Physics - Gravitation Question with Solution | TestHub

Ionization energy corresponding to ionization potential $=-13.6\mathrm{eV}$ Photon energy incident $=12.1\mathrm{eV}$ So, the energy of electron in excited state $$=-13.6+12.1=-1.5\mathrm{eV}$$ $$\begin{array}{l}\text{ ie, }{E}_n=-\frac{13.6}{n^2}\mathrm{eV}\\ -1.5=\frac{-13.6}{n^2}\\ \Rightarrow n^2=\frac{-13.6}{-1.5}\approx 9\\ \therefore n=3\end{array}$$ ie, energy of electron in excited state corresponds to third orbit. The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1 st and 2nd to 1 st. Thus, 3 spectral lines are emitted.