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(A) According to Rydberg's formula for hydrogen-like ions,the wavelength $\lambda$ of the emitted radiation is given by:$\frac{1}{\lambda} = R Z^2 \le

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$9:4$

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(A) According to Rydberg's formula for hydrogen-like ions,the wavelength $\lambda$ of the emitted radiation is given by:$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$Since the transition is from $n_2 = 2$ to $n_1 = 1$ for both ions,the term $\left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \left[ 1 - \frac{1}{4} \right] = \frac{3}{4}$ is constant.Therefore,$\lambda \propto \frac{1}{Z^2}$.For Helium $(He^+)$,$Z_{He} = 2$,so $\lambda_{He} \propto \frac{1}{2^2} = \frac{1}{4}$.For Lithium $(Li^{++})$,$Z_{Li} = 3$,so $\lambda_{Li} \propto \frac{1}{3^2} = \frac{1}{9}$.The ratio of wavelengths is $\frac{\lambda_{He}}{\lambda_{Li}} = \frac{1/4}{1/9} = \frac{9}{4}$.Thus,the ratio is $9:4$.

The ratio of wavelengths for the transition of electrons from the $2^{nd}$ orbit to the $1^{st}$ orbit of Helium $(He^+)$ and Lithium $(Li^{++})$ is (Atomic number of Helium = $2$,Atomic number of Lithium = $3$).

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