3.8 Nuclear physics (A-level only) — revision question pack

8 specification points · notes, questions, answers and worked methods

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.

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Answer all questions in the spaces provided.

3.8.1.1 · Rutherford scattering

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.A reversal requires a very large force and momentum change.
2.Electrostatic repulsion must therefore act near concentrated positive charge.
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

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
State why the metal foil used for an alpha-particle scattering experiment must be very thin.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A small fraction of alpha particles directed at a thin platinum foil are deflected through angles greater than $90^\circ$ . Explain what this reveals about the atom.
[3 marks]
Total for this question: 3
Explain why an alpha particle is deflected before reaching a gold nucleus and why the deflection increases as it gets closer.
[2 marks]
Total for this question: 2
Treat a gold nucleus as fixed. An alpha particle approaches it head-on, stops momentarily and then travels back along its original path. Explain the energy transfers during this motion and why the observation is evidence for a positively charged nucleus.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Explain why the large-angle deflections cannot be caused by the atomic electrons, and what the size of the alpha particle's momentum change shows about the mass and charge concentration of the scattering centre.
[3 marks]
Total for this question: 3
Two gold foils are tested using identical alpha-particle beams. Foil X is extremely thin, whereas foil Y contains many more atomic layers. A much larger fraction of particles is deflected through more than $90^\circ$ by Y. A student concludes that the nuclei in Y are larger. Evaluate this conclusion. Explain why the fraction deflected through more than $90^\circ$ is expected to be proportional to thickness only for a very thin foil.
[4 marks]
Total for this question: 4
A student makes three claims about alpha-particle scattering: straight trajectories prove that no force acts anywhere inside an atom; the rarity of backward scattering by itself proves that nearly all atomic mass is in the nucleus; and a diffuse sphere containing the same total positive charge could reverse an alpha particle. Evaluate all three claims and give the corrected inferences.
[5 marks]
Total for this question: 5
A beam of alpha particles is directed in turn at two very thin foils containing the same number of target nuclei per unit area. Foil L contains nuclei with a low proton number and foil H contains nuclei with a much higher proton number. Predict how replacing L with H affects the fraction of alpha particles scattered through large angles and the fraction that is essentially undeflected. Explain both predictions.
[4 marks]
Total for this question: 4

3.8.1.2 · Alpha, beta and gamma radiation

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, $I=k/x^2$ , using source-to-detector distance and background-corrected count rate.
Required practical 12 tests this by plotting corrected rate against $1/x^2$ .
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 $220\,\text{s}^{-1}$ at $2.0\,\text{m}$ from a gamma source; background is $20\,\text{s}^{-1}$ . Predict the reading at $4.0\,\text{m}$ .

1.Subtract background: source rate is $200\,\text{s}^{-1}$ .
2.Doubling distance reduces source rate by $2^2$ , giving $50\,\text{s}^{-1}$ .
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

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
Explain why beta radiation is suitable for controlling the thickness of aluminium foil, whereas gamma radiation is not.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gamma detector records $920\,\text{s}^{-1}$ at $0.20\,\text{m}$ from a point source. The background rate is $20\,\text{s}^{-1}$ . Calculate the detector reading expected at $0.50\,\text{m}$ .
[3 marks]
Total for this question: 3
A gamma detector records $410\,\text{s}^{-1}$ at $0.30\,\text{m}$ from a point source and $110\,\text{s}^{-1}$ at $0.60\,\text{m}$ . The background count rate is $10\,\text{s}^{-1}$ . Show that the corrected count rates are consistent with the inverse-square law.
[2 marks]
Total for this question: 2
A source emits beta and gamma radiation. In $60\,\text{s}$ a detector records $1240$ counts with no absorber and $400$ counts with aluminium that absorbs all the beta radiation but negligible gamma radiation. A separate background reading is $40$ counts in $60\,\text{s}$ . Calculate the percentage of the background-corrected source count caused by beta radiation.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Discuss the use of a gamma-emitting tracer rather than an alpha-emitting tracer for diagnostic imaging inside a patient.
[5 marks]
Total for this question: 5
Background-corrected count rates from a point gamma source are $784\,\text{s}^{-1}$ and $400\,\text{s}^{-1}$ when a ruler reads $0.20\,\text{m}$ and $0.30\,\text{m}$ respectively. The true source-to-detector distance is $d=x+a$ , where $x$ is the ruler reading and $a$ is a constant offset. Assuming $C=k/d^2$ , determine $a$ and predict the corrected count rate at $x=0.50\,\text{m}$ . Explain the physical origin of the offset.
[5 marks]
Total for this question: 5
A detector records $168\,\text{s}^{-1}$ at $0.240\,\text{m}$ from an isotropic point gamma source. The background count rate is $8.0\,\text{s}^{-1}$ . Predict the detector reading at $0.400\,\text{m}$ . Explain why a source emitting photons at rate $R$ gives the background-corrected detector count rate $C=RS/(4\pi x^2)$ when the detector has small sensitive area $S$ and every photon reaching the detector is counted.
[4 marks]
Total for this question: 4
A student records $5080$ counts in $100\,\text{s}$ at $0.250\,\text{m}$ from a point gamma source and $1352$ counts in $100\,\text{s}$ at $0.500\,\text{m}$ . A background measurement gives $240$ counts in $300\,\text{s}$ . For a count $N$ , take the random counting uncertainty to be $\sqrt{N}$ . Deduce whether the measurements support inverse-square behaviour, allowing for random counting uncertainty.
[5 marks]
Total for this question: 5

3.8.1.3 · Radioactive decay

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 $\lambda$ . Hence $\Delta N/\Delta t=-\lambda N$ , $N=N_0e^{-\lambda t}$ and $A=\lambda N=A_0e^{-\lambda t}$ , where $1\,\text{Bq}=1\,\text{s}^{-1}$ .
Half-life satisfies $T_{1/2}=\ln2/\lambda$ . It can be found from repeated halving on a decay curve or from the gradient $-\lambda$ of a $\ln A$ or $\ln N$ 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 $6.0\,\text{h}$ . Calculate its decay constant in $\text{s}^{-1}$ .

1. $T_{1/2}=6.0\times3600=2.16\times10^4\,\text{s}$ .
2. $\lambda=\ln2/T_{1/2}$ .
3. $\lambda=0.693/(2.16\times10^4)$ .

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

An isotope has a half-life of $6.0\,\text{h}$ . Calculate its decay constant in $\text{s}^{-1}$ .
[2 marks]
Total for this question: 2
State what is meant by saying that radioactive decay is random but has a constant decay probability.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A source initially has activity $480\,\text{Bq}$ and decay constant $1.8\times10^{-4}\,\text{s}^{-1}$ . Determine its activity after $2.40\times10^3\,\text{s}$ .
[3 marks]
Total for this question: 3
The activity of a radioactive sample falls from $720\,\text{Bq}$ to $90\,\text{Bq}$ in $18\,\text{days}$ . Determine its half-life to two significant figures.
[2 marks]
Total for this question: 2
A graph of $\ln(A/\text{Bq})$ against time is a straight line. Its value falls from $7.20$ to $5.40$ during $8.00\,\text{h}$ . Determine the decay constant in $\text{h}^{-1}$ and the half-life in hours.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A pure sample has activity $860\,\text{Bq}$ at a time $5.0\,\text{h}$ after it was prepared. Its half-life is $3.0\,\text{h}$ . Calculate the number of radioactive nuclei present when the sample was prepared.
[5 marks]
Total for this question: 5
A $2.40\,\mu\text{g}$ pure sample of an isotope with molar mass $60.0\,\text{g mol}^{-1}$ has activity $1.60\times10^9\,\text{Bq}$ . Using $N_\text{A}=6.02\times10^{23}\,\text{mol}^{-1}$ , calculate the half-life of the isotope to three significant figures.
[4 marks]
Total for this question: 4
A sample contains two radionuclides P and Q with half-lives $2.00\,\text{h}$ and $6.00\,\text{h}$ respectively. Their combined activity is initially $1200\,\text{Bq}$ and is $225\,\text{Bq}$ after $6.00\,\text{h}$ . Determine the initial activity of each radionuclide and the time after preparation when their activities are equal.
[5 marks]
Total for this question: 5
A radionuclide has activity $4.80\times10^5\,\text{Bq}$ and half-life $9.60\,\text{days}$ at the start of an observation. Determine the number of nuclei that decay during the next $3.20\,\text{days}$ .
[4 marks]
Total for this question: 4
A sealed radionuclide source initially has activity $3.60\times10^{12}\,\text{Bq}$ and half-life $18.0\,\text{h}$ . Each decay deposits $0.740\,\text{MeV}$ as thermal energy in the source. Calculate its initial thermal power, the time for this power to fall below $0.120\,\text{W}$ , and the energy deposited during the first $54.0\,\text{h}$ . Use $1\,\text{MeV}=1.602\times10^{-13}\,\text{J}$ .
[6 marks]
Total for this question: 6

3.8.1.4 · Nuclear instability

Explanation

On an $N$ – $Z$ graph, light stable nuclei lie near $N=Z$ , while stable heavy nuclei need more neutrons than protons. A neutron-rich nucleus may undergo $\beta^-$ decay, giving $N-1$ and $Z+1$ .
A proton-rich nucleus may undergo $\beta^+$ decay or electron capture, each giving $N+1$ and $Z-1$ . Alpha decay reduces $N$ and $Z$ by $2$ each.
Nuclear excited states lose energy by gamma emission without changing $N$ , $Z$ or $A$ ; 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 $\beta^-$ decay. State how its position changes on an $N$ – $Z$ graph.

