Preview

Rutherford Scattering

Discover the nuclear model of the atom through alpha particle scattering

1909, Manchester. Hans Geiger and Ernest Marsden are firing alpha particles — helium nuclei, about 5 MeV apiece — at a gold foil only hundreds of atoms thick, counting tiny flashes on a zinc sulfide screen. Almost every alpha sails straight through, exactly as expected: J.J. Thomson's plum-pudding atom, its positive charge spread softly through the whole atomic volume, could only muster tiny deflections. Then the flashes start appearing in impossible places. Some alphas bounce sideways. A few — about one in eight thousand — come straight back. Rutherford's own verdict: it was 'as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.' Two years of hard thinking later he had the only explanation that fits: the atom's positive charge, and nearly all its mass, sits in a nucleus one hundred thousand times smaller than the atom itself. Most alphas miss that speck by enormous margins — the atom is mostly empty space. But aim close, and the bare Coulomb repulsion between +2e and +79e becomes a wall of field: the alpha climbs it, stops, and turns around without ever touching anything. In this lab you fly those trajectories yourself — every one a hyperbola carved by the 1/r² force — and you price the turnaround with a single line of energy bookkeeping: all kinetic energy becomes electric potential energy at the distance of closest approach. Aim dead-center first. The shell really does come back.

What you'll be able to do

  • Price the head-on turnaround with pure energy bookkeeping: at the Head-On preset (Z=79, E=5 MeV, b=0) the alpha converts all kinetic energy into Coulomb potential energy, KE = k·2Ze²/d, giving d = 45.50 fm from infinity — and because the panel's launcher sits at 80 fm (already A/80 = 2.84 MeV deep in the field, A = 227.5144 MeV·fm), the displayed Closest approach row stops at 29.0 fm. Both numbers dwarf the ~10⁻¹⁵ m nuclear scale (OpenStax CP2e §30.2): the turnaround happens ~29 nuclear scales from the center, without contact — the atom is mostly empty space
  • Read the scattering angle as a monotone function of impact parameter: cot(θ/2) = 2E·b/(k·2Ze²), so the ghost-fan ladder b = 0/3/6/10/16/25/40 fm prices theory angles 180.0/165.0/150.5/132.5/109.8/84.6/59.3° — strictly decreasing from dead-center backscatter to the glancing whisper, the exact relation the θ-vs-b chart board draws and the HTML's f3 sidebar teaches
  • Certify the energy lever at fixed aim: tan(θ/2) ∝ 1/E — sweep Alpha Energy 1 → 20 MeV at b = 100 fm and the theory angle falls 97.4° → 6.5°; doubling E exactly halves tan(θ/2) (2.275 → 1.138 at b=10, the HTML quiz Q3 feedback's own arithmetic). Faster particles spend less time in the field and are LESS deflected — the counter-intuitive direction is the testable one

Formulas

cot(θ/2)=2bEkkqαqAu\cot(\theta/2) = \frac{2bE_k}{k q_\alpha q_{Au}}
Rutherford scattering formula
dclosest=kqαqAuEkd_{closest} = \frac{kq_\alpha q_{Au}}{E_k}
Distance of closest approach

Make a prediction

Load the Head-On Collision preset: a 5 MeV alpha particle aimed dead-center (b = 0) at a gold nucleus, Z = 79. Nothing but the Coulomb field stands in its way. What does the alpha do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Check

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Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It slows to a dead stop tens of femtometers from the center — never touching the nucleus — and turns straight back at θ = 180°; at the turnaround all 5 MeV of kinetic energy sits in the Coulomb field as potential energy

