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(A) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$,wher

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$5 : 9$

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(A) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $R$ is the Rydberg constant.For the Balmer series,$n_1 = 2$.For the maximum wavelength $(\lambda_{\max})$,the transition is from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{5}{36} ight] \Rightarrow \lambda_{\max} = \frac{36}{5R}$.For the minimum wavelength $(\lambda_{\min})$,the transition is from $n_2 = \infty$ to $n_1 = 2$:$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{2^2} - \frac{1}{\infty^2} ight] = R \left[ \frac{1}{4} - 0 ight] = \frac{R}{4} \Rightarrow \lambda_{\min} = \frac{4}{R}$.The ratio of minimum to maximum wavelength is $\frac{\lambda_{\min}}{\lambda_{\max}} = \frac{4/R}{36/5R} = \frac{4}{1} 	imes \frac{5}{36} = \frac{5}{9}$.

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

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