Chemistry - Atomic Structure Question with Solution | TestHub

$\begin{array}{rl}& \text{ Orbital angular momentum }=\sqrt{\ell (\ell +1)}\frac{\mathrm{h}}{2\pi }\\ & \therefore \text{ For }2\mathrm{s}\text{ orbital }:\ell =0\\ & \text{ Orbital angular momentum }=0\\ & \therefore \text{ For }2\mathrm{p}\text{ orbital }:\ell =1\\ & \begin{array}{rl}\text{ Orbital angular momentum }& =\sqrt{1(1+2)}\frac{\mathrm{h}}{2\pi }\\ & =\sqrt{2}\frac{\mathrm{h}}{2\pi }\end{array}\end{array}$