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 -- The Smoke Detector
Total: 12 marks
Most homes in Leeds have smoke detectors containing a tiny source of americium-241 (₀₂₄₁₉₅Am is written ²⁴¹Am with proton number 95). The americium ionises the air between two electrodes inside a small chamber, and a battery drives a tiny current across the gap.
(a)[3]
Explain how the detector senses smoke and triggers the alarm.
Model Answer -- 1(a)
The alpha particles ionise the air in the chamber, so a small (ionisation) current flows between the electrodes [1]
Smoke particles entering the chamber absorb the alpha particles / capture the ions [1]
The current falls, and this drop is detected electronically and sounds the alarm [1]
⚠ If you missed marks here: The alarm is triggered by a current DECREASING, not increasing — the smoke interrupts the steady ionisation current. All three steps are needed: ionisation current flows; smoke absorbs alphas/ions; current drops and is detected.
Mark 1 -- alpha ionises air giving small current (1 mark)
Mark 2 -- smoke absorbs alpha particles or ions (1 mark)
Mark 3 -- current falls and drop triggers alarm (1 mark)
(b)[3]
The americium-241 in a detector is manufactured from plutonium-241 (proton number 94), which decays to americium by beta-minus emission. Explain why the americium formed still has nucleon number 241, state its proton number, and state what an alpha particle (the radiation the americium itself emits) consists of.
Model Answer -- 1(b)
Beta decay does not change the nucleon number: a neutron becomes a proton, so the total count of nucleons stays 241 [1]
The proton number rises by one: 94 + 1 = 95 (americium) [1]
An alpha particle is 2 protons + 2 neutrons (a helium nucleus, ⁴₂He) [1]
⚠ If you missed marks here: In beta-minus decay a NEUTRON turns into a proton (ejecting the electron), so the nucleon total is untouched — 241 stays 241 — while the proton number climbs from 94 to 95. And an alpha is a helium NUCLEUS (2p + 2n), not a helium atom: no electrons.
Mark 1 -- nucleon number unchanged in beta decay explained (1 mark)
Mark 2 -- proton number 95 stated (1 mark)
Mark 3 -- alpha is 2 protons 2 neutrons helium nucleus (1 mark)
(c)[2]
Americium-241 has a half-life of 432 years. Explain why this makes it suitable for a smoke detector.
Model Answer -- 1(c)
The activity stays (almost) constant over the lifetime of the detector (10–20 years is a tiny fraction of 432 years) [1]
So the ionisation current does not fade / the source never needs replacing [1]
⚠ If you missed marks here: Link the half-life to the DEVICE: 432 years means the activity barely changes over the detector’s ~10-year life, so the standing current stays steady and the alarm threshold stays valid. "Long half-life is safer" is the wrong direction — suitability here is about constancy, not safety.
Mark 1 -- activity nearly constant over detector lifetime (1 mark)
Mark 2 -- current does not fade so source never replaced (1 mark)
(d)[2]
Explain why an alpha source is chosen rather than a beta or gamma source.
Model Answer -- 1(d)
Alpha is the most strongly ionising radiation, so it produces a good ionisation current from a tiny source [1]
Beta and gamma are weakly ionising (and would pass straight through smoke and the chamber), so smoke would barely change the current [1]
⚠ If you missed marks here: Two sides needed: alpha ionises strongly (big effect from a tiny, safe source) AND beta/gamma ionise weakly so smoke would make almost no difference to their current. Ionising power, not penetrating power, is the working property here.
Mark 1 -- alpha most strongly ionising giving good current (1 mark)
Mark 2 -- beta gamma weakly ionising so smoke barely changes current (1 mark)
(e)[2]
A worried customer asks whether the americium is dangerous to her family. Use the properties of alpha radiation to explain why the detector is safe in normal use.
Model Answer -- 1(e)
Alpha particles have a very short range — they are stopped by a few centimetres of air / the plastic casing / the dead outer layer of skin [1]
So no radiation escapes the detector; it is only hazardous if the source were swallowed/inhaled (which the sealed casing prevents) [1]
⚠ If you missed marks here: The same property that makes alpha strongly ionising makes it easy to stop: a few cm of air or the casing absorbs it completely. A complete answer also names the real (avoided) hazard — alpha INSIDE the body — and why the sealed source prevents it.
