Question

What transition in the hydrogen spectrum have the same wavelength as Balmer transition $n=4$ to $n=2$ of ${He}^{+}$ spectrum?

Answer 1

For ${He}^{+}$ ion,
$\frac{1}{\lambda }=Z^{2}R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$={\left(2\right)}^{2}R[\frac{1}{2^2}-\frac{1}{4^2}]=\frac{3R}{4}$
For hydrogen atom,
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
So, $\frac{3R}{4}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
or $\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}$
i.e., $n_1=1$ and $n_2=2$.