Step 1: State the Paschen series formula.
The Paschen series spectral line wavelength \( \lambda \) is calculated using:\[\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n^2} \right), \quad n = 4, 5, 6, \dots\]Key values are:\begin{itemize} \( R \): Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n \): Principal quantum number (\( n > 3 \)).\end{itemize}Step 2: Compute the wavelength for \( n=4 \).
When \( n = 4 \):\[\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right).\]Simplifying the expression:\[\frac{1}{\lambda_1} = R \left( \frac{1}{9} - \frac{1}{16} \right).\]Calculating the difference:\[\frac{1}{\lambda_1} = R \cdot \frac{16 - 9}{144} = R \cdot \frac{7}{144}.\]Substituting \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):\[\frac{1}{\lambda_1} = 1.097 \times 10^7 \cdot \frac{7}{144}.\]Resulting in:\[\frac{1}{\lambda_1} \approx 5.33 \times 10^5 \, \text{m}^{-1}.\]Converting to wavelength:\[\lambda_1 = \frac{1}{5.33 \times 10^5} \approx 18800 \, \text{Å}.\]Step 3: Determine the short wavelength limit (\( n = \infty \)).
For \( n = \infty \), the formula simplifies to:\[\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{3^2} \right).\]Using \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):\[\frac{1}{\lambda_{\text{min}}} = 1.097 \times 10^7 \cdot \frac{1}{9}.\]Resulting in:\[\frac{1}{\lambda_{\text{min}}} = \frac{1.097 \times 10^7}{9} \approx 1.22 \times 10^6 \, \text{m}^{-1}.\]Converting to wavelength:\[\lambda_{\text{min}} = \frac{1}{1.22 \times 10^6} \approx 8.225 \times 10^{-7} \, \text{m}.\]Expressing in angstroms:\[\lambda_{\text{min}} = 8225 \, \text{Å}.\]The short wavelength limit for the Paschen series is \( 8225 \, \text{Å} \).