The orbital angular momentum of an electron in a d-orbital is:

Options

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

Explanation

To find the orbital angular momentum of an electron in a d-orbital, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Orbital Angular Momentum**:
- The orbital angular momentum (L) of an electron is related to its motion around the nucleus and is given by the formula:
\[ L = \sqrt{l(l + 1)} \cdot \hbar \]
where \( l \) is the azimuthal quantum number and \( \hbar = \frac{h}{2\pi} \) (with \( h \) being Planck's constant).

2. **Identify the Azimuthal Quantum Number for d-Orbitals**:
- For different types of orbitals, the azimuthal quantum number \( l \) takes specific values:
- \( s \) orbital: \( l = 0 \)
- \( p \) orbital: \( l = 1 \)
- \( d \) orbital: \( l = 2 \)
- \( f \) orbital: \( l = 3 \)
- Since we are dealing with a d-orbital, we have \( l = 2 \).

3. **Substitute the Value of l into the Formula**:
- Now we substitute \( l = 2 \) into the formula for angular momentum:
\[ L = \sqrt{2(2 + 1)} \cdot \hbar = \sqrt{2 \cdot 3} \cdot \hbar = \sqrt{6} \cdot \hbar \]

4. **Express in Terms of h**:
- Since \( \hbar = \frac{h}{2\pi} \), we can rewrite the angular momentum:
\[ L = \sqrt{6} \cdot \frac{h}{2\pi} = \frac{\sqrt{6}h}{2\pi} \]

5. **Final Answer**:
- Therefore, the orbital angular momentum of an electron in a d-orbital is:
\[ L = \frac{\sqrt{6}h}{2\pi} \]

### Conclusion:
The correct answer is \( \frac{\sqrt{6}h}{2\pi} \).