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 given that the separation energy from its third excited state is 2.25 times the separation energy of a hydrogen atom from its first excited state.

### Step-by-Step Solution:

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

2. **For the Hydrogen-like Ion**:
- The ion is in its third excited state, which means \( n = 4 \).
- Therefore, the separation energy \( BE_1 \) for the hydrogen-like ion can be expressed as:
\[ BE_1 = \frac{Z^2}{4^2} \times 13.6 = \frac{Z^2}{16} \times 13.6 \]

3. **For the Hydrogen Atom**:
- The hydrogen atom is in its first excited state, which means \( n = 2 \).
- Thus, the separation energy \( BE_2 \) for the hydrogen atom can be expressed as:
\[ BE_2 = \frac{1^2}{2^2} \times 13.6 = \frac{1}{4} \times 13.6 \]

4. **Setting Up the Relationship**:
According to the problem, the separation energy of the hydrogen-like ion is 2.25 times that of the hydrogen atom:
\[ BE_1 = 2.25 \times BE_2 \]
Substituting the expressions for \( BE_1 \) and \( BE_2 \):
\[ \frac{Z^2}{16} \times 13.6 = 2.25 \times \left( \frac{1}{4} \times 13.6 \right) \]

5. **Simplifying the Equation**:
We can cancel \( 13.6 \) from both sides:
\[ \frac{Z^2}{16} = 2.25 \times \frac{1}{4} \]
Simplifying the right side:
\[ \frac{Z^2}{16} = \frac{2.25}{4} = \frac{9}{16} \]

6. **Solving for \( Z^2 \)**:
Now, multiply both sides by 16:
\[ Z^2 = 9 \]

7. **Finding \( Z \)**:
Taking the square root of both sides:
\[ Z = 3 \]

### Conclusion:
The atomic number of the hydrogen-like ion is \( Z = 3 \).