If the first excitation energy of a hydrogen - like atom is 27.3 eV , then ionization energy of this atom will be

Answer

To find the ionization energy of a hydrogen-like atom given that its first excitation energy is 27.3 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Excitation Energy**: The first excitation energy refers to the energy required to move an electron from the ground state (n=1) to the first excited state (n=2). For hydrogen-like atoms, the energy levels can be described by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
where \(Z\) is the atomic number.

2. **Calculating the Energy Levels**:
- The energy of the ground state (n=1) is:
\[ E_1 = -\frac{13.6 \, Z^2}{1^2} = -13.6 \, Z^2 \text{ eV} \]
- The energy of the first excited state (n=2) is:
\[ E_2 = -\frac{13.6 \, Z^2}{2^2} = -\frac{13.6 \, Z^2}{4} \text{ eV} \]

3. **Finding the Excitation Energy**: The excitation energy from n=1 to n=2 is given by:
\[ E_{\text{excitation}} = E_2 - E_1 = \left(-\frac{13.6 \, Z^2}{4}\right) - \left(-13.6 \, Z^2\right) \]
Simplifying this gives:
\[ E_{\text{excitation}} = -\frac{13.6 \, Z^2}{4} + 13.6 \, Z^2 = 13.6 \, Z^2 \left(1 - \frac{1}{4}\right) = 13.6 \, Z^2 \cdot \frac{3}{4} \]
Thus,
\[ E_{\text{excitation}} = \frac{3 \cdot 13.6 \, Z^2}{4} = 10.2 \, Z^2 \text{ eV} \]

4. **Setting Up the Equation**: We know from the problem statement that the first excitation energy is 27.3 eV:
\[ 10.2 \, Z^2 = 27.3 \]

5. **Solving for Z**:
\[ Z^2 = \frac{27.3}{10.2} = 2.676 \implies Z = \sqrt{2.676} \approx 1.64 \]

6. **Calculating Ionization Energy**: The ionization energy \(E_{\text{ionization}}\) for hydrogen-like atoms is given by:
\[ E_{\text{ionization}} = 13.6 \, Z^2 \text{ eV} \]
Substituting \(Z^2\):
\[ E_{\text{ionization}} = 13.6 \cdot 2.676 \approx 36.4 \text{ eV} \]

### Final Answer:
The ionization energy of the hydrogen-like atom is **36.4 eV**.