Skip to content

AQA A-level Physics revision notes

Nuclear physics (A-level only)

Section 3.8
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
8 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.8

Checked against AQA 7408 section 3.8. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Data and formulae booklet provided · calculator allowed in every paper

Open the printable pack
3.8.1.1

Rutherford scattering

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Rutherford scattering changed the atomic model through linked observations and inferences. Most alpha particles crossed thin metal foil with little deflection, showing that atoms are mostly empty space.
  • Some deflected through small angles because positive alpha particles were repelled by positive charge. Very few scattered through large angles or backwards, requiring a strong force concentrated in a tiny region.
  • The nuclear model therefore places nearly all atomic mass and all positive charge in a nucleus far smaller than the atom.
  • A diffuse positive-charge model could not provide the intense local force needed to reverse an alpha particle.
  • Explanations must connect each observation to its specific inference.
Most alpha particles pass through foil, while rare close approaches produce large-angle scattering.
Worked example

State the inference from the rare observation that an alpha particle returns towards its source.

  1. 1.A reversal requires a very large force and momentum change.
  2. 2.Electrostatic repulsion must therefore act near concentrated positive charge.
  3. 3.Its rarity shows that this massive positive region occupies little atomic volume.

Answer: Nearly all mass and positive charge are concentrated in a very small nucleus.

Common mistakes

  • Don't list scattering observations without linking them to atomic structure.
  • Don't say electrons cause the large deflections of positive alpha particles.
  • Don't infer that the nucleus occupies most of the atom.

Exam tip

Write each scattering observation followed immediately by “therefore” and the matching inference.

Tier 1 · Easy

ORIGINAL

State what is inferred from the observation that most alpha particles cross a very thin gold foil without changing direction.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A small fraction of alpha particles directed at a thin platinum foil are deflected through angles greater than 9090^\circ. Explain what this reveals about the atom.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a scattering experiment, most alpha particles continue straight through a metal foil, some are deflected slightly, and about one in twelve thousand returns towards the source. Explain how these observations support the nuclear model rather than a model with positive charge spread throughout the atom.

[5 marks]

Total for this question: 5

Your progress and exam materials

This section: Evidence from your answers: 0/8 secureYour confidence: 0 self-rated secureTracker status: 0/8 secure, 0 shaky, 8 unseen

Overall: Evidence from your answers: 0/147 secureYour confidence: 0 self-rated secureTracker status: 0/147 secure, 0 shaky, 147 unseen

Progress is saved on this device for guests and accounts right now; cross-device account sync is not live yet.

3.8.1.2

Alpha, beta and gamma radiation

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Alpha is strongly ionising with short range and low penetration; beta has intermediate ionisation and is absorbed by thin aluminium; gamma is weakly ionising but highly penetrating and needs thick dense shielding. Absorption experiments identify radiation and support thickness control: beta suits paper or aluminium foil, while gamma can monitor steel.
  • For a point gamma source, I=k/x2I=k/x^2, using source-to-detector distance and background-corrected count rate.
  • Required practical 12 tests this by plotting corrected rate against 1/x21/x^2.
  • Background originates from sources including cosmic rays and rocks and must be measured separately.
  • Safe handling balances time, distance and shielding, while medical uses require a justified risk–benefit comparison.
Alpha, beta and gamma have increasing penetration and require different absorbers.
Worked example

A detector reads 220s1220\,\text{s}^{-1} at 2.0m2.0\,\text{m} from a gamma source; background is 20s120\,\text{s}^{-1}. Predict the reading at 4.0m4.0\,\text{m}.

  1. 1.Subtract background: source rate is 200s1200\,\text{s}^{-1}.
  2. 2.Doubling distance reduces source rate by 222^2, giving 50s150\,\text{s}^{-1}.
  3. 3.Add background to obtain the detector reading.

Answer: The detector reading is 70 s⁻¹.

Common mistakes

  • Don't apply the inverse-square law to a count rate that still includes background.
  • Don't measure distance from the detector casing rather than the source position.
  • Don't call the most penetrating radiation the most ionising.

Exam tip

Subtract background before analysis and add it back only when a total detector reading is requested.

Tier 1 · Easy

ORIGINAL

A radioactive source produces radiation that is stopped by a sheet of paper. Identify the radiation and state one relative hazard when it is inside the body.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A gamma detector records 920s1920\,\text{s}^{-1} at 0.20m0.20\,\text{m} from a point source. The background rate is 20s120\,\text{s}^{-1}. Calculate the detector reading expected at 0.50m0.50\,\text{m}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Describe an experiment to test the inverse-square law for a sealed gamma source. Your method must explain how background radiation is handled and how the data are analysed.

