Question

Calculate the wavelength of the first line and the series limit for the Lyman series for hydrogen. $(R_H=109678{cm}^{-1})$.

Answer 1

For Lyman series, $n_1=1$.
For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., $n_2=\infty$.
So, $\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{{\infty }^2}]=R_H$
$\lambda =\frac{1}{109678}=9.117\times {10}^{-6}cm$
$=911.7\mathring{A}$
For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., $n_2=2$.
So, $\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{2^2}]=\frac{3}{4}{R}_H$
or $\lambda =\frac{4}{3}\times \frac{1}{R_H}=\frac{4}{3\times 109678}=1215.7\times {10}^{-8}cm$
$=1215.7\mathring{A}$.