The ionisation energy of `H` is `13.6 eV`. Calculate the ionization energy of `Li^(2+)` ions.

Options

A. `54.4 eV`
B. `122.4eV`
C. `244.8eV`
D. `108.8eV`

Explanation

To calculate the ionization energy of \( \text{Li}^{2+} \) ions, we can use the formula for the ionization energy of hydrogen-like ions:

\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]

Where:
- \( E_n \) is the ionization energy,
- \( Z \) is the atomic number of the ion,
- \( n \) is the principal quantum number (for the ground state, \( n = 1 \)).

### Step-by-Step Solution:

1. **Identify the Atomic Number (Z) of Li**:
- Lithium (Li) has an atomic number \( Z = 3 \).

2. **Determine the Value of n**:
- For the ground state of \( \text{Li}^{2+} \), \( n = 1 \).

3. **Substitute Values into the Formula**:
- Plugging in the values into the formula:
\[ E_1 = -\frac{3^2 \cdot 13.6 \, \text{eV}}{1^2} \]

4. **Calculate \( Z^2 \)**:
- \( Z^2 = 3^2 = 9 \).

5. **Calculate the Ionization Energy**:
- Now substitute \( Z^2 \) into the equation:
\[ E_1 = -\frac{9 \cdot 13.6 \, \text{eV}}{1} \]
\[ E_1 = -122.4 \, \text{eV} \]

6. **Interpret the Result**:
- The ionization energy is typically expressed as a positive value, so we take the absolute value:
\[ \text{Ionization Energy of } \text{Li}^{2+} = 122.4 \, \text{eV} \]

### Final Answer:
The ionization energy of \( \text{Li}^{2+} \) ions is \( 122.4 \, \text{eV} \).