3.8 Nuclear physics (A-level only) — printable coverage pack

3.8.1.1 · Rutherford scattering

In Rutherford scattering, most alpha particles pass through a thin metal foil with little or no deflection, showing that atoms are mostly empty space.
A small proportion are deflected through large angles because the positive alpha particles experience electrostatic repulsion near a concentrated positive nucleus.
The very rare backward deflections show that nearly all the mass and positive charge occupy a region much smaller than the atom.
Exam answers must link each observation to an inference. A common error is merely to list the observations without explaining how they contradict a diffuse-charge model.

Tier 1 · Easy

1. State what is inferred from the observation that most alpha particles cross a very thin gold foil without changing direction.[1 mark]
Answer
- The atom is mostly empty space.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- The atom contains a very small, positively charged nucleus in which most of its mass is concentrated.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- The observations imply an atom that is mostly empty space with its positive charge and nearly all its mass concentrated in a tiny nucleus.
Method: 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.

3.8.1.2 · Alpha, beta and gamma radiation

Alpha radiation is strongly ionising and has a short range; beta radiation is moderately ionising and is absorbed by a few millimetres of aluminium; gamma radiation is weakly ionising and needs thick, dense shielding.
For a point gamma source, intensity or background-corrected count rate follows $I=k/x^2$ when distance $x$ is measured from the source.
Subtract the background count rate before testing an inverse-square relationship, then add it back only if a predicted detector reading is required.
Absorption measurements can identify radiation and enable thickness monitoring: beta can monitor thin aluminium or paper, whereas gamma is used for thicker steel.
A common error is to apply the inverse-square law to a raw count rate that still includes background radiation.

Tier 1 · Easy

1. 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]
Answer
- Alpha radiation; it is especially hazardous inside the body because it is strongly ionising.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $164\,\text{s}^{-1}$
Method: 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 $144+20=164\,\text{s}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- Measure background and repeated count rates at several source-detector distances, subtract background, and test whether corrected count rate is proportional to $1/x^2$ .
Method: 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.

3.8.1.3 · Radioactive decay

Radioactive decay is random for an individual nucleus, but each nucleus of a given isotope has the same constant decay probability per unit time, represented by $\lambda$ .
The number of undecayed nuclei follows $N=N_0e^{-\lambda t}$ and activity follows $A=\lambda N=A_0e^{-\lambda t}$ ; activity is measured in becquerels, where $1\,\text{Bq}=1\,\text{s}^{-1}$ .
Half-life and decay constant are related by $T_{1/2}=\ln 2/\lambda$ ; time and $\lambda$ must use reciprocal units.
On a graph of $\ln A$ or $\ln N$ against $t$ , the gradient is $-\lambda$ . A common error is to treat a decay curve as linear or to omit the minus sign.
When activity is used to find a number of nuclei, rearrange $A=\lambda N$ only after converting the half-life into seconds if activity is in Bq.

Tier 1 · Easy

1. An isotope has a half-life of $6.0\,\text{h}$ . Calculate its decay constant in $\text{s}^{-1}$ .[2 marks]
Answer
- $3.2\times10^{-5}\,\text{s}^{-1}$
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $3.1\times10^2\,\text{Bq}$
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $4.3\times10^7$ nuclei
Method: 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.

3.8.1.4 · Nuclear instability

The $N$ - $Z$ graph has a band of stable nuclei: light stable nuclei have approximately $N=Z$ , while heavier stable nuclei require $N>Z$ .
A neutron-rich nucleus tends to undergo beta-minus decay, changing a neutron into a proton so that $N$ decreases by $1$ and $Z$ increases by $1$ .
A proton-rich nucleus can undergo beta-plus decay or electron capture; in either case $Z$ decreases by $1$ and $N$ increases by $1$ .
Alpha decay changes $A$ by $-4$ and $Z$ by $-2$ . Gamma emission changes neither $A$ nor $Z$ because it de-excites the nucleus.
In nuclear equations, conserve nucleon number and charge separately. A common error is to change $A$ during beta decay.

Tier 1 · Easy

1. 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]
Answer
- Beta-minus decay; $N$ decreases by $1$ and $Z$ increases by $1$ .
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- Beta-plus decay: $^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+{}^{0}_{+1}\mathrm{e}+\nu_{\mathrm{e}}$ .
Method: 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 $^{18}_{9}\mathrm{F}\rightarrow{}^{18}_{8}\mathrm{O}+{}^{0}_{+1}\mathrm{e}+\nu_{\mathrm{e}}$ .

Tier 3 · Hard

1. 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]
Answer
- $^{90}_{38}\mathrm{Sr}\rightarrow{}^{90}_{39}\mathrm{Y}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ , then $^{90}_{39}\mathrm{Y}\rightarrow{}^{90}_{40}\mathrm{Zr}+{}^{0}_{-1}\mathrm{e}+\bar{\nu}_{\mathrm{e}}$ ; each step has $N-1$ and $Z+1$ .
Method: 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 $90$ .

