Discussion
For the carbon monoxide (which can be considered a rigid diatomic molecule) for the transition J = 0 to J = 1 (where J is the rotational quantum number), the wave number \(\bar{\nu} = 3.842\ cm^{-1}\). The length of CO bond is :
(Given : Mass of carbon atom is \(19.921 \times 10^{-27}\) kg, mass of oxygen atom is \(26.561 \times 10^{-27}\) kg, Planck's constant \(h = 6.626 \times 10^{-34}\) Js, speed of light \(c = 3 \times 10^{8}\) m/s.)
Answer: C
The bond length is 1.131 Å.
For a rigid rotor the \(J = 0 \to 1\) line lies at \(\bar{\nu} = 2B\), so \(B = 1.921\ cm^{-1} = 192.1\ m^{-1}\). With \(B = h/(8\pi^{2}cI)\) and \(I = \mu r^{2}\):
\[ \begin{aligned} I &= \frac{h}{8\pi^{2}cB} = 1.456 \times 10^{-46}\ \text{kg m}^{2} \\ \mu &= \frac{19.921 \times 26.561}{46.482} \times 10^{-27} \\ &= 1.138 \times 10^{-26}\ \text{kg} \\ r &= \sqrt{I/\mu} = 1.131 \times 10^{-10}\ \text{m} \end{aligned} \]
The key step is halving the line position: using \(\bar{\nu}\) itself as \(B\) gives \(r\) smaller by \(\sqrt{2}\), about 0.80 Å, which is not an option, while the other listed lengths (0.956, 1.483 and 1.762 Å) do not follow from the given masses and line position.
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