Find the largest and shortest wavelengths in the Lyman series for hydrogen. 

The transition equation for Lyman series is given by,

$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right),n=2,3,\dots$ .

The largest wavelength is corresponding to \(n=2\), so

$$\begin{gathered} \frac{1}{{\lambda }_{\max }}=1.097\times {10}^7\left(\frac{1}{1}-\frac{1}{4}\right) \\ \Rightarrow \frac{1}{{\lambda }_{\max }}=0.823\times {10}^7 \\ \Rightarrow {\lambda }_{\max }=1.2154\times {10}^{-7}m \\ \Rightarrow {\lambda }_{\max }=1215 \stackrel{\circ }{A} \end{gathered}$$

The shortest wavelength corresponds to $n\to \infty$, so

$$\begin{gathered} \frac{1}{{\lambda }_{\min }}=1.097\times {10}^7\left(\frac{1}{1}-\frac{1}{\infty }\right) \\ \Rightarrow {\lambda }_{\min }=0.911\times {10}^{-7}m=911 \stackrel{\circ }{A} \end{gathered}$$

Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum.