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 determine the orbital angular momentum of a 'p' electron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Orbital Angular Momentum**:
The formula for the orbital angular momentum \( L \) of an electron in an orbital 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. **Identify the Value of \( L \) for a 'p' Electron**:
For a 'p' electron, the azimuthal quantum number \( L \) is equal to 1.

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

4. **Simplify the Expression**:
Calculate \( 1(1 + 1) \):
\[ 1(1 + 1) = 1 \cdot 2 = 2 \]
Therefore, the expression becomes:
\[ L = \sqrt{2} \cdot \frac{h}{2\pi} \]

5. **Express the Result in Terms of \( \hbar \)**:
Recall that \( \hbar = \frac{h}{2\pi} \). Thus, we can rewrite the expression:
\[ L = \sqrt{2} \cdot \hbar \]

6. **Final Answer**:
The orbital angular momentum for a 'p' electron is:
\[ L = \sqrt{2} \hbar \]

### Summary:
The orbital angular momentum of a 'p' electron is \( \sqrt{2} \hbar \).