What are the values of the orbital angular momentum of an electron in the orbitals `1s,3s,3d` and `2p`:-

Options

A. `0,0sqrt(6h),sqrt(2h)`
B. `1,1sqrt(4h),sqrt(2h)`
C. `0,1sqrt(6h),sqrt(3h)`
D. `0,0sqrt(20h),sqrt(6)`

Explanation

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:

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

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

### Step-by-Step Solution:

1. **Identify the azimuthal quantum number \( l \) 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: \( l = 1 \)

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

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

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

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

3. **Summarize the results:**
- **1s orbital:** \( L_{1s} = 0 \)
- **3s orbital:** \( L_{3s} = 0 \)
- **3d orbital:** \( L_{3d} = \sqrt{6} \cdot \frac{h}{2\pi} \)
- **2p orbital:** \( L_{2p} = \sqrt{2} \cdot \frac{h}{2\pi} \)

### Final Answer:
- The values of the orbital angular momentum for the orbitals are:
- 1s: 0
- 3s: 0
- 3d: \( \sqrt{6} \cdot \frac{h}{2\pi} \)
- 2p: \( \sqrt{2} \cdot \frac{h}{2\pi} \)