Question 144: Atoms & Nuclei

Question

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 $n^{	ext {th }}$ orbit will be proportional to,

Accepted Answer

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 $n^{	ext{th}}$ orbit is proportional to $\sqrt{n}$.

Answer

$\sqrt{n}$

Answer

$n^{2}$

Answer

$\frac{1}{n}$

Answer

$n$

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