A 12·5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.

Given: $\mathrm{\Delta E}=12.5\mathrm{eV}$
Considering the electron to jump from $\mathrm{n}=\mathrm{l} \mathrm{to} \mathrm{n}=\mathrm{n} \mathrm{level}$
$\begin{array}{l}\mathrm{\Delta E}={\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_{\mathrm{l}}\\ \therefore 12.5=-\frac{13.6}{{\mathrm{n}}^2}-\left(\frac{-13.6}{{\mathrm{l}}^2}\right)\\ 12.5=13.6\left(1-\frac{1}{{\mathrm{n}}^2}\right)\\ 1-\frac{12.5}{13.6}=\frac{1}{{\mathrm{n}}^2}\\ \frac{1.1}{13.6}=\frac{1}{{\mathrm{n}}^2}\\ \frac{13.6}{1.1}={\mathrm{n}}^2\\ 12.36={\mathrm{n}}^2\\ \mathrm{n}=3.5\\ \mathrm{n}=3^{\mathrm{rd}}\end{array}$
For $\mathrm{n}=3\text{ to }\mathrm{n}=1$
$\begin{array}{l}\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{3^2}\right.\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(1-\frac{1}{9}\right)\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\times \frac{8}{9}\\ \mathrm{\lambda }=102.55\mathrm{nm}\\ \text{ For }\mathrm{n}=2\mathrm{ton}=1,\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{3}{4}\right)\\ \mathrm{\lambda }=121.54\mathrm{nm}\end{array}$
$\begin{array}{l}\text{ For }\mathrm{n}=2\text{ to }\mathrm{n}=1,\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{3}{4}\right)\\ \mathrm{\lambda }=121.54\mathrm{nm}\\ \text{ For }\mathrm{n}=3\text{ to }\mathrm{n}=2,\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left(\frac{1}{4}-\frac{1}{9}\right)\\ \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\times \frac{5}{36}\\ \mathrm{\lambda }=656.33\mathrm{nm}\end{array}$
Hence, $\mathrm{\lambda }=102.5\mathrm{nm}$ and $121.5\mathrm{nm}$ $\to$ Lyman series and $\mathrm{\lambda }=656.33\mathrm{nm}$ $\to$Balmer series.