Q.An electron in hydrogen atom stays in its second orbit for $10^{-8}$ s. How many revolutions will it make around the nucleus in that time?

From Bohr's two results, $L = nh/2\pi \propto n$ and $r_n \propto n^2$, eliminating $n$ gives the true relation $L \propto \sqrt{r}$ — which is not literally one of the four options offered, so this needs an honest note rather than a forced pick.

In the Bohr model of the hydrogen-like atom, two standard results follow from the quantisation postulate and the Coulomb force providing centripetal force:

1. Bohr's angular-momentum quantisation: $L = mvr = \dfrac{nh}{2\pi}$, so $L \propto n$.

2. The orbit radius: balancing $\dfrac{mv^2}{r} = \dfrac{kZe^2}{r^2}$ together with the quantisation condition gives $r_n = \dfrac{n^2h^2\varepsilon_0}{\pi m Z e^2}$, so $r \propto n^2$.

Eliminating the quantum number $n$ between these two: since $r \propto n^2$, we have $n \propto \sqrt{r}$, and since $L \propto n$, substituting gives

$L \propto \sqrt{r} \quad \text{i.e.} \quad L^2 \propto r$

This is the physically correct relationship — angular momentum grows only as the SQUARE ROOT of the orbit radius, not linearly, inversely, or as its square. None of the four printed choices (L∝r, L∝r⁻¹, L∝r², independent of r) is exactly this. Being honest about that mismatch rather than picking a wrong option with false confidence: if a single choice must be marked, (a) L ∝ r is the least-wrong pick only in the sense that it is also a directly increasing relationship (like the true $\sqrt{r}$ dependence), unlike the inverse or r² options — but the correct physics answer is $L \propto \sqrt{r}$.

Bohr orbit radius scales as $n^2$.

The radius of the $n^{th}$ Bohr orbit is $r_n = n^2 r_1$ (radius $\propto n^2$). So

$\frac{r_8}{r_4} = \left(\frac{8}{4}\right)^2 = 2^2 = 4$

The eighth orbit's radius is 4 times the fourth orbit's radius.