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 \( n = 4 \) and \( m = -3 \), we can follow these steps:

### Step 1: Identify the Principal Quantum Number (n) and Magnetic Quantum Number (m)
The principal quantum number \( n \) is given as 4, and the magnetic quantum number \( m \) is given as -3.

### Step 2: Determine the Azimuthal Quantum Number (l)
The azimuthal quantum number \( l \) can take values from 0 to \( n-1 \). Therefore, for \( n = 4 \), the possible values of \( l \) are 0, 1, 2, and 3.

The magnetic quantum number \( m \) can take values from \( -l \) to \( +l \). Since \( m = -3 \), this means \( l \) must be at least 3 (because \( m \) can take values from \( -l \) to \( +l \)). Thus, we conclude that \( l = 3 \).

### Step 3: Use the Formula for Orbital Angular Momentum
The formula for orbital angular momentum \( L \) is given by:
\[ L = \sqrt{l(l + 1)} \frac{h}{2\pi} \]
Substituting \( l = 3 \) into the formula:
\[ L = \sqrt{3(3 + 1)} \frac{h}{2\pi} \]

### Step 4: Calculate the Value
Now, calculate \( l(l + 1) \):
\[ l(l + 1) = 3 \times 4 = 12 \]
Thus, we have:
\[ L = \sqrt{12} \frac{h}{2\pi} \]
We can simplify \( \sqrt{12} \):
\[ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \]
So, substituting this back, we get:
\[ L = 2\sqrt{3} \frac{h}{2\pi} \]

### Step 5: Simplify the Expression
The \( 2 \) in the numerator and denominator cancels out:
\[ L = \sqrt{3} \frac{h}{\pi} \]

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

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