The angular momentum of an electron is zero. In which orbital may it be present?

Options

A. 2s
B. 2p
C. 3d
D. 4f

Explanation

To determine in which orbital an electron may be present when its angular momentum is zero, we can follow these steps:

### Step 1: Understand Angular Momentum
The angular momentum (L) of an electron in an orbital is given by the formula:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
where:
- \( l \) is the azimuthal quantum number (also known as the angular momentum quantum number).
- \( h \) is Planck's constant.

### Step 2: Identify the Values of l for Different Orbitals
The azimuthal quantum number \( l \) takes different values depending on the type of orbital:
- For an s orbital, \( l = 0 \)
- For a p orbital, \( l = 1 \)
- For a d orbital, \( l = 2 \)
- For an f orbital, \( l = 3 \)

### Step 3: Calculate Angular Momentum for Each Orbital
Using the formula for angular momentum, we can calculate it for each type of orbital:
- For s orbital (\( l = 0 \)):
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} = 0 \]
- For p orbital (\( l = 1 \)):
\[ L = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} = \sqrt{2} \cdot \frac{h}{2\pi} \neq 0 \]
- For d orbital (\( l = 2 \)):
\[ L = \sqrt{2(2 + 1)} \cdot \frac{h}{2\pi} = \sqrt{6} \cdot \frac{h}{2\pi} \neq 0 \]
- For f orbital (\( l = 3 \)):
\[ L = \sqrt{3(3 + 1)} \cdot \frac{h}{2\pi} = \sqrt{12} \cdot \frac{h}{2\pi} \neq 0 \]

### Step 4: Conclusion
The only orbital where the angular momentum is zero is the s orbital, where \( l = 0 \).

Thus, the electron may be present in an **s orbital**.

### Final Answer
The electron may be present in an **s orbital**.

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