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

Options

A. Option1 11,802 Angstroms
B. Option2 13,802 Angstroms
C. Option3 12,502 Angstroms
D. Option4 10,000 Angstroms

Answer

To solve the problem, we need to find the difference between the maximum and minimum wavelengths for the Paschen series, given that the difference for the Lyman series is 340 Angstroms.

### Step-by-Step Solution:

1. **Identify the Series and Constants:**
- For the Lyman series, the electron transitions from n1 = 1 to n2 = 2 (for maximum wavelength) and n2 = ∞ (for minimum wavelength).
- For the Paschen series, the electron transitions from n1 = 3 to n2 = 4 (for maximum wavelength) and n2 = ∞ (for minimum wavelength).

2. **Calculate Maximum Wavelength for Lyman Series:**
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Therefore,
\[ \lambda_{\text{max}} = \frac{4}{3R} \]

3. **Calculate Minimum Wavelength for Lyman Series:**
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
Therefore,
\[ \lambda_{\text{min}} = \frac{1}{R} \]

4. **Find the Difference for Lyman Series:**
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{4}{3R} - \frac{1}{R} = \frac{4 - 3}{3R} = \frac{1}{3R} \]
According to the problem, this difference is given as 340 Angstroms:
\[ \frac{1}{3R} = 340 \text{ Angstroms} \]
Thus,
\[ R = \frac{1}{3 \times 340} \text{ Angstroms} \]

5. **Calculate Maximum Wavelength for Paschen Series:**
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Therefore,
\[ \lambda_{\text{max}} = \frac{144}{7R} \]

6. **Calculate Minimum Wavelength for Paschen Series:**
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{9} \right) \]
Therefore,
\[ \lambda_{\text{min}} = \frac{9}{R} \]

7. **Find the Difference for Paschen Series:**
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{144}{7R} - \frac{9}{R} = \frac{144 - 63}{7R} = \frac{81}{7R} \]

8. **Substituting R:**
Using \( R = \frac{1}{3 \times 340} \):
\[ \lambda_{\text{max}} - \lambda_{\text{min}} = \frac{81}{7 \times \left( \frac{1}{3 \times 340} \right)} = \frac{81 \times 3 \times 340}{7} = \frac{81 \times 1020}{7} \]
Calculating this gives:
\[ \lambda_{\text{max}} - \lambda_{\text{min}} \approx 11802 \text{ Angstroms} \]

### Final Answer:
The difference between the maximum and minimum wavelengths for the Paschen series is approximately **11802 Angstroms**.