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The ratio of wavelengths for transition of electrons from $2^{\text {nd }}$ orbit to $1^{\text {st }}$ orbit of Helium $\left(\mathrm{He}^{++}\right)$and Lithium $\left(\mathrm{Li}^{++}\right)$is (Atomic number of Helium $=2$, Atomic number of Lithium = 3)
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The correct answer is:
$9:4$
Using Rydberg's Formula,
$\begin{aligned}
& \quad \frac{1}{\lambda}=R_H Z^2\left[\frac{1}{n^2}-\frac{1}{m^2}\right] \\
& \Rightarrow \lambda \propto \frac{1}{Z^2} \\
& \therefore \quad \lambda_{\mathrm{Li}}: \lambda_{\mathrm{He}}=9: 4
\end{aligned}$
$\begin{aligned}
& \quad \frac{1}{\lambda}=R_H Z^2\left[\frac{1}{n^2}-\frac{1}{m^2}\right] \\
& \Rightarrow \lambda \propto \frac{1}{Z^2} \\
& \therefore \quad \lambda_{\mathrm{Li}}: \lambda_{\mathrm{He}}=9: 4
\end{aligned}$
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