An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus head-on. Taking Z = 79 and e 2 /(4 0 ) = 1.44 MeV fm, what is the distance of closest approach?

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An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus head-on. Taking Z = 79 and e<sup>2</sup>/(4πε<sub>0</sub>) = 1.44 MeV fm, what is the distance of closest approach?

Answer: 45.5 fm.

A 45.5 fm
B 22.8 fm
C 91.0 fm
D 114 fm

Correct answer: A. 45.5 fm

Explanation: At closest approach, the initial kinetic energy is converted into electrostatic potential energy. Thus K = 1.44 × (2 × 79)/r. With K = 5.0 MeV, r = 45.5 fm.

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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An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus head-on. Taking Z = 79 and e<sup>2</sup>/(4&pi;&epsilon;<sub>0</sub>) = 1.44 MeV fm, what is the distance of closest approach?

Concept context

More Atoms questions

An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus head-on. Taking Z = 79 and e<sup>2</sup>/(4πε<sub>0</sub>) = 1.44 MeV fm, what is the distance of closest approach?