These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- Sellafield Nuclear Waste Management
Total: 12 marks
The Sellafield nuclear reprocessing plant in Cumbria, England, stores radioactive waste from decommissioned nuclear reactors. Different types of waste emit different radiation.
(a)[3]
State the three types of radiation emitted by radioactive materials. For each, give its symbol and state its relative charge.
Model Answer -- Part (a)
Alpha (α) radiation, charge +2
Beta (β) radiation, charge -1
Gamma (γ) radiation, charge 0 (neutral)
Award yourself marks:
⚠ If you missed marks here: The classic slip is mixing up the charges — α is +2 because it is 2 protons + 2 neutrons (a helium nucleus), β is −1 because it is an electron, and γ is 0 because it is an electromagnetic wave, not a particle. If you wrote β as +1 or forgot to state gamma's charge at all, that is where the marks went.
(b)[3]
High-level waste at Sellafield emits all three types of radiation. Explain what shielding material would be needed to stop each type of radiation.
Model Answer -- Part (b)
Alpha stopped by paper or a few centimetres of air
Beta stopped by a few millimetres of aluminium sheet
Gamma reduced by thick lead or concrete shielding
Award yourself marks:
⚠ If you missed marks here: Watch the wording on gamma — it is only reduced by thick lead or concrete, never completely "stopped", and saying aluminium blocks gamma loses the mark. Also make sure you matched each radiation to its own material: paper/air for α, a few mm of aluminium for β.
(c)[3]
Some waste at Sellafield contains strontium-90, which has a half-life of 29 years. The initial activity of a waste sample is 3200 Bq. Calculate the activity after 116 years.
Model Answer -- Part (c)
Number of half-lives = 116 ÷ 29 = 4 half-lives
Halving sequence: 3200 → 1600 → 800 → 400 → 200
Answer: activity = 200 Bq
Award yourself marks:
⚠ If you missed marks here: First find the number of half-lives: 116 ÷ 29 = 4 exactly. If you got 400 Bq you only halved three times; if you divided 3200 by 4 to get 800 Bq, remember each half-life halves the activity, so 4 half-lives means dividing by 24 = 16, giving 200 Bq. Always show the halving chain 3200 → 1600 → 800 → 400 → 200 for the working mark.
(d)[3]
Explain two safety precautions that workers at Sellafield must follow when handling radioactive waste.
Model Answer -- Part (d)
Use long-handled tongs or remote handling to increase distance from source, reducing radiation exposure
Wear protective clothing or lead aprons to shield from radiation
Limit time of exposure and wear dosimeter badges to monitor total radiation dose received
(Any two precautions for 2 marks + 1 mark for explaining why)
Award yourself marks:
⚠ If you missed marks here: The word "explain" is the trap — most students list two precautions (tongs, protective clothing) but never say why they work, which is the third mark. Link each one to reducing the radiation dose: tongs increase distance from the source, shielding absorbs radiation, limiting time cuts total exposure.
Question 2 -- Alpha Scattering at Cambridge Laboratory
Total: 11 marks
In 1909, Geiger and Marsden, working under Rutherford at the University of Manchester (later continued at Cambridge), fired alpha particles at a thin gold foil. Their results led to the nuclear model of the atom.
(a)[3]
Describe the experimental setup for the alpha scattering experiment. Refer to the diagram above in your answer.
Model Answer -- Part (a)
Alpha particles from a radioactive source are directed at a thin gold foil
A moveable detector (zinc sulfide screen) surrounds the foil to detect scattered alpha particles
Experiment is done in a vacuum to prevent alpha particles being absorbed by air
Award yourself marks:
⚠ If you missed marks here: The two most-forgotten points are the vacuum and the moveable detector. Alpha particles travel only a few cm in air, so without a vacuum they never reach the foil — say that, don't just write "in a vacuum". And the detector (zinc sulfide screen) must be able to move all around the foil to catch particles scattered at every angle.
(b)[4]
State the three main observations from the experiment and explain what each tells us about the structure of the atom.
