The value of $(n_{1} + n_{2}) and (n_{2}^{2} - n_{1}^{2})$ for He⁺ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted radiation when electrons jump from n₂ to n₁is

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$\frac{9}{32 R_{H}}$

$\frac{32 R_{H}}{9}$

$\frac{32}{9 R_{H}}$

$\frac{9 R_{H}}{32}$

answer is B.

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Detailed Solution

$(n_{2}^{2} - n_{1}^{2}) = 8; (n_{1} + n_{2}) = 4 (n_{1} + n_{2}) (n_{1} - n_{2}) = 8 (n_{1} - n_{2}) = 8 / 4 = 2$

Hence n₁=1, n₂=3

$\bar{ν} = \frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} 2^{2} (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = \frac{32 R_{H}}{9} λ = \frac{9}{32 R_{H}}$

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