If the first ionisation energy of H-atom is 13.6 eV, then the second ionisation energy of He-atom is :

Options

A. 27.2 eV
B. 40.8 eV
C. 54.4 eV
D. 108.8 eV

Explanation

To find the second ionization energy of the helium atom (He), given that the first ionization energy of the hydrogen atom (H) is 13.6 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Ionization Energy Formula**:
The ionization energy (IE) for hydrogen-like atoms can be expressed using the formula:
\[ \text{IE} = 13.6 \times \frac{Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Identify Parameters for Helium**:
- For helium (He), the atomic number \( Z = 2 \).
- The first ionization energy corresponds to removing one electron from the ground state, so \( n = 1 \).

3. **Calculate First Ionization Energy of Helium**:
Using the formula for the first ionization energy of He:
\[ \text{First IE of He} = 13.6 \times \frac{2^2}{1^2} = 13.6 \times 4 = 54.4 \text{ eV} \]

4. **Consider the Second Ionization Energy**:
After the first ionization, helium becomes a positively charged ion (He\(^+\)) with one electron remaining. The second ionization energy refers to removing this remaining electron.

5. **Use the Formula for He\(^+\)**:
Now, we treat He\(^+\) as a hydrogen-like atom (since it has only one electron). For He\(^+\):
- \( Z = 2 \) (still the atomic number of helium)
- \( n = 1 \) (the electron is still in the ground state)

Therefore, the second ionization energy (removing the second electron from He\(^+\)) is given by:
\[ \text{Second IE of He} = 13.6 \times \frac{2^2}{1^2} = 13.6 \times 4 = 54.4 \text{ eV} \]

6. **Final Calculation**:
Since the second ionization energy is calculated as:
\[ \text{Second IE of He} = 54.4 \text{ eV} \]

### Conclusion:
Thus, the second ionization energy of the helium atom is **54.4 eV**.