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,

<p>According to Bohr&#39;s quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:</p> <p>$L = n\hbar$</p> <p>Where $n$ is an integer and $\hbar$ is the reduced Planck&#39;s constant. The angular momentum $L$ can also be expressed as:</p> <p>$L = mvr$</p> <p>Where $m$ is the mass of the particle, $v$ is its linear velocity, and $r$ is the radius of the orbit.</p> <p>Now, we are given the potential energy $U = \frac{1}{2} m\omega^2r^2$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:</p> <p>$m\frac{v^2}{r} = -\frac{\mathrm{d}U}{\mathrm{d}r} = -m\omega^2r$</p> <p>We can simplify this equation to get the relation between $v$ and $r$:</p> <p>$v^2 = \omega^2r^2$</p> <p>Now, let&#39;s combine the equations for angular momentum and the relation between $v$ and $r$:</p> <p>$n\hbar = mvr = m\sqrt{\omega^2r^2}r = m\omega r^2$</p> <p>We can now solve for the radius $r$ in terms of $n$:</p> <p>$r^2 = \frac{n\hbar}{m\omega}$</p> <p>Taking the square root of both sides, we get:</p> <p>$r \propto \sqrt{n}$</p> <p>So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.</p>