Difference between the wave length of 1^st line of Balmer series and last line of Paschen series for ${Li}^{+ 2} ion is \frac{1}{{xR}_{H}}$ then ‘x’ is

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answer is 5.

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Detailed Solution

$\frac{1}{λ_{1}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) 9; \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{α^{2}}) 9$
On solving $λ_{1} - λ_{\infty} = \frac{1}{5 R}$

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