Question

The time period of revolution of $e^{-}$ in ground state orbit of hydrogen atom is $T_{\lambda }$, then frequency of $e^{-}$ in second excited state of $L{i}^{++}$ ion is

• A. $\frac{1}{3T}$
• B. $\frac{1}{8T}$
• C. $\frac{9}{T}$
• D. $\frac{9}{8T}$

Answer 1

The correct option is D $\frac{9}{8T}$

We know that
$r=0.53\frac{n^2}{z}$

$v=\frac{c}{137}\frac{z}{n}$

$\therefore T=\frac{2\pi r}{v}\Rightarrow T\propto \frac{n^3}{z^2}$

Now time period of electron in ground state in case of hydrogen atom,
$T_1\propto \frac{1}{(1{)}^2}$

For $L{i}^{++}$ atom in second excited state
($n=2,z=3$),
$T_2\propto \frac{8}{9}$
$\frac{T_1}{T_2}=\frac{9}{8}\Rightarrow {\nu }_{Li}.T=\frac{9}{8}$
${\nu }_{Li}=\frac{9}{8T}$