The energy of an atom (or ion) in the ground state is `- 54.4 eV ` .It may be?

Options

A. Hydrogen
B. deuterium
C. `He^(+)`
D. `Li^(++)`

Answer

To solve the problem, we need to determine the atomic number (Z) of the atom or ion based on the given energy of the ground state, which is -54.4 eV. We will use the formula for the energy of an electron in the ground state of a hydrogen-like atom.

### Step-by-Step Solution:

1. **Understand the Formula**: The energy of an electron in the ground state of a hydrogen-like atom is given by the formula:
\[ E = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where:
- \( E \) is the energy of the atom,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)).

2. **Substitute the Known Values**: Since the energy of the atom in the ground state is given as -54.4 eV, we can substitute this into the formula:
\[ -54.4 = -\frac{13.6 \, Z^2}{1^2} \]

3. **Simplify the Equation**: The equation simplifies to:
\[ -54.4 = -13.6 \, Z^2 \]
We can remove the negative signs from both sides:
\[ 54.4 = 13.6 \, Z^2 \]

4. **Solve for \( Z^2 \)**: To isolate \( Z^2 \), divide both sides by 13.6:
\[ Z^2 = \frac{54.4}{13.6} \]

5. **Calculate \( Z^2 \)**: Performing the division:
\[ Z^2 = 4 \]

6. **Find \( Z \)**: Taking the square root of both sides gives us:
\[ Z = 2 \]

7. **Conclusion**: The atomic number \( Z \) is 2, which corresponds to the element Helium (He).

### Final Answer:
The atom or ion with an energy of -54.4 eV in the ground state may be Helium (He), with an atomic number \( Z = 2 \).

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