Inside the atom: plum pudding, a gold foil and the discovery of the nucleus
By the end of this lesson you will be able to
- Outline the plum-pudding and nuclear models of the atom and explain which evidence made scientists replace one with the other.
- Describe the Geiger–Marsden experiment and explain each of its three observations in terms of the nucleus.
- Calculate sizes, ratios and densities for the atom and the nucleus in scientific notation.
- State the relative mass, charge and position of the proton, neutron and electron.
- Use nuclide notation to determine the numbers of protons, neutrons and electrons in atoms and ions.
- Define isotopes and state one similarity and one difference between isotopes in the exact words examiners accept.
You met this before
- All matter is made of atoms, far too small to see: a typical atom is about 1 × 10⁻¹⁰ m across, so ten million of them side by side make one millimetre.
- J. J. Thomson discovered the electron in 1897: a tiny, negatively charged particle about 1 800 times lighter than a hydrogen atom, found inside every atom.
- Opposite charges attract and like charges repel; the closer two charges are, the stronger the force.
- An atom is electrically neutral, so the positive charge inside it exactly balances the negative charge of its electrons.
- Electrons sit in shells around the centre of the atom: 2, then 8, then 8 for the first twenty elements.
An oxygen atom has 8 electrons. How much positive charge must the rest of the atom carry, and where did Thomson think that positive charge was?
Reveal
The atom is neutral, so the positive charge must be +8 (in units of the electron charge) to balance the 8 electrons.
Thomson placed it in a uniform, soft “pudding” of positive charge filling the whole atom, with the electrons dotted through it like raisins. This lesson is the story of why that picture had to go.
Revise it in full: Year 2 · Atoms and their structure · Lesson 2 — Thomson plum pudding: the glowing tube that split the atom · Year 2 · Atoms and their structure · Lesson 3 — Rutherford surprise: the gold foil and the empty atom · Year 2 · Atoms and their structure · Lesson 4 — Bohr and the modern atom: shells, colours and the neutron
The puzzle
In 1909 a twenty-year-old student, Ernest Marsden, sat in a dark room in Manchester counting tiny green flashes through a microscope. Each flash was one alpha particle landing on a screen after being fired at a sheet of gold thinner than a soap bubble.
His supervisor, Hans Geiger, had asked him to check something that everybody “knew” would not happen: whether any alpha particles bounced back off the gold.
An alpha particle is about 7 300 times heavier than an electron and travels at roughly 15 000 km every second. Firing it at gold leaf should be like firing a bullet through tissue paper. Yet about one alpha particle in every 8 000 came back.
Rutherford called it “almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you”. By the end of this lesson you will know exactly why it happened — and what it tells you about the atoms you are made of.
Suppose the positive charge of a gold atom is spread evenly through the whole atom, as Thomson believed. Predict what happens to fast alpha particles fired at a thin gold foil. Then predict what would have to be true inside the atom for a few of them to bounce straight back.
Compare your answer
Spread-out charge exerts only weak forces — the pulls and pushes from every side almost cancel — so every alpha particle should go straight through, deflected by at most a fraction of a degree.
To turn a heavy, fast particle round, you need a very strong force, and a strong electric force needs a lot of charge packed into a very small space, so that the alpha can get very close to it. Only a tiny, massive, concentrated centre could do that. That is the nucleus.
The science — structured notes
1.1 Models change when evidence demands it
A scientific model is a simplified picture that explains what has been observed and predicts what has not. A model is kept while its predictions come true and replaced when a careful measurement contradicts it. The atom is the classic example: five models in about 130 years, each one forced by a new experiment.
