Q. The different lines in the Lyman series have their wavelengths faying between :

Solution:

First line of Lymen series
$\lambda_{1}=\frac{12431}{10.2}=1218.7\, \mathring{A}$
last time of Lymen series
$\lambda_{2}=\frac{12431}{13.6}=914\, \mathring{A}$