The ratio of minimum wavelength of Balmer ser | vedclass

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(C) For hydrogen atom,the Rydberg formula is $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer series,$n_1 = 2$.

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$9$:$100$

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(C) For hydrogen atom,the Rydberg formula is $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$. The minimum wavelength corresponds to $n_2 = \infty$.$\frac{1}{\lambda_{	ext{min, Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = \frac{R}{4} \Rightarrow \lambda_{	ext{min, Balmer}} = \frac{4}{R}$.For the Brackett series,$n_1 = 4$. The maximum wavelength corresponds to $n_2 = 5$.$\frac{1}{\lambda_{	ext{max, Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} ight) = R \left( \frac{1}{16} - \frac{1}{25} ight) = R \left( \frac{25 - 16}{400} ight) = \frac{9R}{400}$.$\lambda_{	ext{max, Brackett}} = \frac{400}{9R}$.The ratio is $\frac{\lambda_{	ext{min, Balmer}}}{\lambda_{	ext{max, Brackett}}} = \frac{4/R}{400/9R} = \frac{4}{R} 	imes \frac{9R}{400} = \frac{36}{400} = \frac{9}{100}$.

The ratio of minimum wavelength of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is

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