| Model | Who and when | The picture | The evidence behind it |
|---|---|---|---|
| Solid sphere | John Dalton, 1803–1808 | Atoms are tiny indivisible balls; each element has its own kind | Elements combine in fixed ratios by mass |
| Plum pudding | J. J. Thomson, 1904 | A sphere of spread-out positive charge with electrons embedded in it | Cathode rays are streams of electrons (1897), found in every element — so atoms have parts |
| Nuclear | Ernest Rutherford, 1911 | A tiny, dense, positive nucleus holds almost all the mass; electrons move in the empty space around it | Geiger and Marsden’s alpha scattering (1909) |
| Shells (planetary) | Niels Bohr, 1913 | Electrons can occupy only certain energy levels (shells) | The sharp line spectra of hydrogen |
| Protons and neutrons | Rutherford 1919; James Chadwick 1932 | The nucleus contains protons and uncharged neutrons | Protons knocked out of nitrogen; neutral radiation from beryllium knocking protons out of paraffin wax |
Notice the pattern: each model kept what the previous one got right. Thomson kept Dalton’s elements but opened the atom; Rutherford kept Thomson’s electrons but moved the positive charge. That is how science normally progresses.
1.2 The Geiger–Marsden experiment
Alpha particles are emitted by some radioactive nuclei. Each is a helium nucleus — 2 protons and 2 neutrons, charge +2e — moving at about 5 % of the speed of light. Geiger and Marsden used them as bullets to probe the inside of atoms.
The design details matter, and examiners ask about them:
- Vacuum. Alpha particles ionise air strongly and stop after 3–7 cm, so the chamber was pumped out.
- Very thin foil. Gold can be beaten thinner than almost any other metal, so each alpha meets only a few hundred layers of atoms and is usually scattered by one nucleus at most.
- Narrow beam. A slit in the lead block makes all the alphas start in the same direction, so any change of direction was caused by the foil.
- Counting by eye. Each alpha hitting the zinc sulfide screen makes one faint flash (a scintillation). The observers sat in the dark for up to half an hour first to let their eyes adapt, and could count reliably for only about a minute at a time.
Explain why the experiment would have failed if the chamber had been full of air, and why a thick gold block would have been useless.
Reveal answer
In air each alpha particle collides with and ionises thousands of air molecules, losing energy until it stops after a few centimetres — it would never reach the screen at large angles, so nothing could be counted.
In a thick block every alpha would be stopped inside the gold (its range in gold is only about 0.01 mm), and those that did get out would have been deflected many times, so no single scattering event could be interpreted.
1.3 Three observations, three conclusions
| Observation | Conclusion about the atom |
|---|---|
| Almost all alpha particles passed straight through the foil, undeflected. | The atom is mostly empty space. |
| A small fraction were deflected through small or moderate angles. | The centre of the atom is positively charged: it repels the positive alphas that pass near it. |
| About 1 in 8 000 was deflected through more than 90° — turned back. | The positive charge and almost all the mass are concentrated in a very small, dense nucleus; only a massive, concentrated charge could exert a force big enough to reverse a fast, heavy alpha. |
Rutherford did more than describe: he calculated how many alphas the nuclear atom predicts at each angle. The count should be proportional to 1 ÷ sin⁴(θ/2), where θ is the scattering angle, and proportional to the square of the nuclear charge. In 1913 Geiger and Marsden tested every prediction and found them right — a model that makes numerical predictions and passes them is far stronger than one that merely explains.
A student writes: “Most alpha particles went straight through because the gold atoms are very small, and some bounced back because they hit electrons.” Identify both errors and rewrite the answer.
Reveal answer
The size of the atom is not the reason: the alphas pass through the atoms, so the reason is that each atom is mostly empty space. And electrons cannot turn an alpha round — an alpha is about 7 300 times heavier.
Rewrite: “Most alpha particles passed straight through because the atom is mostly empty space. A few were turned back because they approached a tiny, dense, positively charged nucleus, which contains almost all the mass of the atom and repelled them strongly.”
1.4 How small, how dense? Sizes in scientific notation
From the angles and numbers of scattered alphas Rutherford estimated that the gold nucleus is smaller than about 3 × 10⁻¹⁴ m. Modern measurements give a radius of about 7 × 10⁻¹⁵ m for gold, while the gold atom has a radius of about 1.4 × 10⁻¹⁰ m. The atom is about twenty thousand times wider than its nucleus.
