PhysicsISC Class 12 · Atoms

Rutherford's Nuclear Model: Alpha-Particle Scattering and the Distance of Closest Approach

Updated 15 October 2026 · Fact-checked

In the Geiger-Marsden experiment most alpha particles passed straight through gold foil, but a few bounced back. Rutherford concluded that the atom's positive charge and mass sit in a tiny nucleus. For a head-on alpha particle of kinetic energy K, the distance of closest approach is $r_0 = \frac{2Ze^2}{4\pi\varepsilon_0K}$, about $10^{-14}$ m.

Understand Rutherford's Model and Distance of Closest Approach

Background. Thomson's "plum pudding" model spread the positive charge through the whole atom. A beam of fast, positively charged alpha particles (helium nuclei, charge +2e) should then pass through thin foil with only tiny deflections.

The Geiger-Marsden experiment. A narrow beam of alpha particles from a radioactive source (inside a lead block) fell on a very thin gold foil (about $1 0^{- 7}$ m thick, so each particle meets few atoms). A movable zinc sulphide screen and microscope counted the flashes produced by scattered alpha particles at different angles θ. The apparatus was in vacuum.

Observations.
1. Most alpha particles went straight through, undeviated or nearly so.
2. Some were deflected through small angles.
3. Very few (roughly 1 in 8000) were deflected by more than 90°; a very small number came almost straight back.

Conclusions (Rutherford's nuclear model).

Most of the atom is empty space (observation 1).
All the positive charge and almost all the mass are concentrated in a very small central nucleus; only a strong, concentrated charge can turn a heavy, fast alpha particle back (observation 3).
Electrons revolve round the nucleus at comparatively large distances; the atom is neutral overall.
The nucleus is about $1 0^{- 14}$ to $1 0^{- 15}$ m across, while the atom is about $1 0^{- 10}$ m: the nucleus is some $1 0^{4}$ to $1 0^{5}$ times smaller. (The mathematical theory of scattering is not required.)

Distance of closest approach. For a head-on collision, the alpha particle slows down as it approaches the nucleus (charge Ze), stops momentarily at distance $r_{0}$ , and returns along its path. At that point all its kinetic energy has become electrostatic potential energy:
$K = \frac{1}{2} m v^{2} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{r _{0}} \Rightarrow r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K} .$
Since the alpha particle does not touch the nucleus (at these energies), $r_{0}$ is an upper limit on the nuclear radius. Handy value: $\frac{e ^{2}}{4 π ε _{0}} = 1.44$ MeV fm, so $r_{0} (fm) = \frac{2 Z \times 1.44}{K ( MeV )}$ , where 1 fm (fermi) = $1 0^{- 15}$ m.

$r_{0} \propto Z$ and $r_{0} \propto \frac{1}{K}$ .
For a general projectile of charge q, $r_{0} = \frac{q Z e}{4 π ε _{0} K}$ .

Impact parameter (qualitative). The perpendicular distance between the alpha particle's initial line of motion and the nucleus is the impact parameter b. Small b (near head-on) gives large deflection; large b gives small deflection. Few particles have very small b, which is why large-angle scattering is rare.

Atomic structure terms. Atomic number Z = number of protons; neutron number N; mass number A = Z + N. For $_{79}^{197}$ Au, Z = 79, N = 118.

Drawbacks of Rutherford's model. (1) Stability: an orbiting electron is accelerating and, by classical theory, should radiate, lose energy and spiral into the nucleus within about $1 0^{- 8}$ s. (2) Spectra: the spiralling electron would emit a continuous range of frequencies, but atoms emit line spectra. Bohr's model addressed both.

Rutherford's Model and Distance of Closest Approach: formulas and key facts

Closest approach: $r_{0} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$; Head-on alpha particle.
Energy balance: $\frac{1}{2} m v^{2} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{r _{0}}$; K converts to U.
Handy form: $r_{0} (fm) = \frac{2.88 Z}{K ( MeV )}$; Uses 1.44 MeV fm.
General projectile: $r_{0} = \frac{q Z e}{4 π ε _{0} K}$; Charge q.
Mass number: $A = Z + N$; Protons plus neutrons.

