If ' $\lambda_1$ ' and ' $\lambda_2$ ' are the wavelengths of the first line of the Lyman and Paschen series…
If ' $\lambda_1$ ' and ' $\lambda_2$ ' are the wavelengths of the first line of the Lyman and Paschen series respectively, then $\lambda_2: \lambda_1$ is
$3: 1$
$30: 1$
$50: 7$
$108: 7$
Solution
Using Rydberg's formula,
$\begin{aligned}
\frac{1}{\lambda} & =\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\
\frac{1}{\lambda_1} & =\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{4}\right] \\
& \ldots\left(\because \mathrm{n}_1=1 \text { and } \mathrm{n}_2=2 \text { for Lyman series }\right) \\
& =\frac{3 \mathrm{R}}{4} \\
\frac{1}{\lambda_2} & =\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right] \\
& \ldots\left(\because \mathrm{n}_1=3 \text { and } \mathrm{n}_2=4 \text { for Paschen series }\right) \\
& =\frac{7 \mathrm{R}}{144}
\end{aligned}$
$\begin{array}{ll}\therefore & \frac{1}{\lambda_1} \times \frac{\lambda_2}{1}=\frac{3 R}{4} \times \frac{144}{7 R} \\ \therefore & \frac{\lambda_2}{\lambda_1}=\frac{108}{7}\end{array}$