For `d` electron, the orbital angular momentum is

Options

A. `sqrt(6)((h)/(2pi))`
B. `sqrt(2)((h)/(2pi))`
C. `((h)/(2pi))`
D. `2((h)/(2pi))`

Explanation

To find the orbital angular momentum for a `d` electron, we can follow these steps:

### Step 1: Understand the formula for orbital angular momentum
The formula for the orbital angular momentum \( L \) is given by:
\[ L = \sqrt{L(L + 1)} \cdot \frac{h}{2\pi} \]
where \( L \) is the azimuthal quantum number and \( h \) is Planck's constant.

### Step 2: Identify the azimuthal quantum number for `d` orbitals
For `d` orbitals, the azimuthal quantum number \( L \) is equal to 2. This is because:
- \( s \) orbitals correspond to \( L = 0 \)
- \( p \) orbitals correspond to \( L = 1 \)
- \( d \) orbitals correspond to \( L = 2 \)

### Step 3: Substitute the value of \( L \) into the formula
Now substituting \( L = 2 \) into the formula:
\[ L = \sqrt{2(2 + 1)} \cdot \frac{h}{2\pi} \]

### Step 4: Simplify the expression
Calculating the expression inside the square root:
\[ 2(2 + 1) = 2 \cdot 3 = 6 \]
So, we have:
\[ L = \sqrt{6} \cdot \frac{h}{2\pi} \]

### Step 5: Final expression for orbital angular momentum
Thus, the orbital angular momentum for a `d` electron is:
\[ L = \sqrt{6} \cdot \frac{h}{2\pi} \]