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The longest wavelength associated with Paschen series is : (Given $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}$ unit)
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Verified Answer
The correct answer is:
$1.876 \times 10^{-6} \mathrm{~m}$
For longest wavelength in Paschen's series:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
For longest $\mathrm{n}_1=3$
$\mathrm{n}_2=4$
$\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{16-9}{16 \times 9}\right] \\ & \Rightarrow \lambda=\frac{16 \times 9}{7 \mathrm{R}}=\frac{16 \times 9}{7 \times 1.097 \times 10^7} \\ & \lambda=1.876 \times 10^{-6} \mathrm{~m}\end{aligned}$
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
For longest $\mathrm{n}_1=3$
$\mathrm{n}_2=4$
$\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{16-9}{16 \times 9}\right] \\ & \Rightarrow \lambda=\frac{16 \times 9}{7 \mathrm{R}}=\frac{16 \times 9}{7 \times 1.097 \times 10^7} \\ & \lambda=1.876 \times 10^{-6} \mathrm{~m}\end{aligned}$
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