Q. The time of revolution of an electron around a nucleus of charge $Ze$ in $n^{th}$ Bohr orbit is directly proportional to
A) $n$
B) $\frac{n^{3}}{Z^{2}}$
C) $\frac{n^{2}}{Z}$
D) $\frac{Z}{n}$
Explanation
(b) $T = \frac{2\pi r}{v};r$ = radius of $n^{th}$ orbit $= \frac{n^{2}h^{2}}{\pi MZe^{2}}$
$v =$ speed of $e^{-}$ in $n^{th}$ orbit $= \frac{ze^{2}}{2\varepsilon_{0}nh}$
$\therefore T = \frac{4\varepsilon_{0}^{2}n^{3}h^{3}}{mZ^{2}e^{4}} \Rightarrow T \propto \frac{n^{3}}{Z^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.