Physics · Atomic and quantum physics · Grade 11

Rutherford's experiment

Why do most alpha particles pass straight through gold foil, while some bounce back?

The situation

1909, Manchester. Under the direction of Ernest Rutherford, his assistant Hans Geiger and the student Ernest Marsden aimed a beam of alpha particles at a very thin metal foil and, in a dark room, counted the tiny flashes on a ZnS screen one by one through a microscope. According to the Thomson model of the time, an atom is a positively charged “pudding” with the electrons scattered in it like raisins, so a heavy, fast alpha particle should pass almost straight through. But some particles were deflected through large angles, and a few even bounced back. What is inside an atom, and how can the charge and size of the nucleus be found by counting flashes?

Topic

Atomic structure: Rutherford's experiment. Firing alpha particles at gold foil: the atom is almost empty

Key ideas

The atomic nucleus
Impact parameter
Rutherford's formula

Formulas

\tan \frac{θ}{2} = \frac{d}{2 b}, d = \frac{2 Z e^{2}}{4 π ε_{0} E}

Impact parameter $b$ and scattering angle $θ$ ; $d$ is the distance of closest approach to the nucleus in a head-on collision ( $E$ is the kinetic energy of the α-particle, $Z$ is the charge of the nucleus).

\frac{d σ}{d Ω} = {(\frac{d}{4})}^{2} \frac{1}{\sin^{4} (θ / 2)}

Rutherford formula — the differential scattering cross-section.

N (θ) = I_{0} n t Δ Ω \frac{d σ}{d Ω} \propto \frac{Z^{2} t}{E^{2} \sin^{4} (θ / 2)}

Flashes counted by the microscope: $I_{0}$ is the α flux, $n$ is the concentration of atoms, $t$ is the foil thickness, $Δ Ω$ is the solid angle of the screen.

r_{\min} = \frac{d}{2} (1 + \frac{1}{\sin (θ / 2)})

Distance of closest approach to the nucleus on a hyperbolic trajectory.

\frac{d σ}{d Ω} = | f (θ) |^{2}, f_{B} (θ) = - \frac{m}{2 π ℏ^{2}} \int V (𝐫) e^{i 𝐪 \cdot 𝐫} d^{3} r

Quantum scattering: amplitude $f$ and the Born approximation ( $ℏ q = 2 p \sin (θ / 2)$ is the change in momentum).

Theory

The Thomson model and its prediction

J. J. Thomson imagined the atom as a uniform sphere of positive charge of radius ≈ $10^{- 10}$ m, with the electrons inside it like “raisins in a pudding” (the Thomson model). In such an atom the electric field is weak: a 7.7 MeV α-particle passing through a gold atom is deflected by at most $θ_{1} = d / (2 R) \approx 0.005 °$ . The electrons are about 7300 times lighter than an α-particle — they can hardly turn it at all. In the foil a particle crosses ≈ 2000 atoms and the deflections add up randomly: on average $Θ \approx 0.8 \sqrt{N} θ_{1} \approx 0.2 °$ . Check in Thomson atom mode: at angles above 5° there is not a single flash.

The Geiger–Marsden experiment: the atom has a nucleus

In 1909–1913 Geiger and Marsden aimed α-particles from a radioactive source at a thin foil in a vacuum chamber and counted the flashes on a ZnS screen that rotated around the foil. Most particles passed through almost undeflected, but a very small fraction came back at angles greater than 90°. Rutherford's conclusion (1911): the positive charge of the atom and almost all of its mass are concentrated in a very small nucleus, with the electrons around it. The radius of the gold nucleus is ≈ 7 fm and that of the atom ≈ 0.15 nm — the atom is almost empty. The nuclear charge $Z e$ equals the atomic number of the element; Chadwick (1920) measured it by exactly this method — repeat it yourself with the unknown foil X in Nuclear model mode.

