Answer: 4pi eps 0 hbar²/(m e e²).
- A hbar/(m<sub>e</sub> c)
- B 4pi eps<sub>0</sub> hbar²/(m<sub>e</sub> e²)
- C e²/(m<sub>e</sub> c²)
- D m<sub>e</sub> e²/(4pi eps<sub>0</sub> hbar²)
Correct answer: B. 4pi eps<sub>0</sub> hbar²/(m<sub>e</sub> e²)
Explanation: a₀ = (4pi eps<sub>0</sub> hbar²)/(m<sub>e</sub> e²) = 0.529 Angstrom. It sets the scale of atomic orbitals.
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.