The wavelength of the first line in balmer series in the hydrogen spectrum' is `lambda`. The wavelength of the second line is `(y lambda)/(27)`. Find `y`.

Answer

To solve the problem, we need to find the value of \( y \) given that the wavelength of the first line in the Balmer series of the hydrogen spectrum is \( \lambda \) and the wavelength of the second line is \( \frac{y \lambda}{27} \).

### Step-by-step solution:

1. **Understand the Balmer series**: The Balmer series corresponds to electronic transitions in a hydrogen atom where the electron falls to the second energy level (n=2) from higher energy levels (n=3, 4, 5, ...). The first line (H-alpha) corresponds to the transition from n=3 to n=2, and the second line (H-beta) corresponds to the transition from n=4 to n=2.

2. **Use the Rydberg formula**: The wavelength \( \lambda \) of the emitted light 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. **Calculate the wavelength for the first line (H-alpha)**:
- For the transition from n=3 to n=2:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Therefore, the wavelength \( \lambda \) is:
\[ \lambda = \frac{36}{5R} \]

4. **Calculate the wavelength for the second line (H-beta)**:
- For the transition from n=4 to n=2:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
- Finding a common denominator (16):
\[ \frac{1}{\lambda_2} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]
- Therefore, the wavelength \( \lambda_2 \) is:
\[ \lambda_2 = \frac{16}{3R} \]

5. **Relate the two wavelengths**: We know that:
\[ \lambda_2 = \frac{y \lambda}{27} \]
Substituting \( \lambda \) from step 3:
\[ \frac{16}{3R} = \frac{y \cdot \frac{36}{5R}}{27} \]

6. **Simplify the equation**:
- Cross-multiplying gives:
\[ 16 \cdot 27 = y \cdot \frac{36}{5} \cdot 3 \]
- Simplifying further:
\[ 432 = y \cdot \frac{108}{5} \]
- Rearranging gives:
\[ y = \frac{432 \cdot 5}{108} = \frac{2160}{108} = 20 \]

7. **Final answer**: Thus, the value of \( y \) is:
\[ y = 20 \]