MCQ
The orbital angular momentum of an electron in $2s$-orbital is
- A$\frac{1}{2}\frac{h}{{2\pi }}$
- B$\frac{h}{{2\pi }}$
- C$\sqrt 2 \frac{h}{{2\pi }}$
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Answer
Correct option: D.
Zero
d
(d) Since $s-$ orbital have $l = 0$
Angular momentum = $\sqrt {l\,(l + 1)} \times \frac{h}{{2\pi }}$ = $0 \times \frac{h}{{2\pi }} = 0$ .
(d) Since $s-$ orbital have $l = 0$
Angular momentum = $\sqrt {l\,(l + 1)} \times \frac{h}{{2\pi }}$ = $0 \times \frac{h}{{2\pi }} = 0$ .
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