A hydrogen - like neutral species in some excited state A, on absorbing a photon of energy 3.066 eV get excited to a new state B. When the electron from state B returns back, photons of a maximum ten different wavelengths can be observed in which some photons are of energy smaller than 3.066 eV, some are of equal energy and only four photons are having energy greater than 3.066 eV. The ionization energy of this atom is

Options

A. 14.6 eV
B. 3.066 eV
C. 6.132 eV
D. 9.2 eV

Answer

To solve the problem, we need to determine the ionization energy of a hydrogen-like neutral species based on the information provided regarding its excited states and photon emissions.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a hydrogen-like atom in an excited state A.
- It absorbs a photon of energy \(3.066 \, \text{eV}\) and gets excited to a new state B.
- When the electron returns from state B, it emits photons of 10 different wavelengths, with some energies less than, equal to, and greater than \(3.066 \, \text{eV}\).

2. **Determining the Number of Excited States**:
- The number of different wavelengths corresponds to the number of transitions possible between excited states.
- The formula for the number of transitions (wavelengths) from \(n\) excited states is given by:
\[ \text{Number of transitions} = \frac{n(n-1)}{2} \]
- Given that there are 10 different wavelengths, we set up the equation:
\[ \frac{n(n-1)}{2} = 10 \]
- Multiplying both sides by 2:
\[ n(n-1) = 20 \]
- Rearranging gives us:
\[ n^2 - n - 20 = 0 \]

3. **Solving the Quadratic Equation**:
- We can factor the quadratic:
\[ (n - 5)(n + 4) = 0 \]
- This gives us two solutions: \(n = 5\) or \(n = -4\). Since \(n\) must be positive, we have:
\[ n = 5 \]

4. **Identifying the Energy Levels**:
- The lowest excited state (ground state) is \(n = 1\).
- The excited state A must be \(n = 2\) (since it is the first excited state).
- The new state B is \(n = 5\).

5. **Calculating the Energy Difference**:
- The energy of the states is given by:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \]
- The energy difference when transitioning from \(n = 2\) to \(n = 5\) is:
\[ E_2 - E_5 = \left(-\frac{13.6 \, Z^2}{2^2}\right) - \left(-\frac{13.6 \, Z^2}{5^2}\right) \]
- This simplifies to:
\[ E_2 - E_5 = -\frac{13.6 \, Z^2}{4} + \frac{13.6 \, Z^2}{25} \]
- Finding a common denominator (100):
\[ E_2 - E_5 = \frac{-13.6 \times 25 + 13.6 \times 4}{100} Z^2 = \frac{13.6 \times (-25 + 4)}{100} Z^2 = \frac{13.6 \times -21}{100} Z^2 \]
- Setting this equal to the absorbed energy \(3.066 \, \text{eV}\):
\[ \frac{13.6 \times -21}{100} Z^2 = 3.066 \]
- Solving for \(Z^2\):
\[ Z^2 = \frac{3.066 \times 100}{13.6 \times -21} = 14.6 \]

6. **Calculating Ionization Energy**:
- The ionization energy corresponds to the transition from \(n = 1\) to \(n = \infty\):
\[ E_{\text{ionization}} = E_1 - E_{\infty} = -\frac{13.6 \, Z^2}{1^2} - 0 = -13.6 \times Z^2 \]
- Substituting \(Z^2 = 14.6\):
\[ E_{\text{ionization}} = -13.6 \times 14.6 = 14.6 \, \text{eV} \]

### Conclusion:
The ionization energy of the atom is \( \boxed{14.6 \, \text{eV}} \).