Which transition in the hydrogen spectrum would have the same wavelength as the Balmer-type transition from to of the spectrum? [2023]

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Solution

Q.
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer-type transition from $n = 4$ to $n = 2$ of the ${He}^{+}$ spectrum [2023]

1 $n = 1$ to $n = 3$

2 $n = 1$ to $n = 2$

3 $n = 2$ to $n = 1$

4 $n = 3$ to $n = 4$

Ans.
(3)

$λ_{H} = λ_{{He}^{+}}$

$R_{H} \times {(1)}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} \times {(2)}^{2} (\frac{1}{{(2)}^{2}} - \frac{1}{{(4)}^{2}})$

$(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = (\frac{4}{4}) - (\frac{4}{16})$

$\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} = \frac{1}{1} - \frac{1}{4}$

$n_{1} = 1, n_{2} = 2 for H-atom$

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