4.4 Atomic structure — revision question pack

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

Checked against AQA 8463 section 4.4. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

How this checking works

(physics only) means this content belongs to AQA GCSE Physics (8463), the separate-science qualification, but not AQA Combined Science: Trilogy (8464). It is not an exam tier: (HT only) separately marks Higher-tier content.

Loading your tier…

Answer all questions in the spaces provided.

4.4.1.1 · The structure of an atom

Explanation

An atom has a tiny, positively charged nucleus containing protons and neutrons, with negatively charged electrons arranged at different energy levels around it.
Compare scales using a ratio: an atom has radius about $1\times10^{-10}\,\mathrm{m}$ , while its nucleus has less than $1/10\,000$ of the atom's radius.
Most atomic mass is concentrated in the nucleus; absorbing electromagnetic radiation can move an electron to a higher energy level, while emitting it can move the electron lower.
A common error is to draw the nucleus as most of the atom: it contains most of the mass but occupies only a very small central region.

A tiny central nucleus with electrons occupying energy levels around it.

Worked example

An atom has radius $1.0\times10^{-10}\,\mathrm{m}$ and its nucleus has radius $8.0\times10^{-15}\,\mathrm{m}$ . Calculate how many times larger the atom's radius is.

1.Calculate $(1.0\times10^{-10})/(8.0\times10^{-15})=0.125\times10^5=1.25\times10^4$ . The atom's radius is therefore $12\,500$ times the nucleus's radius.

Answer: Use atom radius divided by nucleus radius. $1.25\times10^4$ times

Common mistakes

Don't draw the nucleus as most of the atom: it contains most of the mass but occupies only a very small central region.
Don't fall into the trap of drawing electrons inside the nucleus rather than in shells around it.

Exam tip

For atomic structure, give particle charge, relative mass and location precisely.

Tier 1 · Easy

Explain why an atom is described as mostly empty space even though nearly all its mass is concentrated at its centre.
[2 marks]
Total for this question: 2
A model labels a tiny central region as containing almost all the atom's mass, while electrons occupy the much larger surrounding region. State which label describes the nucleus and give one reason the model should show mostly empty space.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe the positions and charges of the three subatomic particles in an atom, and state where nearly all the atom's mass is found.
[4 marks]
Total for this question: 4
An electron begins on energy level 2. It absorbs electromagnetic radiation and moves to level 4, then emits less energy than it absorbed. Decide whether its final energy level can be level 1, 2 or 3, and explain your choice.
[3 marks]
Total for this question: 3
A table describes particle $P$ as having charge $+1$ , relative mass $1$ and a position in the nucleus; $Q$ has charge $0$ , relative mass $1$ and is also in the nucleus; $R$ has charge $-1$ , very small relative mass and occupies energy levels. Identify $P$ , $Q$ and $R$ , then explain why losing one $R$ changes an atom's charge but hardly changes its mass.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An atom absorbs electromagnetic radiation and later emits electromagnetic radiation. Explain what can happen to one of its electrons in the two changes.
[2 marks]
Total for this question: 2
A display model gives an atom a radius of $84\,\mathrm{mm}$ and its nucleus a radius of $0.42\,\mathrm{mm}$ . In a real atom the nucleus has less than $1/10\,000$ of the atom's radius. Calculate the upper limit for a scale-model nucleus radius, then evaluate the display model.
[5 marks]
Total for this question: 5
An atom contains $7$ protons, $8$ neutrons and $7$ electrons. Take the relative masses of a proton and neutron as $1$ and the electron mass as negligible. Calculate the fraction of the atom's listed particles that are in the nucleus, estimate the fraction of its mass in the nucleus, and explain why these fractions do not show that the nucleus occupies most of the atom's volume.
[5 marks]
Total for this question: 5
A row contains $2.5\times10^{6}$ identical atoms, each with radius $1.2\times10^{-10}\,\mathrm{m}$ . Assume neighbouring atoms touch. Calculate the length of the row. The radius of each nucleus is less than $1/10\,000$ of its atom radius. Calculate the upper limit for the combined length of all the nuclear diameters in the row.
[4 marks]
Total for this question: 4
An atom has electron energy levels at $0$ , $3$ , $7$ and $12$ energy units above its lowest level. An electron starts at the second level, absorbs radiation carrying $9$ energy units and then emits radiation carrying $5$ energy units. Determine its level after each transfer and explain whether the electron has gained or lost energy overall.
[5 marks]
Total for this question: 5

4.4.1.2 · Mass number, atomic number and isotopes

Explanation

Atomic number is the number of protons; mass number is the total number of protons and neutrons, so neutron number is mass number minus atomic number.
For a neutral atom, electron number equals proton number; for a positive ion, subtract the positive charge from the proton number to find its electrons.
Isotopes are atoms of the same element with the same proton number but different neutron numbers, so their atomic numbers match but their mass numbers differ.
A common error is to change the nucleus when an ion forms: losing outer electrons changes the charge, not the atomic number or mass number.

Worked example

A neutral atom contains $17$ protons and $20$ neutrons. State its atomic number, mass number and number of electrons.

1.The atomic number equals the proton number, so it is $17$ . Add protons and neutrons for the mass number: $17+20=37$ . A neutral atom has equal proton and electron numbers, so it has $17$ electrons.

Answer: Atomic number $=17$ Mass number $=37$ Number of electrons $=17$

Common mistakes

Don't change the nucleus when an ion forms: losing outer electrons changes the charge, not the atomic number or mass number.
Don't fall into the trap of calculating neutron number by adding atomic number and mass number.

Exam tip

Use neutron number = mass number − atomic number and keep isotope notation consistent.

Tier 1 · Easy

Two atoms have the same number of protons but different numbers of neutrons. State the relationship between the atoms and explain why they are the same element.
[2 marks]
Total for this question: 2
A card describes a magnesium ion as having 12 protons, 12 neutrons and 14 electrons, with charge $2+$ . Assume the proton, neutron and charge entries are correct. Identify the incorrect entry and give the correct electron number.
[3 marks]
Total for this question: 3

Tier 2 · Standard

For the ion ${}^{27}_{13}\mathrm{Al}^{3+}$ , determine the numbers of protons, neutrons and electrons.
[3 marks]
Total for this question: 3
Particle $X$ has 17 protons, 18 neutrons and 17 electrons. Particle $Y$ has 17 protons, 20 neutrons and 18 electrons. Explain the relationship between $X$ and $Y$ , and determine the charge of $Y$ .
[4 marks]
Total for this question: 4
A packet contains five identical ions. Together the ions contain $50$ electrons. Each ion has charge $2+$ . Determine the atomic number of the element. One ion has mass number $26$ ; also determine its neutron number.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two isotopes of element $Q$ have mass numbers $63$ and $65$ . An ion of the first isotope has charge $2+$ and contains $27$ electrons. Determine the atomic number of $Q$ , the neutron number of each isotope, and explain why both are the same element.
[5 marks]
Total for this question: 5
A record states that an ion has 27 protons, 32 neutrons and 25 electrons. Without calculating any particle counts, identify which count entry must change to describe a different element and which must change to describe another isotope of the same element. Explain your choices.
[3 marks]
Total for this question: 3
A box contains one neutral atom of isotope $X$ and one neutral atom of isotope $Y$ of the same element. Together they contain $28$ protons and $32$ neutrons. $Y$ has two more neutrons than $X$ . Determine the atomic number and mass number of each isotope, then state the total number of electrons in the box.
[5 marks]
Total for this question: 5
A particle has $17$ protons, $20$ neutrons and $18$ electrons. Write its full symbol. A second particle has the symbol ${}^{35}_{17}\mathrm{Cl}$ . Explain why the particles are isotopes and why only one is charged.
[5 marks]
Total for this question: 5
A sample contains $50$ neutral boron atoms. Boron has atomic number $5$ , each atom is either boron-10 or boron-11, and the sum of all their mass numbers is $546$ . Determine the number of atoms of each isotope and the total number of electrons in the sample.
[5 marks]
Total for this question: 5

4.4.1.3 · The development of the model of the atom (common content with chemistry)

Explanation

Atoms were first treated as indivisible spheres; discovery of the electron led to the plum pudding model, with electrons embedded in a ball of positive charge.
Use scattering evidence in order: most alpha particles passed through, so atoms are mostly empty space; a few were strongly deflected, so charge and most mass occupy a tiny nucleus.
Bohr proposed electrons at specific distances, later evidence identified protons in the nucleus, and Chadwick's work provided evidence for neutrons.
A common error is to say Rutherford expected every alpha particle to rebound: the key comparison is between the observed pattern and the plum pudding model's prediction of only small deflections.

Worked example

Put these developments in chronological order: the nuclear model, the plum pudding model, evidence for the neutron, and electrons at specific distances from the nucleus.

1.The electron discovery produced the plum pudding model. Alpha scattering then produced the nuclear model. Bohr next placed electrons at specific distances, and Chadwick's neutron evidence came later.

Answer: Plum pudding model, nuclear model, electrons at specific distances, evidence for the neutron

Common mistakes

Don't say Rutherford expected every alpha particle to rebound: the key comparison is between the observed pattern and the plum pudding model's prediction of only small deflections.
Don't fall into the trap of describing Rutherford scattering without linking the observations to the nuclear model.

Exam tip

For model-development questions, link each new observation to the change it forced in the model.

Tier 1 · Easy

In the alpha-scattering experiment, most alpha particles passed straight through the metal foil. What did this observation show about the structure of an atom?
[1 mark]
Total for this question: 1
The plum pudding model predicted that an alpha particle crossing thin foil would experience only a small deflection. A few particles instead travelled back towards the source. Explain why this result counted against that model.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In an alpha-scattering investigation, nearly all particles cross a thin metal sheet without changing direction, while a very small fraction turn through large angles. Explain two conclusions that caused the plum pudding model to be replaced.
[4 marks]
Total for this question: 4
Model $A$ spreads positive charge through the atom and predicts no deflection above $90^\circ$ . Model $B$ concentrates positive charge in a tiny nucleus and predicts very few such deflections. A test records 19 960 particles with no measurable deflection, 36 below $90^\circ$ and 4 above $90^\circ$ . A student chooses $A$ because most particles went straight through. Evaluate the choice.
[4 marks]
Total for this question: 4
An early nuclear model contains protons and electrons but no neutral nuclear particle. Scientists then find atoms with the same proton number but extra nuclear mass. Identify the later discovery that helps explain the extra mass and describe the resulting change to the model.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A student says, 'Once the nucleus was proposed, the atomic model was complete.' Use later changes to the model to evaluate this statement.
[6 marks]
Total for this question: 6
Model $A$ spreads positive charge through the atom and predicts zero large-angle deflections. Model $B$ concentrates positive charge in a tiny nucleus and predicts rare large-angle deflections. Repeat runs under the same conditions record 8 and 7 large-angle deflections. Explain why the repeat evidence strengthens the rejection of $A$ and supports $B$ , despite the small counts.
[5 marks]
Total for this question: 5
A revision card states: (1) the plum pudding model was rejected because most alpha particles were deflected; (2) Bohr's model was accepted because it was simpler; (3) the neutron was discovered before the nuclear model. Identify and correct each error.
[5 marks]
Total for this question: 5
In an alpha-particle scattering experiment, a thin metal foil is replaced by a foil of the same metal that is twice as thick. Predict how the number of large-angle deflections changes, explain your prediction, and state what should still happen to the large majority of alpha particles.
[4 marks]
Total for this question: 4

4.4.2.1 · Radioactive decay and nuclear radiation

Explanation

An unstable nucleus decays at random; activity is its decay rate in becquerels ( $\mathrm{Bq}$ ), while count rate is the number of decays recorded each second by a detector such as a Geiger-Muller tube.
Identify radiation by composition and properties: alpha is two protons plus two neutrons, beta is a fast electron from the nucleus, gamma is electromagnetic radiation, and a neutron may also be emitted.
Alpha has the shortest range and greatest ionising power, beta is intermediate, and gamma is the most penetrating and least ionising of the three.
A common error is to call beta an orbital electron or gamma a charged particle: beta forms when a neutron changes into a proton, while gamma has no charge or mass.

Relative penetration of alpha, beta and gamma radiation.

Worked example

Name the radiation described in each case: (i) two protons and two neutrons, (ii) electromagnetic radiation from a nucleus, (iii) a fast electron emitted when a neutron changes.

1.Match composition before using penetration: a helium nucleus is alpha, an electromagnetic wave from the nucleus is gamma, and the nuclear electron produced in a neutron-to-proton change is beta.

Answer: (i) alpha (ii) gamma (iii) beta

Common mistakes

Don't call beta an orbital electron or gamma a charged particle: beta forms when a neutron changes into a proton, while gamma has no charge or mass.
Don't fall into the trap of saying gamma radiation is a charged particle.

Exam tip

Compare alpha, beta and gamma by ionising power, penetration and range.

