The ionization energy of H-atom is `13.6eV`. Calculate the is ionization energy of `Li^(+2)`ion-

Options

A. 122.4 eV
B. 40.8 eV
C. 30.6 eV
D. 13.6 eV

Answer

To calculate the ionization energy of the \( Li^{2+} \) ion, we will use the formula for ionization energy, which is given by:

\[ IE = \frac{13.6 \, \text{eV} \times Z^2}{n^2} \]

where:
- \( IE \) is the ionization energy,
- \( Z \) is the atomic number of the element,
- \( n \) is the principal quantum number of the electron being removed.

### Step-by-Step Solution:

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

2. **Determine the Principal Quantum Number (n):**
- For the \( Li^{2+} \) ion, we are removing an electron from the 1s orbital. Therefore, the principal quantum number \( n = 1 \).

3. **Substitute Values into the Formula:**
- Now, substitute \( Z = 3 \) and \( n = 1 \) into the ionization energy formula:
\[ IE = \frac{13.6 \, \text{eV} \times (3)^2}{(1)^2} \]

4. **Calculate \( Z^2 \) and \( n^2 \):**
- \( Z^2 = 3^2 = 9 \)
- \( n^2 = 1^2 = 1 \)

5. **Plug in the Values:**
\[ IE = \frac{13.6 \, \text{eV} \times 9}{1} = 13.6 \, \text{eV} \times 9 \]

6. **Perform the Multiplication:**
\[ IE = 122.4 \, \text{eV} \]

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

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