MCQ
The orbital angular momentum of an electron in $2s$ orbitals is
- A$ + \frac{1}{2}.\frac{h}{{2\prod }}$
- ✓$0$
- C$\frac{h}{{2\prod }}$
- D$\sqrt {2.} \frac{h}{{2\prod }}$
$L =\frac{ h }{2 \pi} \sqrt{1(1+1)}$
For $2 s$ orbital, $1=0$ so $L=\frac{h}{2 \pi} \sqrt{0(0+1)}=0$
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