Q.In which of the following systems will the radius of the first orbit of the electron be smallest? (A) hydrogen (B) singly ionized helium (C) deuteron (D) tritium

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.