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

Solution:

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

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

Case 1: Minimum wavelength in the Lyman series:

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

Then, $\frac{1}{\mathrm{\lambda }}=109677\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$

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

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

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

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

Case 2: Maximum wavelength in the Lyman series:

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

Then, wavelength of line is

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

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

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

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

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