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Answer: Shortest \(\approx 911.6\) Å; longest \(\approx 1215\) Å.

For the Lyman series, \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{1^2} - \dfrac{1}{n^2}\right)\), \(n = 2, 3, 4, \dots\)

Shortest wavelength (series limit, \(n = \infty\)):

\[\frac{1}{\lambda_s} = R_H \Rightarrow \lambda_s = \frac{1}{1.097\times10^{7}} = 9.116\times10^{-8}\ \text{m} = 911.6\ \text{Å}\]

Longest wavelength (\(n = 2\)):

\[\frac{1}{\lambda_l} = R_H\left(1 - \frac{1}{4}\right) = \frac{3R_H}{4} \Rightarrow \lambda_l = \frac{4}{3\times1.097\times10^{7}} = 1.215\times10^{-7}\ \text{m} = 1215\ \text{Å}\]
Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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