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International GCSE Physics · Pearson Edexcel 4PH1 · Topic 7
278 free, high-quality past-paper-style questions on Radioactivity and particles for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (b) Radioactivity; (c) Fission and Fusion. For when you have finished the past papers, or want to practise one topic at a time.Notes 7(b) →Notes 7(c) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A nucleus of sodium is represented as 2311Na.
How many neutrons are in this nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| The mass (nucleon) number 23 is the total number of protons plus neutrons. The atomic (proton) number is 11. Number of neutrons = 23 − 11 = 12. 11 is the number of protons, 23 is the total number of nucleons, and 34 comes from wrongly adding the two numbers. | ✓ 1 | Answer B |
Answer: B
Exam tip: The top number counts every particle in the nucleus. Subtract the bottom number to get the neutrons.
Neon gas used in advertising signs contains two main isotopes, neon-20 and neon-22. The atomic number of neon is 10.
State what is meant by the word isotopes.
An atom of neon is electrically neutral. Explain why.
State the number of protons and the number of neutrons in a nucleus of neon-22.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Atoms (nuclei) of the same element / with the same number of protons | ✓ 1 | |
| with different numbers of neutrons | ✓ 1 | |
| Part (b) | ||
| The number of electrons (10) equals the number of protons, so the negative and positive charges cancel | ✓ 1 | |
| Part (c) | ||
| Protons = 10 | ✓ 1 | |
| Neutrons = 22 − 10 = 12 | ✓ 1 | |
Common error: Isotopes have the same number of protons. Do not say they have the same mass number, because that is the thing that differs.
A research reactor produces medical isotopes by fission. One possible fission reaction is
23592U + 10n → 14054Xe + 9438Sr + x 10n
What is the value of x?
| Marking point | Mark | Notes |
|---|---|---|
| Nucleon numbers must balance. The left-hand side totals 235 + 1 = 236. The right-hand side totals 140 + 94 + x, so x = 236 − 234 = 2. Option 1 comes from forgetting the incoming neutron. Option 3 comes from assuming fission always releases three neutrons. Option 0 comes from balancing proton numbers instead: these already balance (92 = 54 + 38) because neutrons carry no charge. | ✓ 1 | Answer B |
Answer: B
Exam tip: Always add the absorbed neutron to the left-hand side before balancing nucleon numbers in a fission equation.
The crew of a nuclear submarine lower the control rods further into the reactor core. What happens?
| Marking point | Mark | Notes |
|---|---|---|
| Control rods absorb neutrons. Inserting them further removes more neutrons from the chain reaction, so fewer neutrons go on to cause fission. The rate of fission falls and the power output decreases. Slowing neutrons is the job of the moderator, not the control rods. Shielding the crew from gamma radiation is the job of the concrete and steel shielding. | ✓ 1 | Answer B |
Answer: B
Common error: Do not mix up the moderator (slows neutrons down) with the control rods (absorb neutrons).
An experimental fusion reactor heats a mixture of deuterium and tritium, two isotopes of hydrogen. The reaction is
21H + 31H → 42He + Z
Identify particle Z.
State two conditions needed inside the reactor for fusion to take place.
Suggest one reason why there are no fusion power stations supplying electricity yet.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A neutron (10n) . | ✓ 1 | |
| Part (b) | ||
| Very high temperature | ✓ 1 | |
| very high pressure / density | ✓ 1 | |
| Part (c) | ||
| Any one from: it is very difficult to reach and maintain the high temperatures and pressures needed / containing the hot plasma is difficult / more energy is currently put in than is obtained . | ✓ 1 | |
Exam tip: Check that both nucleon numbers and proton numbers balance: 2 + 3 = 4 + 1 and 1 + 1 = 2 + 0.
Many household smoke detectors contain a small source of americium-241, 24195Am, which decays by emitting an alpha particle.
Which of these is the nucleus formed by this decay?
| Marking point | Mark | Notes |
|---|---|---|
| An alpha particle is 42He, made of 2 protons and 2 neutrons. The mass number falls by 4 (241 − 4 = 237) and the atomic number falls by 2 (95 − 2 = 93). Atomic number 93 is neptunium, so the product is 23793Np. 24194Pu treats the change like a single proton loss. 23795Am forgets that the proton number changes. 23993Np takes only 2 from the mass number. | ✓ 1 | Answer A |
Answer: A
Exam tip: Check that both the top numbers and the bottom numbers balance on each side of a nuclear equation. If the atomic number changes, the element symbol must change too.
Iodine-131, 13153I, is used in hospitals to treat an overactive thyroid gland.
State the number of protons and the number of neutrons in a nucleus of iodine-131.
Iodine-127 is a stable form of iodine. Iodine-127 and iodine-131 are isotopes.
Explain what is meant by isotopes.
Iodine-131 decays by beta emission to form xenon (Xe).
Write a balanced nuclear equation for this decay.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 53 protons | ✓ 1 | |
| 78 neutrons (131 − 53) | ✓ 1 | |
| Part (b) | ||
| Atoms or nuclei with the same number of protons / the same atomic number / the same element | ✓ 1 | |
| but different numbers of neutrons / different mass numbers | ✓ 1 | |
| Part (c) | ||
| 13153I → 13154Xe + 0−1e | — | |
| Xenon shown with mass number 131 | ✓ 1 | |
| Xenon shown with atomic number 54 | ✓ 1 | |
| Beta particle shown correctly as 0−1e or 0−1β | ✓ 1 | |
Common error: In beta decay the mass number stays the same and the atomic number goes UP by 1, because a neutron changes into a proton. A common error is to subtract 1 instead.
A homeowner is worried that a new granite kitchen worktop is radioactive. A technician uses a Geiger–Müller tube and counter to investigate.
With no worktop nearby, the background count rate is 21 counts per minute.
With the tube held against the worktop, the counter records 290 counts in 5 minutes.
Name two sources of background radiation.
Explain why the technician counts for 5 minutes rather than for only 10 seconds.
Calculate the count rate due to the worktop alone.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any two from: radon gas (from the ground or buildings) / rocks or soil / cosmic rays (from space) / food and drink / medical uses such as X-rays / nuclear industry or fallout | ✓ 2 | |
| Part (b) | ||
| Radioactive decay is random, so the count varies from one moment to the next | ✓ 1 | |
| Counting for longer gives a more reliable or representative average count rate | ✓ 1 | |
| Part (c) | ||
| Measured count rate = 290 ÷ 5 = 58 counts per minute or background in 5 minutes = 21 × 5 = 105 counts | ✓ 1 | |
| Subtract the background: 58 − 21 or 290 − 105 = 185 counts in 5 minutes | ✓ 1 | |
| = 37 counts per minute | ✓ 1 | |
Common error: Put both readings on the same time basis before subtracting the background. Subtracting 21 from the 5-minute total of 290 gives a wrong answer.
Astronauts on a space station wear badges that record their exposure to ionising radiation. Several types of radiation can reach them, including alpha particles, beta particles, gamma rays and neutrons.
State what an alpha particle consists of.
Describe what a beta particle is and where it comes from.
State what gamma radiation is.
State which of alpha, beta and gamma radiation is the most strongly ionising.
Suggest why a beam of neutrons is not deflected as it passes between two charged metal plates.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2 protons and 2 neutrons / a helium nucleus | ✓ 1 | |
| Part (b) | ||
| A high-speed (fast-moving) electron | ✓ 1 | |
| emitted from the nucleus when a neutron changes into a proton | ✓ 1 | |
| Part (c) | ||
| High-frequency electromagnetic radiation (or wave) emitted from the nucleus | ✓ 1 | |
| Part (d) | ||
| Alpha | ✓ 1 | |
| Part (e) | ||
| Neutrons have no charge, so the electric field exerts no force on them | ✓ 1 | |
Common error: A beta particle does not come from the electron shells around the atom. It is created in the nucleus when a neutron changes into a proton.
Iodine-131 is used in hospitals to treat thyroid conditions. A nucleus of iodine-131 has a proton number of 53 and decays by emitting a beta particle (β−).
Which row gives the proton number and the mass number of the new nucleus formed?
| Marking point | Mark | Notes |
|---|---|---|
| In β− decay a neutron in the nucleus changes into a proton and an electron, and the electron is emitted. The proton number therefore rises by 1, from 53 to 54. The total number of nucleons does not change, so the mass number stays at 131. Option A has the change in proton number the wrong way round. Option B treats the electron as if it carried away one unit of mass number. Option D shows the changes for alpha decay. | ✓ 1 | Answer C |
Answer: C
Common error: Beta emission increases the proton number by 1. It does not decrease it. The mass number is unchanged because a neutron has become a proton.
A household smoke alarm contains a small sample of americium-241, which emits alpha particles. The alpha particles ionise the air between two plates so that a small current flows. Smoke entering the alarm reduces this current and sets off the alarm.
Which statement best explains why an alpha source is suitable for this use?
| Marking point | Mark | Notes |
|---|---|---|
| Alpha particles cause a lot of ionisation over a short distance, so they produce a measurable current in the air gap. Because their range in air is only a few centimetres and they are absorbed by the casing, people near the alarm are not exposed. Option A is wrong because alpha particles are the least penetrating radiation, and smoke absorbs them, which is what triggers the alarm. Option C is wrong because alpha particles carry a charge of +2. Option D is wrong because americium-241 has a long half-life (over 400 years), so the alarm keeps working for many years. | ✓ 1 | Answer B |
Answer: B
Exam tip: When you explain the choice of a source, refer to both ionising ability and penetrating power, and link each one to the job the source has to do.
Radon-222 is a radioactive gas that seeps out of some rocks into homes. It is a major source of background radiation. A radon-222 nucleus is written as \({}^{222}_{86}\text{Rn}\).
State the number of protons and the number of neutrons in a nucleus of radon-222.
Radon-219 is another isotope of radon. State what is meant by the term isotopes.
Radon-222 decays by emitting an alpha particle to form an isotope of polonium (Po).
Complete the nuclear equation, giving the mass numbers and proton numbers of the products.
\({}^{222}_{86}\text{Rn} \rightarrow \text{Po} + \alpha\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| protons = 86 ; neutrons = 222 − 86 = 136 | ✓ 2 | |
| Part (b) | ||
| Atoms or nuclei of the same element with the same number of protons / same proton number ; but a different number of neutrons / different mass number | ✓ 2 | |
| Part (c) | ||
| Polonium mass number 218 ; polonium proton number 84 ; alpha particle written as \({}^{4}_{2}\alpha\) or \({}^{4}_{2}\text{He}\) | ✓ 3 | |
Common error: In a nuclear equation the mass numbers must add up to the same total on both sides, and so must the proton numbers. Check both totals before moving on.
A geologist uses a Geiger–Müller tube and counter to check whether a granite worktop is radioactive.
Before testing the granite, the geologist measures the background radiation. State two sources of background radiation.
With no granite present, the counter records 45 counts in 3 minutes.
With the granite in place, it records 1260 counts in 5 minutes.
Calculate the count rate due to the granite alone, in counts per minute.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| First valid source, e.g. radon gas / cosmic rays (radiation from space) / rocks, soil or building materials / food and drink / medical sources such as X-rays / nuclear industry or weapons-testing fallout ; a second, different valid source from the same list | ✓ 2 | |
| Part (b) | ||
| Count rate with granite = 1260 ÷ 5 = 252 counts/min ; background count rate = 45 ÷ 3 = 15 counts/min ; corrected count rate = 252 − 15 = 237 counts/min | ✓ 3 | |
Common error: Convert both readings to counts per minute before you subtract. The two readings were taken over different times, so subtracting the raw counts gives a wrong answer.
A research reactor at a hospital produces radioactive isotopes for medical scanning. Inside the reactor, uranium-235 nuclei undergo induced fission.
Which particle must a uranium-235 nucleus absorb to start this fission?
| Marking point | Mark | Notes |
|---|---|---|
| Induced fission of uranium-235 starts when the nucleus absorbs a slow-moving neutron. The nucleus becomes unstable and splits into two daughter nuclei. A neutron has no charge, so the positive nucleus does not repel it. A proton or an alpha particle is positively charged and is repelled by the nucleus. An electron is not absorbed into the nucleus to cause fission. | ✓ 1 | Answer B |
Answer: B
Common error: Do not choose a proton just because protons are found in the nucleus. Only neutrons trigger fission, because they are uncharged and are not repelled.
A nuclear submarine is powered by a small fission reactor. The reactor uses uranium-235 as its fuel.
When a uranium-235 nucleus undergoes fission, it splits into two daughter nuclei.
State two other things that are released in each fission.
When the submarine is taken out of service, the used fuel must be stored safely for a very long time.
Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Neutrons (accept 'two or three neutrons'; do not accept 'a neutron' or 'one neutron') | ✓ 1 | |
| Energy (accept heat, kinetic energy of the products or gamma radiation) | ✓ 1 | |
| Part (b) | ||
| The daughter nuclei (fission products) are radioactive and/or have long half-lives, so they keep emitting harmful ionising radiation for many years | ✓ 1 | |
Exam tip: Write 'neutrons' in the plural, ideally with 'two or three'. These extra neutrons are what make a chain reaction possible.
A nuclear reactor at a desalination plant provides energy to turn seawater into drinking water. The reactor core contains uranium fuel rods, a graphite moderator and control rods.
State the purpose of the graphite moderator.
The plant needs less energy at night, so the operators lower the control rods further into the core.
Explain how this reduces the power output of the reactor.
Suggest a suitable material for the control rods.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| It slows down the neutrons so that they are more likely to be absorbed by uranium-235 nuclei and cause fission | ✓ 1 | |
| Part (b) | ||
| The control rods absorb (more) neutrons | ✓ 1 | |
| So there are fewer fissions per second, and energy is released at a lower rate | ✓ 1 | |
| Part (c) | ||
| Boron or cadmium | ✓ 1 | |
Common error: Do not mix up the two jobs. The moderator slows neutrons down; the control rods absorb neutrons. Saying control rods 'slow down the reaction' without mentioning neutron absorption does not earn the mark.
A space telescope studies a young star. Astronomers detect that the star is releasing energy by nuclear fusion in its core.
Describe what happens during nuclear fusion.
Name the element that is formed when hydrogen nuclei fuse in the core of the star.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Two light (small) nuclei join together to form a heavier nucleus | ✓ 1 | |
| Energy is released | ✓ 1 | |
| Part (b) | ||
| Helium | ✓ 1 | |
Exam tip: Write about nuclei, not atoms or molecules. Fusion is a nuclear process: the electrons have been stripped away at these temperatures.
Molybdenum-99 is used in hospitals to produce a tracer for medical scans. It is made as a fission product in a research reactor:
\(^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{99}_{42}\text{Mo} + ^{A}_{50}\text{Sn} + 2\,{}^{1}_{0}\text{n}\)
State the number of protons and the number of neutrons in a nucleus of uranium-235.
Calculate the nucleon number A of the tin (Sn) nucleus.
Molybdenum-99 has a half-life of 66 hours. Suggest why hospitals need new deliveries of it every week.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Protons: 92 | ✓ 1 | |
| Neutrons: 235 − 92 = 143 | ✓ 1 | |
| Part (b) | ||
| 235 + 1 = 99 + A + 2 | ✓ 1 | |
| A = 135 | ✓ 1 | |
| Part (c) | ||
| Its short half-life means the activity falls quickly, so after about a week too little is left to be useful | ✓ 1 | |
Exam tip: Check the bottom numbers as well: 92 = 42 + 50 + 0 confirms the equation balances. Nucleon and proton numbers must both be conserved.
In some regions built on granite, radon gas seeps out of the ground and can collect in cellars and basements. Radon is an alpha emitter.
Radon is one source of background radiation. State two other sources of background radiation.
Radon outside the body is much less of a hazard than radon that has been breathed in. Explain why.
Suggest one way that a householder could reduce the amount of radon in a basement.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any two from: cosmic rays / rocks and soil / food and drink / medical uses such as X-rays / nuclear power or weapons testing | ✓ 2 | |
| Part (b) | ||
| Alpha particles cannot pass through the outer layer of skin / have a very short range | ✓ 1 | |
| but inside the lungs they are strongly ionising and damage nearby cells / DNA, which can cause cancer | ✓ 1 | |
| Part (c) | ||
| Improve ventilation, e.g. fit a fan or open vents / seal cracks in the floor | ✓ 1 | |
Common error: Saying alpha is dangerous because it is 'penetrating' is wrong. Alpha is dangerous inside the body because it is highly ionising.
Boxes of sealed surgical gloves are sterilised by passing them close to a strong cobalt-60 source that emits gamma radiation.
Explain the difference between irradiation and contamination.
The sterilised gloves are safe to use straight away. Explain why.
Explain why a gamma source is used rather than an alpha source.
State one precaution that protects workers at the sterilising plant.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Irradiation is being exposed to radiation from a source outside the object, with no radioactive material transferred to it | ✓ 1 | |
| Contamination is when radioactive material gets onto or into the object | ✓ 1 | |
| Part (b) | ||
| They have been irradiated, not contaminated, so they contain no radioactive material and do not become radioactive | ✓ 1 | |
| Part (c) | ||
| Gamma is very penetrating, so it passes through the sealed packaging | ✓ 1 | |
| and kills bacteria on all of the gloves, whereas alpha would be stopped by the packaging | ✓ 1 | |
| Part (d) | ||
| Any one from: thick concrete or lead shielding / handling the source remotely / keeping their distance from the source / limiting time spent near the source | ✓ 1 | |
Common error: Irradiating an object with gamma rays does not make the object radioactive. Also, a dosimeter badge only measures a worker's dose; it does not protect them, so it is not a protective precaution.
A nucleus of radium-226 (22688Ra) emits an alpha particle. Which nucleus is produced?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: B. An alpha particle (42He) removes 4 from the mass number and 2 from the atomic number: 226 − 4 = 222 and 88 − 2 = 86. A keeps the atomic number at 88, forgetting that two protons leave. C changes only the atomic number and leaves the mass number unchanged. D adds the alpha particle's numbers instead of subtracting them. | ✓ 1 | Answer B |
Answer: B
Common error: In alpha decay both numbers go down (mass by 4, atomic number by 2); do not change only one.
In a pool-type research reactor, control rods containing boron are lowered further into the core. What is the effect?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: B. Boron absorbs neutrons, so fewer neutrons are left to cause further fission and the reaction rate falls. A describes the moderator, not control rods. C describes shielding. D describes the coolant. | ✓ 1 | Answer B |
Answer: B
Exam tip: Link each reactor part to one job: moderator slows neutrons, control rods absorb neutrons, shielding absorbs radiation, coolant removes heat.
A university has a pool-type research reactor. The core sits at the bottom of a deep pool of water inside thick concrete walls.
State the purpose of the moderator in a nuclear reactor.
The reactor has control rods made of boron. Explain how the control rods are used to change the rate of fission.
Explain why concrete walls and a deep pool of water are used around the core.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 slows down (fast) neutrons so they are more likely to cause fission | ✓ 1 | |
| Part (b) | ||
| B1 boron absorbs neutrons | ✓ 1 | |
| B1 lowering the rods in absorbs more neutrons so the rate of fission falls (raising them lets the rate increase) | ✓ 1 | |
| Part (c) | ||
| B1 they absorb radiation (gamma rays and neutrons) from the core | ✓ 1 | |
| B1 so that the radiation reaching people outside is reduced to a safe level | ✓ 1 | |
Exam tip: Name the job and the effect: 'absorbs neutrons, so fewer fissions' earns more than 'controls the reaction'.
Many theatres have smoke detectors that contain a small source of americium-241, which emits alpha particles.
Explain why an alpha source is suitable for use in a smoke detector.
Americium-241 has a half-life of 432 years. Explain why this is useful in a smoke detector.
Suggest why a person should not swallow or inhale the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 alpha particles are strongly ionising, so they ionise the air and produce a small current | ✓ 1 | |
| B1 alpha particles are stopped by a few centimetres of air (or the case), so little radiation escapes | ✓ 1 | |
| Part (b) | ||
| B1 the activity stays almost constant for many years, so the detector does not need a new source | ✓ 1 | |
| Part (c) | ||
| B1 the alpha particles would be very ionising inside the body, damaging cells | ✓ 1 | |
Exam tip: An alpha source is relatively safe outside the body but very dangerous inside it; state which case you mean.
A nucleus emits a beta-minus (β−) particle. Which row describes the change in the nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| A neutron changes into a proton and a fast electron, so proton number rises by 1 and neutron number falls by 1 (option A). B describes beta-plus emission. C describes alpha emission. D describes gamma emission. | ✓ 1 | Answer A |
Answer: A
Common error: Beta decay changes the element because the proton number changes; the nucleon number stays the same.
Neon has atomic number 10. Neon-20 and neon-22 are both found in the solar wind.
State the number of protons and the number of neutrons in a nucleus of neon-22.
Explain why neon-20 and neon-22 are isotopes.
Explain why a neon atom has no overall charge.
A neon-22 ion has a charge of +2. State the number of electrons in this ion.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 10 protons | ✓ 1 | |
| 12 neutrons | ✓ 1 | |
| Part (b) | ||
| same number of protons | ✓ 1 | |
| different numbers of neutrons | ✓ 1 | |
| Part (c) | ||
| it has equal numbers of protons (positive) and electrons (negative) | ✓ 1 | |
| Part (d) | ||
| 10 − 2 = 8 | ✓ 1 | |
Exam tip: The nucleon number is protons plus neutrons; the atomic number is protons only.
Smoke detectors in railway stations use americium-241 (atomic number 95), an alpha emitter with a half-life of 432 years. It decays to an isotope of neptunium (atomic number 93).
Complete the nuclear equation: americium-241 → neptunium-X + alpha particle. Give the nucleon numbers and the proton numbers in the equation.
Explain why the alpha source is safe inside a detector but dangerous if swallowed.
A detector contains 0.32 µg of americium-241. Calculate the mass of americium-241 left after 864 years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particle written as helium with nucleon number 4, proton number 2 | ✓ 1 | |
| neptunium nucleon number 237 | ✓ 1 | |
| neptunium proton number 93 | ✓ 1 | |
| Part (b) | ||
| alpha particles are absorbed by a few cm of air or the detector case so they cannot reach people outside | ✓ 1 | |
| inside the body they are strongly ionising and damage cells | ✓ 1 | |
| Part (c) | ||
| 864 ÷ 432 = 2 half-lives | ✓ 1 | |
| 0.32 ÷ 2 ÷ 2 | ✓ 1 | |
| 0.08 µg | ✓ 1 | |
Exam tip: Check that nucleon numbers and proton numbers each balance on both sides of a nuclear equation.
Technicians on a railway use cobalt-60, a gamma emitter with a half-life of 5.3 years, to check welds in rails for cracks. A source has an initial activity of 3200 GBq.
Explain why gamma rays are used to look for cracks inside a weld.
Calculate the activity of the source after 10.6 years.