1.A neutron changes into a proton.
2. $N$ decreases by $1$ and $Z$ increases by $1$ .
3. $A=N+Z$ 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

A nucleus lies above the band of stability on an $N$ - $Z$ graph. State its likely beta decay mode and the changes in $N$ and $Z$ .
[2 marks]
Total for this question: 2
State the changes in neutron number, proton number and nuclear energy when an excited nucleus emits a gamma photon.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Fluorine-18 is proton-rich and decays to oxygen-18. State the decay mode and complete the equation $^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+\ldots$
[3 marks]
Total for this question: 3
Iodine-123 undergoes electron capture. Complete the nuclear equation, including the captured electron, and explain why a characteristic X-ray may be emitted after the capture.
[3 marks]
Total for this question: 3
An excited nucleus loses $0.180\,\text{MeV}$ in a two-step gamma cascade. The first photon has frequency $2.50\times10^{19}\,\text{Hz}$ . Determine the frequency of the second photon. Use $h=6.63\times10^{-34}\,\text{J s}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Strontium-90 undergoes two successive beta-minus decays, first to yttrium and then to zirconium. Complete the equations $^{90}_{38}\mathrm{Sr}\rightarrow\ldots$ and $^{90}_{39}\mathrm{Y}\rightarrow\ldots$ and explain the movement of each nucleus on an $N$ - $Z$ graph.
[5 marks]
Total for this question: 5
A nucleus has energy levels at $0$ , $0.38\,\text{MeV}$ , $0.91\,\text{MeV}$ and $1.24\,\text{MeV}$ . Every downward transition between these levels is allowed. Determine all possible gamma-photon energies and calculate the shortest possible photon wavelength. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[4 marks]
Total for this question: 4
Tin-120 is stable with $N=70$ and $Z=50$ . Xenon-120 has $N=66$ and $Z=54$ and is proton-rich. Explain why stable heavy nuclei need more neutrons than protons, and deduce how xenon-120 can move towards the band of stability without changing its nucleon number.
[4 marks]
Total for this question: 4
On a graph of $N$ against $Z$ , with $Z$ on the horizontal axis, a nucleus is represented by point P at $(Z,N)=(62,96)$ . Points R, S and T have coordinates $(63,95)$ , $(61,97)$ and $(60,94)$ respectively. For each destination, deduce the possible single-step decay process or processes that can move P there. State the change in nucleon number for each process and explain why gamma emission cannot produce any of the three moves.
[5 marks]
Total for this question: 5
A gamma-ray spectrum from excited nuclei of an isotope $^{A}_{Z}\mathrm{X}$ contains lines at $0.263\,\text{MeV}$ , $0.491\,\text{MeV}$ and $0.754\,\text{MeV}$ . The $0.263\,\text{MeV}$ and $0.491\,\text{MeV}$ photons are sometimes detected in rapid succession from one nucleus, with the $0.263\,\text{MeV}$ photon always emitted first, whereas a $0.754\,\text{MeV}$ photon is detected alone. Deduce the two excited-state energies and the two possible routes to the ground state. Here, $*$ denotes an excited state and $**$ a higher excited state. Complete both nuclear equations for the two-step route, including each photon energy: $^{A}_{Z}\mathrm{X}^{**}\rightarrow\ldots$ and $^{A}_{Z}\mathrm{X}^{*}\rightarrow\ldots$ .
[5 marks]
Total for this question: 5

3.8.1.5 · Nuclear radius

Explanation

Nuclear radii are typically of order $10^{-15}\,\text{m}$ . 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 $R=R_0A^{1/3}$ .
Since spherical volume is proportional to $R^3$ , nuclear volume is proportional to $A$ ; nuclear mass is also approximately proportional to $A$ , 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 $R=R_0A^{1/3}$ with $R_0=1.05\,\text{fm}$ to find the radius of aluminium-27.

1. $A^{1/3}=27^{1/3}=3$ .
2. $R=(1.05)(3)=3.15\,\text{fm}$ .
3.In SI form, multiply by $10^{-15}$ .

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 $R=R_0A^{1/3}$ into $\rho=Am_n/(4\pi R^3/3)$ and cancel $A$ .

Tier 1 · Easy

Use $R=r_0A^{1/3}$ with $r_0=1.05\,\text{fm}$ to calculate the radius of an aluminium-27 nucleus.
[2 marks]
Total for this question: 2
State the order of magnitude of a typical nuclear radius in metres.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A nucleus has radius $4.20\,\text{fm}$ . Using $R=r_0A^{1/3}$ and $r_0=1.05\,\text{fm}$ , determine its nucleon number.
[3 marks]
Total for this question: 3
An alpha particle with kinetic energy $6.40\,\text{MeV}$ approaches a stationary tin nucleus of proton number $50$ head-on. Estimate its closest distance of approach to three significant figures.
[3 marks]
Total for this question: 3
Electron diffraction gives a nuclear radius $R=6.30\times10^{-15}\,\text{m}$ for a nuclide with nucleon number $A=216$ . Determine $R_0$ in $R=R_0A^{1/3}$ and state how the diffraction pattern is used to obtain $R$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Model a nucleus as a sphere with $R=r_0A^{1/3}$ , where $r_0=1.05\,\text{fm}$ . Taking each nucleon to have mass $1.67\times10^{-27}\,\text{kg}$ , calculate the nuclear density and show why your result is independent of $A$ .
[5 marks]
Total for this question: 5
An electron-diffraction experiment is repeated for a nuclide of known nucleon number $A$ . Describe the expected intensity-against-angle graph and explain how it can be used to estimate $R_0$ in $R=R_0A^{1/3}$ .
[3 marks]
Total for this question: 3
A $7.50\,\text{MeV}$ alpha particle approaches a stationary gold-197 nucleus head-on. At closest approach they momentarily have the same velocity. Treat their masses as $4\,\text{u}$ and $197\,\text{u}$ and the gold proton number as $79$ . Use conservation of momentum to determine the kinetic energy still associated with their common motion. Hence calculate the closest separation and the percentage by which it exceeds the value obtained by treating the gold nucleus as fixed.
[5 marks]
Total for this question: 5
Assume nuclear matter is uniform. For a nuclide with nucleon number $A$ , take its spherical radius to satisfy $A=(R/R_0)^3$ and use $R_0=1.05\,\text{fm}$ . Define the surface region as the shell extending $0.650\,\text{fm}$ inwards from the outer radius. Calculate the fraction of nuclear volume in this shell for nuclei with $A=27$ and $A=216$ . Explain why the fractions differ even though both nuclei have the same density.
[5 marks]
Total for this question: 5
Measurements give radii $4.30\,\text{fm}$ and $6.46\,\text{fm}$ for nuclei with nucleon numbers $50$ and $169$ respectively. Assuming $R=kA^n$ , determine $n$ . Calculate the ratio of the density of the second nucleus to that of the first. Discuss whether the measurements support constant nuclear density.
[5 marks]
Total for this question: 5

3.8.1.6 · Mass and energy

Explanation

Every energy change has an equivalent mass change through $E=mc^2$ . 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, $1\,\text{u}=931.5\,\text{MeV}$ . 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 $0.00320\,\text{u}$ . Find the energy released using $1\,\text{u}=931.5\,\text{MeV}$ .

1.The positive mass defect means final rest mass is lower.
2. $E=(0.00320)(931.5)\,\text{MeV}$ .
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

