If the wavelength of photon emitted due to transition of electron from third orbit to first orbit in a hydrogen atom is `lambda` then the wavelength of photon emitted due to transition of electron from fourth orbit to second orbit will be

Options

A. `(128)/(27) lambda`
B. `(25)/(9) lambda`
C. `(36)/(7) lambda`
D. `(125)/(11) lambda`

Explanation

To solve the problem, we need to find the wavelength of the photon emitted during the transition of an electron from the fourth orbit to the second orbit in a hydrogen atom, given that the wavelength of the photon emitted during the transition from the third orbit to the first orbit is λ.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Wavelength for the transition from the third orbit (n=3) to the first orbit (n=1) is given as λ.

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

3. **Calculate λ1 for the Transition from n=3 to n=1:**
- For the transition from n=3 to n=1:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]
- Therefore, we have:
\[ \frac{1}{\lambda_1} = R \cdot \frac{8}{9} \]
- Rearranging gives:
\[ \lambda_1 = \frac{9}{8R} \]

4. **Calculate λ2 for the Transition from n=4 to n=2:**
- Now, for the transition from n=4 to n=2:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]
- Therefore, we have:
\[ \frac{1}{\lambda_2} = R \cdot \frac{3}{16} \]
- Rearranging gives:
\[ \lambda_2 = \frac{16}{3R} \]

5. **Find the Ratio of λ2 to λ1:**
- Now, we can find the ratio of λ2 to λ1:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{16}{3R}}{\frac{9}{8R}} = \frac{16 \cdot 8}{3 \cdot 9} = \frac{128}{27} \]
- Thus, we can express λ2 in terms of λ:
\[ \lambda_2 = \frac{128}{27} \lambda_1 \]

6. **Substituting λ1:**
- Since λ1 is given as λ:
\[ \lambda_2 = \frac{128}{27} \lambda \]

### Final Answer:
\[ \lambda_2 = \frac{128}{27} \lambda \]