If $$a_{\circ}$$ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength $$(\lambda)$$ of the electron present in the second orbit of hydrogen atom? [n : any integer]

We need to find the de-Broglie wavelength of the electron in the second orbit of hydrogen atom in terms of the Bohr radius $$a_0$$.

We start by recalling the Bohr model relations. From angular momentum quantization: $$mv_n r_n = n\hbar$$, where $$\hbar = h/(2\pi)$$. The Bohr radius of the $$n$$th orbit is given by $$r_n = n^2 a_0$$.

Next, from $$m v_n r_n = n\hbar = \frac{nh}{2\pi}$$, we have $$ m v_n = \frac{nh}{2\pi r_n} = \frac{nh}{2\pi \cdot n^2 a_0} = \frac{h}{2\pi n a_0} $$. This gives the momentum of the electron in the $$n$$th orbit.

Substituting this into the de-Broglie relation $$\lambda = \frac{h}{mv_n}$$ leads to $$ \lambda = \frac{h}{\frac{h}{2\pi n a_0}} = 2\pi n a_0 $$.

Therefore, for the second orbit where $$n = 2$$, $$ \lambda = 2\pi \times 2 \times a_0 = 4\pi a_0 $$. This can also be written as $$\frac{8\pi a_0}{2} = \frac{8\pi a_0}{n}$$.

Looking at the options, Option A gives $$\frac{8\pi a_0}{n}$$, which for $$n = 2$$ becomes $$\frac{8\pi a_0}{2} = 4\pi a_0$$, matching the result.

The correct answer is Option A: $$\frac{8\pi a_0}{n}$$.