A nuclear reaction has a mass defect of $0.00320\,\text{u}$ . Use $1\,\text{u}=931.5\,\text{MeV}$ to calculate the energy released.
[2 marks]
Total for this question: 2
A fully charged battery releases $4.5\,\text{MJ}$ of energy. Calculate its decrease in mass, giving the answer to two significant figures.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In one fission event, uranium-235 absorbs a neutron and produces barium-141, krypton-92 and three neutrons. The relevant masses are $235.0439\,\text{u}$ , $140.9144\,\text{u}$ , $91.9262\,\text{u}$ and $1.0087\,\text{u}$ for a neutron. Use $1\,\text{u}=931.5\,\text{MeV}$ to calculate the energy released.
[4 marks]
Total for this question: 4
The nuclear mass of helium-4 is $4.00151\,\text{u}$ . Using proton mass $1.00728\,\text{u}$ , neutron mass $1.00867\,\text{u}$ and $1\,\text{u}=931.5\,\text{MeV}$ , calculate its average binding energy per nucleon to three significant figures.
[4 marks]
Total for this question: 4
Carbon-12 has an average binding energy of $7.68\,\text{MeV}$ per nucleon. The proton and neutron masses are $1.00728\,\text{u}$ and $1.00867\,\text{u}$ respectively. Use $1\,\text{u}=931.5\,\text{MeV}/c^2$ to determine the mass of the carbon-12 nucleus.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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 $2.01410\,\text{u}$ , $3.01605\,\text{u}$ and $4.00260\,\text{u}$ ; the neutron mass is $1.00867\,\text{u}$ . Use $1\,\text{u}=931.5\,\text{MeV}$ and $1\,\text{MeV}=1.602\times10^{-13}\,\text{J}$ . 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
A fusion event consumes reactants of total mass $5.0000\,\text{u}$ and produces products of total mass $4.9800\,\text{u}$ . Use $1\,\text{u}=931.5\,\text{MeV}/c^2$ , $1\,\text{MeV}=1.60\times10^{-13}\,\text{J}$ and $1\,\text{u}=1.66\times10^{-27}\,\text{kg}$ . A chemical fuel has a specific energy of $4.20\times10^7\,\text{J kg}^{-1}$ . Calculate the mass defect in u, the energy released per event in MeV and in J, the mass of reactants consumed per event in kg, the specific energy of the fusion fuel, and the factor by which this exceeds the chemical fuel's specific energy.
[6 marks]
Total for this question: 6
A gamma photon removes a neutron from an oxygen-16 atom, producing oxygen-15. The atomic masses of oxygen-16 and oxygen-15 are $15.994915\,\text{u}$ and $15.003066\,\text{u}$ , and the neutron mass is $1.008665\,\text{u}$ . Neglect recoil energy. Determine the minimum photon frequency. Use $1\,\text{u}=931.5\,\text{MeV}/c^2$ , $1\,\text{MeV}=1.602\times10^{-13}\,\text{J}$ and $h=6.63\times10^{-34}\,\text{J s}$ .
[5 marks]
Total for this question: 5
A stationary nucleus X undergoes alpha decay to nucleus Y. The atomic masses of X, Y and helium-4 are $222.014620\,\text{u}$ , $218.004910\,\text{u}$ and $4.002603\,\text{u}$ respectively. Calculate the energy released and, assuming no gamma emission, the kinetic energy of each product. Use $1\,\text{u}=931.5\,\text{MeV}/c^2$ , take the product masses to be proportional to the stated atomic masses and neglect relativistic effects.
[6 marks]
Total for this question: 6
A reactor transfers thermal energy at a steady rate of $3.20\,\text{GW}$ . The mean energy transferred as thermal energy per fission is $202\,\text{MeV}$ . Calculate the fission rate and the corresponding decrease in rest mass during $24.0\,\text{h}$ . Use $1\,\text{MeV}=1.602\times10^{-13}\,\text{J}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5

3.8.1.7 · Induced fission

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 $2.5$ neutrons and $40\%$ cause another fission. Classify the chain reaction.

1.Mean continuing neutrons per fission are $2.5(0.40)$ .
2.The multiplication factor is $1.0$ .
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

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
State what is meant by the critical mass of a fissile material.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Each fission in a reactor releases an average of $2.5$ neutrons. If $40\%$ 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
Two samples contain the same mass of fissile material. One is a compact sphere and the other is a thin sheet. Explain why the sphere is more likely to sustain a chain reaction.
[3 marks]
Total for this question: 3
In a reactor, each fission causes on average $1.015$ further fissions. One generation contains $4.00\times10^{12}$ fissions. Determine the number of fissions in the generation that occurs $30$ generations later. Explain how the control rods should be adjusted to restore steady power.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
In a simplified head-on collision model, a $2.0\,\text{MeV}$ neutron retains a fraction $(11/13)^2$ of its kinetic energy in every collision with a stationary carbon-12 nucleus. Determine the minimum number of collisions needed to reduce its energy below $0.025\,\text{eV}$ .
[5 marks]
Total for this question: 5
A neutron of mass $m$ and speed $u$ makes a one-dimensional elastic collision with a stationary moderator nucleus of mass $A m$ . Show that the fraction of the neutron's kinetic energy remaining is $((A-1)/(A+1))^2$ . Hence calculate the percentage of its energy lost in one collision with hydrogen-1 and with deuterium, and suggest one moderator property that this model does not include.
[5 marks]
Total for this question: 5
Explain why the neutrons released by fission must be slowed before they are likely to induce further fission in uranium-235, and state two possible fates of a neutron that is not slowed.
[3 marks]
Total for this question: 3
A reactor uses a separate solid moderator, requires its coolant to remain in one phase and has a coolant outlet temperature of $620\,\text{K}$ . Coolant L has high specific heat capacity but boils at $560\,\text{K}$ at the operating pressure and absorbs neutrons moderately. Coolant M is a gas with low neutron absorption and chemical stability but needs a high flow rate. Coolant N is a liquid with low neutron absorption and excellent heat transfer but reacts violently with air and water. Evaluate the three coolants and justify a choice for the reactor.
[6 marks]
Total for this question: 6

3.8.1.8 · Safety aspects

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.Inserted rods absorb neutrons and end the induced chain reaction.
2.Existing fission products remain radioactive.
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

State two ways in which worker exposure is reduced while spent reactor fuel is moved.
[2 marks]
Total for this question: 2
State two requirements of a container used for long-term storage of high-level radioactive waste.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Explain why high-level radioactive waste can be immobilised in glass and placed in a deep geological store.
[3 marks]
Total for this question: 3
During remote handling, a spent-fuel flask is found to be damaged while it remains behind a radiation shield. Explain why the shield reduces one hazard but does not make the situation safe, and suggest one additional control.
[3 marks]
Total for this question: 3

Tier 3 · Hard

During remote handling, a detector reads $365\,\text{s}^{-1}$ at $1.5\,\text{m}$ from a compact gamma source. Background is $40\,\text{s}^{-1}$ . Assuming inverse-square behaviour, determine the distance at which the source contribution is $13\,\text{s}^{-1}$ and state the detector reading there. Explain why this distance alone is not a complete safety measure.
[5 marks]
Total for this question: 5
Discuss a proposal to replace a fossil-fuel power station with a nuclear power station at the same site.
[5 marks]
Total for this question: 5
A proposal would move freshly removed spent reactor fuel directly into a sealed dry store, without first using a cooling pond. Explain why pond storage is needed at first and why the dry store cannot accept the fuel until its decay-heat output has fallen.
[5 marks]
Total for this question: 5
Discuss how remote handling, shielding, continued decay-heat cooling and storage reduce the risks during removal and storage of spent fuel from a reactor, and give a reasoned conclusion on the balance of risk and benefit.
[6 marks]
Total for this question: 6
A waste package contains radionuclides X and Y with initial activities $6.40\times10^{13}\,\text{Bq}$ and $1.10\times10^{12}\,\text{Bq}$ and half-lives $2.40\,\text{years}$ and $28.0\,\text{years}$ respectively. Show that the combined activity after $12.0\,\text{years}$ is about $2.8\times10^{12}\,\text{Bq}$ . Explain why this activity does not remove the need for shielding and secure containment.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.8.1.1 · Rutherford scattering

Tier 1 · Easy

Mark scheme for 3.8.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The atom is mostly empty space.	Most alpha particles encounter no concentrated matter or charge capable of exerting a large force, so most of the atomic volume must be empty space.	1
02.1	An alpha particle should be unlikely to undergo more than one deflection.	A thin foil makes multiple scattering unlikely, so the measured change in direction can be attributed to one close encounter.	1

Tier 2 · Standard

Mark scheme for 3.8.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The atom contains a very small, positively charged nucleus in which most of its mass is concentrated.	An alpha particle is positively charged, so a large deflection requires a strong repulsive electrostatic force. Such a strong force acts only when the alpha particle passes very close to a concentrated positive charge. Because large deflections are rare, this charged region is very small; the large change in momentum also shows that it contains most of the atomic mass.	3
02.1	The positive alpha particle is repelled by the positive nucleus through a non-contact electrostatic force that increases as their separation decreases.	Like positive charges repel through an electrostatic force that acts at a distance. The force varies as $1/r^2$ , so it becomes stronger and changes the alpha particle's momentum more rapidly as the separation $r$ decreases.	2
03.1	The alpha particle's kinetic energy is transferred to electric potential energy as it approaches. At closest approach its kinetic energy is zero; electrostatic repulsion from concentrated positive charge then accelerates it away.	The alpha particle and nucleus are both positively charged. Work is done against their repulsive electrostatic force as the alpha particle approaches, so its kinetic energy decreases while electric potential energy increases. At the turning point the alpha particle is instantaneously at rest. The same repulsive force then transfers electric potential energy back to kinetic energy and reverses the alpha particle. A reversal without contact is evidence that positive charge is concentrated in the nucleus.	3

