Chemistry - Atomic Structure Question with Solution | TestHub

For $\mathrm{s}$-subshell $\mathrm{l}=0$

Orbital angular momentum $\left(\mathrm{mvr}\right)=\frac{\mathrm{h}}{2\mathrm{\pi }}\sqrt{\mathrm{l}(\mathrm{l}+1)}$

For $2\mathrm{s}$ orbital, $\mathrm{l}$ (azimuthal quantum number) $=0$
$\therefore$ orbital angular momentum $=\frac{\mathrm{h}}{2\mathrm{\pi }}\sqrt{0(0+1)}$

$=\frac{\mathrm{h}}{2\mathrm{\pi }}\sqrt{0}=0$

Thus, the orbital angular momentum of an electron in $2\mathrm{s}$ -orbital is Zero.