The energy required to remove the electron from a singly ionized Helium atom is `2.2` times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Options

A. Option1 `109 eV`
B. Option2 `34 eV`
C. Option3 `79 eV`
D. Option4 `20 eV`

Answer

To solve the problem, we need to find the total energy required to completely ionize a helium atom, which has two electrons. The energy required to remove the first electron from a neutral helium atom is denoted as \( E_1 \), and the energy required to remove the second electron from a singly ionized helium atom (which has one electron left) is denoted as \( E_2 \). According to the problem, \( E_2 \) is 2.2 times \( E_1 \).

### Step-by-Step Solution:

1. **Understanding the Ionization Energies**:
- Let \( E_1 \) be the energy required to remove the first electron from a neutral helium atom.
- Let \( E_2 \) be the energy required to remove the second electron from the singly ionized helium atom.
- According to the problem, we have the relationship:
\[ E_2 = 2.2 \times E_1 \]

2. **Using the Formula for Ionization Energy**:
- The ionization energy for hydrogen-like atoms is given by the formula:
\[ E = \frac{13.6 \, Z^2}{n^2} \text{ eV} \]
- For helium, which has \( Z = 2 \) and for the first electron (ground state, \( n = 1 \)):
\[ E_1 = \frac{13.6 \times 2^2}{1^2} = \frac{13.6 \times 4}{1} = 54.4 \text{ eV} \]

3. **Calculating \( E_2 \)**:
- Now, using the relationship \( E_2 = 2.2 \times E_1 \):
\[ E_2 = 2.2 \times 54.4 \text{ eV} \]
- Calculating \( E_2 \):
\[ E_2 = 119.68 \text{ eV} \]

4. **Finding Total Energy Required for Complete Ionization**:
- The total energy required to completely ionize the helium atom is the sum of the energies required to remove both electrons:
\[ E_{\text{total}} = E_1 + E_2 \]
- Substituting the values:
\[ E_{\text{total}} = 54.4 \text{ eV} + 119.68 \text{ eV} = 174.08 \text{ eV} \]

5. **Final Answer**:
- The total energy required to ionize the helium atom completely is approximately:
\[ E_{\text{total}} \approx 174.08 \text{ eV} \]