The orbital angular momentum of a p-electron is given as :

Options

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

Answer

To find 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 orbital angular momentum (L) of an electron in an orbital is given by the formula:
\[ L = \sqrt{l(l + 1)} \frac{h}{2\pi} \]
where \( l \) is the azimuthal quantum number and \( h \) is Planck's constant.

2. **Identify the Azimuthal Quantum Number for p-electron**:
For different subshells, the values of \( l \) are:
- For s subshell: \( l = 0 \)
- For p subshell: \( l = 1 \)
- For d subshell: \( l = 2 \)
- For f subshell: \( l = 3 \)

Since we are dealing with a p-electron, we have:
\[ l = 1 \]

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

4. **Calculate the Expression Inside the Square Root**:
Simplifying the expression inside the square root:
\[ L = \sqrt{1 \cdot 2} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} \]

5. **Final Expression for Orbital Angular Momentum**:
We can express this as:
\[ L = \frac{h}{\sqrt{2} \cdot 2\pi} \]
or simply:
\[ L = \frac{h}{\sqrt{2} \cdot 2\pi} \]

6. **Conclusion**:
Thus, the orbital angular momentum of a p-electron is:
\[ L = \frac{h}{\sqrt{2} \cdot 2\pi} \]
The correct option is \( \frac{h}{\sqrt{2} \cdot 2\pi} \).