For an electron, with n=3 has only one radial node.
The orbital angular momentum of the electron will be :

Options

A. `0`
B. `sqrt(6) h/(2pi)`
C. `sqrt(2) h/(2pi)`
D. `3(h/(2pi))`

Explanation

To solve the problem step by step, we will follow the concepts of quantum mechanics related to electron configurations and angular momentum.

### Step 1: Understand the given information
We are given that an electron is in the n=3 energy level and has one radial node. We need to determine the orbital angular momentum of this electron.

### Step 2: Use the formula for radial nodes
The formula for the number of radial nodes (R) in an orbital is given by:
\[ R = n - l - 1 \]
Where:
- \( n \) = principal quantum number
- \( l \) = azimuthal quantum number (orbital angular momentum quantum number)

### Step 3: Substitute the known values
We know:
- \( n = 3 \)
- \( R = 1 \) (as given in the problem)

Substituting these values into the radial node formula:
\[ 1 = 3 - l - 1 \]

### Step 4: Solve for \( l \)
Rearranging the equation:
\[ 1 = 3 - l - 1 \]
\[ 1 + 1 = 3 - l \]
\[ 2 = 3 - l \]
\[ l = 3 - 2 \]
\[ l = 1 \]

### Step 5: Identify the type of orbital
The azimuthal quantum number \( l = 1 \) corresponds to the p-orbital. Therefore, the electron is in a p-orbital.

### Step 6: Calculate the orbital angular momentum
The formula for the orbital angular momentum \( L \) is given by:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
Substituting \( l = 1 \):
\[ L = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{1 \cdot 2} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{2} \cdot \frac{h}{2\pi} \]

### Step 7: Conclusion
Thus, the orbital angular momentum of the electron is:
\[ L = \sqrt{2} \cdot \frac{h}{2\pi} \]

### Step 8: Identify the correct option
From the options provided, we check which one corresponds to the calculated angular momentum. According to the solution, option C is correct.

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