The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively

For hydrogen like species , use the following formula to calculate the wavelength Of radiation of spectral series lines.

$\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Case 1: Minimum wavelength in the Lyman series:

${\mathrm{n}}_1=1 {\mathrm{n}}_2=2$

Then, $\frac{1}{\mathrm{\lambda }}=1096877\left[\frac{1}{I^2}-\frac{1}{2^2}\right]$

$\frac{1}{\mathrm{\lambda }}=R\left[1-\frac{1}{4}\right]=\frac{3R}{4}$

$\Rightarrow \mathrm{\lambda }=\frac{4}{3\mathrm{R}}=\frac{4}{3\times 109677{\mathrm{cm}}^{-1}}$

$\therefore \mathrm{\lambda } =1.215\times {10}^{-5}\mathrm{cm}\left(\mathrm{or}\right)121.5\mathrm{nm}$

$\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Case 2: Maximum wavelength in the Lyman series:

${\mathrm{n}}_1=1 {\mathrm{n}}_2=\infty$

Then, wavelength of line is

$\frac{1}{\mathrm{\lambda }}109677\left[\frac{1}{{\left(1\right)}^2}-\frac{1}{{(\infty )}^2}\right]$

$\frac{1}{\mathrm{\lambda }}=109677\left[1-0\right]=109677c{m}^{-1}$

$\Rightarrow \mathrm{\lambda }=\frac{1}{109677{\mathrm{cm}}^{-1}}=9.117\times {10}^{-6} \mathrm{cm}$

$\therefore \mathrm{\lambda }=91.2 \mathrm{nm}$

$\mathrm{Note}:1\mathrm{nm}={10}^{-7} \mathrm{cm}$