Models and evidence, revisited: three atoms in sixteen years
By the end of this lesson you will be able to
- Outline the Thomson (1897), Rutherford (1911) and Bohr (1913) models of the atom as a sequence of claim → evidence → model change, and state what each model kept from the one before.
- Describe the proton, neutron and electron by relative mass, relative charge and location, and calculate the numbers of each in any atom or ion from its atomic number, mass number and charge.
- Explain why the atom is described as a system, and why a model of a system can be useful before it is complete.
- Interpret alpha-scattering data quantitatively: estimate the size of the nucleus from the fraction of particles turned back.
- Evaluate a model against evidence: decide what must change and what can stay.
The puzzle
In 1897 the atom was the smallest thing there was: a hard, indivisible ball, exactly as Dalton had drawn it. By 1913 it had a tiny nucleus, electrons in fixed shells, and a size that could be written down. The atom did not change in those sixteen years — gold was gold and hydrogen was hydrogen. What changed was three experiments and three people willing to redraw the picture. If the atom never changed, in what sense was each of the three models “right”?
A model predicts that alpha particles fired at a gold foil will all pass through with, at most, a tiny nudge. In the experiment, about 1 in 8 000 comes straight back. Predict which is the scientific response: (i) the experiment must be faulty, so repeat it until it agrees; (ii) throw the whole model away and start again; (iii) find the one feature of the model that the result contradicts, change that, and keep the rest.
Compare your answer
(iii) — and that is exactly what Rutherford did. Repeating the experiment (Geiger and Marsden did, many times) kept giving the same 1 in 8 000, so the result stood. Throwing everything away would have discarded what still worked: the electron, its charge, its tiny mass. The result contradicted only one claim — that the positive charge and mass were spread softly through the whole atom — so that claim was replaced by a nucleus, and everything else stayed. A model is a system of claims; evidence usually breaks one part of it, not all of it.
The science — structured notes
1.1 A model is an idea expressed precisely enough to be wrong
Nobody has ever seen an atom with their eyes and nobody ever will: it is a thousand times smaller than the shortest wavelength of visible light. Everything you know about atoms comes from models — pictures, equations and rules that express an idea about what the atom is like, and that predict what an experiment will measure. If the prediction fails, the model must change. That loop — claim → evidence → model change — is the story of this lesson, and it ran three times in sixteen years.
The key concept of this unit is Systems: an atom is a system of parts (protons, neutrons, electrons) whose interactions produce everything the atom does. A good model names the parts, says where they are, and says how they interact. A model can be a useful system before it is a complete one: Thomson’s atom had no nucleus and still explained static electricity correctly.
1.2 1897 — Thomson: the atom has parts
| Claim before | Evidence | Model change | What was kept |
|---|---|---|---|
| Dalton (1803): atoms are indivisible and identical for each element; nothing smaller exists | Cathode rays bent by electric and magnetic fields towards the positive plate → negative particles; charge-to-mass ratio about 1 800 times that of a hydrogen ion; identical from every metal and every gas in the tube | Thomson (1897; model published 1904): a negatively charged particle about 11 800 of the mass of a hydrogen atom exists inside every atom. The atom is a ball of positive charge with electrons dotted through it — the “plum pudding” | Atoms are the units of elements; each element has its own mass; mass is conserved |
Notice the shape of the argument. The evidence did not say “there is an electron”; it said “something with a charge-to-mass ratio 1 800 times larger than any ion comes out of every material”. Thomson chose the interpretation — same charge as an ion, far smaller mass — and Millikan’s oil drops (1909–1913) later measured the charge directly, 1.6 × 10⁻¹⁹ C, and confirmed it. The plum pudding was then the simplest arrangement that kept the atom neutral: negative dots in a positive sea.