A gold atom has radius 1.4 × 10⁻¹⁰ m and its nucleus has radius 7.0 × 10⁻¹⁵ m. (1) Calculate how many times wider the atom is. (2) If the nucleus were a pea 5.0 mm across, how wide would the atom be? (3) What fraction of the atom’s volume does the nucleus fill?
(1) ratio = atom radius ÷ nucleus radius
ratio = 1.4 × 10⁻¹⁰ m ÷ 7.0 × 10⁻¹⁵ m
ratio = 2.0 × 10⁴ (twenty thousand)
(2) atom width = 5.0 mm × 2.0 × 10⁴
atom width = 1.0 × 10⁵ mm
atom width = 100 m — a football pitch around a pea
(3) volume ∝ radius³, so fraction = (1 ÷ 2.0 × 10⁴)³
fraction = 1 ÷ 8.0 × 10¹²
fraction = 1.25 × 10⁻¹³
Only about a ten-million-millionth (10⁻¹³) of the atom is nucleus — which is why almost every alpha misses.
The gold nucleus has a mass of 3.27 × 10⁻²⁵ kg and a radius of 7.0 × 10⁻¹⁵ m. (1) Calculate the density of nuclear matter. (2) Compare it with solid gold, 1.93 × 10⁴ kg/m³. (3) A neutron star is made of matter this dense: find the mass of one teaspoon (5.0 cm³) of it.
(1) volume V = (4 ÷ 3)πr³
V = (4 ÷ 3) × π × (7.0 × 10⁻¹⁵ m)³
V = 1.44 × 10⁻⁴² m³
density ρ = m ÷ V
ρ = 3.27 × 10⁻²⁵ kg ÷ 1.44 × 10⁻⁴² m³
ρ = 2.3 × 10¹⁷ kg/m³
(2) ratio = 2.3 × 10¹⁷ ÷ 1.93 × 10⁴
ratio ≈ 1.2 × 10¹³ times denser than gold
(3) 5.0 cm³ = 5.0 × 10⁻⁶ m³
mass = 2.3 × 10¹⁷ kg/m³ × 5.0 × 10⁻⁶ m³
mass = 1.1 × 10¹² kg — about a billion tonnes
Solid gold is “dense” only because its nuclei are dense; the rest of the metal is the empty space around them.
A beam delivers 2.0 × 10⁵ alpha particles per minute to a gold foil, and 1 in 8 000 is turned back through more than 90°. Calculate how many are turned back in one hour.
Reveal answer
per minute = 2.0 × 10⁵ ÷ 8 000
per minute = 25
per hour = 25 × 60
per hour = 1 500 (1.5 × 10³)
Geiger and Marsden’s sources were weaker and their screen was small, so they saw only a few flashes per minute at large angles — the patience behind the discovery.
The gold foil is replaced by aluminium foil of the same thickness. Aluminium has 13 protons per nucleus; gold has 79. Predict how the count at 90° changes, using the fact that the count is proportional to the square of the nuclear charge.
Reveal answer
factor = (13 ÷ 79)²
factor = 0.165²
factor = 0.027
The count falls to about one thirty-seventh: from about 21 to about 0.6 per minute. A smaller nuclear charge repels less strongly, so fewer alphas are turned through large angles. (Aluminium atoms are also packed differently, but the nuclear charge is the main effect.) You can test this in the lab below.
1.5 Protons, neutrons, electrons and nuclide notation
Bohr (1913) showed that electrons occupy fixed energy levels. Rutherford (1919) knocked hydrogen nuclei — which he named protons — out of nitrogen with alpha particles. Chadwick (1932) found the neutron, which explained why nuclei weigh about twice as much as their protons alone. Protons and neutrons together are called nucleons.
| Particle | Relative mass | Relative charge | Where |
|---|---|---|---|
| Proton | 1 | +1 | nucleus |
| Neutron | 1 | 0 | nucleus |
| Electron | about 1 ÷ 1 840 (almost 0) | −1 | shells around the nucleus |
A particular nucleus — a nuclide — is written ᴬZX, where X is the chemical symbol:
- Z, the atomic number (proton number): the number of protons. It fixes the element and the nuclear charge.
