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An ionic atom equivalent to hydrogen atom has wavelength equal to 1/4 of the wavelengths of hydrogen lines. The ion will be

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${Li}^{+ +}$

${He}^{+}$

${Ne}^{9 +}$

${Na}^{10 +}$

answer is A.

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Detailed Solution

$\bar{ν} \propto \frac{1}{λ} \propto Z^{2} \Rightarrow {λZ}^{2} =$ constant $\Rightarrow λ = \frac{λ}{4} Z^{2} \Rightarrow Z = 2$

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