Suggest two precautions for the technician using the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma rays pass through the metal | ✓ 1 | |
| more reach the detector through a crack than through solid metal, so cracks show up | ✓ 1 | |
| Part (b) | ||
| 10.6 ÷ 5.3 = 2 half-lives | ✓ 1 | |
| 3200 ÷ 4 = 800 GBq | ✓ 1 | |
| Part (c) | ||
| stay at a distance or behind lead/concrete shielding | ✓ 1 | |
| limit the time of exposure and wear a dosimeter badge | ✓ 1 | |
Exam tip: Convert the time given into a whole number of half-lives before halving the activity.
A smoke detector contains a small source of americium-241, which emits alpha particles. Which statement explains why an alpha source is suitable for this use?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: strong ionisation of the air in the gap gives a small current, and alpha radiation is absorbed within a few cm of air, so the casing keeps it in. B is wrong because alpha is the least penetrating radiation. C is wrong because alpha is the most strongly ionising. D is wrong because a short half-life would mean the source decays away quickly; americium-241 has a long half-life and that is why it rarely needs replacing. | ✓ 1 | Answer A |
Answer: A
Common error: Do not link penetrating power and ionising power the wrong way round: the more strongly a radiation ionises, the less far it penetrates.
Iodine-131 is used in hospitals to treat an overactive thyroid. A nucleus of iodine-131 is written 13153I.
State the number of protons, neutrons and electrons in a neutral atom of iodine-131.
Iodine-127 is the stable isotope of iodine. Explain why iodine-127 and iodine-131 are isotopes of the same element.
Iodine-131 decays by beta-minus emission to form an isotope of xenon (Xe). Complete the nuclear equation: 13153I → ……Xe + ……e.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 53 protons | ✓ 1 | |
| 78 neutrons (131 − 53) | ✓ 1 | |
| 53 electrons | ✓ 1 | |
| Part (b) | ||
| same number of protons (53) / same proton number | ✓ 1 | |
| different number of neutrons (74 and 78) / different mass numbers | ✓ 1 | |
| Part (c) | ||
| xenon mass number 131 | ✓ 1 | |
| xenon proton number 54 | ✓ 1 | |
| beta particle written as 0−1e | ✓ 1 | |
Common error: In a beta-minus decay the mass number stays the same and the proton number goes up by 1, because a neutron changes into a proton.
A student tests three radioactive sources, X, Y and Z, with a Geiger–Müller tube a short distance away. Each source emits only one type of radiation.
Source X: the count rate falls to background when a sheet of paper is placed in front of the source.
Source Y: paper has almost no effect, but a 3 mm aluminium sheet reduces the count rate to background.
Source Z: aluminium has almost no effect, but several centimetres of lead reduce the count rate greatly.
Identify the type of radiation emitted by each of the sources X, Y and Z.
State one safety precaution the student should take when handling the sources.
Source X is swallowed accidentally. Explain why this is more dangerous than if source Y were swallowed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| X is alpha | ✓ 1 | |
| Y is beta | ✓ 1 | |
| Z is gamma | ✓ 1 | |
| Part (b) | ||
| any one of: handle with tongs; keep sources in a lead-lined box when not in use; do not point at people; keep a distance / limit the time of exposure | ✓ 1 | |
| Part (c) | ||
| alpha radiation is strongly ionising | ✓ 1 | |
| inside the body it is absorbed by tissue close to the source, causing more damage than the less ionising beta radiation | ✓ 1 | |
Exam tip: Alpha is stopped by paper, beta by a few mm of aluminium, and gamma is reduced only by thick lead or concrete. Learn these as a set.
A nucleus of radium-226 contains 88 protons. It decays by emitting an alpha particle. Which row describes the nucleus that is formed?
| Marking point | Mark | Notes |
|---|---|---|
| An alpha particle is a helium nucleus (2 protons, 4 nucleons), so the proton number falls by 2 (88 to 86) and the nucleon number falls by 4 (226 to 222): B is correct. A forgets that the protons are lost. C subtracts 4 from the proton number instead of 2. D describes beta-minus decay, where a neutron becomes a proton. | ✓ 1 | Answer B |
Answer: B
Common error: Alpha decay lowers the nucleon number by 4 but the proton number by only 2.
A theatre foyer has a smoke detector containing a small sealed source of americium-241, which emits alpha particles.
Which type of radiation, alpha, beta or gamma, is (i) stopped by a sheet of paper, (ii) an electromagnetic wave, (iii) a stream of fast-moving electrons?
Explain why an alpha source is suitable for a smoke detector in a public building.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 alpha | ✓ 1 | |
| B1 gamma | ✓ 1 | |
| B1 beta | ✓ 1 | |
| Part (b) | ||
| B1 alpha particles are strongly ionising, so they ionise the air in the detector to give a small current | ✓ 1 | |
| B1 alpha particles are stopped by a few centimetres of air / the casing, so little radiation escapes and the source is safe to have in the building | ✓ 1 | |
Exam tip: For each type of radiation, learn its nature, how ionising it is, and what stops it.
A technician at a shipyard checks a radiation detector before using a sealed source. With no source present, the detector records 42 counts in 6 minutes.
State two natural sources of background radiation.
Calculate the background count rate in counts per minute.
With the source in place, the count rate is 52 counts per minute. Calculate the count rate due to the source alone.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 first correct source, e.g. radon gas / rocks / cosmic rays / food and drink | ✓ 1 | |
| B1 second correct source | ✓ 1 | |
| Part (b) | ||
| M1 for 42 ÷ 6 | ✓ 1 | |
| A1 for 7 (counts per minute) | ✓ 1 | |
| Part (c) | ||
| M1 for 52 − 7, using their background count rate (ECF from (b)) | ✓ 1 | |
| A1 for 45 (counts per minute) | ✓ 1 | |
Exam tip: Always subtract the background count rate from a reading before using it in a calculation.
An atom of radium-226 has 88 protons.
Calculate the number of neutrons in a nucleus of radium-226.
A different atom of radium has 140 neutrons in its nucleus. State the number of protons in this atom and explain why it is still radium.
Explain why a radium atom has no overall electric charge.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 226 − 88 | ✓ 1 | |
| A1 for 138 | ✓ 1 | |
| Part (b) | ||
| B1 88 protons | ✓ 1 | |
| B1 the number of protons defines the element, so atoms with the same proton number are the same element (different isotopes) | ✓ 1 | |
| Part (c) | ||
| B1 it has equal numbers of protons (positive) and electrons (negative) | ✓ 1 | |
Common error: The nucleon number counts protons plus neutrons; subtract the proton number to find neutrons.
A jeweller makes a silver clasp for a handbag. One atom of silver is represented by the symbol 10747Ag.
How many neutrons are in the nucleus of this atom?
| Marking point | Mark | Notes |
|---|---|---|
| Neutrons = mass number − atomic number = 107 − 47 = 60, so B is correct. A is the number of protons. C is the mass number (protons plus neutrons together). D adds the two numbers instead of subtracting. | ✓ 1 | Answer B |
Answer: B
Exam tip: In nuclide notation the top number is the total of protons and neutrons; the bottom number is protons only.
The table gives the numbers of protons and neutrons in four nuclides, A to D, found in a fish-feed supplement.
A: 30 protons, 34 neutrons
B: 31 protons, 33 neutrons
C: 30 protons, 36 neutrons
D: 32 protons, 34 neutrons
Identify which two nuclides are isotopes of the same element. Give a reason for your answer.
Calculate the mass number of nuclide D.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A and C identified | ✓ 1 | |
| Same number of protons (30) but different numbers of neutrons (34 and 36) | ✓ 1 | |
| Part (b) | ||
| Mass number = protons + neutrons = 32 + 34 = 66 | ✓ 1 | |
Common error: Isotopes match in protons, not in total nucleons. Check the proton numbers first.
A carbon fibre frame is used in a robot arm. Carbon atoms in the frame are neutral.
State where nearly all of the mass of an atom is found.
Explain why a carbon atom with 6 protons is neutral.
An atom of carbon-14 has 6 protons. Calculate the number of neutrons in the nucleus.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| In the nucleus | ✓ 1 | |
| Part (b) | ||
| It has 6 electrons, the same number as protons | ✓ 1 | |
| Proton charge is positive and electron charge is negative, equal in size, so the charges cancel | ✓ 1 | |
| Part (c) | ||
| Neutrons = mass number − protons = 14 − 6 | ✓ 1 | |
| 8 | ✓ 1 | |
Exam tip: Use the words 'equal and opposite charges' when explaining why atoms are neutral.
A fishery worker handles a sealed gamma source used in a flow gauge.
State two precautions the worker should take when using the source.
Explain why an alpha source outside the body is less dangerous than the same source inside the body.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use tongs or long handles (keep a distance from the source) | ✓ 1 | |
| Limit the time of exposure, or store the source in a lead-lined container | ✓ 1 | |
| Part (b) | ||
| Alpha radiation is stopped by skin or a few centimetres of air | ✓ 1 | |
| Inside the body it is strongly ionising and can damage cells directly | ✓ 1 | |
Exam tip: Precautions are marked for what they do: reducing time, increasing distance or adding shielding.
A marine biologist on a research boat takes a radiation detector out to sea. Far from any radioactive source, the detector still gives a small reading. Which of these is a natural source of this reading?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: cosmic rays reach the Earth from space and are part of natural background radiation. A and D are man-made sources. B is also man-made, even though the fallout is now spread throughout the environment. | ✓ 1 | Answer C |
Answer: C
Common error: Background radiation has natural and man-made sources; do not assume that anything found in the environment is natural.
A team of marine biologists measures the background radiation on a research vessel before testing shellfish samples. With no sample present, a detector records 450 counts in 30 minutes. With a sample of shellfish placed beside it, the detector records 1710 counts in 30 minutes.
State two natural sources of background radiation.
Suggest why the biologists counted for 30 minutes rather than for 1 minute.
Calculate the background count rate in counts per minute.
Calculate the count rate due to the shellfish sample alone.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any one from: radon gas, cosmic rays, rocks and soil, food and drink, building materials | ✓ 1 | |
| Any different second source from the same list | ✓ 1 | |
| Part (b) | ||
| Radioactive decay is random, so counting for longer reduces the effect of random variation (gives a more reliable average) | ✓ 1 | |
| Part (c) | ||
| Rate = counts ÷ time = 450 ÷ 30 | ✓ 1 | |
| 15 counts per minute | ✓ 1 | |
| Part (d) | ||
| Count rate with sample = 1710 ÷ 30 = 57 counts per minute | ✓ 1 | |
| 57 − 15 = 42 counts per minute (ECF from (c)) | ✓ 1 | |
Common error: Remember to subtract the background rate from the reading with the sample, not just quote the reading with the sample.
A marine biologist injects a small amount of a radioactive tracer into a fish to follow its movements. The tracer has a half-life of 6.0 hours. Its initial activity is 960 Bq.
State what is meant by the term half-life.
Calculate the activity of the tracer 24 hours after it was injected.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The time taken for the activity (or the number of undecayed nuclei) of a sample to halve | ✓ 1 | |
| Part (b) | ||
| 24 hours is 4 half-lives (24 ÷ 6.0) | ✓ 1 | |
| 960 → 480 → 240 → 120 → 60 (or 960 ÷ 24) | ✓ 1 | |
| 60 Bq | ✓ 1 | |
Exam tip: Convert the time into a whole number of half-lives first, then halve the activity that many times.
Bone contains calcium-40 and a smaller amount of calcium-44. Both nuclei are calcium nuclei. Which row describes the two nuclei?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: both are calcium, so both have 20 protons, and the nucleon numbers 40 and 44 differ because the neutron numbers (20 and 24) differ. B swaps the roles of protons and neutrons. C is wrong because the nucleon numbers are 40 and 44, not equal. D would make the nuclei identical, which they are not. | ✓ 1 | Answer A |
Answer: A
Exam tip: Isotopes of one element always have the same proton number; only the neutron number (and so the nucleon number) changes.
In a pool-type research reactor the boron control rods are lowered further into the core. What is the effect?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: boron absorbs neutrons, so fewer are left to cause further fission and the reaction slows. B confuses control rods with the moderator, which slows neutrons. C is wrong because control rods remove neutrons and do not make any. D reverses the cause: fewer neutrons absorbed would make the fission rate rise. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link each reactor part to its job: moderator slows neutrons, control rods absorb neutrons, shielding absorbs radiation.
Potassium-40 is a radioactive isotope found in small amounts in the human body. A nucleus of potassium-40 is written 4019K.
State the number of protons and the number of neutrons in a nucleus of potassium-40.
Potassium-40 can decay by beta-minus emission to form calcium. The decay is shown below.
4019K → AZCa + 0−1e
State the value of A and the value of Z.
Explain why potassium-40 and calcium-40 are not isotopes of the same element.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 19 protons | ✓ 1 | |
| 21 neutrons | ✓ 1 | |
| Part (b) | ||
| A = 40 | ✓ 1 | |
| Z = 20 | ✓ 1 | |
| Part (c) | ||
| isotopes of an element have the same number of protons | ✓ 1 | |
| potassium has 19 protons but calcium has 20 protons (so they are different elements) | ✓ 1 | |
Exam tip: In a nuclear equation the top numbers must add up to the same total on both sides, and so must the bottom numbers.
A student measures the count rate from a radioactive source. After the background count rate has been subtracted, the readings are 540 counts per minute with nothing between the source and the detector, 538 counts per minute with a sheet of paper in the way and 2 counts per minute with 3 mm of aluminium in the way. Which type of radiation does the source emit?
| Marking point | Mark | Notes |
|---|---|---|
| The paper makes almost no difference, so no alpha particles are present (alpha would be stopped by paper). The aluminium removes almost everything, which beta particles allow but gamma rays would not (gamma would pass through 3 mm of aluminium with little loss). So the source emits beta only. 'Alpha only' and 'alpha and gamma' fail because paper does not reduce the count; 'gamma only' fails because the aluminium cuts the count to almost zero. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check the thinnest absorber first: if paper has no effect, alpha is absent.
Radiation passing through the steel wall of a pipe is used to produce an image of a weld on film on the far side. The inspection will be repeated over several months. Which source is most suitable?
| Marking point | Mark | Notes |
|---|---|---|
| The radiation must pass through the steel, so a gamma emitter is needed, and the half-life must be long so that the activity stays high for months. An alpha emitter or a beta emitter would be absorbed by the steel before reaching the film. A gamma emitter with a half-life of 40 seconds would have decayed away almost completely within minutes. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a source that must work over a long time, check both the type of radiation and the half-life.
Radium-226 was once used in luminous paint on the dials of watches. A radium-226 nucleus has 88 protons.
State the number of neutrons in a radium-226 nucleus.
Radium-226 decays by emitting an alpha particle to form radon. Determine the nucleon number and the proton number of the radon nucleus.
Explain why alpha particles from the paint on a watch worn on the wrist are much less dangerous than alpha particles from paint that has been swallowed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 226 − 88 = 138 neutrons | ✓ 1 | |
| Part (b) | ||
| nucleon number 226 − 4 = 222 | ✓ 1 | |
| proton number 88 − 2 = 86 | ✓ 1 | |
| Part (c) | ||
| alpha particles have a very short range and are absorbed by the watch glass or the outer layer of skin | ✓ 1 | |
| inside the body they ionise cells in nearby tissue, which can cause damage | ✓ 1 | |
Common error: In alpha decay the nucleon number falls by 4 and the proton number falls by 2.
In a textile factory the thickness of a plastic film used to back a fabric is monitored. A beta source is placed below the film and a detector above it.
Explain why a beta source is more suitable than an alpha source or a gamma source.
The film becomes thicker. State and explain how the count rate changes.
Suggest why a source with a long half-life is chosen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particles would be completely absorbed by the film | ✓ 1 | |
| gamma rays would pass through almost unchanged, so the count rate would hardly depend on the thickness (beta is partly absorbed) | ✓ 1 | |
| Part (b) | ||
| the count rate decreases | ✓ 1 | |
| more beta particles are absorbed in the thicker film | ✓ 1 | |
| Part (c) | ||
| its activity stays almost constant, so the gauge does not need frequent recalibration or replacement | ✓ 1 | |
Exam tip: Choose the radiation whose penetration is partly, not fully or not at all, reduced by the material.
Which row lists alpha, beta and gamma radiation in order of increasing ionising ability?
| Marking point | Mark | Notes |
|---|---|---|
| Gamma rays ionise weakly, beta particles moderately and alpha particles most strongly because they are heavy, slow and doubly charged. The other rows reverse the order or put beta at the wrong end of it. | ✓ 1 | Answer C |
Answer: C
Common error: The most ionising radiation is the least penetrating.
What is a beta-minus particle?
| Marking point | Mark | Notes |
|---|---|---|
| A beta-minus particle is an electron created in the nucleus when a neutron changes into a proton. An electron from the outer shell would only ionise the atom and is not radiation from the nucleus. A helium nucleus is an alpha particle. A high-energy photon from the nucleus is a gamma ray. | ✓ 1 | Answer A |
Answer: A
Common error: Beta particles come from the nucleus, not from the electron shells.
A marine biology laboratory keeps three sealed sources: one emits alpha particles, one beta particles and one gamma rays.
State which source gives the most strongly ionising radiation.
State which radiation is stopped by a thin sheet of paper.
Explain why the radiation that is most strongly ionising has the shortest range in air.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha | ✓ 1 | |
| Part (b) | ||
| alpha | ✓ 1 | |
| Part (c) | ||
| it ionises many atoms along its path, losing energy at each ionisation | ✓ 1 | |
| so it quickly runs out of energy and stops after a short distance | ✓ 1 | |
Common error: The radiation that ionises most is stopped most easily.
A marine biologist tracks a sea turtle using a small tag that emits beta-minus particles. A detector is held outside the turtle.
State what a beta-minus particle is.
Explain why a tag that emits alpha particles would be unsuitable.
Suggest why a tag with a half-life of several months is chosen rather than one of a few seconds.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a fast-moving electron | ✓ 1 | |
| Part (b) | ||
| alpha particles have a very short range and are absorbed by the turtle's tissue | ✓ 1 | |
| so none would reach the detector outside the turtle | ✓ 1 | |
| Part (c) | ||
| the activity must stay high enough to be detected for the whole time the turtle is tracked | ✓ 1 | |
Exam tip: Link the penetration of the radiation to where the detector is placed.
A potassium-40 nucleus is written 4019K. It is one of the radioactive nuclei found in the human body. How many neutrons are in this nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| Neutrons = nucleon number − proton number = 40 − 19 = 21, so this is correct. 19 is the number of protons. 40 is the nucleon number, which counts protons and neutrons together. 59 comes from adding the two numbers instead of subtracting. | ✓ 1 | Answer B |
Answer: B
Common error: Nucleon number is the total of protons and neutrons, so subtract the proton number to find neutrons.
In a school laboratory a Geiger–Müller tube is switched on with no radioactive source nearby. It still records a few counts every minute. Which statement gives the best explanation?
| Marking point | Mark | Notes |
|---|---|---|
| Radioactive material in rocks, soil and air, together with cosmic rays, is always present, so a few counts per minute is normal and this is correct. A working tube is expected to record background counts, so it is not faulty. Sources stored away do not leave radioactivity behind in the room. A detector tube does not emit radiation of its own. | ✓ 1 | Answer A |
Answer: A
Exam tip: Always measure the background count rate first and subtract it from later readings.
Potassium and calcium nuclei are both found in the human body. Potassium-40 is 4019K, a radioactive isotope. It can decay by beta-minus emission into calcium-40, which has proton number 20.
State the number of protons and the number of neutrons in a potassium-40 nucleus.
State the number of protons and the number of neutrons in a calcium-40 nucleus.
Explain why the proton number increases by one when a nucleus emits a beta-minus particle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 19 protons | ✓ 1 | |
| 21 neutrons | ✓ 1 | |
| Part (b) | ||
| 20 protons | ✓ 1 | |
| 20 neutrons, since 40 − 20 = 20 | ✓ 1 | |
| Part (c) | ||
| a neutron in the nucleus changes into a proton | ✓ 1 | |
| a fast-moving electron (the beta particle) is emitted | ✓ 1 | |
Common error: Beta decay does not remove an electron from the atom's shells; the beta particle is created inside the nucleus.
A factory makes foil bags for roasted coffee. A beta source on one side of the moving foil and a detector on the other side are used to monitor the thickness of the foil.
Explain why a beta source is more suitable than an alpha source or a gamma source.
State how the count rate at the detector changes if the foil becomes thicker.
Suggest why a source with a long half-life is chosen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particles would be stopped completely by the foil, so the detector would always read zero | ✓ 1 | |
| gamma rays would pass through almost unaffected, so the count rate would hardly change with thickness | ✓ 1 | |
| beta particles are partly absorbed, so the count rate changes with thickness | ✓ 1 | |
| Part (b) | ||
| the count rate decreases | ✓ 1 | |
| Part (c) | ||
| the activity stays almost constant, so the source does not need frequent replacing or recalibrating | ✓ 1 | |
Exam tip: For a thickness gauge the radiation must be partly, not fully, absorbed by the material.
A hospital receives a radioactive tracer with a half-life of 6 hours for use in a sports-injury scan. It is stored for 18 hours before it is used. What fraction of the original radioactive nuclei have decayed during storage?
| Marking point | Mark | Notes |
|---|---|---|
| 18 hours is 3 half-lives, so the number of undecayed nuclei is halved three times: 1/2 × 1/2 × 1/2 = 1/8 remain, which means 7/8 have decayed. The option 1/8 is the fraction that remains, not the fraction that has decayed. The option 1/3 comes from dividing 6 by 18. The option 3/8 comes from wrongly assuming each half-life removes 1/8 of the original nuclei. | ✓ 1 | Answer D |
Answer: D
Exam tip: Read whether the question asks for the fraction that has decayed or the fraction that remains, and check by adding the two fractions to make 1.
Neptunium-237 (23793Np) is a radioactive isotope found in nuclear waste. It decays through a series of steps that eventually ends at stable bismuth-209 (20983Bi).
State the number of protons and the number of neutrons in a nucleus of neptunium-237.
The first step is alpha decay. Give the proton number and the nucleon number of the nucleus formed.
The next step is beta-minus decay of the nucleus formed in the first step. Give the proton number and the nucleon number of the nucleus formed.
