ExamBro
ExamBro
MHT CET · Physics · Atomic Physics

The shortest wavelength for Lyman series is \(912 Å\). The longest wavelength in Paschen series is

  1. A \(1216 Å\)
  2. B \(3646 Å\)
  3. C \(18760 Å\)
  4. D \(8208 Å\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(18760 Å\)

Step-by-step Solution

Detailed explanation

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
\(
\begin{aligned}
& \frac{1}{\lambda_{\mathrm{p}}}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\
& =\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]=\mathrm{R} \cdot \frac{7}{144} \\
& \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \\
& \therefore \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{L}}}=\frac{144}{7} \mathrm{R}=\frac{144}{7} \\
& \therefore \lambda_{\mathrm{p}}=\frac{144}{7 \mathrm{R}} \cdot \lambda_{\mathrm{L}}=\frac{144}{7} \times 912=18760 Å
\end{aligned}
\)
Same subject
Explore more questions on app