Background radiation, hazards and uses of nuclear radiation (physics only) – GCSE Physics exam questions with mark scheme

1Grade 4–5FoundationTriple only5 marks

The pie chart shows the sources of the average yearly radiation dose received by a person living in the UK.

The average yearly dose is 2.7 mSv.

(a) Determine the percentage of the dose that comes from medical sources. [1]

(b) Give the number of one source in the pie chart that is man-made. [1]

(d) Write your answer to part (c) in sieverts. [1]

(a) 15(%)(1)
(b) 5 / medical or 6 / weapons fallout and nuclear industry accept either(1)
(c) 0.48 × 2.7 allow 48 ÷ 100 × 2.7; allow 1.296 seen(1)
(c) 1.3 (mSv) correct answer with no working gains 2 marks(1)
(d) 0.0013 Sv allow 1.3 × 10⁻³ Sv; allow ecf from (c) ÷ 1000; allow 0.001296 Sv(1)
Answer: (a) 15%. (b) 5 (medical) or 6 (fallout/nuclear industry). (c) 1.3 mSv. (d) 0.0013 Sv.
Examiner tip: To go from mSv to Sv you divide by 1000; students often multiply and get 1300 Sv, a fatal dose.

2Grade 4–5Triple only6 marks

The table shows the average yearly radiation dose received by three people.

Person	Yearly dose in mSv
office worker living in London	2.3
office worker living in Cornwall	6.9
airline pilot living in London	5.1

(a) Calculate how many times greater the dose of the office worker in Cornwall is than the dose of the office worker in London. [2]

(b) Much of the ground in Cornwall is made of granite.

Suggest why people living in Cornwall receive a greater radiation dose than people living in London. [2]

(a) 6.9 ÷ 2.3(1)
(a) 3 (times) allow 3.0; correct answer with no working gains 2 marks(1)
(b) granite (rock) contains radioactive elements / isotopes allow granite is radioactive(1)
(b) which release (more) radon gas (into homes / the air) allow which emit more (gamma) radiation(1)
(c) the pilot spends a lot of time at high altitude(1)
(c) where there is less atmosphere above to absorb cosmic rays, so the pilot is exposed to more cosmic rays allow cosmic radiation is stronger higher up(1)
Answer: (a) 3 times. (b) Granite contains radioactive isotopes that release radon gas. (c) Pilot spends time at high altitude where less atmosphere absorbs cosmic rays.
Examiner tip: For (c) you must link altitude to cosmic rays; "pilots fly near the Sun" or "planes are radioactive" gains nothing.

3Grade 4–5Triple only5 marks

Two radioactive sources, A and B, have the same activity at the start.

Source A has a half-life of 5 minutes.
Source B has a half-life of 5000 years.

(a) Calculate the number of half-lives of source A that pass in 1 hour. [1]

(b) Explain which source is the greater hazard 1 hour after the start. [2]

Explain why this waste must be stored safely for thousands of years. [2]

(a) 12 allow 60 ÷ 5 = 12(1)
(b) source B(1)
(b) the activity of A has fallen to a very small value (1/4096 of the start) but B has (almost) the same activity as at the start allow A has (almost) all decayed but B has hardly decayed; allow B will keep emitting radiation for a long time(1)
(c) the activity / amount of radiation emitted falls (very) slowly allow it stays radioactive for a long time(1)
(c) so the waste will remain hazardous / could irradiate or contaminate people or the environment for thousands of years allow it takes thousands of years for the activity to fall to a safe level(1)
Answer: (a) 12. (b) B: A's activity has dropped to about 1/4096 of its start value, B's is almost unchanged. (c) Activity falls very slowly, so the waste stays dangerous for thousands of years.
Examiner tip: A long half-life does not mean a source is more strongly ionising; the hazard is that it keeps emitting radiation for a long time.

4Grade 6–7Triple only6 marks

Iodine-131 is used to treat an overactive thyroid gland. A patient swallows a capsule containing iodine-131. The iodine is absorbed by the thyroid gland.

Iodine-131 emits beta and gamma radiation. It has a half-life of 8 days.

(a) Explain how the iodine-131 controls the overactive thyroid gland. [2]

(b) The activity of the iodine-131 in the patient is 400 MBq just after the capsule is swallowed.

Calculate the activity after 32 days. Assume none of the iodine leaves the patient's body. [2]

Explain why. [2]

(a) the radiation (from the iodine) is ionising / damages cells(1)
(a) so it destroys (some of the) unwanted thyroid tissue / cells, while little of the rest of the body is affected because the iodine is concentrated in the thyroid allow kills thyroid cells so the gland is less active(1)
(b) 32 ÷ 8 = 4 half-lives allow 400 → 200 → 100 → 50 → 25 shown(1)
(b) 25 MBq correct answer with no working gains 2 marks(1)
(c) (gamma) radiation from the iodine passes out of the patient's body, so the children would be irradiated allow the patient is a source of radiation; allow the children could be contaminated by radioactive iodine in the patient's saliva / sweat / urine(1)
(c) the radiation could damage (the DNA of) the children's cells / cause cancer / mutation allow children are more sensitive to radiation; ignore 'children could become radioactive'(1)
Answer: (a) Radiation ionises/damages thyroid cells, destroying some of the unwanted tissue. (b) 25 MBq. (c) Gamma radiation leaving the patient could irradiate the children and damage their cells.
Examiner tip: In (c) avoid saying the children would "become radioactive": being irradiated does not make something radioactive.

