The angular momentum of electron in `Li^(2+)` is found to be `14(h/11)` . Calculated the potential energy (in eV) of system .

Options

A. `13.6xx(3/8)^2`
B. `-13.6xx(3/8)^2`
C. `-2xx13.6xx(3/8)^2`
D. `-2xx13.6xx(8/3)^2`

Answer

To solve the problem, we need to calculate the potential energy of the electron in the lithium ion \( \text{Li}^{2+} \) given its angular momentum.

### Step-by-Step Solution:

1. **Identify the given angular momentum**:
The angular momentum \( L \) of the electron is given as:
\[ L = \frac{14h}{11} \]

2. **Use the formula for angular momentum**:
The angular momentum of an electron in a hydrogen-like atom is given by:
\[ L = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number.

3. **Set the two expressions for angular momentum equal**:
Equating the two expressions for angular momentum:
\[ \frac{14h}{11} = \frac{nh}{2\pi} \]

4. **Cancel \( h \) from both sides**:
This simplifies to:
\[ \frac{14}{11} = \frac{n}{2\pi} \]

5. **Cross-multiply to solve for \( n \)**:
Cross-multiplying gives:
\[ 14 \cdot 2\pi = 11n \]
Thus,
\[ n = \frac{14 \cdot 2\pi}{11} \]

6. **Calculate the value of \( n \)**:
Approximating \( \pi \) as 3.14:
\[ n \approx \frac{14 \cdot 6.28}{11} \approx \frac{87.92}{11} \approx 7.99 \]
Rounding this, we get:
\[ n \approx 8 \]

7. **Calculate the potential energy (PE)**:
The formula for potential energy in a hydrogen-like atom is:
\[ PE = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For \( \text{Li}^{2+} \), the atomic number \( Z = 3 \) and \( n = 8 \):
\[ PE = -\frac{3^2 \cdot 13.6}{8^2} \]
\[ PE = -\frac{9 \cdot 13.6}{64} \]
\[ PE = -\frac{122.4}{64} \approx -1.915625 \, \text{eV} \]

8. **Final Result**:
The potential energy of the system is approximately:
\[ PE \approx -1.92 \, \text{eV} \]