Calculate the total number of alpha particles and the total number of beta-minus particles emitted in the whole series from neptunium-237 to bismuth-209.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| protons = 93 | ✓ 1 | |
| neutrons = 237 − 93 = 144 | ✓ 1 | |
| Part (b) | ||
| proton number = 91 | ✓ 1 | |
| nucleon number = 233 | ✓ 1 | |
| Part (c) | ||
| proton number = 92 | ✓ 1 | |
| nucleon number = 233 | ✓ 1 | |
| (ECF from (b)) | — | |
| Part (d) | ||
| nucleon number falls by 237 − 209 = 28 | ✓ 1 | |
| only alpha emission changes the nucleon number, so 28 ÷ 4 = 7 alpha particles | ✓ 1 | |
| 7 alpha particles lower the proton number from 93 to 93 − 14 = 79 | ✓ 1 | |
| 83 − 79 = 4 beta-minus particles, since each raises the proton number by 1 | ✓ 1 | |
Common error: Beta-minus emission does not change the nucleon number, so use the alpha emissions alone to account for the fall in nucleon number.
Gamma radiography is used to check the quality of welds in steel railway track. A sealed source containing iridium-192, which has a half-life of 74 days, is placed on one side of the weld and a detector is placed on the other side.
State why gamma radiation, rather than alpha radiation or beta radiation, is used for this test.
State one safety precaution for workers who handle the source.
The source is replaced after about 150 days. Explain why.
Explain why a source with a half-life of a few seconds would be unsuitable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma radiation is highly penetrating, so it can pass through the steel | ✓ 1 | |
| alpha radiation (and beta radiation) would be absorbed by the steel and would not reach the detector | ✓ 1 | |
| Part (b) | ||
| any one of: use long-handled tongs; keep the source in a lead-lined container when it is not in use; stand behind shielding; keep the exposure time short; keep as far from the source as possible | ✓ 1 | |
| Part (c) | ||
| the activity falls as the nuclei decay | ✓ 1 | |
| after about two half-lives the activity is about a quarter of its original value, so much longer exposure times would be needed to form a clear image | ✓ 1 | |
| Part (d) | ||
| it would decay away before it could be delivered and used | ✓ 1 | |
| its activity would fall too fast during the test to give a steady, reliable image | ✓ 1 | |
Exam tip: Link each type of radiation to what stops it: paper stops alpha, a few millimetres of aluminium stops beta, and thick lead or concrete is needed to reduce gamma.
Which statement correctly describes a beta-minus particle?
| Marking point | Mark | Notes |
|---|---|---|
| A beta-minus particle is a fast-moving electron created in the nucleus when a neutron changes into a proton, and it is then emitted from the nucleus, so the third option is correct. The first option describes an orbital electron, which is not what beta radiation is. The second option describes an alpha particle. The fourth option describes gamma radiation. | ✓ 1 | Answer C |
Answer: C
Common error: Do not confuse the electron in a beta particle with the electrons in the shells of the atom: beta particles come from the nucleus.
A hospital laboratory stores a sample of the isotope iodine-131, written as 13153I. How many protons and neutrons are in one nucleus of this isotope?
| Marking point | Mark | Notes |
|---|---|---|
| The lower number, 53, is the number of protons. The neutron number is the nucleon number minus the proton number, 131 − 53 = 78, so the first option is correct. The second option swaps the two particles. The third option uses the nucleon number as the neutron number. The fourth option uses the nucleon number as the proton number and the proton number as the neutron number. | ✓ 1 | Answer A |
Answer: A
Common error: The top number counts all the nucleons (protons plus neutrons), so subtract the bottom number to get the neutrons.
A beekeeper fits a smoke alarm in a honey-processing shed. The alarm contains a small sample of americium-241, which emits alpha particles and has a half-life of 432 years.
Explain why an alpha source, rather than a gamma source, is suitable for a smoke alarm.
Explain why it is safe to have the alarm inside the shed.
Explain why the beekeeper does not need to replace the source for many years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Alpha particles are strongly ionising, so they make the air in the detector conduct a small current | ✓ 1 | |
| Alpha particles travel only a few centimetres in air, so smoke particles absorbing them changes the current (while gamma rays would pass straight through) | ✓ 1 | |
| Part (b) | ||
| The alpha particles cannot leave the casing / are absorbed by the casing or a few centimetres of air | ✓ 1 | |
| Part (c) | ||
| The half-life is very long, so the activity falls only very slowly | ✓ 1 | |
Common error: Alpha particles are the most ionising but the least penetrating; do not describe them as the most penetrating.
A nucleus of polonium-218 has 84 protons. It decays by emitting an alpha particle. Which row describes the nucleus that is formed?
| Marking point | Mark | Notes |
|---|---|---|
| An alpha particle is a helium nucleus with 4 nucleons and 2 protons, so the nucleon number falls by 4 (218 to 214) and the proton number falls by 2 (84 to 82). 218 nucleons wrongly leaves the nucleon number unchanged. 216 nucleons wrongly removes only 2 nucleons. 86 protons wrongly adds protons instead of removing them. | ✓ 1 | Answer B |
Answer: B
Common error: Alpha emission reduces the nucleon number by 4, not by 2: the particle contains two protons and two neutrons.
Radium-226 was once used in luminous paint for the dials of old railway station clocks. A nucleus of radium-226 has 88 protons. It decays by emitting an alpha particle to form a nucleus of radon (Rn).
Complete the nuclear equation.
22688Ra → ……Rn + 42He
Radon-222 also decays by alpha emission, forming a nucleus of polonium-218 with 84 protons. Determine the number of protons and the number of neutrons in a nucleus of polonium-218.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mass number of radon is 222 | ✓ 1 | |
| atomic number of radon is 86 | ✓ 1 | |
| Part (b) | ||
| 84 protons | ✓ 1 | |
| 218 − 84 = 134 neutrons | ✓ 1 | |
Exam tip: In a decay equation, the top numbers must add up to the same total on both sides, and so must the bottom numbers.
Helium-3 and helium-4 are both found in stars. A helium-3 nucleus is written 32He and a helium-4 nucleus is written 42He.
State the number of protons, neutrons and electrons in a neutral atom of helium-3.
Explain why helium-3 and helium-4 are isotopes of the same element.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| protons 2 | ✓ 1 | |
| neutrons 1 | ✓ 1 | |
| electrons 2 | ✓ 1 | |
| Part (b) | ||
| same number of protons (same atomic number) but different numbers of neutrons | ✓ 1 | |
Common error: Isotopes differ in neutron number, not in proton number: changing the proton number would change the element.
Archaeologists compare carbon-14 (146C) with carbon-12 (126C). Which statement about these two isotopes is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Both have 6 protons (atomic number 6), so both are carbon. Carbon-12 has 12 − 6 = 6 neutrons and carbon-14 has 14 − 6 = 8 neutrons, so carbon-14 has two more neutrons. The first and third options are wrong because the proton number defines the element and is the same. The fourth is wrong because a neutral atom has as many electrons as protons, 6 in each. | ✓ 1 | Answer B |
Answer: B
Common error: Isotopes differ only in neutron number: the nucleon number changes but the proton number, and so the element, does not.
Carbon has two isotopes found in a wooden artefact: carbon-12 (126C) and carbon-14 (146C).
State the number of protons and the number of neutrons in a carbon-14 nucleus.
Explain why carbon-12 and carbon-14 are called isotopes of the same element.
State the number of electrons in a neutral carbon-14 atom.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 6 protons | ✓ 1 | |
| 8 neutrons | ✓ 1 | |
| Part (b) | ||
| same number of protons (6) but different number of neutrons | ✓ 1 | |
| Part (c) | ||
| 6 | ✓ 1 | |
Exam tip: The number of electrons in a neutral atom equals the number of protons, which is the atomic number.
The human body contains small amounts of radioactive potassium-40 (4019K). Calcium-40 (4020Ca) is also present in bones.
State the number of protons and the number of neutrons in a nucleus of potassium-40.
Calculate the number of neutrons in a nucleus of calcium-40.
Explain why potassium-40 and calcium-40 are not isotopes of each other.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 19 protons | ✓ 1 | |
| 21 neutrons (40 − 19) | ✓ 1 | |
| Part (b) | ||
| 20 neutrons (40 − 20) | ✓ 1 | |
| Part (c) | ||
| they have different numbers of protons, so they are different elements | ✓ 1 | |
Common error: Isotopes must have the same proton number; two nuclei with the same nucleon number but different proton numbers are different elements.
A cycle manufacturer checks the welds on a steel bicycle frame using gamma radiation.
Explain why gamma radiation is used rather than alpha radiation.
State one safety precaution for the person handling the source.
State why the radiation is described as ionising.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma is very penetrating and passes through the metal, whereas alpha is stopped by thin material | ✓ 1 | |
| Part (b) | ||
| use tongs or long handles / keep the source in a lead-lined container / limit time of exposure / keep distance | ✓ 1 | |
| Part (c) | ||
| it can remove electrons from atoms to form ions | ✓ 1 | |
Exam tip: For safety questions, link the precaution to a reason such as distance, time or shielding.
A detector placed near a radioactive sample records 440 counts/min. The background count rate is 40 counts/min. The half-life of the sample is 2 hours.
What count rate will the detector record 6 hours later?
| Marking point | Mark | Notes |
|---|---|---|
| First find the count rate due to the sample alone: 440 − 40 = 400 counts/min. Six hours is 3 half-lives, so this falls to 400 → 200 → 100 → 50 counts/min. The background is still present, so the detector records 50 + 40 = 90 counts/min. 50 forgets to add the background back on. 55 halves the uncorrected 440 three times. 147 divides by 3 instead of halving three times. | ✓ 1 | Answer C |
Answer: C
Exam tip: Only the sample's count rate halves. Background stays the same, so subtract it first and add it back at the end if the question asks for the measured reading.
A researcher measures the count rate from a new radioactive isotope every 10 minutes. The background count rate is 30 counts/min. Her measured readings are:
0 min: 830 counts/min
10 min: 596 counts/min
20 min: 430 counts/min
30 min: 313 counts/min
40 min: 230 counts/min
Explain why the researcher must subtract the background count rate from each reading.
Determine the half-life of the isotope. Show your working.
Calculate the time from the start until the corrected count rate falls to 25 counts/min.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| So that only the radiation from the sample is used / background does not halve, so including it would give the wrong half-life | ✓ 1 | |
| Part (b) | ||
| Corrected readings: 800 at 0 min, 400 at 20 min (and 200 at 40 min) | ✓ 1 | |
| Half-life = 20 min | ✓ 1 | |
| Part (c) | ||
| 800 → 400 → 200 → 100 → 50 → 25 is 5 half-lives | ✓ 1 | |
| 5 × 20 = 100 min | ✓ 1 | |
Exam tip: Write down the corrected values before you look for the half-life. The measured readings never quite halve.
Engineers designing a fusion reactor must heat hydrogen nuclei to over 100 million °C. Why is such a high temperature needed?
| Marking point | Mark | Notes |
|---|---|---|
| Hydrogen nuclei are positively charged, so they repel each other. Only at extremely high temperatures do they move fast enough, with enough kinetic energy, to get close enough to fuse. Melting and boiling are changes of state and are irrelevant at these temperatures. Fusion joins nuclei together rather than splitting them. Gravity between individual nuclei is negligible at any temperature. | ✓ 1 | Answer B |
Answer: B
Common error: Writing 'the nuclei repel' is not enough. State that the nuclei are positively charged and need high kinetic energy (high speed) to get close enough to fuse.
A nuclear power station supplies a steel works that uses electric-arc furnaces. The power station has an electrical power output of 1.2×109 W and an efficiency of 36%. Each fission of a uranium-235 nucleus releases 3.2×10−11 J of energy.
Calculate the rate at which energy is released by fission in the reactor core.
Calculate the number of uranium-235 nuclei that undergo fission each second.
The steel works suddenly needs less electrical power. Explain how the operators reduce the rate of fission in the reactor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Input power = useful output ÷ efficiency = 1.2×109 ÷ 0.36 | ✓ 1 | |
| = 3.3×109 W (accept 3.33×109 W) | ✓ 1 | |
| Part (b) | ||
| Number per second = 3.33×109 ÷ 3.2×10−11 (allow their value from the first part) | ✓ 1 | |
| = 1.04×1020 per second (accept 1.0×1020 or 1.03×1020) | ✓ 1 | |
| Part (c) | ||
| The control rods are lowered further into the core . | ✓ 1 | |
| They absorb more neutrons, so fewer neutrons go on to cause further fissions . | ✓ 1 | |
Exam tip: Efficiency links useful output to total input. Work out the total energy released by fission each second before dividing by the energy per fission.
During the start-up of a reactor, engineers model the chain reaction. In their model, one fission occurs in the first generation. Each fission then causes exactly two further fissions in the next generation.
Each fission of uranium-235 releases about three neutrons. Suggest two reasons why not all of these neutrons go on to cause another fission.
Calculate the number of fissions in the 20th generation.
State the condition needed for the reactor to give a steady power output.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Acceptable reasons: neutrons escape from the core; neutrons are absorbed by the control rods; neutrons are absorbed by other nuclei, e.g. U-238, the moderator or the coolant; neutrons are moving too fast to be captured by U-235 nuclei. | — | |
| First valid reason | ✓ 1 | |
| Second, different valid reason | ✓ 1 | |
| Part (b) | ||
| Number = 219 | ✓ 1 | |
| = 524 288 ≈ 5.2×105 | ✓ 1 | |
| Part (c) | ||
| On average, exactly one neutron from each fission goes on to cause another fission. Also accept: reproduction factor k = 1 | ✓ 1 | |
Common error: The first generation has 2⁰ = 1 fission, so the 20th generation has 2¹⁹, not 2²⁰.
Molybdenum-99 is used in hospitals to produce a tracer for medical scans. It is made by the fission of uranium-235 in a research reactor:
23592U + 10n → 9942Mo + A50Sn + 2 10n
Calculate the nucleon number A of the tin nucleus.
Explain why the fission products from the reactor must be handled carefully.
Molybdenum-99 has a half-life of 66 hours. A hospital receives a sample containing 40 μg of molybdenum-99. Calculate the mass of molybdenum-99 remaining 198 hours after delivery.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 235 + 1 = 99 + A + 2 | ✓ 1 | |
| A = 135 | ✓ 1 | |
| Part (b) | ||
| Fission products are radioactive / emit ionising radiation . The radiation can damage cells or DNA / cause cancer, so workers must be shielded or handled remotely . | ✓ 2 | |
| Part (c) | ||
| Number of half-lives = 198 ÷ 66 = 3 | ✓ 1 | |
| 40 → 20 → 10 → 5 | ✓ 1 | |
| mass remaining = 5.0 μg | ✓ 1 | |
Exam tip: Find the number of half-lives first, then halve the mass that many times. Show each step of the halving.
A hospital physicist measures a sample of sodium-24. Its activity falls from 3200 Bq to 400 Bq in 45 hours.
What is the half-life of sodium-24?
| Marking point | Mark | Notes |
|---|---|---|
| Halve the activity step by step: 3200 → 1600 → 800 → 400 Bq. That is 3 half-lives. The half-life is 45 ÷ 3 = 15 hours. 11.25 hours comes from counting the 4 values listed instead of the 3 halvings between them. 5.6 hours comes from dividing 45 by the ratio 8. 22.5 hours comes from assuming only 2 half-lives. | ✓ 1 | Answer C |
Answer: C
Common error: Count the halvings (the arrows), not the numbers in the chain. Going from 3200 to 400 involves three halvings, even though four values are written down.
A space probe sent to the outer planets is powered by a radioisotope generator. Heat from a radioactive source, sealed inside a thick metal casing, is converted into electricity. The probe must work for several decades.
Which source is most suitable?
| Marking point | Mark | Notes |
|---|---|---|
| Alpha particles are very strongly ionising and have a short range. Almost all of their energy is absorbed inside the source and its casing, so they heat it well, and little radiation escapes to damage the electronics. A half-life of 88 years means the heat output falls only slightly over a few decades. A 3-day half-life would leave almost no activity within weeks. Gamma rays mostly pass straight out of the source, so they deposit little heat and could damage instruments. A 2-minute half-life would be useless almost at once. | ✓ 1 | Answer A |
Answer: A
Exam tip: When choosing a source, justify two things separately: the type of radiation (penetration or ionisation) and the half-life (how long the activity is needed for).
A factory makes aluminium foil for food packaging by passing aluminium sheet between rollers. A radioactive source is placed on one side of the foil and a detector on the other. The count rate from the detector is used to control the gap between the rollers automatically.
The source emits beta radiation.
Explain why a beta source is used rather than an alpha source or a gamma source.
The count rate suddenly rises above its normal value.
State what has happened to the foil, and describe how the rollers should be adjusted.
Suggest why the source chosen has a long half-life.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Alpha would be completely absorbed by the foil, so nothing would reach the detector | ✓ 1 | |
| Gamma would pass through almost unaffected, so a change in thickness would give almost no change in count rate | ✓ 1 | |
| Beta is partly absorbed, so the count rate changes noticeably when the thickness changes | ✓ 1 | |
| Part (b) | ||
| The foil has become thinner | ✓ 1 | |
| Move the rollers further apart / reduce the pressure of the rollers | ✓ 1 | |
| Part (c) | ||
| The count rate will not fall noticeably because of decay, so a drop is not mistaken for the foil getting thicker / the source does not need replacing often | ✓ 1 | |
Exam tip: For a 'why not alpha or gamma' question, give one reason for each rejected radiation and then say why the chosen one works. Each of these points can earn a separate mark.
Fluorine-18 is used in PET scans to detect cancers. It has a half-life of 110 minutes. It is made in a cyclotron and then driven to a hospital. The journey takes 5 hours 30 minutes.
State what is meant by the half-life of a radioactive isotope.
Calculate the number of half-lives that pass during the journey.
The hospital needs the sample to have an activity of 450 MBq when it arrives.
Calculate the minimum activity the sample must have when it leaves the cyclotron.
Suggest why it is better for a hospital using fluorine-18 to be close to a cyclotron.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The time taken for the number of undecayed nuclei, or the activity, of a sample to fall to half its original value | ✓ 1 | |
| Part (b) | ||
| 5 h 30 min = 330 minutes, so 330 ÷ 110 | ✓ 1 | |
| = 3 half-lives | ✓ 1 | |
| Part (c) | ||
| Activity is halved 3 times, so the starting activity = 450 × 2 × 2 × 2 | ✓ 1 | |
| = 3600 MBq | ✓ 1 | |
| Part (d) | ||
| Less of the sample decays on the way, so a much smaller sample needs to be made, or more useful activity arrives | ✓ 1 | |
Exam tip: When you work backwards to an earlier activity, multiply by 2 for each half-life. Check that your answer is larger than the final activity.
Film studios use anti-static brushes to remove dust from camera lenses. The brush contains a small sealed strip of polonium-210, which emits alpha particles. Static charge on the lens attracts dust.
Explain how the alpha particles remove the static charge from the lens.
Explain why the brush is safe to use normally, even though polonium-210 is very harmful if swallowed.
Polonium-210 has a half-life of 138 days.
Calculate the fraction of the original polonium-210 nuclei that remain after 414 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Alpha particles are strongly ionising and ionise the air near the lens | ✓ 1 | |
| This produces both positive and negative ions | ✓ 1 | |
| Ions with the opposite charge are attracted to the lens and neutralise its charge, so the dust is no longer attracted | ✓ 1 | |
| Part (b) | ||
| Outside the body, alpha particles are absorbed by the sealed casing, by a few centimetres of air or by the outer layer of dead skin, so they cannot reach living cells | ✓ 1 | |
| Inside the body nothing shields the cells, so the strongly ionising alpha particles damage living tissue | ✓ 1 | |
| Part (c) | ||
| 414 ÷ 138 = 3 half-lives | ✓ 1 | |
| Fraction remaining = (1/2)3 = 1/8 | ✓ 1 | |
Common error: Saying that one-third or one-sixth of the nuclei remain after 3 half-lives is wrong. Each half-life halves what is left, so the fraction remaining is 1/2 × 1/2 × 1/2 = 1/8.
Technetium-99m is a gamma-emitting tracer used in medical scans. It has a half-life of 6 hours.
A hospital receives a sample with an activity of 800 MBq at 8 am on Monday.
What is the activity of the sample at 2 am on Tuesday?
| Marking point | Mark | Notes |
|---|---|---|
| From 8 am to 2 am the next day is 18 hours. This is 18 ÷ 6 = 3 half-lives. The activity halves three times: 800 → 400 → 200 → 100 MBq. Option C comes from counting the time as only 12 hours. Option D comes from dividing 800 by 3 instead of halving three times. Option A comes from counting a full 24 hours, which is four half-lives. | ✓ 1 | Answer B |
Answer: B
Exam tip: Work out the elapsed time carefully before anything else. Then halve once for each half-life. Never divide by the number of half-lives.
A factory makes aluminium foil for food packaging. The foil passes between a radioactive source above it and a detector below it. The detector reading is used to control the gap between a pair of rollers that press the foil to the correct thickness.
The source used emits beta particles. Explain why a beta source is used rather than an alpha source or a gamma source.
During one shift the count rate at the detector increases. State what has happened to the thickness of the foil and suggest what the control system should do to the rollers.
Suggest why the source chosen has a long half-life.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Alpha particles would be completely absorbed by the foil, so no change in reading would be detected ; gamma rays would pass through almost unaffected, so small changes in thickness would not change the reading | ✓ 2 | |
| Part (b) | ||
| The foil has become thinner ; the rollers should be moved further apart / the pressure on the rollers reduced | ✓ 2 | |
| Part (c) | ||
| So that the count rate does not fall noticeably because of decay, which would otherwise be mistaken for a change in thickness / so the source does not need replacing often | ✓ 1 | |
Exam tip: For thickness control, link penetrating power to the material. The radiation must be partly absorbed, so that a small change in thickness causes a measurable change in count rate.
Fluorine-18 is a radioactive isotope used as a tracer in PET scans in hospitals. A technician measures the corrected count rate from a sample at different times.
time = 0 min, count rate = 1600 counts/s
time = 55 min, count rate = 1131 counts/s
time = 110 min, count rate = 800 counts/s
time = 220 min, count rate = 400 counts/s
Use the data to determine the half-life of fluorine-18. Show how the data supports your answer.
Calculate the fraction of the original fluorine-18 nuclei that remain undecayed after 440 minutes.
Explain why a tracer with a half-life of this length is suitable for use inside a patient.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Count rate halves from 1600 to 800 counts/s in 110 min, so half-life = 110 min ; check: 800 to 400 counts/s also takes 110 min (from 110 min to 220 min) | ✓ 2 | |
| Part (b) | ||
| 440 ÷ 110 = 4 half-lives ; fraction remaining = (1/2)4 = 1/16 | ✓ 2 | |
| Part (c) | ||
| The activity falls quickly, so the patient receives a low total dose / is not radioactive for long ; but the half-life is long enough for the tracer to be prepared, injected and the scan completed before the activity becomes too low to detect | ✓ 2 | |
Exam tip: When you determine a half-life from data, use two separate halvings to show the value is consistent. A single halving may not earn all the marks.
A patient with an overactive thyroid gland is given a drink containing iodine-131, which emits beta particles and gamma rays. For a few days afterwards the patient is asked to avoid close contact with other people.
