Topic 8 goes inside the nucleus: Rutherford scattering, nuclear radius, accelerators and detectors, and the Standard Model of quarks, leptons and the conservation laws they must obey.
R = r₀A^(1/3) · E = mc² · 1 eV = 1.60 × 10⁻¹⁹ Jthe nucleus is about 10⁻¹⁵ m across — a hundred thousand times smaller than the atom
Scale check: if a nucleus were a 1 mm grain of sand, the atom around it would be a football stadium. Nearly all the mass sits in that grain; the rest is empty space patrolled by electrons.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 8 · Rutherford
Rutherford scattering and the nuclear atom
Geiger and Marsden fired alpha particles at a thin gold foil. Three observations, three conclusions:
Most passed straight through, barely deflected → the atom is mostly empty space.
A few were deflected by large angles → there is a concentrated positive charge to repel them.
About 1 in 8000 bounced back through more than 90° → that charge is in a tiny, massive nucleus.
This demolished the Thomson plum-pudding model, in which a diffuse positive charge could never have produced a large-angle deflection.
R = r₀ A^(1/3) where r₀ ≈ 1.2 fmA = nucleon number · so nuclear volume ∝ A, and nuclear density is CONSTANT
Why R ∝ A^(1/3): volume ∝ R³ and volume ∝ A (each nucleon takes up the same room), so R³ ∝ A. This means all nuclei have essentially the same density, around 10¹⁷ kg m⁻³ — the density of a neutron star.
Calculate
Your turn — nuclear radius
1Calculate the radius of an aluminium-27 nucleus (A = 27), taking r₀ = 1.2 fm. Give your answer in fm.
fm
Hint: R = r₀A^(1/3). The cube root of 27 is exactly 3, so R = 1.2 × 3.
Topic 8 · the electronvolt
The electronvolt and particle accelerators
1 eV = 1.60 × 10⁻¹⁹ Jthe energy gained by ONE electron accelerated through a pd of ONE volt: E = QV
Joules are absurdly large for single particles, so we use eV, keV, MeV and GeV.
Linear accelerator (linac): alternating pds on a line of drift tubes, each of which must get longer as the particle speeds up so that it always arrives in step with the field.
Cyclotron: two D-shaped electrodes in a uniform magnetic field. The magnetic force provides the centripetal force, so r = mv/(BQ) — the radius grows as the particle speeds up, giving a spiral. It is accelerated by the electric field in the gap each half-turn.
Higher energy → shorter de Broglie wavelength (λ = h/p) → finer probing of nuclear structure. That is why bigger accelerators see smaller things.
Detection: charged particles are tracked in magnetic fields — the curvature of the track gives the momentum, and the direction of curvature gives the sign of the charge.
Calculate
Your turn — accelerating an electron
2An electron is accelerated from rest through a potential difference of 5.0 kV. Calculate the kinetic energy it gains. Give your answer as a multiple of 10⁻¹⁶ J — type 8.0 if the answer is 8.0 × 10⁻¹⁶ J. (e = 1.60 × 10⁻¹⁹ C)
× 10⁻¹⁶ J
Hint: E = QV = 1.60 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J.
Calculate
Your turn — the resulting speed
3That electron now has 8.0 × 10⁻¹⁶ J of kinetic energy. Calculate its speed (non-relativistically). Give your answer as a multiple of 10⁷ m s⁻¹ to 2 significant figures. (me = 9.11 × 10⁻³¹ kg)
× 10⁷ m s⁻¹
Hint: ½mv² = 8.0 × 10⁻¹⁶, so v² = (2 × 8.0 × 10⁻¹⁶) ÷ 9.11 × 10⁻³¹ = 1.76 × 10¹⁵. Take the square root.
Topic 8 · E = mc²
Mass–energy equivalence and annihilation
E = mc²mass and energy are the same currency; c² = 9.00 × 10¹⁶ makes the exchange rate brutal
Every particle has an antiparticle with the same mass but opposite charge (and opposite lepton/baryon number). The positron is the electron's antiparticle.
