A hydrogen like species (atomic number Z) is present in a higher excited state of quantum number n. This excited atom can make a transitionn to the first excited state by successive emission of two photons of energies 10.20 eV and 17.0 eV respectively. Altetnatively, the atom from the same excited state can make a transition to the second excited state by successive of two photons of energy 4.25 eV and 5.95 eVv respectively. Determine the value of Z.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the transitions of a hydrogen-like atom with atomic number \( Z \) from a higher excited state to lower energy states by emitting photons.

### Step-by-Step Solution:

1. **Identify the Energy Transitions**:
- The atom transitions from a higher excited state \( n \) to the first excited state \( n = 2 \) by emitting two photons of energies \( E_1 = 10.20 \, \text{eV} \) and \( E_2 = 17.0 \, \text{eV} \).
- The total energy emitted during this transition is:
\[ E_n - E_2 = E_1 + E_2 = 10.20 \, \text{eV} + 17.0 \, \text{eV} = 27.20 \, \text{eV} \]

2. **Transition to the Second Excited State**:
- The atom can also transition from the same excited state \( n \) to the second excited state \( n = 3 \) by emitting two photons of energies \( E_3 = 4.25 \, \text{eV} \) and \( E_4 = 5.95 \, \text{eV} \).
- The total energy emitted during this transition is:
\[ E_n - E_3 = E_3 + E_4 = 4.25 \, \text{eV} + 5.95 \, \text{eV} = 10.20 \, \text{eV} \]

3. **Set Up the Energy Level Equation**:
- For a hydrogen-like atom, the energy levels are given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV} \]
- Therefore, we can express the energy differences as:
\[ E_n - E_2 = -\frac{13.6 Z^2}{n^2} + \frac{13.6 Z^2}{2^2} \]
\[ E_n - E_3 = -\frac{13.6 Z^2}{n^2} + \frac{13.6 Z^2}{3^2} \]

4. **Equate the Energy Differences**:
- From the first transition:
\[ -\frac{13.6 Z^2}{n^2} + \frac{13.6 Z^2}{4} = 27.20 \, \text{eV} \quad \text{(1)} \]
- From the second transition:
\[ -\frac{13.6 Z^2}{n^2} + \frac{13.6 Z^2}{9} = 10.20 \, \text{eV} \quad \text{(2)} \]

5. **Solve the Equations**:
- Rearranging equation (1):
\[ \frac{13.6 Z^2}{4} - \frac{13.6 Z^2}{n^2} = 27.20 \]
\[ \frac{13.6 Z^2}{n^2} = \frac{13.6 Z^2}{4} - 27.20 \]
- Rearranging equation (2):
\[ \frac{13.6 Z^2}{9} - \frac{13.6 Z^2}{n^2} = 10.20 \]
\[ \frac{13.6 Z^2}{n^2} = \frac{13.6 Z^2}{9} - 10.20 \]

6. **Equate the Two Expressions for \( \frac{13.6 Z^2}{n^2} \)**:
- Set the right-hand sides of the two equations equal to each other:
\[ \frac{13.6 Z^2}{4} - 27.20 = \frac{13.6 Z^2}{9} - 10.20 \]

7. **Simplify and Solve for \( Z^2 \)**:
- Rearranging gives:
\[ \frac{13.6 Z^2}{4} - \frac{13.6 Z^2}{9} = 27.20 - 10.20 \]
\[ \frac{13.6 Z^2 (9 - 4)}{36} = 17.00 \]
\[ \frac{13.6 Z^2 \cdot 5}{36} = 17.00 \]
\[ 13.6 Z^2 \cdot 5 = 612 \]
\[ Z^2 = \frac{612}{68} = 9 \]
\[ Z = \sqrt{9} = 3 \]

### Final Answer:
The value of \( Z \) is **3**.