Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $$n^{th}$$ orbital will therefore be proportional to :

The problem states that for a charged particle moving in a circular orbit perpendicular to a magnetic field, the perimeter of the orbit equals an integer number of de-Broglie wavelengths. We need to find how the radius of the nth orbital depends on n.

First, recall the de-Broglie wavelength $$\lambda$$ of a particle is given by $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle.

The condition that the perimeter equals an integer number of de-Broglie wavelengths for the nth orbit means:

$$ 2\pi r_n = n \lambda $$

where $$r_n$$ is the radius of the nth orbit and $$n$$ is a positive integer. Substituting $$\lambda = \frac{h}{p}$$:

$$ 2\pi r_n = n \cdot \frac{h}{p} $$

Rearranging this equation gives:

$$ 2\pi r_n = \frac{n h}{p} \quad \text{(Equation 1)} $$

Now, consider the motion of the charged particle in the magnetic field. The magnetic force provides the centripetal force. For a particle with charge $$q$$ and mass $$m$$ moving with speed $$v$$ perpendicular to a uniform magnetic field $$B$$, the force is $$F = q v B$$. This equals the centripetal force $$\frac{m v^2}{r}$$:

$$ q v B = \frac{m v^2}{r_n} $$

Simplifying by canceling one $$v$$ (assuming $$v \neq 0$$):

$$ q B = \frac{m v}{r_n} $$

Since momentum $$p = m v$$, we can write:

$$ q B = \frac{p}{r_n} $$

Rearranging for $$p$$:

$$ p = q B r_n \quad \text{(Equation 2)} $$

We now have two equations: Equation 1 relates $$r_n$$ and $$p$$, and Equation 2 relates $$p$$ and $$r_n$$. Substitute Equation 2 into Equation 1 to eliminate $$p$$.

Substituting $$p = q B r_n$$ into Equation 1:

$$ 2\pi r_n = \frac{n h}{q B r_n} $$

To solve for $$r_n$$, multiply both sides by $$r_n$$:

$$ 2\pi r_n \cdot r_n = \frac{n h}{q B} $$

$$ 2\pi r_n^2 = \frac{n h}{q B} $$

Divide both sides by $$2\pi$$:

$$ r_n^2 = \frac{n h}{2\pi q B} $$

Taking the square root of both sides:

$$ r_n = \sqrt{ \frac{n h}{2\pi q B} } $$

This can be written as:

$$ r_n = \sqrt{ \frac{h}{2\pi q B} } \cdot \sqrt{n} $$

The term $$\sqrt{ \frac{h}{2\pi q B} }$$ is constant because $$h$$, $$q$$, and $$B$$ are constants. Therefore, $$r_n$$ is proportional to $$\sqrt{n}$$, which is $$n^{1/2}$$.

Hence, the radius of the nth orbital is proportional to $$n^{1/2}$$.

Comparing with the options:

A. $$n^2$$

B. $$n$$

C. $$n^{1/2}$$

D. $$n^{1/4}$$

So, the correct answer is Option C.