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(C) The wavelength $\lambda$ for a transition between energy levels $n_2$ and $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(

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$6.4, 1.2$

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(C) The wavelength $\lambda$ for a transition between energy levels $n_2$ and $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$.For $Li^{2+}$,the atomic number $Z = 3$,so $Z^2 = 9$.$\frac{1}{\lambda_{32}} = R(9) \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = 9R \left(\frac{1}{4} - \frac{1}{9}ight) = 9R \left(\frac{5}{36}ight) = \frac{5R}{4} \implies \lambda_{32} = \frac{4}{5R}$.$\frac{1}{\lambda_{31}} = R(9) \left(\frac{1}{1^2} - \frac{1}{3^2}ight) = 9R \left(1 - \frac{1}{9}ight) = 9R \left(\frac{8}{9}ight) = 8R \implies \lambda_{31} = \frac{1}{8R}$.$\frac{1}{\lambda_{21}} = R(9) \left(\frac{1}{1^2} - \frac{1}{2^2}ight) = 9R \left(1 - \frac{1}{4}ight) = 9R \left(\frac{3}{4}ight) = \frac{27R}{4} \implies \lambda_{21} = \frac{4}{27R}$.Now,calculating the ratios:$\frac{\lambda_{32}}{\lambda_{31}} = \frac{4/5R}{1/8R} = \frac{4}{5} 	imes 8 = \frac{32}{5} = 6.4$.$\frac{\lambda_{21}}{\lambda_{31}} = \frac{4/27R}{1/8R} = \frac{4}{27} 	imes 8 = \frac{32}{27} \approx 1.185 \approx 1.2$.Thus,the ratios are $6.4$ and $1.2$.

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively:

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