The Nuclear Atom

A beam of alpha particles struck thin gold foil. Most particles sailed through it. A tiny fraction bounced almost straight back. That single surprising result overturned the reigning picture of the atom. It forced scientists to accept a new idea. Positive charge and mass are concentrated in an incredibly small nucleus. This article walks through Rutherford's gold-foil experiment. It explains why the "plum pudding" model failed. It then builds the modern nuclear atom: a dense positive nucleus surrounded by a cloud of electrons. These electrons occupy mostly empty space. The reasoning here rests entirely on the properties of the Subatomic Particles: Electron, Proton, Neutron. It also sets up the language of Atomic Number and Mass Number. The story is a clean example of a single well-designed experiment. Such an experiment can demolish a model that seemed reasonable. Before 1909, J. J. Thomson's plum-pudding model imagined the atom differently. It pictured a soft sphere of positive charge with electrons studded through it like raisins. That model predicted alpha particles should pass through foil with only tiny deflections. Rutherford's team expected to confirm this. Instead, they found deflections large enough to imply a concentrated core. Turning that surprise into a quantitative model is the heart of this article. By the end you should be able to describe the experimental setup. You should also explain why most particles pass through undeflected. Use nothing but Coulomb repulsion and geometry to explain why a few scatter backward. Finally, you should estimate the tiny size of the nucleus relative to the atom as a whole.

Key takeaways

Rutherford fired alpha particles (He²⁺, charge +2e) at gold foil only a few thousand atoms thick.
Most alpha particles passed straight through; about 1 in 8000 bounced back at angles greater than 90°.
The plum-pudding model could not explain large-angle scattering — its charge was too spread out to repel an alpha particle strongly.
The only explanation: nearly all atomic mass and positive charge sit in a nucleus roughly $10^{-15}\,\text{m}$ across, some $10^{4}$ – $10^{5}$ times smaller than the atom itself.
The nuclear atom pictures electrons as occupying the vast remaining volume, giving atoms their large apparent size despite negligible electron mass.
Key idea: rare, large-angle scattering of alpha particles reveals a small, dense, positively charged nucleus surrounded by mostly empty space where electrons reside.

The core insight

Deflection angle tells you about charge concentration: a diffuse charge distribution can only nudge a fast-moving alpha particle slightly, but a point-like concentration of charge can turn it back entirely if the alpha particle's aim is close enough to head-on.

The physical picture

Picture an alpha particle — two protons and two neutrons bound together, carrying charge $+2e$ — fired at a sheet of gold foil about $4 \times 10^{-7}\,\text{m}$ thick, roughly 2000 atoms deep. A circular screen coated with zinc sulfide surrounds the foil; every time an alpha particle strikes the screen, it produces a tiny flash of light that an observer counts by eye through a microscope, rotating the detector to sample different angles. If Thomson's plum-pudding model were correct, the positive charge of each gold atom would be smeared over the entire atomic volume, about $10^{-10}\,\text{m}$ across. An approaching alpha particle would feel only a weak, gradually varying force, deflected by at most a degree or two even after passing through thousands of atoms. Rutherford's student Ernest Marsden instead found that roughly 1 in 8000 alpha particles deflected by more than 90°, some coming almost straight back toward the source. Rutherford later remarked that this was "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The resolution requires concentrating essentially all of the atom's positive charge and nearly all its mass into a region far smaller than the atom itself. Only a strong, sharply localized Coulomb repulsion can reverse the momentum of a fast, massive alpha particle, and only a rare, nearly head-on encounter with that concentrated charge produces such a large deflection. Most alpha particles, aimed at random points across the foil, pass nowhere near a nucleus and sail through with negligible deflection — consistent with the overwhelming majority of observed trajectories.

Figure

Rutherford's gold-foil experiment: most α particles pass through; rare close encounters with a nucleus scatter sharply, some backward.

