3D Rutherford Gold Foil Experiment | Rutherford Institute Chemistry

Background: the plum-pudding atom

By 1900 physicists knew that atoms are not indivisible. In 1897 J. J. Thomson had found the electron: a particle with negative charge and less than a thousandth of the mass of the lightest atom. Since atoms are electrically neutral, the matching positive charge — and almost all of the mass — had to be somewhere else. Nobody knew where.

In 1904 Thomson proposed the simplest picture: the positive charge is spread out evenly through the whole atom, a ball about $1 0^{- 10}$ m in size, with the electrons dotted through it like plums in a pudding. Hence the nickname plum-pudding model.

In a plum-pudding atom the electric forces are gentle everywhere. The positive charge is diluted over the whole atom, so it can never push hard on anything passing through, and the electrons are far too light to knock a heavy particle off course. Any fast, heavy, charged particle should go straight through, nudged by only a tiny fraction of a degree — like a cannonball fired through a cloud of mist.

Rutherford had exactly such a particle to test this with. The alpha particle (α), shot out by radioactive atoms, is what we now call a helium nucleus: charge +2 (two protons), about 7,300 times heavier than an electron. The fastest alphas from the radium source used in Manchester carry 7.7 MeV and move at about 19,000 km/s, roughly 6% of the speed of light.

The experiment

The work was done by Hans Geiger and Ernest Marsden, a 20-year-old undergraduate, in Rutherford's laboratory at Manchester between 1909 and 1913.

Source. A small glass tube of radium emanation (radon gas). Its decay products emit alpha particles. A narrow opening lets only a thin beam out towards the foil.
Gold foil. Gold can be beaten into leaf well under a micrometre thick, and its atoms are heavy and carry a large charge (atomic number 79). Even a foil 0.4 µm thick, the one used in the simulation, is about 1,500 atoms thick.
Detector. A small screen coated with zinc sulfide (ZnS). Each alpha that hits it makes one tiny flash of light, a scintillation. Through a microscope fixed to the screen, the observer counted the flashes by eye, in a darkened room. The microscope and screen could be turned around the foil to count at different angles, from 5° to 150°.
Vacuum. Alphas travel only about 7 cm in air, and air molecules deflect them too. The whole apparatus was sealed in a metal box and the air pumped out.

The quantity they measured is the scattering angle $θ$ : the angle between the direction an alpha came in and the direction it leaves the foil. 0° means straight through, 90° means sideways, and 180° means straight back towards the source.

What was expected — and what was seen

Expected (plum-pudding atom)

One gold atom can turn an alpha by about 0.005° at most, half a hundredth of a degree. Crossing 1,500 atoms, the small random nudges add up to a typical deflection of only about a quarter of a degree.

The chance of being turned through more than 90° works out to about $1 0^{- 61000}$ . In other words: never.

Seen

Most alphas did go almost straight through: in a thin gold foil about 99.8% are deflected by less than 10°.

But a few were deflected through large angles, and some came back. For a 0.4 µm gold foil about 1 alpha in 60,000 is turned through more than 90°.

For the curious: where does

1 0^{- 61000}

come from?

Each atom gives the alpha a tiny sideways nudge in a random direction. Random nudges partly cancel, like a drunkard's walk: after many steps you typically end up only about $N$ steps from where you started, not $N$ . For 1,500 gold atoms the simulation's plum-pudding model gives a typical (rms) deflection of 0.24°.

For such a random walk the chance of ending up beyond an angle $θ$ falls like a bell curve, $P (> θ) \approx e^{- (θ /0.2 4^{\circ})^{2}}$ . At 90° the exponent is $(90/0.24)^{2} \approx 141, 000$ , and $e^{- 141000} \approx 1 0^{- 61000}$ : a decimal point followed by 61,000 zeros.

Geiger and Marsden first saw this in 1909. Their famous figure, that about 1 alpha in 8,000 was “reflected”, came from a thick plate of platinum. Alphas entering a thick plate meet many more atoms, so the chance of a big deflection is much higher than in a thin foil. The thin-foil number is smaller, but the puzzle is the same: no plum-pudding atom can turn an alpha around even once.

“It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Ernest Rutherford, recalling the result

Rutherford's explanation (1911)

To turn a fast, heavy alpha around in a single encounter, something inside the atom must push extremely hard. Rutherford realised that this happens if all the positive charge, and nearly all the mass, is packed into a tiny nucleus at the centre, with the electrons spread thinly around it. Most alphas pass far from any nucleus, because the atom is almost empty, and are barely deflected. A very few happen to head almost straight at a nucleus and are flung back.