The Rutherford formula and the scattering cross-section

In the Coulomb field of a nucleus an α-particle moves along a hyperbola (scattering in a central field), and the angle depends only on the impact parameter $b$ : $\tan (θ / 2) = d / (2 b)$ . Particles that fall on the ring between $b$ and $b + d b$ are scattered into the range between $θ$ and $θ + d θ$ ; the area of the ring $d σ = 2 π b d b$ is the scattering cross-section. Dividing by $d Ω = 2 π \sin θ d θ$ , we get $d σ / d Ω = {(d / 4)}^{2} / \sin^{4} (θ / 2)$ . So the number of flashes is $N \propto Z^{2} t / (E^{2} \sin^{4} (θ / 2))$ : if you turn the microscope from 60° to 120°, $N$ falls by a factor of 9. Geiger and Marsden (1913) checked that $N \sin^{4} (θ / 2)$ is constant, and also that $N \propto t$ and $N \propto 1 / E^{2}$ .

Quantum scattering theory (in brief)

In quantum mechanics the incident particle is a plane wave $e^{i k z}$ and the scattered part is a spherical wave $f (θ) e^{i k r} / r$ ; the differential cross-section follows from the scattering amplitude $f$ as $d σ / d Ω = | f |^{2}$ . If the potential is weak, the Born approximation gives $f$ as the Fourier transform of $V (𝐫)$ : for a screened Coulomb potential $V \propto e^{- r / a} / r$ we get $d σ / d Ω \propto {(q^{2} + a^{- 2})}^{- 2}$ , and as $a \to \infty$ exactly the Rutherford formula results. In the partial wave method $f = \frac{1}{k} \sum_{l} (2 l + 1) e^{i δ_{l}} \sin δ_{l} P_{l} (\cos θ)$ : each angular momentum $l \approx k b$ receives its own phase shift $δ_{l}$ . For a 7.7 MeV α-particle $ℏ / p \approx 0.8$ fm, much smaller than $d \approx 30$ fm (Sommerfeld parameter $η \approx 18$ ), so classical trajectories give the right result — watch $l$ and $η$ in Single particle mode.

The planetary model and its crisis

The Rutherford atom resembles the Solar System: electrons orbit the nucleus (the planetary model). But according to classical electrodynamics an accelerating charge radiates (the Larmor formula $P = \frac{2}{3} k e^{2} a^{2} / c^{3}$ ). The electron in a hydrogen atom should lose energy and spiral into the nucleus in about $1.6 \cdot 10^{- 11}$ s, while the frequency of the radiation should grow continuously, giving a continuous spectrum. In experiments, however, atoms are stable and emit a line spectrum. This contradiction was resolved by Bohr's postulates (1913): an electron stays in allowed orbits without radiating, with energy levels $E_{n} = - 13.6 / n^{2}$ eV; the existence of the levels was confirmed by the Franck–Hertz experiment (1914). For details see the “Atomic structure and spectra” and “Orbitals of the hydrogen atom” experiments.

About the model

The nucleus is taken to be stationary (the recoil correction for gold is ≈ 2 %), each particle is scattered once by a single nucleus (thin foil), and the screening by the electron shells and nuclear forces are not taken into account: if $r_{\min}$ reaches the nuclear radius $R \approx 1.2 A^{1 / 3}$ fm, the formula breaks down (such “anomalous scattering” has been observed in light nuclei). The source flux $I_{0} = 2 \cdot 10^{8}$ s⁻¹ and the solid angle seen by the microscope $Δ Ω = 10^{- 3}$ sr are model values. The counting time is sped up 20 times, the picture of the atom is magnified roughly $10^{9}$ – $10^{12}$ times, and the rotation in the planetary model is slowed down.

Did you know?

In his 1936 lecture, recalling the α-particles that bounced back, Rutherford said it was as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it had come back and hit you.
In 1920 James Chadwick measured, from the scattering of α-particles, the nuclear charges of copper, silver and platinum as 29.3, 46.3 and 77.4 elementary charges — very close to their atomic numbers (29, 47, 78).

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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