QJEEM23Ph27c4: JEE Main 2023 Physics - Class 12 JEEM23Ph27c4 Solution: A small particle of mass moves in such a way that its potential …

According to Bohr's quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:

$L = n\hbar$

Where $n$ is an integer and $\hbar$ is the reduced Planck's constant. The angular momentum $L$ can also be expressed as:

$L = mvr$

Where $m$ is the mass of the particle, $v$ is its linear velocity, and $r$ is the radius of the orbit.

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:

$m\frac{v^2}{r} = -\frac{\mathrm{d}U}{\mathrm{d}r} = -m\omega^2r$

We can simplify this equation to get the relation between $v$ and $r$ :

$v^2 = \omega^2r^2$

Now, let's combine the equations for angular momentum and the relation between $v$ and $r$ :

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

We can now solve for the radius $r$ in terms of $n$ :

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

Taking the square root of both sides, we get:

$r \propto \sqrt{n}$

So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$ .