Tier 1 · Easy

State the unit of activity and explain what an activity of $1$ in this unit means.
[2 marks]
Total for this question: 2
A student writes, 'Beta radiation is an electron taken from an atom's outer energy level, and gamma is the most strongly ionising radiation.' Correct both errors.
[4 marks]
Total for this question: 4

Tier 2 · Standard

Radiation $P$ is stopped by card, $Q$ crosses card but is stopped by a thin aluminium sheet, and $R$ crosses both but is reduced by thick lead. Identify $P$ , $Q$ and $R$ , then state which has the greatest ionising power.
[4 marks]
Total for this question: 4
A detector records $18$ counts per minute for background. With a source it records $642$ with no absorber, $634$ behind paper, $260$ behind aluminium and $22$ behind thick lead. Use the data to identify the radiation types emitted by the source.
[4 marks]
Total for this question: 4
A sealed source stays unchanged during a short test. Its activity is $2000\,\mathrm{Bq}$ . A detector records $480$ counts per second with no absorber and $35$ counts per second after thick shielding is inserted. Explain why the detector reading changes while the source activity does not, and distinguish what the two quantities measure.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A sealed source must send radiation through several centimetres of tissue to a target while limiting ionisation of healthy tissue along the path. Compare alpha, beta and gamma, and choose the most suitable radiation.
[5 marks]
Total for this question: 5
Two sealed sources $X$ and $Y$ are known to emit either alpha or beta radiation, one type each. The background count rate is $18$ counts per minute. At source-detector distances of $1$ , $2$ , $4$ , $10$ and $30\,\mathrm{cm}$ , $X$ gives $486$ , $174$ , $18$ , $17$ and $19$ counts per minute, while $Y$ gives $980$ , $720$ , $470$ , $250$ and $96$ counts per minute. Identify the radiation from each source and justify both choices using range in air.
[5 marks]
Total for this question: 5
A sealed source has activity $4.8\,\mathrm{kBq}$ . During $15\,\mathrm{s}$ , a detector records $540$ counts. Its background rate is $4.0$ counts per second. Calculate the corrected count rate and the percentage of the source's decays that are recorded. Explain why the corrected count rate is not the source activity.
[5 marks]
Total for this question: 5
A detector records $20$ counts per minute as background, $1020$ with a source and no absorber, $420$ behind paper and $60$ behind thick lead. The source emits alpha and gamma radiation only. Assume paper absorbs all alpha radiation but no gamma radiation. Determine the corrected alpha and gamma count rates before absorption, calculate the fraction of the corrected count rate due to alpha, and explain the lead reading.
[5 marks]
Total for this question: 5
An alpha source has a constant activity of $3.2\,\mathrm{kBq}$ during a $25\,\mathrm{s}$ interval. Calculate the expected number of alpha particles emitted and the total numbers of protons and neutrons carried by them. Explain why the number emitted in one actual interval may differ slightly from the calculated value.
[4 marks]
Total for this question: 4

4.4.2.2 · Nuclear equations

Explanation

Balance a nuclear equation by making the total mass number and total atomic number equal on both sides.
For alpha emission use ${}^{4}_{2}\alpha$ ; for beta-minus emission use ${}^{0}_{-1}\beta$ , so the daughter's atomic number is one greater while its mass number is unchanged.
For example, after one alpha emission a parent labelled $A,Z$ becomes $A-4,Z-2$ ; gamma emission changes neither number.
A common error is to decrease atomic number in beta-minus decay: the emitted beta has atomic number $-1$ , so the daughter must increase by $1$ to balance.

Worked example

Complete ${}^{218}_{84}X\rightarrow{}^{214}_{82}Y+\,?$ by giving the emitted particle in full nuclear notation.

1.Subtract daughter numbers from parent numbers: $218-214=4$ and $84-82=2$ . The missing radiation is therefore an alpha particle, ${}^{4}_{2}\alpha$ .

Answer: Mass number of the particle $=4$ and atomic number $=2$ . ${}^{4}_{2}\alpha$

Common mistakes

Don't decrease atomic number in beta-minus decay: the emitted beta has atomic number $-1$ , so the daughter must increase by $1$ to balance.
Don't fall into the trap of changing both atomic and mass number for beta-minus decay.

Exam tip

Balance both mass number and atomic number on each side of a nuclear equation.

Tier 1 · Easy

A nucleus emits gamma radiation. State what happens to the nucleus's mass number and atomic number.
[2 marks]
Total for this question: 2
A student completes a beta-minus equation by decreasing the daughter's atomic number by 1. Diagnose the error and state the correct changes to mass number and atomic number.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Complete the beta-minus equation ${}^{131}_{53}X\rightarrow{}^{A}_{Z}Y+{}^{0}_{-1}\beta$ by determining $A$ and $Z$ .
[2 marks]
Total for this question: 2
A student claims that ${}^{222}_{86}R$ can become ${}^{218}_{85}S$ by emitting one alpha particle only. Test the claim by balancing both nuclear numbers, and identify the additional emission needed.
[4 marks]
Total for this question: 4
A decay ledger contains three consecutive entries: ${}^{214}_{84}A\rightarrow{}^{210}_{82}B+{}^{4}_{2}\alpha$ ; ${}^{210}_{82}B\rightarrow{}^{210}_{83}C+{}^{0}_{-1}\beta$ ; ${}^{210}_{83}C\rightarrow{}^{206}_{82}D+{}^{4}_{2}\alpha$ . Check all three entries. Identify the entry that does not balance and correct it.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A nucleus ${}^{240}_{96}M$ emits one alpha particle and then two beta-minus particles. Determine the mass number and atomic number of the final nucleus, showing the change at each stage.
[4 marks]
Total for this question: 4
A decay chain changes ${}^{226}_{88}T$ into ${}^{218}_{85}U$ by alpha and beta-minus emissions only. Determine how many of each emission occurred and justify your answer using conservation of mass number and atomic number.
[5 marks]
Total for this question: 5
A nucleus ${}^{212}_{84}X$ undergoes one alpha emission and one beta-minus emission. Student $A$ applies alpha first; student $B$ applies beta-minus first and claims the different order gives a different final nucleus. Determine the final mass number and atomic number by both routes, then evaluate the claim.
[5 marks]
Total for this question: 5
Cobalt-60 undergoes beta-minus decay to nickel-60, which then emits gamma radiation. The atomic numbers of cobalt and nickel are $27$ and $28$ . Write a balanced nuclear equation for the beta-minus decay, and state what the gamma emission does to the mass number and atomic number of the nickel nucleus.
[4 marks]
Total for this question: 4
Radium-226 has atomic number $88$ and undergoes alpha decay. The possible daughter elements are francium ( $87$ ), radon ( $86$ ), astatine ( $85$ ) and actinium ( $89$ ). Identify the daughter nucleus, write a balanced nuclear equation, and explain why the daughter is a different element.
[5 marks]
Total for this question: 5

4.4.2.3 · Half-lives and the random nature of radioactive decay

Explanation

Half-life is the time for the number of undecayed nuclei, activity or net count rate to fall to half its initial value; individual nuclear decays remain unpredictable.
Subtract background count rate before finding successive halvings, then divide the elapsed time by the number of half-lives.
A fall from $960$ to $120$ is three halvings because $960\rightarrow480\rightarrow240\rightarrow120$ ; Higher tier can express the remaining-to-original ratio as $1:8$ .
A common error is to subtract the same amount in every half-life or to halve a gross detector reading without first removing background.

An exponential activity curve showing one half-life.

Worked example

The activity of a sample falls from $640\,\mathrm{Bq}$ to $160\,\mathrm{Bq}$ in $10$ hours. Determine its half-life.

1.The activity halves twice in the $10$ -hour interval. Divide the total time by two: $10/2=5$ hours.

Answer: $640\rightarrow320\rightarrow160$ is two half-lives. Half-life $=5$ hours

Common mistakes

Don't subtract the same amount in every half-life or to halve a gross detector reading without first removing background.
Don't fall into the trap of saying exactly half the nuclei decay in every small sample.

Exam tip

For half-life, show repeated halving or use two well-separated points on the decay curve.

Tier 1 · Easy

Explain why the exact time at which one particular unstable nucleus will decay cannot be predicted, even when the isotope's half-life is known.
[2 marks]
Total for this question: 2
Gross count rates are $460$ , $260$ and $160$ counts per minute at $0$ , $5$ and $10$ minutes. Background is $60$ counts per minute. A student says the source loses $200$ counts per minute every $5$ minutes. Diagnose the error and determine the half-life.
[4 marks]
Total for this question: 4

Tier 2 · Standard

A detector records $420$ counts per minute beside a source at time zero and $70$ counts per minute $18$ minutes later. Background count rate is $20$ counts per minute. Calculate the source's half-life.
[4 marks]
Total for this question: 4
A smooth gross count-rate graph is interpolated at two times: $590$ counts per minute at $2.5$ minutes and $310$ counts per minute at $8.5$ minutes. The background rate is $30$ counts per minute. Use the corrected readings to determine the half-life.
[4 marks]
Total for this question: 4
A source has a half-life of $4.0$ minutes. A detector gives gross readings of $740$ counts per minute initially and $110$ counts per minute after $12$ minutes. The constant background rate was not recorded. Determine the background rate and the initial corrected count rate.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A detector beside a source reads $830$ counts per minute initially and $130$ counts per minute after $12$ minutes. Background is $30$ counts per minute. Determine the half-life, then explain why repeated one-minute readings taken at the same time would not all be identical.
[6 marks]
Total for this question: 6
Higher only: A detector initially reads $970$ counts per minute beside a source. Background is $10$ counts per minute and the source has a half-life of $3.0$ hours. After $12$ hours, determine the ratio of remaining activity to initial activity and predict the detector reading.
[5 marks]
Total for this question: 5
A sample initially contains $6.4\times10^{8}$ undecayed nuclei. After $20$ days, $4.0\times10^{7}$ undecayed nuclei remain. Determine the half-life, then determine the number of undecayed nuclei remaining after a further $15$ days.
[5 marks]
Total for this question: 5
A source has a half-life of $6.0$ hours. After $18$ hours, a detector gives a gross count rate of $46$ counts per minute; background is $6$ counts per minute. Determine the initial corrected count rate and the initial gross detector reading.
[4 marks]
Total for this question: 4
A source has corrected count rates of $640$ , $320$ , $160$ and $80$ counts per minute at $0$ , $3$ , $6$ and $9$ minutes. The constant background is $20$ counts per minute. Determine the half-life, find the first time at which the gross count rate is below $45$ counts per minute, and explain why the gross count rate can never fall below $20$ counts per minute.
[5 marks]
Total for this question: 5

4.4.2.4 · Radioactive contamination

Explanation

Contamination is the unwanted presence of material containing radioactive atoms; irradiation is exposure to nuclear radiation without transfer of radioactive material.
Compare hazards by asking whether the source can remain on or enter the body, which radiation it emits, and how exposure can be shortened, shielded or kept at a distance.
An alpha contaminant outside the body may be stopped by skin, but the same material inside the body can be especially hazardous because alpha is strongly ionising at short range.
A common error is to say an irradiated object must become radioactive; irradiation stops when exposure ends, whereas contaminating atoms continue to decay until removed or decayed.

Worked example

A wrapped instrument is placed near a sealed gamma source and then removed. No radioactive material touches it. State whether this is contamination or irradiation, and whether the instrument becomes radioactive.

1.The source only exposes the instrument to radiation; no radioactive atoms are transferred. This is irradiation, and the irradiated instrument does not itself become a radioactive source.

Answer: The instrument is irradiated. It does not become radioactive.

Common mistakes

Don't say an irradiated object must become radioactive; irradiation stops when exposure ends, whereas contaminating atoms continue to decay until removed or decayed.
Don't fall into the trap of confusing irradiation with contamination by radioactive material.

Exam tip

State whether the hazard is an external source or radioactive material on or inside the body.

Tier 1 · Easy

Radioactive liquid is spilled onto a worker's glove. Explain why removing the glove reduces the worker's exposure even after the original container has been moved away.
[2 marks]
Total for this question: 2
After a radioactive sample is removed, a bench still gives a count above background. Wiping the bench returns the reading to background. Explain what this evidence shows.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Compare the hazard from an alpha-emitting speck held outside the body with the hazard if the same speck is inhaled. Give a suitable precaution.
[4 marks]
Total for this question: 4
Background is $24$ counts per minute. After an unsealed source is put away, a bench reads $238$ counts per minute. After the bench is wiped it reads $27$ , while the used wipe reads $211$ . Interpret all three readings and state how the wipe should be handled.
[4 marks]
Total for this question: 4
Batch $A$ passes beside a sealed gamma source and returns to background count rate when the source is removed. Batch $B$ passes beneath an unsealed source and still gives a reading above background after the source is removed. State whether each batch has been irradiated or contaminated, and explain which batch can continue exposing a handler.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A small study reports that workers exposed near a sealed radiation source have a higher illness rate. Explain why publishing the method and results for peer review is important before concluding that irradiation caused the illnesses.
[5 marks]
Total for this question: 5
One worker briefly stands near a sealed gamma source. Another handles an unsealed alpha-emitting powder and later finds powder on a sleeve. Compare the two hazards and justify one control measure for each worker.
[6 marks]
Total for this question: 6
A study finds illness in $12$ of $200$ hospital imaging staff and $6$ of $200$ administrative staff at the same hospital. The imaging staff are older on average and a larger fraction smoke. Calculate the illness percentages, explain why the data do not by themselves show that irradiation caused the difference, and suggest two improvements to the investigation.
[6 marks]
Total for this question: 6
Two unsealed solutions each contaminate a bench with initial activity $800\,\mathrm{Bq}$ . Solution $A$ contains nitrogen-16, a beta-minus emitter with half-life $7.1\,\mathrm{s}$ (it also emits gamma radiation). Solution $B$ contains phosphorus-32, a beta-minus emitter with half-life $14.3$ days. After $28.4\,\mathrm{s}$ , calculate the activity from $A$ , give a lower bound for the activity from $B$ , and explain how the contamination should be handled.
[5 marks]
Total for this question: 5

4.4.3.1 · Background radiation (physics only)

Explanation

Background radiation is always present and includes natural sources such as radioactive rocks and cosmic rays, plus man-made fallout from weapons tests and nuclear accidents.
When comparing measurements, allow for location, altitude, surrounding rock and occupation, and subtract a local background count rate when isolating a source's count rate.
For dose, $1000\,\mathrm{mSv}=1\,\mathrm{Sv}$ ; a worker's total dose can be estimated by adding the contributions from different exposures over the stated time.
A common error is to assume a detector should read zero after a test source is removed: background radiation continues to produce counts.