[5 marks]

Total for this question: 5

3.8.1.3

Radioactive decay

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Radioactive decay is random for one nucleus, but every nucleus of an isotope has the same constant probability per unit time, the decay constant λ\lambda. Hence ΔN/Δt=λN\Delta N/\Delta t=-\lambda N, N=N0eλtN=N_0e^{-\lambda t} and A=λN=A0eλtA=\lambda N=A_0e^{-\lambda t}, where 1Bq=1s11\,\text{Bq}=1\,\text{s}^{-1}.
  • Half-life satisfies T1/2=ln2/λT_{1/2}=\ln2/\lambda. It can be found from repeated halving on a decay curve or from the gradient λ-\lambda of a lnA\ln A or lnN\ln N graph.
  • Models may use dice or experimental data.
  • Calculations can require molar mass and Avogadro’s constant.
  • Applications include dating and radioactive-waste storage.
Activity falls exponentially, reaching half its initial value after one half-life.
Worked example

An isotope has half-life 6.0h6.0\,\text{h}. Calculate its decay constant in s1\text{s}^{-1}.

  1. 1.T1/2=6.0×3600=2.16×104sT_{1/2}=6.0\times3600=2.16\times10^4\,\text{s}.
  2. 2.λ=ln2/T1/2\lambda=\ln2/T_{1/2}.
  3. 3.λ=0.693/(2.16×104)\lambda=0.693/(2.16\times10^4).

Answer: The decay constant is 3.2 × 10⁻⁵ s⁻¹.

Common mistakes

  • Don't describe decay as becoming more likely because a nucleus is old.
  • Don't use half-life in hours with activity in becquerels without matching time units.
  • Don't treat the gradient of the curved activity–time graph as the negative decay constant.

Exam tip

On a log graph, include the minus sign when obtaining decay constant from the gradient.

Tier 1 · Easy

ORIGINAL

An isotope has a half-life of 6.0h6.0\,\text{h}. Calculate its decay constant in s1\text{s}^{-1}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A source initially has activity 480Bq480\,\text{Bq} and decay constant 1.8×104s11.8\times10^{-4}\,\text{s}^{-1}. Determine its activity after 2.40×103s2.40\times10^3\,\text{s}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A pure sample has activity 860Bq860\,\text{Bq} at a time 5.0h5.0\,\text{h} after it was prepared. Its half-life is 3.0h3.0\,\text{h}. Calculate the number of radioactive nuclei present when the sample was prepared.

[5 marks]

Total for this question: 5

3.8.1.4

Nuclear instability

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • On an NNZZ graph, light stable nuclei lie near N=ZN=Z, while stable heavy nuclei need more neutrons than protons. A neutron-rich nucleus may undergo β\beta^- decay, giving N1N-1 and Z+1Z+1.
  • A proton-rich nucleus may undergo β+\beta^+ decay or electron capture, each giving N+1N+1 and Z1Z-1. Alpha decay reduces NN and ZZ by 22 each.
  • Nuclear excited states lose energy by gamma emission without changing NN, ZZ or AA; technetium-99m is used as a gamma source in diagnosis.
  • Energy-level diagrams may represent transitions.
  • Nuclear equations must conserve nucleon number and charge.
The band of stability bends above the N equals Z line for heavier nuclei.
Worked example

A neutron-rich nucleus undergoes β\beta^- decay. State how its position changes on an NNZZ graph.

  1. 1.A neutron changes into a proton.
  2. 2.NN decreases by 11 and ZZ increases by 11.
  3. 3.A=N+ZA=N+Z remains unchanged.

Answer: The point moves one unit down and one unit right, generally towards the stability band.

Common mistakes

  • Don't change nucleon number during beta decay.
  • Don't treat gamma emission as loss of a proton or neutron.
  • Don't reverse the neutron-number and proton-number changes for beta-minus decay.

Exam tip

Balance the top and bottom numbers separately before naming a nuclear decay.

Tier 1 · Easy

ORIGINAL

A nucleus lies above the band of stability on an NN-ZZ graph. State its likely beta decay mode and the changes in NN and ZZ.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Fluorine-18 is proton-rich and decays to oxygen-18. State the decay mode and complete the equation 918F818O+^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+\ldots

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Strontium-90 undergoes two successive beta-minus decays, first to yttrium and then to zirconium. Complete the equations 3890Sr^{90}_{38}\mathrm{Sr}\rightarrow\ldots and 3990Y^{90}_{39}\mathrm{Y}\rightarrow\ldots and explain the movement of each nucleus on an NN-ZZ graph.

