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The shortest wavelength for Lyman series is $912 Å$. The longest wavelength in Paschen series is
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Verified Answer
The correct answer is:
$18760 Å$
Shortest wavelength in Lyman series is given by
$$
\begin{aligned}
& \frac{1}{\lambda_{\mathrm{I}}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=\mathrm{R} \\
& \therefore \lambda_{\mathrm{I}}=\frac{1}{\mathrm{R}}
\end{aligned}
$$
The longest wavelength in Paschen series is given by
$$
\begin{aligned}
& \frac{1}{\lambda_{\mathrm{p}}}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\
& =\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]=\mathrm{R} \cdot \frac{7}{144} \\
& \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \\
& \therefore \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{L}}}=\frac{144}{7} \mathrm{R}=\frac{144}{7} \\
& \therefore \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \cdot \lambda_{\mathrm{L}}=\frac{144}{7} \times 912=18760 Å
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{\lambda_{\mathrm{I}}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=\mathrm{R} \\
& \therefore \lambda_{\mathrm{I}}=\frac{1}{\mathrm{R}}
\end{aligned}
$$
The longest wavelength in Paschen series is given by
$$
\begin{aligned}
& \frac{1}{\lambda_{\mathrm{p}}}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\
& =\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]=\mathrm{R} \cdot \frac{7}{144} \\
& \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \\
& \therefore \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{L}}}=\frac{144}{7} \mathrm{R}=\frac{144}{7} \\
& \therefore \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \cdot \lambda_{\mathrm{L}}=\frac{144}{7} \times 912=18760 Å
\end{aligned}
$$
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