How to solve Rutherford's Model and Distance of Closest Approach questions

Closest approach is pure energy conservation; choose SI units or the MeV-fm shortcut and stay with it.

1Write $K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{r _{0}}$ .
2SI route: K in joules (1 MeV = $1.6 \times 1 0^{- 13}$ J), $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ , e = $1.6 \times 1 0^{- 19}$ C.
3Shortcut: $r_{0} = \frac{2 Z \times 1.44}{K}$ fm with K in MeV.
4Ratios: $r_{0} \propto \frac{q Z}{K}$ ; cancel common factors.
5Theory questions: link each observation to its conclusion.

Rutherford's Model and Distance of Closest Approach in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

How empty is an atom?

Suppose you enlarge a gold atom (about $1 0^{- 10}$ m across) until its nucleus (about $1 0^{- 14}$ m) is the size of a 1 cm marble.

How big would the whole atom be?

Try it first, then show the solutionSolution

Scale factor = $\frac{1 0 ^{- 10}}{1 0 ^{- 14}} = 1 0^{4}$ .
Atom size $= 1 0^{4} \times 1$ cm = 100 m: a marble in the middle of a sports ground.

Answer: About 100 m across: the atom is overwhelmingly empty space.

Memory hook: Marble in a stadium.

Radon alphas hitting lead

Imagine alpha particles of 5.5 MeV from radon gas striking lead shielding (Z = 82) head-on.

How close can they get to a lead nucleus?

Try it first, then show the solutionSolution

$r_{0} = \frac{2 \times 82 \times 1.44}{5.5}$ fm.
$r_{0} \approx 42.9$ fm $\approx 4.3 \times 1 0^{- 14}$ m.

Answer: About $4.3 \times 1 0^{- 14}$ m: they are turned back long before touching the nucleus.

Memory hook: Bigger Z pushes back from further away.

Tricks to remember Rutherford's Model and Distance of Closest Approach

Short memory aids. Say each one aloud once, then use it on a question the same day.

Most through, few back
Through → empty space; back → tiny, heavy, positive nucleus.
Use it when: Observation-conclusion pairs.
Two-Z over K
$r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K}$ : the 2 is the alpha's charge.
Use it when: Formula recall.
1.44 MeV fm
$\frac{e ^{2}}{4 π ε _{0}}$ = 1.44 MeV fm.
Use it when: Fast numericals.
Spiral and smear
Rutherford fails twice: electron spirals in; spectrum smears into a continuum.
Use it when: Drawbacks.

Rutherford's Model and Distance of Closest Approach: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: What large-angle scattering shows · 1 mark · MCQBoard level

The scattering of a few alpha particles through more than 90° by gold foil shows that:
(a) electrons are very light
(b) the positive charge and mass of the atom are concentrated in a tiny nucleus
(c) the atom is mostly empty space
(d) alpha particles are negatively charged

Try it first, then show the solutionSolution

Only a strong, concentrated positive charge with large mass can reverse a fast alpha particle.

Answer: (b).

Example 2: Closest approach for a 7.2 MeV alpha particle · 2 marks · Short answerBoard level

An alpha particle of kinetic energy 7.2 MeV approaches a gold nucleus (Z = 79) head-on. Find the distance of closest approach.

Try it first, then show the solutionSolution

$r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K} = \frac{2 \times 79 \times 1.44 MeV fm}{7.2 MeV}$ . (1)
$r_{0} \approx 31.6$ fm $\approx 3.16 \times 1 0^{- 14}$ m. (1)

Answer: ≈ $3.16 \times 1 0^{- 14}$ m.

Example 3: Derivation of closest approach · 3 marks · Long answerHarder board level

Obtain an expression for the distance of closest approach of an alpha particle of kinetic energy K to a nucleus of atomic number Z. Why does it give only an upper limit to the nuclear radius?

Try it first, then show the solutionSolution

In a head-on approach the alpha particle stops momentarily at $r_{0}$ ; its kinetic energy is then all electrostatic potential energy. (1)
$K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{r _{0}}$ , so $r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K}$ . (1)
The alpha particle is turned back before reaching the nuclear surface, so the nucleus must be smaller than $r_{0}$ : an upper limit. (1)

Answer: $r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K}$ ; the nucleus is no larger than this.

