Which transition in hydrogen spectrum would have same wavelength as that in $H e^{+}$ ion for $n = 4$ to $n = 2$ transition?

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Solution

For $H e^{+}$
$R_{H} (\frac{1}{2^{2}} - \frac{1}{4^{2}}) 2^{2} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) 1^{2}$
${(\frac{2}{2})}^{2} - {(\frac{2}{4})}^{2} = {(\frac{1}{n_{1}})}^{2} - {(\frac{1}{n_{2}})}^{2}$
${(\frac{1}{1})}^{2} - {(\frac{1}{2})}^{2} = {(\frac{1}{n_{1}})}^{2} - {(\frac{1}{n_{2}})}^{2}$
$\frac{1}{n_{1}} = \frac{1}{1} \Rightarrow n_{1} = 1$
$\frac{1}{n_{2}} = \frac{1}{2} \Rightarrow n_{2} = 2$
$n_{1} = 1$
$n_{2} = 2$

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