Thomson's plum-pudding model: Question bank — Chemistry deep dive

Symbols and setup you need before the traps

Before any trap uses a symbol, here is exactly what each one means, anchored to one picture.

Definition The objects in the model

$R$ — the radius of the positive sphere (the "pudding"), roughly $1 0^{- 10} m$ , the size of the whole atom. It is the distance from the centre out to the sphere's surface.
$r$ — the distance of the electron from the centre of that sphere. We take $r$ positive measured outward from the centre along the line the electron sits on. So $r = 0$ is dead centre, $r = R$ is the surface.
$N$ — the number of electrons in the atom (a plain counting number: $1$ for hydrogen, $6$ for carbon, and so on). It is the count of "raisins" embedded in the dough.
$Q$ — the total positive charge of the pudding. For a neutral atom with $N$ electrons, $Q = N e$ .
$e$ — the magnitude of one electron's charge, $1.6 \times 1 0^{- 19} C$ . A single electron carries charge $- e$ .
$m_{e}$ — the mass of one electron, $9.1 \times 1 0^{- 31} kg$ . The positive dough is thousands of times heavier, so we treat it as fixed and let only the electron move.
$ε_{0}$ — the constant in Coulomb's law that sets how strong electric forces are.
$E$ — the electric field, a vector. We call it positive when it points radially outward (away from the centre) and negative when it points inward, using the same "outward is positive" convention as $r$ .

Now look at the setup figure below. The yellow outline is the sphere's surface at radius $R$ ; the chalk arrow from the dot at the centre out to the edge is $R$ itself; the blue dot off-centre is the electron; the pink arrow from centre to that dot is the displacement $r$ . Keep this picture in mind — every trap refers back to it.

Figure s01 — The setup: a uniform positive sphere (yellow) of radius $R$ , one electron (blue) at distance $r$ (pink arrow) from the centre. Positive charge $Q = N e$ fills the whole ball.

The Gauss's-law derivation, drawn out

This is the machinery every trap leans on. Follow the two-panel figure step by step as you read.

Figure s02 — Left panel: the dashed blue "Gaussian sphere" of radius $r$ passes through the electron (pink dot); only the charge $q_{in}$ shaded inside it acts, and the pink arrow is the outward field $E$ at that surface. Right panel: the blue curve is $E (r)$ — rising linearly while $r < R$ , then falling off as $1/ r^{2}$ once $r > R$ (dotted line marks $r = R$ ); the pink curve is the restoring force $F = - k r$ , dipping below zero to show it points inward.

Step 1 — Charge density of the pudding. Why? "Uniform" means charge per unit volume is constant, so total charge $Q = e$ divided by total volume gives the density $ρ$ : $ρ = \frac{Q}{\frac{4}{3} π R ^{3}} = \frac{e}{\frac{4}{3} π R ^{3}}$

Step 2 — Charge enclosed inside radius $r$ . Why? Only the dough inside the electron's radius pulls it (the outer shells cancel out). This is the shaded blue region in the left panel of Figure s02. Multiply the density by the volume of the inner ball of radius $r$ : $q_{in} = ρ \cdot \frac{4}{3} π r^{3} = e \frac{\frac{4}{3} π r ^{3}}{\frac{4}{3} π R ^{3}} = e \frac{r ^{3}}{R ^{3}}$

Step 3 — Field from Gauss's law. Why? Draw an imaginary sphere ("Gaussian surface") of radius $r$ through the electron (the dashed circle in Figure s02, left). By symmetry $E$ is the same all over it and points radially outward (our positive direction), so the flux is simply $E$ times the surface area $4 π r^{2}$ . Gauss's law says that flux equals enclosed charge over $ε_{0}$ : $E (r) \cdot 4 π r^{2} = \frac{q _{in}}{ε _{0}} \Rightarrow E (r) = \frac{1}{4 π ε _{0}} \frac{q _{in}}{r ^{2}} = \frac{1}{4 π ε _{0}} \frac{e r}{R ^{3}}$ Because $q_{in} > 0$ , this $E$ is positive (outward-pointing) — matching the pink arrow in Figure s02. The $r^{3}$ from the charge beats the $r^{2}$ in the denominator, leaving $E \propto r$ : the field grows linearly outward, zero at the centre. That is the rising blue line in the right panel of Figure s02.

