Rutherford's gold-foil experiment — nuclear model: Foundations

0. How to read every formula on the parent page

The parent page throws symbols at you: $e$ , $Z$ , $ε_{0}$ , $r_{0}$ , $v$ , $m$ , $\frac{1}{2} m v^{2}$ , $\frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r}$ . If any of those look like alien runes, this page is for you. We define each in plain words → a picture → why the topic needs it, so that when you return to the parent page not a single symbol is a stranger.

Definition The α-particle (our "bullet")

Plain words: the parent page fires α-particles (alpha-particles) at the foil. An α-particle is simply a helium nucleus — a tightly bound bundle of $2$ protons and $2$ neutrons — that has lost its two electrons. Because it kept the two positive protons but none of the balancing electrons, it carries a small amount of positive charge — the exact symbol for "how much" comes in Section 1, where we will write it as $+ 2 e$ . It has a mass of about $4$ atomic mass units. The picture: a small ball marked with two $+$ signs speeding to the right — a compact positive bullet. Why the topic needs it: everything in the experiment is "what happens when this positive bullet meets the positive nucleus?" We must know the bullet is positive (so it is repelled) and fairly heavy (so only a heavier target can bounce it back). From here on, "α" always means this positive helium-nucleus bullet.

1. Charge, and the symbol $e$

The smallest lump of charge nature hands out is called the elementary charge, written $e$ :

Figure 1 (below) shows both cases side by side: on top, two $+ e$ balls flung apart (repulsion); on the bottom, a $+ e$ and a $- e$ ball pulled together (attraction). Read the yellow arrows as "which way the force pushes each ball."

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

2. Atomic number $Z$

3. The Coulomb force — pushing that grows near

This is the single physics law the whole experiment rests on. See Coulomb's Law for its own home page; here we build just what we need.

Figure 2 (below) plots that force against distance $r$ . Follow the yellow curve: on the right (large $r$ ) the force is a gentle sag near the floor; sweep left toward small $r$ and it rockets upward. The pink dot marks "close = huge push," the blue dot "far = gentle push."

Intuition

Why the $1/ r^{2}$ — the spreading-out picture Figure 3 (below) shows the reason with a picture, not hand-waving. Imagine the charge's influence as a fixed number of "force lines" streaming outward equally in all directions from the charge — like paint sprayed uniformly from a point. Those lines have to cover the whole surface of a sphere around the charge. A sphere of radius $r$ has surface area $4 π r^{2}$ , which grows as $r^{2}$ . So the same number of lines is smeared over an area that grows like $r^{2}$ → the crowding of lines (and hence the force) thins out as $1/ r^{2}$ . Double the distance, the sphere's area quadruples, so the force drops to a quarter. The $4 π$ in $\frac{1}{4 π ε _{0}}$ is literally the "surface area of a unit sphere" bookkeeping. Why it matters here: as the α rushes in and $r$ shrinks, the lines get packed ever tighter and the repulsion explodes upward — that ferocious close-range push is exactly what stops and reverses the bullet.

Now the mysterious constant inside that formula:

Definition

$ε_{0}$ (permittivity of free space) Plain words: $ε_{0}$ ("epsilon-nought") is a fixed number of nature called the permittivity of free space — it just sets the strength of electric forces in empty space. It never changes; you plug in its value and move on. In the whole formula it always appears as the bundle $\frac{1}{4 π ε _{0}}$ , whose numerical value is $\frac{1}{4 π ε _{0}} \approx 9 \times 1 0^{9} N\cdotm^{2} / C^{2} .$ Those units ( $N\cdotm^{2} / C^{2}$ ) are just what is needed so that, after multiplying by charges (in C) and dividing by $r^{2}$ (in $m^{2}$ ), the answer lands in newtons (N). The picture: think of $\frac{1}{4 π ε _{0}}$ as a single "conversion knob" that turns "charges and distance" into "newtons of force." Why the topic needs it: every energy and force estimate on the parent page begins by writing $\frac{1}{4 π ε _{0}}$ and then substituting $9 \times 1 0^{9}$ . It is just a number — don't let the Greek letter scare you.

4. Mass $m$ and speed $v$

5. Kinetic energy $\frac{1}{2} m v^{2}$ and potential energy $U$

Energy is the bookkeeping currency that lets the parent page find the closest distance (called $r_{0}$ , defined in Section 6) without knowing the messy details of the collision.

Definition Electrostatic potential energy

$U$ Plain words: potential energy is stored energy that depends on position. For two charges $q_{1}$ and $q_{2}$ (same meanings as in Coulomb's Law) a distance $r$ apart, moving them takes work, and that work is stored as potential energy $U$ : $U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r} .$ (Notice: same ingredients as Coulomb's force, but $r$ instead of $r^{2}$ — energy, not force.) The sign matters — read it carefully:

If the two charges are alike (both $+$ , as with our α and the nucleus), then $q_{1} q_{2} > 0$ , so $U$ is positive and grows larger as $r$ shrinks — like a ball forced up a steeper and steeper hill. This is our case.
If the two charges are opposite (a $+$ and a $-$ , as with a nucleus and an electron), then $q_{1} q_{2} < 0$ , so $U$ is negative — a "valley" they sit down into, which is why an electron stays bound. The picture: for like charges, a ball rolling up a hill (positive, rising $U$ ); for opposite charges, a ball settling into a valley (negative $U$ ). Why the topic needs it: for the head-on α (like charges), at the turning point the α has traded all its positive KE for positive $U$ . Setting KE $= U$ is exactly how the parent page solves for the closest distance $r_{0}$ (defined in Section 6).

