Question

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.

Answer 1

Given : $\Delta E=12.5eV$
Considering the electron to jump from n = 1 to n = n level.
$\Delta E=E_n-E_1$
$\therefore$ $12.5=-\frac{13.6}{n^2}-\left(\frac{-13.6}{1^2}\right)12.5=13.6(1-\frac{1}{n^2})1-\frac{12.5}{13.6}=\frac{1}{n^2}\frac{1.1}{13.6}=\frac{1}{n^2}\frac{13.6}{1.1}=n^{2}12.36=n^{2}n=3.5n=3^{rd}$
$\therefore$
For n = 3 to n = 1,
$\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{n^2})\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{3^2})\frac{1}{\lambda }=1.097\times {10}^7(1-\frac{1}{9})\frac{1}{\lambda }=1.097\times {10}^7\times \frac{8}{9}\lambda =102.55nm$
For n = 2 to n = 1,
$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{1^2}-\frac{1}{2^2})\frac{1}{\lambda }=1.097\times {10}^7\left(\frac{3}{4}\right)\lambda =121.54nm$
For n = 3 to n = 2,
$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{3^2})\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{4}-\frac{1}{9})\frac{1}{\lambda }=1.097\times {10}^7\times \frac{5}{36}\lambda =656.33nm$
Hence , $\lambda =102.5nm$ and 121.5 nm $\to$ Lyman series and $\lambda =656.33nm\to$ Balmer series