NEET 2024 · PhysicsRydberg formulaPrevious Year Question
Match List I with List II: List I (Spectral Lines of Hydrogen for transitions from) (A) n₂ = 3 to n₁ = 2 (B) n₂ = 4 to n₁ = 2 (C) n₂ = 5 to n₁ = 2 (D) n₂ = 6 to n₁ = 2 List II (Wavelengths (nm)) (I) 410.2 (II) 434.1 (III) 656.3 (IV) 486.1 Choose the correct answer from the options given below:
Answer: (A) A-III, B-IV, C-II, D-I. Answer: (a) A-III, B-IV, C-II, D-I Solution: All transitions end at n₁=2 (Balmer).
- A.A-III, B-IV, C-II, D-I✓
- B.A-IV, B-III, C-I, D-II
- C.A-I, B-II, C-III, D-IV
- D.A-II, B-I, C-IV, D-III
Correct Answer
(A) A-III, B-IV, C-II, D-I
Solution & Explanation
Answer: (a) A-III, B-IV, C-II, D-I Solution: All transitions end at n₁=2 (Balmer). Larger jumps give larger energy and shorter wavelength. (A) 3→2 = Hα, longest λ = 656.3 nm (III). (B) 4→2 = Hβ = 486.1 nm (IV). (C) 5→2 = Hγ = 434.1 nm (II). (D) 6→2 = Hδ, shortest = 410.2 nm (I). So A-III, B-IV, C-II, D-I.
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