[5 marks]

Total for this question: 5

3.8.1.5

Nuclear radius

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Nuclear radii are typically of order 1015m10^{-15}\,\text{m}. Closest approach estimates an upper limit by equating an alpha particle’s initial kinetic energy to Coulomb electric potential energy at closest separation.
  • Electron diffraction gives radius more directly: the angular intensity pattern and its minima depend on nuclear size. Experimental data support R=R0A1/3R=R_0A^{1/3}.
  • Since spherical volume is proportional to R3R^3, nuclear volume is proportional to AA; nuclear mass is also approximately proportional to AA, so nuclear density is approximately constant.
  • Calculations require femtometre conversion, order-of-magnitude reasoning and the Coulomb equation.
  • The intensity-against-angle shape for electron diffraction should be recognised.
Nuclear radius increases with the cube root of nucleon number.
Worked example

Use R=R0A1/3R=R_0A^{1/3} with R0=1.05fmR_0=1.05\,\text{fm} to find the radius of aluminium-27.

  1. 1.A1/3=271/3=3A^{1/3}=27^{1/3}=3.
  2. 2.R=(1.05)(3)=3.15fmR=(1.05)(3)=3.15\,\text{fm}.
  3. 3.In SI form, multiply by 101510^{-15}.

Answer: The nuclear radius is 3.15 fm, or 3.15 × 10⁻¹⁵ m.

Common mistakes

  • Don't use nucleon number instead of its cube root in the radius equation.
  • Don't convert femtometres using the wrong power of ten.
  • Don't claim constant radius, rather than constant density, for different nuclei.

Exam tip

To show constant density, substitute R=R0A1/3R=R_0A^{1/3} into ρ=Amn/(4πR3/3)\rho=Am_n/(4\pi R^3/3) and cancel AA.

Tier 1 · Easy

ORIGINAL

Use R=r0A1/3R=r_0A^{1/3} with r0=1.05fmr_0=1.05\,\text{fm} to calculate the radius of an aluminium-27 nucleus.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A nucleus has radius 4.20fm4.20\,\text{fm}. Using R=r0A1/3R=r_0A^{1/3} and r0=1.05fmr_0=1.05\,\text{fm}, determine its nucleon number.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Model a nucleus as a sphere with R=r0A1/3R=r_0A^{1/3}, where r0=1.05fmr_0=1.05\,\text{fm}. Taking each nucleon to have mass 1.67×1027kg1.67\times10^{-27}\,\text{kg}, calculate the nuclear density and show why your result is independent of AA.

[5 marks]

Total for this question: 5

3.8.1.6

Mass and energy

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Every energy change has an equivalent mass change through E=mc2E=mc^2. Binding energy is the energy required to separate a nucleus into free nucleons, and the mass defect between separated constituents and the nucleus represents that energy.
  • For masses in atomic mass units, 1u=931.5MeV1\,\text{u}=931.5\,\text{MeV}. Average binding energy per nucleon measures nuclear stability.
  • Its graph rises steeply for light nuclei, peaks near medium nucleon number and falls slowly for heavy nuclei.
  • Fusion of light nuclei and fission of heavy nuclei release energy because products move towards higher binding energy per nucleon.
  • Calculations should show initial mass, final mass, mass difference and a consistent unit conversion.
Fusion and fission release energy when products have greater binding energy per nucleon.
Worked example

A reaction has mass defect 0.00320u0.00320\,\text{u}. Find the energy released using 1u=931.5MeV1\,\text{u}=931.5\,\text{MeV}.

  1. 1.The positive mass defect means final rest mass is lower.
  2. 2.E=(0.00320)(931.5)MeVE=(0.00320)(931.5)\,\text{MeV}.
  3. 3.Round to the precision of the mass defect.

Answer: The energy released is 2.98 MeV.

Common mistakes

  • Don't subtract initial mass from final mass and report a negative released energy.
  • Don't multiply a mass defect in kilograms by the conversion factor stated for atomic mass units.
  • Don't say a smaller binding energy per nucleon means a more stable nucleus.

Exam tip

Tabulate total reactant and product masses before converting their difference into energy.

Tier 1 · Easy

ORIGINAL

A nuclear reaction has a mass defect of 0.00320u0.00320\,\text{u}. Use 1u=931.5MeV1\,\text{u}=931.5\,\text{MeV} to calculate the energy released.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

In one fission event, uranium-235 absorbs a neutron and produces barium-141, krypton-92 and three neutrons. The relevant masses are 235.0439u235.0439\,\text{u}, 140.9144u140.9144\,\text{u}, 91.9262u91.9262\,\text{u} and 1.0087u1.0087\,\text{u} for a neutron. Use 1u=931.5MeV1\,\text{u}=931.5\,\text{MeV} to calculate the energy released.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A deuterium nucleus and a tritium nucleus fuse to form helium-4 and a neutron. The corresponding atomic masses of deuterium, tritium and helium-4 are 2.01410u2.01410\,\text{u}, 3.01605u3.01605\,\text{u} and 4.00260u4.00260\,\text{u}; the neutron mass is 1.00867u1.00867\,\text{u}. Use 1u=931.5MeV1\,\text{u}=931.5\,\text{MeV} and 1MeV=1.602×1013J1\,\text{MeV}=1.602\times10^{-13}\,\text{J}. Calculate the energy released in both MeV and joules, and explain the release using binding energy per nucleon.