- A, the mass number (nucleon number): the total number of protons and neutrons.
- N, the neutron number: N = A − Z.
- In a neutral atom, the number of electrons equals Z. An ion has gained or lost electrons, never protons.
Copper is written ⁶³29Cu (the May 2021 examination used this atom). (1) State what 29 and 63 tell you. (2) Determine the numbers of protons, neutrons and electrons in a neutral atom. (3) Determine the numbers in a Cu²⁺ ion.
(1) 29 is the number of protons in the nucleus; 63 is the total number of protons and neutrons in the nucleus
(2) protons = Z = 29
neutrons N = A − Z
N = 63 − 29
N = 34
electrons = protons in a neutral atom = 29
(3) Cu²⁺ has lost 2 electrons: electrons = 29 − 2
electrons = 27; still 29 protons and 34 neutrons
Nuclear fuel uses ²³⁵92U. Determine the numbers of protons and neutrons, and the total charge of the nucleus in coulombs (the charge of one proton is 1.6 × 10⁻¹⁹ C).
protons = 92
neutrons = 235 − 92
neutrons = 143
charge = 92 × 1.6 × 10⁻¹⁹ C
charge = 1.5 × 10⁻¹⁷ C
The May 2024 examiners reported that many candidates confused the protons and neutrons of uranium-235. Write N = A − Z every time, even when it feels obvious.
Complete for each nuclide the number of protons, neutrons and electrons in the neutral atom: ⁴2He, ²⁷13Al, ²³⁸92U.
Reveal answer
⁴2He: 2 protons, 2 neutrons, 2 electrons (its nucleus is an alpha particle)
²⁷13Al: 13 protons, 14 neutrons, 13 electrons
²³⁸92U: 92 protons, 146 neutrons, 92 electrons
1.6 Isotopes
Isotopes are atoms of the same element — the same number of protons — with different numbers of neutrons. They have the same number of electrons, so the same chemistry, but different masses, and their nuclei can differ in stability: carbon-12 is stable, carbon-14 is radioactive.
| Isotope | Protons | Neutrons | Stable? | Found in |
|---|---|---|---|---|
| ¹²6C | 6 | 6 | yes | 98.9 % of carbon |
| ¹³6C | 6 | 7 | yes | 1.1 % of carbon |
| ¹⁴6C | 6 | 8 | no — beta emitter, half-life 5 730 years | about 1 atom in 10¹² of carbon |
| ²³⁵92U | 92 | 143 | no — alpha emitter; the fissile fuel | 0.72 % of natural uranium |
| ²³⁸92U | 92 | 146 | no — alpha emitter, half-life 4.5 × 10⁹ years | 99.3 % of natural uranium |
Similarity: the same number of protons (the same nuclear charge).
Difference: a different number of neutrons (C-14 has 8, C-12 has 6); or C-14 is unstable and C-12 is stable.
The relative atomic mass on the periodic table is the average mass of an element’s atoms, weighted by how common each isotope is.
Natural chlorine is 75.0 % chlorine-35 and 25.0 % chlorine-37. Calculate its relative atomic mass.
Ar = (75.0 × 35 + 25.0 × 37) ÷ 100
Ar = (2 625 + 925) ÷ 100
Ar = 35.5
“Tritium, ³H, is a different element from ordinary hydrogen because it is three times as heavy.”
Reveal answer
False. The element is fixed by the number of protons alone. Tritium has one proton, so it is hydrogen; it is an isotope of hydrogen with two extra neutrons. It even forms water — “tritiated water” — that behaves chemically like ordinary water.
Boron has two stable isotopes, boron-10 and boron-11, and its relative atomic mass is 10.81. Calculate the percentage of boron-10 in natural boron.
Reveal answer
Let the fraction of boron-10 be x, so the fraction of boron-11 is 1 − x.