Explain the difference between contamination and irradiation, using this patient as an example.
Describe two precautions that hospital staff could take to reduce their own exposure while caring for this patient.
Explain why ionising radiation can be harmful to living cells.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Contamination: radioactive material gets onto or into a person or object, e.g. the iodine-131 is inside the patient, or traces of it pass to others through body fluids ; irradiation: exposure to radiation from a source outside the body, e.g. a relative sitting near the patient is irradiated by gamma rays without taking in any iodine-131 | ✓ 2 | |
| Part (b) | ||
| First valid precaution, chosen from: keep the time spent near the patient short; keep as large a distance as possible from the patient; use shielding such as lead screens; wear gloves / protective clothing when handling waste or bedding to avoid contamination ; a second, different valid precaution from the same list . Wearing a dosimeter / film badge alone is not credited, as it only monitors dose and does not reduce it. | ✓ 2 | |
| Part (c) | ||
| The radiation ionises atoms or molecules in the cells ; this can damage or kill cells, or damage DNA, causing mutations that may lead to cancer | ✓ 2 | |
Common error: Being irradiated does not make a person radioactive. Only contamination involves radioactive material being transferred to the person.
A student investigates an unknown radioactive source. She places different absorbers between the source and a detector and records the count rate for each one.
no absorber: 610 counts/min
sheet of paper: 604 counts/min
3 mm of aluminium: 182 counts/min
2 cm of lead: 95 counts/min
The background count rate is 30 counts/min.
Deduce which type or types of radiation the source emits. Explain your answer using the data.
Suggest why the small difference between the count rates with no absorber and with paper is not significant.
State one safety precaution the student should take when handling the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| No alpha, because the sheet of paper causes almost no drop in count rate (610 to 604 counts/min) ; beta present, because 3 mm of aluminium causes a large drop (604 to 182 counts/min) ; gamma present, because behind 2 cm of lead, which stops all beta, the count rate (95 counts/min) is still well above the background of 30 counts/min | ✓ 3 | |
| Part (b) | ||
| Radioactive decay is random, so repeated counts vary slightly even under the same conditions; a difference of 6 counts/min is within this variation | ✓ 1 | |
| Part (c) | ||
| Any one from: handle the source with tongs / never point it towards people / store it in its lead-lined box when not in use / limit the time it is out of its box | ✓ 1 | |
Exam tip: Compare each count rate with the background value. A reading still well above background after an absorber that stops beta shows that gamma is present.
Sealed packs of surgical instruments are sterilised by exposing them to gamma rays from a cobalt-60 source. Cobalt-60 has a half-life of 5.3 years.
Explain why gamma radiation is used to sterilise the instruments after they have been sealed in their packs.
A patient worries that the instruments will be radioactive. Explain why they are not.
The plant replaces the cobalt-60 source when its activity has fallen to 25% of its original value. Calculate how long after installation the source is replaced.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Gamma rays are very penetrating, so they pass through the sealed packaging ; they kill bacteria / microorganisms on the instruments, so the instruments stay sterile because the pack is never opened | ✓ 2 | |
| Part (b) | ||
| The instruments are irradiated, not contaminated ; no radioactive material is transferred to them / gamma rays do not make the nuclei of the instruments unstable | ✓ 2 | |
| Part (c) | ||
| 25% is (1/2)2, so 2 half-lives ; time = 2 × 5.3 = 10.6 years | ✓ 2 | |
Exam tip: Turn a percentage into a number of halvings: 50% is one half-life, 25% is two, 12.5% is three and 6.25% is four.
Energy from the Sun reaches a solar farm on Earth. This energy was released in the core of the Sun.
Which statement correctly describes the process that releases this energy?
| Marking point | Mark | Notes |
|---|---|---|
| Stars release energy by nuclear fusion. Light hydrogen nuclei join together to form a heavier helium nucleus, and energy is released. Splitting nuclei is fission, not fusion, so the first and last options are wrong. Nuclei are all positively charged, so they repel each other. The high temperature in the core gives them enough kinetic energy to overcome this repulsion; they do not attract. | ✓ 1 | Answer B |
Answer: B
Exam tip: Fusion means joining light nuclei together and fission means splitting a heavy nucleus. Link stars with fusion of hydrogen into helium.
In a nuclear power station, one fission reaction is:
\(^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{140}_{54}\text{Xe} + \text{X} + 2\,{}^{1}_{0}\text{n}\)
Which of these is nucleus X?
| Marking point | Mark | Notes |
|---|---|---|
| The nucleon (mass) number is conserved. The left side has 235 + 1 = 236 nucleons. The right side has 140 + A + 2(1), so A = 236 − 142 = 94. The proton (atomic) number is also conserved: 92 + 0 = 54 + Z, so Z = 38. X is therefore \(^{94}_{38}\)Sr. The distractors come from common mistakes. 96 ignores the two emitted neutrons. 93 ignores the absorbed neutron. 95 counts only one emitted neutron. | ✓ 1 | Answer C |
Answer: C
Common error: Count every neutron on both sides of the equation. This includes the incoming one and each of the outgoing ones, using the coefficient in front of the neutron symbol.
Engineers are designing a small fission reactor to supply energy to a base on the Moon.
Each fission of a uranium-235 nucleus releases 3.2 × 10−11 J of energy.
The reactor core releases thermal energy at a rate of 40 kW.
Calculate the number of fissions taking place each second in the reactor core.
The reactor supplies 10 kW of electrical power to the base.
Calculate the efficiency of the reactor system.
A night on the Moon lasts about 14 Earth days.
Suggest why a fission reactor is more suitable than solar panels for this base.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Number per second = 40 000 ÷ 3.2 × 10−11 | ✓ 1 | |
| = 1.25 × 1015 | ✓ 1 | |
| Part (b) | ||
| Efficiency = (useful power output ÷ total power input) × 100% = (10 ÷ 40) × 100% | ✓ 1 | |
| = 25% (or 0.25) | ✓ 1 | |
| Part (c) | ||
| There is no sunlight for about 14 days, but the reactor can supply energy continuously | ✓ 1 | |
Exam tip: Convert kW to W (× 1000) before dividing by the energy per fission. The answer then comes out as the number of fissions per second.
In a nuclear power station, the fission of each uranium-235 nucleus releases three neutrons. The station is designed to give a steady power output.
Describe what is meant by a chain reaction in the reactor.
State the average number of neutrons from each fission that must go on to cause another fission for the power output to stay steady.
Suggest two things that could happen to the other neutrons released in each fission.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Neutrons released by one fission are absorbed by other uranium-235 nuclei | ✓ 1 | |
| These nuclei then undergo fission, releasing more neutrons, which cause further fissions | ✓ 1 | |
| Part (b) | ||
| One (accept 1) | ✓ 1 | |
| Part (c) | ||
| Any two different suggestions from: absorbed by the control rods; absorbed by the moderator, coolant or other non-fuel material in the core; absorbed by uranium-238 nuclei without causing fission; escape from the reactor core | — | |
| First correct suggestion | ✓ 1 | |
| Second, different correct suggestion | ✓ 1 | |
Common error: A steady reaction does not need all three neutrons to cause fission. If, on average, more than one per fission did so, the rate would increase rapidly and become uncontrolled.
Phosphorus-32, 3215P, is used to study how plants take up fertiliser. It decays by emitting a beta particle.
Which of these is the nucleus formed?
| Marking point | Mark | Notes |
|---|---|---|
| In beta decay a neutron in the nucleus changes into a proton and an electron, and the electron is emitted. The mass number stays at 32 and the atomic number goes up by 1, from 15 to 16, which gives sulfur-32. Aluminium-28 is what alpha decay would give. Silicon-32 comes from thinking that losing a negative particle lowers the atomic number. Phosphorus-31 treats the beta particle as if it removed a nucleon. | ✓ 1 | Answer A |
Answer: A
Common error: Losing a negative charge makes the nucleus more positive, so the atomic number goes UP by 1 in beta decay.
A geology student tests a small rock sample taken from a disused mine. She first measures the background count rate as 26 counts/min.
She then places different absorbers between the rock and a detector. The detector stays at the same distance from the rock each time. Her results are:
no absorber: 520 counts/min
thin paper: 515 counts/min
5 mm of aluminium: 140 counts/min
5 cm of lead: 27 counts/min
Explain which types of radiation the rock emits. Use the results in your answer.
Suggest why the student keeps the distance between the rock and the detector the same throughout.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| No alpha, because thin paper causes no significant drop in the count rate (520 → 515 counts/min) | ✓ 1 | |
| Beta present, because 5 mm of aluminium causes a large drop in the count rate | ✓ 1 | |
| Gamma present, because the count rate with aluminium is still well above background, and only thick lead reduces it to about background | ✓ 1 | |
| Uses the background to correct the readings, e.g. corrected rates of 494, 489, 114 and 1 counts/min, or states that 27 counts/min is about the same as background | ✓ 1 | |
| Part (b) | ||
| The count rate also depends on distance, so keeping it fixed means any change is caused only by the absorber / it makes it a fair test | ✓ 1 | |
Exam tip: Subtract the background from each reading before comparing them. A corrected count rate that is still well above zero after aluminium is your evidence for gamma.
Some deep-space probes are kept warm by a heater containing plutonium-238, 23894Pu. Plutonium-238 decays by alpha emission to an isotope of uranium (U). Its half-life is 88 years.
Complete the nuclear equation for this decay.
23894Pu → ___U + ___He
When the probe is launched, the heater gives out thermal energy at 2000 W. Calculate how long it takes for this power to fall to 500 W. Assume the power is proportional to the activity.
Explain why an alpha emitter is a good choice for a heater placed near sensitive electronic instruments.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Uranium mass number 234 | ✓ 1 | |
| Uranium atomic number 92 | ✓ 1 | |
| Alpha particle written as 42He | ✓ 1 | |
| Part (b) | ||
| 2000 → 1000 → 500 W is 2 half-lives | ✓ 1 | |
| 2 × 88 = 176 years | ✓ 1 | |
| Part (c) | ||
| Alpha particles have a very short range and are absorbed within the plutonium or its casing | ✓ 1 | |
| so their energy is transferred as heat and very little radiation escapes to damage the instruments | ✓ 1 | |
Exam tip: In any nuclear equation, check that the top numbers balance and the bottom numbers balance on both sides.
A sample of a radioactive gas collected at a volcanic vent has a half-life of 12 hours. It is sealed in a container and stored for 36 hours. What fraction of the nuclei have decayed during storage?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: D. 36 hours is 3 half-lives, so the fraction remaining is (1/2)3 = 1/8 and the fraction decayed is 1 − 1/8 = 7/8. A is the fraction that remains, not the fraction that has decayed. B multiplies 1/8 by 3. C is the fraction decayed after only 2 half-lives. | ✓ 1 | Answer D |
Answer: D
Common error: Read the question carefully: decayed and remaining fractions add up to 1.
A hospital receives a sealed radioactive source by sea. At the start of the voyage its activity is 640 kBq. After 21 days of storage on the ship its activity is 80 kBq.
Work out the fraction of the original activity that remains after 21 days.
State the fraction of the radioactive nuclei that have decayed during the 21 days.
Work out the number of half-lives that have passed during the 21 days.
Calculate the half-life of the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 80 ÷ 640 = 1/8 | ✓ 1 | |
| Part (b) | ||
| B1 7/8 (1 − 1/8) | ✓ 1 | |
| Part (c) | ||
| B1 3 half-lives (1/8 = (1/2)3, or 640 → 320 → 160 → 80) | ✓ 1 | |
| Part (d) | ||
| M1 21 ÷ 3 | ✓ 1 | |
| A1 7 days (7.0 days) (ECF from the number of half-lives in (c)) | ✓ 1 | |
Common error: The activity falls to 1/8 after three half-lives, not four; count the halvings: 640 to 320 to 160 to 80.
Radium-226 (22688Ra) decays through a series of radioactive isotopes. Its first decay gives radon (Rn) and emits an alpha particle.
Determine the mass number and the atomic number of the radon nucleus produced.
Lead-214 (21482Pb) decays by emitting a beta-minus particle to form bismuth (Bi). Write a nuclear equation for this decay.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 mass number 226 − 4 = 222 | ✓ 1 | |
| B1 atomic number 88 − 2 = 86 | ✓ 1 | |
| Part (b) | ||
| B1 bismuth has atomic number 83 | ✓ 1 | |
| B1 mass number stays 214 | ✓ 1 | |
| B1 equation: 21482Pb → 21483Bi + 0−1e (beta-minus particle shown with mass 0 and charge −1) | ✓ 1 | |
Common error: In beta-minus decay the mass number is unchanged and the atomic number goes up by 1.
A shipyard checks welds using a cobalt-60 gamma source. Cobalt-60 has a half-life of 5.3 years.
Explain why a gamma source is used to check welds rather than an alpha source.
The source must be replaced when its activity has fallen to one quarter of its original value. Calculate how long this takes.
State one safety precaution for workers using the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 gamma rays are highly penetrating and pass through the metal | ✓ 1 | |
| B1 alpha particles would be stopped by the metal (or air), so no image of the inside of the weld would be formed | ✓ 1 | |
| Part (b) | ||
| M1 one quarter is 2 half-lives | ✓ 1 | |
| A1 2 × 5.3 = 10.6 years | ✓ 1 | |
| Part (c) | ||
| B1 any one of: keep the source in a lead-lined container when not in use; stay as far away as possible; limit the time of exposure | ✓ 1 | |
Exam tip: Fraction questions: write the sequence 1, 1/2, 1/4, 1/8 to count the half-lives.
A sports scientist uses a radioactive tracer with a half-life of 3 days. Its corrected count rate is 640 counts/min. What is the corrected count rate 9 days later?
| Marking point | Mark | Notes |
|---|---|---|
| 9 days is 3 half-lives, so the rate halves three times: 640 → 320 → 160 → 80 (C). A divides by 3, confusing the number of days with the number of halvings. B halves only twice. D divides by 9. | ✓ 1 | Answer C |
Answer: C
Exam tip: First find how many half-lives have passed, then halve that many times.
A student places absorbers between a source and a detector. The count rate is the same with paper as with no absorber. With 5 mm of aluminium the count rate falls to the background level. What radiation does the source emit?
| Marking point | Mark | Notes |
|---|---|---|
| Paper does not reduce the rate, so no alpha. Aluminium reduces it to background, so the radiation is fully stopped by a few mm of aluminium: beta only (B). A: alpha would be stopped by paper. C: gamma would still give a count above background behind 5 mm of aluminium. D: gamma is hardly reduced by aluminium. | ✓ 1 | Answer B |
Answer: B
Common error: If the count stays above background behind aluminium, some highly penetrating gamma is present.
A hospital uses technetium-99m, which has a half-life of 6.0 hours. A dose has an activity of 400 MBq at 08:00. A scan needs at least 100 MBq at the time of injection.
Calculate the activity of the dose at 14:00.
State why a gamma emitter is chosen for a tracer inside the body rather than an alpha emitter.
Explain why a half-life of 6 hours is better for a tracer than a half-life of 6 days.
Calculate the number of hours after 08:00 at which the activity falls to 100 MBq, and state the latest time of injection.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 400 ÷ 2 = 200 MBq | ✓ 1 | |
| Part (b) | ||
| gamma rays pass out of the body and can be detected outside | ✓ 1 | |
| alpha would be absorbed in the body, doing damage to cells and not being detected | ✓ 1 | |
| Part (c) | ||
| activity falls quickly so little radiation remains in the patient after the scan | ✓ 1 | |
| still long enough to deliver the dose and complete the scan | ✓ 1 | |
| Part (d) | ||
| 400 to 100 is two halvings | ✓ 1 | |
| 2 × 6.0 = 12 hours | ✓ 1 | |
| 20:00 | ✓ 1 | |
Exam tip: Count halvings first, then multiply by the half-life to get a time.
A student investigates a beta source with a Geiger-Müller tube and sheets of aluminium. The background count rate is 20 counts/min. Count rate (counts/min) for thickness of aluminium: 0 mm: 520; 1 mm: 380; 2 mm: 275; 3 mm: 190; 4 mm: 130; 5 mm: 70; 6 mm: 30; 7 mm: 20; 8 mm: 21.
State what causes the background count rate and why it is subtracted from each reading.
Determine the thickness of aluminium needed to stop the beta particles.
Calculate the percentage of the corrected count rate that remains behind 3 mm of aluminium, using the reading at 0 mm as the starting rate.
Explain why a few millimetres of aluminium would not be a suitable shield against gamma rays.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| radiation from rocks, cosmic rays or other natural sources | ✓ 1 | |
| so that only the count from the source is compared | ✓ 1 | |
| Part (b) | ||
| 7 mm | ✓ 1 | |
| Part (c) | ||
| corrected rates 500 and 170 counts/min | ✓ 1 | |
| 170 ÷ 500 × 100 = 34% | ✓ 1 | |
| Part (d) | ||
| gamma rays are much more penetrating than beta particles | ✓ 1 | |
| only a small fraction would be absorbed; thick lead or concrete is needed | ✓ 1 | |
Exam tip: Subtract background from every reading before working out fractions or percentages.
Sports scientists test a radioactive tracer. The corrected count rate from a sample was: 0 min: 640 counts/min; 10 min: 452; 20 min: 320; 30 min: 226; 40 min: 160.
Determine the half-life of the tracer using the data.
Explain why it is not possible to predict when one particular nucleus will decay.
Predict the corrected count rate at 60 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| activity falls from 640 to 320 counts/min | ✓ 1 | |
| this takes 20 minutes | ✓ 1 | |
| Part (b) | ||
| radioactive decay is random | ✓ 1 | |
| it cannot be influenced and is a matter of chance for each nucleus | ✓ 1 | |
| Part (c) | ||
| 60 min is 3 half-lives | ✓ 1 | |
| 640 ÷ 8 = 80 counts/min | ✓ 1 | |
Common error: Half-life is the time for the corrected count rate to halve, wherever you start on the decay curve.
A nucleus of radium-226 has 88 protons and 138 neutrons. It decays by emitting an alpha particle. How many protons and neutrons are in the daughter nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| An alpha particle is 2 protons and 2 neutrons, so both numbers fall by 2: 86 protons and 136 neutrons (radon-222). B only removes the 2 protons and forgets the 2 neutrons, which would give mass number 224. C removes 4 protons, confusing the 4 in the mass number with the proton number. D shows a neutron turning into a proton, which is what happens in beta-minus decay, not alpha decay. | ✓ 1 | Answer A |
Answer: A
Exam tip: Check your answer by adding protons and neutrons: the mass number must fall by 4 for alpha decay (226 to 222 here).
A hospital uses a radioactive tracer to study blood flow. A sample of the tracer has an activity of 960 Bq. After 18 days the activity of the sample has fallen to 120 Bq.
Determine the half-life of the tracer.
Calculate the activity of the sample a further 12 days later, that is 30 days after the start.
Suggest why a tracer that is injected into the body should emit gamma radiation rather than alpha radiation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| number of half-lives: 960 → 480 → 240 → 120 is 3 half-lives | ✓ 1 | |
| half-life = 18 ÷ 3 = 6 days | ✓ 1 | |
| Part (b) | ||
| 12 days is 2 half-lives (ECF from (a)) | ✓ 1 | |
| activity = 120 ÷ 4 = 30 Bq | ✓ 1 | |
| Part (c) | ||
| gamma radiation passes out of the body and can be detected outside | ✓ 1 | |
| alpha would be absorbed in the body (and cause more damage to tissue) / would not reach the detector | ✓ 1 | |
Exam tip: To find a half-life from two activities, count how many times the activity halves, then divide the total time by that number.
A student measures the radiation from a sample of a radioactive isotope. The background count rate, measured with the sample removed, is 18 counts per second. The count rate with the sample present is 418 counts per second at time 0, 218 counts per second after 6 hours and 118 counts per second after 12 hours.
Calculate the corrected count rate (the count rate due to the sample alone) at time 0, after 6 hours and after 12 hours.
Use your corrected count rates to determine the half-life of the isotope.
Predict the count rate that would be measured 18 hours after time 0, including background.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 400, 200 and 100 counts per second (all three correct) | ✓ 1 | |
| Part (b) | ||
| corrected rate halves from 400 to 200 in 6 hours | ✓ 1 | |
| corrected rate halves again from 200 to 100 in the next 6 hours, so the time to halve is constant | ✓ 1 | |
| half-life = 6 hours | ✓ 1 | |
| Part (c) | ||
| 18 hours is 3 half-lives, so corrected rate = 400 ÷ 8 = 50 counts per second | ✓ 1 | |
| add background: 50 + 18 = 68 counts per second | ✓ 1 | |
Common error: Background radiation does not decay with the sample. Subtract it before comparing readings, and add it back if the question asks for the count rate the detector would show.
Radium-226 is a radioactive isotope that is found in some rocks. Radium-226 (proton number 88) decays by alpha emission to radon-222 (Rn). Lead-214 (proton number 82) decays by beta-minus emission to bismuth-214 (Bi).
Complete the equation for the decay of radium-226: 22688Ra → ……Rn + ……He.
Complete the equation for the decay of lead-214: 21482Pb → ……Bi + ……e.
Radon is a gas that can build up inside houses built on rocks containing radium. Explain why breathing in radon is a health risk.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| radon mass number 222 | ✓ 1 | |
| radon proton number 86 | ✓ 1 | |
| alpha particle written as 42He | ✓ 1 | |
| Part (b) | ||
| bismuth mass number 214 | ✓ 1 | |
| bismuth proton number 83 | ✓ 1 | |
| beta particle written as 0−1e | ✓ 1 | |
| Part (c) | ||
| radon emits alpha radiation, which is strongly ionising | ✓ 1 | |
| once inside the lungs the radiation is absorbed by lung tissue, damaging cells / can cause cancer | ✓ 1 | |
Exam tip: Check each equation: the mass numbers on the two sides must add to the same total, and so must the proton numbers.
A student investigates how well aluminium absorbs beta particles. A Geiger–Müller tube is placed 10 cm from a beta source and sheets of aluminium of increasing total thickness are placed between them. With the source removed, the tube records a background count rate of 30 counts per minute. The readings with the source present are:
0.0 mm: 880 counts per minute
0.5 mm: 510 counts per minute
1.0 mm: 270 counts per minute
1.5 mm: 120 counts per minute
2.0 mm: 48 counts per minute
2.5 mm: 31 counts per minute
3.0 mm: 29 counts per minute
3.5 mm: 30 counts per minute
Explain why the student measured the background count rate before starting.
Use the readings to state the smallest thickness of aluminium, from those tested, that stops the beta particles. Explain your answer.
Calculate the percentage fall in the corrected count rate between no aluminium and 1.0 mm of aluminium. Give your answer to 3 significant figures.
