The time of revolution of an electron around a nucleus of charge Ze in nth Bohr orbit is directly proportional to
Answer & explanation
Correct answer: option 2
$T=\frac{2 \pi r}{v}$ ; r = radius of $n^{th}$ orbit = $\frac{n^2 h^2}{\pi m Z e^2}$
v = speed of $e^{-}$ in $n^{th}$ orbit = $\frac{z e^2}{2 \varepsilon_0 n h}$
∴ $T=\frac{4 \varepsilon_0^2 n^3 h^3}{m Z^2 e^4} \Rightarrow T \propto \frac{n^3}{Z^2}$