The wavelength of the first balmer line caused by a transition from the `n = 3` level to the n = 2 level in hydrogen is `lamda_(1)`. The wavelength of the line caused by an electronic transition from `n =5` to `n =3` is

Options

A. `(375)/(128) lamda_(1)`
B. `(125)/(64) lamda_(1)`
C. `(64)/(125) lamda_(1)`
D. `(128)/(378) lamda_(1)`

Explanation

To solve the problem, we need to find the wavelength of the electronic transition from \( n = 5 \) to \( n = 3 \) in hydrogen, given that the wavelength of the first Balmer line (transition from \( n = 3 \) to \( n = 2 \)) is \( \lambda_1 \).

### Step-by-Step Solution:

1. **Understanding the Balmer Series**:
The Balmer series corresponds to electronic transitions in a hydrogen atom where the final energy level \( n_1 \) is 2. The transitions are from higher energy levels \( n_2 \) to \( n_1 \).

2. **Using the Rydberg Formula**:
The wavelength of the emitted light during these transitions can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Calculating \( \lambda_1 \)**:
For the first Balmer line (transition from \( n = 3 \) to \( n = 2 \)):
- \( n_1 = 2 \)
- \( n_2 = 3 \)
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_1} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

4. **Calculating \( \lambda_2 \)**:
Now, for the transition from \( n = 5 \) to \( n = 3 \):
- \( n_1 = 3 \)
- \( n_2 = 5 \)
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{9} - \frac{1}{25} \right) \]
Finding a common denominator (225):
\[ \frac{1}{\lambda_2} = R \left( \frac{25}{225} - \frac{9}{225} \right) = R \left( \frac{16}{225} \right) \]

5. **Finding the Ratio of Wavelengths**:
Now, we can find the ratio of \( \lambda_2 \) to \( \lambda_1 \):
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{1}{R \left( \frac{16}{225} \right)}}{\frac{1}{R \left( \frac{5}{36} \right)}} = \frac{5}{16} \cdot \frac{225}{36} \]
Simplifying this gives:
\[ \frac{\lambda_2}{\lambda_1} = \frac{5 \times 225}{16 \times 36} = \frac{1125}{576} \]

6. **Final Calculation of \( \lambda_2 \)**:
Therefore, we can express \( \lambda_2 \) in terms of \( \lambda_1 \):
\[ \lambda_2 = \lambda_1 \cdot \frac{1125}{576} \]

### Conclusion:
The wavelength of the line caused by the electronic transition from \( n = 5 \) to \( n = 3 \) is given by:
\[ \lambda_2 = \frac{1125}{576} \lambda_1 \]