Model Answer -- Part (b)
Most alpha particles pass straight through → most of the atom is empty space (1 mark for observation + 1 mark for deduction)
Some are deflected at small angles → the nucleus is positively charged and repels the positive alpha particles (1 mark)
Very few (about 1 in 8000) bounce back → the nucleus is very small, dense, and contains most of the mass (1 mark)
Award yourself marks:
⚠ If you missed marks here: This question pays for pairs: each observation must come with its deduction. Writing "most pass straight through" without adding "so the atom is mostly empty space" throws away half the marks. Same pattern for the other two: small deflections → nucleus is positive and repels the positive α; the rare bounce-backs (~1 in 8000) → nucleus is tiny, dense and holds most of the mass.
(c)[2]
Before this experiment, the accepted model was Thomson's "plum pudding" model. Explain why the results of the alpha scattering experiment disproved this model.
Model Answer -- Part (c)
In the plum pudding model, positive charge is spread evenly throughout the atom (not concentrated)
So alpha particles would not experience strong enough repulsion to bounce back / would all pass through with at most small deflections
Award yourself marks:
⚠ If you missed marks here: An incomplete answer just describes the plum pudding model. To disprove it you must make the link: because the positive charge is spread thinly through the whole atom, no single spot is concentrated enough to repel an α particle backwards — yet some did bounce back, so the model must be wrong.
(d)[2]
State two limitations of Rutherford's nuclear model that were later addressed by Bohr.
Model Answer -- Part (d)
Did not explain why electrons don't spiral into the nucleus / no explanation of electron stability in orbits
Did not explain discrete line spectra / energy levels of electrons
Award yourself marks:
⚠ If you missed marks here: Vague answers like "it didn't explain the electrons properly" score nothing. Name the two specific gaps: Rutherford's model could not say why orbiting electrons don't spiral into the nucleus (stability), and it could not explain the discrete line spectra, which needed Bohr's fixed energy levels.
Question 3 -- NHS Cancer Treatment with Radioisotopes
Total: 12 marks
The National Health Service (NHS) uses radioactive isotopes for both diagnosis and treatment of cancer at hospitals across the United Kingdom.
(a)[3]
Technetium-99m is used as a medical tracer for diagnosis. Explain three reasons why it is suitable for this purpose.
Model Answer -- Part (a)
Emits gamma radiation which can be detected outside the body by a gamma camera
Has a short half-life of 6 hours so it does not remain in the body long / reduces radiation dose to patient
'm' means metastable / decays to Tc-99 by gamma emission only / no alpha or beta damage to tissue
Award yourself marks:
⚠ If you missed marks here: Each reason needs its "because": "it emits gamma" only scores when you add that gamma penetrates out of the body so a camera can detect it, and "short half-life of 6 hours" only scores when you link it to a low radiation dose for the patient. The third point most students miss entirely: Tc-99m emits only gamma, so there is no ionising α or β damaging tissue inside the body.
(b)[3]
A patient is injected with 800 MBq of technetium-99m at 08:00. Calculate the activity at 20:00 the same day.
Model Answer -- Part (b)
Time elapsed = 20:00 - 08:00 = 12 hours
Number of half-lives = 12 ÷ 6 = 2 half-lives
Activity: 800 → 400 → 200 MBq
Award yourself marks:
⚠ If you missed marks here: Two half-lives fit in the 12 hours from 08:00 to 20:00 (12 ÷ 6 = 2), so 800 → 400 → 200 MBq. If you got 100 MBq you halved three times; if you got 400 MBq you only halved once. Writing "time elapsed = 12 hours" explicitly earns the first mark, so never skip that line.
(c)[3]
Cobalt-60 is used for external radiotherapy to treat deep tumours. Explain why cobalt-60 is more suitable than a source that emits alpha particles for this purpose.
Model Answer -- Part (c)
Cobalt-60 emits gamma radiation which has high penetrating power
Gamma can reach deep tumours inside the body
Alpha has very low penetrating power / stopped by skin / cannot reach the tumour
Award yourself marks:
⚠ If you missed marks here: Saying gamma is "stronger" or "safer" misses the physics — the key idea is penetration. Gamma passes through the body so it can reach a deep tumour from outside; α is stopped by the skin (or a few cm of air) so it would never arrive. You need both halves of the comparison for full marks, not just the gamma side.
(d)[3]
Cobalt-60 has a half-life of 5.27 years. A hospital receives a cobalt-60 source with an activity of 4000 kBq. Estimate the activity after approximately 16 years. Show your working.
