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)`

Options

A. `0,0,sqrt6ħ,sqrt(2)ħ`
B. `1,1,sqrt(6)ħ,sqrt(2)ħ`
C. `0,1,sqrt(6)ħ,sqrt(3)ħ`
D. `0,0,sqrt(20)ħ,sqrt(6)ħ`

Answer

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

\[ L = \sqrt{l(l + 1)} \cdot \hbar \]

where:
- \( L \) is the orbital angular momentum,
- \( l \) is the azimuthal quantum number,
- \( \hbar \) is the reduced Planck's constant (often expressed as \( h/2\pi \)).

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

### Step 2: Calculate the orbital angular momentum for each orbital

1. **For 1s orbital**:
\[ L_{1s} = \sqrt{0(0 + 1)} \cdot \hbar = \sqrt{0} \cdot \hbar = 0 \]

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

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

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

### Step 3: Compile the results
The values of the orbital angular momentum for the orbitals are:
- **1s**: \( 0 \)
- **3s**: \( 0 \)
- **3d**: \( \sqrt{6} \hbar \)
- **2p**: \( \sqrt{2} \hbar \)

### Final Answer
The values are \( 0, 0, \sqrt{6} \hbar, \sqrt{2} \hbar \).

Thus, the correct option is **(a)**: \( 0, 0, \sqrt{6}h, \sqrt{2}h \).

---