Figure 4 (below) shows the α's hairpin U-turn. Trace the yellow path: it flies in along the bottom with a full tank of KE, slows as the repulsion climbs, halts for an instant at the blue dot (the turning point, $v = 0$ , all energy now stored), then races back out along the top. The double-headed white arrow marks the closest gap, $r_{0}$ .

6. Distance of closest approach $r_{0}$

7. Circular motion: centripetal force and $\frac{m v ^{2}}{r}$

The parent page's Section 4 pictures electrons orbiting. Two ideas make that work; see Centripetal Force and Circular Motion for depth.

Definition Centripetal force — the inward pull that bends a path into a circle

Plain words: anything moving in a circle is constantly turning, and turning means its motion is being changed — that needs a force pointing inward, toward the centre. That inward force is called centripetal. Why exactly $\frac{m v ^{2}}{r}$ (not "because I said so"): picture the electron going round a circle of radius $r$ at speed $v$ . In a short time it sweeps a small arc. Its speed never changes, but the direction of its motion swings around by a small angle. Geometry of that little turn says: the amount of "direction-change per second" (its inward acceleration $a$ ) is bigger when it moves faster (more $v$ ) and bigger when the circle is tighter (smaller $r$ ) — and carefully doing the triangle of arrows gives exactly $a = \frac{v ^{2}}{r} .$ Then force is mass times acceleration ( $F = ma$ ), so the inward force needed is $F_{needed} = m a = \frac{m v ^{2}}{r} .$ The picture: a ball on a string whirled in a circle — the string tugs inward; cut the string and the ball flies off straight. Why the topic needs it: an electron circling the nucleus needs such an inward tug. Rutherford's model says the Coulomb attraction ( $+$ nucleus pulling $-$ electron) supplies it. Setting $\frac{1}{4 π ε _{0}} \frac{Z e \cdot e}{r ^{2}} = \frac{m v ^{2}}{r}$ is where the electron's orbital speed comes from — and this same picture is exactly what Bohr Model of the Atom fixes later.

8. The prerequisite map

The diagram below stacks everything on this page in dependency order. Charge ( $e$ ) and mass/speed ( $m, v$ ) sit at the top as the raw ingredients. Charge feeds the atomic number $Z$ and Coulomb's force; force plus position gives the stored energy $U$ , while mass and speed give the kinetic energy KE. Those two energies meet in the energy-conservation balance (KE $= U$ ), which pins down the closest distance $r_{0}$ . Separately, mass, speed and the Coulomb force meet in the orbital force balance. Both roads arrive at the nuclear model of the atom — so if any top box is shaky, the whole model wobbles.

Read it top-to-bottom: charge and mass at the top feed force and energy; those meet in the energy-conservation balance that gives $r_{0}$ ; that plus the orbital force balance produces the nuclear model.

Equipment checklist

Cover the right side and test yourself. If any answer surprises you, reread that section before returning to the parent page.

What is an α-particle?

A helium nucleus (2 protons + 2 neutrons) stripped of its electrons — a positive bullet of charge

+ 2 e

, mass ~4 u.

What does the symbol

e

mean, and its value?

The elementary charge,

e = 1.6 \times 1 0^{- 19}

C — one proton's worth of positive charge.

What is

Z

for gold and what does it count?

Z = 79

; it counts the protons in the nucleus, giving charge

+ 79 e

State Coulomb's Law in words.

The push/pull between two charges is proportional to both charges and falls off as

1/ r^{2}

— stronger the closer they are.

What do

q_{1}

and

q_{2}

stand for?

The charge on object 1 and the charge on object 2 in Coulomb's Law.

What are the units newton (N) and metre (m)?

N is the unit of force (about an apple's weight); m is the unit of distance (about one stride).

What is

\frac{1}{4 π ε _{0}}

numerically?

About

9 \times 1 0^{9} N\cdotm^{2} / C^{2}

;

ε_{0}

is a fixed constant of nature.

Write kinetic energy and electrostatic potential energy.

KE = \frac{1}{2} m v^{2}

;

U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r}

When is

U

positive and when negative?

Positive for like charges (both

+

, our α–nucleus case); negative for opposite charges (nucleus–electron), which is why electrons stay bound.

Why can we set KE

= U

at closest approach?

Energy is conserved; at the turning point all motion energy has become stored electrostatic energy.

What is

r_{0}

in plain words?

The smallest distance the α-particle reaches before repulsion stops and reverses it — a ceiling on the nucleus's size.

Why is the needed centripetal force

\frac{m v ^{2}}{r}

Turning a circle gives inward acceleration

a = v^{2} / r

; force is

ma

, so

F = m v^{2} / r

For an orbiting electron, what real force plays the centripetal role?

The Coulomb attraction between the positive nucleus and the negative electron.

Recall One-line summary to carry forward

Charge ( $e$ , $Z$ ) → force (Coulomb) → energy (KE, $U$ , conservation) → $r_{0}$ → "atom is mostly empty" = the nuclear model. Every symbol on the parent page is one of these.

Parent topic: Rutherford's gold-foil experiment — nuclear model. Related build-ons: Thomson's Plum Pudding Model, Bohr Model of the Atom, Atomic Spectra (Line Spectra), Discovery of the Proton and Neutron.