Mark 1 -- alpha stopped by few cm air or casing or skin (1 mark)
Mark 2 -- no radiation escapes and sealed source prevents ingestion hazard (1 mark)
Question 2 -- Tracing the Fertiliser
Total: 11 marks
Scientists at Rothamsted Research in Hertfordshire tag a phosphate fertiliser with the radioactive isotope phosphorus-32 (³²P, proton number 15) to trace how quickly plants take up phosphorus. Ordinary, stable phosphorus is phosphorus-31.
(a)[2]
Define the term isotope.
Model Answer -- 2(a)
Atoms of the same element / same number of protons [1]
...with different numbers of neutrons (different nucleon numbers) [1]
⚠ If you missed marks here: Both halves are required: SAME proton number (that is what makes them the same element) but DIFFERENT neutron/nucleon number. "Same element, different mass" scrapes the idea but name the particles to be safe.
Mark 1 -- same element same proton number (1 mark)
Mark 2 -- different number of neutrons (1 mark)
(b)[3]
State the number of protons, neutrons and electrons in a neutral atom of phosphorus-32, and the number of neutrons in phosphorus-31.
Model Answer -- 2(b)
P-32: 15 protons, 15 electrons [1]
P-32 neutrons = 32 − 15 = 17 [1]
P-31 neutrons = 31 − 15 = 16 [1]
⚠ If you missed marks here: Protons = proton number (15); electrons = protons in a NEUTRAL atom (15); neutrons = nucleon number minus proton number (32 − 15 = 17, or 16 for P-31). The two isotopes differ by exactly one neutron — nothing else.
Mark 1 -- 15 protons and 15 electrons (1 mark)
Mark 2 -- 17 neutrons in P-32 (1 mark)
Mark 3 -- 16 neutrons in P-31 (1 mark)
(c)[4]
Phosphorus-32 decays by beta-minus emission to sulfur (S). State the nucleon number and proton number of the sulfur isotope formed, and explain how a nucleus that contains no electrons can emit a beta particle.
Model Answer -- 2(c)
Nucleon number of S = 32 (unchanged) [1]
Proton number of S = 15 + 1 = 16 [1]
In beta decay a NEUTRON in the nucleus changes into a proton and an electron [1]
The electron is created at that moment and immediately ejected as the beta particle (hence proton number rises by one while nucleon number is unchanged) [1]
⚠ If you missed marks here: Beta decay leaves the nucleon number ALONE (32) and raises the proton number by one (15 → 16). The beta particle is not an orbiting electron falling out: a neutron converts into a proton plus an electron, and that new electron is ejected. That conversion is exactly why the proton count rises.
Mark 1 -- sulfur nucleon number 32 unchanged (1 mark)
Mark 2 -- sulfur proton number 16 (1 mark)
Mark 3 -- neutron changes into proton plus electron (1 mark)
Mark 4 -- electron created and ejected as beta particle (1 mark)
(d)[2]
The detectors used in the field always register counts even with no tagged fertiliser nearby, because of background radiation. State two sources of background radiation.
Model Answer -- 2(d)
Any two of: radon gas from rocks/soil [1]
cosmic rays; rocks/building materials; food (e.g. potassium in it); medical sources (X-rays); nuclear weapons testing / industry fallout [1]
⚠ If you missed marks here: Background radiation is everywhere and mostly natural: radon from the ground (the largest share in the UK), cosmic rays, rocks and buildings, food, plus artificial contributions such as medical X-rays. Mobile phones and power lines are NOT ionising radiation and score nothing.
Mark 1 -- one valid source such as radon from rocks (1 mark)
Mark 2 -- second valid source such as cosmic rays or medical (1 mark)
Question 3 -- Finding the Leak
Total: 12 marks
A water company in Reading suspects a leak in a buried water main. Engineers add a small quantity of sodium-24 — a gamma-emitting isotope with a half-life of 15 hours — to the water, then walk the route with a detector: radiation builds up in the soil where water is escaping. The sample added has an initial activity of 640 MBq. The graph shows how the activity of sodium-24 changes with time.
(a)[2]
Use the graph to determine the half-life of sodium-24. Show your method.
Model Answer -- 3(a)
Read the time for the activity to fall from 640 MBq to 320 MBq (or any value to half that value) [1]
Half-life = 15 hours [1]
⚠ If you missed marks here: Pick ANY starting activity and read the time for it to halve — 640 → 320 gives 15 h, and 320 → 160 gives another 15 h (checking with a second pair is what examiners call good practice). If you read the time to reach zero, remember activity never reaches zero.
