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

Before we start, here are the numbers used all over this page, each named in plain words:

Two more names we will use constantly:

The master formula we keep reusing (built by energy conservation in the parent note):

Figure below (s01): a head-on α slowing to a stop at the turning point a distance $r_{0}$ from the nucleus, then reversing. Watch the lavender incoming arrow shorten to zero at the butter-yellow turning point (where KE has fully become $U$ ), then leave as the mint reversed arrow. The double-headed bracket marks $r_{0}$ — the smallest gap the α ever achieves.

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

Level 1 — Recognition

L1.1 In the gold-foil experiment, roughly what fraction of α-particles bounced back at angles greater than $9 0^{\circ}$ , and what does that fact prove about the atom?

Recall Solution

About 1 in 20,000 bounced back. This proves the positive charge and almost all the mass are packed into a tiny, dense central nucleus. If positive charge were spread out (plum pudding), no spot would be strong enough to reverse a fast α-particle.

L1.2 Match each observation to its conclusion. (a) Most pass straight → ? (b) Some deflect a little → ? (c) A very few bounce back → ?

Recall Solution

(a) The atom is mostly empty space.
(b) They passed near (not into) the concentrated charge and felt mild Coulomb repulsion.
(c) A near-head-on hit on the tiny massive nucleus.

L1.3 True or false, with reason: "Rutherford discovered the electron."

Recall Solution

False. Rutherford discovered the nucleus (and later named the proton). The electron was J. J. Thomson (1897); the neutron was Chadwick (1932). See Discovery of the Proton and Neutron.

Level 2 — Application

L2.1 An α-particle with kinetic energy $KE = 5.0 MeV$ is fired head-on at a gold nucleus ( $Z = 79$ ). Find the distance of closest approach $r_{0}$ .

Recall Solution

Why energy conservation? At the turning point $v = 0$ , so all $KE$ has become electric potential energy $U$ . Set $KE = U$ : $KE = k \frac{( 2 e ) ( Z e )}{r _{0}} \Rightarrow r_{0} = k \frac{2 Z e ^{2}}{KE} .$ Convert energy: $KE = 5.0 \times 1.6 \times 1 0^{- 13} = 8.0 \times 1 0^{- 13} J$ . $r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8.0 \times 1 0 ^{- 13}} .$ Numerator: $9 \times 1 0^{9} \times 2 \times 79 \times 2.56 \times 1 0^{- 38} = 3.64 \times 1 0^{- 26}$ . $r_{0} = \frac{3.64 \times 1 0 ^{- 26}}{8.0 \times 1 0 ^{- 13}} \approx 4.55 \times 1 0^{- 14} m .$ So the α gets no closer than about $4.5 \times 1 0^{- 14} m$ .

L2.2 A gold nucleus ( $Z = 79$ ) and an α-particle ( $+ 2 e$ ) sit $2.0 \times 1 0^{- 14} m$ apart. Compute the Coulomb force between them.

Recall Solution

Why Coulomb's Law? It gives the force between two point charges: $F = k q_{1} q_{2} / r^{2}$ . $F = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1.6 \times 1 0 ^{- 19} ) ( 79 \times 1.6 \times 1 0 ^{- 19} )}{( 2.0 \times 1 0 ^{- 14} ) ^{2}} .$ $q_{1} = 3.2 \times 1 0^{- 19}$ , $q_{2} = 1.264 \times 1 0^{- 17}$ , product $= 4.045 \times 1 0^{- 36}$ . $F = \frac{9 \times 1 0 ^{9} \times 4.045 \times 1 0 ^{- 36}}{4.0 \times 1 0 ^{- 28}} \approx 91 N .$ 91 N of repulsion between two specks — that is why the α reverses.

L2.3 For the setup in L2.1, what is the speed $v$ of the α-particle (mass $6.64 \times 1 0^{- 27} kg$ ) far from the nucleus?

Recall Solution

From $KE = \frac{1}{2} m v^{2}$ : $v = \frac{2 KE}{m} = \frac{2 \times 8.0 \times 1 0 ^{- 13}}{6.64 \times 1 0 ^{- 27}} = 2.41 \times 1 0^{14} \approx 1.55 \times 1 0^{7} m/s .$ About $5%$ of the speed of light — genuinely a "bullet."

