Explanation

(d) In Lyman series, wavelength emitted is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ =$\ R\left\lbrack \frac{1}{1^{2}} - \frac{1}{n^{2}} \right\rbrack$
where, $n$ =2,3,4……
and $R =$Rydberg’s constant
$= 1.097\ \times \ 10^{7}m^{- 1}$
For maximum wavelength $n$=2
$\therefore\ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{\max}} = 1.097\ \times 10^{7}\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack$
$\frac{1}{\lambda_{\max}} = 1.097\ \times 10^{7}\left\lbrack \frac{1}{1} - \frac{1}{4} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.097\ \times \ 10^{7} \times \frac{3}{4}$
$\Rightarrow \ \ \ \ \ \ \ \lambda_{\max}$ = $\frac{4}{3.291\ \times 10^{7}}$
$\ = 1216\ \mathring{\mathrm{A}}\ = 121.6\ m$
$\therefore\ \ \ \ \ \ \ \ \lambda_{\max}$ =122nm