Tier 3 · Hard

Mark scheme for 3.8.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The observations imply an atom that is mostly empty space with its positive charge and nearly all its mass concentrated in a tiny nucleus.	Particles travelling straight through show that most of the atom is empty space. Small deflections show repulsion between a positive alpha particle and positive atomic charge. A backward deflection requires a very large force and momentum change, so the positive charge must be highly concentrated and the scattering centre must be massive. The rarity of backward events shows that this region occupies only a tiny fraction of the atomic volume. A diffuse positive charge could not exert the intense, localised force needed for a reversal, so the observations favour the nuclear model.	5
02.1	Electrons are about 7000 times lighter than an alpha particle, so cannot reverse its momentum. A large momentum change requires a massive target with concentrated charge; therefore the scattering centre is a small, dense nucleus.	Atomic electrons have a mass about 7000 times smaller than an alpha particle, so an encounter with an electron cannot produce the observed reversal of the alpha particle's momentum. A large change in momentum needs a scattering centre that is massive, while the strong repulsion requires its positive charge to be concentrated in a very small region. This supports the nuclear model.	3
03.1	The conclusion is invalid because both foils contain gold nuclei of the same size. For a very thin foil, adding layers adds independent opportunities for one close encounter, so the fraction is proportional to thickness. Multiple scattering and attenuation make this relation fail for a thicker foil.	Nuclear size is a property of the nuclide, so extra gold layers do not enlarge a nucleus. When the foil is very thin, the chance of a large deflection is small and each added layer supplies another independent set of nuclei; the single-scattering probability is then proportional to the number of layers and hence to thickness. In a thicker foil, an alpha particle can be deflected more than once, so smaller deflections may combine to exceed $90^\circ$ . Particles may also be stopped or diverted before reaching later layers. These effects prevent a simple proportional relation.	4
04.1	Straight trajectories show that most alpha particles do not pass close to concentrated charge, so the atom is mostly empty space; they do not prove that the force is exactly zero. Rarity establishes that the strongly scattering region is very small, not that it contains most of the mass. The reversal and large momentum change, together with the negligible electron mass, identify a massive scattering centre. A diffuse positive charge cannot provide the intense local repulsion needed for backward scattering.	A straight recorded path may include an undetectably small deflection, so it supports the inference that most atomic volume contains no concentrated matter or charge rather than proving zero force everywhere. The small frequency of backward events measures how rarely an alpha particle approaches the scattering centre closely, so it establishes small size. Mass concentration is instead inferred from the centre's ability to produce a very large alpha-particle momentum change when electrons are far too light to do so. Because the alpha particle is positive, reversal requires a strong, localised repulsive interaction with positive charge. Spreading the same charge over the atomic volume makes the field too weak at any one position to account for the observed reversals.	5
05.1	Foil H produces a greater fraction of large-angle and backward scattering and a smaller fraction of essentially undeflected particles. Its higher- $Z$ nuclei exert a stronger Coulomb repulsion on an alpha particle at the same separation, so encounters that give only a small deflection in L can give a larger deflection in H. Most alpha particles are still essentially undeflected because close approaches to the tiny nuclei are rare.	The alpha particle has charge $+2e$ , while a target nucleus has charge $+Ze$ . Increasing $Z$ strengthens the repulsive electrostatic force at a given separation, so it increases the deflection produced for a given close approach. With the number of target nuclei per unit area fixed, the change is due to nuclear charge rather than foil thickness: more trajectories are scattered through large angles and fewer remain within the undeflected beam. The nucleus still occupies a minute fraction of the atom, so most trajectories do not pass close enough to any nucleus for the stronger force to produce a detectable deflection.	4

3.8.1.2 · Alpha, beta and gamma radiation

Tier 1 · Easy

Mark scheme for 3.8.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Alpha radiation; it is especially hazardous inside the body because it is strongly ionising.	Radiation stopped by paper has the low penetration characteristic of alpha particles. Alpha radiation produces dense ionisation, so an internal source can cause severe local cell damage.	2
02.1	Beta is partly absorbed by the foil, so its detected intensity changes with thickness; gamma is too penetrating for a sensitive change.	The transmitted beta count falls measurably as aluminium thickness increases, providing a feedback signal. Most gamma radiation would pass through a thin foil, so its count would change too little.	2

Tier 2 · Standard

Mark scheme for 3.8.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$164\,\text{s}^{-1}$	First remove the background: $C_1=920-20=900\,\text{s}^{-1}$ . For the source contribution, $C_2=C_1(x_1/x_2)^2=900(0.20/0.50)^2=144\,\text{s}^{-1}$ . The detector also records background, so its expected reading is $144+20=164\,\text{s}^{-1}$ .	3
02.1	The corrected rates are $400\,\text{s}^{-1}$ and $100\,\text{s}^{-1}$ ; doubling the distance reduces the source count rate by a factor of $4$ , as required.	Subtract the background from both readings: $410-10=400\,\text{s}^{-1}$ and $110-10=100\,\text{s}^{-1}$ . The distance ratio is $0.60/0.30=2$ , while the corrected count-rate ratio is $400/100=4=2^2$ , so $I\propto1/x^2$ .	2
03.1	$70\%$ (accept $70.0\%$ )	The background-corrected count with no absorber is $1240-40=1200$ . With aluminium it is $400-40=360$ , which is the gamma contribution. The beta contribution is therefore $1200-360=840$ counts, so the beta percentage is $(840/1200)\times100=70.0\%$ .	3

Tier 3 · Hard

Mark scheme for 3.8.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Measure background and repeated count rates at several source-detector distances, subtract background, and test whether corrected count rate is proportional to $1/x^2$ .	Measure the background count for a long fixed interval with the source removed or well shielded. Place the sealed source and detector at fixed alignment, measure $x$ from the source to the detector, and record counts for the same sufficiently long interval at several distances. Repeat readings to reduce random uncertainty. Convert each result to count rate and subtract the background rate. Plot corrected count rate against $1/x^2$ ; a straight line through the origin within uncertainty supports the inverse-square law. Use tongs, shielding and the shortest practical exposure time throughout.	5
02.1	Gamma can leave the body for external detection and is less ionising than alpha, so a small carefully chosen exposure can provide diagnostic benefit with lower tissue damage.	Gamma radiation is penetrating, allowing photons from the tracer to reach an external detector and form an image. Its relatively weak ionisation reduces energy deposited in tissue for a given passage. Alpha radiation has a very short range in tissue, so it would not reach the detector, and its strong ionisation would cause concentrated damage near the tracer. The gamma exposure still carries an ionisation risk, so the administered activity and exposure time should be minimised. Its use is justified only when the diagnostic information is expected to outweigh that controlled risk.	5
03.1	$a=0.050\,\text{m}$ and the predicted corrected count rate is $162\,\text{s}^{-1}$ . The offset is the distance from a ruler reference point to the source or the detector's sensitive region.	Taking a square root of the count-rate ratio gives $\sqrt{784/400}=1.40=(0.30+a)/(0.20+a)$ . Hence $0.30+a=0.28+1.40a$ , so $a=0.050\,\text{m}$ . Using the first reading, $k=784(0.20+0.050)^2=49.0\,\text{s}^{-1}\text{m}^2$ . At $x=0.50\,\text{m}$ , $d=0.550\,\text{m}$ and $C=49.0/(0.550)^2=162\,\text{s}^{-1}$ to three significant figures. The geometrical distance must be measured between the source and the detector's active region, neither of which need coincide with the ruler's reference points.	5
04.1	$65.6\,\text{s}^{-1}$ . The photons are spread over area $4\pi x^2$ , and the detector intercepts and counts the fraction $S/(4\pi x^2)$ of them.	First subtract background: the source count rate at $0.240\,\text{m}$ is $168-8.0=160\,\text{s}^{-1}$ . The inverse-square prediction at $0.400\,\text{m}$ is $160(0.240/0.400)^2=57.6\,\text{s}^{-1}$ . Adding the background gives a detector reading of $57.6+8.0=65.6\,\text{s}^{-1}$ . At distance $x$ , isotropically emitted photons are spread over a sphere of area $4\pi x^2$ . A small detector intercepts the fraction $S/(4\pi x^2)$ and counts every intercepted photon, so multiplying the source rate $R$ by this fraction gives $C=RS/(4\pi x^2)$ .	4
05.1	Yes. The first result predicts $1330$ total counts at $0.500\,\text{m}$ , compared with the observed $1352$ . The difference of $22$ counts is smaller than the random uncertainty $\sqrt{1352}=36.8$ counts in the second measurement, so the difference is consistent with random counting variation.	The background rate is $240/300=0.800\,\text{s}^{-1}$ , equivalent to $80$ counts in each $100\,\text{s}$ source measurement. The first background-corrected count is therefore $5080-80=5000$ . Doubling the distance should reduce this source count by a factor of four, giving $1250$ source counts and hence $1250+80=1330$ total counts at $0.500\,\text{m}$ . The observed total is $1352$ , only $22$ counts higher. Its random uncertainty is $\sqrt{1352}=36.77$ counts, so the discrepancy is less than one counting uncertainty even before including uncertainty in the first and background measurements. The measurements therefore support inverse-square behaviour; exact equality is not expected because radioactive decay and detection are random.	5

