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Wavelength of emission from Hydrogen atom when electron jumps from fourth orbit to ground state is $λ$ . What is the wavelength of emission coming from He⁺ ion when electron jumps from third orbit to ground state is

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$135 λ / 512$

$147 λ / 412$

$87 λ / 512$

$127 λ / 488$

answer is A.

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Detailed Solution

$\frac{1}{λ} = R (1 - \frac{1}{4^{2}}) = \frac{15 R}{16}$

In $H e^{+}, \frac{1}{λ^{1}} = R {(2)}^{2} (1 - \frac{1}{3^{2}}) = \frac{32 R}{9}$

$∴ \frac{λ^{1}}{λ} = \frac{15 R}{16} \times \frac{9}{32 R} \Rightarrow λ^{1} = \frac{135}{512} λ$

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