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(C) The longest wavelength in the ultraviolet region (Lyman series) occurs for the transition from $n_2 = 2$ to $n_1 = 1$.Using the Rydberg formula: $

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$\frac{27}{4} \lambda_{0}$

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(C) The longest wavelength in the ultraviolet region (Lyman series) occurs for the transition from $n_2 = 2$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{\lambda_{0}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R$.Thus,$R = \frac{4}{3 \lambda_{0}}$.The shortest wavelength in the infrared region (Paschen series) occurs for the transition from $n_2 = \infty$ to $n_1 = 3$.Using the Rydberg formula: $\frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = R \left( \frac{1}{9} ight) = \frac{R}{9}$.Substituting $R = \frac{4}{3 \lambda_{0}}$ into the equation:$\frac{1}{\lambda'} = \frac{1}{9} 	imes \frac{4}{3 \lambda_{0}} = \frac{4}{27 \lambda_{0}}$.Therefore,$\lambda' = \frac{27}{4} \lambda_{0}$.

If the longest wavelength in the ultraviolet region of the hydrogen spectrum is $\lambda_{0}$,then the shortest wavelength in its infrared region is:

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