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 the `Li^(2+)` ion using the given ionization energy of hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ionization Energy of Hydrogen**:
The ionization energy of hydrogen (H) is given as `13.6 eV`.

2. **Identify the Formula for Ionization Energy**:
The formula for calculating the ionization energy (IE) of a hydrogen-like ion is:
\[ \text{IE} = -13.6 \times Z^2 \text{ eV} \]
where \( Z \) is the atomic number of the ion.

3. **Determine the Atomic Number of Lithium**:
The atomic number of lithium (Li) is `3`.

4. **Substitute the Atomic Number into the Formula**:
For the `Li^(2+)` ion, we substitute \( Z = 3 \) into the formula:
\[ \text{IE} = -13.6 \times (3^2) \]

5. **Calculate \( Z^2 \)**:
\[ 3^2 = 9 \]

6. **Calculate the Ionization Energy**:
Now substitute \( Z^2 \) back into the equation:
\[ \text{IE} = -13.6 \times 9 \]
\[ \text{IE} = -122.4 \text{ eV} \]

7. **Final Result**:
The ionization energy of `Li^(2+)` ions is `122.4 eV`.