Rutherford's gold-foil experiment — nuclear model: Worked examples

The scenario matrix

We are almost always juggling one of two equations. Recall them so we never use a symbol unearned:

Formula The two master relations

Closest approach (head-on shot, stops at $r_{0}$ ). The α carries charge $q_{α} = 2 e$ , the nucleus $q_{N} = Z e$ .

Why energy conservation? Nothing is lost in a Coulomb push, so the α's kinetic energy is entirely converted into electrostatic potential energy at the stopping point. That single accounting statement is $all the motion \frac{1}{2} m v^{2} = all the push, stored \frac{1}{4 π ε _{0}} \frac{q _{α} q _{N}}{r _{0}} .$
Why solve for $r_{0}$ ? The question asks "how close?", so we isolate the distance. Multiply both sides by $r_{0}$ , then divide by $\frac{1}{2} m v^{2}$ : $r_{0} = \frac{1}{4 π ε _{0}} \frac{q _{α} q _{N}}{\frac{1}{2} m v ^{2}} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{\frac{1}{2} m v ^{2}} = \frac{1}{4 π ε _{0}} \frac{4 Z e ^{2}}{m v ^{2}} .$ The $\frac{1}{2}$ in the denominator is what turns the $2 Z e^{2}$ on top into $4 Z e^{2}$ .
KE-form (kinetic energy carried whole, so you can plug KE straight in — no need to unpack $\frac{1}{2} m v^{2}$ ): $r_{0} = \frac{1}{4 π ε _{0}} \frac{q _{α} q _{N}}{KE} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{KE} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{KE} .$ Orbit balance (Coulomb pull = centripetal need — why? a circling electron needs an inward force, and Coulomb attraction supplies it): $v = \frac{Z e ^{2}}{4 π ε _{0} m r}$ where $k = \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ , $e = 1.6 \times 1 0^{- 19} C$ , $Z$ = nuclear charge number (protons), $m, v$ = mass and speed of the moving particle.

Here is the full grid of cases. Each row is a distinct "shape" of problem; the last column names the example that kills it.

Cell	Scenario class	What changes / what's tricky	Example
A	Baseline head-on, given KE	plug into $r_{0}$ directly	Ex 1
B	Given speed not KE	must build $\frac{1}{2} m v^{2}$ first	Ex 2
C	Change the nucleus ( $Z ↑$ )	$r_{0} \propto Z$ — bigger repeller, farther stop	Ex 3
D	Change the speed ( $v ↑$ )	$r_{0} \propto 1/ v^{2}$ — faster ⇒ closer	Ex 4
E	Limiting case $v \to \infty$ and $v \to 0$	$r_{0} \to 0$ (touches) / $r_{0} \to \infty$ (never arrives)	Ex 5
F	Degenerate/impossible input	negative KE, wrong-sign charge	Ex 6
G	Orbit-balance side, find $v$	uses the second formula	Ex 7
H	Real-world word problem	translate story → numbers	Ex 8
I	Exam twist: ratio without full data	cancel constants, no calculator	Ex 9
J	Order-of-magnitude sanity	is the nucleus really tiny?	Ex 10
K	Glancing (oblique) hit	impact parameter → deflection angle	Ex 11

Every cell A–K is covered below.

Ex 1 — Cell A: the baseline head-on shot

Step 1 — Convert energy to joules. $KE = 5.0 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} = 8.0 \times 1 0^{- 13} J$ Why this step? Every constant ( $k$ , $e$ ) is in SI, so energy must be in joules, not eV. Mixing units silently is the #1 error.

Step 2 — Use the KE-form of closest approach. $r_{0} = \frac{k ( 2 e ) ( Z e )}{KE} = \frac{k \cdot 2 Z e ^{2}}{KE}$ Why this step? We were given KE directly, so we use the version where KE appears whole — no need to recover $v$ .

Step 3 — Plug in. $r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8.0 \times 1 0 ^{- 13}} \approx 4.55 \times 1 0^{- 14} m$ Why this step? Substituting the actual SI numbers turns the symbolic relation into the physical distance we were asked for — the algebra is worthless until it yields a length.

Ex 2 — Cell B: given speed, not energy

Step 1 — Build the kinetic energy. $KE = \frac{1}{2} m v^{2} = \frac{1}{2} (6.64 \times 1 0^{- 27}) (1.6 \times 1 0^{7})^{2} \approx 8.5 \times 1 0^{- 13} J$ Why this step? The formula's engine is energy. Speed is only raw material — we must forge KE from it first.

Step 2 — Feed it into the KE-form of $r_{0}$ . $r_{0} = \frac{k \cdot 2 Z e ^{2}}{KE} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8.5 \times 1 0 ^{- 13}} \approx 4.28 \times 1 0^{- 14} m$ Why this step? Now that we hold KE as a single number, the whole-KE form takes it directly; using the $4 Z e^{2} / (m v^{2})$ form would force us to re-insert $m$ and $v$ and risks the factor-of-two trap.