3.8.1.3 · Radioactive decay

Tier 1 · Easy

Mark scheme for 3.8.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.2\times10^{-5}\,\text{s}^{-1}$	Convert the half-life: $T_{1/2}=6.0\times3600=2.16\times10^4\,\text{s}$ . Then $\lambda=\ln2/T_{1/2}=0.693/(2.16\times10^4)=3.21\times10^{-5}\,\text{s}^{-1}$ , which is $3.2\times10^{-5}\,\text{s}^{-1}$ to two significant figures.	2
02.1	The time at which a particular nucleus decays cannot be predicted, but each nucleus has the same fixed probability of decaying per unit time.	Random refers to the unpredictable event time for one nucleus. Constant probability means that an undecayed nucleus does not become more or less likely to decay as it ages.	2

Tier 2 · Standard

Mark scheme for 3.8.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.1\times10^2\,\text{Bq}$	Use $A=A_0e^{-\lambda t}$ . The exponent is $-\lambda t=-(1.8\times10^{-4})(2.40\times10^3)=-0.432$ . Hence $A=480e^{-0.432}=311.6\,\text{Bq}$ , giving $3.1\times10^2\,\text{Bq}$ to two significant figures.	3
02.1	$6.0\,\text{days}$	The activity ratio is $720/90=8=2^3$ , so three half-lives have elapsed. Therefore $3T_{1/2}=18\,\text{days}$ and $T_{1/2}=6.0\,\text{days}$ .	2
03.1	$\lambda=0.225\,\text{h}^{-1}$ and $T_{1/2}=3.08\,\text{h}$ .	For $\ln A=\ln A_0-\lambda t$ , the gradient is $-\lambda$ . The gradient is $(5.40-7.20)/8.00=-0.225\,\text{h}^{-1}$ , so $\lambda=0.225\,\text{h}^{-1}$ . Therefore $T_{1/2}=\ln2/\lambda=0.693/0.225=3.08\,\text{h}$ .	3

Tier 3 · Hard

Mark scheme for 3.8.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.3\times10^7$ nuclei	Convert the half-life: $T_{1/2}=3.0\times3600=1.08\times10^4\,\text{s}$ , so $\lambda=\ln2/T_{1/2}=6.42\times10^{-5}\,\text{s}^{-1}$ . Also $t=5.0\times3600=1.80\times10^4\,\text{s}$ . From $A=A_0e^{-\lambda t}$ , $A_0=860e^{(6.42\times10^{-5})(1.80\times10^4)}=2.73\times10^3\,\text{Bq}$ . Finally $A_0=\lambda N_0$ , so $N_0=A_0/\lambda=(2.73\times10^3)/(6.42\times10^{-5})=4.25\times10^7$ , giving $4.3\times10^7$ nuclei.	5
02.1	$1.04\times10^7\,\text{s}$	The sample contains $N=(2.40\times10^{-6}/60.0)(6.02\times10^{23})=2.408\times10^{16}$ nuclei. From $A=\lambda N$ , $\lambda=(1.60\times10^9)/(2.408\times10^{16})=6.64\times10^{-8}\,\text{s}^{-1}$ . Hence $T_{1/2}=\ln2/\lambda=1.04\times10^7\,\text{s}$ to three significant figures.	4
03.1	Initially P has activity $1000\,\text{Bq}$ and Q has activity $200\,\text{Bq}$ ; their activities are equal after $6.97\,\text{h}$ .	Let the initial activity of P be $P_0$ , so $Q_0=1200-P_0$ . After $6.00\,\text{h}$ , P has passed through three half-lives and Q through one, giving $P_0/8+(1200-P_0)/2=225$ . Solving gives $P_0=1000\,\text{Bq}$ and $Q_0=200\,\text{Bq}$ . Equality requires $1000(2^{-t/2})=200(2^{-t/6})$ , so $5=2^{t/3}$ . Hence $t=3\log_2 5=6.97\,\text{h}$ .	5
04.1	$1.18\times10^{11}$ decays (accept $1.17\times10^{11}$ to $1.19\times10^{11}$ ).	In seconds, $T_{1/2}=9.60(86400)=829440\,\text{s}$ , so $\lambda=\ln2/829440=8.3568092\times10^{-7}\,\text{s}^{-1}$ . The interval is $3.20(86400)=276480\,\text{s}$ . The activity at its end is $A_2=(4.80\times10^5)e^{-(8.3568092\times10^{-7})(276480)}=3.8097625\times10^5\,\text{Bq}$ . Since $A=\lambda N$ , the number that decay is $N_1-N_2=(A_1-A_2)/\lambda=(4.80\times10^5-3.8097625\times10^5)/(8.3568092\times10^{-7})=1.1849469\times10^{11}$ , giving $1.18\times10^{11}$ to three significant figures.	4
05.1	The initial power is $0.427\,\text{W}$ , it falls below $0.120\,\text{W}$ after $32.9\,\text{h}$ , and the energy deposited in the first $54.0\,\text{h}$ is $3.49\times10^4\,\text{J}$ (accept $3.48\times10^4$ to $3.50\times10^4\,\text{J}$ ).	The energy per decay is $E_d=(0.740)(1.602\times10^{-13})=1.18548\times10^{-13}\,\text{J}$ , so $P_0=A_0E_d=(3.60\times10^{12})(1.18548\times10^{-13})=0.4267728\,\text{W}$ . Because power is proportional to activity, $0.120=0.4267728e^{-\lambda t}$ with $\lambda=\ln2/18.0=0.038508176\,\text{h}^{-1}$ . Thus $t=\ln(0.4267728/0.120)/0.038508176=32.9478\,\text{h}$ . For the energy, use $\lambda=\ln2/(18.0\times3600)=1.0696716\times10^{-5}\,\text{s}^{-1}$ . In $54.0\,\text{h}$ , three half-lives pass, so the total number of decays is $(7/8)(A_0/\lambda)=2.945\times10^{17}$ . The deposited energy is therefore $(2.945\times10^{17})(1.18548\times10^{-13})=3.49\times10^4\,\text{J}$ .	6

3.8.1.4 · Nuclear instability

Tier 1 · Easy

Mark scheme for 3.8.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Beta-minus decay; $N$ decreases by $1$ and $Z$ increases by $1$ .	Above the stability band the nucleus has an excess of neutrons. In beta-minus decay a neutron changes into a proton, so $N\to N-1$ and $Z\to Z+1$ while $A$ is unchanged.	2
02.1	$N$ and $Z$ are unchanged, while the nucleus moves to a lower energy state.	Gamma emission is a transition between nuclear energy levels rather than a change of nucleon identity. The photon carries away the energy difference, so neutron and proton numbers stay fixed.	2

Tier 2 · Standard

Mark scheme for 3.8.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Beta-plus decay: $^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+{}^{0}_{+1}\mathrm{e}+\nu_{\mathrm{e}}$ .	The nucleon number stays $18$ while the proton number falls from $9$ to $8$ , so a proton changes into a neutron by beta-plus decay. Conserving nucleon number and charge gives $^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+{}^{0}_{+1}\mathrm{e}+\nu_{\mathrm{e}}$ .	3
02.1	$^{123}_{53}\mathrm{I}+{}^{0}_{-1}\mathrm{e}\rightarrow{}^{123}_{52}\mathrm{Te}+\nu_{\mathrm{e}}$ ; an electron falling into the inner-shell vacancy can emit a characteristic X-ray photon.	Write the captured inner-shell electron on the left-hand side. Electron capture changes a proton into a neutron and emits an electron neutrino, so $A$ remains $123$ while $Z$ falls from $53$ to $52$ , giving tellurium-123. The capture leaves a vacancy in an inner atomic shell; when an electron from a higher shell fills it, the energy difference can be emitted as a characteristic X-ray photon.	3
03.1	$1.84\times10^{19}\,\text{Hz}$	The total energy released is $E_{\mathrm{total}}=0.180\times10^6(1.60\times10^{-19})=2.88\times10^{-14}\,\text{J}$ . The first photon carries $E_1=hf_1=(6.63\times10^{-34})(2.50\times10^{19})=1.6575\times10^{-14}\,\text{J}$ . Hence $E_2=2.88\times10^{-14}-1.6575\times10^{-14}=1.2225\times10^{-14}\,\text{J}$ and $f_2=E_2/h=1.84\times10^{19}\,\text{Hz}$ to three significant figures.	3

