How many degenerate orbitals are presentin a subshell if electron associated with that subshell possesses orbital angular momentum `=2 sqrt (3) h` ?

Options

A. No degenerate orbitals
B. Seven degenerates orbitals
C. Three degenerate orbitals
D. No of degenerate orbitals are same as in subshell which possesses `mu_(L) = 3.46 ` B.M.

Answer

To solve the question of how many degenerate orbitals are present in a subshell when the electron associated with that subshell possesses orbital angular momentum equal to \(2 \sqrt{3} h\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Orbital Angular Momentum**:
The orbital angular momentum \(L\) of an electron in a subshell is given by the formula:
\[ L = \sqrt{l(l + 1)} h \]
where \(l\) is the azimuthal quantum number, and \(h\) is Planck's constant.

2. **Setting Up the Equation**:
Given that \(L = 2 \sqrt{3} h\), we can set up the equation:
\[ \sqrt{l(l + 1)} h = 2 \sqrt{3} h \]

3. **Removing Planck's Constant**:
We can divide both sides of the equation by \(h\) (assuming \(h \neq 0\)):
\[ \sqrt{l(l + 1)} = 2 \sqrt{3} \]

4. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ l(l + 1) = (2 \sqrt{3})^2 \]
\[ l(l + 1) = 4 \times 3 = 12 \]

5. **Finding the Value of \(l\)**:
We need to solve the quadratic equation:
\[ l^2 + l - 12 = 0 \]
We can factor this equation:
\[ (l - 3)(l + 4) = 0 \]
This gives us two possible solutions for \(l\):
\[ l = 3 \quad \text{or} \quad l = -4 \]
Since \(l\) cannot be negative, we have:
\[ l = 3 \]

6. **Identifying the Subshell**:
The value \(l = 3\) corresponds to the \(f\) subshell.

7. **Counting the Degenerate Orbitals**:
The number of degenerate orbitals in a subshell is given by the formula \(2l + 1\). For \(l = 3\):
\[ \text{Number of degenerate orbitals} = 2(3) + 1 = 7 \]

### Conclusion:
Thus, the number of degenerate orbitals present in the subshell is **7**.