1.3 1911 — Rutherford: the mass and positive charge are in one tiny place
In Manchester in 1909, Hans Geiger and the undergraduate Ernest Marsden fired alpha particles — fast, positive, 7 000 times heavier than an electron — at gold leaf a few hundred atoms thick and counted the flashes on a zinc sulfide screen. The plum pudding model made a clear prediction: the positive charge is spread so thinly through the atom that no alpha can meet enough of it at once to be turned by more than a fraction of a degree. Most alphas did exactly that. But about 1 in 8 000 turned through more than 90° — some came straight back.
| Claim tested | Evidence | Model change | What was kept |
|---|---|---|---|
| Positive charge and mass are spread evenly through the atom (Thomson) | 1 in 8 000 alphas deflected through more than 90°; the number counted at each angle falls as 1 ÷ sin⁴(θ/2); gold turns back far more than aluminium | Rutherford (1911): all the positive charge and almost all the mass are in a nucleus about 110 000 of the atom’s diameter; electrons occupy the space around it; the atom is mostly empty | The electron, its charge and mass; the atom neutral overall; the atom’s size (about 10⁻¹⁰ m) |
The gold foil in the lab is about 400 atoms thick. About 1 alpha in 8 000 is turned back through more than 90°. Estimate the size of the “target” inside each atom compared with the atom itself.
Chance of being turned back in the whole foil = 18 000. The foil is 400 layers of atoms, so the chance per layer ≈ 18 000 × 400 = 13 200 000.
If an alpha is turned back only when it heads for the target, that chance equals the fraction of the atom’s cross-section that is target: area of targetarea of atom = 13 200 000.
Area scales with radius squared, so radius of targetradius of atom = 1√3 200 000 = 11 800.
A gold atom has a radius of about 1.4 × 10⁻¹⁰ m, so the target radius is about 1.4 × 10⁻¹⁰ ÷ 1 800 ≈ 8 × 10⁻¹⁴ m. The modern value for the radius of the gold nucleus is about 7 × 10⁻¹⁵ m, so this estimate is roughly ten times too large — partly because the electrostatic repulsion reaches well beyond the nuclear surface, so the “target” really is wider than the nucleus, and partly because the calculation crudely assumes every alpha aimed anywhere at that target comes back. Even so, a one-line calculation from a single count puts the nucleus within a power of ten of the right size and thousands of times smaller than the atom. That is what Rutherford meant when he said the result was “as if you fired a 15-inch shell at tissue paper and it came back and hit you.”
1.4 1913 — Bohr: electrons in fixed shells; 1932 — Chadwick: the neutron
Rutherford’s atom had a fatal flaw of its own. According to the physics of the day, an electron circling a nucleus must radiate energy continuously, lose speed and spiral into the nucleus in about a hundredth of a billionth of a second — every atom in the universe should have collapsed, emitting every colour on the way down. Atoms do not collapse, and heated hydrogen does not emit every colour: it emits four sharp lines in the visible spectrum (red, blue-green, blue, violet), at wavelengths Johann Balmer had fitted to a formula in 1885 without knowing why.
| Claim tested | Evidence | Model change | What was kept |
|---|---|---|---|
| Electrons can orbit at any distance and any energy (Rutherford) | Hydrogen emits only certain sharp colours (line spectrum); atoms are stable; the same lines from every hydrogen sample in the universe | Bohr (1913): electrons occupy fixed shells of definite energy; an electron jumps between shells by absorbing or emitting light of exactly the energy difference. Bohr’s formula reproduced Balmer’s lines to four significant figures | The nucleus; the electron; the atom’s size and neutrality |
| The nucleus is made only of protons (Rutherford, 1919) | Beryllium has 4 protons but a mass of 9; alpha particles striking beryllium produce a neutral radiation that knocks protons out of paraffin wax | Chadwick (1932): a neutral particle of mass 1, the neutron, shares the nucleus with the protons | Everything else — the neutron filled a gap rather than breaking a claim |
1.5 The three particles — and counting them
| Particle | Relative mass | Relative charge | Where it is | Discovered |
|---|---|---|---|---|
| Proton | 1 | +1 | Nucleus | Rutherford, 1919 (named 1920) |
| Neutron | 1 | 0 | Nucleus | Chadwick, 1932 |
| Electron | 11 840 (≈ 0) | −1 | Shells around the nucleus | Thomson, 1897 |
Two numbers describe any nucleus. The atomic number (Z) is the number of protons — it decides which element the atom is. The mass number (A) is protons + neutrons. Written as a symbol, the mass number sits top-left and the atomic number bottom-left: ²³₁₁Na. In a neutral atom, electrons = protons. In an ion, the charge tells you how many electrons have been lost (positive) or gained (negative) — the nucleus never changes when an ion forms.
