SOLUTION : After absorbing a photon of energy $12 \cdot 1 eV$ electron jumps from ground state $(n=1)$ to sccond excited state $(n=3)$. Therefore, change in angular momentum.
$\begin{aligned}\Delta L &=L_{3}-L_{1} \quad\left(L=n \frac{h}{2 \pi}\right) \\&=3\left(\frac{h}{2 \pi}\right)-\left(\frac{h}{2 \pi}\right)=\frac{h}{\pi} \\&=\frac{6 \cdot 626 \times 10^{-34}}{3 \cdot 14} J . S \\&= 2 \cdot 11 \times 10^{-34} J . s .\end{aligned}$
