The ionization potential for second `H`e electron is

Options

A. `13.6 eV`
B. `27.2 eV`
C. `54.4 eV`
D. `100 eV`

Explanation

To find the ionization potential for the second electron of helium (He), we can treat it as a hydrogen-like atom. The ionization energy (or potential) for a hydrogen-like atom is given by the formula:

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

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

### Step-by-Step Solution:

1. **Identify the Atomic Number (Z)**:
Helium has an atomic number \( Z = 2 \).

2. **Determine the Principal Quantum Number (n)**:
For the second electron in helium, we consider it to be in the ground state, so \( n = 1 \).

3. **Apply the Formula**:
Substitute \( Z \) and \( n \) into the ionization energy formula:
\[ E = -\frac{2^2 \cdot 13.6 \, \text{eV}}{1^2} \]

4. **Calculate**:
\[ E = -\frac{4 \cdot 13.6 \, \text{eV}}{1} = -54.4 \, \text{eV} \]
Since we are interested in the ionization potential (which is a positive value), we take the absolute value:
\[ \text{Ionization Potential} = 54.4 \, \text{eV} \]

5. **Final Result**:
The ionization potential for the second electron of helium is \( 54.4 \, \text{eV} \).