Worked example

Classify each source of background radiation as natural or man-made: cosmic rays, radioactive rock, and fallout from a nuclear weapons test.

1.Cosmic radiation arrives from space and radioactivity occurs naturally in rock. Fallout is produced by human nuclear weapons testing, so it is man-made.

Answer: Cosmic rays: natural Radioactive rock: natural Weapons-test fallout: man-made

Common mistakes

Don't assume a detector should read zero after a test source is removed: background radiation continues to produce counts.
Don't fall into the trap of assuming background count is zero when no source is present.

Exam tip

Subtract background count from the measured count before interpreting source activity.

Tier 1 · Easy

A detector still records counts after a test source has been removed. Give one source of these counts and explain why the reading is not necessarily a detector fault.
[2 marks]
Total for this question: 2
An office worker and a hospital radiographer live in the same area. Explain why the radiographer's occupation may increase their radiation dose, and classify the extra source as natural or man-made.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A detector averages $18$ counts per minute at sea level on sedimentary ground and $31$ counts per minute at a high-altitude site on granite. Suggest two reasons for the difference and explain why neither reading should be treated as zero-source error.
[4 marks]
Total for this question: 4
Five one-minute background counts are $24$ , $22$ , $23$ , $75$ and $21$ . Treat $75$ as an anomaly. A later gross reading beside a test source is $183$ counts per minute. Calculate the mean background and the source's corrected count rate.
[4 marks]
Total for this question: 4
A person's annual radiation dose is $2.70\,\mathrm{mSv}$ : radon $1.30\,\mathrm{mSv}$ , ground and buildings $0.35\,\mathrm{mSv}$ , food and drink $0.25\,\mathrm{mSv}$ , cosmic rays $0.33\,\mathrm{mSv}$ , medical sources $0.41\,\mathrm{mSv}$ and other man-made sources $0.06\,\mathrm{mSv}$ . Calculate the percentage of the total that comes from man-made sources and name the largest natural contributor.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A cave guide receives $0.006\,\mathrm{mSv}$ per working week from surrounding rock and works $48$ weeks. An airline worker receives $0.009\,\mathrm{mSv}$ per working week from additional cosmic radiation for $48$ weeks. Calculate each annual occupational dose, compare them, and express the larger in sieverts.
[4 marks]
Total for this question: 4
A worker's annual occupational dose is $0.84\,\mathrm{mSv}$ . Work near radioactive rock contributes $0.015\,\mathrm{mSv}$ per week for $36$ weeks. The rest comes from flights at $0.0050\,\mathrm{mSv}$ per hour. Calculate the flight time and express the total annual dose in sieverts.
[5 marks]
Total for this question: 5
Five one-minute background readings at site $P$ are $18$ , $21$ , $19$ , $22$ and $20$ counts. At site $Q$ they are $21$ , $23$ , $19$ , $22$ and $20$ . Calculate the mean for each site and evaluate the claim that the background at $Q$ is definitely higher.
[5 marks]
Total for this question: 5
The same sealed source is measured at two sites. Site $P$ has background $24$ counts per minute and a gross reading of $324$ ; high-altitude site $Q$ has background $45$ and a gross reading of $345$ . Calculate both corrected count rates and explain why the gross readings differ although the source is unchanged.
[4 marks]
Total for this question: 4
A student measures background radiation twice: first for $10$ seconds and records $3$ counts, then for $5$ minutes and records $105$ counts. With a source present, the detector records a gross count of $90$ in $10$ seconds. Calculate both background rates, use the longer background measurement to calculate the corrected source count rate, and give two reasons why the longer background measurement is better.
[5 marks]
Total for this question: 5

4.4.3.2 · Different half-lives of radioactive isotopes (physics only)

Explanation

Radioactive isotopes span a very wide range of half-lives, so the duration and rate of a hazard depend on the isotope present.
Compare hazards using both activity and persistence: a short half-life means rapid decay and a quickly falling hazard, while a long half-life can leave material radioactive for much longer.
For equal numbers of unstable nuclei, a shorter-half-life isotope undergoes decays more rapidly at first; a longer-half-life isotope generally creates the longer waste-management problem.
A common error is to call either short or long half-life always safer: risk also depends on quantity, radiation type, route into the body and exposure time.

Worked example

Two contaminants have half-lives of $6$ hours and $24$ years. Which contaminant can remain a disposal hazard for longer? Explain your choice.

1.A longer half-life means fewer successive halvings occur in a fixed time. The $24$ -year isotope therefore remains radioactive over a much longer storage period.

Answer: The isotope with the $24$ -year half-life Its activity falls much more slowly, so radioactive material persists for longer.

Common mistakes

Don't call either short or long half-life always safer: risk also depends on quantity, radiation type, route into the body and exposure time.
Don't fall into the trap of saying a long half-life always means a high activity.

Exam tip

A half-life comparison must distinguish activity from how quickly activity decreases.

Tier 1 · Easy

Samples $P$ and $Q$ contain equal numbers of unstable nuclei and emit the same type of radiation. $P$ has the shorter half-life. Which sample has the greater initial activity? Explain.
[2 marks]
Total for this question: 2
A student says, 'A long-half-life isotope is always more dangerous than a short-half-life isotope.' Give two reasons why the half-life alone is not enough to support this conclusion.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Samples $A$ and $B$ initially have the same activity and emit the same type of radiation. $A$ has half-life $3$ hours; $B$ has half-life $40$ years. Compare how their hazards change after the samples are securely stored.
[4 marks]
Total for this question: 4
For otherwise comparable samples in stored waste, isotope $P$ has half-life $3.0\times10^2\,\mathrm{s}$ , isotope $Q$ has half-life $8.0$ days and isotope $R$ has half-life $4.5\times10^9$ years. Rank the isotopes from the shortest to the longest time for which they remain a significant hazard, and state which isotope dominates long-term disposal planning. Use the data in your answer.
[4 marks]
Total for this question: 4
A waste sample contains isotope $X$ with a half-life of $1$ day and isotope $Y$ with a half-life of $20$ years. Its total activity falls steeply during the first week and then continues to fall much more slowly. Explain the two parts of this pattern and identify which isotope controls the long-term storage problem.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Equal numbers of nuclei of isotopes $C$ and $D$ are spilled. $C$ has half-life $2.0\times10^2\,\mathrm{s}$ and $D$ has half-life $6.0\times10^7\,\mathrm{s}$ . Compare the likely initial and long-term hazards, stating why half-life alone cannot determine the total risk.
[5 marks]
Total for this question: 5
Waste samples $U$ and $V$ emit the same radiation and are kept in identical containers. $U$ begins at $900\,\mathrm{Bq}$ with a half-life of $2$ days; $V$ begins at $300\,\mathrm{Bq}$ with a half-life of $30$ years. Decide which needs greater immediate shielding and which needs longer secure storage, then explain why neither sample is simply 'safer'.
[5 marks]
Total for this question: 5
A monitoring device must retain more than half its initial source activity throughout $12$ years. Candidate $A$ has half-life $8$ years, $B$ has half-life $15$ years and $C$ has half-life $40$ years. Choose the suitable candidate that creates the shorter long-term storage problem and explain why each of the other two is not chosen.
[5 marks]
Total for this question: 5
Samples $A$ and $B$ emit the same radiation. $A$ starts at $960\,\mathrm{Bq}$ and has a half-life of $2$ days; $B$ starts at $120\,\mathrm{Bq}$ and has a half-life of $8$ days. Calculate both activities after $8$ days and explain which sample presents the greater activity hazard initially and after longer storage.
[5 marks]
Total for this question: 5
A room contains a source with activity $1600\,\mathrm{Bq}$ and half-life $3$ days. The room may reopen only when the activity is below $120\,\mathrm{Bq}$ . Determine the earliest reopening time and explain how a longer half-life would affect it.
[5 marks]
Total for this question: 5

4.4.3.3 · Uses of nuclear radiation (physics only)

Explanation

A medical tracer for exploring an organ should be detectable outside the body, so a penetrating radiation such as gamma is useful, and its half-life should limit the time the patient remains radioactive.
For controlling or destroying unwanted tissue, direct radiation at the target or place a suitable source close to it while limiting dose to healthy cells.
Evaluate a use by comparing diagnostic or treatment benefit with absorbed dose, radiation type, half-life, exposure time and the consequences of not carrying out the procedure.
A common error is to discuss only usefulness: every evaluation needs a linked risk, such as ionisation damaging healthy cells, and a way the exposure is controlled.

Worked example

Give two reasons why a gamma-emitting isotope can be suitable as a tracer for exploring an internal organ.

1.A tracer must be detected without surgery, which requires radiation able to leave the body. Gamma is penetrating and relatively weakly ionising, giving the two linked advantages.

Answer: Gamma can penetrate out of the body to an external detector. Gamma is less ionising than alpha or beta, so it causes less cell damage for a comparable exposure.

Common mistakes

Don't discuss only usefulness: every evaluation needs a linked risk, such as ionisation damaging healthy cells, and a way the exposure is controlled.
Don't fall into the trap of choosing a radiation source without considering penetration and half-life.

Exam tip

Justify a medical or industrial source using penetration, ionisation and half-life together.

Tier 1 · Easy

Explain why a medical tracer's half-life should be long enough for an investigation but not unnecessarily long.
[2 marks]
Total for this question: 2
A radiotherapy machine directs a narrow gamma beam at a tumour from several different angles. Explain how this arrangement can damage the tumour while limiting harm to surrounding healthy tissue.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A doctor proposes directing nuclear radiation at a tumour to destroy unwanted tissue. A smaller dose will also reach nearby healthy tissue, and without treatment the tumour is likely to grow. Evaluate this use of radiation.
[4 marks]
Total for this question: 4
Three possible tracers are tested through a tissue model. $A$ gives no reading outside the model and remains active for hours. $B$ gives a clear external reading and its activity falls substantially during the next day. $C$ gives a clear reading but changes very little over several years. Choose the best tracer and justify the rejection of the others.
[4 marks]
Total for this question: 4
After a gamma-emitting tracer is introduced, identical external detectors monitor two regions that should receive similar blood flow. Corrected count rates in region $P$ are $80$ , $210$ and $120$ counts per minute after $1$ , $3$ and $5$ minutes; region $Q$ gives $24$ , $31$ and $26$ . Interpret the pattern and explain why gamma radiation is suitable for this test.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A hospital needs an internal tracer that will be measured by a detector outside the body during a four-hour investigation. Candidate $J$ emits alpha and has half-life $12$ years; $K$ emits gamma and has half-life $6$ hours; $L$ emits gamma and has half-life $30$ years. Choose the best candidate and justify why the other two are less suitable.
[6 marks]
Total for this question: 6
A non-radiation scan detects $62\%$ of tumours of a particular type. Tracer scan $P$ detects $91\%$ , gives the patient $0.8\,\mathrm{mSv}$ and uses a six-hour-half-life isotope. Tracer scan $Q$ detects $93\%$ , gives $7.4\,\mathrm{mSv}$ and uses a six-year-half-life isotope. Evaluate the evidence and recommend one scan.
[6 marks]
Total for this question: 6
Two radiotherapy plans use the same type of ionising radiation. Plan $R$ gives a tumour $60$ dose units and nearby healthy tissue $24$ units. Plan $S$ gives the tumour $54$ units and healthy tissue $9$ units. The tumour needs at least $50$ units. Calculate the tumour-to-healthy-tissue dose ratio for each plan, then recommend a plan and justify the remaining risk.
[5 marks]
Total for this question: 5
Technetium-99m emits gamma radiation and has a half-life of about $6.0$ hours. A tracer has activity $640\,\mathrm{MBq}$ when administered. Calculate its activity after $12$ hours and explain two features that make technetium-99m suitable for exploring an internal organ.
[4 marks]
Total for this question: 4
To control or destroy unwanted tissue such as a tumour, a doctor may place a suitable source close to it. Give one advantage and one disadvantage of this internal treatment, and explain why a short-range emitter can be suitable.
[5 marks]
Total for this question: 5

4.4.4.1 · Nuclear fission (physics only)

Explanation

Fission is the splitting of a large, unstable nucleus; it usually begins when the nucleus absorbs a neutron.
A fission event produces two smaller nuclei of roughly equal size, two or three neutrons and gamma rays, with released energy appearing as kinetic energy of the products.
Emitted neutrons can trigger further fissions: limiting how many continue gives a controlled reactor chain reaction, while continued multiplication gives an uncontrolled release.
A common error is to describe fission as two small nuclei joining; that is fusion, whereas fission starts with one large nucleus splitting.

A neutron-induced fission event releasing two smaller nuclei and three neutrons.

Worked example

State what usually starts a fission event and name two products other than the two smaller nuclei.

1.Begin with neutron absorption by the large unstable nucleus. After splitting, list products beyond the two daughter nuclei: emitted neutrons, gamma radiation and released kinetic energy.

Answer: A large unstable nucleus absorbs a neutron. Two or three neutrons are emitted. Gamma rays are emitted; energy or kinetic energy is also an acceptable second product.

Common mistakes

Don't describe fission as two small nuclei joining; that is fusion, whereas fission starts with one large nucleus splitting.
Don't fall into the trap of saying a neutron is created from nothing in a fission chain reaction.

Exam tip

In fission, identify neutron absorption, nucleus splitting, released energy and emitted neutrons.

