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The wavelength of first line of Balmer series is 6563Å. The wavelength of first line of Lyman series will be
Options
(a) 1215.4Å
(b) 2500Å
(c) 7500Å
(d) 600Å
Correct Answer:
1215.4Å
Explanation:
λ(ʟyman) / λ(вalmer) = [(1/2²) – (1/3²)] / [(1/1²) – (1/2)²] = 5 / 27
λ(ʟyman) = (5 / 27) × λ(вalmer) = (5 / 27) × 6563 = 1215.4 Å
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Topics: Atoms and Nuclei
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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