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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2023
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  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(\sqrt{n}\)

Step-by-step Solution

Detailed explanation

\(U=\frac{1}{2} m \omega^2 r^2\)

\(F=-\frac{d v}{d r}=-m \omega^2 r\)

Now \(m \omega^2 r=\frac{m v^2}{r} \Rightarrow v=\omega v^{\prime}\)

\(m vr=\frac{n h}{2 \pi} \ldots (ii)\)

From \((i)\) and \((ii)\)

\(m \omega r^2=\frac{n h}{2 \pi}\)

\(\Rightarrow r \propto \sqrt{n}\)

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