It is found that the excitation frequency from ground to the excited state of rotation for the $\,CO$ molecule is close to \[\dfrac{4}{\pi } \times {10^{11}}\,Hz\] . Then the moment of inertia of CO molecule about its centre of mass is close to ( Take \[h = 2\pi \times {10^{ - 34}}\,Js\])
A. \[2.76 \times {10^{ - 46}}\,kg{m^2}\]
B. \[1.87 \times {10^{ - 46}}\,kg{m^2}\]
C. \[4.67 \times {10^{ - 47}}\,kg{m^2}\]
D. \[1.17 \times {10^{ - 47}}\,kg{m^2}\]

Hint: The change in kinetic energy of the CO molecule is equal to the Bohr’ quantization condition $\dfrac{{nh}}{{2\pi }}.$

Complete step by step solution:
Let $I$ be the moment of inertia of a CO molecule about its centre of mass. Then, the rotational kinetic energy is given by \[\dfrac{1}{2}I{\omega ^2}\] , Where \[\omega \] is the angular velocity. According to Bohr’s quantization theory, an electron can revolve only in circular orbits and the angular momentum of the revolving electron is an integral multiple of $\dfrac{h}{{2\pi }}$, where $h$ is the planck’s constant.
$\therefore I\omega = \dfrac{{nh}}{{2\pi }}$
Therefore, rotational kinetic energy = \[\dfrac{1}{2}I{\left( {\dfrac{{nh}}{{2\pi I}}} \right)^2} = \dfrac{{{n^2}{h^2}}}{{8{\pi ^2}I}}\]
As the electron is excited from $n = 2\,to\,n = 1$
$\therefore \Delta KE = \dfrac{{{h^2}}}{{8{\pi ^2}I}}\left( {{2^2} - {1^2}} \right) = \dfrac{{3{h^2}}}{{8{\pi ^2}I}}$
This excitation energy is equal to$h\nu $, where $\nu $is the frequency of the light emitted. Thus,
$h\nu = \dfrac{{3{h^2}}}{{8{\pi ^2}I}} \Rightarrow I = \dfrac{{3h}}{{8{\pi ^2}\nu }}$
Putting the values of $h\,and\,\nu $in the above equation, we get
\[I = \dfrac{{3 \times 2\pi \times {{10}^{ - 34}}}}{{8{\pi ^2} \times \dfrac{4}{\pi } \times {{10}^{11}}}} \Rightarrow I = \dfrac{{3 \times {{10}^{ - 45}}}}{{16}} = 1.87 \times {10^{ - 46}}\]
Hence, the moment of inertia of the CO molecule about its centre of mass is \[1.87 \times {10^{ - 46}}\,kg{m^2}\]

Thus, the correct option is (B).

Note: In 1913 Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus. The energy of an electron depends on the size of the orbit. Radiation can occur only when the electron jumps from one orbit to another. He postulated that the angular momentum of the electron is quantized. Because of the quantization, the electron orbits have fixed sizes and energies. With his model, Bohr explained how electrons could jump from one orbit to another only by emitting or absorbing energy in fixed quanta. For example, if an electron jumps one orbit closer to the nucleus, it must emit energy equal to the difference of the energies of the two orbits.