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(A) The wavelength of spectral lines for a hydrogen atom is given by the Rydberg formula:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^

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$\frac{9}{5}$

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(A) The wavelength of spectral lines for a hydrogen atom is given by the Rydberg formula:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \ldots$.For the minimum wavelength (series limit),$n_2 = \infty$:$\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}$$\lambda_{\min} = \frac{4}{R}$For the maximum wavelength,$n_2 = 3$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9 - 4}{36} ight) = \frac{5R}{36}$$\lambda_{\max} = \frac{36}{5R}$Now,the ratio of maximum to minimum wavelength is:$\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 a hydrogen atom is

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