Mark 1 -- method shown reading time to halve from graph (1 mark)
Mark 2 -- half-life 15 hours (1 mark)
(b)[3]
The company assures customers that the radioactivity will be insignificant within a few days. Calculate the activity 60 hours after the sodium-24 is added.
⚠ If you missed marks here: Count half-lives first: 60 h ÷ 15 h = 4. Then halve four times (divide by 2⁴ = 16), giving 40 MBq. The classic error is dividing by 4 (two halvings) — write the halving chain out and it cannot go wrong.
Mark 1 -- 60 hours identified as 4 half-lives (1 mark)
Mark 2 -- halving chain or divide by 16 shown (1 mark)
Mark 3 -- activity 40 MBq calculated (1 mark)
(c)[3]
Explain why an isotope that emits gamma radiation with a fairly short half-life is a good choice for this leak test.
Model Answer -- 3(c)
Gamma is very penetrating, so it passes up through the soil and can be detected at the surface (alpha or beta would be absorbed by the ground) [1]
The build-up of activity in the wet soil pinpoints the leak without digging up the whole road [1]
The short half-life means the activity in the water supply falls to an insignificant level within days of the test [1]
⚠ If you missed marks here: Match each property to the job: penetration to GET OUT through a metre of soil (alpha and beta cannot), localisation of the leak by where counts peak, and a short half-life so the drinking water is quickly safe again. An alpha emitter would be undetectable from the surface AND linger in the supply if long-lived — wrong on both counts.
Mark 1 -- gamma escapes through soil detected at surface (1 mark)
Mark 2 -- activity build-up locates leak without excavation (1 mark)
Mark 3 -- short half-life so water safe within days (1 mark)
(d)[2]
Above the suspected leak the detector reads 470 counts per minute. Well away from the pipe it reads 20 counts per minute. Calculate the count rate due to the tracer alone, and state what causes the 20 counts per minute.
The 20 counts/min is background radiation (radon, cosmic rays, rocks etc.) [1]
⚠ If you missed marks here: Always SUBTRACT background before using a count rate: 470 − 20 = 450 counts/min from the tracer. The background never switches off — the detector counts radon, cosmic rays and the ground itself even far from the pipe.
Mark 1 -- corrected rate 450 counts per minute (1 mark)
Mark 2 -- 20 counts identified as background radiation (1 mark)
(e)[2]
A trainee checks the tracer’s half-life from detector readings but forgets to subtract the background. Explain why the half-life obtained is wrong, and in which direction.
Model Answer -- 3(e)
The measured counts are all too high by a constant amount, so the time for the READING to halve is not the time for the ACTIVITY to halve [1]
The apparent half-life comes out too long (the reading can never fall below the background level) [1]
⚠ If you missed marks here: Adding a constant to every reading stretches the apparent halving time: 470 falls to 235 only when the true tracer rate falls from 450 to 215, which takes MORE than one true half-life. The reading also flattens out at 20 and never halves again — a clean signature of forgotten background.
Mark 1 -- constant offset means reading halves later than activity (1 mark)
Mark 2 -- apparent half-life too long stated (1 mark)
Question 4 -- Identifying the Radiation
Total: 11 marks
A physics teacher in Bristol demonstrates three sealed radioactive sources. A Geiger counter 2 cm from source X reads 520 counts/min (background already subtracted). Absorbers are then placed between source X and the counter in turn:
absorber
count rate / counts per min
none
520
sheet of paper
518
3 mm aluminium
210
5 cm lead
2
(a)[3]
State the nature of each of the three types of nuclear radiation (alpha, beta, gamma).
Model Answer -- 4(a)
Alpha: a helium nucleus — 2 protons and 2 neutrons, charge +2 [1]
Beta: a (fast) electron emitted from the nucleus, charge −1 [1]
Gamma: a high-frequency electromagnetic wave, no charge and no mass [1]
⚠ If you missed marks here: Nature means WHAT IT IS: alpha = helium nucleus (2p+2n), beta = electron from the nucleus, gamma = electromagnetic wave. Charges +2, −1, 0 respectively. Calling gamma a "particle" or beta an "orbiting electron" loses the mark.
Mark 1 -- alpha helium nucleus charge plus 2 (1 mark)
Mark 2 -- beta electron from nucleus charge minus 1 (1 mark)
Mark 3 -- gamma electromagnetic wave uncharged (1 mark)
(b)[3]
The three radiations pass through a strong magnetic field. Compare how alpha, beta and gamma are deflected, and explain the differences.
