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(A) The potential energy is given by $U = \frac{1}{2} m \omega^2 r^2$.The force acting on the particle is $F = -\frac{dU}{dr} = -m \omega^2 r$. The ma

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$\sqrt{n}$

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(A) The potential energy is given by $U = \frac{1}{2} m \omega^2 r^2$.The force acting on the particle is $F = -\frac{dU}{dr} = -m \omega^2 r$. The magnitude of this force provides the necessary centripetal force for circular motion:$m \omega^2 r = \frac{m v^2}{r} \implies v^2 = \omega^2 r^2 \implies v = \omega r$  --- $(i)$According to Bohr's quantization condition for angular momentum:$mvr = \frac{nh}{2\pi}$  --- $(ii)$Substituting the value of $v$ from $(i)$ into $(ii)$:$m(\omega r)r = \frac{nh}{2\pi}$$m \omega r^2 = \frac{nh}{2\pi}$Solving for $r^2$:$r^2 = \frac{nh}{2\pi m \omega}$Since $h, m, \omega$ are constants,we have:$r^2 \propto n \implies r \propto \sqrt{n}$.

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