The momentum of an electron revolving in $$n^{th}$$ orbit is given by: (Symbols have their usual meanings)

We need to find the momentum of an electron revolving in the $$n^{th}$$ orbit.

Bohr’s quantization condition states that the angular momentum of an electron in the $$n^{th}$$ orbit is quantized:

$$L = mvr = \frac{nh}{2\pi}$$

Since the linear momentum is $$p = mv$$, substituting into the angular momentum relation $$mvr = \frac{nh}{2\pi}$$ yields

$$p \cdot r = \frac{nh}{2\pi}$$

and hence

$$p = \frac{nh}{2\pi r}$$

Hence, the correct answer is Option A.