Step 4 — Force on the electron. Why? The electron has charge $- e$ , so $F = (- e) E$ . Since $E$ points outward but the charge is negative, the force $F$ points inward — opposite to the displacement $r$ . That flip of direction is what the minus sign records, and it is exactly restoring. In our "outward is positive" convention that reads as a negative $F$ , the pink curve dipping below zero in the right panel of Figure s02: $F = - e E = - \frac{e ^{2}}{4 π ε _{0} R ^{3}} r = - k r, k = \frac{e ^{2}}{4 π ε _{0} R ^{3}}$

True or false — justify

Each line: a claim, then the honest verdict with the reason that makes it true or false.

The positive "pudding" is a single point at the atom's centre.

False — the whole point of Thomson's picture is that the positive charge is smeared uniformly over the entire sphere of radius

R

; a central point is Rutherford's nucleus, which came later.

In the plum-pudding model the atom is electrically neutral overall.

True — the diffuse positive charge

Q = N e

is built to exactly cancel the total negative charge

- N e

of all

N

embedded electrons, so from outside the atom shows no charge.

Inside a uniform positive sphere the electric field grows in proportion to the distance

r

from the centre.

True — by Gauss's law only the enclosed charge

q_{in} \propto r^{3}

acts, so

E \propto r^{3} / r^{2} = r

; the field is zero at the centre and rises linearly.

An electron displaced from the centre of the pudding feels a force that pushes it further out.

False — the force is restoring: with

r

measured outward,

F = - k r

points back toward the centre, which is exactly why the electron oscillates instead of flying away.

The electron inside the pudding executes simple harmonic motion.

True — the restoring force is linear in displacement,

F = - k r

with

k = \frac{e ^{2}}{4 π ε _{0} R ^{3}}

, which is Hooke's law form and hence SHM.

The Geiger–Marsden gold-foil result confirmed the plum-pudding model.

False — it disproved it; the large-angle backscattering of

α

-particles could only happen if the positive charge were concentrated in a tiny dense nucleus, not smeared out.

Electrons in Thomson's atom travel in planetary orbits around a nucleus.

False — they sit embedded and static (at most gently oscillating) in the positive dough; orbits belong to the Bohr model, not to Thomson.

Thomson's model could predict light of roughly the right frequency for atomic spectra.

True — the oscillating electron radiates at

ν = ω /2 π \sim 1 0^{15} Hz

for

R \approx 1 0^{- 10} m

, an order-of-magnitude match with visible light — its headline success.

The name "watermelon model" refers to a different model from plum-pudding.

False — watermelon, raisin-bread and plum-pudding are three names for the same Thomson model, seeds/raisins/plums all playing the electron role.

Outside the whole sphere (

r > R

), the field of the pudding still goes like

1/ r^{2}

True — from outside, a uniform sphere behaves as if all its charge

Q

sat at the centre, so ordinary Coulomb

1/ r^{2}

applies; only inside (

r < R

) does the

E \propto r

rule take over.

Spot the error

Each line quotes flawed reasoning; the reveal names the exact broken step.

"Move the electron outward and the pull weakens because it is farther from the charge, so

F \propto 1/ r^{2}

The error is treating all the charge as fixed at the centre; moving out actually encloses more positive charge (

q_{in} \propto r^{3}

), and this outweighs the

1/ r^{2}

dilution, giving

F \propto r

, not

1/ r^{2}

"The pudding for a carbon atom has charge

- 6 e

so it cancels the six electrons."

Sign error — electrons are negative (

- 6 e

total), so the pudding must be

Q = + 6 e

(positive) to cancel them and make the atom neutral.

"Since some

α

-particles bounced back, a diffuse positive cloud must give strong deflections."

The conclusion is backwards: a diffuse cloud gives only tiny deflections; the strong backscattering is exactly what a diffuse cloud cannot produce, which is why it forced the nuclear model.

"The electron at the very centre feels the strongest inward force, so it can't rest there."

