HookOne alpha in eight thousand came straight back
In 1909, in a Manchester basement, Hans Geiger and a twenty-year-old student called Ernest Marsden spent weeks sitting in the dark, counting faint flashes on a zinc sulphide screen — one flash per alpha particle. They were firing alphas at gold foil a few ten-thousandths of a millimetre thick and expected every one to punch through with barely a nudge, because the accepted 'plum pudding' atom smeared its positive charge across the whole atomic volume, far too dilute to push back hard. Most alphas obliged. But roughly one in every 8,000 turned through more than 90° — a few came almost straight back. Rutherford later compared it to firing a naval shell at tissue paper and having it rebound: on the old model, the deflecting forces available were thousands of times too small.
The 1911 verdict rebuilt the atom: nearly all of its mass, and all of its positive charge, sits in a nucleus roughly 10,000 times smaller than the atom — a dense speck in mostly empty space. Section 3.8 lives inside that speck. You will size it by two independent methods, map which proton–neutron combinations survive and which fall apart, put numbers on the falling-apart with the exponential decay law, audit the energy ledger with \(E=mc^2\), and finish at the working end of the subject: fission reactors, the systems that keep them safe, and the waste that outlives everyone who made it.
ModelScattering — reading a nucleus you cannot see
AQA rewards you for pairing each observation with its inference, so learn them as couples. Most alphas passed straight through → the atom is mostly empty space. A small fraction deflected through large angles → the charge and mass are concentrated in a tiny volume, because only an intense, localised electric field bends a 5 MeV projectile. A few rebounded → the scatterer is positive (it repels the positive alpha) and far more massive than the alpha itself — a light target would be knocked away, not turn the projectile around.
The scattering geometry also hands you a ruler. In a head-on approach, the alpha decelerates as electric potential energy grows, stopping momentarily at the distance of closest approach, where all its kinetic energy has become potential energy: \(E_k = \frac{Q_{nucleus}\,q_{\alpha}}{4\pi\varepsilon_0 r}\). Solve for \(r\) and you have an estimate of nuclear scale from nothing but the alpha's energy and the target's proton number. Be precise about its status: the alpha stops short of the surface, so closest approach is an upper limit on the radius — and a more energetic alpha gets closer, so the 'answer' depends on the probe.
A 5.0 MeV alpha (charge \(2e\)) approaches a gold nucleus (\(Z = 79\)) head-on. Convert the energy first: \(E_k = 5.0 \times 1.60 \times 10^{-13} = 8.0 \times 10^{-13}\ \text{J}\).
\[r = \frac{Q q}{4\pi\varepsilon_0 E_k} = \frac{8.99 \times 10^{9} \times (79 \times 1.60 \times 10^{-19}) \times (2 \times 1.60 \times 10^{-19})}{8.0 \times 10^{-13}}\]
\[r = \frac{8.99 \times 10^{9} \times 1.264 \times 10^{-17} \times 3.20 \times 10^{-19}}{8.0 \times 10^{-13}} = 4.5 \times 10^{-14}\ \text{m}\]
About 45 fm. Electron diffraction puts gold's true radius nearer 7 fm, so the 5 MeV alpha stalls a long way out — exactly why you must call this an upper limit. The two habitual slips: forgetting the alpha's charge is \(2e\), and leaving the energy in MeV.
DataAlpha, beta, gamma — reach, ionising power, identification
Alpha is a helium nucleus: charge \(+2e\), mass 4 u, ejected at around 5% of light speed. Being heavy, slow and doubly charged, it ionises densely — tens of thousands of ion pairs per millimetre of air — and therefore exhausts itself fast: a few centimetres of air, a sheet of paper, or the dead outer layer of your skin stops it. Beta-minus is a fast electron from the nucleus: lightly ionising, stopped by a few millimetres of aluminium or about a metre of air. Gamma is a high-energy photon: no charge, no mass, very weakly ionising, never fully stopped — several centimetres of lead only halves the intensity, and in air it simply dilutes with the inverse-square law, \(I \propto 1/x^2\).
