If lambda_1 and lambda_2 are the wavelengths | vedclass

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

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$20$:$7$

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(A) The wavelength $\lambda$ of a spectral line in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant.Frequency $f$ is given by $f = \frac{c}{\lambda} = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first member of the Balmer series,$n_1 = 2$ and $n_2 = 3$. Thus,$f_1 = Rc \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = Rc \left( \frac{1}{4} - \frac{1}{9} ight) = Rc \left( \frac{9-4}{36} ight) = \frac{5Rc}{36}$.For the first member of the Paschen series,$n_1 = 3$ and $n_2 = 4$. Thus,$f_2 = Rc \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = Rc \left( \frac{1}{9} - \frac{1}{16} ight) = Rc \left( \frac{16-9}{144} ight) = \frac{7Rc}{144}$.The ratio of frequencies is $\frac{f_1}{f_2} = \frac{5Rc/36}{7Rc/144} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$.

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first member of the Balmer and Paschen series in a hydrogen atom,respectively,then the ratio of the respective frequencies,$f_1 / f_2$,is:

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