A group of hydrogen atom are prepered in `n = 4` states list the wavelength that are emitted as the atoms make transition and return to `n = 2` states

Explanation

To solve the problem of finding the wavelengths emitted by hydrogen atoms transitioning from the n = 4 state to the n = 2 state, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Possible Transitions**:
The hydrogen atom in the n = 4 state can transition to the n = 2 state through two possible paths:
- Directly from n = 4 to n = 2 (let's call this wavelength λ_c).
- First from n = 4 to n = 3 (λ_a), and then from n = 3 to n = 2 (λ_b).

2. **Use the Rydberg Formula**:
The Rydberg formula for the wavelength of emitted light during transitions in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Calculate λ_a (Transition from n = 4 to n = 3)**:
- Here, \( n_1 = 3 \) and \( n_2 = 4 \).
\[ \frac{1}{\lambda_a} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
- Substitute \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \frac{1}{\lambda_a} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right) \]
- Calculate \( \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \):
\[ \frac{1}{\lambda_a} = 1.097 \times 10^7 \times \frac{7}{144} \]
- Calculate \( \lambda_a \):
\[ \lambda_a = \frac{144}{1.097 \times 10^7 \times 7} \approx 1875 \, \text{nm} \]

4. **Calculate λ_b (Transition from n = 3 to n = 2)**:
- Here, \( n_1 = 2 \) and \( n_2 = 3 \).
\[ \frac{1}{\lambda_b} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- Substitute \( R = 1.097 \times 10^7 \):
\[ \frac{1}{\lambda_b} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Calculate \( \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \):
\[ \frac{1}{\lambda_b} = 1.097 \times 10^7 \times \frac{5}{36} \]
- Calculate \( \lambda_b \):
\[ \lambda_b = \frac{36}{1.097 \times 10^7 \times 5} \approx 656 \, \text{nm} \]

5. **Calculate λ_c (Transition from n = 4 to n = 2)**:
- Here, \( n_1 = 2 \) and \( n_2 = 4 \).
\[ \frac{1}{\lambda_c} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
- Substitute \( R = 1.097 \times 10^7 \):
\[ \frac{1}{\lambda_c} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \]
- Calculate \( \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \):
\[ \frac{1}{\lambda_c} = 1.097 \times 10^7 \times \frac{3}{16} \]
- Calculate \( \lambda_c \):
\[ \lambda_c = \frac{16}{1.097 \times 10^7 \times 3} \approx 487 \, \text{nm} \]

### Summary of Wavelengths:
- \( \lambda_a \approx 1875 \, \text{nm} \) (n = 4 to n = 3)
- \( \lambda_b \approx 656 \, \text{nm} \) (n = 3 to n = 2)
- \( \lambda_c \approx 487 \, \text{nm} \) (n = 4 to n = 2)