The orbital angular momentum of an electron in 3s-orbital is :

Options

A. `+(1)/(2).(h)/(2pi)`
B. zero
C. `(h)/(2pi)`
D. `sqrt2.(h)/(2pi)`

Explanation

To find the orbital angular momentum of an electron in the 3s orbital, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Quantum Numbers**:
- The principal quantum number (n) for the 3s orbital is 3.
- The azimuthal quantum number (l) for an s orbital is always 0.

2. **Understand the Formula for Orbital Angular Momentum**:
- The formula for the orbital angular momentum (L) of an electron is given by:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
where \( h \) is Planck's constant.

3. **Substitute the Value of l**:
- Since we have identified that for the s orbital, \( l = 0 \):
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} \]

4. **Calculate the Orbital Angular Momentum**:
- Simplifying the equation:
\[ L = \sqrt{0} \cdot \frac{h}{2\pi} = 0 \]

5. **Conclusion**:
- Therefore, the orbital angular momentum of an electron in the 3s orbital is:
\[ L = 0 \]

### Final Answer:
The orbital angular momentum of an electron in the 3s orbital is **0**.

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