In the line spectra of the hydrogen atom,the | vedclass

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(A) The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant. L

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$10553$

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(A) The wavelength $\lambda$ 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. Let $C = \frac{1}{R}$. Then $\lambda = C \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)^{-1}$.For the Lyman series $(n_1 = 1)$:Shortest wavelength $\lambda_{L,s} = C \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight)^{-1} = C$.Longest wavelength $\lambda_{L,l} = C \left( \frac{1}{1^2} - \frac{1}{2^2} ight)^{-1} = \frac{4C}{3}$.Difference $\Delta \lambda_L = \frac{4C}{3} - C = \frac{C}{3} = 304\,\mathring{A} \implies C = 912\,\mathring{A}$.For the Paschen series $(n_1 = 3)$:Shortest wavelength $\lambda_{P,s} = C \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight)^{-1} = 9C$.Longest wavelength $\lambda_{P,l} = C \left( \frac{1}{3^2} - \frac{1}{4^2} ight)^{-1} = C \left( \frac{1}{9} - \frac{1}{16} ight)^{-1} = C \left( \frac{7}{144} ight)^{-1} = \frac{144C}{7}$.Difference $\Delta \lambda_P = \frac{144C}{7} - 9C = \frac{144C - 63C}{7} = \frac{81C}{7}$.Substituting $C = 912\,\mathring{A}$:$\Delta \lambda_P = \frac{81 	imes 912}{7} \approx 10553.14\,\mathring{A}$.

In the line spectra of the hydrogen atom,the difference between the largest and the shortest wavelengths of the Lyman series is $304\,\mathring{A}$. The corresponding difference for the Paschen series in $\mathring{A}$ is:

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