Tier 1 · Easy

Explain how neutrons released by one fission event can produce a chain reaction.
[2 marks]
Total for this question: 2
A trainee says that withdrawing control rods reduces reactor power because fewer neutrons can reach fuel nuclei. Correct the trainee's reasoning.
[3 marks]
Total for this question: 3

Tier 2 · Standard

In a simplified chain reaction, every fission releases three neutrons and every released neutron causes one new fission. Starting with one fission in generation 1, calculate the numbers of fissions in generations 2, 3 and 4.
[3 marks]
Total for this question: 3
A reactor log gives the mean number of neutrons from each fission that cause another fission: $0.7$ in state $R$ , $1.0$ in state $S$ and $1.4$ in state $T$ . Predict how reactor power changes in each state and identify the state for steady operation.
[4 marks]
Total for this question: 4
A fission event produces two fast-moving smaller nuclei and gamma rays. Explain how the released energy can be transferred to a reactor's coolant.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Sketch and label a chain-reaction diagram beginning with a neutron absorbed by one large unstable nucleus. Show that fission releasing two neutrons can cause the next generation, then explain how a controlled reactor prevents the number of fissions increasing each generation.
[5 marks]
Total for this question: 5
Each fission in a reactor releases three neutrons. With the control rods inserted, two neutrons per fission are absorbed or escape and one causes another fission. Starting from $800$ fissions in one generation, calculate the next generation. The rods are then withdrawn so that two neutrons per fission continue the chain; calculate the following two generations and explain the power change.
[5 marks]
Total for this question: 5
In a reactor using uranium-235 fuel, each fission releases $3.2\times10^{-11}\,\mathrm{J}$ to the coolant. The coolant receives $1.6\times10^{6}\,\mathrm{J}$ during $0.50\,\mathrm{s}$ . Calculate the number of fissions and the power transferred to the coolant. Explain what must happen to the continuing-neutron number for this power to remain steady.
[5 marks]
Total for this question: 5
A reactor averages $2.5$ neutrons released per fission. For $4.0\times10^{15}$ fissions, $4.0\times10^{15}$ neutrons cause further fission, $3.2\times10^{15}$ are absorbed by control rods and the rest escape or are absorbed elsewhere. Calculate the total neutrons released and the number in the final category, then explain why the reactor power is steady.
[5 marks]
Total for this question: 5
Uranium-235 fission releases about $3.2\times10^{-11}\,\mathrm{J}$ per event. A reactor produces $2.5\times10^{17}$ fissions, and $35\%$ of the released energy becomes electrical output. Calculate the total released energy, the electrical output and the energy transferred by other routes, then state how a controlled chain reaction differs from an uncontrolled chain reaction.
[5 marks]
Total for this question: 5

4.4.4.2 · Nuclear fusion (physics only)

Explanation

Fusion is the joining of two light nuclei to form a heavier nucleus.
Identify fusion from the pattern of two small nuclear reactants becoming one larger nuclear product, rather than from the presence of radiation alone.
The combined mass of the nuclear product can be slightly less than that of the starting nuclei; the mass difference is converted into energy carried by radiation.
A common error is to call any energy-releasing nuclear process fusion: fission splits one large nucleus, while fusion joins two light nuclei.

Worked example

Complete the definition: nuclear fusion is the joining of two ______ nuclei to make a ______ nucleus.

1.Fusion begins with two light nuclei and combines them into a nucleus heavier than either starting nucleus.

Answer: light heavier

Common mistakes

Don't call any energy-releasing nuclear process fusion: fission splits one large nucleus, while fusion joins two light nuclei.
Don't fall into the trap of confusing fusion of light nuclei with fission of a heavy nucleus.

Exam tip

For fusion, name the light nuclei, the heavier product and the need for very high temperature.

Tier 1 · Easy

In nuclear fusion the product nucleus has slightly less mass than the two original nuclei. State what happens to this missing mass, and name the condition needed for fusion to occur.
[2 marks]
Total for this question: 2
In deuterium-tritium fusion, the deuterium and tritium nuclei have a total mass of $8.35\times10^{-27}\,\mathrm{kg}$ . The helium nucleus and neutron produced have a total mass of $8.32\times10^{-27}\,\mathrm{kg}$ . State what the comparison shows without calculating the released energy.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Compare nuclear fusion with nuclear fission in terms of the nuclei before and after each process, and state one energy feature shared by them.
[4 marks]
Total for this question: 4
Record $R$ shows one large nucleus absorbing a neutron and producing two smaller nuclei. Record $S$ shows two light nuclei producing one heavier nucleus and radiation. Identify the fusion record and use both the nuclear pattern and the radiation to justify the choice.
[4 marks]
Total for this question: 4
A chamber initially contains $1.8\times10^{20}$ deuterium nuclei. After a pulse, $1.0\times10^{20}$ deuterium nuclei remain and $4.0\times10^{19}$ heavier nuclei have formed. Each heavier nucleus formed when two deuterium nuclei joined. Determine the number of fusion events and the fraction of the starting deuterium nuclei that fused, then explain why the observation is fusion.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An experiment shows two light nuclei combining into one heavier nucleus while radiation leaves the reaction. The measured mass of the heavier nucleus is slightly smaller than the total mass of the two starting nuclei. Explain why the observations support fusion and account for the mass difference.
[5 marks]
Total for this question: 5
In a fusion event, two starting nuclei have a total mass of $6.70\times10^{-27}\,\mathrm{kg}$ and the product nucleus has mass $6.65\times10^{-27}\,\mathrm{kg}$ . Calculate the mass lost, explain its link to the released energy, and briefly explain why very high temperature and pressure help fusion occur.
[6 marks]
Total for this question: 6
An experimental fusion pulse produces $7.5\times10^{14}$ fusion events. Each event releases $2.8\times10^{-12}\,\mathrm{J}$ , while creating the conditions for the pulse requires $4.5\times10^3\,\mathrm{J}$ . Calculate the released energy and the net energy output, then evaluate the statement, 'The pulse released fusion energy, so it supplied net useful energy.'
[5 marks]
Total for this question: 5
A uranium-235 fission releases about $3.2\times10^{-11}\,\mathrm{J}$ , while one deuterium-tritium fusion event releases about $2.8\times10^{-12}\,\mathrm{J}$ . Calculate how many times greater the energy per fission event is. For equal power, compare the numbers of events per second needed, then compare the nuclear changes in fission and fusion.
[4 marks]
Total for this question: 4
A fusion chamber contains $6.0\times10^{18}$ deuterium nuclei and $4.0\times10^{18}$ tritium nuclei. Each deuterium-tritium event uses one nucleus of each isotope, produces a helium nucleus and a neutron, and releases $2.8\times10^{-12}\,\mathrm{J}$ . Determine the maximum number of fusion events, the number of deuterium nuclei left and the total energy released. Explain why the reaction is fusion.
[5 marks]
Total for this question: 5

Answer key

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

4.4.1.1 · The structure of an atom

Tier 1 · Easy

Mark scheme for 4.4.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The nucleus contains nearly all the mass but occupies only a tiny central region. The electrons occupy energy levels much farther from the nucleus, leaving most of the atom's volume empty.	Separate mass from volume. Protons and neutrons concentrate the mass in the tiny nucleus, while the much larger region containing the electrons is mostly empty space.	2
Total Question 1			2
02.1	The tiny central region is the nucleus. The nucleus occupies only a very small fraction of the atom's volume, while the electrons are much farther out.	Separate the region containing mass from the region occupying volume. Protons and neutrons place almost all the mass in the tiny nucleus, but the electron levels extend across the much larger atom.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Protons are positive and are in the nucleus. Neutrons have no charge and are in the nucleus. Electrons are negative and occupy energy levels around the nucleus. Nearly all the mass is in the nucleus.	Organise the description by particle. Place protons and neutrons together in the central nucleus, then place electrons at energy levels outside it. Finish by linking the proton and neutron masses to the concentration of atomic mass in the nucleus.	4
Total Question 1			4
02.1	Its final energy level is 3. Absorption moved the electron to a higher energy level. Emitting less energy than was absorbed lowers its energy from level 4 but leaves it above its starting level 2.	Track the net energy transfer. The electron gains energy in moving from level 2 to level 4. It gives back only part of that gain, so it must finish below level 4 but above level 2: level 3.	3
Total Question 2			3
03.1	$P$ is a proton and $Q$ is a neutron. $R$ is an electron. Losing a negative electron leaves the atom with one more positive charge than negative charge. An electron has a very small relative mass, so losing one changes the atom's mass by a negligible amount.	Match each row using charge, relative mass and location. Then treat charge and mass separately: removing one negative particle changes the net charge by $+1$ , while removing a particle with very small relative mass barely changes the total mass.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	On absorption, the electron can move to a higher energy level, further from the nucleus. On emission, the electron can move to a lower energy level, closer to the nucleus.	Track the direction of energy transfer. Absorbing radiation raises the electron's energy, so it can move to a level further from the nucleus. Emitting radiation lowers its energy, so it can move to a level closer to the nucleus.	2
Total Question 1			2
02.1	The nucleus radius must be less than $84/10\,000=0.0084\,\mathrm{mm}$ . The displayed nucleus is too large. $0.42/0.0084=50$ It is more than $50$ times the permitted radius. The model therefore exaggerates the fraction of the atom occupied by the nucleus.	Apply the stated radius limit to the $84\,\mathrm{mm}$ atom: $84\div10\,000=0.0084\,\mathrm{mm}$ . The real scale requires a value below this upper limit. Since $0.42/0.0084=50$ , the displayed nucleus is more than $50$ times too large.	5
Total Question 2			5
03.1	There are $7+8=15$ particles in the nucleus and $22$ listed particles in total. The particle fraction in the nucleus is $15/22=0.682$ , or about $68\%$ . The nucleus has relative mass $7+8=15$ and the electrons add negligible mass. The estimated fraction of the atom's mass in the nucleus is therefore about $100\%$ . Particle number and mass do not determine occupied volume: the nucleus is tiny and the electron energy levels extend through a much larger, mostly empty region.	Count nuclear particles separately from all listed particles: $15/22=0.6818\ldots$ . For mass, the protons and neutrons contribute approximately all $15$ relative mass units because the seven electron masses are negligible. Finish by separating mass concentration from physical size.	5
Total Question 3			5
04.1	Each atom has diameter $2(1.2\times10^{-10})=2.4\times10^{-10}\,\mathrm{m}$ . Row length $=(2.5\times10^{6})(2.4\times10^{-10})=6.0\times10^{-4}\,\mathrm{m}$ . Each nucleus has radius less than $1.2\times10^{-14}\,\mathrm{m}$ , so each nuclear diameter is less than $2.4\times10^{-14}\,\mathrm{m}$ . The combined nuclear-diameter length is less than $(2.5\times10^{6})(2.4\times10^{-14})=6.0\times10^{-8}\,\mathrm{m}$ .	Double the atomic radius before multiplying by the number of touching atoms. Apply the less-than- $1/10\,000$ scale to the nuclear radius, double again for nuclear diameter, and multiply by the same number of atoms. Keep the strict inequality because the specification gives an upper limit, not an equality.	4
Total Question 4			4
05.1	The second level has energy $3$ units. After absorption, its energy is $3+9=12$ units, so it moves to the fourth level. After emission, its energy is $12-5=7$ units, so it moves to the third level. The electron finishes $7-3=4$ energy units above its starting energy. It has gained energy overall because it absorbed more energy than it emitted.	Use the supplied level energies as an energy ledger. Add the absorbed radiation energy, match the result to a higher level, then subtract the emitted energy and match again. Compare the final and starting values to determine the overall transfer.	5
Total Question 5			5

