In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in

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Question

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Brackett series is

Accepted Answer

Correct Answer: B — 4λStep-by-step Solution:Shortest wavelength corresponds to n → ∞ (series limit).Balmer series (n₁ = 2): 1/λB = R(1/2² − 0) = R/4 → λB = 4/R = λ (given)Brackett series (n₁ = 4): 1/λBr = R(1/4² − 0) = R/16 → λBr = 16/RRatio: λBr/λB = (16/R)/(4/R) = 4∴ λBr = 4λWhy other options are wrong:A (2λ): This would be the Paschen series limit (n₁=3): λ = 9/R = 9λ/4 ≈ 2.25λ — not exactly 2λ.C (9λ): Paschen series limit = 9/R = 9λ/4, not 9λ.D (16λ): This would require 1/λ = R/64, not seen in standard series.Key Concepts:Rydberg formula: 1/λ = RH(1/n₁² − 1/n₂²)Series limits (shortest λ, n₂ → ∞): 1/λmin = R/n₁²  → λmin = n₁²/RLyman: n₁=1, λmin=1/R; Balmer: n₁=2, λmin=4/R; Paschen: n₁=3, λmin=9/R; Brackett: n₁=4, λmin=16/R

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Brackett series is

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In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Brackett series is

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