Answer: De Broglie standing wave condition: n lambda = 2 pi r.
- A Classical mechanics alone, with little quantum assumption needed
- B De Broglie standing wave condition: n lambda = 2 pi r
- C Conservation of energy alone, with little wave condition imposed
- D Coulomb force balance between the electron and the nucleus alone
Correct answer: B. De Broglie standing wave condition: n lambda = 2 pi r
Explanation: Standing wave condition: for stable orbit, circumference = n wavelengths. 2pi r = n lambda = n h/mv. This gives L = mvr = nh/2pi.
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.