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(B) 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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$\frac{7}{20}$

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(B) 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 longest wavelength line,the transition occurs between adjacent energy levels,i.e.,$n_2 = n_1 + 1$.For the Balmer series,$n_1 = 2$,so $n_2 = 3$. The longest wavelength $\lambda_{BL}$ is:$\frac{1}{\lambda_{BL}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight) \implies \lambda_{BL} = \frac{36}{5R} \quad ... (A)$For the Paschen series,$n_1 = 3$,so $n_2 = 4$. The longest wavelength $\lambda_{PL}$ is:$\frac{1}{\lambda_{PL}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{7}{144} ight) \implies \lambda_{PL} = \frac{144}{7R} \quad ... (B)$Taking the ratio of the longest wavelengths:$\frac{\lambda_{BL}}{\lambda_{PL}} = \frac{36}{5R} 	imes \frac{7R}{144} = \frac{36 	imes 7}{5 	imes 144} = \frac{7}{5 	imes 4} = \frac{7}{20}$.

The ratio of longest wavelength lines in the Balmer and Paschen series of hydrogen spectrum is

Similar Questions

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