Question

Calculate the shortest and longest wavelengths in the hydrogen spectrum of the Lyman series.

Answer 1

Given Data:

The series is of Lyman Series.

So, ${\mathrm{n}}_1$= 1

Applying Rydberg Formula:-

For the shortest wavelength in the Lyman series, emission will be maximum.

So, ${\mathrm{n}}_2=\infty$

By Rydberg's Formula:

$\frac{1}{\mathrm{\lambda }}=R_H\left[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right]$ where ${\mathrm{R}}_{\mathrm{H}}$= 109678 ${\mathrm{cm}}^{-1}$

On putting ${\mathrm{n}}_1$=1 and ${\mathrm{n}}_2=\infty$, we get

$\frac{1}{\mathrm{\lambda }}=109678$${\mathrm{cm}}^{-1}$

$\mathrm{\lambda }=\frac{1}{109678}$${\mathrm{cm}}^{-1}$

$\mathrm{\lambda }$ = 9.117 x ${10}^{-6}$ ${\mathrm{m}}^{-1}$

So,

$\mathrm{\lambda }$ = 911.7 $\mathrm{A}°$

For longest wavelength in Lyman series, ${\mathrm{n}}_2=2$

So, Applying

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

Putting ${\mathrm{n}}_1$=1 and ${\mathrm{n}}_2=2$, we get

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\frac{3}{4}x\frac{1}{R_H} \\ \lambda =\frac{4}{3}x{R}_H \end{gathered}$$

Also, ${\mathrm{R}}_{\mathrm{H}}$ = 109678 ${\mathrm{cm}}^{-1}$

On solving,

$\mathrm{\lambda }$ = 1215.7 x ${10}^{-8}$ ${\mathrm{cm}}^{-1}$

So, $\mathrm{\lambda }$ = 1215.7 $\mathrm{A}°$

Therefore, the shortest wavelength and longest wavelength in the hydrogen spectrum of Lyman Series is 911.7$\mathrm{A}°$ and 1215.7$\mathrm{A}°$ respectively.