Tier 3 · Hard

Mark scheme for 3.8.1.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$^{90}_{38}\mathrm{Sr}\rightarrow{}^{90}_{39}\mathrm{Y}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ , then $^{90}_{39}\mathrm{Y}\rightarrow{}^{90}_{40}\mathrm{Zr}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ ; each step has $N-1$ and $Z+1$ .	For the first decay, conserve $A=90$ and increase the proton number by one: $^{90}_{38}\mathrm{Sr}\rightarrow{}^{90}_{39}\mathrm{Y}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ . Repeating the same change gives $^{90}_{39}\mathrm{Y}\rightarrow{}^{90}_{40}\mathrm{Zr}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ . In each beta-minus event a neutron becomes a proton, so the point moves one unit down in $N$ and one unit right in $Z$ , towards the band of stability; $A=N+Z$ remains $90$ .	5
02.1	$0.33\,\text{MeV}$ , $0.38\,\text{MeV}$ , $0.53\,\text{MeV}$ , $0.86\,\text{MeV}$ , $0.91\,\text{MeV}$ and $1.24\,\text{MeV}$ ; the shortest wavelength is $1.00\times10^{-12}\,\text{m}$ .	Each photon energy is the difference between an upper and a lower level. The six differences are $0.38$ , $0.91$ , $0.53$ , $1.24$ , $0.86$ and $0.33\,\text{MeV}$ . The largest energy gives the shortest wavelength: $E=1.24\times10^6(1.60\times10^{-19})=1.984\times10^{-13}\,\text{J}$ , so $\lambda=hc/E=1.00\times10^{-12}\,\text{m}$ to three significant figures.	4
03.1	Neutrons add strong nuclear attraction without adding electrostatic repulsion. Xenon-120 can undergo $\beta^+$ decay or electron capture, giving $N\rightarrow N+1$ and $Z\rightarrow Z-1$ while $A$ remains $120$ .	The strong nuclear force is attractive and short-ranged, whereas electrostatic repulsion acts between all the protons. As proton number increases, the repulsion becomes more significant. Extra neutrons increase the attractive nuclear interaction without adding electrostatic repulsion, so a stable heavy nucleus needs $N>Z$ . Xenon-120 has too many protons relative to neutrons, so either beta-plus decay or electron capture can change a proton into a neutron. Its point then moves one unit left and one unit up on the $N$ - $Z$ graph: $Z\rightarrow Z-1$ , $N\rightarrow N+1$ and $A$ is unchanged.	4
04.1	P to R is beta-minus decay; P to S is beta-plus decay or electron capture; P to T is alpha decay. The nucleon number is unchanged for either beta process and decreases by $4$ for alpha decay. Gamma emission changes neither $N$ nor $Z$ .	For P to R, the point moves right and down: $Z$ rises by one and $N$ falls by one, so a neutron changes into a proton by beta-minus decay and $A=N+Z$ stays $158$ . For P to S, the point moves left and up: $Z$ falls by one and $N$ rises by one, so a proton changes into a neutron by beta-plus decay or electron capture; again $A=158$ . For P to T, both $Z$ and $N$ fall by two, identifying alpha emission and reducing $A$ from $158$ to $154$ . A gamma photon removes excitation energy only, leaving both coordinates unchanged, so it cannot move the point to R, S or T.	5
05.1	The excited states are $0.491\,\text{MeV}$ and $0.754\,\text{MeV}$ above the ground state. The nucleus either emits one $0.754\,\text{MeV}$ photon directly or emits $0.263\,\text{MeV}$ and then $0.491\,\text{MeV}$ in a cascade: $^{A}_{Z}\mathrm{X}^{*}\rightarrow{}^{A}_{Z}\mathrm{X}^{}+\gamma(0.263\,\text{MeV})$ and $^{A}_{Z}\mathrm{X}^{*}\rightarrow{}^{A}_{Z}\mathrm{X}+\gamma(0.491\,\text{MeV})$ .	Rapid successive detection shows that the $0.263\,\text{MeV}$ and $0.491\,\text{MeV}$ photons form one cascade. Their energies add to $0.754\,\text{MeV}$ , so the upper state is $0.754\,\text{MeV}$ above ground. The second photon ends at the ground state, placing the intermediate state at $0.491\,\text{MeV}$ ; the upper-to-intermediate gap is $0.754-0.491=0.263\,\text{MeV}$ . The isolated $0.754\,\text{MeV}$ photon is the direct upper-to-ground transition. Gamma emission changes neither $A$ nor $Z$ , so both cascade equations retain the same isotope $^{A}_{Z}\mathrm{X}$ .	5

3.8.1.5 · Nuclear radius

Tier 1 · Easy

Mark scheme for 3.8.1.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.15\,\text{fm}$	For aluminium-27, $A=27$ and $27^{1/3}=3$ . Therefore $R=(1.05\,\text{fm})(3)=3.15\,\text{fm}$ .	2
02.1	$10^{-15}\,\text{m}$	Nuclear radii are a few femtometres, and $1\,\text{fm}=10^{-15}\,\text{m}$ .	1

Tier 2 · Standard

Mark scheme for 3.8.1.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$A=64$	Rearrange to $A=(R/r_0)^3$ . Hence $A=(4.20/1.05)^3=4.00^3=64$ .	3
02.1	$2.25\times10^{-14}\,\text{m}$	At closest approach, the initial kinetic energy equals the electric potential energy: $6.40\times10^6(1.60\times10^{-19})=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(50e)}{r}$ . Using $e=1.60\times10^{-19}\,\text{C}$ and $\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}$ gives $r=2.25\times10^{-14}\,\text{m}$ to three significant figures.	3
03.1	$R_0=1.05\times10^{-15}\,\text{m}$ ; $R$ is obtained from the angle of the first diffraction minimum.	Since $216^{1/3}=6$ , $R_0=R/A^{1/3}=(6.30\times10^{-15})/6=1.05\times10^{-15}\,\text{m}$ . In electron diffraction, the measured angle of the first minimum is used to determine the nuclear radius $R$ .	3

Tier 3 · Hard

Mark scheme for 3.8.1.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.44\times10^{17}\,\text{kg m}^{-3}$	The nuclear mass is $m=A(1.67\times10^{-27})\,\text{kg}$ . Its volume is $V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi(r_0A^{1/3})^3=\frac{4}{3}\pi r_0^3A$ . Therefore $\rho=m/V=\frac{3(1.67\times10^{-27})}{4\pi(1.05\times10^{-15})^3}$ . The factor $A$ cancels, giving $\rho=3.44\times10^{17}\,\text{kg m}^{-3}$ , independent of nucleon number.	5
02.1	The graph has a central maximum followed by minima and smaller secondary maxima. The angle of the first minimum is used to estimate $R$ , then $R_0$ is found from $R/A^{1/3}$ .	Electron diffraction produces a strong central maximum, a first minimum and further alternating maxima and minima whose intensity decreases away from the centre. For electrons of known wavelength, the measured angle of the first minimum determines the nuclear radius using a relation between the first-minimum angle and $R$ given in the question. Repeating this for a known $A$ allows $R_0=R/A^{1/3}$ to be estimated.	3
03.1	The common-motion kinetic energy is $0.149\,\text{MeV}$ , the closest separation is $3.09\times10^{-14}\,\text{m}$ , and this is $2.03\%$ greater than the fixed-nucleus value.	If the alpha particle initially has speed $u$ , momentum conservation gives the common speed $v=4u/(4+197)$ . The common-motion kinetic energy is therefore $\frac12(201\,\text{u})v^2=(4/201)(\frac12(4\,\text{u})u^2)=(4/201)(7.50)=0.149\,\text{MeV}$ . The maximum electric potential energy is $7.50-0.149=7.35\,\text{MeV}$ . Using $7.35\times10^6e=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(79e)}{r}$ gives $r=3.092\times10^{-14}\,\text{m}$ . The fixed-nucleus calculation gives $3.031\times10^{-14}\,\text{m}$ , so using unrounded values the increase is $2.03\%$ .	5
04.1	The shell contains $50.0\%$ of the $A=27$ nuclear volume and $27.9\%$ of the $A=216$ nuclear volume (accept $49.8$ – $50.2\%$ and $27.7$ – $28.1\%$ ). The fixed shell thickness is a larger fraction of the smaller radius; constant density concerns mass per total volume, not the proportion lying near the surface.	The radii are $R_{27}=1.05(27^{1/3})=3.15\,\text{fm}$ and $R_{216}=1.05(216^{1/3})=6.30\,\text{fm}$ . For shell thickness $d$ , the shell fraction is $f=[R^3-(R-d)^3]/R^3=1-(1-d/R)^3$ . Hence $f_{27}=1-(1-0.650/3.15)^3=0.500$ , or $50.0\%$ , and $f_{216}=1-(1-0.650/6.30)^3=0.279$ , or $27.9\%$ . Although $A/R^3$ is constant, the surface-area-to-volume ratio decreases as radius increases, so a shell of fixed thickness occupies a smaller fraction of the larger nucleus.	5
05.1	$n=0.334$ and $\rho_{169}/\rho_{50}=0.997$ ; both results support $R\propto A^{1/3}$ and approximately constant nuclear density.	Taking ratios gives $6.46/4.30=(169/50)^n$ , so $n=\ln(6.46/4.30)/\ln(169/50)=0.3342002$ , or $0.334$ . Nuclear density is proportional to $A/R^3$ , hence $\rho_{169}/\rho_{50}=(169/50)/(6.46/4.30)^3=0.9968378$ , or $0.997$ . The exponent agrees with $1/3$ to the precision of the measurements and the densities differ by only about $0.3\%$ , supporting the constant-density model. Measured values of the nuclear-radius constant lie between about $1.05$ and $1.4\,\text{fm}$ .	5