Energy conservation runs the whole show: the alpha climbs the Coulomb hill until all its kinetic energy has become electric potential energy, KE = k·2Ze²/d, which prices the stop at d = 45.5 fm for a 5 MeV alpha arriving from far away. This panel's launcher sits at 80 fm — already 2.84 MeV deep in the field — so the displayed Closest approach row stops at 29.0 fm; same law, honest launch pad. Either way the stop happens ~29 nuclear scales from a ~10⁻¹⁵ m nucleus: nothing touches anything, and then the field hands every joule back, sending the alpha out the way it came — θ = 180.0° on the panel. Option B is Thomson's plum-pudding atom making its own prediction: 'The incident massive alpha particles would suffer only small deflections in such a model' (OpenStax College Physics 2e §30.2, verbatim). Geiger and Marsden watched about 1 in 8000 alphas come straight back — a result the soft uniform charge cannot produce and the point nucleus produces on demand, every time you fire this preset. Option C is the hard-ball misreading: the alpha reverses course while still tens of femtometers away, turned by a 1/r² field that acts at a distance — 'These particles interact with matter mostly via the Coulomb force' (CP2e §30.2). No wall, no contact, no exception to energy conservation — just a very small, very charged nucleus.

Quiz (0/3)

An alpha particle aimed directly at a gold nucleus (Z=79). What is the closest approach distance?

Why did most alpha particles pass through the foil almost undeflected?

How did Rutherford estimate the nucleus size from the maximum scattering angle?

You can now

  • Price the head-on turnaround with pure energy bookkeeping: at the Head-On preset (Z=79, E=5 MeV, b=0) the alpha converts all kinetic energy into Coulomb potential energy, KE = k·2Ze²/d, giving d = 45.50 fm from infinity — and because the panel's launcher sits at 80 fm (already A/80 = 2.84 MeV deep in the field, A = 227.5144 MeV·fm), the displayed Closest approach row stops at 29.0 fm. Both numbers dwarf the ~10⁻¹⁵ m nuclear scale (OpenStax CP2e §30.2): the turnaround happens ~29 nuclear scales from the center, without contact — the atom is mostly empty space
  • Read the scattering angle as a monotone function of impact parameter: cot(θ/2) = 2E·b/(k·2Ze²), so the ghost-fan ladder b = 0/3/6/10/16/25/40 fm prices theory angles 180.0/165.0/150.5/132.5/109.8/84.6/59.3° — strictly decreasing from dead-center backscatter to the glancing whisper, the exact relation the θ-vs-b chart board draws and the HTML's f3 sidebar teaches
  • Certify the energy lever at fixed aim: tan(θ/2) ∝ 1/E — sweep Alpha Energy 1 → 20 MeV at b = 100 fm and the theory angle falls 97.4° → 6.5°; doubling E exactly halves tan(θ/2) (2.275 → 1.138 at b=10, the HTML quiz Q3 feedback's own arithmetic). Faster particles spend less time in the field and are LESS deflected — the counter-intuitive direction is the testable one

Step-by-step

  1. Alpha particles launch automatically toward the gold nucleus while a ghost fan of trajectories shows the whole scattering family at once.
  2. Decrease the impact parameter to aim closer to the center and watch the deflection climb toward a full 180° bounce-back; raise the alpha energy and see the deflection shrink instead.
  3. Read the scattering angle and closest-approach rows in the data panel, and tour the three presets from head-on collision to glancing shot.

Key formulas

  • cot(θ/2)=2bEkkqαqAu\cot(\theta/2) = \frac{2bE_k}{k q_\alpha q_{Au}}Rutherford scattering formula
  • dclosest=kqαqAuEkd_{closest} = \frac{kq_\alpha q_{Au}}{E_k}Distance of closest approach

Frequently asked questions

An alpha particle aimed directly at a gold nucleus (Z=79). What is the closest approach distance?
D = k×q_α×q_Au/KE; use KE=5MeV=8×10⁻¹³J; q_α=3.2×10⁻¹⁹C; q_Au=79e.
Why did most alpha particles pass through the foil almost undeflected?
The nucleus is tiny compared to atomic size; most alphas pass far from any nucleus.
How did Rutherford estimate the nucleus size from the maximum scattering angle?
Closest approach for direct hits gives upper bound on nuclear radius.