If `He^(+)` ion is in its first excited state then its ionization energy is

Options

A. `13.6eV`
B. `48.8eV`
C. if `f(x)={{:(1|pix|+1,xle5),(b|x-pi|+3,xgt5):}`
D. `-13.6eV`

Answer

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

### Step 1: Understand Ionization Energy
Ionization energy is the energy required to remove an electron from an atom or ion in a given state. For a hydrogen-like atom, the formula for ionization energy can be expressed as:
\[ IE = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where:
- \( IE \) is the ionization energy,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number of the electron's state.

### Step 2: Identify the Parameters
For the `He^(+)` ion:
- The atomic number \( Z = 2 \) (since helium has 2 protons).
- The first excited state corresponds to \( n = 2 \).

### Step 3: Substitute Values into the Formula
Now, we substitute the values of \( Z \) and \( n \) into the ionization energy formula:
\[ IE = \frac{13.6 \, \text{eV} \cdot (2)^2}{(2)^2} \]

### Step 4: Simplify the Equation
Calculating the values:
\[ IE = \frac{13.6 \, \text{eV} \cdot 4}{4} = 13.6 \, \text{eV} \]

### Step 5: Conclusion
Thus, the ionization energy of the `He^(+)` ion in its first excited state is:
\[ \text{Ionization Energy} = 13.6 \, \text{eV} \]