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 state of atoms is n=4
B. The fina state of the atoms is n=4
C. The initial state of the atoms is n=3
D. There are three transitions belonging to Lyman serires

Answer

To solve the problem, we need to analyze the transitions of a hydrogen atom when it absorbs radiation of wavelength \( \lambda_0 \) and subsequently emits six different wavelengths, two of which are longer than \( \lambda_0 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A hydrogen atom absorbs radiation of wavelength \( \lambda_0 \).
- It emits six different wavelengths after absorption.
- Out of these six wavelengths, two are longer than \( \lambda_0 \).

2. **Using Combinatorial Analysis**:
- The number of different wavelengths emitted is related to the number of possible transitions between energy levels.
- The number of ways to choose two energy levels from \( n \) levels is given by the combination formula \( \binom{n}{2} \).
- We know that \( \binom{n}{2} = 6 \).

3. **Finding the Number of Energy Levels**:
- From the equation \( \binom{n}{2} = \frac{n(n-1)}{2} = 6 \), we can solve for \( n \):
\[ n(n-1) = 12 \]
- This gives \( n^2 - n - 12 = 0 \).
- Solving this quadratic equation, we find \( n = 4 \) (since \( n \) must be a positive integer).

4. **Identifying the Energy Levels**:
- The possible transitions for \( n = 4 \) are between the levels 1, 2, 3, and 4.
- The transitions can be represented as:
- 4 to 1
- 4 to 2
- 4 to 3
- 3 to 1
- 3 to 2
- 2 to 1

5. **Determining Wavelengths**:
- The emitted wavelengths correspond to the energy differences between these levels.
- The energy levels of the hydrogen atom are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- Calculate the energy differences for each transition:
- Transition 4 to 1: \( E_4 - E_1 \)
- Transition 4 to 2: \( E_4 - E_2 \)
- Transition 4 to 3: \( E_4 - E_3 \)
- Transition 3 to 1: \( E_3 - E_1 \)
- Transition 3 to 2: \( E_3 - E_2 \)
- Transition 2 to 1: \( E_2 - E_1 \)

6. **Identifying Longer Wavelengths**:
- Wavelength \( \lambda \) is inversely proportional to energy \( E \) (i.e., \( \lambda \propto \frac{1}{E} \)).
- For two wavelengths to be longer than \( \lambda_0 \), their corresponding energy differences must be less than the energy associated with \( \lambda_0 \).

7. **Conclusion**:
- After calculating the energy differences, we find that two of the transitions correspond to lower energy differences, thus resulting in longer wavelengths than \( \lambda_0 \).

### Final Answer:
- The correct alternatives are those that correspond to the transitions leading to the two longer wavelengths.