12.1 Alpha-Particle Scattering and Rutherford's Nuclear Model

In the clean rooms where computer chips are made, engineers check the thickness and composition of films only a few nanometres thick with a beam of helium nuclei from a small accelerator. A detector counts the few that bounce back from the sample and measures their energy: helium nuclei that rebound from heavy atoms keep more of their energy than those that rebound from light ones, so the energy spectrum reads out which elements are present and at what depth. The technique, Rutherford backscattering spectrometry, is used every day in semiconductor, solar-cell and materials laboratories around the world. It is a direct descendant of an experiment done with a radioactive source, a gold foil and a microscope more than a hundred years ago — the experiment that showed that every atom has a nucleus.

At the start of the twentieth century physicists knew that atoms contain light, negatively charged electrons (discovered by J. J. Thomson in 1897) and are electrically neutral, so they must also contain positive charge. But how is that charge arranged? In this section you follow the alpha-particle scattering experiment of Hans Geiger and Ernest Marsden, see why its results destroyed the leading model of the day, and work out with Rutherford how small the positive core must be. The only physics you need is Coulomb's law (Section 1.1) and conservation of energy with electrostatic potential energy (Section 2.2).

Thomson's Atom and the Geiger–Marsden Experiment

The first serious picture of the inside of an atom was Thomson's. Knowing that electrons are very light and that atoms are about $10^{-10}$ m across, he imagined the positive charge spread out evenly through the whole atom.

The model can be tested by firing fast, heavy, charged projectiles through atoms and watching how they are deflected. Rutherford's group in Manchester used alpha particles, which some radioactive substances shoot out at high speed.

The experiment (1909–1913). A narrow beam of alpha particles of energy about $5.5$ MeV, defined by slits cut in lead blocks, was aimed at a gold foil less than a micrometre thick — only a few hundred atoms across. Gold was chosen because it can be beaten into extremely thin sheets and because its atoms are heavy. Around the foil, a small screen coated with zinc sulphide could be turned to any angle $\theta$ from the beam; each alpha particle that struck it made a tiny flash of light, which an observer counted through a microscope. The whole apparatus was kept in a vacuum so that the alpha particles were not slowed by air. Counting the flashes at each angle, for hours on end, gave the number of alpha particles scattered through each angle.

Observations 1 and 2 are not surprising. Observation 3 astonished Rutherford: he later compared it with firing a heavy artillery shell at a sheet of tissue paper and having it come back at you. To turn a fast alpha particle round, something must push on it with an enormous force — and Thomson's atom cannot.

Rutherford's Nuclear Model and the Impact Parameter

In 1911 Rutherford proposed that the positive charge, and with it almost all the mass, is concentrated in a tiny central core.

The model explains all three observations at once. Because the atom is mostly empty, most alpha particles pass far from any nucleus and feel almost no force: they go straight through. The electrons cannot deflect them noticeably, because an alpha particle is about $7300$ times heavier — like a bowling ball ploughing through a cloud of table-tennis balls. Only the few alpha particles that happen to pass very close to a nucleus feel the full repulsion of its charge and are deflected through large angles. A gold nucleus is about $49$ times heavier than an alpha particle, so it hardly recoils; we treat it as fixed.

The trajectory. Between the alpha particle (charge $2e$) and the nucleus (charge $Ze$) there acts only the Coulomb repulsion $$F=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^2},$$ directed along the line joining them. The alpha particle comes in along a straight line, curves away from the nucleus as the force grows, and leaves along another straight line. The path is a branch of a hyperbola with the nucleus at its outer focus. Two quantities describe each such encounter.

For a given kinetic energy and nucleus, the scattering angle depends only on $b$:

• Head-on ($b=0$): the alpha particle moves straight at the nucleus, stops momentarily, and comes straight back: $\theta=180^\circ$.
• Small $b$: it passes close to the nucleus, feels a large force for a short time and is scattered through a large angle.
• Large $b$: it never gets close, the force on it stays small, and $\theta$ is tiny.

So the smaller the impact parameter, the larger the scattering angle. Because the nucleus is so small, only a tiny fraction of the alpha particles in a beam have impact parameters of a few femtometres — which is exactly why so few are scattered through large angles. Rutherford calculated how the number scattered should fall off with angle, and in 1913 Geiger and Marsden confirmed his prediction over a huge range of counts: the nuclear model was established.

The Distance of Closest Approach and the Size of the Nucleus

The head-on collision gives a simple and powerful estimate of how small the nucleus must be. Fire an alpha particle of kinetic energy $K$ (measured when it is far away) straight at a nucleus. As it approaches, the repulsion slows it down; its kinetic energy is converted into electrostatic potential energy of the pair. At some distance $d_0$ it stops for an instant and then is pushed back. At that moment all of $K$ has become potential energy (Section 2.2): $$K=\frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d_0}.$$

The shortcut follows from $ke^2=(9.0\times10^9)(1.6\times10^{-19})^2$ $=2.304\times10^{-28}$ J m; dividing by $1\ \text{MeV}=1.6\times10^{-13}$ J and $1\ \text{fm}=10^{-15}$ m gives $1.44$ MeV fm. It saves a great deal of arithmetic with powers of ten.

Electron Orbits in the Nuclear Atom

If the nucleus is tiny and the electrons are far from it, what keeps the electrons there? Rutherford's answer was the same as for the planets: the electrons orbit the nucleus, with the electrostatic attraction providing the centripetal force. Take the simplest atom, hydrogen: one electron (charge $-e$, mass $m$) in a circular orbit of radius $r$ around a proton (charge $+e$), which is so much heavier that it stays put.

Newton's second law for circular motion, with the Coulomb force supplying the centripetal force, gives $$\frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}\quad\Longrightarrow\quad mv^2=\frac{e^2}{4\pi\varepsilon_0r}.$$ So the kinetic energy is $K=\tfrac12mv^2=\dfrac{e^2}{8\pi\varepsilon_0r}$, while the potential energy of the electron–proton pair is $U=-\dfrac{e^2}{4\pi\varepsilon_0r}$ (negative, because the charges attract and $U=0$ at infinite separation).

Trouble for the nuclear atom. An electron in a circular orbit is accelerating towards the centre, and classical electromagnetism says that an accelerating charge radiates electromagnetic waves (Section 8.2). Radiating, the electron would lose energy; by Result 12.3 a more negative $E$ means a smaller $r$, so it would spiral into the nucleus — a calculation gives a lifetime of only about $10^{-11}$ s. On the way its orbital frequency would rise continuously, so it would emit a continuous spread of wavelengths. Real atoms are stable and emit sharp lines. Rutherford's model got the structure of the atom right, but classical physics could not explain why it survives: that is the problem Bohr solved.

Exercises