The energy of an atom or ion in the first excited state is -13.6 eV. It may be

Options

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

Explanation

To solve the problem, we need to identify which atom or ion corresponds to the given energy of -13.6 eV in its first excited state. We will use the formula for the energy levels of hydrogen-like atoms.

### Step-by-Step Solution:

1. **Understand the Energy Formula**:
The energy of an electron in the nth energy level of a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{Z^2}{n^2} \times 13.6 \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Identify the Principal Quantum Number**:
Since we are looking for the first excited state, we set \( n = 2 \).

3. **Set Up the Equation**:
We know the energy in the first excited state is given as:
\[ E_2 = -13.6 \text{ eV} \]
Therefore, we can write:
\[ -\frac{Z^2}{2^2} \times 13.6 = -13.6 \]

4. **Simplify the Equation**:
Cancelling the negative signs and multiplying both sides by \( 4 \) (since \( 2^2 = 4 \)):
\[ Z^2 \times 13.6 = 54.4 \]
Now, divide both sides by \( 13.6 \):
\[ Z^2 = \frac{54.4}{13.6} = 4 \]

5. **Solve for Z**:
Taking the square root of both sides:
\[ Z = 2 \]

6. **Identify the Atom/Ion**:
The atomic number \( Z = 2 \) corresponds to Helium. Since we are looking for an ion, we consider \( \text{He}^+ \) (Helium ion).

7. **Conclusion**:
Therefore, the atom or ion in the first excited state with energy -13.6 eV is \( \text{He}^+ \).

### Final Answer:
The correct answer is **Helium ion (He⁺)**.

---