3.8.1.6 · Mass and energy

Tier 1 · Easy

Mark scheme for 3.8.1.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.98\,\text{MeV}$	Convert the mass defect directly: $E=(0.00320\,\text{u})(931.5\,\text{MeV per u})=2.9808\,\text{MeV}$ . To three significant figures, $E=2.98\,\text{MeV}$ .	2
02.1	$5.0\times10^{-11}\,\text{kg}$	Use $E=mc^2$ , so $m=E/c^2=(4.5\times10^6)/(3.00\times10^8)^2=5.0\times10^{-11}\,\text{kg}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.8.1.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$173\,\text{MeV}$	The initial mass is $235.0439+1.0087=236.0526\,\text{u}$ . The final mass is $140.9144+91.9262+3(1.0087)=235.8667\,\text{u}$ . The mass defect is $236.0526-235.8667=0.1859\,\text{u}$ . Therefore $E=(0.1859)(931.5)=173.2\,\text{MeV}$ , which is $173\,\text{MeV}$ to the precision of the mass data.	4
02.1	$7.08\,\text{MeV per nucleon}$	The mass of two free protons and two free neutrons is $2(1.00728)+2(1.00867)=4.03190\,\text{u}$ . The mass defect is $4.03190-4.00151=0.03039\,\text{u}$ , so the total binding energy is $(0.03039)(931.5)=28.31\,\text{MeV}$ . Dividing by four gives $7.08\,\text{MeV per nucleon}$ to three significant figures.	4
03.1	$11.997\,\text{u}$ (accept $12.0\,\text{u}$ to three significant figures)	The total binding energy is $12(7.68)=92.16\,\text{MeV}$ , corresponding to a mass defect of $92.16/931.5=0.09894\,\text{u}$ . Six separate protons and six separate neutrons have mass $6(1.00728)+6(1.00867)=12.09570\,\text{u}$ , so the nuclear mass is $12.09570-0.09894=11.99676\,\text{u}$ , which is $11.997\,\text{u}$ to five significant figures. The mass of the nucleus is less than the mass of its separated nucleons by the mass defect.	4

Tier 3 · Hard

Mark scheme for 3.8.1.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$17.6\,\text{MeV}=2.82\times10^{-12}\,\text{J}$ ; the helium product is more tightly bound per nucleon.	The initial mass is $2.01410+3.01605=5.03015\,\text{u}$ . The final mass is $4.00260+1.00867=5.01127\,\text{u}$ . Hence $\Delta m=5.03015-5.01127=0.01888\,\text{u}$ . Using the atomic-mass conversion, $E=(0.01888)(931.5)=17.5867\,\text{MeV}=17.6\,\text{MeV}$ . In joules, $E=(17.5867)(1.602\times10^{-13})=2.82\times10^{-12}\,\text{J}$ . The products have a greater average binding energy per nucleon, so their lower total rest mass corresponds to the released energy.	6
02.1	$\Delta m=0.0200\,\text{u}$ ; $E=18.63\,\text{MeV}=2.9808\times10^{-12}\,\text{J}$ ; mass consumed $=8.30\times10^{-27}\,\text{kg}$ ; specific energy $=3.59\times10^{14}\,\text{J kg}^{-1}$ ; comparison factor $=8.55\times10^6$ (accept $8.55\times10^6$ to $8.57\times10^6$ from rounded intermediates)	The mass defect is $\Delta m=5.0000-4.9800=0.0200\,\text{u}$ . The energy released per event is $E=(0.0200)(931.5)=18.63\,\text{MeV}$ . Converting this unrounded value to joules gives $E=(18.63)(1.60\times10^{-13})=2.9808\times10^{-12}\,\text{J}$ , or $2.98\times10^{-12}\,\text{J}$ to three significant figures. The reactant mass consumed per event is $m=(5.0000)(1.66\times10^{-27})=8.30\times10^{-27}\,\text{kg}$ . Therefore the specific energy, using the unrounded energy, is $E/m=(2.9808\times10^{-12})/(8.30\times10^{-27})=3.5913253\times10^{14}\,\text{J kg}^{-1}$ , or $3.59\times10^{14}\,\text{J kg}^{-1}$ . Finally, the unrounded comparison is $(3.5913253\times10^{14})/(4.20\times10^7)=8.5507745\times10^6$ , giving $8.55\times10^6$ ; accept values from $8.55\times10^6$ to $8.57\times10^6$ when rounded intermediates are carried forward.	6
03.1	$3.78\times10^{21}\,\text{Hz}$	The atomic electrons cancel because both atoms are oxygen. The separated products have mass $15.003066+1.008665=16.011731\,\text{u}$ , so the required mass increase is $16.011731-15.994915=0.016816\,\text{u}$ . The threshold energy, with recoil neglected, is $(0.016816)(931.5)=15.6641\,\text{MeV}=15.6641(1.602\times10^{-13})=2.509\times10^{-12}\,\text{J}$ . Therefore $f=E/h=(2.509\times10^{-12})/(6.63\times10^{-34})=3.78\times10^{21}\,\text{Hz}$ to three significant figures.	5
04.1	$6.620\,\text{MeV}$ is released; the alpha particle receives $6.501\,\text{MeV}$ and Y receives $0.1194\,\text{MeV}$ (accept values differing by the final displayed rounding).	Atomic electrons balance because the daughter atom and helium atom together have the parent's electron number. The mass defect is $\Delta m=222.014620-218.004910-4.002603=0.007107\,\text{u}$ , so $Q=(0.007107)(931.5)=6.6201705\,\text{MeV}$ . Momentum conservation gives the products equal momentum magnitudes. Since $K=p^2/(2m)$ , $K_\alpha/K_Y=m_Y/m_\alpha$ . Therefore $K_\alpha=Qm_Y/(m_Y+m_\alpha)=6.6201705(218.004910)/(218.004910+4.002603)=6.5008146\,\text{MeV}$ . The daughter receives $K_Y=Q-K_\alpha=0.1193559\,\text{MeV}$ . Their sum is the unrounded released energy.	6
05.1	The fission rate is $9.89\times10^{19}\,\text{s}^{-1}$ and the rest-mass decrease in $24.0\,\text{h}$ is $3.07\times10^{-3}\,\text{kg}$ .	The energy transferred per fission is $(202)(1.602\times10^{-13})=3.23604\times10^{-11}\,\text{J}$ . Therefore the fission rate is $(3.20\times10^9)/(3.23604\times10^{-11})=9.88863\times10^{19}\,\text{s}^{-1}$ , or $9.89\times10^{19}\,\text{s}^{-1}$ . In $24.0\,\text{h}$ the transferred energy is $E=Pt=(3.20\times10^9)(24.0\times3600)=2.7648\times10^{14}\,\text{J}$ . From $E=\Delta mc^2$ , $\Delta m=(2.7648\times10^{14})/(3.00\times10^8)^2=3.072\times10^{-3}\,\text{kg}$ , giving $3.07\times10^{-3}\,\text{kg}$ to three significant figures.	5

3.8.1.7 · Induced fission

Tier 1 · Easy

Mark scheme for 3.8.1.7 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The moderator slows neutrons; the control rods absorb neutrons to regulate the chain reaction.	The moderator reduces neutron kinetic energy so that thermal neutrons are more likely to induce fission. Control rods remove neutrons from the chain reaction by absorption, controlling the rate of fission.	2
02.1	It is the minimum mass, for a particular arrangement, that can sustain a chain reaction.	Below the critical mass too many neutrons escape without causing another fission. At the critical mass, one neutron per fission causes a further fission on average.	2

Tier 2 · Standard

Mark scheme for 3.8.1.7 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$k=1.0$ ; the reactor is critical.	The mean number of next-generation fissions per current fission is $k=(2.5)(0.40)=1.0$ . One fission therefore replaces itself with one further fission on average, so the chain reaction is critical and steady.	3
02.1	The sphere has a smaller surface-area-to-volume ratio, so fewer fission neutrons escape and more can induce further fissions.	For a fixed volume, a sphere has the least surface area. A smaller fraction of its fissile nuclei lie close to a surface, reducing neutron leakage. More released neutrons therefore remain available to be absorbed by fissile nuclei, making a self-sustaining chain reaction more likely than in the thin sheet.	3
03.1	$6.25\times10^{12}$ fissions; insert the control rods further so that more neutrons are absorbed and each fission causes one further fission on average.	The number in the later generation is $(4.00\times10^{12})(1.015)^{30}=6.252\times10^{12}$ , or $6.25\times10^{12}$ to three significant figures. Inserting the control rods further increases neutron absorption. This reduces the mean number of further fissions caused by each fission to one, giving steady power.	3

