The Nuclear Model of the Atom
In 1909, Hans Geiger and Ernest Marsden at Manchester University, under Ernest Rutherford's direction, fired 8 MeV alpha particles at gold foil 10⁻⁷ m thick. Most passed straight through, but 1 in 20,000 rebounded at angles greater than 90°. Rutherford calculated that such deflections required a nuclear radius of less than 10⁻¹⁴ m, meaning the nucleus occupies only 10⁻¹⁵ of the atom's volume and the atom is otherwise empty space. Rutherford published the nuclear model of the atom in 1911.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Imagine firing tiny positively charged particles at a thin gold foil. Most pass straight through, but a few bounce back at large angles.
Before reading on, answer:
- What does it mean that most particles pass straight through?
- What does it mean that some particles bounce back?
- What model of atomic structure would explain both observations?
Warm-up: In Thomson's "plum pudding" model, the positive charge in an atom is:
Know, Rutherford's Model
- Tiny dense nucleus (~$10^{-15}$ m)
- Electrons orbit at large distance (~$10^{-10}$ m)
- Mostly empty space
- Electron evidence: cathode rays, Thomson, Millikan
Understand, Scattering Evidence
- Geiger-Marsden experiment
- Most particles undeflected
- Large-angle scattering explained
- How cathode rays revealed the electron
Can Do, Apply Nuclear Physics
- Calculate closest approach ($r_{min}$)
- Analyse scattering data
- Evaluate atomic models critically
- Combine Thomson's $e/m$ and Millikan's $e$ to find $m_e$
Core Content
How Rutherford discovered the nucleus
In 1909, Hans Geiger and Ernest Marsden, under Rutherford's direction, fired alpha particles (helium nuclei, $^4_2\text{He}$, charge $+2e$, mass $\approx 4$ u) at a very thin gold foil (~$10^{-7}$ m thick). They observed three key results:
- Most alpha particles passed straight through with little or no deflection. This indicated that atoms are mostly empty space.
- Some alpha particles were deflected through small angles ($< 10°$). This suggested the positive charge was concentrated, not spread evenly.
- A very few alpha particles (~1 in 8,000) were deflected by more than 90°, with some bouncing almost straight back. This required a very small, very dense, positively charged centre capable of exerting enormous repulsive Coulomb force.
Rutherford described this as "almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The results were completely incompatible with Thomson's "plum pudding" model, where positive charge was spread throughout the atom like a diffuse cloud.
Rutherford proposed the nuclear model:
- Almost all the atom's mass is concentrated in a tiny nucleus (~$10^{-15}$ m).
- The nucleus is positively charged.
- Electrons orbit the nucleus at relatively large distances (~$10^{-10}$ m).
- Most of the atom is empty space.
Figure 1, Alpha particle scattering showing mostly straight-through, some small deflection, and rare large-angle scattering. The fraction scattered by >90° was ~1 in 8,000.
Why would the plum pudding model predict only small-angle scattering? Why does large-angle scattering require a concentrated positive charge?
The Geiger-Marsden experiment (1909) fired alpha particles (+2e) at gold foil. Most passed straight through (atom is mostly empty space); ~1 in 8,000 scattered back (>90°), disproving Thomson's plum pudding model. Rutherford proposed a tiny dense positive nucleus (~10⁻¹⁵ m) with electrons orbiting at ~10⁻¹⁰ m.
Write this in your own words, exam questions often ask you to describe what the results implied about atomic structure.
In the Geiger-Marsden experiment, the fact that most alpha particles passed straight through the gold foil indicates that:
Using energy conservation to probe the nucleus
We just saw that large-angle scattering in the Geiger-Marsden experiment proved atoms have a tiny dense positive nucleus. That raises a question: can we use the scattering data to actually measure how big the nucleus is? This card answers it → yes, by setting kinetic energy equal to Coulomb potential energy at closest approach, we get an upper limit on nuclear radius.
When an alpha particle approaches a nucleus head-on, its kinetic energy is converted to electrical potential energy. At the closest approach, all kinetic energy has become potential energy:
$$E_k = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q_1 q_2}{r}$$For an alpha particle ($q_1 = +2e$) approaching a nucleus with atomic number $Z$ ($q_2 = +Ze$):
$$r_{min} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{2Ze^2}{E_k}$$This gives an upper limit on the nuclear radius, the actual nucleus must be smaller than $r_{min}$ because the alpha particle turns around before reaching the nuclear surface. Rutherford's calculations gave nuclear radii on the order of $10^{-15}$ m (femtometres), about 100,000 times smaller than the atomic radius (~$10^{-10}$ m).