[6 marks]

Total for this question: 6

3.8.1.7

Induced fission

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A fissile nucleus absorbs a slow thermal neutron, becomes unstable and splits, releasing energy and more neutrons. A chain reaction is critical when one neutron per fission, on average, causes another fission; critical mass is the minimum arrangement able to sustain it.
  • A moderator slows fast neutrons by elastic collisions, working best with light nuclei and low neutron absorption; water or graphite can be used. Control rods such as boron or cadmium absorb neutrons and regulate reaction rate.
  • Coolant removes thermal energy and needs suitable heat capacity, flow behaviour, stability and low neutron absorption.
  • Material choices follow these functions.
  • A mechanical collision model explains efficient neutron energy transfer to light moderator nuclei.
A thermal reactor uses moderator, control rods and coolant around fissile fuel.
Worked example

Each fission releases 2.52.5 neutrons and 40%40\% cause another fission. Classify the chain reaction.

  1. 1.Mean continuing neutrons per fission are 2.5(0.40)2.5(0.40).
  2. 2.The multiplication factor is 1.01.0.
  3. 3.One fission replaces itself with one further fission on average.

Answer: The reactor is critical and the chain reaction is steady.

Common mistakes

  • Don't say the moderator absorbs neutrons instead of slowing them.
  • Don't say control rods remove heat rather than neutrons.
  • Don't choose heavy moderator nuclei even though elastic energy transfer is then less effective.

Exam tip

For each reactor component, state its function first and then link a material property to that function.

Tier 1 · Easy

ORIGINAL

State the function of the moderator and the function of the control rods in a thermal nuclear reactor.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Each fission in a reactor releases an average of 2.52.5 neutrons. If 40%40\% of these neutrons induce another fission, calculate the multiplication factor and state whether the reactor is subcritical, critical or supercritical.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Explain how induced fission becomes a controlled chain reaction in a thermal reactor. Include the roles and suitable material properties of the moderator, control rods and coolant.

[5 marks]

Total for this question: 5

3.8.1.8

Safety aspects

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Nuclear safety combines suitable fuel containment, remote handling, shielding and emergency shutdown. Remote tools increase distance from intense sources; shielding reduces external exposure and containment limits contamination.
  • Rapid insertion of neutron-absorbing rods stops the fission chain reaction, but radioactive products continue producing decay heat, so cooling remains essential.
  • Waste production, remote handling and storage depend on activity and half-life: highly active waste needs shielding, secure containment and long-term management.
  • Time, distance and shielding reduce exposure but do not substitute for contamination controls.
  • A balanced judgement compares specific accident, waste and radiation risks with benefits such as reliable low-carbon electricity, rather than claiming either zero risk or no benefit.
Containment, shielding and distance work together to reduce exposure during remote handling.
Worked example

Explain why coolant must continue circulating immediately after emergency control rods stop fission.

  1. 1.Inserted rods absorb neutrons and end the induced chain reaction.
  2. 2.Existing fission products remain radioactive.
  3. 3.Their decay releases thermal energy that must be removed.

Answer: Continued cooling removes decay heat and prevents fuel or containment overheating.

Common mistakes

  • Don't claim emergency shutdown instantly stops all heat production.
  • Don't treat shielding as protection against radioactive contamination entering the body.
  • Don't give a risk–benefit conclusion without comparing specific risks and benefits.

Exam tip

In an evaluation, pair each stated nuclear risk with the control used and a specific societal benefit.

Tier 1 · Easy

ORIGINAL

State two ways in which worker exposure is reduced while spent reactor fuel is moved.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain why a reactor still needs coolant circulation immediately after an emergency shutdown has fully inserted the control rods.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

During remote handling, a detector reads 365s1365\,\text{s}^{-1} at 1.5m1.5\,\text{m} from a compact gamma source. Background is 40s140\,\text{s}^{-1}. Assuming inverse-square behaviour, determine the distance at which the source contribution is 13s113\,\text{s}^{-1} and state the detector reading there. Explain why this distance alone is not a complete safety measure.

[5 marks]

Total for this question: 5

Want help turning these notes into marks?

Bring a tricky specification point or a recent answer, and we can work through the method and exam wording together.