To identify an unknown source, insert absorbers in sequence and watch the count rate: a drop with paper means alpha present; a drop with 3 mm of aluminium means beta; whatever penetrates centimetres of lead is gamma. (Deflection in a magnetic field works too — alphas and betas curve opposite ways, gamma goes straight — but absorbers are the bench method.) The danger ranking inverts with location: outside the body alpha is the least hazardous; inhaled or ingested it is the worst, dumping all its energy into a few cells — polonium-210, the alpha emitter used in the 2006 London poisoning of Alexander Litvinenko, was harmless outside its victim.
Every measurement floats on background radiation: radon from the ground (roughly half the UK's average annual dose of about 2.7 mSv, and several times higher over Cornish granite), cosmic rays, rocks and buildings, medical exposures, even potassium-40 in food. Measure it with the source absent and subtract it — no analysis in this section is valid without that step.
DataRP12 — testing the inverse-square law for gamma
The method: record background counts with the source locked away (several 60 s counts, averaged). Then mount a sealed gamma source (cobalt-60 or caesium-137) in its holder facing a Geiger–Müller tube along a metre rule, and log counts over a fixed interval at a series of separations \(d\). Subtract background from every reading. Independent variable: distance; dependent: corrected count rate \(C\); controls: counting interval, the same source and tube, tube voltage.
Two error sources dominate, one systematic and one random. Systematic: you cannot see the true origin — the active material sits recessed inside its capsule and the detection region sits inside the GM tube — so the true separation is \(d + d_0\) for some unknown offset. The fix is the plot: if \(C = \frac{k}{(d+d_0)^2}\) then \(\frac{1}{\sqrt{C}} = \frac{d + d_0}{\sqrt{k}}\), so graph \(1/\sqrt{C}\) against \(d\): a straight line is the evidence for inverse-square behaviour, and the intercept measures \(d_0\) instead of letting it wreck the result. Random: decay is genuinely random, and a total of \(N\) counts carries an uncertainty of about \(\sqrt{N}\) — 100 counts is a 10% measurement, 10,000 counts a 1% one — so count for longer where the rate is low. Safety is marked too: handle with tongs, keep the source pointing away, maximise distance and minimise exposure time.
A student measures 435 counts per minute at \(d = 0.25\ \text{m}\) and 131 per minute at \(d = 0.50\ \text{m}\), with background 30 per minute.
Corrected rates: \(C_1 = 405\), \(C_2 = 101\ \text{min}^{-1}\). Doubling the distance should quarter the rate: \(C_1/C_2 = 405/101 = 4.0\) — inverse square confirmed. Cross-check with the constancy test: \(C_1 d_1^2 = 405 \times 0.0625 = 25.3\); \(C_2 d_2^2 = 101 \times 0.25 = 25.3\). Consistent.
Now skip the background subtraction: \(435/131 = 3.3\), and the law appears to fail. A constant added to both readings destroys the proportionality — which is why 'subtract background' is not bookkeeping but the difference between confirming and wrongly rejecting the physics.
ModelThe decay clock — randomness with a reliable average
Radioactive decay is random — you cannot say when a given nucleus will go, only quote a fixed probability per unit time, the decay constant \(\lambda\) — and spontaneous: temperature, pressure and chemical bonding change nothing, which is why decay makes an incorruptible clock. With vast numbers the statistics become certainty: activity \(A = \lambda N\) (in becquerels), the population obeys \(N = N_0 e^{-\lambda t}\), and activity and measured count rate fall by the same law. The half-life is \(T_{1/2} = \frac{\ln 2}{\lambda}\), and in log form \(\ln N = \ln N_0 - \lambda t\): a straight line of gradient \(-\lambda\), the standard graphical route.
Half-life is a design parameter. Technetium-99m (\(T_{1/2} = 6\) hours) is injected for tens of millions of hospital scans a year precisely because it images in the afternoon and is effectively gone by the weekend. Americium-241 (432 years) sits in smoke detectors because its activity is essentially constant over the product's life while its alphas travel only centimetres. Carbon-14 (5,730 years) dates anything organic back to about 50,000 years: living tissue exchanges carbon with the air, holding its \(^{14}\text{C}\) fraction steady, and death stops the exchange and starts the clock. For rocks, slower clocks — uranium-238's 4.5-billion-year chain among them — do the same job on geological time.
