For a dynamically stable orbit in a hydrogen atom, according to Rutherford’s atomic model the relation between the orbit radius (r) and the electron velocity (v) is:

$r=\frac{e^2}{4 \pi \varepsilon_0 m v^2}$

The correct answer is Option (3) → $r=\frac{e^2}{4 \pi \varepsilon_0 m v^2}$

According to Rutherford’s model

Centripetal force on electron = Electrostatic force of attraction between electron and proton

∴  $\frac{m v^2}{r}=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$

$r=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{m v^2}$