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(D) The shortest wavelength $(\lambda_{\min})$ in the entire hydrogen emission spectrum occurs for the transition from $n = \infty$ to $n = 1$ (Lyman

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$900/11$

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(D) The shortest wavelength $(\lambda_{\min})$ in the entire hydrogen emission spectrum occurs for the transition from $n = \infty$ to $n = 1$ (Lyman series limit).Using the Rydberg formula: $\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.Thus, $\lambda_{\min} = \frac{1}{R}$.The longest wavelength $(\lambda_{\max})$ among the five series (Lyman, Balmer, Paschen, Brackett, Pfund) occurs for the transition from $n = 6$ to $n = 5$ (the last line of the Pfund series).Using the Rydberg formula: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) = R \left( \frac{1}{25} - \frac{1}{36} \right) = R \left( \frac{36 - 25}{900} \right) = \frac{11R}{900}$.Thus, $\lambda_{\max} = \frac{900}{11R}$.The ratio of the longest wavelength to the shortest wavelength is $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{900/11R}{1/R} = \frac{900}{11}$.

The ratio of the longest wavelength to the shortest wavelength observed in the five spectral series of the emission spectrum of hydrogen is

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