4.4.1.2 · Mass number, atomic number and isotopes

Tier 1 · Easy

Mark scheme for 4.4.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	They are isotopes of the same element. Element identity is fixed by proton number, which is the same in both atoms.	Use proton number to identify the element. Equal proton numbers mean the atoms are the same element; different neutron numbers make them isotopes.	2
Total Question 1			2
02.1	The electron entry is incorrect. A $2+$ ion has lost two electrons. It should have $12-2=10$ electrons.	Positive ion charge records missing electrons, not extra electrons. Starting from 12 electrons in the neutral atom, subtract two to obtain 10.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.4.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$13$ protons $14$ neutrons $10$ electrons	The lower number gives $13$ protons. The neutron number is $27-13=14$ . A $3+$ ion has lost three electrons, so its electron number is $13-3=10$ .	3
Total Question 1			3
02.1	$X$ is an atom and $Y$ is an ion of another isotope of the same element. They have the same 17 protons. They have different neutron numbers, 18 and 20. $Y$ has one more electron than proton, so its charge is $1-$ .	Use proton number to decide element identity and neutron number to decide whether the particles are isotopes. Then compare $Y$ 's 18 negative electrons with its 17 positive protons to obtain one net negative charge.	4
Total Question 2			4
03.1	Each ion has $50/5=10$ electrons. A $2+$ ion has two more protons than electrons. The atomic number is $12$ . The neutron number is $26-12=14$ .	First undo the aggregation to obtain 10 electrons per ion. A $2+$ charge means two electrons are missing, so the proton number is $10+2=12$ . Subtract this atomic number from the mass number to obtain $14$ neutrons.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Atomic number $=29$ The mass- $63$ isotope has $34$ neutrons. The mass- $65$ isotope has $36$ neutrons. Both have $29$ protons, so they are the same element. Their different neutron numbers make them isotopes.	A $2+$ ion has two fewer electrons than protons, so the proton number is $27+2=29$ . Hence the atomic number is $29$ . The neutron numbers are $63-29=34$ and $65-29=36$ . Element identity depends on proton number, which is unchanged between the two atoms; only the neutron number differs.	5
Total Question 1			5
02.1	The proton-number entry must change to describe a different element because proton number fixes element identity. The neutron-number entry must change, while the proton number remains 27, to describe another isotope of the same element. Changing only the electron-number entry would change the ion's charge, not its element or isotope.	Classify what each given count controls. Proton number identifies the element, neutron number distinguishes isotopes of that element, and electron number determines the ion's charge.	3
Total Question 2			3
03.1	Each atom has $28/2=14$ protons, so both isotopes have atomic number $14$ . If $X$ has $n$ neutrons, $Y$ has $n+2$ , so $2n+2=32$ and $n=15$ . $X$ has mass number $14+15=29$ . $Y$ has mass number $14+17=31$ . The atoms are neutral, so they contain $14+14=28$ electrons in total.	Because the atoms are isotopes of the same element, split the 28 protons equally. Use the two-neutron difference with the total: $n+(n+2)=32$ . Add each neutron number to 14 for the two mass numbers, then use neutrality to match the total electron and proton counts.	5
Total Question 3			5
04.1	The first particle has mass number $17+20=37$ . It has one more electron than proton, so its full symbol is ${}^{37}_{17}\mathrm{Cl}^{-}$ . They are isotopes because they have the same proton number but different neutron numbers. The first particle is charged because its $18$ electrons outnumber its $17$ protons by one. ${}^{35}_{17}\mathrm{Cl}$ is neutral because no charge is shown, so it has equal numbers of protons and electrons.	Add protons and neutrons to obtain mass number $37$ , use the one excess electron to obtain charge $1-$ , and place both around chlorine's proton number $17$ . Compare the proton and neutron numbers with chlorine-35, then use electron balance to explain the charge difference.	5
Total Question 4			5
05.1	Let the number of boron-10 atoms be $x$ , so the number of boron-11 atoms is $50-x$ . $10x+11(50-x)=546$ $x=4$ , so there are $4$ boron-10 atoms and $46$ boron-11 atoms. Each neutral boron atom has $5$ electrons because its atomic number is $5$ . The sample contains $50\times5=250$ electrons.	Use the two possible mass numbers and the fixed total of $50$ atoms to form $10x+11(50-x)=546$ . Then use neutrality to equate the electron number of each atom to boron's proton number before scaling to the whole sample.	5
Total Question 5			5

4.4.1.3 · The development of the model of the atom (common content with chemistry)

Tier 1 · Easy

Mark scheme for 4.4.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Most of the atom is empty space — equivalently, the nucleus occupies only a very small fraction of the atom's volume.	A particle travelling straight through has encountered very little matter. Because this happened to most alpha particles, most of each atom must be empty space.	1
Total Question 1			1
02.1	The observed large deflections did not match the prediction of only small deflections. They required positive charge and mass to be concentrated in a small nucleus rather than spread through the atom.	Compare prediction with observation. Diffuse positive charge cannot exert the large repulsive force needed to send an alpha particle backwards, so the evidence favours a concentrated nucleus.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Most particles passing straight through shows that most of an atom is empty space. Large deflections of a small fraction show that the positive charge is concentrated. The deflections also show that most mass is concentrated in a tiny central region. This evidence supports a small charged nucleus rather than spread-out positive charge.	Link each observation to one structural inference. Unchanged paths need very little matter in most of the atom. Rare, large changes of direction require a concentrated region capable of a strong interaction. Together these contradict diffuse positive charge and support the nuclear model.	4
Total Question 1			4
02.1	The choice of $A$ is not supported because four particles were deflected above $90^\circ$ , while $A$ predicts none. Those rare large deflections support the concentrated positive nucleus in $B$ . Their rarity is consistent with the nucleus occupying only a tiny fraction of the atom. The 19 960 undeflected particles show that most of the atom is empty space, which is also consistent with $B$ .	Compare the observation with each model's prediction rather than voting by the most common outcome. The four large deflections falsify $A$ 's zero prediction, while both the rare large paths and the common straight paths fit a tiny nucleus in a mostly empty atom.	4
Total Question 2			4
03.1	The later discovery was the neutron. Neutrons are neutral particles in the nucleus. They add nuclear mass without adding positive charge or changing the proton number. The model was revised so that the nucleus contains both protons and neutrons.	Use the two constraints together: the new particle must be in the nucleus to account for nuclear mass, but it must be uncharged so the proton number and positive charge can stay fixed. Chadwick's neutron evidence supplied that missing component.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The nuclear model explained scattering by concentrating positive charge and mass at the centre. It did not yet specify electrons at fixed distances or energy levels. Bohr adapted it by placing electrons at specific distances, supported by agreement between calculations and observations. Later evidence showed nuclear positive charge is made from whole units called protons. Chadwick's work supplied evidence for neutral particles in the nucleus. Therefore the model remained provisional and changed when new evidence improved its explanations.	Evaluate by separating what the first nuclear model explained from what later evidence added. It accounted for scattering, but Bohr's electron arrangement, the proton and the neutron were later refinements. The sequence shows that a scientific model can be useful without being complete and is revised when evidence supports a better description.	6
Total Question 1			6
02.1	Both runs contradict model $A$ because it predicts zero large-angle deflections but non-zero counts were observed. The similar counts of 8 and 7 show that the result is repeatable. Repeatability makes a one-off anomaly or chance observation less plausible. A tiny concentrated positive nucleus can exert the large repulsive force needed for a large deflection. The small counts are consistent with model $B$ because only the rare alpha particles passing close to the tiny nucleus are deflected through large angles.	Compare the repeated observations directly with the models' predictions. Two similar non-zero results reject a zero prediction more strongly than a single run, while their rarity fits the small target presented by a tiny nucleus and the large deflections follow from electrostatic repulsion.	5
Total Question 2			5
03.1	Statement (1) is wrong because most alpha particles passed straight through the foil. The few large-angle deflections were what the plum pudding model could not explain. Statement (2) is wrong: Bohr's model was accepted because his calculations agreed with observations. Statement (3) is wrong: the neutron was discovered after the nuclear model. Chadwick's neutron discovery came about 20 years after the nuclear model was proposed.	Audit each statement against the evidence and chronology. Most alpha particles passed through, while the rare large-angle deflections contradicted diffuse positive charge. Bohr's calculations matched observations, and Chadwick discovered the neutron about two decades after the nuclear model was proposed.	5
Total Question 3			5
04.1	There should be about twice as many large-angle deflections. The foil is twice as thick, so the alpha particles pass through about twice as many nuclei. The large majority should still pass through without being deflected because nuclei occupy only a tiny fraction of the foil's volume. This supports the conclusion that an atom is mostly empty space.	Doubling the thickness approximately doubles the number of nuclei encountered and therefore the chance of a close approach that causes a large-angle deflection. Even so, nuclei fill such a small fraction of the foil that most alpha particles should remain undeflected, showing that atoms are mostly empty space.	4
Total Question 4			4

4.4.2.1 · Radioactive decay and nuclear radiation

Tier 1 · Easy

Mark scheme for 4.4.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The unit is the becquerel ( $\mathrm{Bq}$ ). $1\,\mathrm{Bq}$ means one nuclear decay per second.	Activity is the rate at which unstable nuclei decay. Its unit is the becquerel, and one becquerel is one decay each second.	2
Total Question 1			2
02.1	A beta particle is a fast electron emitted from the nucleus. It is produced when a neutron changes into a proton. Gamma is the least ionising of alpha, beta and gamma. Alpha is the most strongly ionising of the three.	Correct composition and property separately. Beta is created in a nuclear change rather than removed from an electron shell. The ionising order is alpha, beta, gamma from greatest to least.	4
Total Question 2			4

Tier 2 · Standard

Mark scheme for 4.4.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$P$ is alpha. $Q$ is beta. $R$ is gamma. Alpha, $P$ , has the greatest ionising power.	Order the radiations by penetration. Card stopping the least penetrating identifies alpha; aluminium stopping the intermediate radiation identifies beta; substantial transmission until lead identifies gamma. Ionising power follows the reverse order, so alpha is greatest.	4
Total Question 1			4
02.1	The source does not emit a detectable alpha component because paper causes almost no decrease. The large decrease from $634$ to $260$ shows that beta is present and is stopped by aluminium. The reading of $260$ remains well above the background of $18$ , so a penetrating component remains after aluminium. The fall to near background behind lead identifies that component as gamma; the source emits beta and gamma.	Compare each reading with both the previous absorber and the $18$ -count background. Paper removes no substantial component, aluminium removes a large intermediate-penetration component, and lead removes the remaining penetrating component.	4
Total Question 2			4
03.1	Activity is the rate of nuclear decay in the source, so $2000\,\mathrm{Bq}$ means $2000$ decays per second. Count rate is the number of events recorded each second by the detector. The shielding absorbs much of the radiation before it reaches the detector, so the count rate falls from $480$ to $35$ counts per second. The shielding does not change how rapidly the unstable nuclei decay, so the source activity is unchanged during the stated test.	Assign each number to the correct system. Becquerels describe all decays occurring in the source; detector count rate describes only recorded events. Inserting an absorber changes transmission to the detector without changing the nuclei in the sealed source.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Alpha would be absorbed over a very short distance and is strongly ionising. Beta has an intermediate range and may not penetrate to the target. Gamma is much more penetrating, so it can reach the target. Gamma is the least ionising of the three, reducing ionisation along the path for a given exposure. Gamma is the most suitable of the three, with dose and shielding still controlled.	Test each option against both requirements. Alpha fails the penetration requirement and has high ionising power. Beta penetrates further but may still be absorbed before the target. Gamma best reaches a deep target and has lower ionising power, although exposure must still be managed because it passes through tissue.	5
Total Question 1			5
02.1	$X$ emits alpha radiation. $X$ falls from $486$ counts per minute at $1\,\mathrm{cm}$ to the $18$ -count background by about $4\,\mathrm{cm}$ . Alpha has a range of only a few centimetres in air. $Y$ emits beta radiation. $Y$ is still $96$ counts per minute at $30\,\mathrm{cm}$ , far above the $18$ -count background, so its range in air is much greater than $X$ 's; beta has a range of tens of centimetres in air.	Compare each series with the stated background as distance increases. $X$ becomes indistinguishable from background after only a few centimetres, matching alpha. $Y$ falls with distance but stays far above background at $30\,\mathrm{cm}$ , matching beta's much longer range in air.	5
Total Question 2			5
03.1	The gross count rate is $540/15=36$ counts per second. The corrected count rate is $36-4.0=32$ counts per second. $4.8\,\mathrm{kBq}=4800\,\mathrm{Bq}=4800$ decays per second. The recorded percentage is $(32/4800)\times100=0.67\%$ . Activity counts all nuclear decays in the source, whereas the detector records only radiation that reaches it and is detected.	Convert the 15-second total to a rate and subtract the background: $540/15-4.0=32\,\mathrm{s}^{-1}$ . Convert kilobecquerels to becquerels, then compare detected events with all decays: $32/4800\times100=0.666\ldots\%$ , giving $0.67\%$ .	5
Total Question 3			5
04.1	The corrected rate with no absorber is $1020-20=1000$ counts per minute. Behind paper the corrected rate is $420-20=400$ counts per minute, so the original gamma component is $400$ counts per minute. The alpha component is $1000-400=600$ counts per minute. The alpha fraction is $600/1000=0.60$ , or $60\%$ . Behind lead the corrected rate is only $60-20=40$ counts per minute because thick lead absorbs most, but not necessarily all, of the penetrating gamma radiation.	Subtract the same background from each gross reading. The paper reading isolates the gamma component under the stated assumption; subtract it from the unshielded corrected rate to obtain alpha. Compare the remaining lead count with both background and the initial gamma component.	5
Total Question 4			5
05.1	$3.2\,\mathrm{kBq}=3200$ decays per second. Expected number of alpha particles emitted $=3200\times25=80\,000$ . Each alpha particle contains two protons and two neutrons, so the emitted particles carry $160\,000$ protons and $160\,000$ neutrons. Radioactive decay is random, so $80\,000$ is an expected value and the number emitted in one actual $25\,\mathrm{s}$ interval may fluctuate around it.	Convert kilobecquerels to decays per second and multiply by the duration to obtain an expected decay count. One alpha particle is emitted per stated decay and contains two protons plus two neutrons. Keep the statistical prediction separate from the random number in a single interval.	4
Total Question 5			4