Atoms of one element can differ in their neutrons. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, so they share an atomic number and differ in mass number: ³⁵₁₇Cl has 18 neutrons and ³⁷₁₇Cl has 20, and both are chlorine because both have 17 protons. Because the proton number is unchanged, so is the number and arrangement of electrons — and since chemistry is done by electrons, isotopes of an element react in the same way; only the mass differs, which is why the relative atomic mass of chlorine is 35.5 rather than a whole number. A few isotopes are unstable and radioactive, which is where the tracers in the PET scan below come from.
Give the numbers of protons, neutrons and electrons in ²⁷₁₃Al, in the ion Al³⁺, and in the ion ³²₁₆S²⁻.
²⁷Al: protons = Z = 13; neutrons = A − Z = 27 − 13 = 14; electrons = 13 (neutral).
Al³⁺: the 3+ charge means three electrons have been lost. Protons 13, neutrons 14 (unchanged — the nucleus is untouched), electrons 13 − 3 = 10. Check: 13 positive, 10 negative → 3+.
³²S²⁻: protons 16; neutrons 32 − 16 = 16; the 2− charge means two electrons gained: electrons 16 + 2 = 18. Check: 16 − 18 = −2.
Notice that Al³⁺ and S²⁻ have 10 and 18 electrons — the same as the noble gases neon and argon. Lesson 5.2 explains why that is no coincidence.
1.6 The scale of the system
An atom’s diameter is about 1 × 10⁻¹⁰ m; its nucleus’s diameter about 1 × 10⁻¹⁴ m. Express the nucleus as a fraction of the atom, by length and by volume, and find where the mass is.
By length: 1 × 10⁻¹⁴1 × 10⁻¹⁰ = 110 000. Scale the atom up to the National Stadium in Gaborone, about 200 m across: the nucleus becomes 200 m ÷ 10 000 = 2 cm — a marble on the centre spot, with the electrons somewhere in the stands and nothing at all in between.
By volume: volume scales with the cube of length, so the nucleus occupies 110 000³ = 110¹² of the atom — one part in a million million. The atom is 99.9999999999 % empty.
By mass: in a carbon-12 atom the nucleus has 12 units (6 protons, 6 neutrons) and the six electrons together have 6 × 11 840 ≈ 1307 of a unit. The nucleus holds about 99.97 % of the mass in 110¹² of the volume. A teaspoon of pure nuclei would have a mass of about a thousand million tonnes — which is what a neutron star is made of.
Complete the particle counts.
(a) ⁴⁰₂₀Ca and Ca²⁺.
(b) ¹⁹₉F⁻.
(c) An ion has 12 protons, 12 neutrons and 10 electrons. Write its symbol with mass number, atomic number and charge.
Reveal answer
(a) Ca: 20 p, 20 n, 20 e. Ca²⁺: 20 p, 20 n, 18 e — two electrons lost, nucleus unchanged.
(b) F⁻: 9 p, 10 n (19 − 9), 10 e (9 + 1 gained).
(c) 12 protons makes it magnesium; mass number 12 + 12 = 24; 12 positive against 10 negative gives 2+: ²⁴₁₂Mg²⁺.
Geiger and Marsden swapped the gold foil for aluminium of the same thickness and found far fewer alphas turned back at every angle. Explain this with the nuclear model, and state what it told Rutherford about the nucleus.
Reveal answer
An alpha is turned back by the electrostatic repulsion of the nucleus, and the size of that repulsion depends on the nuclear charge. Aluminium’s nucleus carries 13 positive charges against gold’s 79, so an alpha passing at the same distance feels a much weaker push and is deflected through a smaller angle; only alphas aimed even more precisely at the centre turn back, and there are fewer of those. The count rose in proportion to the square of the nuclear charge, which told Rutherford that the charge on the nucleus — not its mass — is what deflects, and that the charge is different for each element: the beginnings of the atomic number.
“Rutherford’s experiment proved that Thomson’s model was completely wrong.”
