Find the first excitation potential of `He^(+)` ion (a)Find the ionization potential of `Li^(++)` ion

Explanation

To solve the given problem, we need to find the first excitation potential of the `He^(+)` ion and the ionization potential of the `Li^(++)` ion.

### Step 1: Find the first excitation potential of `He^(+)` ion

1. **Understand the formula for potential energy of hydrogen-like atoms**:
The potential energy \( V \) of a hydrogen-like atom in the \( n \)th state is given by:
\[ V = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number.

2. **Identify the transition for excitation**:
For the first excitation of `He^(+)`, the transition is from \( n = 1 \) to \( n = 2 \).

3. **Calculate the potential energy for \( n = 1 \)**:
\[ V_1 = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \text{ eV} \]

4. **Calculate the potential energy for \( n = 2 \)**:
\[ V_2 = -\frac{13.6 \times 2^2}{2^2} = -\frac{13.6 \times 4}{4} = -13.6 \text{ eV} \]

5. **Find the first excitation potential**:
The first excitation potential is the difference in potential energy between these two states:
\[ \text{Excitation Potential} = V_2 - V_1 = -13.6 - (-54.4) = 40.8 \text{ eV} \]

### Step 2: Find the ionization potential of `Li^(++)` ion

1. **Use the same formula for ionization potential**:
The ionization potential for hydrogen-like atoms is also given by:
\[ \text{Ionization Potential} = -V_1 = \frac{13.6 \, Z^2}{1^2} \]

2. **Identify \( Z \) for `Li^(++)`**:
For `Li^(++)`, \( Z = 3 \).

3. **Calculate the ionization potential**:
\[ \text{Ionization Potential} = 13.6 \times 3^2 = 13.6 \times 9 = 122.4 \text{ eV} \]

### Final Answers:
- The first excitation potential of `He^(+)` is **40.8 eV**.
- The ionization potential of `Li^(++)` is **122.4 eV**.

---