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 find the potential energy of the electron in the Li²⁺ ion given its angular momentum. Let's break down the solution step by step.

### Step 1: Understand the Angular Momentum Formula
The angular momentum \( L \) of an electron in a hydrogen-like atom is given by the formula:
\[ L = \frac{n h}{2\pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

### Step 2: Set Up the Equation
We are given that the angular momentum of the electron in \( \text{Li}^{2+} \) is \( \frac{14h}{11} \). We can set up the equation:
\[ \frac{n h}{2\pi} = \frac{14h}{11} \]

### Step 3: Cancel \( h \) from Both Sides
Since \( h \) appears on both sides of the equation, we can cancel it out:
\[ \frac{n}{2\pi} = \frac{14}{11} \]

### Step 4: Solve for \( n \)
Now, we can solve for \( n \):
\[ n = \frac{14}{11} \times 2\pi \]
Using \( \pi \approx 3.14 \):
\[ n \approx \frac{14 \times 6.28}{11} \approx \frac{87.92}{11} \approx 7.99 \]
Rounding \( n \) gives us:
\[ n = 8 \]

### Step 5: Calculate the Potential Energy
The potential energy \( U \) of an electron in a hydrogen-like atom is given by the formula:
\[ U = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number. For lithium (\( \text{Li} \)), \( Z = 3 \).

### Step 6: Substitute the Values
Now, substituting \( Z = 3 \) and \( n = 8 \) into the potential energy formula:
\[ U = -\frac{13.6 \times 3^2}{8^2} \]
Calculating \( 3^2 = 9 \) and \( 8^2 = 64 \):
\[ U = -\frac{13.6 \times 9}{64} \]

### Step 7: Calculate the Final Value
Now, calculate:
\[ U = -\frac{122.4}{64} \approx -1.915625 \text{ eV} \]
Rounding gives us:
\[ U \approx -1.92 \text{ eV} \]

### Final Answer
Thus, the potential energy of the system is approximately:
\[ \boxed{-1.92 \text{ eV}} \]

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