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Atoms & Nuclei

JEE Main level, single correct, medium

150 of 239

Question 150 of 239

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

Show answer and solution

Answer: (C) $6.4,1.2$

\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}

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