If the ratio of energies of electron in the excited states of $\mathrm{H}$ and $\mathrm{Li}^{2+}$ is $1: 9$,…
If the ratio of energies of electron in the excited states of $\mathrm{H}$ and $\mathrm{Li}^{2+}$ is $1: 9$, the radius ratio of electron in the same excited states of $\mathrm{H}$ and $\mathrm{Li}^{2+}$ is
$9: 1$
$3: 1$
$1: 9$
$1: 3$
Solution
$\because$ For same excited state
$
r \propto \frac{1}{Z} \text { and } E \propto Z^2
$
Where, $r=$ radius of element
$Z=$ atomic number of element
' $\mathrm{Z}$ ' for $(\mathrm{H})$ and $\left(\mathrm{Li}^{2+}\right)$ are 1 and 3 respectively thus,
$
\begin{aligned}
& \frac{r_{(\mathrm{H})}}{r_{\left(\mathrm{L}^{2+}\right)}}=\frac{Z_{\left(\mathrm{Li}^{2+}\right)}}{Z(\mathrm{H})} \\
& \frac{r_{(\mathrm{H})}}{r\left(\mathrm{Li}^{2+}\right)}=\frac{3}{1}=3: 1
\end{aligned}
$