The angular momentum of electron in 'd' orbital is equal to:

Options

A. `sqrt(2)h`
B. `2sqrt(3)h`
C. `0h`
D. `sqrt(6)h`

Explanation

To solve the question regarding the angular momentum of an electron in a 'd' orbital, we need to understand the quantum mechanics behind electron orbitals.

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Quantum Mechanics**:
- In quantum mechanics, the angular momentum of an electron in an orbital is quantized. For any orbital, the angular momentum \( L \) can be expressed using the formula:
\[ L = \sqrt{l(l+1)} \hbar \]
where \( l \) is the azimuthal quantum number and \( \hbar \) is the reduced Planck's constant.

2. **Identifying the Azimuthal Quantum Number for 'd' Orbitals**:
- The 'd' orbitals correspond to the azimuthal quantum number \( l = 2 \). This is because:
- For 's' orbitals, \( l = 0 \)
- For 'p' orbitals, \( l = 1 \)
- For 'd' orbitals, \( l = 2 \)
- For 'f' orbitals, \( l = 3 \)

3. **Calculating the Angular Momentum for 'd' Orbitals**:
- Now, substituting \( l = 2 \) into the angular momentum formula:
\[ L = \sqrt{2(2+1)} \hbar = \sqrt{2 \times 3} \hbar = \sqrt{6} \hbar \]

4. **Conclusion**:
- Therefore, the angular momentum of an electron in a 'd' orbital is:
\[ L = \sqrt{6} \hbar \]

### Final Answer:
The angular momentum of an electron in a 'd' orbital is \( \sqrt{6} \hbar \).

---