4.4.2.2 · Nuclear equations

Tier 1 · Easy

Mark scheme for 4.4.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Its mass number is unchanged. Its atomic number is unchanged.	Gamma radiation carries no mass number and no atomic number. Both totals therefore remain unchanged in a balanced nuclear equation.	2
Total Question 1			2
02.1	The daughter's atomic number should increase by 1, not decrease. The mass number is unchanged. The emitted beta particle has atomic number $-1$ , so the daughter must gain 1 in atomic number for the equation to balance.	Balance both number columns. A beta-minus particle contributes zero to mass number and $-1$ to atomic number, forcing the daughter to keep the parent's mass number and have an atomic number one greater.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.4.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$A=131$ $Z=54$	Mass number balances as $131=A+0$ , so $A=131$ . Atomic number balances as $53=Z+(-1)$ , so $Z=54$ . Beta-minus emission therefore leaves mass number unchanged and raises the daughter's atomic number by one.	2
Total Question 1			2
02.1	One alpha emission changes the mass number from $222$ to $218$ . It changes the atomic number from $86$ to $84$ . The stated final atomic number is $85$ , so alpha emission alone does not balance the equation. One beta-minus emission is also needed to raise atomic number from $84$ to $85$ without changing mass number.	Apply the alpha change first: $(222,86)\rightarrow(218,84)$ . Compare this intermediate nucleus with the claimed $(218,85)$ product. The remaining $+1$ atomic-number change is supplied by beta-minus emission.	4
Total Question 2			4
03.1	The first entry balances: $214=210+4$ and $84=82+2$ . The second entry balances: $210=210+0$ and $82=83+(-1)$ . The third entry has the correct mass number because $210=206+4$ , but its atomic numbers do not balance because $83\ne82+2$ . The daughter in the third entry must have atomic number $81$ , so it should be ${}^{206}_{81}D$ .	Audit mass number and atomic number separately for every row. The first two satisfy both conservation checks. In the alpha row, the daughter atomic number must be $83-2=81$ while its mass number remains $210-4=206$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	After alpha emission: mass number $236$ and atomic number $94$ . Each beta-minus emission leaves mass number unchanged. Two beta-minus emissions raise atomic number from $94$ to $96$ . Final nucleus: ${}^{236}_{96}N$ for any suitable daughter symbol $N$ .	Alpha emission changes $(A,Z)$ from $(240,96)$ to $(236,94)$ . Each beta-minus emission changes $Z$ by $+1$ without changing $A$ . Two beta emissions therefore give $(236,96)$ . The final element name is not needed; balanced numbers are sufficient.	4
Total Question 1			4
02.1	The mass number decreases by $226-218=8$ . Each alpha emission reduces mass number by 4, so two alpha emissions occurred. Those two alpha emissions reduce atomic number from $88$ to $84$ . The final atomic number is $85$ . One beta-minus emission raises $84$ to $85$ , so the chain contains two alpha emissions and one beta-minus emission.	Use mass number to fix the alpha count first: an 8-unit decrease requires two alphas. Their combined atomic-number change is $-4$ , giving 84. One beta-minus change of $+1$ then reaches 85 without affecting mass number.	5
Total Question 2			5
03.1	Alpha first gives ${}^{208}_{82}$ because mass number falls by 4 and atomic number by 2. The following beta-minus emission gives ${}^{208}_{83}$ . Beta-minus first gives ${}^{212}_{85}$ because mass number is unchanged and atomic number rises by 1. The following alpha emission also gives ${}^{208}_{83}$ . The claim is false: the changes $-4$ in mass number and $-2+1=-1$ in atomic number give the same final nucleus in either order.	Track both routes explicitly. Route $A$ is $(212,84)\rightarrow(208,82)\rightarrow(208,83)$ . Route $B$ is $(212,84)\rightarrow(212,85)\rightarrow(208,83)$ . The net changes are independent of the order.	5
Total Question 3			5
04.1	Beta-minus equation: ${}^{60}_{27}\mathrm{Co}\rightarrow{}^{60}_{28}\mathrm{Ni}+{}^{0}_{-1}\beta$ . The beta-minus equation balances because $60=60+0$ and $27=28+(-1)$ . Beta-minus emission leaves the mass number unchanged and raises the daughter's atomic number by one. Gamma emission changes neither mass number nor atomic number.	Balance the mass-number and atomic-number columns independently. A beta-minus particle contributes $(0,-1)$ , so the daughter must be ${}^{60}_{28}\mathrm{Ni}$ . Gamma emission removes energy without changing either nuclear number.	4
Total Question 4			4
05.1	Alpha decay reduces the mass number by $4$ , giving $226-4=222$ . It reduces the atomic number by $2$ , giving $88-2=86$ . The element with atomic number $86$ is radon, so the daughter is ${}^{222}_{86}\mathrm{Rn}$ . The balanced equation is ${}^{226}_{88}\mathrm{Ra}\rightarrow{}^{222}_{86}\mathrm{Rn}+{}^{4}_{2}\alpha$ . The daughter is a different element because it has a different proton number from radium.	Apply the alpha changes $A-4$ and $Z-2$ to radium-226, then use the supplied element list to identify atomic number $86$ as radon. Check that both mass number and atomic number balance across the equation.	5
Total Question 5			5

4.4.2.3 · Half-lives and the random nature of radioactive decay

Tier 1 · Easy

Mark scheme for 4.4.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Radioactive decay is random. Half-life predicts the behaviour of a large number of nuclei, not the decay time of one nucleus.	Distinguish an individual event from a population trend. A particular decay is unpredictable, while half-life describes the statistically predictable fall for a large sample.	2
Total Question 1			2
02.1	The decrease is not constant: the gross rate falls by 200 then by 100. Subtract 60 before testing for halving. The net rates are $400$ , $200$ and $100$ counts per minute, so they halve in each $5$ -minute interval. The half-life is $5$ minutes.	First test the student's constant-decrease claim against the gross intervals: the drops are $200$ and $100$ , not equal. Then subtract $60$ from every gross reading to obtain $400\rightarrow200\rightarrow100$ ; this repeated halving gives a $5$ -minute half-life.	4
Total Question 2			4

Tier 2 · Standard

Mark scheme for 4.4.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Initial net count rate $=400$ counts per minute. Final net count rate $=50$ counts per minute. $400\rightarrow200\rightarrow100\rightarrow50$ is three half-lives. Half-life $=6$ minutes	Remove background from both readings: $420-20=400$ and $70-20=50$ . The net count rate halves three times over $18$ minutes. Therefore the half-life is $18/3=6$ minutes.	4
Total Question 1			4
02.1	First corrected rate $=590-30=560$ counts per minute Second corrected rate $=310-30=280$ counts per minute The corrected rate has halved between the two interpolated points. Half-life $=8.5-2.5=6.0$ minutes	Correct both graph readings using the same background: $560$ and $280$ counts per minute. Because the second is exactly half the first, the elapsed $6.0$ minutes is one half-life.	4
Total Question 2			4
03.1	$12$ minutes is three half-lives. Let the background be $B$ : $(740-B)/8=110-B$ . $B=20$ counts per minute. Initial corrected rate $=740-20=720$ counts per minute.	Let the background be $B$ . Three half-lives require $(740-B)/8=110-B$ . Hence $740-B=880-8B$ , so $7B=140$ and $B=20\,\mathrm{counts\,min^{-1}}$ . The initial corrected rate is $720\,\mathrm{counts\,min^{-1}}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Initial net count rate $=800$ counts per minute. Final net count rate $=100$ counts per minute. $800\rightarrow400\rightarrow200\rightarrow100$ is three half-lives. Half-life $=12/3=4$ minutes. The decay of each individual nucleus is random. The number of decays recorded in equal short intervals therefore fluctuates around an expected value.	Subtract background to obtain $830-30=800$ and $130-30=100$ counts per minute. This is three halvings in $12$ minutes, so the half-life is $4$ minutes. Half-life describes the statistical behaviour of many nuclei; random individual decays make short repeat counts vary.	6
Total Question 1			6
02.1	Initial net rate $=970-10=960$ counts per minute $12/3.0=4$ half-lives Remaining-to-initial activity ratio $=1:16$ Final net rate $=960/16=60$ counts per minute Predicted detector reading $=60+10=70$ counts per minute	Remove background before applying decay. Four half-lives give $960\rightarrow480\rightarrow240\rightarrow120\rightarrow60$ , so the net activity is one sixteenth of its starting value. Restore the $10$ -count background to predict the gross reading.	5
Total Question 2			5
03.1	The number of undecayed nuclei falls by a factor of $(6.4\times10^{8})/(4.0\times10^{7})=16$ . A factor of $16=2^{4}$ means four half-lives have elapsed. $20/4=5.0$ , so the half-life is $5.0$ days. A further $15$ days is three more half-lives. $4.0\times10^{7}/8=5.0\times10^{6}$ undecayed nuclei remain.	Compare the initial and final populations: the decrease is by a factor of $16=2^4$ , so four half-lives fit into $20$ days. Divide to obtain $5.0$ days. A further $15$ days is $15/5.0=3$ more half-lives, so the population halves three more times: $4.0\times10^{7}/2^{3}=5.0\times10^{6}$ undecayed nuclei.	5
Total Question 3			5
04.1	The corrected rate after $18$ hours is $46-6=40$ counts per minute. $18/6.0=3$ half-lives have elapsed. Working backwards through three halvings gives an initial corrected rate of $40\times2^{3}=320$ counts per minute. The initial gross reading was $320+6=326$ counts per minute.	Remove background before reversing the decay. Three half-lives have elapsed, so double the final corrected rate three times. Restore the unchanged background only after recovering the source's initial corrected rate.	4
Total Question 4			4
05.1	The corrected count rate halves every $3$ minutes, so the half-life is $3$ minutes. At $12$ minutes the corrected rate is $40$ counts per minute, so the gross rate is $40+20=60$ counts per minute, still above $45$ . At $15$ minutes the corrected rate is $20$ counts per minute, so the gross rate is $20+20=40$ counts per minute. The gross count rate first falls below $45$ counts per minute at $15$ minutes. The gross rate can never fall below $20$ counts per minute because the background count continues even as the source count approaches zero.	Read the $3$ -minute halving interval directly from the corrected data. Extend the sequence to $40$ at $12$ minutes and $20$ at $15$ minutes, restoring the $20$ -count background each time. The background is a continuing detector contribution, so it sets the lower limit for the gross rate.	5
Total Question 5			5

4.4.2.4 · Radioactive contamination

Tier 1 · Easy

Mark scheme for 4.4.2.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The glove is contaminated because radioactive material is present on it. Removing the glove removes the contaminating atoms, which would otherwise continue to emit radiation.	Identify whether radioactive material was transferred. The liquid contains radioactive atoms, so moving the container is not enough; removing the contaminated glove takes the continuing source away from the body.	2
Total Question 1			2
02.1	Radioactive material had been transferred to the bench, so the bench was contaminated. Wiping removed the radioactive atoms; irradiation alone would not leave removable radioactive material behind.	Use the persistence and removability of the source. A raised reading after the sample has gone, followed by a fall after cleaning, is evidence of radioactive material on the surface.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Outside the body, alpha has a very short range and is stopped by skin. If inhaled, the source remains close to living cells. Alpha's high ionising power can then cause substantial cell damage. Prevent inhalation or spread, for example by using sealed containment and protective handling.	Keep the radiation properties fixed and change only source position. Skin shields an external alpha source, but inhalation removes that shielding and creates continuing internal exposure. A precaution should prevent transfer of the radioactive material, such as sealed containment.	4
Total Question 1			4
02.1	The $238$ reading well above background shows contamination remained on the bench. The fall to $27$ , close to the $24$ background, shows that wiping removed most of it. The wipe reading of $211$ shows that radioactive material was transferred onto the wipe. The wipe should be treated as radioactive waste and contained for safe disposal.	Compare each survey reading with the measured background. The high bench reading locates contamination, the cleaned reading tests whether removal worked, and the high wipe reading tracks where the radioactive material went.	4
Total Question 2			4
03.1	$A$ was irradiated because it was exposed without radioactive material being transferred. $B$ was contaminated because radioactive material remained on it. $A$ no longer exposes the handler once the external source is removed. $B$ can continue exposing the handler because the contaminating atoms keep decaying until they are removed or decay away.	Use persistence after source removal as the discriminator. Irradiation ends when the external source is absent; contamination leaves radioactive atoms on the object, so radiation continues to be emitted.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.2.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Other scientists can check how exposure and illness were measured. They can identify uncontrolled variables or alternative causes. They can check the analysis and whether the sample is sufficient. The study can be repeated or compared with independent evidence. A reproducible association supports a conclusion more strongly than one unreviewed result and still does not by itself prove causation.	Treat the reported association as evidence to test, not an automatic cause. Peer reviewers inspect the procedure, controls, sample and analysis. Publication also permits repetition. Agreement across well-controlled studies makes the conclusion more reliable; weaknesses or failure to reproduce it reduce confidence.	5
Total Question 1			5
02.1	The first worker is irradiated but not contaminated because the sealed source transfers no radioactive material. Gamma can penetrate the body and ionise cells. Their dose can be reduced using shielding, greater distance or shorter exposure time. The second worker is contaminated because radioactive powder is present on the sleeve. The powder could enter the body, where strongly ionising alpha radiation would damage nearby tissue. Removing and containing the sleeve, and preventing inhalation or ingestion, controls the continuing contamination hazard.	First decide whether radioactive atoms were transferred. Then match controls to the hazard: time, distance and shielding reduce irradiation from a sealed source, while containment and removal prevent an unsealed contaminant remaining on or entering the body.	6
Total Question 2			6
03.1	The imaging-staff illness percentage is $(12/200)\times100=6\%$ . The administrative-staff illness percentage is $(6/200)\times100=3\%$ . The study shows an association, but it does not establish that radiation caused the illnesses. Age and smoking are confounding variables that could produce some or all of the difference. Compare groups matched for age and smoking, or adjust the analysis for those variables. Use measured individual radiation doses and a larger sample, so that any dose-response pattern can be tested.	Convert each count to a percentage using the equal group size. Then separate correlation from causation: staff role changes alongside age and smoking. A stronger design controls those variables, records individual dose and uses a larger sample to test for a dose-response pattern.	6
Total Question 3			6
04.1	$28.4/7.1=4$ half-lives for nitrogen-16, so the activity from $A$ is $800/2^{4}=50\,\mathrm{Bq}$ . $28.4\,\mathrm{s}$ is far less than the $14.3$ -day half-life of phosphorus-32, so less than one half-life has passed and the activity from $B$ remains greater than $400\,\mathrm{Bq}$ . Both benches are contaminated because unwanted radioactive atoms from unsealed solutions are present on them; this is not irradiation alone. $B$ creates the more persistent hazard because its contaminating atoms decay much more slowly. Both spills should be removed using controlled decontamination, and the cleaning materials should be contained and handled as radioactive waste because the atoms are transferred rather than made non-radioactive by wiping.	Use whole half-lives for the short-lived contaminant and a bound for the long-lived one: before one half-life its activity must remain above half its initial value. Then distinguish decay from removal. Waiting reduces activity at different rates, whereas decontamination transfers the radioactive material into controlled waste.	5
Total Question 4			5