Ex 3 — Cell C: swap the nucleus (why $r_{0} \propto Z$ )

Step 1 — Notice only $Z$ changed. $\frac{r _{0}^{Al}}{r _{0}^{Au}} = \frac{Z _{Al}}{Z _{Au}} = \frac{13}{79}$ Why this step? In $r_{0} = \frac{k 2 Z e ^{2}}{KE}$ , everything except $Z$ is identical, so a ratio cancels all the messy constants — no re-plugging.

Step 2 — Scale Ex 1's answer. $r_{0}^{Al} = 4.55 \times 1 0^{- 14} \times \frac{13}{79} \approx 7.49 \times 1 0^{- 15} m$ Why this step? Since $r_{0} \propto Z$ , multiplying the known gold answer by the charge ratio gives the aluminium answer instantly — faster and less error-prone than a fresh full substitution.

Ex 4 — Cell D: double the speed (why $r_{0} \propto 1/ v^{2}$ )

Step 1 — Locate $v$ in the formula. $r_{0} = \frac{k \cdot 4 Z e ^{2}}{m v ^{2}} \Rightarrow r_{0} \propto \frac{1}{v ^{2}}$ Why this step? Here we deliberately use the velocity form because the question is about $v$ itself; KE $= \frac{1}{2} m v^{2} \propto v^{2}$ , so $r_{0} \propto 1/ v^{2}$ . Squaring matters.

Step 2 — Apply the doubling. $v \to 2 v \Rightarrow r_{0} \to \frac{r _{0}}{2 ^{2}} = \frac{r _{0}}{4}$ Why this step? Because $r_{0}$ depends on $v^{- 2}$ , replacing $v$ by $2 v$ divides $r_{0}$ by $2^{2} = 4$ — the square is what turns a ×2 in speed into a ÷4 in distance.

Ex 5 — Cell E: the two limits (figure)

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

Step 1 — Read the curve $r_{0} = \frac{k 4 Z e ^{2}}{m v ^{2}}$ . Look at the red curve (the legend labels it $r_{0} \propto 1/ v^{2}$ ): it plunges toward the horizontal axis on the right and shoots up along the vertical axis on the left. The two dashed asymptotes drawn on the figure are exactly the two limits below. Why this step? A single picture of $r_{0}$ against $v$ lets us see both limits at once as the ends of one curve, instead of trusting two separate algebraic manipulations blindly.

(a) $v \to \infty$ : denominator $\to \infty$ , so $r_{0} \to 0$ . Why? An infinitely fast α has infinite KE — the Coulomb hill can never stop it, so it reaches (in the point-charge idealisation) $r_{0} = 0$ , the nucleus itself. In reality it would smash into the nucleus and this classical formula breaks (nuclear forces take over).

(b) $v \to 0$ : denominator $\to 0$ , so $r_{0} \to \infty$ . Why? A crawling α has almost no KE; even far away the repulsion turns it back. It "stops" at a huge distance — effectively never reaches the nucleus at all.

Ex 6 — Cell F: degenerate / impossible inputs

Case F1 — negative KE.

Step 1 — Substitute the bad number. With $KE = - 3 MeV = - 4.8 \times 1 0^{- 13}$ J, $r_{0} = \frac{k 2 Z e ^{2}}{KE} \Rightarrow KE < 0 \Rightarrow r_{0} < 0.$ Why this step? We deliberately push the impossible value through to expose how the formula reacts, rather than rejecting it by hand — the algebra itself flags the nonsense.

Step 2 — Evaluate the magnitude. $r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{- 4.8 \times 1 0 ^{- 13}} \approx - 7.58 \times 1 0^{- 14} m .$ Why this step? Putting an actual number on it shows the output is a negative length — physically impossible, since kinetic energy is $\frac{1}{2} m v^{2} \geq 0$ . The negative sign is the formula screaming "reject this input."

Case F2 — attractive instead of repulsive. If the projectile charge were $- 2 e$ (attraction to the $+ Z e$ nucleus), the "closest approach = stopping point" idea collapses entirely: attraction speeds the particle up as it approaches, so it never stops on the way in — there is no turning point. The head-on-stop derivation assumed repulsion. Wrong sign ⇒ wrong physics, not just a wrong number.

Ex 7 — Cell G: the orbit-balance side

Step 1 — Choose the right formula. This is an orbit, not a stopping shot, so we use $v = \frac{Z e ^{2}}{4 π ε _{0} m r} = \frac{k Z e ^{2}}{m r} .$ Why this step? Closest-approach is about a particle brought to rest; here the electron is balanced in circular motion — Centripetal Force and Circular Motion with Coulomb pull as the centripetal supplier (Coulomb's Law).

Step 2 — Plug in. $v = \frac{( 9 \times 1 0 ^{9} ) ( 1 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 9.1 \times 1 0 ^{- 31} ) ( 5.3 \times 1 0 ^{- 11} )} \approx 2.2 \times 1 0^{6} m s^{- 1}$ Why this step? Inserting the SI values under the root converts the balance condition into the concrete orbital speed the question demands; the square root is essential because $v$ appeared squared in the force balance.

