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(C) The wavelength of spectral lines in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1

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(C) The wavelength of spectral lines in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Paschen series,$n_1 = 3$ and $n_2 = 4, 5, 6, \dots$.The first wavelength (maximum wavelength) corresponds to the transition from $n_2 = 4$ to $n_1 = 3$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = \frac{7R}{144}$.So,$\lambda_{\max} = \frac{144}{7R} = 18,800 \, \mathring A$.The minimum wavelength (series limit) corresponds to the transition from $n_2 = \infty$ to $n_1 = 3$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = \frac{R}{9}$.So,$\lambda_{\min} = \frac{9}{R}$.Taking the ratio: $\frac{\lambda_{\min}}{\lambda_{\max}} = \frac{9/R}{144/7R} = \frac{9}{R} 	imes \frac{7R}{144} = \frac{63}{144} = \frac{7}{16}$.Therefore,$\lambda_{\min} = \frac{7}{16} 	imes \lambda_{\max} = \frac{7}{16} 	imes 18,800 = 8,225 \, \mathring A$.

The first wavelength of the Paschen series in hydrogen is $18,800 \, \mathring A$. What is the minimum wavelength of the Paschen series in $\mathring A$?

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