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Hydrogen (¹H), Deuterium (¹H²), singly ionised Helium (²He⁴⁺), and doubly ionised lithium (³Li⁶⁺) all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ₁, λ₂, λ₃ and λ₄ respectively then approximately which one of the following is correct?

1. 4λ₁ = 2λ₂ = 2λ₃ = λ₄
2. λ₁ = 2λ₂ = 2λ₃ = λ₄
3. λ₁ = λ₂ = 4λ₃ = 9λ₄
4. λ₁ = 2λ₂ = 3λ₃ = 4λ₄

Correct answer: λ₁ = λ₂ = 4λ₃ = 9λ₄

Solution

For n=2->1, 1/lambda is proportional to Z^2*(3/4), so lambda ~ 1/Z^2. H and D both have Z=1 (same wavelength), He+ has Z=2 (1/4), Li2+ has Z=3 (1/9). Hence lambda1 = lambda2 = 4*lambda3 = 9*lambda4.

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