Rutherford's gold-foil experiment — nuclear model: Visual walkthrough

Step 1 — What is an α-particle, and what is it flying at?

WHAT. Picture a single fast bullet — the α-particle. It is a helium nucleus: two protons stuck together, so its electric charge is $+ 2 e$ , where $e$ is the elementary charge (the amount of charge on one proton, a fixed number of nature, $e = 1.6 \times 1 0^{- 19}$ coulombs — a coulomb is just the unit we measure charge in). It heads straight at a gold atom's core, the nucleus, which has charge $+ Z e$ . Here $Z$ is the atomic number — literally how many protons are in that nucleus. Gold has $Z = 79$ .

WHY these two charges matter. Both are positive. Two positive charges push each other apart — like two north poles of magnets, but for electricity. That push is the whole story of what follows.

PICTURE. The blue bullet ( $+ 2 e$ ) approaches the yellow nucleus ( $+ Z e$ ) head-on along a straight line.

Figure — Rutherford's gold-foil experiment — nuclear model

Step 2 — Why the α-particle carries energy of motion

WHAT. A moving object stores kinetic energy — "energy of motion". Its size is $KE = fixed number \frac{1}{2} mass of α m speed, squared v^{2} .$

WHY this exact form. The $v^{2}$ (speed times itself) says: double the speed and the stored energy quadruples. A fast α therefore carries a lot of energy — that is its "budget" for pushing into the nucleus. The $m$ says a heavier bullet at the same speed carries more.

PICTURE. A speed bar for the α: full speed far away, full kinetic-energy tank.

Step 3 — Why the push stores energy: potential energy

WHAT. When two positive charges are near each other, the system stores electrostatic potential energy — call it $U$ . It is the energy hidden in the arrangement of the charges, ready to be released as motion the instant you let go. For two charges $q_{1}$ and $q_{2}$ a distance $r$ apart: $U = a constant, call it k \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2} the two charges}{their separation r} .$

WHY this shape. Look at the $r$ in the bottom. As $r$ shrinks (charges closer), $U$ grows — the arrangement becomes more "loaded", like a spring squeezed tighter. As $r \to 0$ , $U \to \infty$ : you can never fully close the gap between two like charges, because that would need infinite energy. This is exactly Coulomb's Law expressed as stored energy instead of as a force.

The clump $\frac{1}{4 π ε _{0}}$ is just a conversion factor of nature — $ε_{0}$ (the "permittivity of free space") sets how strongly charges talk to each other in empty space. To keep arithmetic sane we bundle it: $k = \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} (in SI units) .$

PICTURE. The potential-energy hill: $U$ rises steeply as the blue bullet nears the yellow core.

Step 4 — The trade: motion turns into the push (energy conservation)

WHAT. As the α climbs the potential-energy hill, it slows down. Its kinetic-energy tank drains exactly into potential energy — nothing leaks away (no friction in vacuum, the Coulomb interaction is clean). This is conservation of energy: $KE + U = constant, always .$

WHY this is the master idea. We don't need to track the messy force at every instant. We only compare two snapshots — far away and the turning point — because the total is the same in both. That is why energy conservation is the right tool here rather than solving the motion step by step.

PICTURE. Two tanks swapping: the KE tank empties as the $U$ tank fills, their sum flat.

Step 5 — The turning point: define $r_{0}$ and set the two snapshots equal

WHAT. Call the closest separation the α ever reaches $r_{0}$ — the distance of closest approach. At that instant the speed is zero (the bullet is reversing, like a ball thrown straight up at the peak of its flight). Equate the total energy of Snapshot A and Snapshot B: $all KE, at start \frac{1}{2} m v^{2} = all PE, at r_{0} k \frac{( 2 e ) ( Z e )}{r _{0}} .$

WHY the right side looks like that. Plug the two real charges into the potential-energy formula: $q_{1} = 2 e$ (the α), $q_{2} = Z e$ (the nucleus), separation $= r_{0}$ . Left side is the α's starting fuel; right side is where all that fuel has gone.

PICTURE. The bullet frozen at $r_{0}$ , velocity arrow shrunk to zero, hill height = full starting KE.