Model Answer -- 4(b)
Alpha and beta are deflected in OPPOSITE directions (opposite charges) [1]
Beta is deflected much MORE than alpha (far smaller mass, despite smaller charge) [1]
Gamma is not deflected at all (uncharged) [1]
⚠ If you missed marks here: Three comparisons: opposite directions for alpha and beta (charges +2 and −1); beta bends far more because an electron is thousands of times lighter than a helium nucleus; gamma sails straight through because only charged particles feel a magnetic force.
Mark 1 -- alpha and beta deflect opposite ways (1 mark)
Mark 2 -- beta deflected much more due to small mass (1 mark)
Mark 3 -- gamma undeflected because uncharged (1 mark)
(c)[3]
Use the table to deduce which types of radiation source X emits. Justify your answer using the data.
Model Answer -- 4(c)
Paper makes (almost) no difference (520 → 518), so NO alpha is present [1]
3 mm aluminium cuts the count substantially (518 → 210), so BETA is present [1]
A count (210) remains after aluminium but is removed by thick lead (→ 2), so GAMMA is also present; X emits beta and gamma [1]
⚠ If you missed marks here: Work absorber by absorber: paper stops only alpha — no change, so no alpha. Aluminium stops beta — a big drop, so beta present. Whatever survives aluminium but dies in lead is gamma. Quote the numbers; "the count went down" without saying WHICH absorber caused WHICH drop earns little.
Mark 1 -- no alpha because paper changes nothing (1 mark)
Mark 2 -- beta present because aluminium halves count (1 mark)
Mark 3 -- gamma present because lead removes the rest (1 mark)
(d)[2]
State two precautions the teacher should take when handling the sources.
Model Answer -- 4(d)
Handle with tongs / keep at arm’s length; never point at anyone; minimise handling time [1]
Store in a lead-lined box when not in use; wash hands / no eating; keep class at a distance [1]
⚠ If you missed marks here: The three levers are always distance, time and shielding: tongs (distance), quick demonstrations (time), lead-lined storage (shielding). "Wear gloves" alone is weak — gloves stop contamination, not gamma.
Mark 1 -- one valid precaution such as tongs or distance (1 mark)
Mark 2 -- second valid precaution such as lead storage or short time (1 mark)
Question 5 -- Fission and Fusion
Total: 12 marks
In a nuclear reactor, a uranium-235 nucleus (proton number 92) absorbs a neutron and splits. One possible reaction produces barium-144 (proton number 56), an isotope of krypton (Kr, proton number 36), and three neutrons.
(a)[3]
Determine the nucleon number of the krypton isotope produced. Show how you used the conservation of nucleon number and proton number.
⚠ If you missed marks here: Do not forget the absorbed neutron on the LEFT (236 nucleons in total) and all THREE released neutrons on the right: 236 − 144 − 3 = 89. The proton check (56 + 36 = 92) costs nothing and catches most slips.
Mark 1 -- total nucleons 236 counted including absorbed neutron (1 mark)
Mark 2 -- krypton nucleon number 89 calculated (1 mark)
Mark 3 -- proton numbers checked 56 plus 36 equals 92 (1 mark)
(b)[3]
The three neutrons released can cause further fissions. Explain how this leads to a chain reaction, and why a minimum amount of uranium is needed to sustain it.
Model Answer -- 5(b)
Each released neutron may be absorbed by another U-235 nucleus, causing it to split and release more neutrons [1]
The number of fissions can grow (each fission triggers more than one new fission) — a chain reaction [1]
In a small piece too many neutrons escape through the surface before being absorbed, so the chain dies out; enough uranium (a critical amount) is needed for the reaction to sustain itself [1]
⚠ If you missed marks here: The chain reaction is fission → neutrons → more fissions → more neutrons. The minimum-mass idea is about neutron ESCAPE: in a small lump most neutrons leave through the surface without being absorbed, so on average less than one new fission follows each fission and the chain fizzles.
Mark 1 -- released neutrons cause further fissions (1 mark)
Mark 2 -- multiplying fissions described as chain reaction (1 mark)
Mark 3 -- small mass loses neutrons through surface so minimum needed (1 mark)
(c)[2]
Describe how the energy released by fission appears in the reactor, and how it is used to generate electricity.
Model Answer -- 5(c)
The energy is carried as kinetic energy of the fission fragments and neutrons, which heats the reactor core / coolant [1]
The heat boils water to steam, which drives a turbine that turns a generator [1]
⚠ If you missed marks here: The energy leaves the nucleus as KINETIC energy of the fast-moving fragments — which is heating, once they collide with surrounding atoms. From there it is the ordinary power-station chain: heat → steam → turbine → generator.