Level 3 — Analysis

L3.1 Two α-particles hit gold: one with $KE = 5.0 MeV$ , one with $KE = 10.0 MeV$ . Without a calculator, which gets closer, and by what factor do their $r_{0}$ differ?

Recall Solution

$r_{0} \propto 1/ KE$ (KE is in the denominator). Doubling KE halves $r_{0}$ . The $10 MeV$ α gets closer, with $r_{0}$ exactly half that of the $5 MeV$ one. Check: $4.55 \times 1 0^{- 14} m \to 2.28 \times 1 0^{- 14} m$ .

L3.2 The same $5.0 MeV$ α is fired at a silver nucleus ( $Z = 47$ ) instead of gold ( $Z = 79$ ). Does $r_{0}$ increase or decrease, and by what factor?

Recall Solution

$r_{0} \propto Z$ (nuclear charge is in the numerator). Smaller $Z$ ⇒ weaker repulsion ⇒ the α penetrates closer ⇒ smaller $r_{0}$ . Factor $= 47/79 \approx 0.595$ . So $r_{0}^{Ag} = 0.595 \times 4.55 \times 1 0^{- 14} \approx 2.71 \times 1 0^{- 14} m$ .

L3.3 Explain quantitatively why an α-particle (mass $4 u$ ) bounces backward off a gold nucleus (mass $197 u$ ) but would not bounce backward off an electron (mass $\approx 1/1836 u$ ).

Recall Solution

Why mass ratio matters: in a head-on elastic collision, a light projectile reverses only if it hits something heavier than itself (like a ball off a wall). Off a lighter target it just shoves the target forward and keeps going.

Gold: $197 ≫ 4$ ⇒ the α reverses. ✓
Electron: $m_{e} \approx 0.0005 u ≪ 4$ ⇒ the α barely notices it, plowing on. The electron flies off; the α is undeflected. This is exactly the parent-note "bowling ball vs pebble" argument — and it's why backscatter proves the mass is concentrated with the positive charge, not spread among light electrons.

Figure below (s02): two side-by-side collisions. On the left, the light α (lavender) meets a heavy $197 u$ nucleus (coral) and rebounds as the mint arrow — reversal. On the right, the α (coral arrow) meets a near-massless electron (small lavender dot) and plows straight on, barely deflected. The contrast is the mass-concentration argument.

Level 4 — Synthesis

L4.1 A classical electron orbits a hydrogen nucleus ( $Z = 1$ ) at radius $r = 5.3 \times 1 0^{- 11} m$ . Using the balance "Coulomb attraction = centripetal force," find its orbital speed $v$ . (Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .)

Recall Solution

Why this balance? A circling electron needs an inward pull (centripetal force, from Centripetal Force and Circular Motion); Coulomb attraction supplies it: $k \frac{Z e ^{2}}{r ^{2}} = \frac{m _{e} v ^{2}}{r} \Rightarrow v = \frac{k Z e ^{2}}{m _{e} r} .$ $v = \frac{( 9 \times 1 0 ^{9} ) ( 1 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 9.11 \times 1 0 ^{- 31} ) ( 5.3 \times 1 0 ^{- 11} )} .$ Numerator: $9 \times 1 0^{9} \times 2.56 \times 1 0^{- 38} = 2.304 \times 1 0^{- 28}$ . Denominator: $9.11 \times 1 0^{- 31} \times 5.3 \times 1 0^{- 11} = 4.83 \times 1 0^{- 41}$ . $v = 4.77 \times 1 0^{12} \approx 2.19 \times 1 0^{6} m/s .$ About $0.7%$ of light speed — consistent with the Bohr Model of the Atom value.

L4.2 For that same electron, compute the centripetal acceleration $a = v^{2} / r$ . This is the acceleration that (classically) forces the electron to radiate and spiral in.

Recall Solution

$a = \frac{v ^{2}}{r} = \frac{( 2.19 \times 1 0 ^{6} ) ^{2}}{5.3 \times 1 0 ^{- 11}} = \frac{4.80 \times 1 0 ^{12}}{5.3 \times 1 0 ^{- 11}} \approx 9.05 \times 1 0^{22} m/s^{2} .$ An enormous acceleration. Classical electromagnetism says any accelerating charge radiates, so this electron should lose energy and spiral into the nucleus — the fatal flaw. Only quantization (Bohr Model of the Atom) rescues stability and explains the line spectra.