Annihilation: a particle meets its antiparticle and both vanish, their entire rest energy appearing as two gamma photons emitted in opposite directions (momentum must be conserved — one photon alone could not do it).
Pair production: the reverse. A photon of at least 2mc² creates a particle–antiparticle pair. It must occur near a nucleus so that momentum can be conserved.
Worked example — electron–positron annihilation
Rest energy of each = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 9.11 × 10⁻³¹ × 9.00 × 10¹⁶ = 8.20 × 10⁻¹⁴ J
Total energy released = 2 × 8.20 × 10⁻¹⁴ = 1.64 × 10⁻¹³ J, shared as two photons of 8.20 × 10⁻¹⁴ J each (= 0.511 MeV each).
Calculate
Your turn — annihilation photon
4An electron and a positron, both at rest, annihilate. Two identical gamma photons are produced. Calculate the energy of one photon. Give your answer as a multiple of 10⁻¹⁴ J to 2 significant figures. (me = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m s⁻¹)
× 10⁻¹⁴ J
Hint: Each photon carries the rest energy of one particle: E = mc² = 9.11 × 10⁻³¹ × 9.00 × 10¹⁶.
Calculate
Your turn — rest energy in MeV
5Calculate the rest energy of a proton (m = 1.67 × 10⁻²⁷ kg). Give your answer in MeV to 3 significant figures. (c = 3.00 × 10⁸ m s⁻¹, 1 MeV = 1.60 × 10⁻¹³ J)
MeV
Hint: E = mc² = 1.67 × 10⁻²⁷ × 9.00 × 10¹⁶ = 1.503 × 10⁻¹⁰ J. Divide by 1.60 × 10⁻¹³ J per MeV.
Topic 8 · the Standard Model
Hadrons, leptons and quarks
Every particle is either a hadron (feels the strong nuclear force, made of quarks) or a lepton (fundamental, does not feel the strong force).
Baryons — hadrons made of three quarks. Proton = uud, neutron = udd. Baryon number B = +1.
Mesons — hadrons made of a quark and an antiquark. Pions and kaons. Baryon number B = 0.
Leptons — electron, muon, tau and their three neutrinos. Lepton number L = +1 (and −1 for antileptons).
β⁺ decay: p → n + e⁺ + νeat the quark level: u → d + e⁺ + νₑ — a WEAK interaction
Why the antineutrino? Beta particles emerge with a continuous spectrum of energies, not a single value. Pauli proposed an unseen particle carrying away the missing energy and momentum — and conserving lepton number. That particle is the (anti)neutrino.
Quick check
What Rutherford proved
?In the Geiger–Marsden experiment, a small number of alpha particles were deflected through more than 90°. What did this show?
Quick check
Quark content
?A neutron is composed of which quarks?
Quick check
Check the conservation laws
?Which conservation law forbids the decay p → n + e⁺ from happening for a free proton in isolation, even though charge, baryon number and lepton number all balance?
Quick check
Annihilation
?Why must electron–positron annihilation produce two gamma photons rather than one?
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Recap
The big ideas to know
Rutherford: most alphas pass through (empty space); a few backscatter (small, dense, positive nucleus)
Nuclear radius: R = r₀A^(1/3) with r₀ ≈ 1.2 fm → nuclear density is the same for all nuclei
Electronvolt: 1 eV = 1.60 × 10⁻¹⁹ J; accelerators reach MeV and GeV
E = mc²: annihilation gives two back-to-back photons; pair production needs at least 2mc²
Hadrons: baryons (3 quarks, e.g. p = uud, n = udd) and mesons (quark + antiquark)
Leptons: electron, muon, tau and their neutrinos — fundamental, no strong force
Conservation: charge, baryon number, lepton number, energy, momentum — and strangeness in strong interactions only
That is Edexcel Topic 8 — from the gold-foil experiment to the Standard Model. Press Finish to see your score.
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