10x + 11(1 − x) = 10.81
11 − x = 10.81
x = 0.19
boron-10 = 19 % (boron-11 is 81 %)
Boron-10 absorbs neutrons strongly, which is why boron is used in the control rods of nuclear reactors (Lesson 5).
Try it — Geiger and Marsden on your screen: counts against scattering angle
The lab below is the 1909 apparatus: an alpha source in a vacuum chamber, a thin foil and a detector on a movable arm that counts the alpha particles arriving at each angle in one minute. You set the angle (the independent variable, 10° to 150°) and can change the foil metal and thickness. Radioactive emission is random, so repeated counts differ — the scatter is real physics, not a fault.
Common mistakes to avoid
- “The alpha particles hit electrons and bounced back.” Electrons are about 7 300 times lighter than alphas; only the massive nucleus can reverse them.
- Pairing the wrong observation with the wrong conclusion. Straight through → empty space; deflected → positive nucleus; bounced back → small, dense, massive nucleus.
- Saying the nucleus “attracts” the alpha particle. Both are positive: it repels.
- “The atom is empty.” It is mostly empty space; the electrons and the nucleus are real.
- Mixing up A and Z. The bigger number is the mass number (top); the smaller is the atomic number (bottom). Neutrons = A − Z.
- “Same atomic number, different mass number” for isotopes. Examiners want the particles: same number of protons, different number of neutrons.
- Changing the number of protons to make an ion. Ions gain or lose electrons; changing protons makes a different element.
- Forgetting the vacuum and the thin foil when asked to describe or evaluate the apparatus.
Study smart: how this is examined
Study routine. Today: write the observation–conclusion table from memory and draw the three alpha paths (straight, deflected, back) past a nucleus. In two days: do the proton–neutron–electron counts for ten nuclides and ions, then the pea and teaspoon EXAMPLES with the book closed. In a week: run the scattering lab as a full investigation, process the product count × sin⁴(θ/2), and take the quiz.
How we found out — flashes in the dark
J. J. Thomson measured the charge-to-mass ratio of cathode rays in Cambridge in 1897 and showed that they were particles far lighter than any atom — electrons. Since atoms are neutral, something positive had to balance them, and in 1904 Thomson proposed the “plum pudding”: electrons embedded in a sphere of spread-out positive charge.
Ernest Rutherford, a New Zealander working in Manchester, had already shown that radioactivity gave off two kinds of rays, which he named alpha and beta (1899), and with Thomas Royds proved in 1908–1909 that alpha particles are helium nuclei. He noticed that a beam of alphas passing through thin mica became slightly blurred — something inside matter could deflect them.
In 1909 his assistant Hans Geiger and the undergraduate Ernest Marsden looked for alphas reflected from metal foils. They found that about 1 in 8 000 came back through more than 90°. Two years later, in 1911, Rutherford published the explanation: all the positive charge and nearly all the mass sit in a tiny central nucleus.
He calculated how the number of alphas should depend on the angle, the foil thickness, the nuclear charge and the alphas’ speed. Geiger and Marsden tested all four predictions over the next two years, counting many thousands of flashes by eye, and published in 1913 that every one held.
The model still had a problem — orbiting electrons should radiate energy and spiral into the nucleus. Niels Bohr, who had worked with Rutherford, solved it in 1913 with fixed energy levels.
In 1919 Rutherford fired alpha particles into nitrogen and detected the protons knocked out of its nuclei (Patrick Blackett showed in 1925 that the nitrogen becomes oxygen). In 1932 James Chadwick identified the neutron. Marsden later moved to New Zealand and led its scientific research for decades.
Geiger and Marsden spent two years testing Rutherford’s numerical predictions after he had already published his explanation. Explain why this second stage was scientifically important.
Reveal answer
An explanation that fits one surprising result could be a coincidence; other models might explain back-scattering too. A model that predicts how many alphas arrive at each angle, and how that number changes with thickness, nuclear charge and speed, can be proved wrong by any one measurement.
When all four predictions were confirmed, the evidence for the nuclear model became very strong, and other scientists accepted it. Testable, quantitative predictions are what make a model scientific.