Later experiments confirmed that nuclear radius scales with mass number:
$$R = R_0 A^{1/3}$$where $R_0 \approx 1.2$ fm and $A$ is the mass number (protons + neutrons). Since volume $\propto R^3 \propto A$ and mass $\propto A$, this implies nuclear density is approximately constant for all nuclei, nucleons are packed at roughly the same density throughout the periodic table.
$r_{min} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{2Ze^2}{E_k}$, closest approach distance (head-on)
$R = R_0 A^{1/3}$, nuclear radius ($R_0 \approx 1.2$ fm)
$F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q_1 q_2}{r^2}$, Coulomb force
$E_p = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q_1 q_2}{r}$, electrical potential energy
$k = \dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^9$ N·m²/C²; $e = 1.6\times10^{-19}$ C; 1 MeV $= 1.6\times10^{-13}$ J
Figure 2, At closest approach, all kinetic energy converts to electrical potential energy. The alpha particle momentarily stops before being repelled back.
An alpha particle with kinetic energy 5.0 MeV approaches a gold nucleus ($Z = 79$) head-on. Calculate the distance of closest approach.
Given: $e = 1.6\times10^{-19}$ C, $k = 8.99\times10^9$ N·m²/C², 1 MeV $= 1.6\times10^{-13}$ J
Closest approach formula: r_min = k·2Ze²/E_k (set E_k = E_p at r_min). This is an upper limit on nuclear radius, the actual nucleus is smaller. R = R₀A^(1/3) with R₀ ≈ 1.2 fm shows constant nuclear density. Convert 1 MeV = 1.6×10⁻¹³ J before calculating. Nuclear radius ~10⁻¹⁵ m is 100,000× smaller than atomic radius ~10⁻¹⁰ m.
Note the key formula and the MeV conversion, both will appear in HSC calculations.
The distance of closest approach $r_{min}$ is an upper limit on the nuclear radius.
The nuclear radius formula $R = R_0 A^{1/3}$ implies that larger nuclei are less dense than smaller ones.
Increasing the kinetic energy of the alpha particle decreases the closest approach distance.
What each model explains, and what it cannot
We just saw how Rutherford used closest approach calculations to pin down the nuclear radius at ~10⁻¹⁵ m. That raises a question: if Rutherford's nuclear model was such a success, why did physicists immediately need to replace it? This card answers it → classical electromagnetism predicts orbiting electrons would radiate energy and spiral into the nucleus, Rutherford's model cannot explain stable atoms.
Scientific models are judged by their ability to explain existing observations and predict new ones. The transition from Thomson's to Rutherford's model is a textbook example of how unexpected experimental evidence forces a paradigm shift.
Figure 3, Thomson's plum pudding model (left) predicts only small-angle alpha scattering. Rutherford's nuclear model (right) explains large-angle scattering via strong Coulomb repulsion from the dense nucleus.
Rutherford's nuclear model explained the scattering data brilliantly, but it had a major unresolved problem: according to classical electromagnetism, an accelerating charge radiates energy. An orbiting electron would continuously radiate, lose energy, and spiral into the nucleus in a fraction of a second. This required a new theory, Bohr's quantum model, to explain why electrons remain in stable orbits.
A common exam trap: confusing closest approach with nuclear radius. The closest approach is an upper limit, the alpha particle turns around before reaching the actual nuclear surface because Coulomb repulsion stops it. The actual nuclear radius is smaller. When calculating closest approach, convert MeV to joules: $1\text{ MeV} = 1.6\times10^{-13}$ J. Remember that $r_{min} \propto Z/E_k$, higher energy alphas or lower-Z targets give smaller closest approach distances. Also note that Rutherford's model could not explain why electrons don't spiral into the nucleus (this required Bohr's quantum model).
Thomson's plum pudding model predicts only small-angle scattering, it is disproved by large-angle results. Rutherford's nuclear model (1911) explains large-angle scattering but fails classically: an orbiting electron radiates energy and should spiral into the nucleus in ~10⁻¹¹ s. This instability required Bohr's quantum model (1913) to resolve with quantised orbits.