Step 6 — Solve for $r_{0}$ and read what it means

WHAT. Rearrange the equality to isolate $r_{0}$ . Multiply both sides by $r_{0}$ , then divide by $\frac{1}{2} m v^{2}$ : $r_{0} = k \frac{2 ( 2 e ) ( Z e )}{m v ^{2}} = \frac{1}{4 π ε _{0}} \frac{4 Z e ^{2}}{m v ^{2}}$

Term by term:

$4 Z e^{2}$ ::: the two charges multiplied out — $(2 e) (Z e) = 2 Z e^{2}$ , and the leftover $\times 2$ from clearing the $\frac{1}{2}$ gives $4 Z e^{2}$ . Bigger $Z$ ⇒ stronger push ⇒ the α is stopped farther out.
$m v^{2}$ (the denominator) ::: the α's energy budget. A faster or heavier α punches closer in, making $r_{0}$ smaller.
$k = 1/4 π ε_{0}$ ::: the universal strength of the electric interaction.

WHY this number matters. $r_{0}$ is an upper bound on the size of the nucleus: the α clearly never touched anything solid out there, so the actual nucleus must be even smaller than $r_{0}$ . Compare $r_{0} \sim 1 0^{- 14}$ m to the whole atom at $\sim 1 0^{- 10}$ m — the core is ten-thousand times smaller across. That gaping emptiness is the nuclear model, and it feeds directly into the Bohr Model of the Atom and Centripetal Force and Circular Motion balance in the parent note.

PICTURE. A ruler comparing $r_{0}$ to the full atom — a dot inside a stadium.

Step 7 — Edge & degenerate cases (so nothing surprises you)

WHAT / WHY, case by case:

Not head-on (a glancing shot). If the α is aimed off-center, it never reaches $r_{0}$ — it swerves away sooner, curving along a hyperbola, and deflects by only a small angle. Our formula is the extreme case (perfectly head-on) giving the smallest possible closest approach. Every other path stays farther out.
Very slow α ( $v$ tiny). The denominator $m v^{2}$ shrinks ⇒ $r_{0}$ grows huge: a feeble α is turned back far away, never sampling the nucleus at all. Useless as a probe.
Very fast α (imaginary future experiment). $r_{0}$ shrinks toward the true nuclear radius. Push hard enough and the α would reach the nucleus — at which point Coulomb's inverse-square law breaks down and nuclear forces take over. That is how later physicists measured real nuclear sizes.
Neutral or negative target ( $Z \to 0$ or opposite sign). With $Z = 0$ there is no charge, no push, $r_{0}$ is meaningless — the α sails on. With an attracting target the formula's sign flips and there is no turning point at all; the α is pulled in, not out. This is why backscatter specifically demands a concentrated positive core.

PICTURE. Three trajectories — head-on (stops at $r_{0}$ ), glancing (swerves early), and slow (turns far away).

The one-picture summary

Everything above is one sentence: the α trades all its motion-energy for the arrangement-energy of two like charges, and the separation where the trade completes is $r_{0}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Imagine rolling a magnet up a ramp toward another magnet that's facing it the wrong way, so they shove each other apart. You roll it fast — it has lots of "roll energy." As it climbs toward the other magnet, the shove gets stronger and stronger, and the magnet slows. At some spot it stops dead — all its roll energy has become "shove energy" stored between the two magnets — and then it comes rolling right back at you.

The α-particle is that first magnet; the gold nucleus is the second; the "shove" is the electric repulsion between two positive charges. The stopping spot is $r_{0}$ . We didn't need to watch every inch of the motion — we just said "the roll energy at the start equals the shove energy at the stopping spot," set the two equal, and solved. Out popped a distance about ten-thousand times smaller than a whole atom. That tiny stopping distance is our proof that all the atom's positive charge is crammed into a speck at the center — the nucleus — with vast emptiness all around it.

Recall Quick self-check

Why can we ignore the messy force and just equate two snapshots? ::: Energy is conserved, so KE+U is the same far away and at the turning point.
At $r_{0}$ , what is zero — the speed or the force? ::: The speed. The force is at its maximum there.
Which way does $r_{0}$ move if the α is faster? ::: Smaller — more energy punches closer in.
Is $r_{0}$ the nucleus's true size? ::: No — it's an upper bound; the real nucleus is smaller.

Parent: Rutherford's gold-foil experiment — nuclear model · Prerequisites: Coulomb's Law, Centripetal Force and Circular Motion · Where it leads: Bohr Model of the Atom, Atomic Spectra (Line Spectra), Discovery of the Proton and Neutron.