Exercises — Thomson's plum-pudding model

Pudding = the sphere of uniform positive charge that fills the whole atom. Raisins = the electrons, embedded inside that sphere. Nucleus? No. The positive charge is diffuse (spread out), not lumped in the centre. A central lump is Rutherford's later idea.

The atom is neutral, so the pudding's positive charge must exactly cancel the 7 electrons' negative charge. Each electron carries $-e$, so 7 of them carry $-7e$. The pudding therefore carries $\boxed{+7e}$, spread uniformly through the sphere.

False. $1/r^2$ is how the field behaves outside a charge. Inside a uniform sphere, only the charge enclosed within radius $r$ acts, and that enclosed charge grows like $r^3$. Dividing $r^3$ (charge) by $r^2$ (the $1/r^2$ spreading) leaves $E\propto r$ — the field grows linearly, starting from zero at the centre.

What we do: plug numbers into the derived $k$. Why: $k$ tells us how stiff the "restoring spring" felt by the electron is. $k=9.0\times 10^9\cdot \frac{(1.6\times 10^{-19}{)}^2}{(1.0\times 10^{-10}{)}^3}=9.0\times 10^9\cdot \frac{2.56\times 10^{-38}}{1.0\times 10^{-30}}$ $=9.0\times 10^9\cdot 2.56\times 10^{-8}=230\text{N/m}.$

Why $\omega =\sqrt{k/m_e}$: this is the standard result for any mass on a Hooke's-law spring — a stiffer spring (bigger $k$) or lighter mass (smaller $m_e$) means faster wiggling. $\omega =\sqrt{\frac{230}{9.1\times 10^{-31}}}=\sqrt{2.53\times 10^{32}}\approx 1.6\times 10^{16}\text{rad/s}.$ $\nu =\frac{\omega }{2\pi }=\frac{1.6\times 10^{16}}{6.283}\approx 2.5\times 10^{15}\text{Hz}.$ This lands near the visible/ultraviolet band — the model's headline "success".

Since $E\propto r$ inside, the ratio is simply $\frac{E(0.5R)}{E(R)}=\frac{0.5R}{R}=\mathrm{0.5.}$ Halfway out, the field is exactly half the surface value. (Look at figure s02 — the field is a straight ramp from $0$ at the centre to its peak at $R$.)

Predict: bigger atom = softer spring = slower wiggle, so $\nu$ should drop. Prove: $\omega =\sqrt{\frac{e^2}{4\pi {\epsilon }_0{m}_e{R}^3}}\propto R^{-3/2}$. So $\frac{{\nu }_B}{{\nu }_A}={\left(\frac{R_B}{R_A}\right)}^{-3/2}=2^{-3/2}=\frac{1}{2\sqrt{2}}\approx \mathrm{0.354.}$ Doubling the radius drops the frequency to about 35% of the original — a big slowdown, because the $R^3$ inside the square root is so sensitive.

Gauss's law says field $=\frac{1}{4\pi {\epsilon }_0}\frac{(\text{charge enclosed})}{r^2}$.

• Point charge: enclosed charge is a fixed $e$ (never changes). So $E=\frac{1}{4\pi {\epsilon }_0}\frac{e}{r^2}\propto \frac{1}{r^2}$.
• Uniform sphere: enclosed charge grows as you move out: $q_{\text{in}}=e{r}^3/R^3$. Then $E=\frac{1}{4\pi {\epsilon }_0}\frac{e{r}^3/R^3}{r^2}=\frac{1}{4\pi {\epsilon }_0}\frac{er}{R^3}\propto r.$ The flipping fact: as $r$ grows, more charge joins in ($r^3$) faster than the $1/r^2$ dilution weakens it. Net exponent $3-2=+1$. That is the whole reason a restoring (spring-like) force exists — and thus why SHM appears at all.

$F=-kr$, so $F=0$ only at $r=0$ — the exact centre. There the electron feels no pull and can sit at rest (stable equilibrium). If nudged to some small $r$, the force $-kr$ points straight back to centre, overshoots, and the electron oscillates back and forth through the centre: simple harmonic motion. (Figure s01: the arrow always aims inward, for a displacement in any direction.)

