PASSAGE - I : The electron with the highest energy in an atom has the quantum numbers `n=4,l=1,m=0,s=1//2`
The orbital angular momentum of the electron is

Options

A. zero
B. `1.4ħ`
C. `4.5ħ`
D. `0.87ħ`

Explanation

To find the orbital angular momentum of the electron with the given quantum numbers \( n=4, l=1, m=0, s=\frac{1}{2} \), we can use the formula for orbital angular momentum:

### Step-by-Step Solution:

1. **Identify the Quantum Number \( l \)**:
- The azimuthal quantum number \( l \) is given as \( l = 1 \).

2. **Use the Orbital Angular Momentum Formula**:
- The formula for the orbital angular momentum \( L \) is:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
- Here, \( h \) is the Planck's constant.

3. **Substitute the Value of \( l \)**:
- Substitute \( l = 1 \) into the formula:
\[ L = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} \]

4. **Calculate \( l(l + 1) \)**:
- Calculate \( l(l + 1) \):
\[ l(l + 1) = 1 \cdot (1 + 1) = 1 \cdot 2 = 2 \]

5. **Take the Square Root**:
- Now, take the square root:
\[ L = \sqrt{2} \cdot \frac{h}{2\pi} \]

6. **Express in Terms of \( \hbar \)**:
- We can express \( \frac{h}{2\pi} \) as \( \hbar \):
\[ L = \sqrt{2} \cdot \hbar \]

7. **Final Answer**:
- Therefore, the orbital angular momentum of the electron is:
\[ L = \sqrt{2} \hbar \]