At the centre the enclosed charge is zero, so

E = 0

and

F = 0

; the centre is the stable equilibrium, the one place the electron feels no net force.

ω = k / m_{e}

uses the mass of the whole atom."

It uses the electron's mass

m_{e}

, because the heavy positive dough is treated as fixed and the electron is the only object that moves.

"Because the atom is neutral, there is no electric field anywhere inside it."

Neutrality of the whole atom does not cancel the field at interior points; at radius

r

only the enclosed positive charge

q_{in}

acts and it is uncancelled, producing

E \propto r

pulling on the electron.

Why questions

Why does a restoring force appear at all, rather than a constant or an outward force?

Because displacement outward encloses more attracting positive charge and displacement inward encloses less, so the force always opposes the displacement — the defining feature that makes oscillation possible.

Why did Thomson assume the positive charge was spread out instead of concentrated?

Nothing in 1904 hinted at concentration, so the simplest neutral arrangement consistent with the newly found electrons was uniform positive charge filling the atom.

Why is Gauss's law the right tool for finding the field inside the sphere?

The sphere is perfectly symmetric, so at radius

r

the field has the same size everywhere and points radially; choosing an imaginary sphere of radius

r

as the Gaussian surface turns the flux into

E \cdot 4 π r^{2}

, which equals

q_{in} / ε_{0}

— this sidesteps the hard Coulomb sum over every piece of dough and gives

E (r) = \frac{1}{4 π ε _{0}} \frac{q _{in}}{r ^{2}}

in one line.

Why did the model's spectral prediction feel convincing at first?

The single oscillation frequency

ν \sim 1 0^{15} Hz

landed in the visible/UV range of real atomic spectra, so the model appeared to explain where light comes from.

Why can a single oscillation frequency never fully explain real spectra?

Real atoms emit many discrete spectral lines, whereas one embedded electron gives essentially one frequency — a mismatch the Bohr model later addressed with quantized orbits.

Why does the discovery in the cathode-ray tube logically force some positive charge to exist?

It proved matter contains negatives, yet bulk matter is neutral, so an equal amount of positive charge must be present to balance the books.

Edge cases

What is the force on an electron placed exactly at the centre (

r = 0

Zero — the enclosed charge is zero there, so

E = 0

and

F = - k r = 0

; it is the stable equilibrium point.

What happens to the oscillation frequency

ω

as the atomic radius

R

grows very large?

Since

ω = e^{2} / (4 π ε_{0} m_{e} R^{3})

ω \to 0

R \to \infty

— a bigger, more dilute pudding gives a weaker spring and slower wiggle.

What happens to

ω

R \to 0

(charge crammed tighter)?

ω \to \infty

because

k \propto 1/ R^{3}

blows up; a tiny dense sphere would demand impossibly high radiation frequencies — an early hint the diffuse picture had limits.

If the electron sits right at the surface (

r = R

), what force does it feel?

The maximum restoring force,

F = - \frac{e ^{2}}{4 π ε _{0} R ^{3}} R = - \frac{e ^{2}}{4 π ε _{0} R ^{2}}

, which matches the ordinary Coulomb value for charge

e

at distance

R

— inside and outside formulas agree exactly at the boundary

r = R

In this model, is there any empty space inside the atom?

No — the positive dough fills the whole volume and electrons are embedded in it; the notion of an atom being "mostly empty" is a later, nuclear-model idea.

For a hydrogen-like pudding with a single electron, where does it rest?

Precisely at the centre, the unique point where the inward pull vanishes; with several electrons they spread into a symmetric arrangement that balances mutual repulsion against the central pull.

Recall One-line self-test

Which single fact, if you remember it, defends against half these traps? ::: "Positive charge is diffuse, so inside the sphere $E \propto r$ (restoring, SHM) — not a central lump and not $1/ r^{2}$ ."

Connections

Parent topic — full derivation these traps are built on.
Gauss's law — why only enclosed charge counts.
Simple Harmonic Motion — the $F = - k r$ oscillation.
Geiger–Marsden gold foil experiment — the disproof several traps target.
Rutherford's nuclear model / Bohr model — the successors students confuse with this model.