The ratio of wavelengths for transition of electrons from 2 nd orbit to 1 st orbit of Helium (He⁺) and Lithium (Li⁺⁺) is (Atomic number of Helium = 2, Atomic number of Lithium = 3) (a) 9:4 (b) 9:36 (c) 4:9 (d) 2:3

Answer: (a). For hydrogen-like ions (1)/(λ) = RZ²((1)/(nf²)-(1)/(ni²)); same transition gives λ ∝ 1/Z². λHe⁺λLi⁺⁺ = ZLi²ZHe² = 9/4 So the answer is (a) 9:4.

The ratio of wavelengths for transition of electrons from 2 nd orbit… | MHT CET 2023 (9 May, Shift 1)

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Transitions, Photon Emission & Excitation

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

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Answer: (a)

For hydrogen-like ions $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ ; same transition gives $\lambda \propto \frac{1}{Z^2}$ .
$\frac{\lambda_{He^+}}{\lambda_{Li^{++}}} = \frac{Z_{Li}^2}{Z_{He}^2} = \frac{9}{4}$
So the answer is (a) $9:4$ .

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