JEE Mains · Physics · STD 12 - 12. atoms
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is _______.
- A \(5: 36\)
- B \(5: 27\)
- C \(3: 4\)
- D \(27: 5\)
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 \(\mathrm {Balmer}\) \(\frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36}\)…
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