In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is `340Å` . The corresponding difference of the Paschan series in `Å` is ...........

Answer

To solve the problem, we need to find the difference between the largest and shortest wavelengths of the Paschen series of the hydrogen atom, given that the difference for the Lyman series is 340 Å.

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Lyman series corresponds to electronic transitions from higher energy levels (n ≥ 2) to the first energy level (n = 1).
- The Paschen series corresponds to transitions from higher energy levels (n ≥ 4) to the third energy level (n = 3).

2. **Finding the Wavelengths**:
- The wavelength (λ) 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 (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Calculating the Shortest and Longest Wavelengths for Lyman Series**:
- For the Lyman series:
- Shortest wavelength (n2 = ∞ to n1 = 1):
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \]
\(\lambda_{min} = \frac{1}{R}\)
- Longest wavelength (n2 = 2 to n1 = 1):
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \]
\(\lambda_{max} = \frac{4}{3R}\)

4. **Difference in Wavelengths for Lyman Series**:
- The difference between the longest and shortest wavelengths for the Lyman series is:
\[ \Delta \lambda_{Lyman} = \lambda_{max} - \lambda_{min} = \frac{4}{3R} - \frac{1}{R} = \frac{4 - 3}{3R} = \frac{1}{3R} \]
- Given that this difference is 340 Å:
\[ \frac{1}{3R} = 340 \, \text{Å} \implies R = \frac{1}{3 \times 340} \, \text{Å}^{-1} \]

5. **Calculating the Difference for Paschen Series**:
- For the Paschen series:
- Shortest wavelength (n2 = ∞ to n1 = 3):
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{9} \right) \implies \lambda_{min} = \frac{9}{R} \]
- Longest wavelength (n2 = 4 to n1 = 3):
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \]
\(\lambda_{max} = \frac{144}{7R}\)

6. **Difference in Wavelengths for Paschen Series**:
- The difference between the longest and shortest wavelengths for the Paschen series is:
\[ \Delta \lambda_{Paschen} = \lambda_{max} - \lambda_{min} = \frac{144}{7R} - \frac{9}{R} = \frac{144 - 63}{7R} = \frac{81}{7R} \]

7. **Substituting R**:
- From earlier, we have \( R = \frac{1}{3 \times 340} \):
\[ \Delta \lambda_{Paschen} = \frac{81}{7 \times \frac{1}{3 \times 340}} = \frac{81 \times 3 \times 340}{7} = \frac{81 \times 1020}{7} = \frac{82620}{7} \approx 11803 \, \text{Å} \]

### Final Answer:
The corresponding difference of the Paschen series is approximately \( 11803 \, \text{Å} \).