5Grade 6–7HigherTriple only6 marks

After a nuclear accident, a sample of soil was found to contain iodine-131 and caesium-137.

The graph shows how the activity of each isotope in the sample changed during the first 40 days.

(a) Use the graph to determine the half-life of iodine-131. Show on the graph how you got your answer. [2]

(b) Use the graph to determine the time when the activities of the two isotopes are equal. [1]

Explain why, one year after the accident, caesium-137 is a much greater hazard than iodine-131, even though iodine-131 had the greater activity at the start. [3]

(a) reads a value and its half from the iodine curve, e.g. 800 → 400 kBq allow any correct pair, e.g. 400 → 200 or 200 → 100(1)
(a) 8 days allow 7.5–8.5 days(1)
(b) 16 days allow 15–17 days(1)
(c) one year is about 46 half-lives of iodine-131, so its activity has become negligible / almost zero allow iodine has almost all decayed(1)
(c) one year is only a small fraction of the half-life of caesium-137, so its activity has hardly changed (still about 200 kBq) allow caesium has hardly decayed(1)
(c) so caesium will continue to emit radiation / the soil stays contaminated for many years (decades) allow land unsafe to farm or live on for a long time(1)
Answer: (a) 8 days. (b) about 16 days. (c) After 1 year iodine has gone through ~46 half-lives (activity negligible); caesium's activity is almost unchanged (~195 kBq) and stays high for decades.
Examiner tip: In (a) you must show the halving on the graph and use a starting value from the curve; quoting "8 days" from memory without working loses the method mark.

6Grade 4–5Triple only5 marks

The diagram shows how gamma radiation can be used to treat a tumour inside a patient's body.

Three beams of gamma radiation are aimed at the tumour from different directions.

(a) Explain why gamma radiation is used rather than alpha radiation. [2]

(b) Explain why three weak beams from different directions are used rather than one strong beam. [2]

(a) gamma can pass through the body / skin to reach the tumour allow gamma is (very) penetrating(1)
(a) alpha would be absorbed by the skin / would not reach the tumour allow alpha has a range of only a few cm in air / is stopped by skin(1)
(b) the tumour receives the radiation from all three beams, so it receives a large dose (enough to kill the cancer cells)(1)
(b) healthy tissue is only in the path of one beam, so it receives a smaller dose / less damage allow fewer healthy cells are killed(1)
(c) to reduce the radiation dose the radiographer receives (as they treat many patients each day) allow so they are not exposed to / irradiated by the gamma radiation; ignore 'so they don't get radiation poisoning'(1)
Answer: (a) Gamma passes through the body to reach the tumour; alpha is stopped by skin. (b) Tumour gets dose from all beams, each region of healthy tissue only from one. (c) To reduce the radiographer's own dose.
Examiner tip: In (b) you need both halves: what happens to the tumour and what happens to the healthy tissue.

7Grade 6–7HigherTriple only7 marks

A doctor recommends that a patient has a bone scan. A gamma-emitting tracer is injected and a camera outside the body detects the gamma radiation.

The table gives some information.

Information	Value
dose received from the bone scan	4.0 mSv
average yearly dose from background radiation in the UK	2.7 mSv
estimated risk of a fatal cancer per sievert of dose	1 in 20

(a) Calculate the number of years of average background radiation that give the same dose as the bone scan. Give your answer to 2 significant figures. [2]

(b) Calculate the estimated risk of a fatal cancer caused by the bone scan. Give your answer in the form '1 in ...'. [3]

Use your answers to evaluate the patient's decision. [2]

(a) 4.0 ÷ 2.7 allow 1.48 seen(1)
(a) 1.5 (years) correct answer with no working gains 2 marks(1)
(b) 4.0 mSv = 0.0040 Sv(1)
(b) risk = 0.0040 × 1/20 (= 0.0002) allow 0.0040 × 0.05; allow 0.02%(1)
(b) 1 in 5000 allow ecf for one error in unit conversion; correct answer with no working gains 3 marks(1)
(c) the risk is very small (1 in 5000) / the dose is about the same as 1.5 years of background radiation that the patient receives anyway(1)
(c) the benefit of diagnosing / treating a bone problem is (much) greater than the small risk, so the decision is not sensible allow a conclusion either way only if it is justified by weighing risk against benefit(1)
Answer: (a) 1.5 years. (b) 0.0040 Sv × 1/20 = 0.0002 → 1 in 5000. (c) Risk is tiny (≈1.5 years of background) compared with the benefit of a diagnosis, so refusing is not justified.
Examiner tip: Forgetting to convert 4.0 mSv to 0.0040 Sv before using a 'per sievert' risk gives 1 in 5, a common and costly slip.