Example 4: Energy needed to come closer · 2 marks · Short answerHarder board level

What minimum kinetic energy must an alpha particle have to approach a gold nucleus (Z = 79) head-on to within 10 fm? Give your answer in MeV.

Try it first, then show the solutionSolution

$K = \frac{2 Z \times 1.44}{r _{0}} = \frac{2 \times 79 \times 1.44}{10}$ MeV. (1)
K ≈ 22.8 MeV. (1)

Answer: ≈ 22.8 MeV.

Example 5: Proton or alpha particle? · 3 marks · Short answerHOTS / competency

A proton and an alpha particle approach the same nucleus head-on. Find the ratio of their distances of closest approach $\frac{r _{α}}{r _{p}}$ when (a) they have the same kinetic energy, and (b) they have the same speed (take $m_{α} = 4 m_{p}$ ).

Try it first, then show the solutionSolution

$r_{0} = \frac{q Z e}{4 π ε _{0} K} \propto \frac{q}{K}$ . (1)
(a) Same K: $\frac{r _{α}}{r _{p}} = \frac{2 e}{e} = 2$ . (1)
(b) Same speed: $K_{α} = 4 K_{p}$ , so $\frac{r _{α}}{r _{p}} = \frac{2}{4} = \frac{1}{2}$ . (1)

Answer: (a) 2; (b) $\frac{1}{2}$ .

Example 6: Re-running Geiger and Marsden · 4 marks · Case-basedHOTS / competency

Suppose a class repeats the Geiger-Marsden experiment with 5.5 MeV alpha particles and gold foil.
(i) Why must the foil be very thin and the apparatus evacuated? (1 mark)
(ii) State the key observation and the conclusion drawn from it. (1 mark)
(iii) Find the distance of closest approach for a head-on collision (Z = 79). (1 mark)
(iv) State two reasons the resulting nuclear model was incomplete. (1 mark)

Try it first, then show the solutionSolution

(i) Thin foil: each alpha particle is scattered by essentially one nucleus. Vacuum: air would slow and scatter the alpha particles.
(ii) A very few alpha particles are deflected by more than 90°: the positive charge and mass are in a tiny nucleus.
(iii) $r_{0} = \frac{2 \times 79 \times 1.44}{5.5} \approx 41.4$ fm $\approx 4.1 \times 1 0^{- 14}$ m.
(iv) It cannot explain the stability of atoms (accelerating electrons should radiate and spiral in) or their line spectra.

Answer: (i) single scattering, no air; (ii) large-angle scattering, nucleus; (iii) ≈ $4.1 \times 1 0^{- 14}$ m; (iv) stability and line spectra.

Where students lose marks in Rutherford's Model and Distance of Closest Approach

The slips examiners see most often, and what to write instead.

Mistake: Charge of the alpha particle taken as e
Students forget the alpha particle carries +2e.
Wrong working
$r_{0} = \frac{Z e ^{2}}{4 π ε _{0} K}$ .
Right working
$r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K}$ .
Do this: Alpha = helium nucleus, charge 2e.
Usually costs: 1 mark.
Mistake: MeV not converted to joules
Mixing SI constants with MeV gives nonsense.
Do this: 1 MeV = $1.6 \times 1 0^{- 13}$ J, or use the 1.44 MeV fm shortcut.
Usually costs: 1 mark.
Mistake: Calling $r_{0}$ the nuclear radius
The particle is stopped before contact.
Do this: Say $r_{0}$ is an upper limit (estimate) of the nuclear size.
Usually costs: ½ to 1 mark.
Mistake: Observations without conclusions
Lists of observations score little alone.
Do this: Pair each observation with its conclusion.
Usually costs: 1 mark.

Exam tips

Draw the scattering set-up: source in lead block, collimator, foil, ZnS screen and microscope.
Give the size comparison: atom about $1 0^{- 10}$ m, nucleus about $1 0^{- 14}$ to $1 0^{- 15}$ m.
Quote 1 fm = $1 0^{- 15}$ m.
Use the syllabus form $U = \frac{( 2 e ) ( Z e )}{4 π ε _{0} r _{0}}$ in derivations.