4.4.3.1 · Background radiation (physics only)

Tier 1 · Easy

Mark scheme for 4.4.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A valid source is cosmic rays, radioactive rocks or another source of background radiation. Background radiation is always present, so a non-zero count is expected without the test source.	Separate the test source from radiation already in the surroundings. Natural and man-made background sources continue to reach the detector when the test source is absent.	2
Total Question 1			2
02.1	The radiographer may be exposed to additional medical radiation, such as X-rays, at work. Medical exposure is a man-made source of background radiation.	Hold location constant and compare occupations. Working with medical radiation can add to a person's dose, and this exposure arises from human activity rather than a natural source.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Greater altitude can increase exposure to cosmic rays. Granite can contain more radioactive material than the other ground. Both locations receive real natural background radiation. Therefore non-zero readings without a test source are expected, not automatically detector error.	Use one location factor from above the site and one from below it: cosmic-ray exposure can rise with altitude, and local geology can change rock radiation. These sources exist without any test source, so a non-zero count is physically expected.	4
Total Question 1			4
02.1	Total of retained background readings $=24+22+23+21=90$ Mean background $=90/4=22.5$ counts per minute Corrected source rate $=183-22.5$ Corrected source rate $=160.5$ counts per minute	Exclude only the identified anomaly, then average the four retained readings. Subtract this local mean background from the gross source measurement: $183-22.5=160.5$ counts per minute.	4
Total Question 2			4
03.1	Man-made dose $=0.41+0.06=0.47\,\mathrm{mSv}$ . $(0.47/2.70)\times100=17\%$ . Radon is the largest natural contributor.	Add the medical and other man-made contributions, divide by the total dose and multiply by $100$ . Compare the listed natural contributions directly; radon is the largest.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Cave-guide dose $=0.288\,\mathrm{mSv}$ Airline-worker dose $=0.432\,\mathrm{mSv}$ The airline worker's dose is $0.144\,\mathrm{mSv}$ greater. $0.432\,\mathrm{mSv}=4.32\times10^{-4}\,\mathrm{Sv}$	Multiply weekly dose by $48$ : $0.006\times48=0.288\,\mathrm{mSv}$ and $0.009\times48=0.432\,\mathrm{mSv}$ . The difference is $0.432-0.288=0.144\,\mathrm{mSv}$ . Divide millisieverts by $1000$ to obtain $0.000432\,\mathrm{Sv}=4.32\times10^{-4}\,\mathrm{Sv}$ .	4
Total Question 1			4
02.1	Rock dose $=0.015\times36=0.54\,\mathrm{mSv}$ Flight dose $=0.84-0.54=0.30\,\mathrm{mSv}$ Flight time $=0.30/0.0050=60$ hours $0.84\,\mathrm{mSv}=0.84/1000\,\mathrm{Sv}$ Total annual dose $=8.4\times10^{-4}\,\mathrm{Sv}$	Find the known rock contribution before solving the missing flight duration. The remaining $0.30\,\mathrm{mSv}$ divided by the hourly dose gives $60$ hours. Convert millisieverts to sieverts by dividing by $1000$ .	5
Total Question 2			5
03.1	The mean at $P$ is $(18+21+19+22+20)/5=20$ counts per minute. The mean at $Q$ is $(21+23+19+22+20)/5=21$ counts per minute. The mean for $Q$ is only $1$ count per minute higher. The ranges overlap: $P$ spans $18$ to $22$ and $Q$ spans $19$ to $23$ , so random variation could account for the difference. The data do not show that $Q$ is definitely higher; longer counting intervals or more repeats are needed.	Find both means, then judge the size of their difference against the observed spread rather than using the means alone. Because radioactive counts fluctuate randomly and the ranges overlap substantially, the small difference is not decisive with these data.	5
Total Question 3			5
04.1	At $P$ , the corrected rate is $324-24=300$ counts per minute. At $Q$ , the corrected rate is $345-45=300$ counts per minute. The source produces the same corrected count rate at both sites. The higher gross reading at $Q$ comes from its higher background, which can be explained by greater cosmic-ray exposure at high altitude.	Treat each gross reading as source plus local background. Subtract the background measured at that site rather than using one common correction, then link the remaining gross difference to the location-dependent cosmic contribution.	4
Total Question 4			4
05.1	The short background measurement gives $(3/10)\times60=18$ counts per minute, while the longer one gives $105/5=21$ counts per minute. The gross source rate is $(90/10)\times60=540$ counts per minute. Using the longer background measurement, the corrected count rate is $540-21=519$ counts per minute. The longer measurement records more counts, so individual random fluctuations have less effect on the calculated rate. It therefore gives a better estimate of the mean background count rate.	Convert each observation to counts per minute before comparing them. Use the five-minute background rate, $21$ counts per minute, to correct the $540$ -count gross rate. Longer counting increases the total count and reduces the relative effect of random variation.	5
Total Question 5			5

4.4.3.2 · Different half-lives of radioactive isotopes (physics only)

Tier 1 · Easy

Mark scheme for 4.4.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Sample $P$ has the greater initial activity. With the shorter half-life, a larger fraction of its nuclei decay each second.	The samples begin with equal numbers of unstable nuclei, so compare their decay rates. The shorter-half-life sample loses nuclei more quickly and therefore has the greater activity initially.	2
Total Question 1			2
02.1	Risk also depends on the activity or amount of radioactive material present. It also depends on factors such as radiation type, route into the body, exposure time, shielding or containment.	Treat half-life as one hazard factor rather than a complete risk measure. A valid critique adds the initial decay rate or quantity and at least one exposure factor that changes the dose received.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Their initial hazards from activity and radiation type are similar under the stated conditions. $A$ loses activity much faster because its half-life is short. $A$ therefore becomes a much smaller hazard relatively soon. $B$ retains significant activity for many years and needs longer secure storage.	The given equal activity and radiation type make the initial comparison fair. Then apply half-life: many $3$ -hour halvings occur in a short storage time, but very little of a $40$ -year half-life passes. Thus $A$ declines rapidly while $B$ remains a persistent hazard.	4
Total Question 1			4
02.1	$P$ remains significant for the shortest time because $3.0\times10^2\,\mathrm{s}$ is only $300\,\mathrm{s}$ . $Q$ is intermediate because $8.0$ days is far longer than $300\,\mathrm{s}$ but far shorter than billions of years. $R$ remains significant for the longest time because its half-life is $4.5\times10^9$ years. $R$ dominates long-term disposal planning because its activity decreases over a timescale of billions of years, much more slowly than the other two isotopes.	Read each standard-form value with its unit before ranking the timescales: $300$ seconds, $8.0$ days and $4.5$ billion years. The longest half-life gives the slowest decrease in activity and therefore controls the longest-term storage decision.	4
Total Question 2			4
03.1	$X$ decays rapidly because its half-life is only $1$ day. Its rapidly decreasing contribution causes much of the steep initial fall. $Y$ changes by only a small fraction during the first week because its half-life is $20$ years, so the later decline is much slower. $Y$ controls the long-term storage problem because it remains radioactive for far longer.	Treat the measured total as the sum of two components with very different timescales. The short-lived component changes strongly at first and soon becomes much smaller; the long-lived component changes little over a week and therefore dominates persistence.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$C$ has the shorter half-life and so decays more rapidly initially for equal numbers of nuclei. Its activity and associated hazard also fall rapidly. $D$ decays more slowly initially. $D$ persists as a radioactive contaminant for much longer. Total risk also depends on radiation type, amount, exposure route and whether the material is removed or contained.	Equal numbers allow a qualitative decay-rate comparison: the population with the shorter half-life loses a larger fraction per second, so $C$ is initially more active but fades quickly. $D$ has the longer-lived waste hazard. Do not declare one universally more dangerous because ionising power, penetration, contamination route and control measures are not specified.	5
Total Question 1			5
02.1	$U$ needs greater immediate shielding because its initial activity is higher. $U$ 's activity falls relatively quickly because its half-life is short. $V$ needs longer secure storage because its activity decreases much more slowly. $V$ begins with the lower activity but remains a source for far longer. Overall risk also depends on amount, exposure route, distance, shielding and containment, so the samples trade immediate activity against persistence.	Use the measured starting activities for the immediate hazard and the half-lives for persistence. This separates the short-term control decision from the long-term storage decision and avoids treating one variable as a complete risk ranking.	5
Total Question 2			5
03.1	$A$ is unsuitable because $12$ years exceeds its $8$ -year half-life, so less than half its activity remains. $B$ is suitable because $12$ years is less than its $15$ -year half-life, so more than half remains. $C$ is also active enough because its half-life is $40$ years. $B$ creates the shorter long-term storage problem because it decays more quickly than $C$ after use. $B$ is therefore the best candidate: it meets the operating requirement without the unnecessary persistence of $C$ .	Use the definition of half-life as a threshold. A source retains more than half before one full half-life has elapsed. Both $B$ and $C$ pass the 12-year constraint, so apply the second criterion and choose the shorter of their half-lives.	5
Total Question 3			5
04.1	For $A$ , $8$ days is four half-lives, so its activity is $960/2^{4}=60\,\mathrm{Bq}$ . For $B$ , $8$ days is one half-life, so its activity is $120/2=60\,\mathrm{Bq}$ . $A$ presents the greater activity hazard initially because $960\,\mathrm{Bq}>120\,\mathrm{Bq}$ for the same radiation type. The activity hazards are equal after $8$ days because both activities are $60\,\mathrm{Bq}$ . After longer storage, $B$ presents the greater activity hazard because its longer half-life makes its activity decrease more slowly.	Count the number of half-lives separately for each sample over the same storage time. Use the calculated activities for the immediate comparison, then use the different decay rates to extend the comparison beyond the crossover time.	5
Total Question 4			5
05.1	Successive activities are $1600\rightarrow800\rightarrow400\rightarrow200\rightarrow100\,\mathrm{Bq}$ . Each step takes $3$ days. After $9$ days the activity is $200\,\mathrm{Bq}$ , which is still above $120\,\mathrm{Bq}$ . After $12$ days the activity is $100\,\mathrm{Bq}$ , so the earliest reopening time is $12$ days. A longer half-life would make each halving take longer, so the activity would take longer to fall below the threshold and reopening would be later.	Halve the activity in three-day steps until it is strictly below the threshold. Check the previous step to prove that the first acceptable value is $100\,\mathrm{Bq}$ at $12$ days, not $200\,\mathrm{Bq}$ at $9$ days.	5
Total Question 5			5

4.4.3.3 · Uses of nuclear radiation (physics only)

Tier 1 · Easy

Mark scheme for 4.4.3.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It must remain active long enough to be detected throughout the investigation. It should then decay quickly enough to limit the time the patient remains exposed to radiation.	Apply both timing requirements. Too short a half-life can make the tracer undetectable before the test ends; too long a half-life keeps the patient radioactive for longer than needed.	2
Total Question 1			2
02.1	Ionising gamma radiation damages or kills cells in the tumour. The beams overlap at the tumour, giving it the greatest total exposure. Different healthy regions lie in each beam path, so no one healthy region receives the full tumour exposure.	Follow where the radiation paths overlap. Each path contributes to the target dose, while rotating the direction spreads the unavoidable entry and exit exposure across different healthy tissue.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.4.3.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Ionisation can damage or kill tumour cells, controlling or destroying the unwanted tissue. Radiation reaching healthy tissue can damage healthy cells and may increase later health risks. A focused beam and controlled exposure can increase tumour dose while limiting healthy-tissue dose. Treatment is justified if the benefit of controlling the likely tumour growth outweighs the controlled risk to healthy tissue.	Link the intended effect and side effect to ionisation. Radiation can kill unwanted tumour cells but can also harm nearby healthy cells. Then use the supplied consequence of no treatment and a control measure, such as careful aiming and limited exposure, to reach a balanced judgement.	4
Total Question 1			4
02.1	$B$ is the best tracer. Its radiation penetrates the tissue sufficiently to reach an external detector. $A$ cannot be detected outside the body, so it fails the diagnostic requirement. $C$ remains radioactive for unnecessarily long, while $B$ reduces the continuing exposure after the investigation.	Infer penetration from the external detector and persistence from the activity observations. A useful tracer must satisfy both: it must be detectable during the test and should not remain a source for much longer than needed.	4
Total Question 2			4
03.1	The large rise and later fall at $P$ shows the tracer arriving and then moving on from region $P$ . The readings at $Q$ stay much lower than at $P$ . Under the stated equal-flow expectation, this suggests reduced tracer arrival at $Q$ , so a blockage or reduced blood flow may be present. Gamma radiation is penetrating enough to leave the body and reach the external detectors.	Compare the time profiles rather than a single reading. The pronounced peak at $P$ is evidence of tracer passage, while the consistently low $Q$ profile is evidence of reduced arrival under otherwise comparable conditions. Link external detection to gamma penetration.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.3.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$K$ is the best candidate. $K$ emits penetrating gamma radiation that can reach an external detector. $K$ remains active for the four-hour investigation. Its $6$ -hour half-life limits how long the patient remains a source after the test. $J$ 's alpha radiation is poorly penetrating and strongly ionising, and its long half-life prolongs contamination risk. $L$ is detectable but its $30$ -year half-life exposes the patient and creates a disposal hazard for unnecessarily long.	Apply three criteria: detectability, sufficient duration and rapid removal of hazard. Alpha from $J$ is unsuitable for external detection and is strongly ionising internally. Both $K$ and $L$ emit detectable gamma, but $K$ lasts long enough for four hours and then decays relatively quickly. $L$ persists for decades without adding diagnostic benefit.	6
Total Question 1			6
02.1	$P$ detects substantially more tumours than the non-radiation scan. $Q$ improves detection only slightly compared with $P$ , by 2 percentage points. $Q$ gives a much larger radiation dose than $P$ . $Q$ 's six-year half-life creates a much longer continuing radiation hazard than $P$ 's six-hour half-life. $P$ offers a large diagnostic benefit with the lower dose and shorter-lived source. $P$ is the better choice from the supplied evidence, while the medical benefit must still be judged against the ionisation risk for the patient.	Compare benefit and risk rather than selecting the highest detection percentage automatically. $P$ gains 29 percentage points over no radiation; $Q$ gains only 2 more than $P$ but carries a much higher dose and far longer persistence.	6
Total Question 2			6
03.1	For $R$ , the ratio is $60/24=2.5$ . For $S$ , the ratio is $54/9=6.0$ . $S$ still meets the minimum tumour dose because $54>50$ units. $S$ is preferred because it gives a much larger target-to-healthy-tissue ratio and a lower healthy-tissue dose. Healthy cells still receive ionising radiation, so treatment time and aiming must remain carefully controlled.	Calculate both benefit-to-harm ratios using the tumour dose as numerator. Do not choose on ratio alone: first verify that the $54$ -unit tumour dose meets the stated minimum. Then compare the unavoidable healthy-tissue exposure and retain a linked ionisation risk.	5
Total Question 3			5
04.1	$12$ hours is two half-lives. The activity falls $640\rightarrow320\rightarrow160\,\mathrm{MBq}$ , so it is $160\,\mathrm{MBq}$ after $12$ hours. Gamma radiation is penetrating enough to leave the body and be recorded by an external detector. The approximately $6$ -hour half-life allows detection during an investigation but makes the activity fall substantially afterwards, limiting the duration of exposure.	Apply two successive halvings to the stated activity. Then link the radiation type to external detection and the half-life to the competing timing requirements: useful activity during the test and a reduced continuing exposure afterwards.	4
Total Question 4			4
05.1	Placing the source close to the tumour concentrates the radiation dose in the unwanted tissue. Less healthy tissue lies in the radiation's path than for a beam sent through the body from outside. Placement and removal may require an invasive procedure. The patient carries a radioactive source during treatment, so exposure time and access must be controlled. A short-range emitter deposits its energy within the tumour while sparing more distant healthy tissue.	Compare the radiation path for a nearby internal source with an external beam, then retain the practical cost of putting radioactive material inside the patient. Link short range directly to local energy deposition and reduced dose farther from the tumour.	5
Total Question 5			5

