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Hydrogen ( $_{1} H^{1}$ ), deuterium (₁H²), singly ionised helium (₂He⁴)⁺ and doubly ionised lithium (₃Li⁶)⁺⁺ and doubly ionised lithium (₃Li⁶)⁺⁺ all have one electron around THE nucleus.
Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are $λ_{1}, λ_{2}, λ_{3}$ and $λ_{4}$ respectively, then approximately WHICH of the
following is correct?

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$λ_{1} = 2 λ_{2} = 3 λ_{3} = 4 λ_{4}$

$4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

$λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

$λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

answer is C.

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Detailed Solution

$\frac{1}{λ} = R Z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

$\frac{1}{λ_{1}} = R (1)^{2} (3 / 4), \frac{1}{λ_{2}} = R (1)^{2} (3 / 4)$

$\frac{1}{λ_{3}} = R (2)^{2} (3 / 4), \frac{1}{λ_{4}} = R 3^{2} (3 / 4)$

$\frac{1}{λ_{1}} = \frac{1}{λ_{2}} = \frac{1}{4 λ_{3}} = \frac{1}{9 λ_{4}}$

$\Rightarrow λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

SO, option (c) is correct.

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