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Q.

Shortest wavelength in Balmer series of ${He}^{+}$ ion will be

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a

$\frac{4}{3 R_{H}}$

b

$\frac{15}{16 R_{H}}$

c

$\frac{4}{R_{H}}$

d

$\frac{1}{R_{H}}$

answer is B.

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Detailed Solution

Shortest wavelength means maximum wave number
$∴ n_{2} = \infty$
$= \frac{1}{λ} = R_{H} (2^{2}) [\frac{1}{2^{2}} - \frac{1}{\infty}]$
$= R_{H}$
$λ = \frac{1}{R_{H}}$

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Shortest wavelength in Balmer series of He+ ion will be