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 of finding the difference between the largest and shortest wavelengths of the Paschen series in the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Lyman series corresponds to transitions where the final energy level (n_final) is 1.
- The Paschen series corresponds to transitions where the final energy level (n_final) is 3.

2. **Wavelength Formula**:
- The wavelength (λ) of the emitted light can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]
- Here, \( R \) is the Rydberg constant.

3. **Finding λ_min and λ_max for Lyman Series**:
- For the Lyman series (n_final = 1):
- The shortest wavelength (λ_min) occurs when n_initial is at its minimum, which is 2:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
\[ \lambda_{min} = \frac{4}{3R} \]
- The longest wavelength (λ_max) occurs when n_initial approaches infinity:
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{1^2} - 0 \right) = R \]
\[ \lambda_{max} = \frac{1}{R} \]

4. **Calculating the Difference for Lyman Series**:
- The difference between the largest and shortest wavelengths for the Lyman series is given as 340 Å:
\[ \lambda_{max} - \lambda_{min} = \frac{1}{R} - \frac{4}{3R} \]
\[ = \frac{3}{3R} - \frac{4}{3R} = \frac{-1}{3R} \]
- Therefore, we can set up the equation:
\[ \frac{1}{3R} = 340 \Rightarrow R = \frac{1}{3 \times 340} \]

5. **Finding λ_min and λ_max for Paschen Series**:
- For the Paschen series (n_final = 3):
- The shortest wavelength (λ_min) occurs when n_initial is 4:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \]
\[ \lambda_{min} = \frac{144}{7R} \]
- The longest wavelength (λ_max) occurs when n_initial approaches infinity:
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \]
\[ \lambda_{max} = \frac{9}{R} \]

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

7. **Finding the Corresponding Difference**:
- We know that the difference for the Lyman series is 340 Å, so we can set up the ratio:
\[ \frac{\lambda_{max} - \lambda_{min} \text{ (Paschen)}}{\lambda_{max} - \lambda_{min} \text{ (Lyman)}} = \frac{-81/7R}{-1/3R} \]
\[ = \frac{81 \times 3}{7} = \frac{243}{7} \]
- Therefore, the difference for the Paschen series is:
\[ \lambda_{max} - \lambda_{min} \text{ (Paschen)} = 340 \times \frac{243}{7} \approx 1176.43 \text{ Å} \]

### Final Answer:
The corresponding difference of the Paschen series is approximately **1176.43 Å**.