Reveal answer
False. The experiment contradicted one claim — that positive charge and mass are spread through the whole atom. Thomson’s electron, its charge, its tiny mass and the neutrality of the atom all survived and are part of every model since. Evidence usually falsifies part of a system of claims; the skill is to identify which part.
Suppose the plum pudding model had been correct. Describe what the count on the detector would have looked like at 10°, 90° and 150°, compared with what was actually seen.
Reveal answer
With charge spread softly through the atom, an alpha can never meet a concentrated push, so every alpha would pass through with a deflection of a fraction of a degree. The detector at 10° would already read almost nothing, and at 90° and 150° it would read exactly zero, run after run. What was seen was a count that fell steeply with angle but never reached zero — thousands per minute near the beam, a handful per minute at 150° — and that “handful, not zero” is the entire case for the nucleus.
Try it — the gold-foil experiment, angle by angle
Move the detector round the arc from 10° to 150° and count the flashes for one minute at each angle. Plot the count against angle: how steeply does it fall? Then change the foil to aluminium and to a thicker gold foil, and see what each does to the count at the same angle. This is the experiment that broke the plum pudding — and the data are the proof.
Suppose your one-minute counts with thin gold were: 10° → 90 500; 30° → 1 160; 50° → 164; 70° → 48; 90° → 21; 110° → 12; 130° → 8; 150° → 6.
(a) Describe the relationship, quoting values.
(b) Calculate the fraction of the 10° count that reaches the detector at 90°, as a stacked fraction in its simplest form, and explain what it says about the atom.
(c) Explain why the count at 150° is small but never zero.
Reveal answer
(a) The count falls extremely steeply as the angle increases: from 90 500 per minute at 10° to 1 160 at 30° — a factor of 78 for the first 20° — then to 21 at 90° and only 6 at 150°. The fall is fastest at small angles and flattens at large angles; it is a curve, not a straight line, and it never reaches zero.
(b) 2190 500 ≈ 14 300. Only about one alpha in four thousand of those that pass close to the beam is turned through a right angle or more: almost all of the atom offers nothing to push against, and the part that does is tiny.
(c) A few alphas in every million are aimed almost exactly at a nucleus; they feel the full repulsion of 79 positive charges concentrated at a point and are turned almost straight back. Because there are many nuclei in the foil and many alphas per minute, a small steady count is seen at every large angle — it is small because the nucleus is tiny, and never zero because the nucleus exists.
Common mistakes to avoid
- “The new model proved the old one wrong” — say which claim the evidence contradicted and what was kept.
- Changing protons to make an ion — ions differ from atoms only in electrons; the nucleus is untouched.
- Neutrons = mass number — neutrons = mass number − atomic number.
- Electron mass “zero” — it is 11 840 of a proton: negligible for mass number, not for chemistry.
- “Most alphas bounced back” — most passed straight through; about 1 in 8 000 turned through more than 90°, and that tiny fraction is the evidence.
- Bohr’s shells as planets — they are energy levels; the electron does not trace a path we can follow (Lesson 5.5).
How we found out — the number that put the table in order
In 1913 a 25-year-old in Rutherford’s Manchester laboratory, Henry Moseley, fired electrons at targets of different metals and measured the X-rays they gave off. He found that the X-ray frequency rose in perfectly regular steps from element to element — not with the atomic mass Mendeleev had used, but with a whole number that increased by exactly one at each step. That number was the charge on the nucleus: the atomic number. For the first time the periodic table had a physical basis: the elements are numbered by their protons, and Mendeleev’s awkward reversals (tellurium before iodine, though heavier) were explained at once. Moseley’s data also showed gaps at 43, 61, 72 and 75 — elements not yet discovered, three of which were found within the next twenty-five years exactly where he said. He volunteered for the army in 1914 and was killed by a sniper at Gallipoli in August 1915, aged 27; the British government afterwards stopped sending scientists to the front.
