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Physics · Ch 15 — Structure of Atoms and Nuclei

Rutherford's Atomic Model

15.4

Rutherford's Atomic Model

Rutherford reasoned carefully from the Geiger-Marsden results. In Thomson's plum-pudding model, the positive charge is spread over a sphere the size of the whole atom, so its charge density is very low everywhere, and even a head-on alpha particle would experience only a weak, gradually varying force -- nowhere near enough to reverse its direction. The observed large-angle scattering, rare as it was, could therefore only happen if an alpha particle occasionally passed extremely close to a small, concentrated lump of positive charge and mass, strong enough to repel it back the way it came.

From the fraction of particles scattered at each angle, Rutherford could work out how small and how massive this central charge had to be: about 10 fm across (1 fm = 10−1510^{-15} m), which is roughly 10−510^{-5} times the size of the whole atom -- meaning its VOLUME is only about 10−1510^{-15} of the atom's volume. He called this tiny, dense, positively charged core the nucleus. …

Misc Ex.15.2Kinetic energy of an alpha particle at closest approach to a gold nucleus

Worked out. An incident alpha particle (radius 1.80 fm) moving directly toward a target gold nucleus (radius 6.23 fm, Z = 79) is assumed to stop exactly at the point where the two surfaces touch, i.e. at a centre-to-centre separation equal to the sum of the two radii. At that instant all of the alpha particle's initial kinetic energy has converted into electrostatic potential energy of the two positive charges at that separation, so equating K=14πϵ0(2e)(Ze)r1+r2K=\frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{r_1+r_2} gives the alpha particle's original kinetic energy directly from Coulomb's law, without needing any details of the actual (curved) scattering trajectory -- only the distance of closest approach matters. The calculation gives about 28.3 MeV, illustrating why alpha sources used in scattering experiments need energies …