A doubly ionised $\mathrm{Li}$ atom is excited from its ground $\operatorname{state}(n=1)$ to $n=3$ state.…
A doubly ionised $\mathrm{Li}$ atom is excited from its ground $\operatorname{state}(n=1)$ to $n=3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratio $\lambda_{32} / \lambda_{31}$ and $\lambda_{21} / \lambda_{31}$ are, respectively
$8.1,0.67$
$8.1,1.2$
$6.4,1.2$
$6.4,0.67$
Solution
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ where $R=$ Rydberg
$
\begin{aligned}
& \text { constant } \\
& \frac{1}{\lambda_{32}}=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \\
& \Rightarrow \lambda_{32}=\frac{36}{5}
\end{aligned}
$
Similarly solving for $\lambda_{31}$ and $\lambda_{21}$
$
\begin{aligned}
& \lambda_{31}=\frac{9}{8} \text { and } \lambda_{21}=\frac{4}{3} \\
& \therefore \quad \frac{\lambda_{32}}{\lambda_{31}}=6.4 \text { and } \frac{\lambda_{21}}{\lambda_{31}} \simeq 1.2
\end{aligned}
$