Mark 1 -- energy as kinetic energy of fragments heating core (1 mark)
Mark 2 -- steam turbine generator chain described (1 mark)
(d)[4]
Nuclear fusion is a different nuclear process. State what happens in fusion, explain why extremely high temperatures are needed for it to occur, and give one place where fusion happens naturally.
Model Answer -- 5(d)
Fusion: two light nuclei join/combine to form a heavier nucleus (releasing energy) [1]
Nuclei are all positively charged, so they repel each other (electrostatic repulsion) [1]
Only at extremely high temperature do the nuclei move fast enough / have enough kinetic energy to overcome the repulsion and get close enough to fuse [1]
Fusion occurs naturally in the Sun (and other stars) [1]
⚠ If you missed marks here: Fusion JOINS light nuclei (fission splits heavy ones — do not mix the verbs). The high temperature is not for "melting": both nuclei are positive and repel, so they must collide at enormous speed to touch. The Sun does this with hydrogen in its core.
Mark 1 -- light nuclei combine into heavier nucleus (1 mark)
Mark 2 -- positive nuclei repel electrostatically (1 mark)
Mark 3 -- high temperature gives speed to overcome repulsion (1 mark)
Mark 4 -- occurs naturally in the Sun or stars (1 mark)
Question 6 -- Sterilising Surgical Instruments
Total: 12 marks
A plant near Glasgow sterilises sealed packs of surgical instruments by passing them, on a conveyor, close to a powerful cobalt-60 gamma source. Cobalt-60 has a half-life of 5.3 years. A new source has an activity of 240 kBq per gram of cobalt.
(a)[2]
Explain why gamma radiation is suitable for sterilising instruments that are already sealed inside their plastic packs.
Model Answer -- 6(a)
Gamma is highly penetrating, so it passes through the packaging and reaches every surface of the instruments [1]
It kills the bacteria/microbes inside without the pack ever being opened (so the contents stay sterile until use) [1]
⚠ If you missed marks here: The whole commercial point is sterilising THROUGH the sealed pack: only gamma penetrates the plastic. Alpha or beta would be absorbed by the packaging and sterilise nothing.
Mark 1 -- gamma penetrates the sealed packaging (1 mark)
Mark 2 -- kills microbes inside so pack stays sterile (1 mark)
(b)[3]
The dose of radiation each pack receives depends on the source and on the conveyor speed. Explain why, as the cobalt-60 source ages, the conveyor must be run more slowly to sterilise the packs properly.
Model Answer -- 6(b)
The dose a pack receives depends on the number of gamma photons absorbed, which grows with EXPOSURE TIME [1]
As the source decays its activity falls, so it emits fewer gamma photons per second [1]
To deliver the same total dose from a weaker source, each pack must spend longer in the beam — hence a slower conveyor [1]
⚠ If you missed marks here: Dose = rate of exposure × time. The activity (and so the photon rate) falls as the source decays, so the time must rise to compensate — the conveyor slows in step with the decay curve. This is also why plants track their source’s age so carefully in part (c).
Mark 1 -- dose depends on photons absorbed over exposure time (1 mark)
Mark 2 -- ageing source emits fewer photons per second (1 mark)
Mark 3 -- slower conveyor restores same total dose (1 mark)
(c)[3]
Calculate the activity per gram of the cobalt-60 source 15.9 years after installation.
Model Answer -- 6(c)
15.9 / 5.3 = 3 half-lives [1]
240 → 120 → 60 → 30 [1]
Activity = 240 / 2³ = 30 kBq per gram [1]
⚠ If you missed marks here: 15.9 years is exactly 3 half-lives of 5.3 years. Halve three times: 240 → 120 → 60 → 30 kBq/g. Dividing 240 by 3 (=80) is the error the halving chain protects you from.
Mark 1 -- three half-lives identified (1 mark)
Mark 2 -- halving chain shown (1 mark)
Mark 3 -- activity 30 kBq per gram calculated (1 mark)
(d)[2]
Radioactive decay is described as random. State what random means here, and why the half-life is still a reliable quantity.
Model Answer -- 6(d)
It is impossible to predict WHEN any particular nucleus will decay (or which one will decay next) [1]
But with enormous numbers of nuclei the AVERAGE behaviour is predictable, so the time for half of them to decay is constant [1]
⚠ If you missed marks here: Random applies to the individual nucleus; half-life applies to the population. One coin flip is unpredictable, but half of a million coins reliably land heads — same statistics, which is why the half-life is dependable while a single decay is not.