Where it is used now
- Rutherford backscattering in industry. Beams of helium ions are still fired at thin films, exactly as in 1909; the energy of the bounced-back ions reveals which elements are present and at what depth. Chip makers use it to check layers only nanometres thick.
- Ion-beam analysis of heritage objects. The AGLAE accelerator beneath the Louvre in Paris analyses paintings and jewellery without taking a sample. iThemba LABS in South Africa uses ion beams in materials science and medical research.
- Finding the quarks. In the late 1960s, electrons fired at protons at the Stanford Linear Accelerator were scattered through unexpectedly large angles — Geiger and Marsden’s surprise again, one level deeper — revealing that protons contain point-like quarks.
- Isotope fingerprints. The ratio of hydrogen-2 to hydrogen-1 and of oxygen-18 to oxygen-16 in water depends on climate, so isotope ratios in ice cores record past temperatures, and in food or ivory can reveal where it came from.
- Heavy water. Water made with deuterium is used as the moderator in CANDU nuclear reactors (Lesson 5).
Science now — and the scientist you need to be
What is happening in 2026
The question Geiger and Marsden asked — what is inside? — is still being asked with scattering experiments. At Brookhaven National Laboratory in the United States the Electron-Ion Collider is being built, planned to start in the 2030s, to map how quarks and gluons produce the mass and spin of protons and nuclei.
Meanwhile, electron microscopes combined with computer reconstruction (“ptychography”) now resolve individual atoms in materials, and machine-learning tools help interpret the enormous data sets from particle detectors and from microscopes.
Careers grow from here: materials science, semiconductor engineering, nuclear medicine, geochemistry, forensic science and data science all use isotopes or scattering every day. The skill that links them is the one Rutherford used: reasoning from indirect evidence to a model you cannot see, then testing its predictions.
The mindshift
What you need to be able to do
Explain each gold-foil observation with its conclusion. Handle numbers from 10⁻¹⁹ to 10¹⁷ in scientific notation without losing a power of ten. Read ᴬZX and count particles instantly. Describe isotopes in terms of protons and neutrons. Judge a model by whether its predictions match the data.
See an example
A typical first answer says the alphas bounced off “the nucleus” but does not say why the nucleus could do it. Add: the nucleus is positive (repulsion) and contains almost all the mass of the atom in a tiny volume (so the force is huge at close range and the nucleus hardly recoils).
For uranium-235: 235 − 92 = 143 neutrons. If an answer ever gives 92 neutrons, it has confused Z with N.
Key terms
| Model | A simplified representation that explains observations and makes testable predictions. |
| Plum-pudding model | Thomson’s 1904 model: electrons embedded in a sphere of uniformly spread positive charge. |
| Nuclear model | Rutherford’s 1911 model: a tiny, dense, positive nucleus holding almost all the mass, surrounded by electrons in mostly empty space. |
| Alpha particle | A helium nucleus, ⁴2He: two protons and two neutrons, charge +2. |
| Nucleus | The central part of an atom, containing protons and neutrons; radius about 10⁻¹⁵ to 10⁻¹⁴ m. |
| Nucleon | A proton or a neutron. |
| Atomic number, Z | The number of protons in the nucleus (proton number). |
| Mass number, A | The total number of protons and neutrons in the nucleus (nucleon number). |
| Nuclide | A particular kind of nucleus with a stated A and Z, written ᴬZX. |
| Isotopes | Atoms with the same number of protons but different numbers of neutrons. |
Think like a scientist
Criterion C A student repeated the gold-foil experiment with a modern alpha source, a thin gold foil in a vacuum chamber and a solid-state detector on a rotating arm. She counted the alpha particles arriving in one minute, three times at each angle.
| Angle θ / ° | Count trial 1 | Count trial 2 | Count trial 3 | Mean count per minute |
|---|---|---|---|---|
| 30 | 1 141 | 1 198 | 1 150 | |
| 50 | 159 | 170 | 162 | |
| 70 | 47 | 81 | 46 | |
| 90 | 21 | 19 | 23 | |
| 110 | 12 | 11 | 12 | |
| 150 | 6 | 5 | 7 |
Values of sin⁴(θ/2): 30°: 0.00449; 50°: 0.0319; 70°: 0.108; 90°: 0.250; 110°: 0.450; 150°: 0.871.