Suggest one way to improve the reliability of the readings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| so it can be subtracted from the readings to find the count rate due to the source | ✓ 1 | |
| Part (b) | ||
| 2.5 mm | ✓ 1 | |
| count rate has fallen to the background level (about 30) and does not fall further with extra thickness | ✓ 1 | |
| Part (c) | ||
| corrected rates: 880 − 30 = 850 and 270 − 30 = 240 counts per minute | ✓ 1 | |
| fall = 850 − 240 = 610 | ✓ 1 | |
| percentage = 610 ÷ 850 × 100 = 71.8% | ✓ 1 | |
| Part (d) | ||
| repeat each reading | ✓ 1 | |
| and calculate a mean / count for a longer time | ✓ 1 | |
Exam tip: Count rates are random, so small differences close to the background level are not significant. Repeating and averaging makes it clear when the count has really reached background.
A factory makes plastic film for packaging. The thickness of the film is monitored by passing it between a beta source and a detector. The source is strontium-90, which has a half-life of 29 years. The source has an activity of 3.2 MBq when new.
Explain why a beta source is more suitable than an alpha source or a gamma source for monitoring thin plastic film.
Suggest why a source with a long half-life is chosen.
Calculate the activity of the source 58 years after it is new.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha would be stopped completely by the film (so nothing reaches the detector) | ✓ 1 | |
| gamma would pass through almost unchanged, so changes in thickness would not be noticed | ✓ 1 | |
| beta is partly absorbed, so the count rate at the detector changes with the thickness | ✓ 1 | |
| Part (b) | ||
| the activity stays almost constant, so the readings remain reliable | ✓ 1 | |
| the source does not need to be replaced often | ✓ 1 | |
| Part (c) | ||
| 58 years is 2 half-lives | ✓ 1 | |
| 3.2 ÷ 4 = 0.8 MBq | ✓ 1 | |
Exam tip: When a question gives a time that is an exact multiple of the half-life, find the number of half-lives first, then halve that many times.
A volcanologist measures a higher count rate in a granite cave than on a boat at sea, using the same detector with no source present. Which is the best explanation?
| Marking point | Mark | Notes |
|---|---|---|
| Natural rocks such as granite contain radioactive isotopes, and radon gas seeping from them accumulates in poorly ventilated caves, so A is correct. B is wrong because cosmic rays do not make ordinary rock radioactive. C is wrong because water only absorbs part of the radiation and there is still a background count at sea. D is wrong because caves are full of air, and alpha particles only travel a few centimetres in air. | ✓ 1 | Answer A |
Answer: A
Exam tip: Background radiation is always present, so a detector will never read zero in normal surroundings.
A radioactive sample used on a ship has an activity of 640 Bq. After 27 hours its activity is 80 Bq. What is the half-life of the sample?
| Marking point | Mark | Notes |
|---|---|---|
| 640 Bq to 80 Bq is a factor of 8, which is 3 half-lives (640 to 320 to 160 to 80). So the half-life is 27 ÷ 3 = 9.0 h, so B is correct. A divides 27 h by the factor 8 instead of the number of half-lives. C assumes 2 half-lives. D takes the whole time as one half-life. | ✓ 1 | Answer B |
Answer: B
Common error: Count how many times the activity halves; do not divide by the overall decay factor.
Radium-226 undergoes alpha decay to form radon-222. Radon is a radioactive gas that can seep from volcanic rock into the cellars of buildings.
Complete the decay equation.
22688Ra → XYRn + 42He
Give the values of X and Y.
Radon-222 also decays by alpha emission to form polonium. State the number of protons in the polonium nucleus formed.
Explain why radon gas that is breathed in is dangerous.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 X = 222 | ✓ 1 | |
| B1 Y = 86 | ✓ 1 | |
| Part (b) | ||
| B1 for 84 | ✓ 1 | |
| Part (c) | ||
| B1 alpha particles from radon in the lungs are strongly ionising | ✓ 1 | |
| B1 and are then in direct contact with living tissue, so they can damage cells or DNA and cause cancer | ✓ 1 | |
Common error: A nuclear equation must balance for both nucleon number (top) and proton number (bottom).
The table shows the average sources of the annual radiation dose for a person in the UK. The average total dose is 2.7 mSv per year.
| Source | Share of total dose |
| Radon gas | 50% |
| Medical | 14% |
| Ground and buildings | 14% |
| Food and drink | 12% |
| Cosmic rays | 10% |
State which source in the table gives the greatest dose.
Calculate the average annual dose from cosmic rays.
Explain why aircrew receive a larger dose from cosmic rays than people who stay at ground level.
State one source in the table that is artificial.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 radon gas | ✓ 1 | |
| Part (b) | ||
| M1 for 0.10 × 2.7 | ✓ 1 | |
| A1 for 0.27 (mSv) | ✓ 1 | |
| Part (c) | ||
| B1 at high altitude there is less atmosphere above them | ✓ 1 | |
| B1 so less of the cosmic radiation is absorbed before it reaches them | ✓ 1 | |
| Part (d) | ||
| B1 medical | ✓ 1 | |
Exam tip: Convert a percentage to a decimal fraction before multiplying: 10% is 0.10.
A fishery adds a radioactive tracer to a tank. The count rate from a water sample falls from 640 counts/min to 80 counts/min in 18 hours. Background radiation is negligible.
What is the half-life of the tracer?
| Marking point | Mark | Notes |
|---|---|---|
| 640 → 320 → 160 → 80 is three halvings, so 3 half-lives = 18 hours and the half-life is 18 ÷ 3 = 6 hours (C). A divides 18 by 8, the overall factor of the fall, instead of by the number of halvings. B divides by 4, counting the four values listed instead of the three halvings. D divides by 2, assuming only one halving. | ✓ 1 | Answer C |
Answer: C
Common error: Count the number of times the value is halved, not the number of readings or the overall factor of the fall.
A fashion designer uses carbon-12 and carbon-14 in a study of old cotton fibres. Which statement about neutral atoms of carbon-12 and carbon-14 is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Isotopes of one element have the same proton number (6) so a neutral atom has the same number of electrons (6); only the neutron number differs (6 and 8), so A is correct. B and C reverse which particle number changes. In C, mass numbers are 12 and 14, so they are not the same. D wrongly links extra mass to extra electrons; the extra mass comes from neutrons. | ✓ 1 | Answer A |
Answer: A
Common error: The element is fixed by the number of protons; extra mass in an isotope comes from extra neutrons, not electrons.
A fish farm injects a radioactive tracer into a tank. The activity of a sample of the water falls from 2400 Bq to 300 Bq in 36 hours.
Calculate the number of half-lives that have passed in 36 hours.
Calculate the half-life of the tracer.
Calculate the activity of the sample a further 24 hours later.
Explain why a tracer with a half-life of 12 years would be unsuitable here.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2400 → 1200 → 600 → 300, so 3 half-lives | ✓ 1 | |
| Part (b) | ||
| Half-life = 36 ÷ 3 | ✓ 1 | |
| 12 hours | ✓ 1 | |
| Part (c) | ||
| 24 hours is 2 half-lives (ECF from (b)) | ✓ 1 | |
| 300 ÷ 4 = 75 Bq | ✓ 1 | |
| Part (d) | ||
| It would stay radioactive for a very long time, so the water and fish would remain contaminated | ✓ 1 | |
Exam tip: Write down the sequence of halved values; it shows the number of half-lives and earns method marks even if the final step slips.
A student in a robotics lab measures the count rate from a small radioactive sample using a Geiger counter. With the sample removed, the count rate is 20 counts/min. The readings with the sample are:
time 0 min: 820 counts/min
time 15 min: 420 counts/min
time 30 min: 220 counts/min
time 45 min: 120 counts/min
Explain why the student needs the reading with the sample removed.
Calculate the corrected count rate at each time.
Determine the half-life of the sample.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Background radiation (from rocks, air and cosmic rays) is always detected | ✓ 1 | |
| It must be subtracted so that only the count from the sample is used | ✓ 1 | |
| Part (b) | ||
| 800, 400, 200 and 100 counts/min | ✓ 1 | |
| Part (c) | ||
| Time for the corrected count rate to halve (for example 800 to 400) | ✓ 1 | |
| 15 minutes | ✓ 1 | |
Common error: Subtract the background from every reading before looking for the halving time; otherwise the readings do not halve properly.
A logistics company tests an unlabelled sealed source before it is shipped. With the source removed, the detector reads 50 counts per minute. The table shows the readings with the source in place.
No absorber: 346 counts per minute
One sheet of paper: 344 counts per minute
3 mm of aluminium: 52 counts per minute
5 cm of lead: 50 counts per minute
What does the source emit?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct. Subtracting the background of 50 gives 296, 294, 2 and 0 counts per minute. Paper makes almost no difference, so there is no alpha. Thin aluminium removes almost all the radiation, so it is beta, and with 5 cm of lead the corrected reading is zero, so there is no gamma. A is wrong because alpha would be stopped by paper. C is wrong because gamma would pass through 3 mm of aluminium. D is wrong because it comes from not subtracting the background, which makes 52 and 50 look like a remaining gamma signal. | ✓ 1 | Answer B |
Answer: B
Exam tip: Always subtract the background count rate before deciding how much radiation an absorber has stopped.
In a simulation of a radiation experiment, a radioactive source is placed in front of a detector and sheets of aluminium of different thickness are placed between them. With the source removed, the detector reads 12 counts per minute. The table gives the readings with the source in place.
Thickness of aluminium / mm: 0, 1, 2, 3, 4, 5
Count rate / counts per minute: 412, 329, 264, 212, 171, 138
The source emits only one type of radiation. State which type it is and give one reason for your answer.
Determine the thickness of aluminium that halves the count rate from the source.
Each further thickness of aluminium equal to your answer to (b) halves the corrected count rate again. Calculate the count rate the detector would show with 6 mm of aluminium.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Beta (particles) | ✓ 1 | |
| Count rate falls steadily with thickness of aluminium: alpha would be stopped completely by the first thin sheet, whereas gamma would be hardly reduced by a few millimetres | ✓ 1 | |
| Part (b) | ||
| Corrected count rate with no aluminium = 412 − 12 = 400 counts per minute, so half is 200 counts per minute | ✓ 1 | |
| 212 − 12 = 200 at 3 mm, so thickness = 3 mm | ✓ 1 | |
| Part (c) | ||
| 6 mm is two halvings (two thicknesses of 3 mm), ECF from (b) | ✓ 1 | |
| Corrected count rate = 400 ÷ 4 = 100 counts per minute | ✓ 1 | |
| Add background: 100 + 12 = 112 counts per minute | ✓ 1 | |
Exam tip: In count-rate tables, subtract the background reading from every value before comparing them.
At a port, a gamma source and a detector are used to scan sealed shipping containers. The detector is on the opposite side of the container from the source. With no container in place, the detector reads 210 counts per minute. With a loaded container in place, it reads 60 counts per minute. With the source removed, it reads 10 counts per minute.
Explain why gamma radiation is used for this scan rather than alpha radiation.
State one safety precaution for workers who handle the source.
Calculate the fraction of the gamma count rate from the source that passes through the container and cargo.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Gamma radiation is very penetrating, so it can pass through the container and cargo (alpha would be stopped by the container wall) | ✓ 1 | |
| Part (b) | ||
| Any one from: store in a lead-lined container, use tongs or remote handling, keep the source pointing away from people, limit time of exposure, wear a radiation badge | ✓ 1 | |
| Part (c) | ||
| Subtract background: 210 − 10 = 200 and 60 − 10 = 50 | ✓ 1 | |
| Fraction = 50 ÷ 200 | ✓ 1 | |
| 0.25 (one quarter) | ✓ 1 | |
Exam tip: Subtract the background from both readings before forming a fraction, otherwise the result will be wrong.
A potassium-40 nucleus decays by emitting a beta-minus particle and becomes a calcium-40 nucleus. Which row describes the change in the nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: inside the nucleus a neutron changes into a proton and an electron (the beta-minus particle) is emitted, so protons go from 19 to 20 while the nucleon number stays 40. A describes alpha decay. B has the proton number changing in the wrong direction (as if a proton became a neutron). C wrongly treats the emitted particle as a lost nucleon. | ✓ 1 | Answer D |
Answer: D
Common error: Do not describe beta-minus emission as an orbital electron leaving; the electron is created in the nucleus when a neutron becomes a proton.
Which statement explains why a fusion reactor must reach a temperature of many millions of degrees?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: nuclei are positively charged and repel, so only very fast-moving (very hot) nuclei approach close enough to fuse. B describes the moderator in a fission reactor. C is irrelevant: a change of state does not make nuclei fuse. D is wrong because fusion does not require the fuel to become radioactive first. | ✓ 1 | Answer A |
Answer: A
Common error: High temperature in fusion is needed to overcome repulsion between nuclei, not to start a chain reaction.
A space probe is powered by a generator containing a radioactive isotope that emits alpha particles. The isotope has a half-life of 22 years. The thermal power produced by the generator is proportional to the activity of the isotope. At launch, the thermal power is 1600 W.
Explain why an alpha emitter is suitable as a heat source inside a sealed generator.
Calculate the thermal power produced 66 years after launch.
The probe needs a thermal power of at least 400 W to keep all its instruments working. Calculate the number of years after launch for which all the instruments can be used.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particles have a short range / are easily absorbed by the generator casing | ✓ 1 | |
| so all of their kinetic energy is transferred to thermal energy inside the generator | ✓ 1 | |
| Part (b) | ||
| 66 years is 3 half-lives | ✓ 1 | |
| power = 1600 / 23 | ✓ 1 | |
| 200 W | ✓ 1 | |
| Part (c) | ||
| the power halves twice (1600 → 800 → 400) | ✓ 1 | |
| 2 × 22 = 44 years | ✓ 1 | |
Common error: In half-life questions count how many halvings fit into the time; do not multiply the starting value by the number of half-lives.
A factory that roasts coffee rolls aluminium foil for its packaging. A source of beta-minus radiation is placed on one side of the moving foil and a radiation detector is placed on the other side. The detector is connected to a counter that is used to monitor the thickness of the foil.
State what a beta-minus particle is.
Explain how a nucleus produces a beta-minus particle.
The foil becomes thicker. Describe the change in the count rate and explain it.
Explain why a source of alpha radiation or a source of gamma radiation would not be suitable for this gauge.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a high-speed electron emitted from the nucleus of an atom | ✓ 1 | |
| Part (b) | ||
| a neutron in the nucleus changes into a proton | ✓ 1 | |
| an electron is created and ejected from the nucleus at high speed (accept an antineutrino is also emitted) | ✓ 1 | |
| Part (c) | ||
| the count rate decreases | ✓ 1 | |
| because more of the beta particles are absorbed by the thicker foil | ✓ 1 | |
| Part (d) | ||
| alpha particles would be absorbed by the foil (or air) so none reach the detector | ✓ 1 | |
| gamma rays pass through the foil almost unchanged so the count rate would hardly change with thickness | ✓ 1 | |
Exam tip: When asked to explain a choice of radiation, compare penetrating power with the material it must pass through.
A hospital uses a radioactive isotope of calcium as a tracer to study bone growth. The table shows the corrected count rate from a sample of the isotope.
| Time / days | Corrected count rate / counts per minute |
|---|---|
| 0 | 960 |
| 6 | 480 |
| 12 | 240 |
Determine the half-life of the isotope.
Predict the corrected count rate 24 days after the start.
Explain why an isotope with a half-life of several years would be unsuitable as a medical tracer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the count rate halves from 960 to 480 in the first 6 days | ✓ 1 | |
| half-life = 6 days | ✓ 1 | |
| Part (b) | ||
| 24 days is 4 half-lives | ✓ 1 | |
| 960 / 24 = 60 counts per minute | ✓ 1 | |
| Part (c) | ||
| any valid reason, for example: the patient would be irradiated for a very long time | ✓ 1 | |
| a second valid reason, for example: few nuclei decay each second so the signal would be weak | ✓ 1 | |
Common error: Take the half-life from the time for the count rate to halve, not from the time for it to reach zero.
Before a bridge repair, a technician tests a sealed source found in a storeroom. A Geiger-Müller tube measures the corrected count rate with different materials placed between the source and the tube.
| Material | Corrected count rate / counts per minute |
|---|---|
| none | 412 |
| sheet of paper | 409 |
| 3 mm of aluminium | 4 |
| 50 mm of lead | 2 |
Deduce the type of radiation emitted by the source. Give reasons for your answer.
Calculate the percentage decrease in the corrected count rate when the 3 mm sheet of aluminium is placed between the source and the tube.
Suggest one change to the procedure that would make the readings more reliable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the paper hardly reduces the count rate so the source is not an alpha emitter | ✓ 1 | |
| the aluminium absorbs almost all of the radiation, which is consistent with beta particles | ✓ 1 | |
| the count rate after aluminium is close to zero so there is no significant gamma radiation | ✓ 1 | |
| Part (b) | ||
| (412 − 4) / 412 × 100 | ✓ 1 | |
| 99.0 % | ✓ 1 | |
| Part (c) | ||
| repeat each reading and take an average (or count for a longer time) | ✓ 1 | |
Exam tip: For a table of absorber results, state which radiation is ruled out by each result as well as which one is present.
A potassium-40 nucleus (19 protons, 21 neutrons) decays into a calcium-40 nucleus (20 protons, 20 neutrons) by emitting a beta-minus particle. Which change inside the nucleus accounts for this?
| Marking point | Mark | Notes |
|---|---|---|
| The number of protons rises by one and the number of neutrons falls by one while the total stays at 40, so a neutron has changed into a proton and a fast electron (the beta particle) is created and emitted. A proton changing into a neutron would lower the proton number. Ejecting a neutron would reduce the nucleon number to 39. Emitting two protons and two neutrons is alpha decay, which would reduce the nucleon number to 36. | ✓ 1 | Answer D |
Answer: D
Common error: A beta particle is not in the nucleus beforehand; it is created when a neutron changes into a proton.
A student investigates a radioactive source using a Geiger-Müller tube. The count rates (in counts per minute) are: source with no absorber 840; source with a sheet of paper 836; source with 2 mm of aluminium 212; source with 20 mm of lead 118. With the source removed, the tube records 36 counts per minute.
State why the student measured the count rate with the source removed.
Deduce which types of radiation the source emits. Use the corrected count rates in your answer.
Calculate the percentage of the corrected count rate removed by the aluminium. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| to find the background count rate, caused by radiation already present (for example from rocks or cosmic rays), so that it can be subtracted from each reading | ✓ 1 | |
| Part (b) | ||
| not alpha: paper changes the corrected rate very little (800 compared with 804) | ✓ 1 | |
| beta is present: 2 mm of aluminium cuts the corrected rate greatly (804 to 176) | ✓ 1 | |
| gamma is present: a significant corrected rate remains after the aluminium, and it is reduced further by lead (82) | ✓ 1 | |
| Part (c) | ||
| M1 corrected rates 840 − 36 = 804 and 212 − 36 = 176 | ✓ 1 | |
| A1 (804 − 176) ÷ 804 × 100 = 78.1 % | ✓ 1 | |
Exam tip: Subtract the background from every reading before you compare or take ratios.
In a seafood factory a robot arm loads packaged fish fillets onto a conveyor that carries them past a sealed cobalt-60 source. The source emits gamma rays, which kill bacteria on the fish.
Suggest why a gamma emitter is used rather than a beta emitter.
Explain why the fillets do not become radioactive.
State one precaution that protects the factory workers from the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma rays pass through the packaging and the fish, whereas beta particles would be absorbed by the packaging | ✓ 1 | |
| Part (b) | ||
| the source is sealed, so no radioactive material gets onto the fish (they are irradiated, not contaminated) | ✓ 1 | |
| the gamma rays ionise atoms but do not make the nuclei of the atoms in the fish unstable | ✓ 1 | |
| Part (c) | ||
| any one of: loading and conveyor run by robot or remote control; thick concrete or lead shielding around the source; keeping people at a distance; limiting the time spent near the source | ✓ 1 | |
Common error: Irradiated objects are not radioactive; they are only radioactive if radioactive material gets on or in them.
A gamma source is used to inspect welds in a pipeline. At installation the source has an activity of 3.2 TBq. Its half-life is 74 days.
Explain why a gamma source, rather than an alpha source, is suitable for inspecting welds inside a steel pipeline.
State one reason why the source must be replaced after a long time.
Calculate the activity of the source 222 days after installation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma rays can pass through the steel wall to reach the film or detector | ✓ 1 | |
| alpha particles would be absorbed by a few centimetres of air or by the pipe wall | ✓ 1 | |
| Part (b) | ||
| its activity has fallen, so much longer exposure times would be needed to form an image | ✓ 1 | |
| Part (c) | ||
| M1 222 days is 3 half-lives (222 ÷ 74 = 3) | ✓ 1 | |
| A1 3.2 ÷ 2 ÷ 2 ÷ 2 = 0.40 TBq | ✓ 1 | |
Exam tip: For half-life questions, first find the number of half-lives, then halve that many times.
Potassium-40 occurs naturally in the human body. A potassium-40 nucleus has 19 protons and a nucleon number of 40. It can decay by beta-minus emission to form calcium-40, which has 20 protons.
Determine the number of neutrons in a potassium-40 nucleus.
Explain how the numbers of protons, neutrons and nucleons change when potassium-40 decays to calcium-40.
A bone sample contains 6.4×1012 potassium-40 nuclei. The half-life of potassium-40 is 1.25×109 years. Calculate the number of potassium-40 nuclei that remain after 3.75×109 years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 40 − 19 = 21 neutrons | ✓ 1 | |
| Part (b) | ||
| a neutron changes into a proton (and an electron is emitted), so protons increase from 19 to 20 and neutrons decrease from 21 to 20 | ✓ 1 | |
| the total number of nucleons stays at 40 | ✓ 1 | |
| Part (c) | ||
| M1 3.75×109 years is 3 half-lives | ✓ 1 | |
| A1 6.4×1012 ÷ 8 = 8.0×1011 nuclei | ✓ 1 | |
Common error: Beta decay keeps the nucleon number the same; only the proton number changes.
A fisheries scientist fits small beta-emitting tags to fish. A detector held close to a tag gives a corrected count rate of 1280 counts per minute at release and 320 counts per minute 18 days later.
Explain why a beta emitter is used for the tag rather than an alpha emitter.
Determine the half-life of the tag.
Calculate the corrected count rate 45 days after release. Use a half-life of 9.0 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particles would be absorbed by the tissue of the fish, whereas beta particles can pass through it to reach a detector outside | ✓ 1 | |
| Part (b) | ||
| M1 1280 → 640 → 320, so 18 days is 2 half-lives | ✓ 1 | |
| A1 18 ÷ 2 = 9.0 days | ✓ 1 | |
| Part (c) | ||
| M1 45 days is 5 half-lives | ✓ 1 | |
| A1 1280 ÷ 25 = 40 counts per minute | ✓ 1 | |
Exam tip: Write the halving sequence out (1280, 640, 320) to check the number of half-lives.