This combines SHM energy with our $k$. Amplitude $A=r_0=0.4\times 10^{-10}=4.0\times 10^{-11}\text{m}$. (b) Where: speed is maximum where all spring energy has become kinetic — at the centre ($r=0$), where $F=0$. (a) How fast: for SHM, $v_{\max }=\omega A$ with $\omega =\sqrt{k/m_e}=1.59\times 10^{16}\text{rad/s}$. $v_{\max }=1.59\times 10^{16}\times 4.0\times 10^{-11}\approx 6.4\times 10^5\text{m/s}.$ (Reassuringly this is well under the speed of light $3\times 10^8$ m/s, so non-relativistic SHM is fine.)

Invert $\nu =\frac{1}{2\pi }\sqrt{\frac{e^2}{4\pi {\epsilon }_0{m}_e{R}^3}}$ to solve for $R$: $R^3=\frac{e^2}{4\pi {\epsilon }_0{m}_e(2\pi \nu {)}^2}.$ Numbers: numerator $=9.0\times 10^9\times (1.6\times 10^{-19}{)}^2=2.30\times 10^{-28}$. $(2\pi \nu {)}^2=(2\pi \cdot 4.6\times 10^{14}{)}^2=(2.89\times 10^{15}{)}^2=8.35\times 10^{30}$. Denominator $=9.1\times 10^{-31}\times 8.35\times 10^{30}=7.60$. $R^3=\frac{2.30\times 10^{-28}}{7.60}=3.03\times 10^{-29}\Rightarrow R\approx 3.1\times 10^{-10}\text{m}.$ That is about 3 Ångström — the right order of magnitude for an atom. So the model plausibly explains visible emission from atom-sized objects: exactly why it briefly felt convincing.

The restoring force on one electron comes from the pudding's field, not from the other electrons (which we've ignored). So $k$ scales with the pudding's total charge $Q$: repeating the derivation with $Q=6e$ gives $k=\frac{6e^2}{4\pi {\epsilon }_0{R}^3}$ — six times the single-unit case. $k=6\times 9.0\times 10^9\times \frac{(1.6\times 10^{-19}{)}^2}{(10^{-10}{)}^3}=6\times 230=1.38\times 10^3\text{N/m}.$ (Careful: the field generating the force uses the pudding charge $Q$; the charge being pushed is the one electron $-e$. Both appear — hence $Qe=6e^2$.)

Both use $E=\frac{1}{4\pi {\epsilon }_0}\frac{e}{r^2}$; only the closest-approach distance $r$ differs.

• Pudding surface: $r=R=10^{-10}\text{m}$.
• Point nucleus: $r=10^{-14}\text{m}$ (an $\alpha$ can dive right up to it, since there's now empty space). The field ratio is $\frac{E_{\text{nuc}}}{E_{\text{pud}}}={\left(\frac{10^{-10}}{10^{-14}}\right)}^2=(10^4{)}^2=10^8.$ The concentrated nucleus delivers a field one hundred million times stronger at close range. A force that huge, applied over the tiny time an $\alpha$ skims past, can reverse the $\alpha$'s motion (>90° backscatter). The diffuse pudding, whose field never exceeds the weak surface value, can only nudge $\alpha$'s by fractions of a degree. That $10^8$ gap is the reason plum-pudding died. → Geiger–Marsden gold foil experiment, Rutherford's nuclear model.

At equilibrium the inward pudding pull balances the outward electron–electron push, for the electron at $+d$: $\underset{\text{pudding pulls in}}{\underbrace{\frac{2e^2}{4\pi {\epsilon }_0{R}^3}d}}=\underset{\text{other electron pushes out}}{\underbrace{\frac{e^2}{4\pi {\epsilon }_0(2d{)}^2}}}.$ Cancel $\frac{e^2}{4\pi {\epsilon }_0}$ from both sides: $\frac{2d}{R^3}=\frac{1}{4d^2}\Rightarrow 8d^3=R^3\Rightarrow d=\frac{R}{2}.$ So each electron sits at $d=R/2$, i.e. the separation $2d=R$. The electrons settle exactly one radius apart, straddling the centre — a genuine stable arrangement, which is precisely the kind of ordered electron layout Thomson hoped would explain the periodic table.

The restoring law $F=-kr$ has one fixed $k$, hence exactly one natural frequency $\nu =\frac{1}{2\pi }\sqrt{k/m_e}$ — no matter the amplitude (that's the defining property of SHM: frequency is amplitude-independent). So the model predicts essentially a single spectral frequency, but hydrogen shows a whole discrete ladder of lines. One spring cannot make a spectrum. This mismatch pointed beyond classical mechanics altogether — toward quantised energy levels, resolved by the Bohr model and later quantum mechanics. So even before the gold-foil disproof, the spectra were quietly warning that a single classical oscillator was too poor a model.