In centre-symmetrical system, the orbital angular momentum, a measure of the momentum of a particle travelling around the nucleus, is quantised. Its magnitude is

Options

A. `sqrt(l(l+1))h/(2pi)`
B. `sqrt(l(l-1))h/(2pi)`
C. `sqrt(S(S+1))h/(2pi)`
D. `sqrt(S(S-1))h/(2pi)`

Answer

To solve the question regarding the quantization of orbital angular momentum in a central symmetrical system, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Orbital Angular Momentum**:
- In quantum mechanics, the orbital angular momentum \( L \) of an electron in an atom is quantized. It is related to the orbital quantum number \( l \).

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

3. **Rewriting the Formula**:
- The expression can also be written using \( \hbar \) (h-bar), where \( \hbar = \frac{h}{2\pi} \):
\[ L = \sqrt{l(l + 1)} \cdot \hbar \]

4. **Identifying the Correct Option**:
- The question provides multiple options for the magnitude of orbital angular momentum. We need to identify which option corresponds to the derived formula:
- Option A: \( \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \) (This matches our formula)
- Option B: \( \sqrt{l - 1} \cdot \frac{h}{2\pi} \) (Incorrect)
- Option C: \( \sqrt{l + 1} \cdot \frac{h}{2\pi} \) (Incorrect)
- Option D: \( \sqrt{s - 1} \cdot \frac{h}{2\pi} \) (Incorrect)

5. **Conclusion**:
- The correct answer is Option A: \( \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \).