When a particle as an electron exist in a box, three quantum numbers are necessary to describe the spatial distributioni fo electrons in atoms. The describe an electron in an atom completely, a fourth quantum number, `m_(s)` called the spin quantum number must be specified. This is because every electron has magnetic moment associated with it which is quantized in one of two possible orientations. Electrons having the same spin strongly repel each other and tend to occupy different region of space. This result of a fundamental law of nature is known as Pauli exclusion principle.
The orbital angular momentum of an electron in 2s orbital is

Options

A. `+1/2 h/(2pi)`
B. zero at the nucleus
C. `h/(2pi)`
D. `sqrt(2)h/(2pi)`

Answer

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

### Step-by-Step Solution:

1. **Identify the Quantum Numbers**:
- For the 2s orbital, we need to identify the principal quantum number (n) and the azimuthal quantum number (l).
- The principal quantum number (n) for the 2s orbital is 2.
- The azimuthal quantum number (l) for s orbitals is always 0.

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

3. **Substitute the Value of l**:
- Since we have determined that \(l = 0\) for the 2s orbital, we can substitute this value into the formula:
\[ L = \sqrt{0(0 + 1)} \frac{h}{2\pi} \]

4. **Calculate the Orbital Angular Momentum**:
- Simplifying the expression:
\[ L = \sqrt{0} \frac{h}{2\pi} = 0 \]
- Therefore, the orbital angular momentum of an electron in the 2s orbital is 0.

5. **Conclusion**:
- The final answer is that the orbital angular momentum of an electron in the 2s orbital is **0**.