Jot down both models' key features and the one fatal flaw in Rutherford's, this compare-and-contrast is a regular HSC question.
An alpha particle with $E_k = 7.7$ MeV approaches a gold nucleus ($Z = 79$) head-on. Using $r_{min} = k \cdot 2Ze^2/E_k$ with $k = 8.99\times10^9$ N·m²/C², $e = 1.6\times10^{-19}$ C, and 1 MeV $= 1.6\times10^{-13}$ J, the distance of closest approach in metres (to 2 sig figs) is approximately $2.96\times10^{-14}$ m. Enter the coefficient (the first two sig figs): _____×10⁻¹⁴ m.
Three of these statements about Rutherford's nuclear model are correct. Pick the odd one out.
The first direct evidence that atoms contain charged particles
We just saw that Rutherford's nuclear model places a tiny positive nucleus at the centre of a mostly empty atom, orbited by electrons. That raises a question: how did physicists know electrons existed at all, decades before the nucleus was found? This card answers it → cathode-ray experiments in the 1890s isolated the electron and measured its properties, the very evidence NESA requires you to model.
A cathode ray is a stream of negatively charged particles, electrons, emitted from the negative electrode (cathode) when a high voltage is applied across a gas at very low pressure in a sealed tube. In the late 1800s, physicists studied these glowing discharges intensively to work out what the rays actually were.
Careful experiments revealed a consistent set of properties:
- They travel in straight lines. An object placed in the beam casts a sharp shadow on the far end of the tube.
- They are deflected by both electric and magnetic fields. A charged ray must respond to both, which a beam of light (electromagnetic radiation) would not do in the same way.
- They carry energy and momentum. A small paddle wheel placed in the tube spins as the rays strike it, showing the rays transfer mechanical momentum, evidence they are particles with mass.
- They are deflected toward the positive plate. Because the rays bend toward the positive electrode in an electric field, they must carry negative charge.
These observations settled a long-running wave-versus-particle debate. Some physicists argued cathode rays were a form of wave in the ether; others argued they were charged particles. The deflection by electric and magnetic fields, together with the paddle-wheel momentum transfer, decided the matter in favour of particles.
Most strikingly, the same particle appeared regardless of the gas filling the tube or the metal used for the electrodes. This universality implied that electrons are a fundamental constituent of all matter, not a property of one particular substance. J.J. Thomson is credited with identifying this particle, the electron, in 1897.
Figure 4, A cathode-ray tube. Electrons stream from the cathode, pass through the anode slit, and deflect toward the positive plate in an electric field, confirming they are negatively charged particles.
If the same particle is produced no matter which gas fills the tube or which metal forms the electrodes, what does that tell you about where electrons come from? Why does the paddle-wheel result favour the particle model over the wave model?
Cathode rays are streams of electrons from the cathode in a low-pressure discharge tube. They travel in straight lines (cast shadows), are deflected by electric and magnetic fields, carry energy and momentum (spin a paddle wheel), and bend toward the positive plate, so they carry negative charge. The same particle appears regardless of the gas or electrode metal, so the electron is a universal constituent of all matter, resolving the wave-versus-particle debate in favour of particles.
List the four cathode-ray properties in your own words, NESA asks you to model the experimental evidence for the existence and properties of the electron.
Cathode rays are deflected toward the positively charged plate in an electric field. This observation shows that cathode rays:
Measuring the electron's charge, mass and the quantisation of charge
We just saw that cathode rays are negatively charged particles common to all matter. That raises a question: how big is an electron's charge, and how heavy is it? This card answers it → Thomson measured the charge-to-mass ratio, Millikan measured the charge itself, and together they pin down the electron mass.
In 1897, J.J. Thomson measured the charge-to-mass ratio ($e/m$) of the electron using crossed electric and magnetic fields inside a cathode-ray tube.
His method works in two stages:
- Balance the fields to find the speed. An electric field (strength $E$) and a magnetic field (strength $B$) are applied at right angles so their forces on the beam act in opposite directions. The field strengths are adjusted until the beam passes through undeflected. When the electric force balances the magnetic force, $qE = qvB$, so the speed is
- Remove the magnetic field and measure the deflection. With only the electric field acting, the beam curves. Measuring how far the beam deflects lets Thomson calculate $e/m$ from the geometry of the tube and the known fields.
