Question 190: 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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