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MHT CET 2025 (21 Apr, Shift 2) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from MHT CET 2025 (21 Apr, Shift 2)

Q40MHT CET 2025 · 21 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( h=h = Planck's constant, I=I = moment of inertia of diatomic molecule)

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Answer: (a)

  1. Quantisation: L=nh2πL = \frac{nh}{2\pi}, so KE=L22I=n2h28π2IKE = \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}.
  2. For n=3n = 3: KE=9h28π2IKE = \frac{9h^2}{8\pi^2 I}
  3. So the answer is (a) 9h28π2I\frac{9h^2}{8\pi^2 I}.
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