Formula 1 The push between alpha and nucleus (Coulomb's law)

F = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( 79 e )}{r ^{2}}

In simple words

Two positive charges push each other apart, and the push grows fast as they get closer: halve the distance and the push is 4 times stronger; ten times closer, 100 times stronger. Because all of the nucleus's charge sits in one tiny point, an alpha can get extremely close to it, so the push can become enormous. In a plum pudding the charge is smeared over the whole atom and the push never gets big.

$F$: the push (force) on the alpha
$r$: distance between the alpha and the centre of the nucleus
$2 e, 79 e$: the charges: the alpha has 2 protons, a gold nucleus 79
$\frac{1}{4 π ε _{0}}$: Coulomb's constant, a fixed number of nature that sets how strong electric forces are

Worked example

At $r = 30$ fm (the closest a 7.7 MeV alpha gets, see below) the push is about 40 newtons, the weight of a 4 kg bag of shopping, acting on a particle that weighs $7 \times 1 0^{- 27}$ kg. At the edge of a plum-pudding atom ( $r \approx 150, 000$ fm), the strongest push it could ever give is $(150, 000/30)^{2} =$ 25 million times weaker: about 1.6 millionths of a newton.

Formula 2 How close a head-on alpha gets: the distance of closest approach $d$

d = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( 79 e )}{K} \approx \frac{2 \times 79 \times 1.44 MeV fm}{7.69 MeV} \approx 29.6 fm

In simple words

Imagine rolling a ball straight up a hill. It slows down, stops for an instant where all of its speed has been turned into height, then rolls back down. For an alpha fired straight at a nucleus, the “hill” is the electric push and the “height” is electrical energy. $d$ is the point where it stops and turns back. A faster ball climbs higher up the hill: double the alpha's energy and it gets twice as close ( $d$ halves, to 14.8 fm).

$d$: closest distance for an alpha aimed dead-centre
$K$: the alpha's energy of motion: 7.69 MeV for the fastest radium alphas
$2 \times 79$: the two charges multiplied together, counted in protons
1.44 MeV fm: a handy value of $e^{2} /4 π ε_{0}$ for nuclear sizes
fm: femtometre, $1 0^{- 15}$ m (a millionth of a billionth of a metre)

Worked example

$2 \times 79 = 158$ ; $158 \times 1.44 = 227.5$ MeV fm; divide by 7.69 MeV: $d \approx$ 29.6 fm. The atom's radius is about 150,000 fm, so a head-on alpha gets 5,000 times closer to the centre than the atom's edge. If the nucleus were bigger than 30 fm, these alphas would hit it, and the simple Coulomb law would fail. It doesn't, so the nucleus is smaller than that.

For the curious: deriving

d

At the turning point all of the kinetic energy has become electrical potential energy:

K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( 79 e )}{d} ⟹ d = \frac{1}{4 π ε _{0}} \frac{2 \cdot 79 e ^{2}}{K} .

Strictly, the gold nucleus recoils a little (it is 49 times heavier than the alpha), so only the energy of relative motion counts. That gives 30.2 fm instead of 29.6 fm. The simulation includes this recoil.

Under an inverse-square repulsion each alpha follows a curved path, a hyperbola. Its fate is set by the impact parameter $b$ : how far off-centre it is aimed, measured from the line through the nucleus.

Formula 3 Aim versus turn: impact parameter $b$ and deflection $θ$

b = \frac{d}{2} cot \frac{θ}{2}

In simple words

The closer the aim, the bigger the turn. Think of rolling marbles past a smooth, steep bump. Aim far to the side and the marble barely bends; aim close and it swings round sharply; aim dead-centre and it rolls straight back. For small turns there is a simple rule: aim 10 times closer, turn 10 times more ( $θ \approx 57. 3^{\circ} \times d / b$ ).

