A small particle of mass $$m$$ moves in such a way that its potential energy $$U = \dfrac{1}{2}m\omega^2 r^2$$ 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$$^{th}$$ orbit will be proportional to

Given: $$U = \frac{1}{2}m\omega^2 r^2$$, where $$\omega$$ is constant.

The force is: $$F = -\frac{dU}{dr} = -m\omega^2 r$$ (directed toward the origin).

For a circular orbit of radius $$r$$, this force provides the centripetal force:

$$ m\omega^2 r = \frac{mv^2}{r} \implies v = \omega r $$

Applying Bohr's quantization condition for angular momentum:

$$ mvr = n\hbar $$

$$ m(\omega r)r = n\hbar $$

$$ m\omega r^2 = n\hbar $$

Solving for $$r$$:

$$ r^2 = \frac{n\hbar}{m\omega} \implies r = \sqrt{\frac{n\hbar}{m\omega}} $$

Therefore, $$r \propto \sqrt{n}$$.

The correct answer is $$\sqrt{n}$$.