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(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \r

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$9$:$5$

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(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots$.Maximum wavelength $(\lambda_{max})$ occurs for the transition $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda_{max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight) \implies \lambda_{max} = \frac{36}{5R}$.Minimum wavelength $(\lambda_{min})$ occurs for the transition $n_2 = \infty$ to $n_1 = 2$:$\frac{1}{\lambda_{min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = R \left( \frac{1}{4} - 0 ight) = \frac{R}{4} \implies \lambda_{min} = \frac{4}{R}$.The ratio of maximum to minimum wavelength is:$\frac{\lambda_{max}}{\lambda_{min}} = \frac{36/5R}{4/R} = \frac{36}{5R} 	imes \frac{R}{4} = \frac{9}{5}$.

The ratio of maximum to minimum wavelength in the Balmer series of the hydrogen atom is

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