Hydrogen (H), deuterium (D), singly ionized helium ( ${He}^{+}$ ) and doubly ionized lithium (Li) all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of emitted radiations are $λ_{1}, λ_{2}, λ_{3} and λ_{4}$ respectively. Then approximately

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$λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

$4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

$λ_{1} = 2 λ_{2} = 2 \sqrt{2} λ_{3} = 3 \sqrt{2} λ_{4}$

$λ_{1} = λ_{2} = 2 λ_{3} = 3 \sqrt{2} λ_{4}$

answer is A.

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Detailed Solution

$\begin{array}{l} Using ΔE \propto Z^{2} (∴ n_{1} and n_{2} are same) \\ \Rightarrow \frac{hc}{λ} \propto Z^{2} \Rightarrow {λZ}^{2} = constant \\ \Rightarrow λ_{1} Z_{1}^{2} = λ_{2} Z_{2}^{2} = λ_{3} Z_{3}^{2} = λ_{4} Z_{4}^{2} \\ \Rightarrow λ_{1} \times 1 = λ_{2} \times 1^{2} = λ_{3} \times 2^{2} = λ_{4} \times 3^{3} \\ \Rightarrow λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4} \end{array}$