The orbital angular momentum corresponding to n = 4 and m = -3 is :

Options

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

Explanation

To find the orbital angular momentum corresponding to the principal quantum number \( n = 4 \) and magnetic quantum number \( m = -3 \), we will follow these steps:

### Step 1: Identify the azimuthal quantum number \( l \)
The azimuthal quantum number \( l \) can take values from \( 0 \) to \( n-1 \). For \( n = 4 \):
\[ l = 0, 1, 2, 3 \]
Thus, the possible values of \( l \) are \( 0, 1, 2, \) and \( 3 \).

### Step 2: Verify the magnetic quantum number \( m \)
The magnetic quantum number \( m \) can take values from \( -l \) to \( +l \). Since we have \( m = -3 \), we need to check if this value is valid for \( l = 3 \):
\[ m = -l, -l + 1, ..., l \]
For \( l = 3 \):
\[ m = -3, -2, -1, 0, 1, 2, 3 \]
Since \( m = -3 \) is within this range, it is valid.

### Step 3: 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 \( l = 3 \):
\[ L = \sqrt{3(3 + 1)} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{3 \cdot 4} \cdot \frac{h}{2\pi} \]
\[ L = \sqrt{12} \cdot \frac{h}{2\pi} \]
\[ L = 2\sqrt{3} \cdot \frac{h}{2\pi} \]
\[ L = \frac{\sqrt{3}h}{\pi} \]

### Final Answer
Thus, the orbital angular momentum corresponding to \( n = 4 \) and \( m = -3 \) is:
\[ L = \frac{\sqrt{3}h}{\pi} \]

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