Which is the shortest wavelength line in the Lyman series of the...

For Lyman series, $\frac{1}{\lambda}=R$\left(\frac{1}{(1)^2}-\frac{1}{n^2}\right)$hence, for$\lambda$to be smallest,$n$should be greatest (i.e.,$\infty$)$\therefore \quad \frac{1}{\lambda_{\min }}=1.097 \times 10^{-2}\left(1-\frac{1}{\infty^2}\right)$$\begin{aligned} & =1.097 \times 10^{-2} \\ \text{ or } \quad \lambda_{\min } & =\frac{1}{1.097 \times 10^{-2}}=91.16 \mathrm{~nm}\end{aligned}$$