Thomson's result was a charge-to-mass ratio of about
$$\dfrac{e}{m} \approx 1.76\times10^{11}\ \text{C/kg}$$This value was enormous, far larger than the charge-to-mass ratio of a hydrogen ion, which told Thomson the electron is either very highly charged or, as turned out to be the case, extremely light.
Thomson's experiment gives the ratio $e/m$, but not $e$ and $m$ separately. The missing piece came from Robert Millikan's oil-drop experiment (1909-1913). Millikan sprayed tiny charged oil droplets between two horizontal plates and adjusted the voltage until the upward electric force balanced gravity, holding a droplet stationary. By repeating this for thousands of droplets, he found that the charge on a droplet was always a whole-number multiple of a single smallest value:
$$e = 1.6\times10^{-19}\ \text{C}$$This showed that electric charge is quantised, it comes in discrete packets of $e$, the elementary charge. Combining Millikan's $e$ with Thomson's $e/m$ gives the electron mass directly:
$$m_e = \dfrac{e}{(e/m)} = \dfrac{1.6\times10^{-19}}{1.76\times10^{11}} \approx 9.1\times10^{-31}\ \text{kg}$$This tiny mass, roughly $1/1836$ of a proton's mass, confirmed that the electron is by far the lightest constituent of the atom.
Figure 5, Left: Thomson balances electric and magnetic forces so the beam is undeflected, giving v = E/B and then e/m. Right: Millikan balances a charged oil drop against gravity to show charge comes in multiples of e.
Use Thomson's $e/m \approx 1.76\times10^{11}$ C/kg and Millikan's $e \approx 1.6\times10^{-19}$ C to find the electron mass.
$m_e = e \div (e/m) = (1.6\times10^{-19}) \div (1.76\times10^{11}) \approx 9.1\times10^{-31}$ kg. Notice that neither experiment alone gives the mass, you need both results together.
Thomson measured the electron's charge-to-mass ratio with crossed E and B fields: balance them so the beam is undeflected to get v = E/B, then remove B and measure the deflection, giving e/m ≈ 1.76×10¹¹ C/kg. Millikan's oil-drop experiment showed charge is quantised in multiples of e = 1.6×10⁻¹⁹ C. Combining the two gives the electron mass m_e = e ÷ (e/m) ≈ 9.1×10⁻³¹ kg, about 1/1836 of a proton.
Memorise the three numbers and how they connect: e/m from Thomson, e from Millikan, and m_e from combining them.
Why is Millikan's oil-drop experiment needed in addition to Thomson's charge-to-mass experiment to find the electron's mass?
Activities
Practice using $r_{min} = k \cdot 2Ze^2/E_k$
- An alpha particle with $E_k = 5.0$ MeV approaches a gold nucleus ($Z = 79$) head-on. Calculate the distance of closest approach in metres. ($k = 8.99\times10^9$ N·m²/C², $e = 1.6\times10^{-19}$ C, 1 MeV $= 1.6\times10^{-13}$ J)
- Repeat for a silver target ($Z = 47$). How does the closest approach compare to gold?
- If the alpha particle's kinetic energy is doubled, what happens to $r_{min}$? Explain using the formula.
- The nuclear radius formula gives $R = R_0 A^{1/3}$ with $R_0 = 1.2$ fm. Calculate the radius of a gold nucleus ($A = 197$) and compare it to your answer in Q1.
Compare models and explain experimental evidence
- Describe two observations from the Geiger-Marsden experiment and explain how each supports Rutherford's nuclear model over Thomson's plum pudding model.
- Explain why Rutherford's model was a significant advance over Thomson's model, even though it could not explain stable electron orbits.
- A student argues: "Because most alpha particles passed straight through, the nucleus must be large." Identify the flaw in this reasoning.
- The nuclear density of gold is approximately $2.3\times10^{17}$ kg/m³. Compare this to the density of water ($1000$ kg/m³). What does this tell us about how matter is distributed in an atom?
A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 4(3 marks) 1. An alpha particle with kinetic energy 5.0 MeV approaches a silver nucleus ($Z = 47$) head-on. (a) State the energy equation used to find the closest approach. (b) Calculate the distance of closest approach in metres. (c) Explain why this distance is an upper limit for the nuclear radius.
