For a 'p' electron the orbial angular momentum is

Options

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

Answer

To find the orbital angular momentum for a 'p' electron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Orbital Angular Momentum**:
- Orbital angular momentum (L) is quantized and can be expressed in terms of the azimuthal quantum number (l). According to Bohr's atomic model, the orbital angular momentum is given by the formula:
\[ L = \sqrt{l(l + 1)} \frac{h}{2\pi} \]
- Here, \( h \) is Planck's constant and \( \frac{h}{2\pi} \) is often denoted as \( \hbar \) (h cross).

2. **Identify the Azimuthal Quantum Number for 'p' Electrons**:
- The azimuthal quantum number \( l \) corresponds to different types of orbitals:
- For s orbital, \( l = 0 \)
- For p orbital, \( l = 1 \)
- For d orbital, \( l = 2 \)
- For f orbital, \( l = 3 \)
- Since we are dealing with a 'p' electron, we set \( l = 1 \).

3. **Substitute the Value of l into the Formula**:
- Now, substitute \( l = 1 \) into the formula for orbital angular momentum:
\[ L = \sqrt{1(1 + 1)} \frac{h}{2\pi} \]

4. **Calculate the Expression**:
- Simplify the expression:
\[ L = \sqrt{1 \cdot 2} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} \]
- This can also be expressed as:
\[ L = \sqrt{2} \hbar \]

5. **Conclusion**:
- Therefore, the orbital angular momentum for a 'p' electron is:
\[ L = \sqrt{2} \hbar \]

### Final Answer:
For a 'p' electron, the orbital angular momentum is \( \sqrt{2} \hbar \).

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