A small particle of mass m moves in such a way that its potential energy U = 1 2mω2r2 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,

JEE Mains - Quantized Circular Orbits in a Harmonic Potential

Question: A small particle of mass \( m \) moves in such a way that its potential energy is \[ U = \frac{1}{2} m\omega^2 r^2, \] where \( \omega \) is constant and \( r \) is the distance from the origin. Assuming Bohr's quantization of angular momentum and a circular orbit, the radius of the \( n \)th orbit will be proportional to ______.
[6-Apr-2023 Shift 2]

Select the correct answer:

(A) \( n \)

(B) \( \sqrt{n} \)

(C) \( n^2 \)

(D) \( \frac{1}{\sqrt{n}} \)

Solution

For a particle in a harmonic oscillator potential, the potential energy is given by: \[ U = \frac{1}{2} m\omega^2 r^2. \]

For a circular orbit, the force required for circular motion is provided by the restoring force: \[ m\omega^2 r = \frac{mv^2}{r} \quad \Longrightarrow \quad v = \omega r. \]

Bohr's quantization condition states that the angular momentum is quantized: \[ mvr = n\hbar. \] Substituting \( v = \omega r \) into the quantization condition, we get: \[ m(\omega r)r = m\omega r^2 = n\hbar. \]

Solving for \( r \): \[ r^2 = \frac{n\hbar}{m\omega} \quad \Longrightarrow \quad r \propto \sqrt{n}. \]

Therefore, the radius of the \( n \)th orbit is proportional to \( \sqrt{n} \).