrm{I}=\mu r^{2} ext { or } r=\sqrt{\frac{I}{\mu}}\ \ ext { Bond distance } r=\sqrt{\frac{1.4567 imes 10^{-46}}{1.14 imes 10^{-26}}}=1.13 imes 10^{-8} \mathrm{~cm}\ \\quad \quad \quad \quad \quad \qquad=1.13 imes 10^{-10} \mathrm{~m}=1.131 \mathrm{~A}^{\circ}=113 \mathrm{pm} \end{array}[/latex]

Question 17.5.1: For CO molecule the rotational spectral lines appear at 3.84......

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