The picture that emerges is the nuclear atom: a nucleus of radius near $10^{-15}\,\text{m}$ to $10^{-14}\,\text{m}$ , holding essentially all the mass and all the positive charge, surrounded by electrons occupying a region roughly $10^{-10}\,\text{m}$ across. The ratio of these two length scales — about 10,000 to 100,000 — is why matter, built almost entirely of empty space at the atomic scale, nonetheless feels solid: electromagnetic forces between the outer electron clouds of neighboring atoms prevent bulk matter from collapsing.

Figure

Scale comparison: nuclear radius versus atomic radius, drawn to logarithmic proportion.

The core relations

The scattering behavior follows from Coulomb repulsion between the alpha particle and the nucleus. Three relations summarize the quantitative reasoning used to interpret the experiment.

Coulomb's law for the repulsive force between the alpha particle (charge $+2e$ ) and a nucleus of atomic number $Z$ (charge $+Ze$ ), separated by distance $r$ :

F = \frac{k \,(2e)(Ze)}{r^2}

where:

$F$ is the repulsive force (N),
$k = 8.99\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is Coulomb's constant,
$e = 1.602\times10^{-19}\,\text{C}$ is the elementary charge,
$Z$ is the atomic number of the target nucleus (79 for gold),
$r$ is the separation between the alpha particle and the nucleus (m).

Distance of closest approach for a head-on collision, found by equating the alpha particle's initial kinetic energy to the electrostatic potential energy at the turning point:

d = \frac{k\,(2e)(Ze)}{E_k}

where:

$d$ is the closest approach distance (m), an upper bound on the nuclear radius,
$E_k$ is the initial kinetic energy of the alpha particle (J),
other symbols as above.

Fraction of particles scattered, relating the observed scattering to the size of the target nucleus (Rutherford scattering formula, angle-dependence form omitted; only the geometric proportionality is needed here):

\frac{N_{\text{scattered}}}{N_{\text{incident}}} \propto \left(\frac{Z\,e^2}{E_k}\right)^{2}

where:

$N_{\text{scattered}}/N_{\text{incident}}$ is the observed fraction of alpha particles deflected beyond a given angle,
the proportionality shows that heavier nuclei (larger $Z$ ) and slower alpha particles (smaller $E_k$ ) scatter more strongly.

Derivation AEstimating the nuclear radius from the distance of closest approach

Goal: show how a measured deflection angle and a known alpha-particle energy give an upper bound on the size of the gold nucleus, using only energy conservation and Coulomb's law.

Step 1. Far from the nucleus, the alpha particle carries only kinetic energy $E_k = \tfrac12 m v^2$ ; the electrostatic potential energy is negligible because $r$ is large.

Step 2. As the alpha particle approaches head-on, it slows down and converts kinetic energy into electrostatic potential energy. At the single instant of closest approach the particle is momentarily at rest (for the head-on case), so all its initial kinetic energy has become potential energy:

E_k = \frac{k\,(2e)(Ze)}{d}

Step 3. Solve for $d$ , the distance of closest approach, by isolating it algebraically:

d = \frac{k\,(2e)(Ze)}{E_k}

Step 4. Insert typical experimental numbers. Rutherford's alpha particles had kinetic energy near $E_k = 7.7\,\text{MeV} = 1.23\times10^{-12}\,\text{J}$ , and gold has $Z = 79$ :

d = \frac{(8.99\times10^{9})(2)(1.602\times10^{-19})(79)(1.602\times10^{-19})}{1.23\times10^{-12}}

Step 5. Evaluate the numerator first:

(8.99\times10^{9})(2)(1.602\times10^{-19})(79)(1.602\times10^{-19}) = 3.64\times10^{-26}\ \text{J}\,\cdot\,\text{m}

Step 6. Divide by the kinetic energy to get $d$ :

d = \frac{3.64\times10^{-26}\ \text{J}\,\cdot\,\text{m}}{1.23\times10^{-12}\ \text{J}} = 2.96\times10^{-14}\,\text{m}

Step 7. Since the alpha particle never got any closer than this without being stopped by an even stronger repulsion than pure Coulomb repulsion predicts, $d \approx 3\times10^{-14}\,\text{m}$ is an upper bound on the gold nuclear radius. This is already some 4000 times smaller than the accepted atomic radius of gold, near $1.4\times10^{-10}\,\text{m}$ , and it is consistent with the modern measured gold nuclear radius of about $7\times10^{-15}\,\text{m}$ (the modern value is smaller because real nuclei also involve short-range nuclear forces not captured by pure Coulomb repulsion at very short range).

