Q51 / 57 · JEE Advanced 2010 (Paper 2) · Physics
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question:
It is found that the excitation frequency from ground to the first excited state of rotation for the $\mathrm{CO}$ molecule is close to $\frac{4}{\pi} \times 10^{11} \mathrm{~Hz}$. Then the moment of inertia of CO molecule about its centre of mass is close to (Take $h=2 \pi \times 10^{-34} J-s$ )
- A$2.76 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2$
- Correct answerB$1.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2$
- C$4.67 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2$
- D$1.17 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2$
Solution
$$
\begin{gathered}
h v=K_2-K_1=\frac{3 h^2}{8 \pi^2 I} \\
\therefore I=\frac{3 h}{8 \pi^2 f}=\frac{3 \times 2 \pi \times 10^{-34} \times \pi}{8 \times \pi^2 \times 4 \times 10^{11}}
\end{gathered}
$$
$$
=1.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2
$$
$\therefore$ The correct answer is (b).
Hence, the correct answer is option: 'B'.