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MHT CET 2026 (13 May, Shift 1) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from MHT CET 2026 (13 May, Shift 1)

Q28MHT CET 2026 · 13 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum

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

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Answer: (b)

  1. Quantisation: L=Iω=nh2πL = I\omega = \frac{nh}{2\pi}.
  2. Rotational KE =L22I=n2h28π2I= \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}.
  3. For n=2n = 2: KE=4h28π2I=h22π2IKE = \frac{4h^2}{8\pi^2 I} = \frac{h^2}{2\pi^2 I}
  4. So the answer is (b) h22Iπ2\frac{h^2}{2I\pi^2}.
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