A bone fragment's carbon-14 activity per gram is 19% of that of living bone. First \(\lambda = \frac{\ln 2}{5730} = 1.21 \times 10^{-4}\ \text{yr}^{-1}\). Then from \(A = A_0 e^{-\lambda t}\):
\[t = \frac{\ln(A_0/A)}{\lambda} = \frac{\ln(1/0.19)}{1.21 \times 10^{-4}} = \frac{1.66}{1.21 \times 10^{-4}} = 13{,}700\ \text{years}\]
End of the last Ice Age. Sanity-check by bracketing: 19% lies between two half-lives (25%) and three (12.5%), so the answer must sit between 11,460 and 17,190 years — it does. Examiners consistently reward that one-line bracket, and it catches sign errors in the logarithm before they cost you the question.
MechanismThe N–Z map — which way an unstable nucleus falls
Plot neutron number \(N\) against proton number \(Z\) for every known nuclide and the stable ones trace a narrow curve: \(N \approx Z\) for light nuclei, bending to roughly \(N \approx 1.5Z\) for heavy ones. The bend is a tug-of-war. Electrostatic repulsion acts between every pair of protons across the whole nucleus, while the strong force saturates — each nucleon only binds its nearest neighbours — so heavy nuclei need extra neutrons as charge-free glue. Beyond \(Z = 83\) no amount of glue suffices and every nuclide is unstable. (Bismuth-209, the traditional last stable nuclide, was finally caught alpha-decaying in 2003 with a half-life of \(2 \times 10^{19}\) years — a billion times the age of the Universe, so 'stable' remains a fair working description.)
Where a nuclide sits relative to the curve dictates its exit route. Above the curve (neutron-rich): \(\beta^-\) decay converts a neutron to a proton (a down quark becomes up), emitting an electron and an antineutrino — the nuclide steps diagonally towards stability. Below the curve (proton-rich): \(\beta^+\) emission converts a proton to a neutron, or the nucleus performs electron capture, swallowing an inner-shell electron; the shell vacancy refills and emits characteristic X-rays, the tell-tale signature. Heavy nuclei shed bulk with \(\alpha\) emission, dropping two protons and two neutrons at once. And after almost any decay the daughter is born excited, shedding the surplus as gamma photons — no change in \(A\) or \(Z\), just the nucleus relaxing. Technetium-99m is exactly this: a metastable excited state whose delayed, nearly pure gamma emission is what makes it imageable and injectable.
ModelNuclear radius — matter at three hundred trillion tonnes per cubic metre
Closest approach was the rough ruler; electron diffraction is the precise one. Accelerate electrons until their de Broglie wavelength \(\lambda = h/p\) shrinks to femtometres — a few hundred MeV — and fire them at a thin target: they diffract around nuclei exactly as light diffracts around a disc, producing minima at angles given by \(\sin\theta \approx 1.22\lambda/d\), where \(d\) is the nuclear diameter. Electrons are leptons, blind to the strong force, so the pattern is a clean electromagnetic map — the method's decisive advantage over alpha scattering, and a favourite two-mark comparison.
The collected results compress into one law: \(R = R_0 A^{1/3}\), with \(R_0\) typically given as about 1.05 fm. Cube it and the meaning appears: \(R^3 \propto A\), volume proportional to nucleon count, so nucleons pack like marbles in a bag and every nucleus has the same density. Hydrogen or uranium, it makes no difference — nuclear matter is incompressible stuff at a fixed, staggering density, and that constancy is precisely what a saturating short-range force predicts.
Estimate the density of nuclear matter using carbon-12. Radius: \(R = R_0 A^{1/3} = 1.05 \times 12^{1/3} = 1.05 \times 2.29 = 2.40\ \text{fm}\). Volume: \(V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi (2.40 \times 10^{-15})^3 = 5.8 \times 10^{-44}\ \text{m}^3\). Mass: \(12 \times 1.66 \times 10^{-27} = 1.99 \times 10^{-26}\ \text{kg}\).
\[\rho = \frac{m}{V} = \frac{1.99 \times 10^{-26}}{5.8 \times 10^{-44}} = 3.4 \times 10^{17}\ \text{kg m}^{-3}\]
A teaspoon of it would weigh about 1.7 billion tonnes. Repeat with gold-197 (\(R = 1.05 \times 197^{1/3} = 6.1\ \text{fm}\)) and the density comes out the same — the point of the whole calculation, and the same density you will meet again if you study neutron stars in the astrophysics option.
