Question

The orbital angular momentum of an electron in 2s-orbital is

• A. $+\frac{1}{2}.\frac{h}{2 \pi}$
• B. zero
• C. $\frac{h}{2 \pi}$
• D. $\sqrt2.\frac{h}{2 \pi}$

Answer 1

Answer: (B)

zero

Answer 2

Expression for orbital angular momentum (L) is
$\, \, \, \, \, \, \, \, L=\sqrt{l(l +1)} \frac{h}{2 \pi}=0 \, \, \, \, \,$ for 2s-electrons
$\because \, \,$ For s-orbital, l= 0.