The orbital angular momentum of an electron of an electron in `2s` orbitals is

Options

A. 4
B. 1
C. 0
D. `(h)/(2pi)`

Explanation

To find the orbital angular momentum of an electron in the 2s orbital, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Quantum Numbers for the 2s Orbital**:
For the 2s orbital:
- The principal quantum number \( n = 2 \).
- The azimuthal quantum number \( l = 0 \) (since 's' orbitals correspond to \( l = 0 \)).

3. **Substitute the Value of \( l \) into the Formula**:
Now, substituting \( l = 0 \) into the formula:
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} \]

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

5. **Conclusion**:
Therefore, the orbital angular momentum of an electron in the 2s orbital is:
\[ L = 0 \]

### Final Answer:
The orbital angular momentum of an electron in the 2s orbital is **0**.

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