$b$: impact parameter: how far to the side of the nucleus the alpha is aimed
$θ$: the angle it is turned through
$d$: closest approach from Formula 2, 29.6 fm
$cot$: cotangent, a calculator button: very large for small angles, 1 at 45°, 0 at 90°

Turned through $θ$	1°	10°	30°	60°	90°	150°	180°
Aim needed, $b$	1,700 fm	170 fm	55 fm	26 fm	15 fm	4 fm	0 (head-on)

Worked example

For a 90° turn: $θ /2 = 4 5^{\circ}$ and $cot 4 5^{\circ} = 1$ , so $b = 29.6/2 \times 1 =$ 14.8 fm. For a 10° turn: $cot 5^{\circ} = 11.43$ , so $b = 14.8 \times 11.43 \approx$ 169 fm. The aim has to be about 11 times closer to get a 90° turn instead of 10°.

For the curious: where the formula comes from

For a repulsive inverse-square force the path is a hyperbola with the nucleus at its outer focus. The incoming and outgoing straight parts (the asymptotes) meet at an angle $18 0^{\circ} - θ$ , and the geometry of the hyperbola gives $tan (θ /2) = d / (2 b)$ . The alpha's closest distance on a path that turns by $θ$ is $r_{min} = \frac{d}{2} (1 + 1/ sin \frac{θ}{2})$ : 36 fm at 90°, 30 fm head-on.

The table includes the small recoil of the gold nucleus (less than 1% here).

To be turned back by more than 90°, an alpha must therefore pass within 15 fm of a nucleus, a target only a ten-thousandth of the atom's radius. How often does that happen?

Formula 4 The chance of bouncing back

P (θ > 9 0^{\circ}) = n t \times π b_{90}^{2} \approx 1.6 \times 1 0^{- 5} \approx \frac{1}{62 , 000}

In simple words

A game of darts. Around every gold nucleus imagine a tiny invisible bullseye of radius $b_{90} = 15$ fm. Any alpha that lands inside one bounces back by more than 90°. So the chance of bouncing back is just the fraction of the foil covered by bullseyes. Blow the foil up to the size of a football pitch: all of its bullseyes together would cover about 0.12 m², less than two sheets of A4 paper.

$n t$: how many gold atoms each square centimetre of foil holds ( $n$ atoms per cm³ × thickness $t$ )
$π b_{90}^{2}$: the area of one bullseye (area of a circle, $π r^{2}$ )
$P$: the chance that one alpha is turned through more than 90°

Worked example

$b_{90} = 14.8$ fm $= 1.48 \times 1 0^{- 12}$ cm, so one bullseye covers $π \times (1.48 \times 1 0^{- 12})^{2} = 6.9 \times 1 0^{- 24}$ cm². The foil holds $2.36 \times 1 0^{18}$ atoms per cm². Multiply: $6.9 \times 1 0^{- 24} \times 2.36 \times 1 0^{18} \approx 1.6 \times 1 0^{- 5}$ , about 1 alpha in 62,000. That is rare, but nowhere near the plum-pudding prediction of “never”.

For the curious: counting the atoms

Gold has a density of 19.3 g/cm³ and 197 g per mole, so one cm³ holds $n = 19.3/197 \times 6.02 \times 1 0^{23} \approx 5.9 \times 1 0^{22}$ atoms. A foil $t = 0.4 μ m = 4 \times 1 0^{- 5}$ cm thick therefore holds $n t \approx 2.36 \times 1 0^{18}$ atoms per cm². The bullseyes cover so little of the foil that they almost never overlap, and an alpha meets at most one of them: that is why a thin foil gives clean single-scattering results.

Rutherford's scattering formula

Rutherford worked out what fraction of the alphas should land in a small detector placed at angle $θ$ . This is the formula Geiger and Marsden put to the test.

Formula 5 How many alphas a detector at angle $θ$ catches

\frac{Δ N}{N _{0}} = n t (\frac{d}{4})^{2} \frac{ΔΩ}{sin ^{4} ( θ /2 )} so N (θ) \propto \frac{1}{sin ^{4} ( θ /2 )}

In simple words

Everything in front, $n t$ , $d$ and $ΔΩ$ , is fixed by the foil, the alphas and the detector. Only the last part changes as you move the detector round: $1/ sin^{4} (θ /2)$ , and it gets small very fast. Rule of thumb: double the angle, about 16 times fewer alphas.

Why 16? To land at half the angle, an alpha's aim must be twice as far out, in a ring of aims that is also wider: about 8 times more alphas. And those alphas are crowded onto a ring of sky only half as big around: another factor of 2. And 8 × 2 = 16.

