If the `1^(st)` ionization energy of H atom is 13.6 eV, then the `2^(nd)` ionization 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), we can use the relationship between the ionization energies of hydrogen (H) and helium based on their atomic numbers and energy levels.

### Step-by-Step Solution:

1. **Understand the Formula**:
The energy of an electron in an atom can be described by the formula:
\[ E = -\frac{k \cdot Z^2}{n^2} \]
where:
- \(E\) is the energy of the electron,
- \(k\) is a constant,
- \(Z\) is the atomic number,
- \(n\) is the principal quantum number (energy level).

2. **Identify the Values for Hydrogen**:
For hydrogen (H):
- Atomic number \(Z_H = 1\)
- First ionization energy \(E_H = 13.6 \, \text{eV}\)
- Principal quantum number for the first ionization \(n_H = 1\)

3. **Identify the Values for Helium**:
For helium (He):
- Atomic number \(Z_{He} = 2\)
- We are looking for the second ionization energy, which corresponds to the same principal quantum number \(n_{He} = 1\) (since we are removing an electron from the first energy level).

4. **Set Up the Ratio of Energies**:
Using the formula for both atoms, we can set up the ratio of the energies:
\[ \frac{E_H}{E_{He}} = \frac{Z_H^2 \cdot n_{He}^2}{Z_{He}^2 \cdot n_H^2} \]
Plugging in the values:
\[ \frac{13.6 \, \text{eV}}{E_{He}} = \frac{1^2 \cdot 1^2}{2^2 \cdot 1^2} \]
Simplifying this gives:
\[ \frac{13.6 \, \text{eV}}{E_{He}} = \frac{1}{4} \]

5. **Solve for \(E_{He}\)**:
Rearranging the equation to find \(E_{He}\):
\[ E_{He} = 4 \cdot 13.6 \, \text{eV} \]
Calculating this gives:
\[ E_{He} = 54.4 \, \text{eV} \]

6. **Conclusion**:
The second ionization energy of the helium atom is \(54.4 \, \text{eV}\).

### Final Answer:
The second ionization energy of He atom is **54.4 eV**.