The maximum wavelength in the Lyman series of the hydrogen spectrum (transition n₂ → 1) corresponds to n₂ =
A{'text': '3', 'label': 'A'}
B{'text': '2', 'label': 'B'}
C{'text': '4', 'label': 'C'}
D{'text': '∞', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': '2', 'label': 'B'}
1. Lyman series: transitions ending at n = 1, emissions in UV.
2. Maximum wavelength = minimum energy = smallest jump.
3. Smallest jump to n = 1 is from n = 2, giving λ_max.
4. This transition emits Lyman-α at 121.6 nm.
_Source: NCERT Class 11 Chemistry, Ch 2 "Structure of Atom", §2.4_
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