The shortest wavelength of H atom in the Lyman series is λ1 . The longest wavelength in the Balmer series of He+ is :

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The shortest wavelength of H atom in the Lyman series is $λ_{1}$ .

The longest wavelength in the Balmer series of He⁺ is :

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$\frac{9 λ_{1}}{5}$

$\frac{5 λ_{1}}{9}$

$\frac{27 λ_{1}}{5}$

$\frac{36 λ_{1}}{5}$

answer is D.

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Detailed Solution

Shortest wavelength → Max energy ( ∞ → 1 ) (Lyman series)

$\frac{1}{λ_{1}} = R_{H} \times 1^{2} [\frac{1}{1^{2}} - 0] R_{H} = \frac{1}{λ_{1}} F o r b a l m e r s e r i e s \frac{1}{λ_{1}} = R_{H} \times 2^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] \frac{1}{λ_{1}} = \frac{5 R_{H}}{9} λ_{1} = = \frac{9}{5 R_{H}} = \frac{9 λ_{1}}{5}$