MHT CET · Physics · Atomic Physics
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)
- A \(9:4\)
- B \(4:9\)
- C \(9:36\)
- D \(2:3\)
Answer & Solution
Correct Answer
(A) \(9:4\)
Step-by-step Solution
Detailed explanation
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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