In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $$n^{th}$$ excited state, is :

In the Bohr model, the electron moves in a circular orbit around the proton. We need to find the magnetic moment when the electron is in the nth excited state. However, note that the options are expressed in terms of $$n$$, and the correct answer matches the expression for the principal quantum number $$n$$. Therefore, we will derive the magnetic moment for the nth orbit (where $$n$$ is the principal quantum number) and interpret the nth excited state as corresponding to principal quantum number $$n$$.

The magnetic moment $$\mu$$ for a current loop is given by $$\mu = I \times A$$, where $$I$$ is the current and $$A$$ is the area of the loop. The electron has charge $$e$$ and moves in a circular orbit of radius $$r$$ with speed $$v$$. The current $$I$$ is the charge passing a point per unit time. The time period $$T$$ for one complete orbit is the circumference divided by the speed, so $$T = \frac{2\pi r}{v}$$. The current is then $$I = \frac{\text{charge}}{\text{time}} = \frac{e}{T} = e \times \frac{v}{2\pi r}$$.

The area $$A$$ of the circular orbit is $$A = \pi r^2$$. Substituting into the magnetic moment formula, we get:

$$\mu = I \times A = \left(e \times \frac{v}{2\pi r}\right) \times \pi r^2 = \frac{e v r}{2}.$$

In the Bohr model, the angular momentum is quantized. For the nth orbit (principal quantum number $$n$$), the angular momentum is $$m v r = n \frac{h}{2\pi}$$, where $$m$$ is the electron mass and $$h$$ is Planck's constant. Solving for $$v r$$, we get:

$$v r = \frac{n h}{2\pi m}.$$

Substituting this into the expression for $$\mu$$:

$$\mu = \frac{e}{2} \times v r = \frac{e}{2} \times \frac{n h}{2\pi m} = \frac{e n h}{4\pi m}.$$

We can rewrite this expression to match the given options. Factor out $$\frac{e}{2m}$$:

$$\mu = \frac{e n h}{4\pi m} = \left(\frac{e}{2m}\right) \times \frac{n h}{2\pi}.$$

This matches option C.

Hence, the correct answer is Option C.