In the visible-region Balmer series, the longest-wavelength line corresponds to the transition:
An = 3 → n = 2
Bn = 5 → n = 2
Cn = 4 → n = 2
Dn = 6 → n = 2
Answer & Solution
Correct answer: A. n = 3 → n = 2
Smallest ΔE among Balmer lines gives longest wavelength: 3 → 2 (Hα).
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