Thank you for reporting, we will resolve it shortly
Q.
Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n=4$ to $n=2$ ) of $He ^{+}$spectrum?
$\because \frac{1}{\lambda}=Z^{2} \cdot R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
$\therefore $ For $He ^{+}$ion, $Z=2$
$\frac{1}{\lambda}=2^{2} \cdot R\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$
$\frac{1}{\lambda}=4 \times R \times \frac{3}{16}=\frac{3}{4} R$
The same value for $H$-atom is possible when electron jumps from $n=2$ to $n=1$, i.e.
$\frac{1}{\lambda}=1 \times R\left[\frac{1}{1}-\frac{1}{4}\right]$
$\Rightarrow \frac{3}{4} R$