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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2

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$1 : 9$

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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$.For deuterium $(D)$,the atomic number $Z_D = 1$. Thus,$\frac{1}{\lambda_D} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.For doubly ionized lithium $(Li^{2+})$,the atomic number $Z_{Li} = 3$. Thus,$\frac{1}{\lambda_{Li}} = R (3)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 9R \left( \frac{3}{4} \right) = \frac{27R}{4}$.Taking the ratio of the wavelengths: $\frac{\lambda_{Li}}{\lambda_D} = \frac{4/27R}{4/3R} = \frac{3}{27} = \frac{1}{9}$.Therefore,the ratio is $1 : 9$.

Find the ratio of the wavelength of the first line of the Lyman series for a doubly ionized lithium atom $(Li^{2+})$ to the wavelength of the first line of the Lyman series for deuterium $(D)$.

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