A `He^(+)` ion is in its first excited state. Its ionization energy is

Options

A. 48.36 eV
B. 54.40 eV
C. 13.60 eV
D. 6.04 eV

Answer

To find the ionization energy of a \( \text{He}^+ \) ion in its first excited state, we can follow these steps:

### Step 1: Understand Ionization Energy
Ionization energy is defined as the minimum energy required to remove an electron from an atom or ion. For a bound electron, this energy is equal to the negative of the total binding energy of that electron.

### Step 2: Identify the State of the Electron
In this problem, the \( \text{He}^+ \) ion is in its first excited state. The first excited state corresponds to the principal quantum number \( n = 2 \).

### Step 3: Use the Formula for Energy Levels
For a hydrogen-like atom, the energy of an electron in a given state is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 4: Substitute Values
For \( \text{He}^+ \):
- The atomic number \( Z = 2 \).
- The electron is in the first excited state, so \( n = 2 \).

Substituting these values into the formula:
\[ E_2 = -\frac{13.6 \, \text{eV} \cdot (2)^2}{(2)^2} = -\frac{13.6 \, \text{eV} \cdot 4}{4} = -13.6 \, \text{eV} \]

### Step 5: Calculate Ionization Energy
The ionization energy is the magnitude of the binding energy:
\[ \text{Ionization Energy} = -E_2 = 13.6 \, \text{eV} \]

### Conclusion
Thus, the ionization energy of the \( \text{He}^+ \) ion in its first excited state is \( 13.6 \, \text{eV} \).

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