1 mark: correct equation · 1 mark: correct $r_{min}$ · 1 mark: upper-limit explanation
AnalyseBand 6(5 marks) 2. (a) Describe the Geiger-Marsden experiment including the apparatus, procedure and the three key observations. (b) Explain how each observation led Rutherford to propose the nuclear model. (c) Outline one limitation of Rutherford's model and identify the theory that resolved it. (d) The nuclear radius formula is $R = R_0 A^{1/3}$ with $R_0 = 1.2$ fm. State what this formula implies about nuclear density across the periodic table.
1 mark: apparatus/procedure · 1 mark: three observations · 1 mark: nuclear model reasoning · 1 mark: limitation + resolution · 1 mark: constant density implication
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer, Model Answers
Q1 (3 marks): (a) At closest approach, all kinetic energy converts to electrical potential energy: $E_k = k \cdot \frac{q_1 q_2}{r_{min}} = k \cdot \frac{2Ze^2}{r_{min}}$ (1 mark). (b) $r_{min} = k \cdot 2Ze^2/E_k = (8.99\times10^9)(2)(47)(1.6\times10^{-19})^2 / (5.0\times1.6\times10^{-13})$. Numerator: $8.99\times10^9 \times 94 \times 2.56\times10^{-38} = 8.99 \times 94 \times 2.56\times10^{-29} \approx 2.162\times10^{-27}$ J·m. Denominator: $8.0\times10^{-13}$ J. So $r_{min} \approx 2.70\times10^{-14}$ m (1 mark). (c) The alpha particle stops at $r_{min}$ due to Coulomb repulsion before it reaches the nuclear surface; the actual nuclear radius must be smaller than $r_{min}$, so it is an upper limit (1 mark).
Q2 (5 marks): (a) Apparatus: radioactive alpha source in a lead box directed at a thin gold foil (~100 nm thick); zinc sulfide detector screens surrounded the foil to detect scattering at all angles. Three observations: (1) most alphas passed straight through with no deflection; (2) a small fraction was deflected through small angles; (3) ~1 in 8,000 was deflected by more than 90°, some bouncing almost directly back (1 mark for apparatus + procedure; 1 mark for all three observations). (b) Observation (1) → atoms are mostly empty space (nucleus very small relative to atom); (2) → positive charge is concentrated, not diffuse; (3) → a tiny, dense, highly positive nucleus can repel alphas strongly enough for large-angle scattering, all of which disprove the plum pudding model (1 mark). (c) Rutherford's model cannot explain why orbiting electrons don't radiate energy and spiral into the nucleus, classical electromagnetism predicts accelerating charges radiate. Resolved by Bohr's quantum model (1913), in which electrons occupy stable, quantised energy levels (1 mark). (d) Since $R \propto A^{1/3}$, volume $V \propto R^3 \propto A$, and mass $m \propto A$, density $= m/V$ is approximately constant, all nuclei have roughly the same density regardless of size (1 mark).
At the start you were asked about the Geiger-Marsden experiment at Manchester University in 1909, firing 8 MeV alpha particles at gold foil and finding that 1 in 20,000 bounced back at more than 90°. This result caused Rutherford to conclude that the nuclear radius must be less than 10⁻¹⁴ m, overturning Thomson's plum-pudding model.
- Did you predict most alpha particles pass straight through the Geiger-Marsden gold foil because atoms are mostly empty space? Correct, the nucleus is tiny (~10⁻¹⁵ m) compared to the atomic size (~10⁻¹⁰ m), leaving vast empty space for alphas to traverse.
- Did you predict the 1-in-20,000 backscatter requires a dense, concentrated positive charge at the atom's centre? Correct, large-angle Coulomb deflection requires a nuclear radius less than 10⁻¹⁴ m, as Rutherford calculated.
- Did you predict a nuclear model with a tiny dense core and orbiting electrons? Correct, this is Rutherford's 1911 nuclear model, the direct consequence of Geiger and Marsden's scattering data.
Extend: A proton (charge $+e$, kinetic energy 3.0 MeV) approaches a lead nucleus ($Z = 82$) head-on. (a) Calculate the distance of closest approach. (b) Compare this to the nuclear radius of lead ($A = 207$, $R_0 = 1.2$ fm). (c) Explain what happens physically if the proton's energy is increased well beyond the Coulomb barrier.
Five timed questions on the nuclear model and Rutherford scattering. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).
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