Alpha-particle scattering experiment

Fire a beam of bullets at a sheet of tissue paper and every bullet punches straight through — nobody is surprised. Now imagine firing those same bullets at tissue paper and watching one bullet in eight thousand bounce clean back at you, as if it had struck a wall of steel hidden somewhere inside the paper. That is almost exactly what happened in 1909 in Ernest Rutherford's laboratory in Manchester, when Hans Geiger and Ernest Marsden fired fast, heavy alpha particles at a foil of gold only a few hundred atoms thick. At the time, the accepted picture of the atom was J. J. Thomson's "plum pudding" model: a soft, spread-out ball of positive charge with electrons studded through it like raisins in a pudding. If that picture were right, an alpha particle — thousands of times heavier than an electron — should barrel through the diffuse positive charge with at most a gentle wobble, the way a cannonball shrugs off a hail of pebbles. Geiger and Marsden set out simply to measure how much wobble there was. What they found instead astonished Rutherford so much that he later said it was "as if you fired a fifteen-inch shell at a piece of tissue paper and it came back and hit you." This article rebuilds that famous experiment from the ground up: the apparatus, the three observations that broke the old model, the chain of reasoning that replaced it with a nucleus, and the one piece of arithmetic — the distance of closest approach — that turns a qualitative story into a hard number for how small that nucleus must be. Along the way you will see exactly why gold, why thin, and why alpha particles, and you will work the same closest-approach calculation twice, once at board level and once with a twist that examiners like to spring in competitive exams.

Key takeaways

Rutherford, Geiger and Marsden fired alpha particles ( $+2e$ , from a radioactive source) at a thin gold foil and counted scintillations on a ZnS screen at every angle $\theta$ .
Observation 1: about 99% of alpha particles passed through almost undeviated — the atom is mostly empty space.
Observation 2: roughly 1 in 8000 alpha particles was scattered through more than $90°$ — something inside the atom is small, heavy and strongly charged.
Observation 3: a very small number bounced back through angles near $180°$ — that something carries a large positive charge concentrated in a tiny volume.
These three observations together killed Thomson's plum-pudding model and led Rutherford to propose the nuclear model: a tiny, dense, positively charged nucleus at the atom's centre, with electrons orbiting far outside it.
The distance of closest approach $r_0$ , found by equating the alpha particle's kinetic energy to the Coulomb potential energy at head-on impact, gives an experimental upper bound on nuclear size — about $10^{-14}$ m, some $10^4$ times smaller than the atom itself.
Key idea: because a tiny fraction of alpha particles bounce back off a thin gold foil while the vast majority sail straight through, the atom's positive charge and nearly all its mass must be concentrated in a nucleus about $10^{-14}$ m across, surrounded by mostly empty space.

The core insight

The rare large-angle scattering of alpha particles off a thin gold foil proves that an atom's positive charge and mass are concentrated in a tiny, dense nucleus surrounded by mostly empty space, and equating kinetic energy to Coulomb potential energy at head-on approach turns that proof into a numerical upper limit on the nucleus's size.

Where it fits in the syllabus

The alpha-particle scattering experiment opens CBSE Class 12 Physics, Unit IX ("Atoms and Nuclei"), and is squarely inside the current 2025-26 prescribed scope: the experiment itself, and the qualitative reasoning that leads to Rutherford's nuclear model, are examinable as described here. You are expected to describe the apparatus, state the three key observations, and reason from each observation to a conclusion about atomic structure — not merely recite facts. The distance-of-closest-approach calculation is standard board-level numerical work built on Coulomb's law and energy conservation, both already familiar from Class 12 Electrostatics. This experiment is purely the experimental foundation; it says nothing about how electrons are arranged or how they move, which is the job of the Bohr model covered later in the same chapter.

The physical picture

Picture the apparatus first, because every conclusion in this topic is read directly off it. A radioactive source emits a narrow, collimated beam of alpha particles — helium nuclei, charge $+2e$ , moving at a few percent of the speed of light — aimed at an extremely thin gold foil, only a few hundred atoms thick. Surrounding the foil is a circular screen coated with zinc sulphide (ZnS), a material that emits a tiny flash of light (a scintillation) wherever a charged particle strikes it. A microscope mounted on the screen can be rotated to any angle $\theta$ measured from the original beam direction, so the experimenters could count, angle by angle, how many alpha particles emerged in each direction after encountering the foil. The entire setup sits inside a vacuum chamber, since alpha particles are easily absorbed by ordinary air.