The neutron completed the count. By 1920 Rutherford had shown that the hydrogen nucleus — the proton — is a component of all nuclei, but the numbers did not add up: helium had 2 units of charge and 4 of mass. He suggested a neutral particle to make up the difference, and for twelve years nobody could find it, because a particle with no charge leaves no track and is bent by no field. In 1932 James Chadwick, working in Cambridge with a few weeks of nearly sleepless experiments, showed that the mysterious neutral radiation from beryllium bombarded by alphas could knock protons out of paraffin wax only if it consisted of particles of about the proton’s mass — a neutral particle, the neutron. The three-particle atom was complete: a system of protons, neutrons and electrons, and every model since has been about how they are arranged.
Moseley’s atomic numbers left gaps at 43 and 61. Explain why a gap in a sequence of measured numbers counts as evidence for an undiscovered element, when Mendeleev’s gaps of 1869 were treated with suspicion.
Reveal answer
Mendeleev’s gaps came from a pattern in properties and masses that he had partly arranged himself, so a gap could be an artefact of his arrangement. Moseley’s X-ray frequencies gave each known element a whole-number position that no one had chosen — it was measured — and the positions ran 42, 44, 45, 46 with nothing at 43. A whole number missing from a measured sequence cannot be explained by a mistake in ordering; either an element with 43 protons exists or the whole nuclear-charge model is wrong. Since the model fitted every other element, the gap was a prediction: technetium (43) was made in 1937 and promethium (61) identified in 1945, both with exactly the X-ray lines Moseley’s rule required.
Where it is used now
- Sorting Botswana’s diamonds. At Jwaneng and Orapa, crushed kimberlite passes through X-ray sorters: diamond, made of light carbon atoms with tiny nuclei, lets X-rays through differently from the heavier surrounding rock, and a jet of air knocks each stone out. The sorter is Moseley’s physics working at industrial speed.
- Seeing atoms. Electron microscopes use Thomson’s particle as a probe; because electrons have a far shorter wavelength than light, they resolve individual atoms — used to inspect the copper and nickel ores from the Kalahari Copperbelt and the coatings on solar panels.
- Identifying elements from their nuclei. Mass spectrometers separate ions by charge-to-mass ratio, exactly as Thomson did with cathode rays; laboratories in Gaborone use them to test water for lead and arsenic and to check the purity of medicines.
- Nuclear medicine. A PET scan detects the radiation from unstable nuclei injected into a patient; the images are maps of where the nuclei went. Chadwick’s neutron, fired at ordinary elements, is how the radioactive tracers are made.
- Smoke detectors. The americium source in a smoke alarm emits alpha particles that ionise air; smoke absorbs them and the current drops. Rutherford’s bullet in every home.
Science now — and the scientist you need to be
What is happening in 2026
The atom is now photographed, not just modelled. Scanning tunnelling microscopes have imaged single atoms since 1981 and can pick them up and place them one at a time; in the last three years electron microscopes have resolved the positions of atoms to a few hundredths of a nanometre and mapped the electron clouds between them. The 2023 Nobel Prize in Physics went to Pierre Agostini, Ferenc Krausz and Anne L’Huillier for light pulses lasting attoseconds — billionths of a billionth of a second — short enough to film an electron leaving an atom, so Bohr’s “jump” between shells is now a measured event with a duration. Quantum computers built from rows of individual atoms held in laser traps passed a thousand atoms in 2024 and are being used to simulate other atoms; and AI models trained on millions of measured spectra now predict a new molecule’s spectrum before it is made. Each of these is evidence being gathered faster than any generation before could, and the models are changing to match: not the picture of a nucleus and shells, which has held for a century, but the fine details of how electrons move and share. Models are still created and modified by evidence — only the evidence arrives quicker.
The mindshift
What you need to be able to do
Criterion A in this unit rewards three habits: count particles correctly from symbols and charges; explain each model change as claim → evidence → change, naming what was kept; and reason with numbers about scale. The examiner reports are blunt that “explain” means a mechanism — for alpha scattering, the mechanism is electrostatic repulsion between two positive charges, and the word “repulsion” is the mark. The Think task below asks for all three habits on an unfamiliar element.
See an example
Column 1, Thomson: claims a positive sphere with electrons in it; evidence, cathode rays from every material; broken by 1 in 8 000 alphas coming back. Column 2, Rutherford: claims a tiny massive nucleus; evidence, the 1 ÷ sin⁴(θ/2) counts; broken by the line spectrum and the collapse problem. Column 3, Bohr: claims fixed shells; evidence, hydrogen’s four visible lines matched to four significant figures; broken (later) by atoms with more than one electron, which the shells could not fully predict. The assistant said the electron “would lose energy as it orbits and spiral in, emitting a continuous rainbow of light, which is not observed” — the mechanism, correctly named.
