The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes $n=2 \righta

Question

The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes $n=2 \rightarrow n=3$ and $n=4 \rightarrow n=6$ transitions, respectively, is

Professor PC Thomas Classes · Answer

(b) : From Rydberg's formula
$
\frac{1}{\lambda}=R_{\mathrm{H}}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}ight]
$
For $n_i=2, n_f=3$
$\frac{1}{\lambda_1}=R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}ight]=R_{\mathrm{H}}\left[\frac{5}{36}ight]$  ...(i)

For $n_i=4, n_f=6$
$\frac{1}{\lambda_2}=R_{\mathrm{H}}\left[\frac{1}{4^2}-\frac{1}{6^2}ight]=R_{\mathrm{H}}\left[\frac{1}{16}-\frac{1}{36}ight]=R_{\mathrm{H}}\left[\frac{5}{144}ight]$  ...(ii)

On dividing eqn. (ii) by (i), we get
$
\begin{aligned}
& \frac{\lambda_1}{\lambda_2}=R_{\mathrm{H}}\left[\frac{5}{144}ight] 	imes \frac{1}{R_{\mathrm{H}}}\left[\frac{36}{5}ight] \
& \frac{\lambda_1}{\lambda_2}=\frac{1}{4}
\end{aligned}
$