Point out the anugular momentum of an electron in,
(a) `4s` orbital (b) `3p` orbital (c ) `4^(th)` orbit )according to Bohr model)

Explanation

To find the angular momentum of an electron in different orbitals, we can use the following principles:

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Quantum Mechanics**:
The angular momentum \( L \) of an electron in an orbital can be calculated using 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. **(a) Angular Momentum of an Electron in the `4s` Orbital**:
- For the `s` orbital, the azimuthal quantum number \( l = 0 \).
- Therefore, the angular momentum \( L \) is:
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} = 0 \]
- **Result**: The angular momentum of an electron in the `4s` orbital is **0**.

3. **(b) Angular Momentum of an Electron in the `3p` Orbital**:
- For the `p` orbital, the azimuthal quantum number \( l = 1 \).
- Therefore, the angular momentum \( L \) is:
\[ L = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} = \sqrt{2} \cdot \frac{h}{2\pi} \]
- **Result**: The angular momentum of an electron in the `3p` orbital is \( \frac{h \sqrt{2}}{2\pi} \).

4. **(c) Angular Momentum of an Electron in the `4th` Orbit (Bohr Model)**:
- According to the Bohr model, the angular momentum \( L \) of an electron in the \( n^{th} \) orbit is given by:
\[ L = n \cdot \frac{h}{2\pi} \]
- For the 4th orbit (\( n = 4 \)):
\[ L = 4 \cdot \frac{h}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi} \]
- **Result**: The angular momentum of an electron in the `4th` orbit is \( \frac{2h}{\pi} \).

### Summary of Results:
- (a) Angular momentum in `4s` orbital: **0**
- (b) Angular momentum in `3p` orbital: \( \frac{h \sqrt{2}}{2\pi} \)
- (c) Angular momentum in `4th` orbit: \( \frac{2h}{\pi} \)