What is the orbital angular momentum of an electron in 'f' orbital ?

Options

A. `(1.5 h)/(pi )`
B. `(sqrt6 h)/(pi)`
C. `(sqrt3 h)/(pi )`
D. `(sqrt3 h)/(2pi)`

Explanation

To find the orbital angular momentum of an electron in the 'f' orbital, we can follow these steps:

### Step 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 (orbital angular momentum quantum number).
- \( h \) is Planck's constant.

### Step 2: Identify the azimuthal quantum number for the 'f' orbital
For the 'f' orbital, the azimuthal quantum number \( L \) is 3. This is because:
- s orbital corresponds to \( L = 0 \)
- p orbital corresponds to \( L = 1 \)
- d orbital corresponds to \( L = 2 \)
- f orbital corresponds to \( L = 3 \)

### Step 3: Substitute the value of L into the formula
Now, we substitute \( L = 3 \) into the formula:
\[ L = \sqrt{3(3 + 1)} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{3 \cdot 4} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{12} \cdot \frac{h}{2\pi} \]

### Step 4: Simplify the expression
Now we simplify \( \sqrt{12} \):
\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]
Thus, we can rewrite the angular momentum as:
\[ L = 2\sqrt{3} \cdot \frac{h}{2\pi} \]

### Step 5: Final expression
Now, we can cancel the 2 in the numerator and denominator:
\[ L = \sqrt{3} \cdot \frac{h}{\pi} \]

### Conclusion
The orbital angular momentum of an electron in the 'f' orbital is:
\[ L = \sqrt{3} \cdot \frac{h}{\pi} \]