(a) Identify the anomalous result, suggest a specific cause, and calculate the mean count for each angle, to the nearest whole count. [3]
(b) Calculate mean count × sin⁴(θ/2) for each angle and state what the results show about Rutherford’s prediction. [3]
(c) Describe the graph you would plot to test whether the count is directly proportional to 1 ÷ sin⁴(θ/2), and the line you would expect. [2]
(d) The spread of the three trials is about 5 % of the mean at 30° but about 30 % of the mean at 150°. Explain why, and suggest how to make the 150° result more reliable. [3]
(e) Write a conclusion for the investigation, using data. [2]
Model answer
(a) Anomaly: 81 at 70°, far above 47 and 46 [1]. Cause: the arm slipped towards a smaller angle during that count, or the counter was not reset from a previous run and added counts — not “a mistake” [1].
Means [1]: 1 163, 164, 47 (46.5 rounded, excluding 81), 21, 12 (11.7), 6.
(b) Products [1 for method, 1 for values]:
1 163 × 0.00449 = 5.2
164 × 0.0319 = 5.2
47 × 0.108 = 5.1
21 × 0.250 = 5.3
12 × 0.450 = 5.4
6 × 0.871 = 5.2
The product is constant within the scatter (5.1 to 5.4), so the count is inversely proportional to sin⁴(θ/2), as the nuclear model predicts [1].
(c) Mean count on the vertical axis against 1 ÷ sin⁴(θ/2) on the horizontal, both labelled with quantity and unit (counts per minute; no unit for the sine term) [1]. A straight line through the origin with gradient about 5.2 counts per minute [1].
(d) Radioactive emission and scattering are random; a count of N typically varies by about √N [1].
At 30°, √1 163 is about 34, 3 % of the count; at 150°, √6 is about 2.4, 40 % of the count, so small counts scatter far more in proportion [1]. Improve by counting for longer at large angles — ten minutes gives about 60 counts and a spread near 13 % — and dividing by the time [1].
(e) As the angle increased from 30° to 150° the count fell from 1 163 to 6 per minute, a factor of about 190 [1], and mean count × sin⁴(θ/2) stayed between 5.1 and 5.4, so the count is inversely proportional to sin⁴(θ/2), supporting the nuclear model [1].
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
In the Geiger–Marsden experiment most alpha particles passed straight through the gold foil, some were deflected and very few were turned back. Explain what each observation shows about the structure of the atom.
Reveal answer
Most passed straight through: the atom is mostly empty space [1].
Some deflected: the nucleus is positively charged and repels the positive alpha particles [1].
Very few turned back: the positive charge and almost all the mass are concentrated in a very small, dense nucleus [1].
Select the correct words. In ⁵⁶26Fe, 26 is the (number of neutrons / number of protons / number of nucleons) and 56 is the (number of neutrons / number of protons / total number of protons and neutrons).
Reveal answer
26 is the number of protons [1]; 56 is the total number of protons and neutrons [1].
State one similarity and one difference between the nuclei of uranium-235 and uranium-238.
Reveal answer
Similarity: the same number of protons, 92 (the same nuclear charge) [1].
Difference: uranium-238 has three more neutrons (146 against 143); or uranium-235 is fissile and uranium-238 is not [1]. (“Same atomic number” and “different mass number” are not accepted.)
Determine the numbers of protons, neutrons and electrons in the ion ³²16S²⁻.
Reveal answer
protons = 16 [1]
neutrons = 32 − 16
neutrons = 16 [1]
electrons = 16 + 2
electrons = 18 (the ion has gained two electrons) [1]
Explain why the Geiger–Marsden apparatus had to be evacuated.
