Question

For d-electron, the orbital angular momentum is

Answer 1

Final Answer : d)
$\frac{ \sqrt{6}h }{2\pi}$

Steps and Understanding:

1) For d-orbital,
Value of Azimuthal Quantum Number, l = 2

2) Orbital angular momentum is given by :
$\sqrt{l \times (l + 1)} \: \: \frac{h}{2\pi} \\ = > \sqrt{2 \times (2 + 1)} \: \: \frac{h}{2\pi} \\ = > \sqrt{6} \frac{h}{2\pi}$

Therefore, The Orbital Angular Momentum of d-electron is
$\boxed { \frac{h}{2\pi} \sqrt{6} }$