Model Answer -- Part (d)
Number of half-lives ≈ 16 ÷ 5.27 ≈ 3 half-lives
Halving: 4000 → 2000 → 1000 → 500 kBq
Answer: approximately 500 kBq
Award yourself marks:
⚠ If you missed marks here: 16 ÷ 5.27 ≈ 3.04, so round to 3 half-lives — if you rounded up to 4 you got 250 kBq, and if you used 16 ÷ 5 ≈ 3.2 then halved 3.2 times you may have written an odd in-between number. This is an "estimate" question: 4000 → 2000 → 1000 → 500 kBq is exactly what the examiner wants.
Question 4 -- UK Nuclear Power Station
Total: 11 marks
Hinkley Point C in Somerset is the UK's newest nuclear power station, using uranium-235 as fuel. It is designed to provide 7% of the UK's electricity.
(a)[2]
Define nuclear fission.
Model Answer -- Part (a)
The splitting of a large / heavy nucleus
Into two (or more) smaller nuclei, releasing energy and neutrons
Award yourself marks:
⚠ If you missed marks here: Two common slips: saying an "atom" splits instead of a large/heavy nucleus, and stopping at "it splits into two smaller nuclei" without adding that energy and neutrons are released. Also be careful not to describe fusion (joining) — fission is splitting.
(b)[3]
Explain how a chain reaction occurs in a nuclear reactor.
Model Answer -- Part (b)
A neutron is absorbed by a uranium-235 nucleus causing it to split (fission)
The fission releases 2 or 3 additional neutrons
These neutrons go on to cause further fission reactions in other U-235 nuclei
Award yourself marks:
⚠ If you missed marks here: The most common gap is skipping the first step: a neutron must be absorbed by the U-235 nucleus before it splits — "the uranium splits and releases neutrons" on its own loses that mark. Then state the number (2 or 3 neutrons released) and close the loop: those neutrons go on to cause further fissions in other U-235 nuclei.
(c)[3]
Write a nuclear equation for one possible fission reaction of uranium-235. Ensure your equation balances for both mass number (A) and atomic number (Z).
Model Answer -- Part (c)
23592U + 10n → 14156Ba + 9236Kr + 310n
Check: A: 235 + 1 = 236 = 141 + 92 + 3(1) = 236 ✓
Check: Z: 92 + 0 = 92 = 56 + 36 + 0 = 92 ✓
Award yourself marks:
⚠ If you missed marks here: The usual errors are forgetting the incoming neutron on the left-hand side (so A totals 235 instead of 236) or not checking the balance at the end. Verify both: top numbers 235 + 1 = 141 + 92 + 3×1 = 236, bottom numbers 92 + 0 = 56 + 36 = 92. Any daughter pair works as long as A and Z balance — but you must show they do.
(d)[3]
Explain the role of control rods in a nuclear reactor. What material are they typically made from and why?
Model Answer -- Part (d)
Control rods absorb neutrons
This controls the rate of the chain reaction / prevents it from going too fast / can shut down reactor
Made from boron or cadmium because these materials are good neutron absorbers
Award yourself marks:
⚠ If you missed marks here: Don't confuse control rods with the moderator — the moderator (graphite/water) slows neutrons, control rods absorb them to control the rate of the chain reaction. Naming the material (boron or cadmium) is a separate mark, and you must say why that material: it is a good neutron absorber.
Question 5 -- Half-life from Experimental Data
Total: 12 marks
A physics student at an Oxford sixth-form college investigates the half-life of a radioactive source using a Geiger-Müller tube and counter.
The student records the following data:
Time / min
0
1
2
3
4
5
6
7
8
Count rate / cpm
420
340
270
220
180
150
130
115
105
Background count rate = 20 counts per minute.
(a)[2]
State what is meant by background radiation and give two sources.
Model Answer -- Part (a)
Radiation that is present in the environment at all times / always present
Sources include: cosmic rays, radon gas from rocks, medical X-rays, nuclear fallout, radioactive isotopes in food (any two)
Award yourself marks:
⚠ If you missed marks here: Background radiation is radiation that is always present in the environment — not "radiation left over from the source", which is a common wrong definition. Then name two specific sources (cosmic rays, radon gas from rocks, medical X-rays, food, fallout); vague answers like "the Sun" or "nature" are too loose to score.
(b)[2]
The background count rate is 20 counts per minute. Explain why the background must be subtracted from the readings.
