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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2010 · Paper 2 · Physics · Atomic Physics

Paragraph: The key feature of Bohr's theory of spectrum of hydrogen...

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 CO\mathrm{CO} molecule is close to 4π×1011 Hz\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=2π×10−34J−sh=2 \pi \times 10^{-34} J-s )

Options

  1. A2.76×10−46 kg−m22.76 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2
  2. B1.87×10−46 kg−m21.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2
  3. C4.67×10−47 kg−m24.67 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2
  4. D1.17×10−47 kg−m21.17 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2

Correct answer

B. 1.87×10−46 kg−m21.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2

Step-by-step solution

hv=K2−K1=3h28π2I∴I=3h8π2f=3×2π×10−34×π8×π2×4×1011=\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 ×\times10^{-46} \mathrm{~kg}-\mathrm{m}^2 ∴\therefore The correct answer is (b).
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