4.4.4.1 · Nuclear fission (physics only)

Tier 1 · Easy

Mark scheme for 4.4.4.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A released neutron can be absorbed by another large unstable nucleus and make it split. That fission releases more neutrons, which can trigger further fissions.	Follow one emitted neutron into the next event. Its absorption causes another fission, and the additional neutrons released allow the sequence to continue.	2
Total Question 1			2
02.1	Control rods absorb neutrons. Withdrawing them causes fewer neutrons to be absorbed by the rods. More neutrons then cause further fissions, so reactor power increases rather than decreases.	Track the neutrons removed by the rods. Pulling absorbers out leaves more neutrons available to reach fuel, so more branches of the chain reaction continue.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.4.4.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Generation 2: $3$ fissions Generation 3: $9$ fissions Generation 4: $27$ fissions	Multiply by three in each generation because every fission supplies three successful neutrons: $1\times3=3$ , then $3\times3=9$ , then $9\times3=27$ .	3
Total Question 1			3
02.1	In $R$ , power decreases because fewer than one continuing fission follows each fission on average. In $S$ , power remains steady because one continuing fission replaces each fission on average. In $T$ , power increases because more than one continuing fission follows each fission on average. $S$ is the state for steady operation.	Compare each continuation number with one. Below one, successive generations shrink; at one, they stay the same size; above one, they grow.	4
Total Question 2			4
03.1	The smaller nuclei carry kinetic energy after the fission event. Their collisions transfer energy to surrounding material as thermal energy. Gamma radiation can also be absorbed and transfer energy, so the coolant is heated.	Follow the energy carriers rather than the chain reaction. The product nuclei transfer kinetic energy through collisions, while absorbed gamma radiation provides another route into the internal energy of the reactor material and coolant.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.4.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The diagram shows an incoming neutron absorbed by a large unstable nucleus. It shows that nucleus splitting into two smaller nuclei and emitting two neutrons. Each emitted neutron is shown travelling to a different unstable nucleus and causing another fission. A controlled reactor prevents excess neutrons from causing further fissions, for example by absorbing them. Keeping about one continuing neutron per fission prevents the reaction rate and energy release from multiplying.	Draw the process as a branching sequence: one neutron enters a large nucleus, the nucleus splits, and two outgoing neutron arrows lead to two further nuclei that split. Label the smaller nuclei and released neutrons. For control, explain that absorbing enough neutrons leaves about one successful neutron per fission, so successive generations do not grow.	5
Total Question 1			5
02.1	With one continuing neutron per fission, the next generation has $800$ fissions. After withdrawal, the first growing generation has $800\times2=1600$ fissions. The following generation has $1600\times2=3200$ fissions. The number of fissions now doubles in each generation. Energy released per generation and reactor power therefore increase, so more neutrons must be absorbed to restore control.	Use only the neutrons that successfully cause another fission. One successor keeps $800$ unchanged. After withdrawal, two successors per fission multiply the generations by two: $800\rightarrow1600\rightarrow3200$ .	5
Total Question 2			5
03.1	Number of fissions $=(1.6\times10^{6})/(3.2\times10^{-11})=5.0\times10^{16}$ . Use $P=E/t$ for the transfer to the coolant. $P=(1.6\times10^{6})/0.50=3.2\times10^{6}\,\mathrm{W}$ . On average, one neutron from each fission must cause one further fission. If more than one continues the chain the power rises, while fewer than one makes it fall.	Divide the total transferred energy by the energy per event to count events. Use $P=E/t$ with time in seconds. A steady chain reaction requires one successful continuing neutron per fission on average.	5
Total Question 3			5
04.1	Total neutrons released $=(2.5)(4.0\times10^{15})=1.0\times10^{16}$ . The stated continuing and control-rod totals account for $4.0\times10^{15}+3.2\times10^{15}=7.2\times10^{15}$ neutrons. The final category contains $1.0\times10^{16}-7.2\times10^{15}=2.8\times10^{15}$ neutrons. Exactly $4.0\times10^{15}$ neutrons cause further fission from $4.0\times10^{15}$ current fissions, so there is one continuing neutron per fission on average. Successive generations therefore contain the same number of fissions, keeping the reaction rate and reactor power steady.	Scale the mean neutron yield by the number of fissions, then complete the neutron budget by subtraction. Compare successful continuing neutrons with current fissions: equality gives one successor per event, the condition for a steady chain reaction.	5
Total Question 4			5
05.1	Total released energy $=(2.5\times10^{17})(3.2\times10^{-11})=8.0\times10^{6}\,\mathrm{J}$ . Electrical output $=0.35(8.0\times10^{6})=2.8\times10^{6}\,\mathrm{J}$ . Energy transferred by other routes $=8.0\times10^{6}-2.8\times10^{6}=5.2\times10^{6}\,\mathrm{J}$ . In a controlled chain reaction, about one neutron from each fission causes a further fission, so the fission rate and power remain steady. In an uncontrolled chain reaction, more than one neutron per fission causes further fission, so the number of fissions grows in each generation, as in a nuclear weapon.	Multiply the real approximate energy per uranium-235 fission by the event count. Apply the supplied energy fraction and subtract from the total. Connect steady output to a controlled neutron population rather than to the energy calculation alone.	5
Total Question 5			5

4.4.4.2 · Nuclear fusion (physics only)

Tier 1 · Easy

Mark scheme for 4.4.4.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The lost mass is converted into energy, which is released as radiation. Fusion needs a very high temperature — equivalently a very high pressure, or enough energy to overcome the electrostatic repulsion between the nuclei.	Fusion converts the mass difference between reactants and product into released energy. It only occurs when the nuclei approach closely enough to fuse, which requires the very high temperatures and pressures found in stars.	2
Total Question 1			2
02.1	The helium nucleus and neutron have less total mass than the deuterium and tritium nuclei. The lost mass is converted into energy released by the fusion event.	Compare the total masses on the two sides of the reaction. The products total $8.32\times10^{-27}\,\mathrm{kg}$ , which is $3.0\times10^{-29}\,\mathrm{kg}$ less than the $8.35\times10^{-27}\,\mathrm{kg}$ reactant total, so the mass difference provides evidence for released energy.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.4.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Fusion joins two light nuclei to form a heavier nucleus. Fission begins with one large unstable nucleus. Fission splits it into two smaller nuclei of roughly equal size. Both processes release energy.	Describe each process by its input and output pattern. Fusion changes two light nuclei into a heavier one; fission changes one large nucleus into two smaller ones. Finish with the shared outcome that energy is released.	4
Total Question 1			4
02.1	$S$ records fusion. It begins with two light nuclei joining together. It produces one nucleus heavier than either starting nucleus. The emitted radiation carries away energy released in the process; $R$ instead describes fission.	Identify the reaction from the change in nuclei rather than from radiation alone. Two light reactants becoming one heavier product is fusion; one large nucleus splitting is fission.	4
Total Question 2			4
03.1	$4.0\times10^{19}$ heavier nuclei formed, so there were $4.0\times10^{19}$ fusion events. The events used $(4.0\times10^{19})\times2=8.0\times10^{19}$ deuterium nuclei, consistent with $1.8\times10^{20}-1.0\times10^{20}=8.0\times10^{19}$ no longer remaining. The fraction that fused is $(8.0\times10^{19})/(1.8\times10^{20})=4/9$ . The observation is fusion because pairs of light deuterium nuclei joined to form heavier nuclei.	Count one event for each heavier product and use two deuterium nuclei per event. Cross-check the $8.0\times10^{19}$ used nuclei against the change in the chamber, then divide by the starting $1.8\times10^{20}$ . The direction from light nuclei to heavier products identifies fusion.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.4.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Two light nuclei are joining, which is the defining pattern of fusion. The joined product is a heavier nucleus than either reactant. The heavier nucleus having less mass shows that not all starting mass remains as nuclear mass. Some of the mass has been converted into energy. The emitted radiation carries energy away from the reaction.	Use the change in number and size of nuclei to identify fusion: two light reactants become one heavier nucleus. Then apply the specified mass-energy principle. The small reduction in total mass corresponds to energy released from the process, and the observed radiation is one way that energy leaves.	5
Total Question 1			5
02.1	Mass lost $=(6.70-6.65)\times10^{-27}\,\mathrm{kg}$ $=0.05\times10^{-27}\,\mathrm{kg}$ $=5.0\times10^{-29}\,\mathrm{kg}$ The product's lower mass shows that some starting mass has been converted into energy. That energy is released by the fusion event, for example as radiation or kinetic energy. The nuclei are positively charged and repel; very high temperature gives them enough kinetic energy to approach, while high pressure keeps many nuclei close and increases collision chance.	Subtract the product mass from the total starting mass and normalise the result: $0.05\times10^{-27}=5.0\times10^{-29}\,\mathrm{kg}$ . Link this mass decrease to energy release, then give only the brief conditions link: kinetic energy and frequent close approaches help positively charged nuclei overcome repulsion and fuse.	6
Total Question 2			6
03.1	Released energy $=(7.5\times10^{14})(2.8\times10^{-12})=2.1\times10^3\,\mathrm{J}$ . Net output $=2.1\times10^3-4.5\times10^3=-2.4\times10^3\,\mathrm{J}$ . The negative result means the pulse used $2.4\times10^3\,\mathrm{J}$ more than it released. Individual fusion events did release energy. The statement is false overall because released fusion energy is not the same as positive net useful energy after the input is included.	Multiply energy per event by the number of events, keeping the powers of ten separate: $7.5\times2.8\times10^2=2.1\times10^3\,\mathrm{J}$ . Subtract the supplied energy and interpret the sign rather than stopping at the fusion-release calculation.	5
Total Question 3			5
04.1	Energy factor $=(3.2\times10^{-11})/(2.8\times10^{-12})=11.4$ , so the fission event releases about $11$ times as much energy as the fusion event using the supplied approximate values. For equal power, the number of events per second is inversely proportional to the energy released per event. About $11$ times as many fusion events per second are therefore needed as fission events. Fusion joins two light nuclei, whereas fission splits one large nucleus.	Divide the fission energy by the fusion energy, keeping the power-of-ten factor. For fixed power, smaller energy per event requires a proportionally larger event rate. Identify fusion by joining two light nuclei and fission by splitting one large nucleus.	4
Total Question 4			4
05.1	Tritium limits the number of events, so the maximum is $4.0\times10^{18}$ fusion events. The events use $4.0\times10^{18}$ deuterium nuclei. Deuterium left $=6.0\times10^{18}-4.0\times10^{18}=2.0\times10^{18}$ nuclei. Energy released $=(4.0\times10^{18})(2.8\times10^{-12})=1.12\times10^{7}\,\mathrm{J}$ , or $1.1\times10^{7}\,\mathrm{J}$ to $2$ significant figures. It is fusion because two light nuclei, one deuterium and one tritium nucleus, join in each event to form a heavier helium nucleus; a neutron and energy are also released.	One event consumes one nucleus of each isotope, so the smaller starting population fixes the event count. Subtract the consumed deuterium, multiply the event count by the real approximate deuterium-tritium energy per event, and identify fusion from the joining of light nuclei.	5
Total Question 5			5