An electron in hydrogen atom after absorbing-an energy photon jumps from energy state `n_(1)` to `n_(2)`. Then it returns to ground state after emitting six different wavelengths in emission spectrum.'The energy of emitted photons is either equal to, less than or grater than the absorbed photons: Then `n_(1)` and `n_(2)` are:

Options

A. `n_(2) = 4, n_(1) = 3`
B. `n_(2) = 5, n_(1) = 3`
C. `n_(2) = 4, n_(1) = 2`
D. `n_(2) = 4, n_(1) = 1`

Explanation

To solve the problem, we need to analyze the transitions of an electron in a hydrogen atom and determine the values of \( n_1 \) and \( n_2 \) based on the information provided.

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
The energy levels of a hydrogen atom are quantized and can be represented by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Photon Absorption**:
When the electron absorbs a photon, it jumps from an initial energy state \( n_1 \) to a higher energy state \( n_2 \). The energy of the absorbed photon is given by:
\[ E_{\text{photon}} = E_{n_2} - E_{n_1} \]

3. **Photon Emission**:
The electron then returns to the ground state (which is \( n = 1 \)) and emits six different wavelengths. Each transition corresponds to a different wavelength, which means the electron must transition through multiple energy levels.

4. **Determining the Energy Levels**:
To emit six different wavelengths, the electron must transition from \( n_2 \) to \( n_1 \) through several intermediate states. The possible transitions can be represented as:
- From \( n_2 \) to \( n_1 \)
- From \( n_2 \) to \( n_1 + 1 \)
- From \( n_2 \) to \( n_1 + 2 \)
- From \( n_2 \) to \( n_1 + 3 \)
- From \( n_2 \) to \( n_1 + 4 \)
- From \( n_2 \) to \( n_1 + 5 \)

5. **Choosing Values for \( n_1 \) and \( n_2 \)**:
- The maximum number of transitions (and hence wavelengths) emitted occurs when \( n_2 \) is at least 4. This is because from \( n = 4 \), the electron can transition to \( n = 1, 2, 3 \) and also take intermediate steps.
- Therefore, we can set \( n_2 = 4 \) and \( n_1 = 2 \). This allows for transitions:
- \( 4 \to 3 \)
- \( 4 \to 2 \)
- \( 4 \to 1 \)
- \( 3 \to 2 \)
- \( 3 \to 1 \)
- \( 2 \to 1 \)

6. **Conclusion**:
Thus, the values of \( n_1 \) and \( n_2 \) are:
\[ n_1 = 2, \quad n_2 = 4 \]

### Final Answer:
- \( n_1 = 2 \)
- \( n_2 = 4 \)