ModelMass defect and binding energy — the nuclear ledger
Weigh a nucleus and it comes up light: less than the summed masses of its separated protons and neutrons. The shortfall is the mass defect \(\Delta m = Zm_p + (A-Z)m_n - m_{nucleus}\), and Einstein's \(E = mc^2\) prices it: assembling the nucleus released \(\Delta m c^2\) of energy — the binding energy — and dismantling it costs exactly that much back. The exam currency conversion sits on your data sheet: 1 u of mass defect is worth 931.5 MeV, which spares you multiplying by \(c^2\) in SI units.
Divide by \(A\) and plot binding energy per nucleon against nucleon number: a steep climb through the light nuclei, a broad peak of about 8.8 MeV near iron-56, then a gentle decline out to uranium. Iron is the most tightly bound arrangement of nucleons in the Universe, and every energy-releasing nuclear reaction is a step towards that peak: fusion climbs it from the left (huge gains per nucleon, but positive nuclei must be slammed together against Coulomb repulsion — hence stellar temperatures), fission steps up from the right (a modest ~0.9 MeV per nucleon, but 235 nucleons at a time). For scale, a chemical bond rearrangement releases a few eV; a single fission releases about 200 MeV — a factor of around \(10^8\), which is the entire case for nuclear power in one ratio.
Find the binding energy per nucleon of helium-4, given \(m_p = 1.00728\) u, \(m_n = 1.00867\) u and a nuclear mass of 4.00151 u.
\[\Delta m = 2(1.00728) + 2(1.00867) - 4.00151 = 4.03190 - 4.00151 = 0.03039\ \text{u}\]
Binding energy: \(0.03039 \times 931.5 = 28.3\ \text{MeV}\). Per nucleon: \(28.3/4 = 7.1\ \text{MeV}\).
That is an extraordinary figure for such a light nucleus — helium-4 is a local spike of stability on the curve — and it is why hydrogen fusion in stars runs to helium and largely stops there, and why alpha particles emerge from heavy nuclei ready-made. Watch the mark scheme's favourite trap: use the nuclear mass, not the atomic mass, or state that you are subtracting the electron masses.
MechanismInduced fission — running a chain reaction on purpose
Uranium-235 rarely splits on its own; it must be induced. Absorb a neutron, form furiously unstable uranium-236, and it tears into two mid-mass fragments (a typical pair: barium and krypton) plus two or three fresh neutrons and about 200 MeV, mostly as fragment kinetic energy. Those fresh neutrons are the loaded dice: if, on average, exactly one per fission goes on to cause another, the reaction is critical and runs at steady power; below one it fizzles; above one it grows exponentially. The critical mass is the minimum lump that sustains the chain — neutron production scales with volume (\(\propto r^3\)) while escape scales with surface (\(\propto r^2\)), so too small a lump leaks too many.
A thermal reactor is three jobs bolted around that arithmetic. The moderator — graphite in Britain's AGRs, water in pressurised-water reactors like Sizewell B — slows the fast fission neutrons to 'thermal' speeds through dozens of elastic collisions with light nuclei, because slow neutrons are enormously more likely to be captured by U-235 and trigger the next fission. The control rods — boron or cadmium, hungry neutron absorbers — slide in and out to hold the average at exactly one, and drop fully home under gravity for an emergency shutdown. The coolant — carbon dioxide gas in an AGR, pressurised water in a PWR — carries gigawatts of heat from the core to boilers, and from there it is Victorian engineering: steam, turbine, generator.
