For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Question

Aicharya · Accepted Answer

$\mathrm {Lyman}$

$\frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4}$
$\lambda_1=\frac{4}{3 R}$ ________...(1)
and $\mathrm {Balmer}$

$\frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36}$
$\lambda_2=\frac{36}{5 R}$
Then, $\frac{\lambda_1}{\lambda_2}=\frac{5}{27}$

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Solution