The ionisation energy of H atom is `13.6 ` eV. What will be the ionisation energy of `He^(o+)` and `Li^(2+)` ions ?

Options

A. 54.4 ev. 122.4 ev
B. 5.44 ev. 12.24 ev
C. 6.82 ev, 13.6 ev
D. 13.4 ev, 122.4 ev

Answer

To find the ionization energy of the `He^(o+)` and `Li^(2+)` ions based on the given ionization energy of the hydrogen atom, we can use the formula that relates ionization energy to the atomic number (Z) of the element.

### Step-by-Step Solution:

1. **Understand the Ionization Energy Formula**:
The ionization energy (IE) of an atom is directly proportional to the square of its atomic number (Z). The formula can be expressed as:
\[ \text{IE} = \text{IE}_H \times Z^2 \]
where \(\text{IE}_H\) is the ionization energy of hydrogen, which is given as 13.6 eV.

2. **Calculate Ionization Energy for Helium (He)**:
- The atomic number (Z) of Helium is 2.
- Using the formula:
\[ \text{IE}_{He} = 13.6 \, \text{eV} \times (2^2) \]
\[ \text{IE}_{He} = 13.6 \, \text{eV} \times 4 = 54.4 \, \text{eV} \]

3. **Calculate Ionization Energy for Lithium (Li)**:
- The atomic number (Z) of Lithium is 3.
- Using the formula:
\[ \text{IE}_{Li} = 13.6 \, \text{eV} \times (3^2) \]
\[ \text{IE}_{Li} = 13.6 \, \text{eV} \times 9 = 122.4 \, \text{eV} \]

4. **Final Results**:
- The ionization energy of `He^(o+)` is **54.4 eV**.
- The ionization energy of `Li^(2+)` is **122.4 eV**.

### Summary:
- Ionization energy of `He^(o+)` = **54.4 eV**
- Ionization energy of `Li^(2+)` = **122.4 eV**