The excitation energy of a hydrogen -like ion in its first excited state is `40.8 eV` Find the energy needed to remain the electron from the ion

Explanation

To solve the problem, we need to find the energy required to remove an electron from a hydrogen-like ion given that the excitation energy of the first excited state is 40.8 eV.

### Step-by-step Solution:

1. **Identify the Ion and its Properties**:
- The problem states that the excitation energy in the first excited state is 40.8 eV.
- For hydrogen-like ions, the excitation energy can be expressed using the formula:
\[ E_n = -R_H Z^2 \left( \frac{1}{n^2} \right) \]
- Here, \( R_H \) is the Rydberg constant for hydrogen, approximately 13.6 eV, and \( Z \) is the atomic number of the ion.

2. **Use the Excitation Energy Formula**:
- The excitation energy from the ground state to the first excited state (n=2) is given by:
\[ E_{excitation} = E_2 - E_1 = -R_H Z^2 \left( \frac{1}{2^2} \right) - \left( -R_H Z^2 \left( \frac{1}{1^2} \right) \right) \]
- Simplifying this gives:
\[ E_{excitation} = R_H Z^2 \left( 1 - \frac{1}{4} \right) = R_H Z^2 \left( \frac{3}{4} \right) \]
- Setting this equal to the given excitation energy:
\[ R_H Z^2 \left( \frac{3}{4} \right) = 40.8 \text{ eV} \]

3. **Solve for Z**:
- Plugging in the value of \( R_H \):
\[ 13.6 Z^2 \left( \frac{3}{4} \right) = 40.8 \]
- Rearranging gives:
\[ Z^2 = \frac{40.8 \times 4}{13.6 \times 3} = \frac{163.2}{40.8} = 4 \]
- Thus, \( Z = 2 \).

4. **Identify the Ion**:
- Since \( Z = 2 \), the ion is \( He^+ \) (helium ion).

5. **Calculate the Ground State Energy**:
- The energy of the ion in the ground state (n=1) is given by:
\[ E_1 = -R_H Z^2 \left( \frac{1}{1^2} \right) = -R_H Z^2 \]
- Substituting the values:
\[ E_1 = -13.6 \times 2^2 = -13.6 \times 4 = -54.4 \text{ eV} \]

6. **Energy Required to Remove the Electron**:
- The energy required to remove the electron from the ion (ionization energy) is the absolute value of the ground state energy:
\[ E_{ionization} = 54.4 \text{ eV} \]

### Final Answer:
The energy needed to remove the electron from the ion is **54.4 eV**.