Figure

A side view of the scattering apparatus showing an alpha source firing a beam at a thin gold foil, with most particles passing straight through, a few deflected at moderate angles, and a rare particle bounced almost straight back toward a circular detector screen

Now read the geometry mathematically. Suppose an alpha particle of charge $+2e$ heads directly at a gold nucleus of charge $+Ze$ (for gold, $Z=79$ ) in a perfectly head-on collision. As it approaches, the repulsive Coulomb force slows it down, converting kinetic energy into electrostatic potential energy, until at some closest distance $r_0$ the particle is momentarily at rest and all its kinetic energy has become potential energy. Energy conservation for this head-on case gives

\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}

where: $m$ and $v$ are the mass and initial speed of the alpha particle far from the nucleus, so $\tfrac12 mv^2$ is its initial kinetic energy; $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is Coulomb's constant; $2e$ and $Ze$ are the charges of the alpha particle and the gold nucleus; and $r_0$ is the distance of closest approach, measured from the centre of the nucleus. Physically, this equation says the alpha particle trades all of its forward motion for stored electrical energy at the single instant it is nearest the nucleus, then the same repulsion flings it straight back out the way it came — a head-on collision is the one case where the particle genuinely stops before reversing. Solving for $r_0$ shows it shrinks as kinetic energy increases and grows with the target's charge $Z$ :

r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{\left(\tfrac12 mv^2\right)}

For real alpha sources (a few MeV of kinetic energy) striking gold, this works out to $r_0$ of order $10^{-14}$ m — some ten thousand times smaller than the accepted size of an atom, about $10^{-10}$ m. Since no alpha particle was ever seen to be deflected by more than this closest-approach calculation would allow, $r_0$ stands as an experimental upper limit on the size of the nucleus, not a precise measurement of it — real gold nuclei, as later scattering experiments with higher-energy particles showed, are in fact several times smaller than this crude estimate.

Derivation ADistance of closest approach in a head-on alpha-nucleus encounter

The goal is to turn the qualitative observation "a few alpha particles bounce almost straight back" into a hard number for how close an alpha particle can get to a nucleus, using nothing beyond Coulomb's law and conservation of energy already familiar from electrostatics. The idea is to isolate the single cleanest trajectory — a particle aimed dead-centre at the nucleus — because along this path the particle's velocity, and hence its kinetic energy, drops smoothly to exactly zero at one well-defined point, letting us equate energies without any messy angle-dependent geometry.

Step 1. Set up energy conservation for the head-on path. Far from the nucleus (formally, at infinite separation) the alpha particle has only kinetic energy $\tfrac12 mv^2$ and negligible potential energy, since the Coulomb potential energy $\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r}$ falls to zero as $r\to\infty$ . As it moves toward the nucleus along the straight line joining them, the repulsive force continuously does negative work on it, converting kinetic energy into potential energy while total mechanical energy stays fixed.

Step 2. Identify the turning point. Because the approach is exactly head-on, the particle's velocity component along the line of approach shrinks steadily; at the point of closest approach it must be instantaneously zero — if it were not, the particle would still be moving toward the nucleus and would not yet be at its closest point. At that instant all of the initial kinetic energy has been converted into potential energy, with none left over as kinetic energy.

Step 3. Write the energy-conservation equation at that instant. Initial kinetic energy equals potential energy at closest approach: $\tfrac12 mv^2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}$ , where $r_0$ is the unknown distance of closest approach.

Step 4. Solve for $r_0$ by rearranging: $r_0 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{\left(\tfrac12 mv^2\right)} = \dfrac{4Ze^2}{4\pi\varepsilon_0\, mv^2}$ . This says $r_0$ is directly proportional to the target's atomic number $Z$ and inversely proportional to the alpha particle's kinetic energy — a faster alpha particle, or a lighter target nucleus, gets closer before turning back.

