Hydrogen atom absorbs radiations of wavelength `lambda_0` and consequently emit radiations of 6 different wavelengths, of which two wavelengths are longer than `lambda_0`. Chosses the correct alternative(s).

Options

A. The final excited staate of the atoms is n=4
B. The initial state of the atoms is n =2
C. The initial state of the atoms is n =3
D. There ae three transitions belonging to Lyman series

Explanation

To solve the problem, we need to analyze the transitions of an electron in a hydrogen atom when it absorbs radiation of wavelength \( \lambda_0 \) and subsequently emits radiation of 6 different wavelengths, with 2 of those wavelengths being longer than \( \lambda_0 \).

### Step-by-Step Solution:

1. **Understanding the Absorption and Emission Process**:
- When a hydrogen atom absorbs a photon of wavelength \( \lambda_0 \), the electron transitions from a lower energy level to a higher energy level. This higher energy level will be denoted as \( n \).

2. **Determining the Number of Emission Wavelengths**:
- The problem states that the atom emits 6 different wavelengths. The number of possible transitions (or wavelengths) from an energy level \( n \) is given by the formula \( \frac{n(n-1)}{2} \).
- Setting this equal to 6, we have:
\[ \frac{n(n-1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]
- Solving the quadratic equation \( n^2 - n - 12 = 0 \), we find that \( n = 4 \) (the positive solution).

3. **Identifying the Energy Levels**:
- The possible energy levels for the transitions are \( n = 1, 2, 3, 4 \). The electron can transition from \( n = 4 \) to lower energy levels \( n = 3, 2, 1 \).

4. **Determining Wavelengths Longer than \( \lambda_0 \)**:
- The energy difference between two levels is inversely proportional to the wavelength of the emitted radiation. Therefore, transitions with smaller energy differences will correspond to longer wavelengths.
- The transitions from \( n = 4 \) to \( n = 3 \) and from \( n = 3 \) to \( n = 2 \) will have smaller energy differences compared to transitions from \( n = 4 \) to \( n = 1 \) or \( n = 2 \) to \( n = 1 \).

5. **Identifying Valid Transitions**:
- The transitions that yield wavelengths longer than \( \lambda_0 \) are:
- \( n = 4 \) to \( n = 3 \) (longer wavelength)
- \( n = 3 \) to \( n = 2 \) (longer wavelength)
- The other transitions (from \( n = 4 \) to \( n = 1 \), \( n = 3 \) to \( n = 1 \), and \( n = 2 \) to \( n = 1 \)) will produce shorter wavelengths.

6. **Conclusion**:
- The final excited state of the atom is \( n = 4 \).
- The initial excited state could be \( n = 2 \) since it can also produce longer wavelengths.
- There are three transitions in the Lyman series (from \( n = 2 \) to \( n = 1 \), \( n = 3 \) to \( n = 1 \), and \( n = 4 \) to \( n = 1 \)).

### Final Answer:
The correct alternatives are:
- The final excited state of the atom is \( n = 4 \).
- The initial excited state of the atom may be \( n = 2 \).
- There are three transitions in the Lyman series.