Ex 8 — Cell H: real-world word problem

Step 1 — Translate the story. "No closer than" = closest approach $r_{0}$ . Given KE $= 7.7$ MeV $= 1.23 \times 1 0^{- 12}$ J. Why this step? Museum English → physics symbol. "Closest" is our $r_{0}$ , and the energy must be in joules to match our constants.

Step 2 — Compute. $r_{0} = \frac{k 2 Z e ^{2}}{KE} = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1.23 \times 1 0 ^{- 12}} \approx 2.96 \times 1 0^{- 14} m .$ Why this step? We use the whole-KE form because energy is what we were given; substituting the numbers turns the claim-checking into an actual length we can compare.

Step 3 — Compare with the claim. $2.96 \times 1 0^{- 14} m = 29.6 fm$ , not $6$ fm. Why this step? Converting to femtometres puts our answer in the exhibit's own units so the comparison is apples-to-apples; the mismatch (≈30 fm vs 6 fm) is a factor of ~5, so the exhibit's specific number is wrong even though the conclusion (a tiny nucleus, $\sim 1 0^{4}$ times smaller than the atom) still stands.

Ex 9 — Cell I: exam twist — ratio with no calculator

Step 1 — Write the ratio symbolically. $\frac{r _{0}^{(1)}}{r _{0}^{(2)}} = \frac{k 2 Z _{1} e ^{2} / KE _{1}}{k 2 Z _{2} e ^{2} / KE _{2}} = \frac{Z _{1}}{Z _{2}} \cdot \frac{KE _{2}}{KE _{1}} .$ Why this step? $k$ , $e$ , the factor $2$ all cancel — an exam wants you to see structure, not grind arithmetic.

Step 2 — Substitute. $= \frac{79}{47} \cdot \frac{4}{2} = \frac{79}{47} \cdot 2 = \frac{158}{47} \approx 3.36.$ Why this step? With only $Z$ and KE surviving, plugging the four small integers in gives the numerical ratio directly — no constants, no calculator needed.

Ex 10 — Cell J: order-of-magnitude sanity

Step 1 — Radius ratio. $\frac{R _{atom}}{r _{0}} = \frac{1.0 \times 1 0 ^{- 10}}{4.55 \times 1 0 ^{- 14}} \approx 2.2 \times 1 0^{3} .$ Why this step? Volume $\propto (radius)^{3}$ , so we must get the linear ratio first before cubing — cubing the raw radii separately would be needless extra work.

Step 2 — Cube it for volume. $(\frac{R _{atom}}{r _{0}})^{3} \approx (2.2 \times 1 0^{3})^{3} \approx 1.06 \times 1 0^{10} .$ Why this step? A sphere's volume scales as the cube of its radius, so cubing the radius ratio converts our length comparison into the volume comparison the question actually asks for.

Ex 11 — Cell K: a glancing (oblique) hit — figure

Step 1 — Understand each symbol on the figure. On the picture, the α comes in along the dashed line; the red segment is $b$ , the perpendicular gap between that line and the nucleus; $θ$ is the angle between the incoming and outgoing directions. Why this step? $cot$ ("cotangent") is $cos / sin$ — it is huge for small angles and small for large angles, so it encodes the physical fact that a small miss-distance gives a big swerve. We must know exactly what $b$ and $θ$ mean geometrically before trusting the formula.

Step 2 — Handle the two ends first (all cases).

Head-on, $b = 0$ : $θ \to 18 0^{\circ}$ , $cot (9 0^{\circ}) = 0$ — consistent, the particle comes straight back (this is the Ex 1 situation).
Far miss, $b \to \infty$ : $θ \to 0^{\circ}$ , $cot (θ /2) \to \infty$ — no deflection, the α sails straight past. Why this step? Checking the extremes confirms the formula behaves sensibly before we trust it on the in-between $9 0^{\circ}$ case.

Step 3 — Put in $θ = 9 0^{\circ}$ . Then $θ /2 = 4 5^{\circ}$ and $cot 4 5^{\circ} = 1$ , so $b = \frac{k Z ( 2 e ) e}{2 KE} = \frac{( 9 \times 1 0 ^{9} ) ( 79 ) ( 2 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2 ( 8.0 \times 1 0 ^{- 13} )} \approx 2.28 \times 1 0^{- 14} m .$ Why this step? At $4 5^{\circ}$ the cotangent is exactly $1$ , which strips the trig away and leaves a clean substitution — the cleanest angle to demonstrate the formula.

Recall Quick self-test (cover answers)

$r_{0}$ depends on speed how? ::: As $1/ v^{2}$ (via KE). Double $Z$ , same KE → $r_{0}$ ? ::: Doubles ( $r_{0} \propto Z$ ). Negative KE gives what and means what? ::: Negative $r_{0}$ = impossible input. Which formula for an orbiting electron's speed? ::: $v = k Z e^{2} / (m r)$ . Why does $v \to 0$ give $r_{0} \to \infty$ ? ::: No KE to climb the Coulomb hill — turns back far away. Impact parameter for a $9 0^{\circ}$ deflection? ::: Half the head-on $r_{0}$ .