Atomic Structure - Result Question 54
####30. The orbital angular momentum of an electron in $2 s$-orbital is
(a) $+\frac{1}{2} \cdot \frac{h}{2 \pi}$
(b) zero
(c) $\frac{h}{2 \pi}$
(d) $\sqrt{2} \cdot \frac{h}{2 \pi}$
(1996, 1M)
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Solution:
- Expression for orbital angular momentum $(L)$ is
$$ L=\sqrt{l(l+1)} \frac{h}{2 \pi}=0 \text { for } 2 s \text {-electrons } $$
$\because \quad$ For $s$-orbital, $l=0$.
