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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 ${He}^{+}$ spectrum

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n = 2 to n = 1

n = 1 to n = 3

n = 3 to n = 4

n = 3 to n = 2

answer is A.

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Detailed Solution

$E = - 13.6 \times Z^{2} \times [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$= - 13.6 \times 2^{2} \times [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$

$= - 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

In hydrogen $n_{1} = 1$ and $n_{2} = 2$ ;

$2 \to 1$ transition

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