What are the values of the orbital angular momentum of an electron in the orbitals `1s,3s,3d` and `2p`:-
(a). `0,0sqrt(6h),sqrt(2h)`
(b). `1,1sqrt(4h),sqrt(2h)`
(c). `0,1sqrt(6h),sqrt(3h)`
(d). `0,0sqrt(20h),sqrt(6)`

Answer

To find the values of the orbital angular momentum of an electron in the orbitals 1s, 3s, 3d, and 2p, we will use the formula for orbital angular momentum:

\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]

Where:
- \(L\) is the orbital angular momentum,
- \(l\) is the azimuthal quantum number,
- \(h\) is Planck's constant.

### Step 1: Identify the azimuthal quantum numbers for each orbital
- For **1s**: \(l = 0\)
- For **3s**: \(l = 0\)
- For **3d**: \(l = 2\)
- For **2p**: \(l = 1\)

### Step 2: Calculate the orbital angular momentum for each orbital
1. **For 1s orbital**:
\[ L_{1s} = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} = \sqrt{0} = 0 \]

2. **For 3s orbital**:
\[ L_{3s} = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} = \sqrt{0} = 0 \]

3. **For 3d orbital**:
\[ L_{3d} = \sqrt{2(2 + 1)} \cdot \frac{h}{2\pi} = \sqrt{6} \cdot \frac{h}{2\pi} = \sqrt{6h} \]

4. **For 2p orbital**:
\[ L_{2p} = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} = \sqrt{2} \cdot \frac{h}{2\pi} = \sqrt{2h} \]

### Step 3: Compile the results
- **1s**: \(0\)
- **3s**: \(0\)
- **3d**: \(\sqrt{6h}\)
- **2p**: \(\sqrt{2h}\)

### Final Result
The values of the orbital angular momentum for the orbitals 1s, 3s, 3d, and 2p are:
\[ 0, 0, \sqrt{6h}, \sqrt{2h} \]

### Conclusion
The correct option is **(a)**: \(0, 0, \sqrt{6h}, \sqrt{2h}\).