Key terms
| Model | A representation of a system — picture, rule or equation — that makes predictions which can be tested. |
| Evidence | Measurements that a model either predicts correctly or fails to predict; the reason models change. |
| Plum pudding model | Thomson’s 1904 atom: electrons dotted through a sphere of positive charge. |
| Nuclear model | Rutherford’s 1911 atom: a tiny, dense, positive nucleus with electrons in the space around it. |
| Shell | A fixed energy level in which electrons sit (Bohr, 1913); electrons jump between shells by absorbing or emitting light. |
| Nucleus | The centre of the atom, about one ten-thousandth of its diameter, containing protons and neutrons and almost all the mass. |
| Proton / neutron / electron | Relative mass 1, 1 and 11 840; charge +1, 0, −1; nucleus, nucleus, shells. |
| Atomic number (Z) | The number of protons; identifies the element and fixes its place in the periodic table (Moseley, 1913). |
| Mass number (A) | Protons + neutrons; neutrons = A − Z. |
| Isotopes | Atoms of the same element with the same number of protons but different numbers of neutrons; identical chemistry, different mass. |
| Ion | An atom that has lost electrons (positive) or gained electrons (negative); the nucleus is unchanged. |
| Alpha particle | A helium nucleus (2 protons, 2 neutrons), charge 2+, used by Rutherford as a probe. |
Think like a scientist
Criterion A Read the information, then answer every part.
Potassium has the symbol ³⁹₁₉K and forms the ion K⁺. In 1913 Geiger and Marsden repeated the alpha-scattering experiment with foils of gold (Z = 79), silver (Z = 47), copper (Z = 29) and aluminium (Z = 13), all of the same thickness, and counted the alphas turned through 90° or more in ten minutes: gold 132, silver 46, copper 18, aluminium 4. In 1932 Chadwick showed that the radiation from beryllium (Z = 4, mass number 9) struck by alphas was a stream of neutral particles with a mass close to that of a proton.
(a) State the numbers of protons, neutrons and electrons in an atom of ³⁹K and in the ion K⁺. [3]
(b) Explain, using the nuclear model, why the count fell from gold to aluminium, and describe how the count changes with nuclear charge, using the data. [3]
(c) Compare the Thomson and Rutherford models of the atom in terms of where the positive charge and mass are. Write a comparison, not two descriptions. [2]
(d) Explain why the beryllium atom required a neutral particle of mass 1 to be added to the model of the nucleus, using its atomic number and mass number. [3]
(e) The plum pudding model was replaced in 1911. Explain why it is still described as good science. [2]
Model answer
(a) ³⁹K atom: 19 protons, 39 − 19 = 20 neutrons, 19 electrons. K⁺: 19 protons and 20 neutrons — the nucleus is unchanged — and 19 − 1 = 18 electrons, because the 1+ charge means one electron has been lost.
(b) An alpha is turned back by electrostatic repulsion between its own positive charge and the positive charge of the nucleus it approaches. A gold nucleus carries 79 positive charges and an aluminium nucleus only 13, so an alpha passing an aluminium nucleus at the same distance feels a much smaller repulsion and is deflected through a smaller angle; fewer alphas are turned through 90° or more. The data show the count rising steeply with nuclear charge: from 4 for aluminium (Z = 13) to 132 for gold (Z = 79). The charge is six times larger and the count about 33 times larger — close to 6² = 36 — so the count rises roughly with the square of the nuclear charge, as Rutherford’s model predicts.
(c) In Thomson’s model the positive charge and the mass are spread evenly through the whole volume of the atom, whereas in Rutherford’s model both are concentrated in a nucleus about one ten-thousandth of the atom’s diameter, so that Rutherford’s atom is almost entirely empty space while Thomson’s is filled throughout.
