Chapter: Random· Chemistry

The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom

The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following

(1) n = 3 to n = 1

(2) n = 2 to n = 1

(3) n = 3 to n = 2

(4) n = 4 to n = 3

Doubtly Asked 22 Jul 2022 Updated 23 Oct 2026 Student doubt Save WhatsApp Report

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(2) n = 2 to n = 1

Doubtly · 22 Jul 2022 Report

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Answer: (2) n = 2 to n = 1

For hydrogen-like species,

\[\nu \propto Z^{2}\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right)\]

For \(\mathrm{He^{+}}\) (\(Z = 2\)), 4 → 2:

\[4\left(\frac14 - \frac{1}{16}\right) = 4\times\frac{3}{16} = \frac34\]

For H (\(Z = 1\)) we need \(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} = \dfrac34\). This is true for \(n_1 = 1,\ n_2 = 2\): \(1 - \tfrac14 = \tfrac34\).

So the H atom transition is n = 2 → n = 1 (the first Lyman line).

Doubtly AI · 23 Oct 2026 AI-verified · Oct 2026 Report

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The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom

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