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JEE Main 2023 (6 Apr, Shift 2) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2023 (6 Apr, Shift 2)

Q75JEE Main 2023 · 6 APR · SHIFT 2Difficult

Bohr Model: Orbits, Speed & Angular Momentum

A small particle of mass mm moves in such a way that its potential energy U=12mω2r2U = \frac{1}{2} m\omega^2 r^2 where ω\omega is constant and rr is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of nnth orbit will be proportional to,

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Answer: (a)

  1. Force F=−dUdr=mω2rF = -\frac{dU}{dr} = m\omega^2 r gives mv2r=mω2r⇒v=ωr\frac{mv^2}{r} = m\omega^2 r \Rightarrow v = \omega r.
  2. Bohr condition: mvr=mωr2=nh2πmvr = m\omega r^2 = \frac{nh}{2\pi}.
  3. So r2∝n⇒r∝nr^2 \propto n \Rightarrow r \propto \sqrt{n}.
  4. So the answer is (a) n\sqrt{n}.
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