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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2025
  • Easy
  • MCQ

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is _______.

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Answer & Solution

Correct Answer

(B) \(5: 27\)

Step-by-step Solution

Detailed explanation

\(\mathrm {Lyman}\)

\(\frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4}\)
\(\lambda_1=\frac{4}{3 R}\) ________...(1)
and \(\mathrm {Balmer}\)

\(\frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36}\)
\(\lambda_2=\frac{36}{5 R}\)
Then, \(\frac{\lambda_1}{\lambda_2}=\frac{5}{27}\)
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