Tier 3 · Hard

Mark scheme for 3.8.1.7 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Fission neutrons are slowed by a low-absorption moderator, excess neutrons are absorbed by adjustable control rods, and a stable coolant removes thermal energy from the core.	Absorption of a thermal neutron makes a fissile nucleus split and emit further neutrons, allowing a chain reaction. A moderator with light nuclei slows fast neutrons efficiently by elastic collisions and should have low neutron absorption. Control rods made from strong neutron absorbers are moved to keep the mean number of further fissions near one. A coolant with good heat-transfer properties carries energy from the core while remaining stable and not absorbing too many neutrons. Together these components maintain a critical, controlled reaction and remove the released energy.	5
02.1	$55$ collisions	The initial energy is $2.0\times10^6\,\text{eV}$ . After $n$ collisions, $E=(2.0\times10^6)(121/169)^n$ . Solving $(2.0\times10^6)(121/169)^n<0.025$ gives $n>54.47$ . Since the number of collisions is an integer, at least $55$ are required; after $54$ the model gives $0.0292\,\text{eV}$ , while after $55$ it gives $0.0209\,\text{eV}$ .	5
03.1	The retained fraction is $((A-1)/(A+1))^2$ ; the neutron loses $100\%$ with hydrogen-1 and $88.9\%$ with deuterium. Neutron absorption probability is also important (accept chemical stability or suitability at reactor temperature).	Let the neutron and moderator speeds after the collision be $v$ and $V$ . Momentum conservation gives $u=v+AV$ , while kinetic-energy conservation gives $u^2=v^2+AV^2$ . Substituting $v=u-AV$ into the energy equation gives $V=2u/(A+1)$ and hence $v=(1-A)u/(A+1)$ . Therefore the neutron retains the fraction $v^2/u^2=((A-1)/(A+1))^2$ . For hydrogen-1, $A=1$ , so the retained fraction is zero and the neutron loses $100\%$ of its kinetic energy in this head-on model. For deuterium, $A=2$ , so the loss is $1-1/9=8/9=88.9\%$ . A practical moderator must also have low neutron absorption; the collision model alone does not test this.	5
04.1	Fission neutrons are fast, whereas a slow thermal neutron is more likely to be absorbed by uranium-235 and induce fission. A neutron that is not slowed may escape from the reactor core or be absorbed by uranium-238 without inducing uranium-235 fission.	Neutrons released in fission have high kinetic energy. Collisions in a moderator reduce this energy, increasing the probability that uranium-235 absorbs a neutron and undergoes further fission. Without sufficient moderation, a neutron can leave the core before interacting or be captured by uranium-238 instead of sustaining the uranium-235 chain reaction.	3
05.1	Coolant L is unsuitable at the stated pressure because its boiling point is below the outlet temperature and it would absorb neutrons that are needed to cause further fissions. M is defensible because it is stable and absorbs few neutrons, although compression and high flow increase pumping demand. N is also defensible because it remains liquid and transfers heat effectively with little neutron loss, provided sealed circuits and secondary heat exchange control its chemical hazard. Credit either M or N when the recommendation follows a balanced comparison.	A coolant must remove thermal energy without disrupting neutron economy or becoming unsafe at the operating temperature. L's high specific heat capacity is useful, but at $620\,\text{K}$ it would boil under the stated pressure; vapour formation would impair predictable cooling, and moderate neutron capture removes neutrons needed for fission. M remains chemically stable and has low absorption, but its low thermal capacity as a gas demands a large mass flow and greater pumping power. N combines low absorption with strong heat transfer and remains liquid, but contact with air or water can create a serious secondary hazard. M can be chosen where chemical simplicity is prioritised; N can be chosen where sealed primary and secondary circuits, leak detection and an isolated heat exchanger adequately control reactivity. A conclusion must link the recommendation to heat removal, neutron absorption and operational safety.	6

3.8.1.8 · Safety aspects

Tier 1 · Easy

Mark scheme for 3.8.1.8 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Use remote handling and place dense shielding around the fuel.	Remote handling increases the worker's distance from the source and avoids direct contact. Dense shielding absorbs radiation before it reaches the worker.	2
02.1	It must prevent radioactive material escaping and resist corrosion or physical damage for a long time.	Secure containment prevents environmental contamination. Long-term chemical and mechanical stability reduces the chance that the barrier fails during storage.	2

Tier 2 · Standard

Mark scheme for 3.8.1.8 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The rods stop the fission chain reaction, but radioactive fission products continue to decay and release heat, so coolant must prevent overheating.	Inserted control rods absorb neutrons, so further induced fissions rapidly cease. The existing fission products remain radioactive and their decay transfers energy to the core. Coolant circulation must therefore continue to remove this decay heat and prevent fuel or containment damage.	3
02.1	Glass limits dispersal, robust containers provide containment, and stable deep rock isolates the waste from people and the surface environment.	Vitrification locks radioactive material into a solid that is difficult to disperse or dissolve. Sealed corrosion-resistant containers add another containment barrier. Stable, low-permeability geology provides shielding, limits groundwater flow and delays radionuclide migration to the surface environment.	3
03.1	The shield reduces external irradiation, but damaged containment may allow radioactive material to escape and cause contamination or internal irradiation. Isolate the area and place the flask in secure secondary containment using remote equipment.	The shield attenuates radiation travelling from the spent fuel towards workers, reducing external exposure. It does not seal a damaged flask, so released radioactive material could contaminate the surroundings or enter the body. Access should be prevented while trained staff use remote equipment to place the flask in secure secondary containment.	3

Tier 3 · Hard

Mark scheme for 3.8.1.8 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$7.5\,\text{m}$ and $53\,\text{s}^{-1}$ ; shielding, exposure-time control and contamination controls are still required.	Remove background at $1.5\,\text{m}$ : $C_1=365-40=325\,\text{s}^{-1}$ . With $C\propto1/x^2$ , $C_1x_1^2=C_2x_2^2$ , so $x_2=1.5\sqrt{325/13}=1.5\sqrt{25}=7.5\,\text{m}$ . The detector reading includes background: $13+40=53\,\text{s}^{-1}$ . Distance reduces external dose but does not eliminate it and does not prevent contamination, so shielding, short exposure times, remote tools and controlled containment are also needed.	5
02.1	Nuclear generation offers reliable electricity with low operational greenhouse-gas emissions, but accident consequences and long-lived waste require effective engineered controls and long-term management.	A nuclear station can deliver large, continuous power output while producing very little carbon dioxide during operation, reducing the climate and air-pollution impacts of fossil-fuel generation. Its small fuel mass also reduces routine fuel transport. However, loss of cooling or containment could release radioactive material, so redundant shutdown, cooling, shielding and emergency systems are necessary. High-level waste remains hazardous and needs secure remote handling and storage for long periods. A justified decision weighs the reduced routine emissions and dependable supply against the probability and severity of accidents, waste obligations, cost and the suitability of the site.	5
03.1	Fresh fuel has high activity and decay-heat output, so circulating water is needed initially for cooling and shielding. The dry store cannot accept the fuel until passive heat removal can keep its temperature within safe limits.	The fission chain reaction has stopped, but freshly removed fuel contains highly active fission products whose decay releases substantial thermal energy. Circulating pond water carries this energy away and its depth provides radiation shielding. During pond storage, short-lived products decay, so the heat output falls. A sealed dry store relies on limited or passive heat transfer; accepting the fuel too early could make its temperature rise enough to damage the fuel or containment. Transfer is safe only after the decay-heat output has fallen below the store's heat-removal capacity.	5
04.1	Remote handling keeps workers farther from the highly radioactive spent fuel, while shielding absorbs radiation and reduces external dose. Radioactive fission products continue to release decay heat after removal, so cooling prevents overheating and damage to the fuel or containment. Shielded, secure storage contains radioactive material and isolates it while activity and heat output fall. Transfer and long-term storage retain risks, but once the fuel is cool enough, moving it into a suitable passive store can reduce reliance on active cooling; the benefit outweighs the controlled transfer risk only if cooling, shielding and containment remain effective.	Freshly removed spent fuel contains intensely radioactive fission products. Remote tools increase the distance between workers and the fuel, reducing dose, and thick shielding attenuates the radiation emitted during lifting and transport. Although induced fission has stopped, radioactive decay continues to transfer energy, so pond water or another cooling system must remove decay heat and prevent overheating, fuel damage and loss of containment. Storage must provide shielding, physical security and durable containment against dispersal or contamination; early pond storage also allows short-lived products to decay before transfer to a dry store with passive heat removal. Removal and transfer introduce exposure, handling and containment risks, and long-lived activity requires continuing management. However, after sufficient cooling, secure passive storage isolates the fuel and reduces dependence on powered cooling. The balance is favourable only while monitoring, shielding, cooling capacity and containment keep the residual risks acceptably low.	6
05.1	After $12.0\,\text{years}$ the combined activity is $2.817\times10^{12}\,\text{Bq}$ , which is about $2.8\times10^{12}\,\text{Bq}$ . This still represents ionising radiation, while long-lived Y remains hazardous. Suitable shielding controls external exposure, and secure containment prevents dispersal, contamination and internal irradiation.	At $12.0\,\text{years}$ , $A_X=(6.40\times10^{13})2^{-12.0/2.40}=(6.40\times10^{13})2^{-5}=2.0000\times10^{12}\,\text{Bq}$ . Also $A_Y=(1.10\times10^{12})2^{-12.0/28.0}=8.1729686\times10^{11}\,\text{Bq}$ . Therefore the combined activity is $A=2.8172969\times10^{12}\,\text{Bq}$ , or about $2.8\times10^{12}\,\text{Bq}$ . The activity is not zero, and activity alone does not specify radiation type or photon energy. A radiation survey and isotope data determine suitable shielding. Durable sealed containment is still required to prevent radionuclides escaping, entering the body or contaminating the environment during the long period for which Y remains active.	6