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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: A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the nnth level ( n=0n=0 is not allowed) is

Options

  1. A1n2(h28π2I)\frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)
  2. B1n(h28π2I)\frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)
  3. Cn(h28π2I)n\left(\frac{h^2}{8 \pi^2 I}\right)
  4. Dn2(h28π2I)n^2\left(\frac{h^2}{8 \pi^2 I}\right)

Correct answer

D. n2(h28π2I)n^2\left(\frac{h^2}{8 \pi^2 I}\right)

Step-by-step solution

L=Iω=nh2π∴ω=nh2πIK=12Iω2=12I(nh2πI)2=n2h28π2I\begin{aligned}L=I \omega& =\frac{n h}{2 \pi} \therefore \omega=\frac{n h}{2 \pi I}\\ K & =\frac{1}{2}I \omega^2=\frac{1}{2}I\left(\frac{n h}{2 \pi I}\right)^2 \\ & =\frac{n^2 h^2}{8 \pi^2 I} \end{aligned}\therefore$ The correct answer is (d).
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