Reveal answer
Alpha particles are strongly ionising and lose their energy in collisions with air molecules [1], so in air they would stop within a few centimetres and never reach the detector — or would be deflected by the air as well as by the foil [1].
Gallium has two stable isotopes, gallium-69 (60.1 %) and gallium-71 (39.9 %). Calculate the relative atomic mass of gallium to three significant figures.
Reveal answer
Ar = (60.1 × 69 + 39.9 × 71) ÷ 100 [1]
Ar = (4 146.9 + 2 832.9) ÷ 100
Ar = 69.798 [1]
Ar = 69.8 [1]
7. An alpha particle with kinetic energy 7.7 MeV is fired straight at a gold nucleus (79 protons). It stops momentarily at distance d from the centre, where all its kinetic energy has become electric potential energy, E = kQq ÷ d.
Use k = 8.99 × 10⁹ N m² C⁻², 1 MeV = 1.6 × 10⁻¹³ J and a proton charge of 1.6 × 10⁻¹⁹ C. Calculate d and explain what it tells you about the size of the gold nucleus.
Reveal answer
E = 7.7 × 1.6 × 10⁻¹³ J
E = 1.23 × 10⁻¹² J [1]
Qq = (79 × 1.6 × 10⁻¹⁹ C) × (2 × 1.6 × 10⁻¹⁹ C)
Qq = 4.04 × 10⁻³⁶ C² [1]
d = kQq ÷ E
d = 8.99 × 10⁹ × 4.04 × 10⁻³⁶ ÷ 1.23 × 10⁻¹²
d = 3.0 × 10⁻¹⁴ m [1]
The alpha still obeyed the electric-force law at this distance, so the nucleus must be no bigger than about 3 × 10⁻¹⁴ m in radius — an upper limit, consistent with the modern value of 7 × 10⁻¹⁵ m [1].
8. A square of gold leaf 1.00 cm by 1.00 cm has a mass of 0.19 mg. Gold has a density of 19.3 g/cm³ and its atoms are about 2.9 × 10⁻¹⁰ m apart. Estimate how many atoms thick the leaf is, and explain why a thin foil was essential to Rutherford’s interpretation.
Reveal answer
thickness = mass ÷ (density × area)
thickness = 0.19 × 10⁻³ g ÷ (19.3 g/cm³ × 1.00 cm²)
thickness = 9.8 × 10⁻⁶ cm
thickness = 9.8 × 10⁻⁸ m [1]
atoms = 9.8 × 10⁻⁸ m ÷ 2.9 × 10⁻¹⁰ m
atoms ≈ 340 [1]
With only a few hundred layers, and each nucleus filling about 10⁻¹³ of its atom, an alpha almost never passes close to more than one nucleus [1]. So each large deflection is a single encounter with one nucleus, which is what Rutherford’s calculation assumed; in a thick foil, many small deflections would add up and hide the pattern [1].
Summary
- Models of the atom changed with evidence: Dalton (solid spheres), Thomson (plum pudding, 1904), Rutherford (nuclear, 1911), Bohr (shells, 1913), Chadwick (neutron, 1932).
- Geiger–Marsden (1909): most alphas straight through (mostly empty space); some deflected (positive nucleus); about 1 in 8 000 turned back (tiny, dense, massive nucleus).
- The nucleus is about 20 000 times smaller than the atom but holds almost all the mass; nuclear matter has a density of about 2 × 10¹⁷ kg/m³.
- Proton: mass 1, charge +1. Neutron: mass 1, charge 0. Electron: mass about 1 ÷ 1 840, charge −1.
- ᴬZX: Z is the number of protons and A the number of protons plus neutrons.
- N = A − Z; in a neutral atom the number of electrons equals Z.
- Isotopes: same number of protons, different number of neutrons — same chemistry, different mass and nuclear stability.
Next time → Unstable nuclei: why some nuclei are stable and others are not, and the nature of alpha, beta and gamma radiation — what each is, what stops it, how it bends in electric and magnetic fields, and why it ionises.