Step 5. Read the physical meaning of the result. Because $r_0$ is computed for the single most extreme case — a dead-centre hit, which is also the rarest kind of encounter — it represents the closest any alpha particle can plausibly get, and hence sets an upper bound on the nucleus's radius: the true nucleus cannot be larger than $r_0$ , or head-on alpha particles would have stopped even farther out. Plugging in typical numbers (a few MeV alpha particles on gold, $Z=79$ ) gives $r_0$ of order $10^{-14}$ m, roughly ten thousand times smaller than the atom's overall radius of about $10^{-10}$ m — the numerical heart of the claim that the nucleus is tiny and dense.

Worked examples

Board level - closest approach for a 5.3 MeV alpha particle on gold

An alpha particle of kinetic energy $5.3$ MeV is fired head-on at a gold nucleus ( $Z=79$ ). Estimate the distance of closest approach $r_0$ .

Step 1. Convert the kinetic energy to joules: $5.3\ \text{MeV} = 5.3\times10^{6}\times1.6\times10^{-19}\ \text{J} = 8.48\times10^{-13}\ \text{J}$ .

Step 2. Write the closest-approach condition: $KE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}$ , so $r_0 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{KE}$ .

Step 3. Substitute $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ , $Z=79$ , $e = 1.6\times10^{-19}$ C: numerator $= 9\times10^{9}\times2\times79\times(1.6\times10^{-19})^2 = 9\times10^{9}\times158\times2.56\times10^{-38} = 3.64\times10^{-26}$ .

Step 4. Divide by the kinetic energy: $r_0 = \dfrac{3.64\times10^{-26}}{8.48\times10^{-13}} \approx 4.3\times10^{-14}\ \text{m}$ .

Answer: $r_0 \approx 4.3\times10^{-14}$ m, about $43$ femtometres — roughly $10^4$ times smaller than the atom's overall size of $\sim10^{-10}$ m, exactly the scale the experiment demands.

Competitive level - comparing closest approach across two targets

A different alpha source of kinetic energy $7.7$ MeV is fired head-on, once at a silver foil ( $Z=47$ ) and once at the same gold foil as above ( $Z=79$ ). Find $r_0$ for silver, and state the ratio $r_0(\text{gold})/r_0(\text{silver})$ without recomputing everything from scratch.

Step 1. Convert kinetic energy: $7.7\ \text{MeV} = 7.7\times10^{6}\times1.6\times10^{-19} = 1.232\times10^{-12}\ \text{J}$ .

Step 2. Apply the same closest-approach formula for silver, $Z=47$ : numerator $= 9\times10^{9}\times2\times47\times(1.6\times10^{-19})^2 = 9\times10^{9}\times94\times2.56\times10^{-38} = 2.166\times10^{-26}$ .

Step 3. Divide by the new kinetic energy: $r_0(\text{silver}) = \dfrac{2.166\times10^{-26}}{1.232\times10^{-12}} \approx 1.76\times10^{-14}\ \text{m}$ .

Step 4. Use the proportionality from the derivation instead of a fresh gold calculation: $r_0 \propto Z/KE$ , so with a different kinetic energy for gold too, first find $r_0(\text{gold})$ at $7.7$ MeV by scaling the board example's result: $r_0(\text{gold, 7.7 MeV}) = 4.3\times10^{-14}\times\dfrac{5.3}{7.7} \approx 2.96\times10^{-14}\ \text{m}$ . The ratio is then $\dfrac{r_0(\text{gold})}{r_0(\text{silver})} = \dfrac{2.96\times10^{-14}}{1.76\times10^{-14}} \approx 1.68$ , matching the expected charge ratio $79/47 \approx 1.68$ exactly, since both were probed at the same kinetic energy.

Answer: $r_0(\text{silver}) \approx 1.76\times10^{-14}$ m at $7.7$ MeV, and gold's closest approach is about $1.68$ times larger than silver's at the same energy, confirming $r_0 \propto Z$ . Timing note: under ninety seconds once the formula is memorised. The efficient route is never to recompute the full fraction twice — extract the proportionality $r_0 \propto Z/KE$ from the first calculation and scale, rather than substituting numbers from scratch for every target and energy combination. The one trap: don't forget to rescale for kinetic energy if it changes between cases, or the "ratio equals $Z$ ratio" shortcut silently gives the wrong answer.

