The angle made by angular momentum vector of an electron with Z-axis is.

Options

A. `cos theta = l//m`
B. `cos theta = sqrt((l)/(m))`
C. `cos theta = sqrt((l + 1)l)/(m)`
D. `cos theta = (m)/(sqrt((l + 1)l))`

Explanation

To find the angle made by the angular momentum vector of an electron with the Z-axis, we can follow these steps:

### Step 1: Understand Angular Momentum in Quantum Mechanics
In quantum mechanics, the angular momentum of an electron is quantized and can be described using quantum numbers. The total angular momentum \( L \) is given by the formula:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
where \( l \) is the azimuthal quantum number and \( h \) is Planck's constant.

### Step 2: Angular Momentum along the Z-axis
The component of angular momentum along the Z-axis, denoted as \( L_z \), is given by:
\[ L_z = m \cdot \frac{h}{2\pi} \]
where \( m \) is the magnetic quantum number.

### Step 3: Relate \( L_z \) to the Total Angular Momentum
The relationship between the total angular momentum \( L \), its component along the Z-axis \( L_z \), and the angle \( \theta \) is given by:
\[ L_z = L \cdot \cos(\theta) \]

### Step 4: Substitute the Expressions
Substituting the expressions for \( L \) and \( L_z \):
\[ m \cdot \frac{h}{2\pi} = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \cdot \cos(\theta) \]

### Step 5: Simplify the Equation
We can cancel \( \frac{h}{2\pi} \) from both sides:
\[ m = \sqrt{l(l + 1)} \cdot \cos(\theta) \]

### Step 6: Solve for \( \cos(\theta) \)
Rearranging the equation gives:
\[ \cos(\theta) = \frac{m}{\sqrt{l(l + 1)}} \]

### Step 7: Find the Angle \( \theta \)
To find the angle \( \theta \), we can take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{m}{\sqrt{l(l + 1)}}\right) \]

### Conclusion
Thus, the angle made by the angular momentum vector of an electron with the Z-axis is:
\[ \theta = \cos^{-1}\left(\frac{m}{\sqrt{l(l + 1)}}\right) \]