For a d electron, the orbital angular momentum is

Question

For a d electron, the orbital angular momentum is (A) √6((h/2 π)) (B) √2((h/2 π)) (C) ((h/2 π)) (D) 2((h/2 π))

Tardigrade Admin · Accepted Answer

The orbital angular momentum is: (h/2 π) √l(l+1) The orbital angular momentum for an electron ind orbital (l=2) is (h/2 π) √l(l+1)=(h/2 π) √2(2+1)=(h/2 π) √6

Q. For a $d$ electron, the orbital angular momentum is

$6 (\frac{h}{2 π})$

$2 (\frac{h}{2 π})$

$(\frac{h}{2 π})$

$2 (\frac{h}{2 π})$

The orbital angular momentum is: $\frac{h}{2 π} l (l + 1)$
The orbital angular momentum for an electron ind orbital $(l = 2)$ is
$\frac{h}{2 π} l (l + 1) = \frac{h}{2 π} 2 (2 + 1) = \frac{h}{2 π} 6$

Q. For a d electron, the orbital angular momentum is

6​(2πh​)

2​(2πh​)

(2πh​)

2(2πh​)

The orbital angular momentum is: 2πh​l(l+1)​ The orbital angular momentum for an electron ind orbital (l=2) is 2πh​l(l+1)​=2πh​2(2+1)​=2πh​6​

Q. For a $d$ electron, the orbital angular momentum is

$6 (\frac{h}{2 π})$

$2 (\frac{h}{2 π})$

$(\frac{h}{2 π})$

$2 (\frac{h}{2 π})$

The orbital angular momentum is: $\frac{h}{2 π} l (l + 1)$
The orbital angular momentum for an electron ind orbital $(l = 2)$ is
$\frac{h}{2 π} l (l + 1) = \frac{h}{2 π} 2 (2 + 1) = \frac{h}{2 π} 6$