Worked examples

Example 1 — Distance of closest approach for a slower alpha particle striking silver. Suppose an alpha particle with kinetic energy $E_k = 5.30\,\text{MeV}$ is fired head-on at a silver nucleus ( $Z = 47$ ). Find the distance of closest approach. Convert energy to joules:

E_k = 5.30\times10^{6}\,\text{eV} \times 1.602\times10^{-19}\,\text{J/eV} = 8.49\times10^{-13}\,\text{J}

Apply the closest-approach relation:

d = \frac{k\,(2e)(Ze)}{E_k} = \frac{(8.99\times10^{9})(2)(1.602\times10^{-19})(47)(1.602\times10^{-19})}{8.49\times10^{-13}}

Evaluate the numerator:

(8.99\times10^{9})(2)(1.602\times10^{-19})(47)(1.602\times10^{-19}) = 2.166\times10^{-26}\ \text{J}\,\cdot\,\text{m}

Divide:

d = \frac{2.166\times10^{-26}}{8.49\times10^{-13}} = 2.55\times10^{-14}\,\text{m}

This is smaller in $Z$ but the particle is also slower than in the earlier gold example, and the two effects partly offset; the resulting distance, $2.55\times10^{-14}\,\text{m}$ , is still roughly four orders of magnitude smaller than a typical atomic radius, confirming again that essentially all positive charge is packed into a region far tinier than the atom.

Limiting case check: if the alpha particle's kinetic energy were doubled to $E_k = 10.60\,\text{MeV}$ while $Z$ stayed at 47, the formula predicts $d$ should be cut exactly in half, since $d \propto 1/E_k$ :

d' = \frac{2.166\times10^{-26}}{1.698\times10^{-12}} = 1.28\times10^{-14}\,\text{m} \approx \frac{2.55\times10^{-14}}{2}

Faster alpha particles get closer before the repulsion turns them around — exactly the physically expected trend.

Example 2 — Comparing scattering rates for two target metals. Using the same alpha-particle energy, compare the expected large-angle scattering fraction from a copper foil ( $Z = 29$ ) to that from a gold foil ( $Z = 79$ ), using the proportionality

\frac{N_{\text{scattered}}}{N_{\text{incident}}} \propto Z^2

(holding $E_k$ fixed, since the fraction depends on $Z^2/E_k^2$ and $E_k$ is unchanged). Form the ratio of the two scattering fractions:

\frac{(N_{\text{scattered}}/N_{\text{incident}})_{\text{gold}}}{(N_{\text{scattered}}/N_{\text{incident}})_{\text{copper}}} = \left(\frac{Z_{\text{Au}}}{Z_{\text{Cu}}}\right)^2 = \left(\frac{79}{29}\right)^2

Evaluate the ratio inside the parentheses first:

\frac{79}{29} = 2.724

Square it:

2.724^2 = 7.42

So gold foil should produce roughly 7.4 times more large-angle scattering events than copper foil under identical beam conditions — a testable, falsifiable prediction that historically matched experiment and pinned down the $Z^2$ dependence of nuclear charge on scattering strength.

Limiting case check: if the two targets had equal atomic number, $Z_{\text{Au}} = Z_{\text{Cu}}$ , the ratio above reduces to exactly 1, meaning identical scattering rates — as it must, since with equal nuclear charge there is no physical reason for one foil to scatter more than the other.

