$\begin{array}{l}\mathrm{A}\mathrm{doubly}\mathrm{ionized}{\mathrm{Li}}^{2+}\mathrm{ion}\text{ in the ground state absorbs 108.8 eV of energy.}\\ \text{ Find the increase in angular momentum of electron. }\left(\mathrm{take}\mathrm{h}=6.63\times {10}^{-34}\mathrm{J}-\mathrm{s}\right)\end{array}$

$\begin{array}{l}{E}_n=\frac{-13.6}{n^2}{Z}^2\\ Z=3,E_n=\frac{-122.4}{n^2}\\ E_1=-122.4eV\\ E_2=-30.6eV\\ E_3=-13.6eV\\ E_3-E_1=-13.6-\left(-122.4\right)=108.8eV\\ \text{Change in angular momentum}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\frac{3h}{2\pi }-\frac{h}{2\pi }\\ =2\left(\frac{h}{2\pi }\right)\\ =2.11\times {10}^{-34}\end{array}$