The orbital angular momentum of electron in 4s orbital of H atom is ……….

Options

A. `(1)/(2). (h )/(2pi)`
B. zero
C. `(h)/( 2 pi)`
D. `(2.5) (h)/(2pi)`

Answer

To find the orbital angular momentum of an electron in the 4s orbital of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Quantum Numbers**:
- The principal quantum number (n) for the 4s orbital is 4.
- The azimuthal quantum number (l) for an s orbital is always 0.

2. **Use the Formula for Orbital Angular Momentum**:
- The formula for orbital angular momentum (L) 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 are dealing with an s orbital, we have \(l = 0\).
- Substitute \(l\) into the formula:
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} \]

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

5. **Conclusion**:
- Therefore, the orbital angular momentum of the electron in the 4s orbital of a hydrogen atom is:
\[ L = 0 \]

### Final Answer:
The orbital angular momentum of the electron in the 4s orbital of a hydrogen atom is **0**.

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