JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2023
- Medium
- 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.
Pick an option to check your answer.
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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