Mark 1 -- cannot predict when a given nucleus decays (1 mark)
Mark 2 -- large numbers make average halving time constant (1 mark)
(e)[2]
When not in use, the cobalt-60 source is lowered into a deep pool of water, and the plant walls are thick concrete. Explain these safety features.
Model Answer -- 6(e)
Gamma is very penetrating, so thick/dense material (several metres of water, thick concrete) is needed to absorb it [1]
This shields workers outside — reducing their dose to a safe level while allowing safe storage between uses [1]
⚠ If you missed marks here: The shielding must match the radiation: a source that shines through plastic packs needs metres of water or thick concrete to stop. Note the logic pair with (a) — the property that makes gamma useful (penetration) is exactly what makes it hard to shield.
Mark 2 -- shielding protects workers reducing dose (1 mark)
Question 7 -- Radon in the Granite
Total: 10 marks
Parts of Cornwall are built on granite, which contains traces of radium-226 (proton number 88). Radium decays to radon-222 (Rn), a radioactive GAS, which can seep out of the ground and collect inside houses.
(a)[3]
Radium-226 decays to radon-222 by emitting one particle. Identify the particle emitted, and show that the nucleon and proton numbers are consistent with your answer.
Model Answer -- 7(a)
The particle is an alpha particle (⁴₂He) [1]
Nucleon numbers: 226 = 222 + 4 ✓ [1]
Proton numbers: 88 = 86 + 2, so radon has proton number 86 ✓ [1]
⚠ If you missed marks here: The nucleon number falls by 4 and the proton number by 2 — the fingerprint of ALPHA decay. Beta would leave the nucleon number unchanged. Show both conservation sums; the numbers are the evidence, not decoration.
Mark 1 -- alpha particle identified (1 mark)
Mark 2 -- nucleon numbers balance 226 equals 222 plus 4 (1 mark)
Radon is an alpha emitter. Explain why breathing in radon gas is far more dangerous than standing near an alpha source outside the body.
Model Answer -- 7(b)
Outside the body, alpha is stopped by air / the dead outer layer of skin, so it cannot reach living cells [1]
Inhaled radon decays INSIDE the lungs, where the alpha particles strike living tissue directly [1]
Alpha is the most strongly ionising radiation, so it does severe damage (can cause cancer) over its short range in the lung tissue [1]
⚠ If you missed marks here: Alpha’s short range cuts both ways: harmless at arm’s length (stopped by skin and air) but devastating when the SOURCE IS INSIDE you — all of that intense ionisation is delivered straight into living lung cells. This inside/outside reversal is the single most examined idea about alpha.
Mark 1 -- external alpha stopped by air or dead skin (1 mark)
Mark 2 -- inhaled radon decays inside lungs on living tissue (1 mark)
Mark 3 -- strong ionisation damages lung cells causing cancer risk (1 mark)
(c)[2]
A radon detector in a Cornish cellar records 47 counts per minute. The UK average background count rate is 12 counts per minute. Calculate the count rate due to radon in the cellar, assuming the other contributions to background match the UK average.
Model Answer -- 7(c)
47 − 12 = 35 counts/min due to radon [1]
(Assumption stated/used: the non-radon background in the cellar equals the 12 counts/min average) [1]
⚠ If you missed marks here: Subtract the ordinary background to isolate the radon: 47 − 12 = 35 counts/min. The stated assumption matters — if the cellar’s granite also raises the non-radon background, 35 is an overestimate of the radon alone.
Mark 1 -- radon count rate 35 per minute calculated (1 mark)
Mark 2 -- assumption about average background used or stated (1 mark)
(d)[2]
Suggest two measures a homeowner can take to reduce the radon level inside the house.
Model Answer -- 7(d)
Increase ventilation / fit a fan or radon sump that pumps air from below the house to the outside [1]
Seal cracks and gaps in floors / install an airtight membrane so the gas cannot seep in [1]
⚠ If you missed marks here: Radon is a GAS, so the fixes are gas fixes: stop it getting in (sealed floors, membrane) and remove what does get in (ventilation, under-floor sump and fan). Lead shielding is useless here — you cannot shield yourself from a gas you are breathing.
Mark 1 -- ventilation or radon sump suggested (1 mark)
Mark 2 -- sealing floors or membrane suggested (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
Tick all mark checkboxes you have earned, then click "Calculate Grade" below.