Model Answer -- Part (b)
Background radiation contributes to the measured count rate / detector picks up background as well as source
Must subtract to find the count rate due to the source alone / to get an accurate value for half-life
Award yourself marks:
⚠ If you missed marks here: "To make it accurate" on its own is not enough — you need the mechanism: the detector records background radiation as well as the source, so the raw readings are 20 cpm too high. Subtracting 20 leaves the count rate due to the source alone, which is what the half-life must be measured from.
(c)[3]
Complete a corrected count rate table by subtracting the background count rate of 20 from each reading.
Model Answer -- Part (c)
Corrected count rates (subtract 20 from each):
Time / min
0
1
2
3
4
5
6
7
8
Corrected / cpm
400
320
250
200
160
130
110
95
85
Award yourself marks:
⚠ If you missed marks here: This is pure care: subtract exactly 20 from every one of the 9 readings (420 → 400 down to 105 → 85). Marks usually go on arithmetic slips near the end of the row (115 − 20 = 95, not 85) or on subtracting from some readings but not all. Double-check the last two values — that is where most errors hide.
(d)[3]
Using the corrected data, determine the half-life of the source. Show your method clearly.
Model Answer -- Part (d)
Corrected count rate halves from 400 to 200
This occurs from t = 0 to t = 3 minutes
Half-life = 3 minutes (can verify: 200 → 100 would be at approximately t = 6, and corrected value at t = 6 is 110, close given experimental uncertainty)
Award yourself marks:
⚠ If you missed marks here: The trap is using the raw data: looking for 420 → 210 gives a wrong half-life because background never halves. Use the corrected values — 400 falls to 200 between t = 0 and t = 3 min, so the half-life is 3 minutes. State the two count values, the two times, and the final answer as separate lines to secure all three marks.
(e)[2]
Predict the corrected count rate after 12 minutes (i.e., 4 half-lives from the start).
Corrected count rate = 25 cpm (or total count rate = 25 + 20 = 45 cpm including background)
Award yourself marks:
⚠ If you missed marks here: Start the halving from the corrected 400, not the raw 420 (which would give a messy 26.25). Four half-lives means halving four times: 400 → 200 → 100 → 50 → 25 cpm — if you got 50 you stopped one halving short. The detector would actually read 25 + 20 = 45 cpm because background is still there, and saying so shows real understanding.
Question 6 -- Industrial Thickness Monitoring
Total: 10 marks
A paper mill in Yorkshire uses radioactive sources to monitor the thickness of paper during production. A radioactive source is placed on one side of the paper and a detector on the other.
(a)[3]
Explain why a beta-emitting source is used rather than an alpha or gamma source for monitoring paper thickness.
Model Answer -- Part (a)
Alpha would be stopped by even thin paper / no radiation reaches detector / cannot distinguish thickness changes
Gamma would pass through paper regardless of thickness / detector reading would barely change
Beta is partially absorbed by paper / amount transmitted depends on thickness / suitable for detecting changes
Award yourself marks:
⚠ If you missed marks here: This is a three-way elimination and each option is a separate mark — only saying "beta is partially absorbed" scores 1 of 3. You must also rule out the other two: α would be completely stopped by even thin paper (detector reads nothing), and γ would sail through regardless of thickness (detector reading barely changes), so neither can detect a thickness change.
(b)[2]
Describe what happens to the detector reading when the paper becomes: (i) too thick, (ii) too thin.
Model Answer -- Part (b)
(i) Too thick: count rate decreases / less radiation passes through the paper
(ii) Too thin: count rate increases / more radiation passes through the paper
Award yourself marks:
⚠ If you missed marks here: Students often reverse the logic under pressure — think "more paper, more absorption": thicker paper absorbs more β, so the count rate falls; thinner paper lets more through, so it rises. Give the direction of change for both (i) and (ii) with a reason, not just one.
(c)[2]
The source used has a half-life of 30 years. Explain why a source with a long half-life is chosen.
Model Answer -- Part (c)
Activity remains approximately constant over the working period / does not change significantly day to day
Source does not need frequent replacement / readings remain reliable and consistent over time
Award yourself marks:
⚠ If you missed marks here: "It lasts longer" is only half the story. The physics mark is that with a 30-year half-life the activity stays approximately constant during use — so a falling count rate must mean the paper thickness changed, not the source decaying. The practical mark is that the source rarely needs replacing.
(d)[3]
A different factory uses gamma radiation to check for cracks inside metal welds on steel pipes. Explain why gamma is suitable for this application but beta would not be.