3.8.1.5 · Nuclear radius

Nuclear radii are typically of order $10^{-15}\,\text{m}$ and follow $R=r_0A^{1/3}$ , where the value of $r_0$ must be taken from the question or experimental data.
The closest approach of an alpha particle can estimate an upper limit for nuclear radius by equating its kinetic energy to electrostatic potential energy.
Electron diffraction gives nuclear size from the angular positions of intensity minima; electrons are suitable because their de Broglie wavelength can be comparable with a nuclear diameter.
Because $R^3\propto A$ , nuclear volume is proportional to nucleon number, providing evidence that nuclear matter has approximately constant density.
Convert femtometres using $1\,\text{fm}=10^{-15}\,\text{m}$ . A common error is to use $A$ rather than $A^{1/3}$ in the radius equation.

Tier 1 · Easy

1. Use $R=r_0A^{1/3}$ with $r_0=1.05\,\text{fm}$ to calculate the radius of an aluminium-27 nucleus.[2 marks]
Answer
- $3.15\,\text{fm}$
Method: For aluminium-27, $A=27$ and $27^{1/3}=3$ . Therefore $R=(1.05\,\text{fm})(3)=3.15\,\text{fm}$ .

Tier 2 · Standard

1. 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]
Answer
- $A=64$
Method: Rearrange to $A=(R/r_0)^3$ . Hence $A=(4.20/1.05)^3=4.00^3=64$ .

Tier 3 · Hard

1. 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]
Answer
- $3.44\times10^{17}\,\text{kg m}^{-3}$
Method: 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.

3.8.1.6 · Mass and energy

Any energy change has an equivalent mass change through $E=mc^2$ ; a decrease in total rest mass appears as released energy.
For nuclear mass differences quoted in atomic mass units, use $1\,\text{u}=931.5\,\text{MeV}$ and show the reactant mass, product mass and mass defect before conversion.
Binding energy is the energy required to separate a nucleus into free nucleons; a larger average binding energy per nucleon means a more tightly bound nucleus.
Fusion of light nuclei and fission of heavy nuclei release energy because the products have a higher average binding energy per nucleon.
Use atomic masses consistently so that electron masses cancel where appropriate. A common error is to multiply a mass defect in kilograms by $931.5$ instead of using $E=mc^2$ .

Tier 1 · Easy

1. 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]
Answer
- $2.98\,\text{MeV}$
Method: Convert the mass defect directly: $E=(0.00320\,\text{u})(931.5\,\text{MeV per u})=2.9808\,\text{MeV}$ . To three significant figures, $E=2.98\,\text{MeV}$ .

Tier 2 · Standard

1. 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]
Answer
- $173\,\text{MeV}$
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $17.6\,\text{MeV}=2.82\times10^{-12}\,\text{J}$ ; the helium product is more tightly bound per nucleon.
Method: 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.

3.8.1.7 · Induced fission

A slow, thermal neutron can be absorbed by a fissile nucleus, making it unstable so that it splits into two smaller nuclei, releases energy and emits further neutrons.
A chain reaction is critical when, on average, one neutron from each fission induces another fission; below this it dies away and above this it grows.
A moderator slows neutrons by elastic collisions and should contain light nuclei while absorbing few neutrons; water and graphite are examples.
Control rods absorb neutrons and are inserted or withdrawn to regulate the reaction; boron and cadmium are suitable because of their high neutron absorption.
A coolant transfers thermal energy from the core and should have suitable thermal properties and chemical stability. A common error is to say the moderator absorbs neutrons rather than slows them.

Tier 1 · Easy

1. State the function of the moderator and the function of the control rods in a thermal nuclear reactor.[2 marks]
Answer
- The moderator slows neutrons; the control rods absorb neutrons to regulate the chain reaction.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $k=1.0$ ; the reactor is critical.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.8.1.8 · Safety aspects

Reactor fuel and radioactive waste are handled remotely to increase distance and reduce the time for which workers are exposed; dense shielding absorbs ionising radiation.
Emergency shutdown inserts neutron-absorbing control rods rapidly, but radioactive decay continues to produce heat, so cooling must continue after fission stops.
Waste is classified and contained according to its activity and half-life; high-activity material requires remote handling, shielding and secure long-term storage.
Inverse-square reasoning can reduce exposure from a compact source, but it does not replace shielding, contamination control or controlled access.
Risk-benefit discussions should compare specific hazards with benefits such as reliable low-carbon electricity. A common error is to state that shutdown immediately removes all heat production.

Tier 1 · Easy

1. State two ways in which worker exposure is reduced while spent reactor fuel is moved.[2 marks]
Answer
- Use remote handling and place dense shielding around the fuel.
Method: Remote handling increases the worker's distance from the source and avoids direct contact. Dense shielding absorbs radiation before it reaches the worker.

Tier 2 · Standard

1. Explain why a reactor still needs coolant circulation immediately after an emergency shutdown has fully inserted the control rods.[3 marks]
Answer
- The rods stop the fission chain reaction, but radioactive fission products continue to decay and release heat, so coolant must prevent overheating.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $7.5\,\text{m}$ and $53\,\text{s}^{-1}$ ; shielding, exposure-time control and contamination controls are still required.
Method: 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.