Separation energy of a hydrogen like ion from its third excited state is 2.25 times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion.

Answer

To solve the problem, we need to find the atomic number \( Z \) of a hydrogen-like ion based on the separation energy from its third excited state and the separation energy of a hydrogen atom from its first excited state.

### Step-by-Step Solution:

1. **Understanding Separation Energy**:
The separation energy \( E_n \) for a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Identify the States**:
- For the hydrogen-like ion, the third excited state corresponds to \( n = 4 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), the second excited state is \( n = 3 \), and the third excited state is \( n = 4 \)).
- For the hydrogen atom, the first excited state corresponds to \( n = 2 \).

3. **Calculate the Separation Energy for Hydrogen Atom**:
For the hydrogen atom (\( Z = 1 \)):
\[ E_{H} = -\frac{1^2 \cdot 13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Calculate the Separation Energy for Hydrogen-like Ion**:
For the hydrogen-like ion in the third excited state (\( n = 4 \)):
\[ E_{ion} = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{4^2} = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{16} \]

5. **Set Up the Equation**:
According to the problem, the separation energy of the hydrogen-like ion from its third excited state is 2.25 times the separation energy of hydrogen from its first excited state:
\[ -\frac{Z^2 \cdot 13.6 \, \text{eV}}{16} = 2.25 \times (-3.4 \, \text{eV}) \]

6. **Simplify the Right Side**:
\[ 2.25 \times (-3.4 \, \text{eV}) = -7.65 \, \text{eV} \]

7. **Set the Equations Equal**:
\[ -\frac{Z^2 \cdot 13.6}{16} = -7.65 \]

8. **Eliminate the Negative Signs**:
\[ \frac{Z^2 \cdot 13.6}{16} = 7.65 \]

9. **Multiply Both Sides by 16**:
\[ Z^2 \cdot 13.6 = 7.65 \times 16 \]

10. **Calculate \( 7.65 \times 16 \)**:
\[ 7.65 \times 16 = 122.4 \]

11. **Solve for \( Z^2 \)**:
\[ Z^2 = \frac{122.4}{13.6} = 9 \]

12. **Find \( Z \)**:
\[ Z = \sqrt{9} = 3 \]

### Final Answer:
The atomic number \( Z \) of the hydrogen-like ion is **3**.