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

Question

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

Aicharya · Accepted Answer

Using Rydberg's Formula, $\begin{aligned} & \quad \frac{1}{\lambda}=R_H Z^2\left[\frac{1}{n^2}-\frac{1}{m^2}ight] \ & \Rightarrow \lambda \propto \frac{1}{Z^2} \ & 	herefore \quad \lambda_{\mathrm{Li}}: \lambda_{\mathrm{He}}=4: 9 \end{aligned}$

The ratio of wavelengths for transition of electrons from $2^{\text {nd }}$ orbit to $1^{\text {st }}$ orbit…

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