Among the following, which transition in the hydrogen spectrum would have the same wavelength as Balmer…
Among the following, which transition in the hydrogen spectrum would have the same wavelength as Balmer transition, $n=4$ to $n=2$ in $\mathrm{He}^{+}$spectrum?
$n=3 \longrightarrow n=1$
$n=3 \longrightarrow n=2$
$n=4 \longrightarrow n=1$
$n=2 \longrightarrow n=1$
Solution
We know that,
$\frac{1}{\lambda}=R_{\mathrm{H}} Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
For $\mathrm{He},(Z=2)$
$\frac{1}{\lambda_{\mathrm{He}}}=R_{\mathrm{H}}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$=R_{\mathrm{H}} 4\left[\frac{1}{4}-\frac{1}{16}\right]=R_{\mathrm{H}} 4 \times \frac{3}{16}$
$\frac{1}{\lambda_{\mathrm{He}}}=\frac{3}{4} R_{\mathrm{H}}$
For option (d), $n=1$ and $n_2=2$
$\frac{1}{\lambda_{\mathrm{H}}}=\frac{R_{\mathrm{H}}}{1}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4} R_{\mathrm{H}}$
So, option (d) is correct.