If an electron is moving in the $$n^{th}$$ orbit of the hydrogen atom, then its velocity ($$v_n$$) for the $$n^{th}$$ orbit is given as:

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit around the nucleus. The key condition is that the angular momentum of the electron is quantised: $$m v_n r_n = \frac{nh}{2\pi}$$, where $$m$$ is the electron mass, $$v_n$$ is the velocity in the $$n^{th}$$ orbit, $$r_n$$ is the radius of the $$n^{th}$$ orbit, $$n$$ is the principal quantum number, and $$h$$ is Planck's constant.

For the electron in a circular orbit, the centripetal force is provided by the Coulomb attraction: $$\frac{mv_n^2}{r_n} = \frac{e^2}{4\pi\varepsilon_0 r_n^2}$$. This simplifies to $$mv_n^2 = \frac{e^2}{4\pi\varepsilon_0 r_n}$$ $$-(1)$$.

From the quantisation condition, $$r_n = \frac{nh}{2\pi m v_n}$$ $$-(2)$$. Substituting $$(2)$$ into $$(1)$$: $$mv_n^2 = \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{2\pi m v_n}{nh}$$.

Cancelling $$m v_n$$ from both sides (since $$v_n \neq 0$$): $$v_n = \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{2\pi}{nh} = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n}$$.

Since $$\frac{e^2}{2\varepsilon_0 h}$$ is a collection of constants, we conclude that $$v_n \propto \frac{1}{n}$$. As the quantum number increases, the electron moves more slowly.

The correct answer is option 2: $$v_n \propto \frac{1}{n}$$.