If lambda_{1} and lambda_{2} are the waveleng | vedclass

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(C) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \rig

Vedclass · Accepted Answer

$7: 135$

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(C) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight)$.For the third member of the Lyman series,$n_{f} = 1$ and $n_{i} = 1 + 3 = 4$. Thus,$\frac{1}{\lambda_{1}} = R \left( \frac{1}{1^{2}} - \frac{1}{4^{2}} ight) = R \left( 1 - \frac{1}{16} ight) = R \left( \frac{15}{16} ight)$.For the first member of the Paschen series,$n_{f} = 3$ and $n_{i} = 3 + 1 = 4$. Thus,$\frac{1}{\lambda_{2}} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16 - 9}{144} ight) = R \left( \frac{7}{144} ight)$.Taking the ratio $\frac{\lambda_{1}}{\lambda_{2}} = \frac{1/\lambda_{2}}{1/\lambda_{1}} = \frac{R(7/144)}{R(15/16)} = \frac{7}{144} 	imes \frac{16}{15} = \frac{7}{9 	imes 15} = \frac{7}{135}$.Therefore,$\lambda_{1} : \lambda_{2} = 7 : 135$.

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of the Lyman series and the first member of the Paschen series respectively,then the value of $\lambda_{1} : \lambda_{2}$ is

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