How much coal is a kilogram of U-235 worth? Number of nuclei: \(N = \frac{1000}{235} \times 6.02 \times 10^{23} = 2.6 \times 10^{24}\). Each fission releases about 200 MeV \(= 200 \times 1.60 \times 10^{-13} = 3.2 \times 10^{-11}\ \text{J}\).
\[E = 2.6 \times 10^{24} \times 3.2 \times 10^{-11} = 8.2 \times 10^{13}\ \text{J}\]
Coal releases about 30 MJ per kilogram, so the equivalent is \(\frac{8.2 \times 10^{13}}{3.0 \times 10^{7}} = 2.7 \times 10^{6}\ \text{kg}\) — roughly 2,700 tonnes of coal per kilogram of fuel. That factor of a few million is the same \(10^8\)-ish ratio between nuclear and chemical energy scales, arriving from the opposite direction.
CaseSafety systems and waste that outlives its makers
Reactor safety is layered containment. The fuel is ceramic, sealed in metal cans; the core sits in a steel pressure vessel; around that stands a metres-thick concrete biological shield that stops the neutrons and gamma cold; spent fuel is moved entirely by remote handling. Emergency shutdown means every control rod dropping fully in — the chain reaction stops in seconds. But the fission fragments are neutron-rich and intensely beta-active, so decay heat continues: several per cent of full thermal power at the instant of shutdown, around one per cent a day later, which for a gigawatt reactor is still megawatts. Cooling must therefore outlive the chain reaction — the lesson of Fukushima in 2011, where the reactors shut down correctly but the tsunami drowned the pumps that removed the heat afterwards.
Waste is sorted by activity. Low-level (gloves, filters, clothing) is compacted and buried in engineered shallow facilities. Intermediate-level (cladding, reactor components) is set in cement inside steel drums. High-level waste — spent fuel and its reprocessing residues — is a few per cent of the volume but around 95% of the radioactivity: it spends years under water in cooling ponds while the decay heat subsides, is then vitrified — melted into borosilicate glass, sealed in stainless steel — and stored, in Britain's case largely at Sellafield, awaiting deep geological disposal. Finland's Onkalo repository, carved 450 m into ancient bedrock, is the world's first; the UK is still siting its own. The timescale is set by \(A = \lambda N\) running both ways: short half-lives burn out fast and fierce, but plutonium-239's 24,100 years means engineering containment for something like ten half-lives — a quarter of a million years, or fifty times the age of Stonehenge.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This is Paper 2 material, and AQA weights the whole qualification roughly 40% AO2 — applying physics to unfamiliar contexts — so expect the decay law dressed up as a hospital tracer, a smoke alarm or a museum bone rather than asked cold. Two bookkeeping habits pay throughout. In every nuclear equation, balance nucleon number and proton number explicitly, and never write a beta decay without its antineutrino (or a beta-plus without its neutrino) — the omission is a routine lost mark. And use the data sheet's currency: 1 u = 931.5 MeV kills the c² arithmetic, and 1 MeV = 1.60 × 10⁻¹³ J converts the other way; candidates who bounce through kilograms lose time and pick up rounding errors.
Decay calculations live in log space. State the equation, take logs cleanly, and quote λ with units (s⁻¹ or yr⁻¹ — and be consistent with the half-life's units). When reading half-life from a graph, take at least two different halvings and average; when asked for evidence that decay is exponential, the creditable phrase is 'constant ratio in equal time intervals' or a straight-line log plot, not 'it curves downwards'. Always subtract background before doing anything else with count data — in RP12 analysis specifically, examiners look for background handling, the recognition that the true distance carries an unknown offset (fixed by plotting 1/√C against d), and a counting-statistics point: total counts of N carry a √N uncertainty, so longer counting at low rates is an accuracy improvement, not an inconvenience.
The 6-markers cluster around the reactor and waste. Structure them component → function → physics: moderator slows neutrons by elastic collision with light nuclei because slow neutrons induce fission far more readily; control rods absorb neutrons to hold exactly one per fission; coolant transfers the thermal power out; shielding, remote handling, ponds, vitrification and geological disposal for the back end — each with its reason attached. And in 'estimate' questions (closest approach, nuclear density), an order-of-magnitude answer with clearly stated assumptions scores; a bald number without the energy-conversion working does not.