A bismuth-213 nucleus has 83 protons. It emits a beta-minus particle and then an alpha particle. What are the nucleon number and the proton number of the nucleus formed?
| Marking point | Mark | Notes |
|---|---|---|
| The alpha particle reduces the nucleon number by 4 (213 to 209) and the proton number by 2. The beta-minus particle raises the proton number by 1 and leaves the nucleon number unchanged. So the proton number is 83 + 1 − 2 = 82. Proton number 84 forgets that alpha emission lowers it. Nucleon number 213 forgets that alpha emission lowers it. Proton number 81 forgets the increase from the beta decay. | ✓ 1 | Answer D |
Answer: D
Exam tip: Do alpha and beta changes separately: alpha changes both numbers, beta changes only the proton number.
A radioactive isotope has a half-life of 6.0 days. A pallet of this isotope is kept in a store for 18 days. What fraction of the original nuclei has decayed by the end of the storage?
| Marking point | Mark | Notes |
|---|---|---|
| 18 days is 3 half-lives, so the fraction remaining is 1/2 × 1/2 × 1/2 = 1/8 and the fraction decayed is 7/8. A fraction of 1/8 is the fraction still present. 3/8 comes from adding three separate eighths and 1/3 from dividing the half-life by the storage time, and neither is correct. | ✓ 1 | Answer B |
Answer: B
Common error: Decide whether the question asks for the fraction remaining or the fraction decayed.
A waste store holds containers contaminated with neptunium-237, which has 93 protons. Its decay series ends at the stable nucleus bismuth-209, which has 83 protons. Only alpha particles and beta-minus particles are emitted in the series.
Calculate the number of alpha particles emitted in the series.
Calculate the number of beta-minus particles emitted in the series.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (237 − 209) ÷ 4 | ✓ 1 | |
| A1 7 alpha particles | ✓ 1 | |
| Part (b) | ||
| M1 7 alpha particles lower the proton number to 93 − 14 = 79 | ✓ 1 | |
| A1 83 − 79 = 4 beta-minus particles | ✓ 1 | |
Exam tip: Use the change in nucleon number to count alpha particles first, then use the proton number for beta particles.
A logistics company stored a sealed radioactive source in a depot for 20 years. After this time, 93.75% of the radioactive nuclei in the source had decayed.
Calculate the fraction of the original nuclei that remained.
Determine the number of half-lives that passed during the 20 years.
Calculate the half-life of the source.
The source is moved to a freezer. State whether this changes its half-life and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1 − 0.9375 = 0.0625 (or 1/16) | ✓ 1 | |
| Part (b) | ||
| 1/2 → 1/4 → 1/8 → 1/16, so 4 half-lives | ✓ 1 | |
| Part (c) | ||
| 20 ÷ 4 = 5.0 years | ✓ 1 | |
| Part (d) | ||
| no, half-life is not affected by temperature or other physical or chemical conditions | ✓ 1 | |
Common error: The fraction remaining is not the same as the fraction decayed.
Radon-222, which has 86 protons, is a radioactive gas that can build up inside buildings. It decays by emitting an alpha particle to form polonium.
Determine the nucleon number and proton number of the polonium nucleus formed.
Explain why radon gas is more hazardous when it is breathed in than when it is outside the body.
A sealed sample of air contains 6.4×105 radon-222 nuclei. The half-life of radon-222 is 3.8 days. Calculate the number of radon-222 nuclei remaining after 15.2 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number 222 − 4 = 218 | ✓ 1 | |
| proton number 86 − 2 = 84 | ✓ 1 | |
| Part (b) | ||
| outside the body, alpha particles are stopped by air or the skin | ✓ 1 | |
| inside the lungs, alpha particles ionise cells directly in the tissue, which can cause damage | ✓ 1 | |
| Part (c) | ||
| M1 15.2 days is 4 half-lives | ✓ 1 | |
| A1 6.4×105 ÷ 16 = 4.0×104 nuclei | ✓ 1 | |
Exam tip: Danger from alpha emitters depends on whether the source is inside or outside the body.
An architect specifies smoke detectors that contain a small sealed source of americium-241, which has 95 protons and a half-life of 432 years. Americium-241 decays by alpha emission to neptunium.
Explain why an alpha source is suitable inside a smoke detector.
Explain why a source with a long half-life is suitable.
Determine the nucleon number and the proton number of the neptunium nucleus formed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha particles ionise the air strongly, so a small electric current flows | ✓ 1 | |
| the particles are absorbed by the casing, so little radiation escapes | ✓ 1 | |
| Part (b) | ||
| its activity stays almost constant, so the detector works for many years | ✓ 1 | |
| Part (c) | ||
| nucleon number 241 − 4 = 237 | ✓ 1 | |
| proton number 95 − 2 = 93 | ✓ 1 | |
Exam tip: Alpha particles are strongly ionising, which is the useful property in a smoke detector.
A technician at a coffee roasting plant investigates radioactivity in a sack of green coffee beans, which contain a little potassium-40. She first measures the background radiation with a Geiger–Müller tube and counts 330 counts in 10 minutes.
State two natural sources of background radiation.
Calculate the background count rate in counts per minute.
With the sack placed beside the tube, the count rate is 71 counts per minute. Calculate the count rate caused by the beans alone.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any one from: radon gas, rocks and soil, cosmic rays, food and drink | ✓ 1 | |
| a second different natural source | ✓ 1 | |
| Part (b) | ||
| 330 ÷ 10 = 33 counts per minute | ✓ 1 | |
| Part (c) | ||
| M1 subtracts the background: 71 − 33 | ✓ 1 | |
| A1 38 counts per minute | ✓ 1 | |
Exam tip: The corrected count rate is always the measured rate minus the background rate.
A space probe carries a small radioactive source used to test its radiation detector. The activity of the source was 4800 Bq at launch. 54 years later its activity was 600 Bq.
State what is meant by the half-life of a radioactive source.
Calculate the half-life of the source.
The probe continues to operate for a further 36 years. Calculate the activity of the source at the end of this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the time taken for the activity (or number of undecayed nuclei) to fall to half of its original value | ✓ 1 | |
| Part (b) | ||
| M1 4800 → 2400 → 1200 → 600, so three half-lives | ✓ 1 | |
| M1 54 ÷ 3 | ✓ 1 | |
| A1 18 years | ✓ 1 | |
| Part (c) | ||
| M1 36 years is two half-lives, so 600 → 300 → 150 | ✓ 1 | |
| A1 150 Bq (ECF from (b)) | ✓ 1 | |
Exam tip: Work out the number of half-lives first, then halve the activity that many times.
A gamma source is used to check the welds in a steel bridge. The source is held on one side of a weld and photographic film is placed on the other side. The source has a half-life of 4.0 years and an activity of 3200 MBq when new.
Explain why a gamma source is used rather than an alpha source.
Calculate the activity of the source after 12 years.
Suggest two safety precautions for the technician using the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha radiation is stopped by (a thin layer of) the steel and cannot reach the film | ✓ 1 | |
| gamma radiation is very penetrating and passes through the steel to the film, where differences in absorption show up faults | ✓ 1 | |
| Part (b) | ||
| M1 12 years is three half-lives | ✓ 1 | |
| A1 3200 → 1600 → 800 → 400 MBq | ✓ 1 | |
| Part (c) | ||
| any one from: handle with long tongs or remotely, keep the source in a lead container when not in use, stand behind a shield | ✓ 1 | |
| any different one from: keep as far away as possible, limit the time of exposure, wear a dose badge | ✓ 1 | |
Common error: Penetrating power and ionising power are opposite: alpha is the most ionising but the least penetrating.
A student measures the activity of a radioactive sample. The background count rate in the laboratory is 20 counts per minute. The table gives the count rate measured with the sample present.
Time = 0 min: 340 counts per minute
Time = 12 min: 180 counts per minute
Time = 24 min: 100 counts per minute
Time = 36 min: 60 counts per minute
State the corrected count rate at time 0.
Determine the half-life of the sample.
Predict the count rate that the student would measure at time 48 min.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 340 − 20 = 320 counts per minute | ✓ 1 | |
| Part (b) | ||
| M1 subtracts the background from the other readings: 160, 80 and 40 counts per minute | ✓ 1 | |
| M1 finds the time for the corrected count rate to halve, for example 320 → 160 in 12 minutes | ✓ 1 | |
| A1 12 minutes | ✓ 1 | |
| Part (c) | ||
| M1 corrected rate halves again: 40 → 20 counts per minute at 48 min | ✓ 1 | |
| A1 add the background: 20 + 20 = 40 counts per minute | ✓ 1 | |
Exam tip: After predicting a corrected count rate, remember to add the background back if the question asks for the measured rate.
The table shows the average annual radiation dose for one person in a region.
Radon gas: 1.30 mSv
Rocks and buildings: 0.35 mSv
Cosmic rays: 0.35 mSv
Food and drink: 0.30 mSv
Medical procedures: 0.40 mSv
Other artificial sources: 0.10 mSv
Calculate the total annual dose.
Calculate the percentage of the total dose that comes from natural sources. Give your answer to 2 significant figures.
A space probe outside the Earth's atmosphere receives a much larger dose from cosmic rays than a person on the ground. Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2.80 mSv | ✓ 1 | |
| Part (b) | ||
| M1 natural sources total 1.30 + 0.35 + 0.35 + 0.30 = 2.30 mSv | ✓ 1 | |
| M1 2.30 ÷ 2.80 × 100 | ✓ 1 | |
| A1 82% (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the atmosphere absorbs most cosmic rays, but there is no atmosphere to shield the probe | ✓ 1 | |
Common error: Medical procedures are an artificial source, so leave them out when you add up the natural sources.
A nucleus of neptunium-237 (23793Np) decays through a series of alpha and beta-minus emissions to become bismuth-209 (20983Bi). How many alpha particles and how many beta-minus particles are emitted in total?
| Marking point | Mark | Notes |
|---|---|---|
| Only alpha emission changes the nucleon number, and each alpha lowers it by 4: (237 − 209) ÷ 4 = 7 alpha particles. These 7 alpha particles lower the proton number from 93 to 79, but bismuth has 83 protons, so 4 beta-minus emissions (each raising the proton number by 1) are needed. The option with 4 alpha and 7 beta-minus swaps the two counts. The option with 7 alpha and 10 beta-minus assumes the whole fall of 10 in proton number comes from beta decay and ignores the charge lost in alpha decay. The option with 28 alpha treats the fall in nucleon number as the number of alpha particles instead of dividing by 4. | ✓ 1 | Answer B |
Answer: B
Common error: Alpha emission changes both the proton number (by 2) and the nucleon number (by 4); work out the alpha count from the nucleon number first.
The count rate from a radioactive source is measured in a laboratory where the background count rate is 20 counts per minute. At the start the measured count rate is 180 counts per minute. Twelve minutes later it is 60 counts per minute. What is the half-life of the source?
| Marking point | Mark | Notes |
|---|---|---|
| Subtracting the background gives corrected count rates of 160 and 40 counts per minute. The count rate has fallen by a factor of 4, which is 2 half-lives, so the half-life is 12 ÷ 2 = 6.0 minutes. The value 3.0 minutes treats the factor of 4 as 4 half-lives. The value 4.0 minutes uses 180 ÷ 60 = 3 as the number of half-lives and forgets to remove the background. The value 12 minutes takes the whole interval as one half-life. | ✓ 1 | Answer C |
Answer: C
Exam tip: Always subtract the background count rate from every reading before looking for halving times.
In one short interval, 4000 uranium-235 nuclei undergo fission in a research reactor that is operating at constant power. On average, each fission releases 2.5 neutrons.
Calculate the total number of neutrons released in this interval.
State how many of these neutrons must go on to cause a fission, and explain your answer.
Of the neutrons that do not cause fission, 1800 escape from the reactor core and 700 are absorbed by uranium-238 nuclei without causing fission. The rest are absorbed by the control rods. Calculate the number absorbed by the control rods.
The control rods are now partly withdrawn. Explain what happens to the power output of the reactor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4000 × 2.5 | ✓ 1 | |
| = 10 000 neutrons | ✓ 1 | |
| Part (b) | ||
| 4000 | ✓ 1 | |
| power is constant, so the number of fissions in the next interval must equal 4000 (on average one neutron per fission causes another fission) | ✓ 1 | |
| Part (c) | ||
| 10 000 − 4000 − 1800 − 700 | ✓ 1 | |
| = 3500 neutrons | ✓ 1 | |
| (ECF from (a) and (b)) | — | |
| Part (d) | ||
| fewer neutrons are absorbed by the control rods | ✓ 1 | |
| so more neutrons cause fission, which means on average more than one neutron from each fission causes another fission | ✓ 1 | |
| the number of fissions per second increases, so the power output increases | ✓ 1 | |
Exam tip: For neutron-balance questions, write the total released first, then subtract each loss in turn so that every step can be checked.
A sports-science laboratory buys a batch of fluorine-18, a radioactive tracer with a half-life of 2.0 hours, for an imaging study of an athlete’s muscles. The batch contains 3.2 × 1012 unstable nuclei when it is made. It is used 8.0 hours after it is made.
Calculate the number of half-lives that pass before the tracer is used.
Calculate the fraction of the original nuclei that decay during these 8.0 hours. Give your answer as a fraction.
Calculate the number of nuclei that decay before the tracer is used.
Suggest two reasons why a tracer with a half-life of a few hours is chosen rather than one with a half-life of several years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8.0 ÷ 2.0 = 4 | ✓ 1 | |
| Part (b) | ||
| fraction remaining = (1/2)4 = 1/16 | ✓ 1 | |
| fraction decayed = 15/16 | ✓ 1 | |
| Part (c) | ||
| 3.2 × 1012 × 15/16 | ✓ 1 | |
| = 3.0 × 1012 nuclei | ✓ 1 | |
| (ECF from (b)) | — | |
| Part (d) | ||
| first valid reason | ✓ 1 | |
| second valid reason | ✓ 1 | |
| Accept any two of: the tracer does not stay radioactive in the body for long, so the dose after the scan is small; a long half-life gives a low activity from the same number of nuclei, so a poor image; the waste decays away quickly so storage and disposal are easier | — | |
Common error: Decayed and remaining fractions are different; check that the two add up to 1.
A student measures the count rate from a small radioactive source at 10-minute intervals. The background count rate, measured separately, is 24 counts per minute.
Time / min: 0, 10, 20, 30, 40
Measured count rate / counts per minute: 424, 224, 124, 74, 49
State what is meant by background radiation and give one source of it.
Show that the half-life of the source is 10 minutes.
Calculate the count rate the student would measure at 50 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| radiation that is always present in the environment, not from the source being studied | ✓ 1 | |
| one source from: radon gas, rocks and soil, cosmic rays, food, medical sources, fallout | ✓ 1 | |
| Part (b) | ||
| corrected count rates are 400, 200, 100, 50 and 25 counts per minute after subtracting 24 | ✓ 1 | |
| the corrected count rate halves in every 10-minute interval, for example 400 to 200 | ✓ 1 | |
| so the half-life is 10 minutes | ✓ 1 | |
| Part (c) | ||
| corrected count rate = 25 ÷ 2 = 12.5 counts per minute | ✓ 1 | |
| add the background: 12.5 + 24 | ✓ 1 | |
| = 36.5 counts per minute | ✓ 1 | |
Common error: Remove the background before finding the half-life, then add it back when predicting a measured count rate.
A hospital uses a sample of technetium-99m with a half-life of 6.0 hours. At 08:00 the sample has an activity of 480 MBq. What is its activity at 02:00 the next morning, 18 hours later?
| Marking point | Mark | Notes |
|---|---|---|
| 18 hours is three half-lives (18 ÷ 6.0 = 3). The activity halves three times: 480 → 240 → 120 → 60 MBq, so the last option is correct. 240 MBq is the value after only one half-life. 160 MBq comes from dividing by 3 instead of halving three times. 120 MBq is the value after only two half-lives. | ✓ 1 | Answer D |
Answer: D
Exam tip: Work out how many half-lives have passed first, then halve the activity that many times.
The space probe Meridian is powered by a generator that uses plutonium-238, 23894Pu. Plutonium-238 emits alpha particles and has a half-life of 88 years. The heat produced in the generator is converted into electrical power.
State the number of protons and the number of neutrons in a nucleus of plutonium-238.
Give two reasons why an alpha-emitting source with a long half-life is suitable for powering a probe.
When the probe is launched the generator releases heat at a rate of 440 W. Calculate the rate at which the generator releases heat 176 years after launch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 94 protons | ✓ 1 | |
| 238 − 94 = 144 neutrons | ✓ 1 | |
| Part (b) | ||
| Any two from: alpha particles are absorbed inside the source or its casing so all of the energy is transferred as heat; little shielding is needed because alpha particles are easily stopped; long half-life so the power output stays high for many years; no need to refuel during the mission | ✓ 2 | |
| Part (c) | ||
| M1 176 ÷ 88 = 2 half-lives | ✓ 1 | |
| M1 power is halved twice: 440 → 220 → 110 | ✓ 1 | |
| A1 110 W | ✓ 1 | |
Common error: A quantity that halves each half-life does not fall by half of the original value each time: halve the remaining value.
A hospital physicist measures the activity of a sample of a radioactive isotope used as a tracer. The table shows her results.
| Time / h | 0 | 3 | 6 | 9 | 12 |
| Activity / MBq | 640 | 453 | 320 | 226 | 160 |
Use the table to determine the half-life of the isotope.
Calculate how long it takes for the activity to fall from 640 MBq to 40 MBq.
The isotope emits gamma rays. Explain why a gamma emitter with a short half-life is a good choice for a tracer injected into a patient.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 identifies that the activity falls from 640 MBq to 320 MBq | ✓ 1 | |
| A1 half-life = 6 h | ✓ 1 | |
| Part (b) | ||
| M1 four halvings are needed: 640 → 320 → 160 → 80 → 40 | ✓ 1 | |
| M1 4 × 6 h | ✓ 1 | |
| A1 24 h | ✓ 1 | |
| Part (c) | ||
| Gamma rays pass out of the body and can be detected outside it | ✓ 1 | |
| Short half-life means the activity falls quickly, so the patient is exposed to radiation for a short time / little radioactive material remains | ✓ 1 | |
Exam tip: Read the half-life from the data by finding the time for a value to halve; you do not need to start at time zero.
A Formula 1 team uses gamma radiography to check welds in a titanium suspension part. A gamma source is placed on one side of the part and photographic film on the other. When the source is new its activity is 1.6 TBq and the film needs an exposure time of 15 s. The source has a half-life of 74 days.
Explain why gamma rays, rather than alpha or beta particles, are used to check the inside of a metal part.
Calculate the activity of the source 222 days after it was new.
The film needs the same total exposure each time, and the exposure is proportional to the activity multiplied by the time. Calculate the exposure time needed 222 days after the source was new.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Gamma rays are highly penetrating and pass through the metal, whereas alpha and beta particles would be absorbed | ✓ 1 | |
| Part (b) | ||
| M1 222 ÷ 74 = 3 half-lives | ✓ 1 | |
| A1 1.6 ÷ 8 = 0.20 TBq | ✓ 1 | |
| Part (c) | ||
| M1 the activity is 8 times smaller so the time is 8 times longer | ✓ 1 | |
| A1 15 × 8 = 120 s | ✓ 1 | |
Exam tip: If two quantities are proportional in this way, a fall in one by a factor of 8 means the other must rise by a factor of 8.
Fluorine-18, 189F, is a radioactive isotope used in hospital scanners. Fluorine-19, 199F, is a stable isotope of the same element. Fluorine-18 has a half-life of 110 minutes.
Explain why fluorine-18 and fluorine-19 are isotopes of the same element.
State the number of electrons in a neutral atom of fluorine-18.
A patient is injected with a sample of fluorine-18 of activity 400 MBq. Calculate the time taken for the activity of the sample to fall to 25 MBq.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Both nuclei have 9 protons | ✓ 1 | |
| They have different numbers of neutrons, 9 and 10 | ✓ 1 | |
| Part (b) | ||
| 9 | ✓ 1 | |
| Part (c) | ||
| M1 four halvings: 400 → 200 → 100 → 50 → 25 | ✓ 1 | |
| M1 4 × 110 minutes | ✓ 1 | |
| A1 440 minutes (7 h 20 min) | ✓ 1 | |
Exam tip: Isotopes always have the same proton number; only the neutron number (and so the nucleon number) differs.
A factory makes thin plastic film. The thickness is monitored continuously by placing a radioactive source on one side of the film and a detector on the other. Which source is most suitable?
| Marking point | Mark | Notes |
|---|---|---|
| Beta particles pass through thin plastic but are partly absorbed, so the detected count rate changes with thickness, and a half-life of years keeps the source activity almost constant. Alpha particles would be stopped completely by the film. Gamma rays would pass through almost unchanged, so changes in thickness would hardly alter the reading. A 20 second half-life would make the source useless within minutes. | ✓ 1 | Answer D |
Answer: D
Exam tip: When choosing a source, check both the type of radiation (penetration) and the half-life (how long the source lasts).
Lead-214 is one of the nuclei formed as radon-222 decays. A nucleus of lead-214 (Pb) has 82 protons and it decays by beta-minus emission to a nucleus of bismuth (Bi).
State the number of neutrons in a nucleus of lead-214.
Explain why the nucleon number does not change in beta-minus decay.
Complete the equation for this decay and state the atomic number of bismuth.
21482Pb → ……Bi + 0−1e
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 214 − 82 = 132 | ✓ 1 | |
| Part (b) | ||
| a neutron changes into a proton | ✓ 1 | |
| and an electron is emitted, so the total number of protons plus neutrons stays the same | ✓ 1 | |
| Part (c) | ||
| mass number of bismuth is 214 | ✓ 1 | |
| atomic number of bismuth is 83 | ✓ 1 | |
Common error: A beta-minus particle is an electron from the nucleus, so the atomic number goes up by 1 while the mass number stays the same.
A company has three radioactive sources available for industrial use.
Source A: alpha emitter, half-life 430 years.
Source B: beta emitter, half-life 28 years.
Source C: gamma emitter, half-life 6.0 hours.
A paper mill wants to monitor the thickness of paper as it leaves the rollers. State which source is most suitable and give two reasons for your choice.
A water company wants to find a leak in a buried pipe by adding a source to the water and searching the ground with a detector. State which source is most suitable and give one reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| source B | ✓ 1 | |
| beta is partly absorbed by paper, so the count rate depends on thickness (alpha would be stopped, gamma would pass through almost unchanged) | ✓ 1 | |
| long half-life so the activity stays almost constant for years | ✓ 1 | |
| Part (b) | ||
| source C | ✓ 1 | |
| reason, any one from: gamma passes through soil and the pipe wall to reach the detector; short half-life means it decays away quickly and leaves little lasting contamination | ✓ 1 | |
Exam tip: Alpha is stopped by paper, beta by a few millimetres of aluminium and gamma only by thick lead or concrete, so match the radiation to the material it must pass through.
