For lyman series `lambda_max` -`lambda_min` = `340 Angstrom` . Find same for paschen series

Options

A. 11,802 Angstroms
B. 13,802 Angstroms
C. 12,502 Angstroms
D. 10,000 Angstroms

Answer

To solve the problem of finding the difference between the maximum and minimum wavelengths for the Paschen series, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Lyman series corresponds to transitions where the electron falls to the n=1 energy level.
- The Paschen series corresponds to transitions where the electron falls to the n=3 energy level.

2. **Identifying Maximum and Minimum Wavelengths**:
- For the Lyman series:
- **Maximum Wavelength (λ_max)**: This occurs for the transition from n=2 to n=1.
- **Minimum Wavelength (λ_min)**: This occurs for the transition from n=∞ to n=1.
- For the Paschen series:
- **Maximum Wavelength (λ_max)**: This occurs for the transition from n=4 to n=3.
- **Minimum Wavelength (λ_min)**: This occurs for the transition from n=∞ to n=3.

3. **Using the Rydberg Formula**:
The Rydberg formula for the wavelength of emitted light is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant.

4. **Calculating λ_min for Paschen Series**:
- For λ_min in the Paschen series (n1=3, n2=∞):
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \]
Therefore,
\[ \lambda_{\text{min}} = \frac{9}{R} \]

5. **Calculating λ_max for Paschen Series**:
- For λ_max in the Paschen series (n1=3, n2=4):
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
To simplify:
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \]
Therefore,
\[ \lambda_{\text{max}} = \frac{144}{7R} \]

6. **Finding the Difference (λ_max - λ_min)**:
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{144}{7R} - \frac{9}{R} \]
Taking a common denominator (7R):
\[ = \frac{144 - 63}{7R} = \frac{81}{7R} \]

7. **Substituting the Value of R**:
From the Lyman series, we know that:
\[ \frac{1}{R} = 340 \text{ Å} \]
Thus,
\[ R = \frac{1}{340} \text{ Å}^{-1} \]
Substituting this into the equation:
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{81}{7 \cdot \frac{1}{340}} = \frac{81 \cdot 340}{7} \]
Calculating this gives:
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{27540}{7} \approx 3934.29 \text{ Å} \]

### Final Answer:
The difference between the maximum and minimum wavelengths for the Paschen series is approximately **3934.29 Å**.