(d) Beryllium’s atomic number is 4, so its nucleus contains 4 protons, giving 4 units of charge and 4 units of mass. Its mass number is 9, so the nucleus must contain 5 more units of mass that carry no charge — otherwise the atom would have a charge of 9+ with only 4 electrons to balance it, and would not be neutral. A particle with mass 1 and charge 0 fills the gap exactly: 4 protons + 5 neutrons = mass 9, charge 4+. Chadwick’s radiation had exactly those properties — it was neutral (undeflected by fields) and it could knock out protons, which needs a particle of similar mass.
(e) The model was built from evidence (the electron’s charge-to-mass ratio, the neutrality of atoms) and made testable predictions — including the prediction that alphas would not be turned back. It was replaced because that prediction was tested and failed, which is the scientific process working, not failing; and its key discovery, the electron, was kept in every later model. A model that makes a prediction precise enough to be proved wrong is good science even when it is wrong.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Give the numbers of protons, neutrons and electrons in ⁵⁶₂₆Fe³⁺.
Reveal answer
Protons 26 [1]; neutrons 56 − 26 = 30 [1]; electrons 26 − 3 = 23, because three electrons are lost to give the 3+ charge [1].
Explain why the observation that most alpha particles passed straight through gold foil, while a few were turned back, led to the nuclear model.
Reveal answer
Most passing straight through means most of the atom contains nothing able to deflect a fast, heavy alpha — it is mostly empty space [1]; a few turned back means somewhere in the atom there is a concentration of positive charge and mass large enough to repel an alpha strongly [1]; because it happens to only about 1 in 8 000, that concentration — the nucleus — must be very small compared with the atom [1].
Explain what evidence Bohr used to place electrons in fixed shells rather than anywhere around the nucleus.
Reveal answer
Hydrogen emits light only at a few sharp wavelengths (a line spectrum), not every colour [1]; if electrons could have any energy they would emit every colour, so the sharp lines show they can only have certain fixed energies, and each line is a jump between two of them [1].
A student writes: “Mg²⁺ has 10 protons because it lost two; the neutron has a charge of +1; and the nucleus is about a tenth of the size of the atom.” Identify and correct three errors.
Reveal answer
Mg²⁺ still has 12 protons — it has lost two electrons, leaving 10 electrons [1]; the neutron has no charge (0); it is the proton that is +1 [1]; the nucleus is about one ten-thousandth of the atom’s diameter, not a tenth [1].
True or false, with a reason: “Because the electron has almost no mass, it can be ignored when describing an atom.”
Reveal answer
False [1]: the electron contributes almost nothing to the mass, so it is ignored in the mass number, but the electrons — especially the outer ones — decide every chemical property of the atom, including what it bonds to and how it conducts [1].
In the gold-foil experiment the foil is made twice as thick. Predict what happens to the count at 90°, explain why, and state what would happen to the count if the beam were fired through air instead of a vacuum.
Reveal answer
The count at 90° roughly doubles [1], because there are twice as many nuclei in the alpha’s path, so twice the chance of a close approach to one — each nucleus acts independently and the atoms are mostly empty, so the extra layers do not block the beam [1]; in air the alphas would be stopped within a few centimetres by collisions with air molecules before reaching the foil or the detector, so the count at every angle would fall to nearly zero — which is why the apparatus is evacuated [1].
Summary
- A model is a system of testable claims; evidence usually breaks one claim, and the next model keeps the rest. Thomson (1897) added the electron; Rutherford (1911) the nucleus; Bohr (1913) the shells; Chadwick (1932) the neutron.
- Proton: mass 1, charge +1, nucleus. Neutron: mass 1, charge 0, nucleus. Electron: mass 11 840, charge −1, shells.
- Protons = atomic number; neutrons = mass number − atomic number; electrons = protons in an atom, protons − charge in an ion. The nucleus never changes when an ion forms.
- The nucleus is about 110 000 of the atom’s diameter and holds 99.97 % of its mass; the count of alphas turned back lets you estimate that from the data.
- Moseley’s atomic number (1913) gave the periodic table its order: elements are numbered by protons, and the outer electrons do the chemistry.
Next time → Electronic configuration: filling the shells 2, 8, 8 for elements 1 to 20, why the group number is written in the outer shell, and why eight is the number atoms want.