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Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n=4 to n=2 of $H e^{+}$ spectrum

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n=1 to n=3

n=1 to n=2

n=3 to n=4

n=2 to n=1

answer is D.

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Detailed Solution

$\bar{ϑ} = R \times Z^{2} (\frac{1}{h_{1}^{2}} - \frac{1}{h_{2}^{2}})$

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