One electron species having ionization enegry of `54.4 eV` is

Options

A. `He^+`
B. H
C. `Be^(2+)`
D. `Be^(3+)`

Answer

To solve the problem, we need to determine which one-electron species has an ionization energy of 54.4 eV. We can use the formula for the ionization energy of a hydrogen-like atom, which is given by:

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

Where:
- \( Z \) is the atomic number of the element,
- \( n \) is the principal quantum number (for the ground state, \( n = 1 \)).

### Step-by-Step Solution:

1. **Identify the Given Ionization Energy**:
We are given that the ionization energy is \( 54.4 \, \text{eV} \).

2. **Set Up the Equation**:
Since we are looking for a one-electron species, we can assume \( n = 1 \). Thus, the equation simplifies to:
\[ \text{Ionization Energy} = 13.6 \, Z^2 \]
Setting this equal to the given ionization energy:
\[ 13.6 \, Z^2 = 54.4 \]

3. **Solve for \( Z^2 \)**:
To find \( Z^2 \), divide both sides by \( 13.6 \):
\[ Z^2 = \frac{54.4}{13.6} \]
Calculating the right side:
\[ Z^2 = 4 \]

4. **Find \( Z \)**:
Taking the square root of both sides gives:
\[ Z = \sqrt{4} = 2 \]

5. **Determine the Element**:
The atomic number \( Z = 2 \) corresponds to Helium (He). Therefore, the one-electron species with an ionization energy of \( 54.4 \, \text{eV} \) is the Helium ion \( \text{He}^+ \).

### Conclusion:
The one-electron species having an ionization energy of \( 54.4 \, \text{eV} \) is **Helium (He)**.