For a d-electron, the orbital angular momentum is :

Options

A. `sqrt6((h)/(2pi))`
B. `sqrt2((h)/(2pi))`
C. `((h)/(2pi))`
D. `2((h)/(2pi))`

Explanation

To determine the orbital angular momentum for a d-electron, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Orbital Angular Momentum**:
The formula for the orbital angular momentum \( L \) is given by:
\[ L = \sqrt{L(L + 1)} \cdot \frac{h}{2\pi} \]
where \( L \) is the azimuthal quantum number and \( h \) is Planck's constant.

2. **Determine the Azimuthal Quantum Number for d-Orbitals**:
For d-orbitals, the azimuthal quantum number \( L \) is equal to 2. This is because:
- s-orbitals have \( L = 0 \)
- p-orbitals have \( L = 1 \)
- d-orbitals have \( L = 2 \)

3. **Substitute the Value of \( L \) into the Formula**:
Now, substituting \( L = 2 \) into the formula:
\[ L = \sqrt{2(2 + 1)} \cdot \frac{h}{2\pi} \]

4. **Calculate the Value Inside the Square Root**:
Calculate \( 2(2 + 1) \):
\[ 2(2 + 1) = 2 \times 3 = 6 \]

5. **Complete the Calculation**:
Now we can substitute this back into the formula:
\[ L = \sqrt{6} \cdot \frac{h}{2\pi} \]

6. **Final Result**:
Thus, the orbital angular momentum for a d-electron is:
\[ L = \sqrt{6} \cdot \frac{h}{2\pi} \]

### Conclusion:
The orbital angular momentum for a d-electron is \( \sqrt{6} \cdot \frac{h}{2\pi} \).