Common traps

Believing most alpha particles were deflected: in fact about 99% went through nearly undeviated, and only about 1 in 8000 was scattered through more than $90°$ — the rare event, not the common one, is what carries the physics.
Calling the large-angle scattering a physical "collision": it is electrostatic (Coulomb) repulsion between two positive charges acting at a distance, not contact between hard surfaces.
Treating $r_0$ as the measured radius of the nucleus: it is only an upper limit, obtained from the single most extreme (head-on) trajectory; the true nucleus is smaller.
Mixing up what the experiment does and does not prove: it establishes only that a small, dense, positively charged nucleus exists — it says nothing about how electrons are arranged or how they move, which the Bohr model addresses separately.
Getting the charge of the alpha particle wrong in the formula: it is $+2e$ (a helium nucleus, two protons and two neutrons), not $+e$ .

Common trapThe nucleus is inferred from a rare event, not from the majority behaviour

The single most common exam slip is reasoning from the wrong observation. The fact that most alpha particles pass through undeviated only tells you the atom is mostly empty space — it says nothing about a nucleus at all. The nucleus is inferred entirely from the rare large-angle and near-180° scattering events. A student who writes "since most particles passed through, the atom has a small nucleus" has the logic backwards: it is precisely because so few particles were deflected strongly that the charged, massive region responsible must be small, occupying only a tiny fraction of the atom's volume, while its ability to occasionally reverse a fast, heavy alpha particle shows that region must carry a large charge and most of the atom's mass.

You should be able to…

Describe the apparatus: alpha source, thin gold foil, circular ZnS scintillation screen, rotatable detector, all inside a vacuum chamber.
State the three key observations (mostly undeviated, about 1 in 8000 beyond $90°$ , a very few near $180°$ ) and connect each one to a conclusion about atomic structure.
Explain why the results rule out Thomson's plum-pudding model and require a small, dense, positively charged nucleus.
Derive and apply the distance-of-closest-approach formula for a head-on alpha-nucleus encounter, and explain why it gives only an upper limit on nuclear size.
Justify the experimental choices: why alpha particles, why gold, and why a thin foil.
State clearly what the experiment does not establish — namely, anything about electron arrangement or energy levels.

Ask the tutor: want to see how the distance of closest approach changes if you swap the alpha particle for a proton, or fire it at a much lighter nucleus like aluminium — ask the AI tutor to walk through the numbers with you.

Why did Rutherford use alpha particles instead of electrons or protons to probe the atom?

Alpha particles were the most energetic, readily available projectile from natural radioactive sources at the time, and being helium nuclei they are relatively heavy and carry a sizeable charge ( $+2e$ ), which makes the Coulomb interaction with a nucleus strong enough to produce a clearly measurable deflection. A lighter particle like an electron would be pushed around far more easily by the atom's individual electrons and would not cleanly probe the central charge the way a heavy, fast alpha particle does.

Why does a thin foil matter so much — would a thick block not show the same thing?

A thin foil, only a few hundred atoms thick, ensures each alpha particle very likely interacts with at most one nucleus on its way through, so the observed scattering angle reflects a single, clean encounter. In a thick block, an alpha particle could be nudged slightly by many nuclei in sequence, and untangling whether a large final deflection came from one strong encounter or many weak ones would make the results impossible to interpret cleanly.

If the nucleus is so small, why does it matter so much for the atom's mass and charge?

Size and mass are different things. The nucleus occupies only about one part in ten thousand of the atom's diameter, yet it holds essentially all of the atom's mass (protons and neutrons are roughly 1800 times heavier than an electron) and all of its positive charge. The electrons, though they define the atom's overall size, contribute almost nothing to its mass — so a tiny nucleus can still dominate both the mass and the charge of the whole atom.

Does this experiment tell us anything about where the electrons are or how they move?

No, and that is a genuinely common point of confusion. The scattering experiment only used positively charged alpha particles interacting with the positively charged nucleus; it says nothing about electron positions, orbits, or energies. Explaining how electrons are arranged around this newly discovered nucleus without the atom collapsing was a separate problem, solved later in the same chapter by Bohr's postulates.