The ratio of te shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:

Options

A. Lyman and Paschen
B. Balmer an Brackett
C. Brackett and Pfund
D. Paschen and Pfund

Answer

To solve the problem, we need to identify the two spectral series of the hydrogen spectrum based on the given ratio of the shortest wavelengths.

### Step-by-Step Solution:

1. **Understanding the Wavelength Formula**:
The wavelength (\( \lambda \)) of spectral lines in hydrogen can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Identifying the Shortest Wavelength**:
The shortest wavelength corresponds to the transition where \( n_2 \) approaches infinity. Thus, for the shortest wavelength, the formula simplifies to:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} \right) \]
This means that the shortest wavelength is inversely proportional to \( n_1^2 \).

3. **Setting Up the Ratio**:
Let’s denote the two spectral series by their lower energy levels:
- For the first series, let \( n_1 = m \).
- For the second series, let \( n_1 = n \).

The ratio of the shortest wavelengths is given as:
\[ \frac{\lambda_m}{\lambda_n} = \frac{n^2}{m^2} = 9 \]

4. **Solving the Ratio**:
From the ratio \( \frac{n^2}{m^2} = 9 \), we can take the square root of both sides:
\[ \frac{n}{m} = 3 \quad \text{or} \quad \frac{m}{n} = \frac{1}{3} \]
This implies that \( n = 3m \).

5. **Identifying the Series**:
The possible values for \( m \) and \( n \) must correspond to the known spectral series:
- Lyman series (UV region): \( n_1 = 1 \)
- Balmer series (visible region): \( n_1 = 2 \)
- Paschen series (IR region): \( n_1 = 3 \)
- Brackett series (IR region): \( n_1 = 4 \)

If we set \( m = 1 \) (Lyman series), then \( n = 3 \) (Paschen series).

6. **Conclusion**:
The two spectral series corresponding to the given ratio of the shortest wavelengths are:
- Lyman series (for \( n_1 = 1 \))
- Paschen series (for \( n_1 = 3 \))

### Final Answer:
The spectral series are the **Lyman series** and the **Paschen series**.