Engineers add a small amount of a radioactive material to the oil in a locomotive axle bearing, to measure how fast the bearing wears. The count rate from the oil sample, corrected for background, is measured over time.
Time / h: 0, 4, 8, 12
Corrected count rate / counts per minute: 800, 400, 200, 100
Determine the half-life of the material.
Predict the corrected count rate after 20 hours.
Explain why the count rate was corrected for background radiation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time for the count rate to fall from 800 to 400 | ✓ 1 | |
| = 4.0 hours | ✓ 1 | |
| Part (b) | ||
| 20 hours is 5 half-lives | ✓ 1 | |
| 800 ÷ 32 = 25 counts per minute | ✓ 1 | |
| Part (c) | ||
| background radiation from the surroundings is also detected, so it must be subtracted to find the count from the oil sample alone | ✓ 1 | |
Exam tip: To find a half-life from a table, choose any count rate and find how long it takes to halve, then check this with a second pair of readings.
A technician uses a source of iridium-192, which emits gamma radiation, to check welds in a steel railway track. The source has a half-life of 74 days and an initial activity of 4.0 × 1011 Bq.
State two safety precautions the technician should take when handling the source.
Explain why a gamma source, rather than an alpha source, is used to check the welds.
Calculate the activity of the source after 222 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any two from: use long tongs or remote handling; store in a lead-lined container; keep exposure time short; keep as far from the source as possible; wear a dose badge | ✓ 2 | |
| Part (b) | ||
| alpha particles are stopped by the steel | ✓ 1 | |
| gamma rays pass through the steel and can be detected or recorded on the other side | ✓ 1 | |
| Part (c) | ||
| 222 days is 3 half-lives | ✓ 1 | |
| 4.0 × 1011 ÷ 8 = 5.0 × 1010 Bq | ✓ 1 | |
Exam tip: Divide the elapsed time by the half-life first; if the result is a whole number, halve the activity that many times.
A sealed source used to check the thickness of metal sheet emits beta-minus particles. What is a beta-minus particle?
| Marking point | Mark | Notes |
|---|---|---|
| In beta-minus decay a neutron in the nucleus changes into a proton and an electron, and the electron is emitted at high speed. A helium nucleus is an alpha particle, a short-wavelength electromagnetic wave is gamma radiation, and an electron knocked out of an inner shell comes from the electron cloud, not from the nucleus, so it is not beta radiation. | ✓ 1 | Answer D |
Answer: D
Common error: A beta particle is an electron, but it comes from the nucleus, not from the electron shells around it.
A technician in a data centre keeps a sealed radioactive source in storage for a time equal to 4 half-lives. What fraction of the original radioactive nuclei have decayed by the end of storage?
| Marking point | Mark | Notes |
|---|---|---|
| After 4 half-lives the fraction remaining is (1/2)4 = 1/16, so the fraction that has decayed is 1 − 1/16 = 15/16. The option 1/16 is the fraction left, not the fraction decayed. The option 1/4 comes from halving only twice. The option 3/4 comes from subtracting 1/4 from 1, which treats 4 half-lives as a quarter remaining. | ✓ 1 | Answer C |
Answer: C
Exam tip: Check whether the question asks for the fraction that has decayed or the fraction that remains.
A technician in a data centre stores a radioactive source used to test shielding. The half-life of the source is 8.0 days. The source is stored for 24 days.
Calculate the number of half-lives in the storage time.
Calculate the fraction of the original nuclei that have not yet decayed after storage.
Calculate the percentage of the original nuclei that have decayed during storage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 ÷ 8.0 = 3 | ✓ 1 | |
| Part (b) | ||
| (1/2)3 or halving three times | ✓ 1 | |
| 1/8 (0.125) | ✓ 1 | |
| Part (c) | ||
| 87.5% (accept 88%) | ✓ 1 | |
Exam tip: Find the number of half-lives first, then halve the amount that many times.
A cyclist's body contains radioactive potassium-40. A scan shows that the potassium-40 in the cyclist's body has an activity of 4000 Bq. The half-life of potassium-40 is 1.25 × 109 years.
State what is meant by an activity of 4000 Bq.
Explain why the activity stays almost constant during a one-hour race.
Calculate the number of decays in the cyclist's body during the one-hour race. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4000 nuclei decay (disintegrations) each second | ✓ 1 | |
| Part (b) | ||
| the half-life is very much longer than one hour, so almost no change in the number of nuclei | ✓ 1 | |
| Part (c) | ||
| 4000 × 3600 (s) | ✓ 1 | |
| 1.44 × 107 | ✓ 1 | |
Common error: Convert the time to seconds before multiplying by an activity in becquerel.
A smoke detector in a data centre contains americium-241, which emits alpha particles. The half-life of americium-241 is 432 years.
State what an alpha particle is.
Explain why the alpha particles from the detector are not a danger outside its casing.
Calculate the fraction of the original americium-241 nuclei remaining after 864 years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a helium nucleus (2 protons and 2 neutrons) | ✓ 1 | |
| Part (b) | ||
| alpha particles are stopped by a few centimetres of air or by thin material such as the casing | ✓ 1 | |
| Part (c) | ||
| 864 ÷ 432 = 2 half-lives, so (1/2)2 | ✓ 1 | |
| 1/4 (0.25) | ✓ 1 | |
Exam tip: Write the number of half-lives explicitly before halving; it makes the method easy to follow.
A technician at a medical isotope supplier checks a new batch of a short-lived isotope. The background count rate has already been subtracted from these readings.
Time 0 min: 960 counts per minute
Time 2 min: 680 counts per minute
Time 4 min: 480 counts per minute
Time 6 min: 340 counts per minute
Time 8 min: 240 counts per minute
Use the data to determine the half-life of the isotope.
Predict the corrected count rate at a time of 16 minutes.
Calculate the time at which the corrected count rate falls to 15 counts per minute.
Suggest why a reading taken at this time would be less reliable than the early readings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The count rate halves, for example from 960 to 480 or from 680 to 340 | ✓ 1 | |
| in 4 minutes, so the half-life = 4 min | ✓ 1 | |
| Part (b) | ||
| 16 ÷ 4 = 4 half-lives | ✓ 1 | |
| 960 ÷ 24 = 960 ÷ 16 = 60 counts per minute | ✓ 1 | |
| Part (c) | ||
| 960 ÷ 15 = 64 = 26, so 6 half-lives | ✓ 1 | |
| 6 × 4 = 24 minutes | ✓ 1 | |
| Part (d) | ||
| Decay is random, so the count rate fluctuates | ✓ 1 | |
| At low count rates these fluctuations, and any error in the background value, are a large fraction of the reading | ✓ 1 | |
Exam tip: Find the half-life using more than one pair of readings, for example 960 → 480 and 680 → 340. If both pairs give the same time, you can be confident in your value.
Engineers inspect the welds that join sections of a steel gas pipeline. A sealed source of iridium-192 is placed inside the pipe at each weld. Photographic film is wrapped around the outside of the weld. Iridium-192 emits gamma radiation and has a half-life of 74 days.
Explain how this method shows whether a weld contains a crack, and why iridium-192 is a suitable source for this inspection.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Gamma radiation is penetrating enough to pass through the thick steel wall and weld | ✓ 1 | |
| Alpha and beta radiation would be absorbed by the steel, so they would not reach the film | ✓ 1 | |
| Radiation reaching the film exposes it, making it darker | ✓ 1 | |
| A crack or air gap has less steel to absorb the gamma radiation, so more radiation passes through at that point | ✓ 1 | |
| The crack therefore shows up as a darker line or patch on the film, compared with the evenly exposed film over a good weld | ✓ 1 | |
| A 74-day half-life means the activity falls very little during a day's or week's work, so exposure times stay consistent and the source does not need frequent replacement | ✓ 1 | |
Exam tip: In a 6-mark explanation, link each step with 'so' or 'because': gamma penetrates, so it reaches the film; a crack absorbs less, so the film is darker there. A list of facts with no links scores less well.
Radon-222, 22286Rn, is a radioactive gas that can collect in the basements of houses. It decays by alpha emission to polonium (Po). This starts a chain of further decays.
Write a balanced nuclear equation for the alpha decay of radon-222.
Starting from one radon-222 nucleus, the chain includes two alpha decays and two beta decays in total.
Calculate the mass number and the atomic number of the nucleus formed after these four decays.
State, with a reason, whether the nucleus formed is the same element as polonium-218.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 22286Rn → 21884Po + 42He | — | |
| Polonium shown with mass number 218 | ✓ 1 | |
| Polonium shown with atomic number 84 | ✓ 1 | |
| Alpha particle shown correctly as 42He or 42α | ✓ 1 | |
| Part (b) | ||
| Mass number = 222 − (2 × 4) = 214, because beta decays do not change the mass number | ✓ 1 | |
| Two alpha decays lower the atomic number by 4 | ✓ 1 | |
| Two beta decays raise it by 2, so atomic number = 86 − 4 + 2 = 84 | ✓ 1 | |
| Part (c) | ||
| Yes, it is also polonium (an isotope of polonium-218) because it has the same atomic number (proton number) of 84 | ✓ 1 | |
Exam tip: For a decay chain, add up all the changes before you calculate. Each alpha decay gives −4 to the mass number and −2 to the atomic number. Each beta decay gives 0 to the mass number and +1 to the atomic number.
Some deep-space probes are powered by a radioisotope generator. The generator uses heat from the decay of plutonium-238, \({}^{238}_{94}\text{Pu}\), which emits alpha particles and has a half-life of 88 years.
At launch, the plutonium in one probe releases thermal energy at a rate of 470 W.
Plutonium-238 decays by alpha emission to an isotope of uranium. State the mass number and the proton number of this uranium nucleus.
Explain why a source that emits alpha particles, rather than beta particles or gamma rays, is a good choice of fuel for a generator on a space probe.
The rate of energy release is proportional to the activity of the plutonium. Calculate how many years after launch the rate of energy release falls to 117.5 W.
The electrical output of the generator falls faster than your calculation in (c) predicts. Suggest one reason why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mass number = 234 ; proton number = 92 | ✓ 2 | |
| Part (b) | ||
| Alpha particles are strongly absorbed (highly ionising, short range) within the fuel or its casing, so their energy is transferred to the fuel as heat ; very little shielding is needed to stop the radiation reaching or damaging the probe's electronics, so the probe's mass stays low | ✓ 2 | |
| Part (c) | ||
| 470 W → 235 W → 117.5 W, or 470 ÷ 117.5 = 4 ; this is 2 half-lives ; time = 2 × 88 = 176 years | ✓ 3 | |
| Part (d) | ||
| Any one valid reason, e.g. the thermocouples or other heat-to-electricity converters deteriorate or become less efficient over time / the electrical connections or insulation degrade / more of the heat is lost from the generator instead of being converted | ✓ 1 | |
Exam tip: When a power or energy rate is proportional to activity, it halves once every half-life. Count the number of halvings needed, then multiply by the half-life.
A hospital wants a radioactive tracer to inject into a patient's bloodstream to check how well the kidneys are working. A detector outside the body will record the radiation as the tracer passes through the kidneys. The scan takes about one hour.
Four isotopes are available:
isotope W: emits alpha, half-life 6 hours
isotope X: emits gamma, half-life 6 hours
isotope Y: emits beta, half-life 12 years
isotope Z: emits gamma, half-life 30 years
Evaluate the suitability of each isotope for this use and state which isotope the hospital should choose.
A patient is injected with isotope X with an activity of 400 MBq. Calculate the activity of the tracer 24 hours later, assuming none has left the body.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W: alpha would not escape the body to reach the detector ; W: alpha is highly ionising, so it would cause serious damage to tissue inside the body ; Y: beta is partly absorbed by tissue, so the signal would be weak, and it would add to the dose ; Y and Z: half-lives of many years mean the patient would stay radioactive for a very long time / receive a large total dose ; X and Z: gamma passes out of the body to the external detector and is weakly ionising ; X is the best choice because its gamma rays can be detected and its 6-hour half-life lasts long enough for a one-hour scan but decays quickly afterwards | ✓ 6 | |
| Part (b) | ||
| 24 ÷ 6 = 4 half-lives ; activity = 400 ÷ 24 = 25 MBq | ✓ 2 | |
Exam tip: In an evaluation, judge each option against every requirement: whether the radiation can be detected, the harm it causes and its half-life. Then end with a clearly justified choice.
A growing coastal city must choose how to generate extra electricity in the future. One option is a nuclear fission power station. Another is a nuclear fusion power station, which may become available in the future.
Evaluate the use of nuclear fission and nuclear fusion for generating the city's electricity.
You should compare the processes, the fuels and the waste produced, and give a conclusion.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Process: fission splits a large nucleus (e.g. uranium-235, after it absorbs a neutron) into two smaller nuclei, while fusion joins light nuclei (isotopes of hydrogen) to form a heavier nucleus | ✓ 1 | |
| Fuel: fission uses uranium (or plutonium), which is a finite resource that must be mined, while fusion uses hydrogen isotopes such as deuterium, which is plentiful, e.g. from seawater | ✓ 1 | |
| Waste: fission products are radioactive with long half-lives, so they need secure long-term storage, while fusion produces much less long-lived radioactive waste | ✓ 1 | |
| Fission is an established technology: the chain reaction can be controlled using control rods (and a moderator), so a station could be built now | ✓ 1 | |
| Fusion needs extremely high temperatures and pressures to overcome the electrostatic repulsion between nuclei, so it is not yet working commercially | ✓ 1 | |
| A conclusion supported by the earlier points, e.g. fission for the near future because fusion is not yet available, with fusion the better long-term option because of its plentiful fuel and less long-lived waste | ✓ 1 | |
| Credit another valid comparison in place of one of the first five points, e.g. neither process releases carbon dioxide while generating electricity. | — | |
Exam tip: 'Evaluate' needs points on both sides and a final judgement. Cover every item the question lists, here the processes, the fuels and the waste. A conclusion earns credit only when it is backed up by the points you have already made.
In a nuclear power station, the fission of uranium-235 releases energy in a reactor core containing fuel rods, a moderator and control rods.
Describe how fission produces a chain reaction and explain how the reaction is kept at a steady rate.
Describe how fission produces a chain reaction and explain how the reaction is kept at a steady rate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A slow-moving neutron is absorbed by a uranium-235 nucleus | ✓ 1 | |
| The nucleus becomes unstable and splits into two smaller daughter nuclei, releasing energy | ✓ 1 | |
| Two or three neutrons are also released | ✓ 1 | |
| These neutrons can go on to be absorbed by further uranium-235 nuclei, causing more fissions, which is a chain reaction | ✓ 1 | |
| The moderator (e.g. graphite or water) slows the neutrons so that they are more likely to be absorbed and cause fission | ✓ 1 | |
| Control rods (e.g. boron) absorb neutrons and are raised or lowered so that on average one neutron from each fission causes another, keeping the rate steady | ✓ 1 | |
Common error: The moderator does not absorb neutrons. It slows them down. Absorbing neutrons is the job of the control rods.
One possible fission of uranium-235 is:
23592U + 10n → 14156Ba + 92ZKr + x 10n
Determine the values of Z and x.
Explain why this reaction can lead to a chain reaction.
Each fission releases 3.2 × 10−11 J. A reactor core produces thermal energy at a rate of 1.6 × 109 W. Calculate the number of fissions taking place each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Z = 92 − 56 = 36 | ✓ 1 | |
| 235 + 1 = 141 + 92 + x, so x = 3 | ✓ 1 | |
| Part (b) | ||
| Each fission releases three neutrons but absorbs only one, so more neutrons are released than are absorbed | ✓ 1 | |
| The released neutrons can be absorbed by other uranium-235 nuclei, causing further fissions | ✓ 1 | |
| Part (c) | ||
| Number per second = energy per second ÷ energy per fission = 1.6 × 109 ÷ 3.2 × 10−11 | ✓ 1 | |
| = 5.0 × 1019 | ✓ 1 | |
Exam tip: A power of 1.6 × 10⁹ W means 1.6 × 10⁹ J are transferred every second. Dividing by the energy per fission gives fissions per second.
A nucleus of radon-222 (22286Rn) undergoes two successive alpha decays. What are the mass number and atomic number of the nucleus that results?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: C. Each alpha decay lowers the mass number by 4 and the atomic number by 2. Two decays: 222 − 8 = 214 and 86 − 4 = 82. A is the nucleus after only one decay. B reduces the mass number by 8 but forgets to reduce the atomic number. D reduces the atomic number but forgets to change the mass number. | ✓ 1 | Answer C |
Answer: C
Common error: Count the decays first, then multiply both changes (−4 and −2) by that number.
Radium-226 (22688Ra) decays through a chain of isotopes, emitting only alpha particles, beta-minus particles and gamma rays, until it reaches stable lead-206 (20682Pb).
Work out the number of alpha particles emitted in the whole chain.
Work out the number of beta-minus particles emitted in the whole chain.
State the change in the nucleus caused by the emission of a gamma ray.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 change in mass number 226 − 206 = 20, and each alpha particle removes 4 | ✓ 1 | |
| A1 20 ÷ 4 = 5 alpha particles | ✓ 1 | |
| Part (b) | ||
| M1 five alpha particles lower the atomic number by 10, from 88 to 78; the final atomic number is 82 | ✓ 1 | |
| A1 82 − 78 = 4 beta-minus particles (each raises the atomic number by 1) | ✓ 1 | |
| Part (c) | ||
| B1 no change in mass number or atomic number (the nucleus loses energy) | ✓ 1 | |
Exam tip: Use the mass number first to find the number of alpha particles, then use the atomic number to find the number of beta particles.
A student tests an unknown source with a detector. The background count rate is 24 counts/min. Count rates in counts/min: no absorber 640; paper 638; 3 mm aluminium 142; 10 mm lead 30.
Calculate the corrected count rate with 3 mm of aluminium.
Deduce which types of radiation the source emits. Give a reason for each type you identify or exclude.
Suggest two safety precautions when handling the source.
Calculate the percentage fall in the corrected count rate when 3 mm of aluminium is placed in front of the source. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 142 − 24 = 118 counts/min | ✓ 1 | |
| Part (b) | ||
| no alpha: paper does not reduce the count rate | ✓ 1 | |
| beta is present: aluminium reduces the count rate a lot | ✓ 1 | |
| gamma is present: count rate behind aluminium is still well above background, and only lead reduces it nearly to background | ✓ 1 | |
| Part (c) | ||
| use tongs and avoid touching it | ✓ 1 | |
| keep it as far away as possible and store it in a lead-lined box when not in use | ✓ 1 | |
| Part (d) | ||
| 616 and 118 counts/min, so fall of 498 | ✓ 1 | |
| 498 ÷ 616 × 100 = 80.8% | ✓ 1 | |
Common error: A reading that equals the background level means the radiation is stopped, not that it is zero before correction.
A radioactive sample is placed near a detector. Without the sample, the detector records a background count rate of 40 counts per minute. With the sample present, the count rate is 840 counts per minute at the start and 140 counts per minute 6 hours later. What is the half-life of the source?
| Marking point | Mark | Notes |
|---|---|---|
| Subtract the background first: 800 counts per minute falls to 100 counts per minute. 800 to 400 to 200 to 100 is three halvings in 6 hours, so the half-life is 2.0 hours (B). A counts the four numbers in the list as four halvings (6 divided by 4). C ignores the background and uses 840 to 140, which is a factor of 6, about 2.6 half-lives. D assumes only two halvings, giving 6 ÷ 2. | ✓ 1 | Answer B |
Answer: B
Common error: Always subtract the background count rate from every reading before looking for halvings.
A laboratory holds a mixture of two radioactive isotopes, A and B, each emitting only gamma radiation. A has a half-life of 3 hours and B has a half-life of 12 hours. At time 0, isotope A has an activity of 800 Bq and isotope B has an activity of 800 Bq.
Calculate the activity of isotope A after 12 hours.
Calculate the activity of isotope B after 12 hours.
Calculate the percentage of the total activity at 12 hours that is due to isotope A. Give your answer to 3 significant figures.
Explain why, after several days, the activity of the mixture is almost entirely due to isotope B.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 12 hours is 4 half-lives of A | ✓ 1 | |
| 800 ÷ 16 = 50 Bq | ✓ 1 | |
| Part (b) | ||
| 12 hours is 1 half-life of B | ✓ 1 | |
| 800 ÷ 2 = 400 Bq | ✓ 1 | |
| Part (c) | ||
| total activity = 50 + 400 = 450 Bq (ECF from (a) and (b)) | ✓ 1 | |
| 50 ÷ 450 × 100 = 11.1% | ✓ 1 | |
| Part (d) | ||
| isotope A has a shorter half-life, so its activity falls much faster | ✓ 1 | |
| isotope B still has a significant activity (decays more slowly) while A has almost disappeared | ✓ 1 | |
Common error: Activities of different sources add, but each one must be halved using its own half-life. Do not use a single half-life for the whole mixture.
A robot-controlled line makes thin metal foil. The foil thickness is checked by a radioactive source on one side and a detector on the other. A separate robot fits a tracer into a buried pipe to find leaks. Which source is best for the tracer in the buried pipe?
| Marking point | Mark | Notes |
|---|---|---|
| A tracer in a buried pipe must be detected through soil, so it must emit penetrating gamma radiation, and it should have a short half-life so it does not stay hazardous: C. A is absorbed by the soil and pipe wall and has a very long half-life. B is only moderately penetrating and stays active for decades. D is too short-lived: the activity would be gone before the pipe could be checked, and beta is absorbed by soil. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a tracer, give two reasons: penetrating radiation to reach the detector, and a half-life short enough to be safe but long enough to complete the job.
A robotics lab tests a radiation sensor with a sample of an unstable isotope. The activity of the sample is 3840 Bq at 09:00 and 120 Bq at 11:30.
Calculate the half-life of the isotope.
Calculate the activity of the sample at 13:30.
A student says: 'After exactly one half-life, half of the nuclei in the sample will have decayed, so I can predict which nuclei these will be.' Explain what is wrong with this statement.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3840 ÷ 120 = 32, which is 5 halvings (5 half-lives) | ✓ 1 | |
| 150 min ÷ 5 = 30 minutes | ✓ 1 | |
| Part (b) | ||
| 13:30 is 270 min after 09:00, which is 9 half-lives (ECF from (a)) | ✓ 1 | |
| 3840 ÷ 29 = 7.5 Bq | ✓ 1 | |
| Part (c) | ||
| Decay is random | ✓ 1 | |
| It is impossible to predict which individual nucleus will decay or when; only the proportion decaying in a large sample is predictable | ✓ 1 | |
Common error: Half-life gives the time for the activity of a large sample to halve; it says nothing about when any one nucleus will decay.
