The wavelength of the H $α$ line in the Balmer series of the hydrogen atom is $λ$ The wavelength
of the same line in doubly ionized lithium is

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$\frac{λ}{9}$

$\frac{λ}{8}$

$\frac{λ}{2}$

$\frac{λ}{27}$

answer is A.

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Detailed Solution

For a given set of n₁and n $(∵ \frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}))$
$∴ \frac{λ_{He}}{λ_{H}} = \frac{Z^{2}}{Z_{He}^{2}} = \frac{1}{3^{2}} \Rightarrow λ_{He} = \frac{λ_{H}}{9} = \frac{λ}{9}$

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