A small particle of mass moves in such a way that its potential energy where is constant and is the distance of the particle from origin. [2023]

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Solution

Q.
A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to. [2023]

1 $\sqrt{n}$

2 $n$

3 $n^{2}$

4 $\frac{1}{n}$

Ans.
(1)

$U = \frac{1}{2} m ω^{2} r^{2}$

$F = - \frac{d U}{d r} = - m ω^{2} r$

$Now m ω^{2} r = \frac{m v^{2}}{r} \Rightarrow v = ω r \dots (i)$

$and m v r = \frac{n h}{2 π} \dots (i i)$

$From (i) and (ii),$

$m ω r^{2} = \frac{n h}{2 π} \Rightarrow r \propto \sqrt{n}$

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