Question

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. $\frac{Z}{n}$

Answer 1

Answer: (B)

$\frac{n^{3}}{Z^{2}}$

Answer 2

$T = \frac{2\pi r}{v}$; r = radius of n^th orbit$= \frac{n^{2}h^{2}}{\pi mZe^{2}}$

v = speed of $e^{-}$ in n^th orbit $= \frac{ze^{2}}{2\varepsilon_{0}nh}$

∴$T = \frac{4\varepsilon_{0}^{2}n^{3}h^{3}}{mZ^{2}e^{4}}$ ⇒ $T \propto \frac{n^{3}}{Z^{2}}$