The ionisation energy of H atom is `13.6 eV` The inoisation energy of `Li^(2+)` law will be

Options

A. 54.4 eV
B. 122.4 eV
C. 13.6 eV
D. 27.2 eV

Explanation

To find the ionization energy of the \( \text{Li}^{2+} \) ion, we can use Bohr's model of the hydrogen-like atom. The formula for the energy of an electron in a hydrogen-like atom is given by:

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

where:
- \( E \) is the energy of the electron,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 1: Identify the parameters for \( \text{Li}^{2+} \)
For \( \text{Li}^{2+} \):
- The atomic number \( Z = 3 \) (since lithium has an atomic number of 3).
- The principal quantum number \( n = 1 \) (because \( \text{Li}^{2+} \) has one electron in the first shell).

### Step 2: Substitute the values into the formula
Now, we can substitute the values into the energy formula:

\[ E = -\frac{13.6 \, \text{eV} \cdot (3)^2}{(1)^2} \]

### Step 3: Calculate the energy
Calculating the above expression:

\[ E = -\frac{13.6 \, \text{eV} \cdot 9}{1} = -122.4 \, \text{eV} \]

### Step 4: Determine the ionization energy
The ionization energy is the energy required to remove the electron from the ion, which is the absolute value of the energy calculated:

\[ \text{Ionization Energy} = +122.4 \, \text{eV} \]

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

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