Practice questions on Rutherford's Model and Distance of Closest Approach

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If the kinetic energy of an alpha particle is doubled, its distance of closest approach to a nucleus:
(a) doubles
(b) halves
(c) is unchanged
(d) becomes four times
Hint
$r_{0} \propto \frac{1}{K}$ .
Answer
(b).
Q2 · 1 mark · Short answerBoard level
For the nucleus $_{79}^{197}$ Au, state Z, N and A.
Hint
A = Z + N.
Answer
Z = 79, N = 118, A = 197.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Only a very small fraction of alpha particles are scattered through large angles by gold foil.
Reason (R): The nucleus is very small, so few alpha particles pass close enough to it.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Impact parameter.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Find the distance of closest approach of a 6 MeV alpha particle to a silver nucleus (Z = 47).
Hint
2 × 47 × 1.44 / 6.
Answer
≈ 22.6 fm ≈ $2.26 \times 1 0^{- 14}$ m.
Q5 · 2 marks · Short answerHarder board level
Using SI units ( $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ N m² C⁻², e = $1.6 \times 1 0^{- 19}$ C), find the closest approach of a 3 MeV alpha particle to gold (Z = 79).
Hint
K = $4.8 \times 1 0^{- 13}$ J.
Answer
≈ $7.58 \times 1 0^{- 14}$ m.
Q6 · 2 marks · Short answerHOTS / competency
A gold nucleus has a radius of roughly 7 fm. What kinetic energy would an alpha particle need to just reach its surface head-on? What new effect might then appear?
Hint
K = 2 × 79 × 1.44 / 7.
Answer
≈ 32.5 MeV. At contact the short-range nuclear force acts and the simple Coulomb picture fails (nuclear reactions can occur).
Q7 · 2 marks · Short answerHOTS / competency
Explain why Thomson's model could not account for the backward scattering of alpha particles.
Hint
Spread-out charge, weak field.
Answer
In Thomson's model the positive charge is spread through the whole atom, so the electric field inside is weak and can give only tiny deflections; a heavy, fast alpha particle cannot be turned back.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose alpha particles of 4 MeV are fired at copper foil (Z = 29).
(i) Find the distance of closest approach. (1 mark)
(ii) Would gold give a larger or smaller value at the same energy? Why? (1 mark)
(iii) What happens to the alpha particle at $r_{0}$ ? (1 mark)
(iv) Give one reason why the observed fraction scattered beyond 90° is so small. (1 mark)
Hint
$r_{0} \propto Z$ .
Answer
(i) ≈ 20.9 fm. (ii) Larger: $r_{0} \propto Z$ and gold has Z = 79. (iii) It stops momentarily (all K converted to U) and returns. (iv) Very few particles have a small enough impact parameter because the nucleus is tiny.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Charge of an alpha particle?Show answer
+2e.
2.Formula for closest approach?Show answer
$r_{0} = \frac{2 Z e ^{2}}{4 π ε _{0} K}$ .
3.Order of nuclear size?Show answer
$1 0^{- 14}$ to $1 0^{- 15}$ m.
4.What did most alpha particles do?Show answer
Pass straight through.
5.One drawback of Rutherford's model?Show answer
It cannot explain atomic stability (or line spectra).

Rutherford's Model and Distance of Closest Approach: frequently asked questions

Why was gold used?

Gold can be beaten into extremely thin foil, and its large Z gives strong scattering.

Why did electrons not deflect the alpha particles?

An electron is about 7000 times lighter than an alpha particle, so it cannot change the alpha particle's path appreciably.

Is the distance of closest approach the same as the impact parameter?

No. r_0 is the nearest distance in a head-on collision; the impact parameter is the sideways offset of the initial path from the nucleus.

What does the ISC 2027 syllabus need here?

Rutherford's nuclear model from the Geiger-Marsden experiment, the nuclear radius from the closest approach using U = \frac{(2e)(Ze)}{4\pi\varepsilon_0r_0}, r_0 \sim 10^{-15} m, and general ideas of Z, N and A. The mathematical theory of scattering is excluded.

Has this been asked recently?

Not as a separate item in the 2024 or 2025 ISC papers, but closest approach is named in the 2027 syllabus and is a standard short numerical.