Limiting cases & common traps

Situation	What actually happens	Common mistake
Alpha particle aimed far from any nucleus	Passes through with negligible deflection	Assuming every particle should deflect somewhat, contradicting the observed straight-through majority
Nearly head-on aim at a nucleus	Large deflection, occasionally back-scattered	Thinking back-scattering requires a solid, continuous positive sphere (plum-pudding), when it actually requires a point-like concentrated charge
Foil made thicker	More scattering events overall, but each nucleus still acts independently	Believing thickness changes the mechanism, rather than just the number of chances for a close encounter
Comparing metals of different $Z$	Scattering fraction scales as $Z^2$	Assuming scattering scales linearly with $Z$
Nucleus radius vs. atomic radius	Nucleus is $10^4$ – $10^5$ times smaller than the atom	Picturing the nucleus and electron cloud as comparable in size, as a simple two-circle diagram might mislead one to think

The chemist behind it

Ernest Rutherford was a New Zealand-born physicist working at the University of Manchester when he directed Hans Geiger and Ernest Marsden's gold-foil experiments between 1908 and 1913. Rutherford had already earned a Nobel Prize in Chemistry in 1908 for earlier work on radioactive decay and the classification of alpha and beta radiation — work that itself depended on understanding the properties of the Subatomic Particles: Electron, Proton, Neutron then only recently identified. He initially expected the gold-foil results to confirm Thomson's plum-pudding model; instead the anomalous back-scattering led him, in 1911, to propose the nuclear model of the atom. Rutherford's reasoning was purely quantitative: he used Coulomb's law and conservation of energy, exactly the relations developed above, to show that only a concentrated positive charge of nuclear dimensions could reproduce the observed scattering pattern. This nuclear picture set the stage for the definitions of atomic number and mass number that quantify exactly how much charge and mass the nucleus carries, covered under Atomic Number and Mass Number.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · Directed the gold-foil scattering experiments at the University of Manchester and proposed the nuclear model of the atom in 1911, building on his earlier Nobel Prize-winning studies of radioactivity.

Ask the tutor: How does the number of protons in this newly discovered nucleus define the identity of an element, and what does that have to do with atomic number and mass number?

Why did most alpha particles pass straight through the gold foil?

Because the atom is mostly empty space. The nucleus occupies only about one part in $10^{12}$ to $10^{15}$ of the atom's volume, so the overwhelming majority of alpha particles never pass close enough to any nucleus to feel a strong force, and they travel through the foil in an almost straight line.

Why does the fraction of large-angle scattering events depend on

Z^2

rather than

Z

The scattering angle depends on the strength of the Coulomb force, which is proportional to the product of the two charges, $(2e)(Ze)$ , so it already carries one factor of $Z$ . The precise Rutherford scattering formula additionally involves the square of the force term in computing scattering probability, producing an overall $Z^2$ dependence in the observed scattering rate. This is exactly why gold, with its very high atomic number, was chosen for the classic experiment — it scatters alpha particles far more dramatically than a lighter metal would.

Does the gold-foil experiment tell us the nucleus contains protons and neutrons?

Not directly. The experiment only established that positive charge and most of the mass are concentrated in a tiny nucleus; it measured the nuclear charge (from the scattering pattern) and inferred a nuclear size. The internal composition of the nucleus in terms of individual protons and neutrons — the particles introduced under Subatomic Particles: Electron, Proton, Neutron — was worked out through later experiments, including the discovery of the proton itself by Rutherford in 1917 and the neutron by Chadwick in 1932.

How was Rutherford so confident the nucleus was small, rather than just very dense?

The kinetic energy of the alpha particles was known from their radioactive source, so the distance of closest approach could be calculated directly from energy conservation, as in the derivation above. Because that calculated distance came out to about $10^{-14}\,\text{m}$ — far smaller than the known atomic radius near $10^{-10}\,\text{m}$ — the positive charge had to be confined within that small distance, regardless of exactly how mass and charge distribute inside it.

Why is a fast, light electron not similarly confined to a tiny volume like the nucleus?

The gold-foil experiment probes positive charge and heavy mass, which is where nearly all of an atom's mass resides; electrons are roughly 7300 times lighter than a proton and do not contribute meaningfully to alpha-particle scattering by momentum transfer in this experiment. Electron behavior and their spatial distribution around the nucleus needed separate experimental and theoretical developments beyond the scope of the gold-foil result itself.