Model Answer -- Part (d)
Gamma has high penetrating power / can pass through thick steel
A crack or void inside the weld allows more gamma through / shows up as a brighter area on detector or film
Beta would be stopped by the steel / cannot penetrate thick metal to detect internal defects
Award yourself marks:
⚠ If you missed marks here: "Gamma detects cracks" is not an explanation — give the mechanism: gamma penetrates thick steel, and where there is a crack or void there is less metal to absorb it, so more gamma reaches the detector or film, showing up as a brighter spot. Then finish the comparison: β is stopped by the steel entirely, so it could never reveal an internal defect.
Question 7 -- Radiocarbon Dating at the British Museum
Total: 12 marks
Archaeologists at the British Museum use radiocarbon dating to determine the age of organic artifacts. Carbon-14 (146C) is a radioactive isotope with a half-life of 5730 years. Living organisms maintain a constant ratio of C-14 to C-12 by exchanging carbon with the environment. When an organism dies, the C-14 decays without being replaced.
(a)[2]
State the type of decay that carbon-14 undergoes and write the decay equation.
Model Answer -- Part (a)
Beta decay (β−)
146C → 147N + 0-1e
Check: A: 14 = 14 + 0 ✓ Z: 6 = 7 + (-1) ✓
Award yourself marks:
⚠ If you missed marks here: In β− decay the nucleon number stays at 14 but the proton number goes up by one (6 → 7, carbon → nitrogen) because a neutron turns into a proton plus an electron. The usual errors are writing Z going down to 5, or giving the electron as 01e instead of 0−1e — check A and Z balance on both sides.
(b)[2]
Explain why living organisms have a constant proportion of carbon-14 but dead organisms do not.
Model Answer -- Part (b)
Living organisms constantly take in / exchange carbon with the environment (through food, photosynthesis, respiration)
When the organism dies, no more carbon is taken in but C-14 continues to decay / the ratio of C-14 to C-12 decreases over time
Award yourself marks:
⚠ If you missed marks here: Both halves are needed: living organisms constantly take in and exchange carbon with the environment (food, photosynthesis), which tops the C-14 back up; after death the intake stops but the decay does not, so the C-14 to C-12 ratio falls. Answers that only mention "C-14 decays" without the replacement idea score just one mark.
(c)[3]
A wooden beam from an Anglo-Saxon settlement is found to contain 1/8 of the original C-14 proportion. Calculate the age of the beam.
Model Answer -- Part (c)
1/8 = (1/2)3 so 3 half-lives have passed
Age = 3 × 5730
Age = 17 190 years
Award yourself marks:
⚠ If you missed marks here: 1/8 means three halvings (1 → 1/2 → 1/4 → 1/8), not eight — if you wrote 8 × 5730 = 45 840 years, that is the trap. Write "1/8 = (1/2)3, so 3 half-lives" as its own line, then 3 × 5730 = 17 190 years.
(d)[3]
Explain why carbon-14 dating is suitable for artifacts up to about 50 000 years old but not suitable for rocks that are millions of years old.
Model Answer -- Part (d)
After about 50 000 years (approximately 8-9 half-lives), the remaining C-14 is too little to measure accurately
The activity becomes indistinguishable from background radiation
For rocks millions of years old, need isotopes with much longer half-lives (e.g., uranium-238 with a 4.5 billion year half-life)
Award yourself marks:
⚠ If you missed marks here: "It's too old" is not physics — say why: after ~50 000 years (8–9 half-lives of 5730 years) less than 1/256 of the C-14 remains, too little to measure, and its tiny activity is swamped by (indistinguishable from) background radiation. The third mark needs the fix: rocks require an isotope with a far longer half-life, e.g. uranium-238 at 4.5 billion years.
(e)[2]
A linen wrapping from an Egyptian mummy at the British Museum has 1/4 of the original C-14 remaining. Calculate its age.
Model Answer -- Part (e)
1/4 = (1/2)2 so 2 half-lives have passed
Age = 2 × 5730 = 11 460 years
Award yourself marks:
⚠ If you missed marks here: 1/4 = (1/2)2, so two half-lives have passed — if you got 22 920 years you multiplied by 4 instead of 2. Age = 2 × 5730 = 11 460 years; show the "2 half-lives" step for the first mark even if the final answer feels obvious.
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Self-Assessment
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