L4.3 Combine ideas: for a $7.7 MeV$ α on gold, compare $r_{0}$ to the actual nuclear radius of gold, $R \approx 7.3 \times 1 0^{- 15} m$ . Can this α "touch" the nucleus?

Recall Solution

$KE = 7.7 \times 1.6 \times 1 0^{- 13} = 1.232 \times 1 0^{- 12} J$ . $r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1.232 \times 1 0 ^{- 12}} = \frac{3.64 \times 1 0 ^{- 26}}{1.232 \times 1 0 ^{- 12}} \approx 2.96 \times 1 0^{- 14} m .$ Since $r_{0} \approx 3.0 \times 1 0^{- 14} m > R \approx 0.73 \times 1 0^{- 14} m$ , the α stops well outside the nucleus. It never touches it — pure Coulomb repulsion turns it around. That's why Rutherford scattering (at these energies) probes charge, not nuclear surface.

Level 5 — Mastery

L5.1 Design argument: You want to measure the nuclear radius of gold directly by making the α just touch the surface ( $r_{0} = R = 7.3 \times 1 0^{- 15} m$ ). What α kinetic energy (in MeV) is required, and why does this experiment fail at lower energies?

Recall Solution

Set $r_{0} = R$ and solve for KE: $KE = k \frac{2 Z e ^{2}}{R} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{7.3 \times 1 0 ^{- 15}} .$ Numerator $= 3.64 \times 1 0^{- 26}$ (same as before). $KE = \frac{3.64 \times 1 0 ^{- 26}}{7.3 \times 1 0 ^{- 15}} = 4.99 \times 1 0^{- 12} J = \frac{4.99 \times 1 0 ^{- 12}}{1.6 \times 1 0 ^{- 13}} \approx 31 MeV .$ You need about 31 MeV. Why lower energies fail: below this, $r_{0} > R$ , so the α turns around before reaching the surface — you only ever measure the charge, never the nuclear size. This is exactly why Rutherford's $\sim 5$ – $8 MeV$ α's could confirm a tiny charged core but not resolve the true nuclear radius. (In reality, above $\sim 30 MeV$ the strong nuclear force also enters and the pure-Coulomb formula breaks — a hint that new physics lives at the surface.)

L5.2 Conceptual synthesis: List, in order, the three things the backscatter observation lets you conclude, and for each name the specific physical principle that forces that conclusion.

Recall Solution

Charge is concentrated ← Coulomb's Law: only a localized $+ Z e$ gives a field strong enough (short range, $\propto 1/ r^{2}$ ) to reverse a fast α.
Mass is concentrated ← elastic collision mechanics: a light projectile reverses only off a heavier target, so the scatterer must far outweigh the α.
The atom is mostly empty ← statistics of straight-through majority: rare backscatter ⇒ the dense target has a minuscule cross-section. Together these are the nuclear model, replacing Thomson's Plum Pudding Model.

L5.3 Critical thinking: A student claims, "If we used a thicker gold foil, more α-particles would bounce back, giving a bigger backscatter fraction — that's a better experiment." Evaluate.

Recall Solution

Partly true, badly flawed. A thicker foil does give more scattering events — but each α now suffers many interactions, so a single deflection can no longer be cleanly linked to one nucleus. Rutherford deliberately used ultra-thin foil (~ $1 0^{- 7}$ m) so that one interaction dominates, making the scattering-angle data interpretable and letting him extract $r_{0}$ and the $1/ sin^{4} (θ /2)$ scattering law. Fix: thin foil = clean single-scattering physics; thick foil = muddled multiple scattering. More bounces ≠ better data.

Recall wrap-up

Recall Quick self-quiz

How does $r_{0}$ scale with KE? ::: Inversely — $r_{0} \propto 1/ KE$ ; more energy burrows deeper (smaller $r_{0}$ ). How does $r_{0}$ scale with nuclear charge? ::: Directly — $r_{0} \propto Z$ ; more charge repels sooner (larger $r_{0}$ ). Why does the α NOT bounce off electrons? ::: Electrons are ~8000× lighter than the α; a light projectile reverses only off a heavier target. What principle gives the electron's orbital speed? ::: Coulomb attraction = centripetal force, so $v = k Z e^{2} / (m_{e} r)$ . Why use ultra-thin foil? ::: To ensure single scattering, so each deflection maps to one nucleus.