The rule is exact only for small angles. Checked with the real formula: 10° → 20° gives 15.8 times fewer, 45° → 90° gives 11.7 times fewer, and 90° → 180° only 4 times fewer. At large angles the curve flattens out.

$Δ N / N_{0}$: fraction of all the alphas fired that land in the detector
$n t, d$: atoms per cm² of foil, and the closest approach (Formulas 2 and 4)
$ΔΩ$: the detector's size as seen from the foil: its area ÷ (distance)², in steradians
$sin^{4} (θ /2)$: sine of half the angle, multiplied by itself four times

Detector at $θ$	1°	10°	20°	30°	60°	90°	120°	180°
Count, compared with 180°	170 million ×	17,300 ×	1,100 ×	223 ×	16 ×	4 ×	1.8 ×	1 ×

Worked example

A screen 2 mm × 2 mm, 2 cm from the foil: $ΔΩ = 4 mm^{2} / (20 mm)^{2} = 0.01$ sr. At 90°: $(d /4)^{2} = (7.4 fm)^{2} = 54.8 fm^{2}$ , $n t = 2.36 \times 1 0^{- 8}$ atoms per fm², and $sin^{4} 4 5^{\circ} = 0.25$ . So $Δ N / N_{0} = 2.36 \times 1 0^{- 8} \times 54.8 \times 0.01/0.25 \approx 5.2 \times 1 0^{- 8}$ : about 1 alpha in 19 million lands on the screen. Move it to 10° and it catches about 1 in 4,500.

For the curious: deriving the

1/ sin^{4} (θ /2)

law

Alphas aimed through the thin ring between $b$ and $b + d b$ (area $2 πb d b$ ) come out between $θ$ and $θ + d θ$ , a ring of sky with solid angle $2 π sin θ d θ$ . The ratio of the two is the cross-section per unit solid angle:

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

Using $sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}$ this simplifies to $d σ / d Ω = (d /4)^{2} / sin^{4} (θ /2)$ . Multiplying by the number of atoms per area, $n t$ , and by the detector's $ΔΩ$ gives Formula 5. The graph on the Experiment page uses the same formula, with the small recoil of the gold nucleus included.

It falls off steeply. A detector at 90° counts about 4,300 times fewer alphas than the same detector at 10°, and one at 10° counts about 10,000 times fewer than one at 1°.
Then it flattens. From 90° to 180°, $sin^{4} (θ /2)$ changes only from $\frac{1}{4}$ to 1, so the count falls by just a factor of 4. Backscattering is rare, but once an alpha comes close enough to be turned through 90°, all backward angles are about equally easy.
It predicts more than the angle. Double the foil thickness and you double the count. Make the nucleus charge twice as big and the count goes up 4 times. Give the alphas half the energy and 4 times as many are scattered (a 5 MeV alpha instead of 7.69 MeV: 2.4 times as many). Geiger and Marsden tested all of these in 1913, and the formula held every time.

Conclusions

The atom has a nucleus. All of its positive charge and more than 99.9% of its mass sit in a tiny central core.
The nucleus is tiny. Even head-on alphas, which get within about 30 fm, still obeyed the pure Coulomb law, so the nucleus must be smaller than about $3 \times 1 0^{- 14}$ m. Today we know a gold nucleus has a radius of about $7 \times 1 0^{- 15}$ m, roughly 20,000 times smaller than the atom. If an atom were a stadium 200 m across, its nucleus would be a pea at the centre spot.
The atom is mostly empty space. That is why almost every alpha sails straight through 1,500 layers of gold.
The nuclear charge can be counted. The scattering rate depends on the square of the nuclear charge. Later measurements (Chadwick, 1920) showed that the charge equals the element's atomic number $Z$ , here 79 for gold. The nuclear atom led directly to Bohr's model of the atom (1913) and to nuclear physics.

Now see it happen

The experiment, live

Stand inside a dome-shaped zinc-sulfide screen while 120,000 alphas a second hit a gold foil. Every flash is one alpha, scattered by true Rutherford physics. Switch the atoms to plum pudding and watch the wide-angle flashes vanish, and see the counts fill a live graph.

Go to the experiment or jump straight to the live graph

Rutherford's gold-foil experiment

Background: the plum-pudding atom

The experiment

What was expected — and what was seen

Expected (plum-pudding atom)

Seen

Rutherford's explanation (1911)

Rutherford's scattering formula

Conclusions

The experiment, live

Two angles name every direction from the foil