8Grade 6–7Triple only6 marks

A doctor wants to check how well blood flows through a patient's kidneys. A radioactive tracer is injected into the patient's blood. A detector outside the body records the radiation as the tracer passes through the kidneys. The scan takes about 1 hour.

The table gives information about four radioactive isotopes.

Isotope	Radiation emitted	Half-life
W	alpha	6 hours
X	gamma	6 hours
Y	gamma	60 days
Z	beta	10 minutes

Evaluate the suitability of each isotope for use as the tracer. [6]

X is the most suitable isotope(1)
gamma can pass out of the body to reach the detector(1)
gamma is the least ionising, so it causes the least damage to the patient's cells(1)
alpha (W) cannot pass out of the body so would not be detected(1)
alpha (W) is strongly ionising so it would damage kidney / body cells(1)
beta (Z) is partly absorbed by the body and is more ionising than gamma(1)
a half-life of 10 minutes (Z) is too short: the activity would fall a lot before / during the 1 hour scan (to 1/64 after 1 hour)(1)
a half-life of 6 hours (X) is long enough for the scan to be completed with enough activity(1)
but short enough that the activity in the patient falls quickly afterwards, so the dose received is small(1)
a half-life of 60 days (Y) is too long: the patient would be irradiated for months / receive a large total dose(1)

Level	Marks	What it looks like
Level 3	5–6	Isotope X is chosen and all four isotopes are evaluated. Both the type of radiation (ability to leave the body and ionising effect) and the half-life (lasting the scan but not irradiating the patient for long) are used in a clear, logical argument.
Level 2	3–4	Isotope X is chosen with reasons, and at least two of the other isotopes are rejected with a reason. Both radiation type and half-life are considered, though some links are incomplete.
Level 1	1–2	Simple relevant statements, e.g. X chosen with one reason, or one isotope rejected with a reason. Little or no linking of ideas.
No relevant content	0	Nothing worthy of credit.

Answer: X: gamma escapes the body and is weakly ionising; a 6-hour half-life lasts the 1-hour scan but decays quickly afterwards. W (alpha) is not detected outside and damages cells; Z decays too quickly and beta is more ionising; Y would irradiate the patient for months.
Examiner tip: To reach Level 3 you must comment on both the type of radiation and the half-life of each isotope, not just pick X.

9Grade 8–9HigherTriple only6 marks

A technician works in an area of a nuclear power station where the radiation dose rate is 0.0080 mSv per hour.

The company sets a limit of 20 mSv per year on the total dose a worker may receive, including background radiation.

The technician also receives 2.4 mSv per year from background radiation.

(a) Calculate the maximum number of hours per year the technician can work in this area. [3]

(b) In one year the technician works 1800 hours in this area.

Calculate the total dose the technician receives in that year. Give your answer in sieverts to 2 significant figures. [2]

(c) Suggest one way the technician's dose could be reduced without reducing the number of hours worked in the area. [1]

(a) dose allowed from the work area = 20 − 2.4 = 17.6 (mSv)(1)
(a) hours = 17.6 ÷ 0.0080(1)
(a) 2200 (hours) 2500 hours (ignoring background) gains 1 mark only; correct answer with no working gains 3 marks(1)
(b) 1800 × 0.0080 + 2.4 = 16.8 (mSv) allow 14.4 + 2.4(1)
(b) 0.017 Sv allow 1.7 × 10⁻² Sv; allow ecf of their mSv value ÷ 1000 to 2 sf(1)
(c) use (more) shielding / lead screens / stand further from the source of radiation allow wear protective clothing that absorbs radiation; ignore 'wear gloves' or 'wear a mask'; do not accept 'wear a film badge' (it measures dose, it does not reduce it)(1)
Answer: (a) 2200 hours. (b) 16.8 mSv = 0.017 Sv. (c) Use shielding or increase distance from the source.
Examiner tip: Read the limit carefully: if it includes background radiation you must subtract 2.4 mSv before dividing, otherwise you get 2500 hours.

10Grade 1–3FoundationTriple only5 marks

Everybody is exposed to background radiation.

(a) What is meant by background radiation? [1]

(b) Some sources of background radiation are listed in the table. Tick (✓) one box in each row to show whether the source is natural or man-made. [2]

Source	Natural	Man-made
cosmic rays from space
fallout from nuclear weapons testing
rocks such as granite
a nuclear power station accident

(d) A radiation worker must not receive a dose of more than 0.020 in this unit in one year. Convert this dose into the matching unit with the prefix milli. [1]

(a) (low-level) radiation that is around us all the time allow radiation from natural and man-made sources in the environment; ignore radiation from the Sun(1)
(b) cosmic rays and rocks both ticked natural(1)
(b) weapons-testing fallout and nuclear accident both ticked man-made(1)
(c) sievert / Sv do not accept becquerel / Bq(1)
(d) 20 mSv allow 20 millisieverts; unit required; allow ecf: 20 m + the unit given in (c)(1)
Answer: (a) Radiation that is around us all the time. (b) Natural: cosmic rays, rocks. Man-made: weapons fallout, nuclear accident. (c) sievert (Sv). (d) 20 mSv.
Examiner tip: Becquerel is the unit of activity, not dose: dose is always measured in sieverts.

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