JEE Mains · Physics · STD 12 - 12. atoms
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.
- A \(\sqrt{n}\)
- B \(n\)
- C \(n^2\)
- D \(\frac{1}{n}\)
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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