Visible light emission from hydrogen atom (Balmer series) involves transitions to:

Visible light emission from hydrogen…

Answer: n=2.

A n=1
B n=2
C n=3
D n=4

Correct answer: B. n=2

Explanation: Balmer series: electrons fall to n=2. These transitions produce visible light (H-alpha, H-beta, etc.).

Comparison of the Thomson model and the Rutherford model of the atom: in the Thomson plum pudding model alpha particles pass almost straight through a diffuse positive sphere, while in the Rutherford model a concentrated central nucleus deflects some alpha particles through large angles, with the lower panels showing the alpha particle source and gold foil and the observed result of wide-angle scattering

The Geiger–Marsden alpha-scattering result: a few alpha particles bounce back at large angles, which rules out Thomson’s diffuse model and demands a tiny, massive, positively charged nucleus. Image: Kurzon, CC BY 3.0, via Wikimedia Commons.

Concept context

Thomson's plum-pudding model, Rutherford's nuclear model from alpha-scattering, Bohr's postulates and hydrogen spectrum, spectral series (Lyman to Pfund), Rydberg formula, and limitations of each model.

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Visible light emission from hydrogen atom (Balmer series) involves transitions to:

Concept context

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