A sample of a radioactive isotope with a half-life of 12 hours is placed near a detector. The detector reads 640 counts per minute at the start. Background radiation contributes 40 counts per minute to this reading. What will the detector read 36 hours later?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct. The count rate from the sample alone is 640 − 40 = 600 counts per minute. 36 hours is three half-lives, so this falls to 600 ÷ 8 = 75 counts per minute. The background of 40 does not decay, so the reading is 75 + 40 = 115. A forgets to add the background back. B halves the whole reading three times, including the background. C uses only two half-lives. | ✓ 1 | Answer D |
Answer: D
Exam tip: Only the count rate from the source halves each half-life; the background stays constant, so subtract it first and add it back afterwards.
A radioactive sample is delivered to a laboratory. A technician measures its count rate every 20 minutes. With the sample removed, the detector reads 24 counts per minute. The table shows the readings with the sample in place.
Time / min: 0, 20, 40, 60
Count rate / counts per minute: 344, 184, 104, 64
Explain why the technician subtracts 24 counts per minute from each reading.
Determine the half-life of the sample.
Predict the reading the detector will give at 100 minutes.
The sample may be handled without special precautions once its corrected count rate is below 2.0 counts per minute. The technician checks only at times that are multiples of 20 minutes. Determine the first time at which handling is allowed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The 24 counts per minute are due to background radiation, not the sample, so they must be removed to find the count rate from the sample | ✓ 1 | |
| Part (b) | ||
| Subtract background: 320, 160, 80, 40 counts per minute | ✓ 1 | |
| The corrected count rate halves from 320 to 160 in the first 20 minutes (and again from 160 to 80 and from 80 to 40) | ✓ 1 | |
| Half-life = 20 minutes | ✓ 1 | |
| Part (c) | ||
| 100 min is 5 half-lives (ECF from (b)) | ✓ 1 | |
| Corrected count rate = 320 ÷ 25 = 10 counts per minute | ✓ 1 | |
| Add background: 10 + 24 = 34 counts per minute | ✓ 1 | |
| Part (d) | ||
| After 7 half-lives (140 min) the rate is 320 ÷ 128 = 2.5, which is not below 2.0 | ✓ 1 | |
| After 8 half-lives the rate is 320 ÷ 256 = 1.25, which is below 2.0 | ✓ 1 | |
| Time = 8 × 20 = 160 minutes | ✓ 1 | |
Common error: The background does not decay, so take it away before looking for halving, and do not halve the raw readings.
A logistics company wants to track the arrival of a batch of fuel travelling along an underground steel pipeline 480 km long. A small amount of radioactive tracer is added to the front of the batch and detectors above ground are used to detect it. The fuel travels at a constant speed of 1.0 m/s. The table shows four possible tracers.
Source A: alpha emitter, half-life 4.0 years
Source B: gamma emitter, half-life 6.0 hours
Source C: beta emitter, half-life 28 years
Source D: gamma emitter, half-life 20 days
Calculate the time taken for the fuel to travel the whole length of the pipeline. Give your answer in days to 2 significant figures.
Evaluate which of the four sources is the most suitable tracer. Use the type of radiation and the half-life in your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Distance = 480 000 m | ✓ 1 | |
| Time = distance ÷ speed = 480 000 ÷ 1.0 = 480 000 s | ✓ 1 | |
| 480 000 ÷ 86 400 = 5.6 days | ✓ 1 | |
| Part (b) | ||
| Radiation must be able to pass through the soil and steel pipe to reach the detectors, so gamma is needed | ✓ 1 | |
| Alpha (A) would be stopped by the pipe wall and soil, and beta (C) would be absorbed by the steel, so neither could be detected above ground | ✓ 1 | |
| B is gamma, but a half-life of 6.0 hours is far too short: the trip takes about 5.6 days, which is over 20 half-lives, so almost none of its activity would be left | ✓ 1 | |
| D is gamma and its half-life of 20 days means that a large fraction of the activity (about 80%) remains after 5.6 days, so it is easily detected | ✓ 1 | |
| D also decays away within a few months, so it does not leave long-lasting contamination, unlike A and C with half-lives of years | ✓ 1 | |
| Conclusion: source D is the most suitable | ✓ 1 | |
Exam tip: In an evaluation, consider both the type of radiation and the half-life for every option before you choose one.
A steel bridge is inspected using gamma radiography. A technician places a sealed source of cobalt-60 on one side of a weld and a photographic film on the other side. Cobalt-60 has a half-life of 5.3 years.
Explain why gamma radiation is used rather than alpha or beta radiation.
The activity of the source is 640 GBq when it is new. Calculate its activity after 15.9 years.
Explain why a half-life of about 5 years is suitable for this source.
State two safety precautions the technician should take when handling the source.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma rays are penetrating enough to pass through steel and affect the film (alpha and beta particles would be absorbed) | ✓ 1 | |
| Part (b) | ||
| 15.9 years is 3 half-lives (15.9 / 5.3 = 3) | ✓ 1 | |
| 640 / 23 = 80 GBq | ✓ 1 | |
| Part (c) | ||
| long enough that the source stays active for several years of use without being replaced | ✓ 1 | |
| not so long that it remains a hazard for thousands of years once it is disposed of | ✓ 1 | |
| Part (d) | ||
| any one: use long tongs / handle it from a distance; store it in a lead container; limit the time of exposure | ✓ 1 | |
| a second different precaution from the list above or wearing a dosimeter badge | ✓ 1 | |
Common error: A half-life of several years does not make a source safe; state precautions about distance, time and shielding.
A student records the count rate from a sample of a radioactive isotope every 10 minutes. The readings, including background, are: 345, 185, 105 and 65 counts per minute at 0, 10, 20 and 30 minutes. The background count rate is 25 counts per minute.
Determine the half-life of the isotope.
Predict the count rate the student would measure at 50 minutes, including background.
Explain why the measured count rate never falls to zero, however long the student waits.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtract background: 320, 160, 80, 40 counts per minute | ✓ 1 | |
| M1 corrected rate halves in each 10 minute interval | ✓ 1 | |
| A1 half-life = 10 minutes | ✓ 1 | |
| Part (b) | ||
| M1 50 minutes is 5 half-lives | ✓ 1 | |
| M1 corrected rate 320 ÷ 32 = 10 counts per minute | ✓ 1 | |
| A1 add background: 10 + 25 = 35 counts per minute | ✓ 1 | |
| Part (c) | ||
| background radiation is always detected | ✓ 1 | |
Common error: Do not halve the raw readings; subtract the background first.
A radioactive sample is placed near a Geiger counter. The reading is 270 counts per minute. The background count rate is 30 counts per minute. After 36 minutes the reading is 60 counts per minute. What is the half-life of the sample?
| Marking point | Mark | Notes |
|---|---|---|
| The corrected count rate falls from 270 − 30 = 240 to 60 − 30 = 30 counts per minute. 240 → 120 → 60 → 30 is three halvings, so the half-life is 36 ÷ 3 = 12 minutes, so this is correct. 18 minutes comes from ignoring the background and judging that the reading halves about twice. 9 minutes treats the ratio 270 : 60, about 4, as the number of half-lives. 4.5 minutes divides 36 by the ratio 240 : 30 = 8, which is not the number of half-lives. | ✓ 1 | Answer C |
Answer: C
Exam tip: Subtract the background count rate from every reading before looking for halvings.
The human body contains potassium-40. For one person, about 4400 of the potassium-40 nuclei decay every second, so the activity is 4400 Bq. About 89% of the decays produce stable calcium-40 nuclei. The half-life of potassium-40 is 1.25 × 109 years.
State what is meant by saying that radioactive decay is random.
Calculate the number of potassium-40 decays in one day (86 400 s). Give your answer to 2 significant figures.
Calculate the number of calcium-40 nuclei formed from these decays in one day. Use the unrounded value from (b) and give your answer to 2 significant figures.
Explain why the activity of potassium-40 in a person's body hardly changes during their lifetime.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it cannot be predicted when a particular nucleus will decay | ✓ 1 | |
| Part (b) | ||
| M1 4400 × 86 400 = 3.8016 × 108 | ✓ 1 | |
| A1 3.8 × 108 | ✓ 1 | |
| Part (c) | ||
| M1 0.89 × 3.8016 × 108 = 3.383 × 108 | ✓ 1 | |
| A1 3.4 × 108 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| the half-life is far longer than a human lifetime | ✓ 1 | |
| so only a tiny fraction of the nuclei decay and the number of nuclei (and the activity) stays almost constant | ✓ 1 | |
Exam tip: Keep the unrounded value on your calculator between parts, and round only the final answer.
A hospital sample contains two radioactive isotopes, P and Q. P has a half-life of 2.0 hours and Q has a half-life of 6.0 hours. At time zero each isotope contributes an activity of 960 kBq to the sample. Ignore any other radiation.
Calculate the activity of P after 12 hours.
Calculate the activity of Q after 12 hours.
Calculate the fraction of the total activity at 12 hours that is due to Q. Give your answer to 3 significant figures.
Calculate the time at which the activity of Q has fallen to 30 kBq.
Explain why, after a long time, almost all of the activity of the sample comes from Q.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 12 hours is 6 half-lives | ✓ 1 | |
| 960 ÷ 26 = 15 kBq | ✓ 1 | |
| Part (b) | ||
| 12 hours is 2 half-lives | ✓ 1 | |
| 960 ÷ 4 = 240 kBq | ✓ 1 | |
| Part (c) | ||
| total activity = 15 + 240 = 255 kBq | ✓ 1 | |
| 240 ÷ 255 = 0.941 | ✓ 1 | |
| (ECF from (a) and (b)) | — | |
| Part (d) | ||
| 960 ÷ 30 = 32 | ✓ 1 | |
| 32 = 25, so 5 half-lives | ✓ 1 | |
| 5 × 6.0 = 30 hours | ✓ 1 | |
| Part (e) | ||
| P has the shorter half-life, so its activity falls more quickly | ✓ 1 | |
| the activity of Q halves more slowly, so it becomes much larger than that of P | ✓ 1 | |
Common error: Each isotope has its own half-life, so find the number of half-lives separately for P and for Q before adding activities.
A research team studies how far bees forage by feeding a colony a small, safe amount of a radioactive tracer. A sample of the colony's honey is tested with a radiation detector. Background radiation gives a count rate of 20 counts/min. The first reading, with the sample present, is 420 counts/min. Nine days later the reading is 70 counts/min.
Calculate the count rate due to the sample alone at the start.
Show that the half-life of the tracer is 3.0 days.
Predict the reading on the detector, with the sample present, 12 days after the first reading.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 420 − 20 = 400 counts/min | ✓ 1 | |
| Part (b) | ||
| Count rate due to the sample after 9 days = 70 − 20 = 50 counts/min | ✓ 1 | |
| 400 → 200 → 100 → 50, which is 3 half-lives | ✓ 1 | |
| 9 ÷ 3 = 3.0 days | ✓ 1 | |
| Part (c) | ||
| M1 12 ÷ 3.0 = 4 half-lives | ✓ 1 | |
| A1 sample count rate = 400 ÷ 16 = 25 counts/min | ✓ 1 | |
| A1 reading = 25 + 20 = 45 counts/min | ✓ 1 | |
Common error: The background count does not decay: take it away before halving and add it back at the end.
In a Formula 1 factory, carbon-fibre sheet is rolled to a constant thickness. A thickness gauge has a strontium-90 beta source on one side of the sheet and a detector on the other side. The detector measures the count rate of beta particles that pass through the sheet. Strontium-90 has a half-life of 29 years.
Explain why a beta source is used in the gauge rather than an alpha source or a gamma source.
The count rate at the detector rises above the value set for the correct thickness. State what this shows about the sheet and what the rollers should do.
Explain why the gauge does not need its source replaced after a few months.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Alpha particles would all be absorbed by the sheet, so the detector would read only background | ✓ 1 | |
| Gamma rays would pass through almost unchanged, so the count rate would hardly change with thickness | ✓ 1 | |
| Beta particles are partly absorbed, so the count rate depends on the thickness | ✓ 1 | |
| Part (b) | ||
| More beta particles are getting through, so the sheet is too thin | ✓ 1 | |
| The rollers should move apart (or press less) to make the sheet thicker | ✓ 1 | |
| Part (c) | ||
| The half-life is very long compared with a few months, so the activity has hardly changed | ✓ 1 | |
Common error: Compare the three radiations by what happens to each in the material: choose the one that is reduced partly, not fully and not hardly at all.
A nucleus of an unstable isotope emits one alpha particle and then two beta-minus particles, one after the other. How does the final nucleus compare with the original nucleus?
| Marking point | Mark | Notes |
|---|---|---|
| The alpha particle lowers the proton number by 2 and the nucleon number by 4. Each beta-minus decay raises the proton number by 1 and leaves the nucleon number unchanged, so two of them raise it by 2. The net change is no change in proton number and a fall of 4 in nucleon number, so the final nucleus is an isotope of the original. Ignoring the beta decays gives a proton number 2 less. Counting the beta decays twice gives 2 more. Treating each beta as removing a proton gives 4 less. | ✓ 1 | Answer C |
Answer: C
Exam tip: Track the nucleon number and the proton number separately after each decay, then add up the changes.
Sports scientists estimate the blood volume of an athlete using sodium-24, which has a half-life of 15 hours. A solution containing sodium-24 with an activity of 2.4 × 106 Bq is injected into the athlete. Assume it mixes evenly through the blood and none leaves the blood. After 30 hours, a 2.0 cm3 sample of blood is taken and found to have an activity of 240 Bq. Assume the sample is measured immediately and ignore background radiation.
Calculate the total activity of the sodium-24 in the athlete's blood after 30 hours.
Calculate the volume of the athlete's blood, in cm3.
Suggest why sodium-24 is a suitable tracer for use in the body.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 30 hours is 2 half-lives | ✓ 1 | |
| 2.4 × 106 ÷ 4 = 6.0 × 105 Bq | ✓ 1 | |
| Part (b) | ||
| activity per cm3 = 240 ÷ 2.0 = 120 Bq/cm3 | ✓ 1 | |
| volume = 6.0 × 105 ÷ 120 (ECF from (a)) | ✓ 1 | |
| = 5000 cm3 | ✓ 1 | |
| Part (c) | ||
| any one from: emits gamma which passes out of the body to be detected; short half-life so the activity soon falls to a safe level | ✓ 1 | |
Common error: The sample activity is for 2.0 cm³ only, so divide by the sample volume to get the activity per cm³ before scaling up.
Radium-226 has 88 protons. After a series of decays it forms stable lead-206, which has 82 protons. Only alpha particles and beta-minus particles are emitted in the series.
Determine the number of alpha particles emitted in the series.
Determine the number of beta-minus particles emitted in the series.
Explain why an alpha emitter such as radium is more hazardous inside the body than outside it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| change in mass number 226 − 206 = 20 | ✓ 1 | |
| 20 ÷ 4 = 5 alpha particles | ✓ 1 | |
| Part (b) | ||
| 5 alpha particles reduce the atomic number from 88 to 78, but 82 is needed | ✓ 1 | |
| so 82 − 78 = 4 beta-minus particles (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| alpha radiation is highly ionising | ✓ 1 | |
| outside the body it is absorbed by air or skin, but inside it is absorbed by living cells, which are damaged | ✓ 1 | |
Exam tip: Use the mass numbers to find the alpha particles first, then use the atomic numbers to find how many beta-minus particles are needed.
Potassium-40 (4019K) is found in the human body and can decay by beta-minus emission to calcium. Which statement describes the calcium nucleus formed?
| Marking point | Mark | Notes |
|---|---|---|
| Potassium-40 has 19 protons and 40 − 19 = 21 neutrons. In beta-minus decay one neutron becomes a proton, so the nucleus has 20 protons and 20 neutrons, and the nucleon number stays 40. The option with 18 protons would need a proton to change into a neutron. The option with 19 protons and 20 neutrons loses a neutron without gaining a proton. The option with 17 protons and 19 neutrons is what alpha emission would give. | ✓ 1 | Answer A |
Answer: A
Common error: In beta-minus decay the nucleon number stays the same and the proton number goes up by one; do not confuse it with alpha emission.
A laboratory stores a sample of iodine-131, which has a half-life of 8.0 days. When the sample is taken out of storage, 93.75% of the original iodine-131 nuclei have decayed.
Calculate the percentage of the original nuclei that remain.
Calculate the number of half-lives that passed during storage.
Calculate the storage time in days.
A second sample, stored for the same time, began with 1.6 × 1012 iodine-131 nuclei. Calculate the number of these nuclei that remain. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 100 − 93.75 = 6.25% | ✓ 1 | |
| Part (b) | ||
| 6.25% = 1/16 = (1/2)4 | ✓ 1 | |
| 4 half-lives | ✓ 1 | |
| Part (c) | ||
| 4 × 8.0 = 32 days | ✓ 1 | |
| Part (d) | ||
| 1.6 × 1012 × (1/2)4 or ÷ 16 | ✓ 1 | |
| 1.0 × 1011 | ✓ 1 | |
Common error: Do not stop at the number of half-lives; the question asks for a time, so multiply by the half-life.
Why must the temperature of a fusion reactor be extremely high?
| Marking point | Mark | Notes |
|---|---|---|
| Nuclei are positively charged and repel one another; only at very high temperatures do they move fast enough to overcome this repulsion and fuse (A). B is wrong because nuclei do not need to be broken up. C describes fission, not fusion. D is wrong because fusion needs no pre-existing radioactivity. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link fusion to charge: all nuclei are positive, so the key idea is overcoming electrostatic repulsion.
Nuclear fusion reactors must heat the fuel to temperatures of over 100 million degrees Celsius. Which statement explains why such high temperatures are needed?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: nuclei are positively charged and repel each other, so they need very high speeds to get close enough to fuse. B is wrong because fusion is not about melting. C is wrong because neutrons are not removed from nuclei to cause fusion. D is wrong because fusion does not rely on nuclei being unstable. | ✓ 1 | Answer A |
Answer: A
Common error: Fusion needs high temperature to overcome repulsion between nuclei; it is not needed to make nuclei radioactive or to split them.
In a nuclear reactor operating at a constant power, each fission releases on average 2.4 neutrons. On average, how many of the neutrons released by one fission go on to cause another fission?
| Marking point | Mark | Notes |
|---|---|---|
| At constant power the number of fissions in each generation stays the same, so on average exactly one neutron from each fission must cause another fission. The other 1.4 neutrons are lost by escaping or by being absorbed (for example in control rods) without causing fission. The value 2.4 assumes every neutron causes a fission, which would make the reaction grow rapidly. The value 1.4 is the number of neutrons lost, not the number that cause fission. The value 0.4 comes from subtracting 2 instead of 1.4. | ✓ 1 | Answer A |
Answer: A
Common error: A steady chain reaction needs one neutron per fission to cause another fission, not all of the neutrons released.
The Sun releases energy by nuclear fusion in its core.
Describe what happens in nuclear fusion.
Explain why fusion happens in the core of the Sun but not at its surface.
Name the element that is the main fuel for fusion in the Sun.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| two light nuclei join together to form a heavier nucleus | ✓ 1 | |
| energy is released | ✓ 1 | |
| Part (b) | ||
| fusion needs very high temperature and pressure so that nuclei can get close enough despite their repulsion | ✓ 1 | |
| the core is hot and under high pressure, whereas the surface is much cooler | ✓ 1 | |
| Part (c) | ||
| hydrogen | ✓ 1 | |
Common error: Fusion joins small nuclei together; do not describe it as splitting.
A patient is injected with technetium-99m so that her kidneys can be imaged with a gamma camera. Technetium-99m emits only gamma radiation and has a half-life of 6 hours.
Describe the difference between contamination and irradiation.
For several hours after the scan, the patient's urine contains some technetium-99m. Nurses wear disposable gloves when handling it.
Explain why.
Explain why technetium-99m is suitable for this use.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Irradiation: being exposed to radiation from a source outside the body; the person does not become radioactive | ✓ 1 | |
| Contamination: radioactive material gets onto the skin or into the body, so it keeps emitting radiation | ✓ 1 | |
| Part (b) | ||
| It stops the radioactive material getting onto their skin or into their body / prevents contamination | ✓ 1 | |
| Material on or in the body would keep exposing their cells to radiation, which can damage cells or DNA and may cause cancer | ✓ 1 | |
| Part (c) | ||
| Gamma radiation passes out of the body to reach the camera, and it is weakly ionising so it causes little damage to cells | ✓ 1 | |
| A 6-hour half-life is long enough to complete the scan, but the activity falls quickly afterwards, so the patient receives only a small dose | ✓ 1 | |
Common error: Do not write that an irradiated object 'becomes radioactive'. Exposure to alpha, beta or gamma radiation does not make an object radioactive; only contamination leaves radioactive material behind.
A fusion research reactor heats hydrogen nuclei to over 100 million °C. The hot nuclei are held away from the reactor walls by strong magnetic fields.
Explain why fusion only happens at very high temperatures.
The reactor also keeps the hot nuclei at a high density (high pressure).
Suggest why this increases the rate of fusion.
Suggest why the hot nuclei must not touch the walls of the reactor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The nuclei are all positively charged | ✓ 1 | |
| So they repel each other electrostatically | ✓ 1 | |
| At a high temperature the nuclei have enough kinetic energy (move fast enough) to overcome the repulsion and get close enough to fuse | ✓ 1 | |
| Part (b) | ||
| The nuclei collide more often, so more fusions happen each second | ✓ 1 | |
| Part (c) | ||
| The nuclei would lose energy and cool so fusion stops, or the walls would be damaged or melted | ✓ 1 | |
Common error: The difficulty is not that 'nuclei are hard to find'. It is the electrostatic repulsion between positive charges, so state the charge and the repulsion explicitly.
Used fuel from a fission reactor contains caesium-137, which has a half-life of 30 years. A sealed container of this waste has an activity of 8.0 × 1012 Bq due to caesium-137.
State why used fuel is more radioactive than fresh uranium fuel.
Calculate how long it takes for the activity from the caesium-137 to fall to 5.0 × 1011 Bq.
Suggest one precaution for storing the container during this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| It contains fission products (daughter nuclei), which are unstable or radioactive | ✓ 1 | |
| Part (b) | ||
| 8.0 × 1012 → 4.0 × 1012 → 2.0 × 1012 → 1.0 × 1012 → 5.0 × 1011 Bq, or ratio 8.0 × 1012 ÷ 5.0 × 1011 = 16 | ✓ 1 | |
| That is 4 half-lives | ✓ 1 | |
| Time = 4 × 30 = 120 years | ✓ 1 | |
| Part (c) | ||
| Any one of: store it behind thick concrete or lead shielding; store it in a secure, sealed site underground or away from people; monitor it for leaks | ✓ 1 | |
Exam tip: Show each halving step. This makes the number of half-lives clear, and the working can still earn marks if the final multiplication goes wrong.
A red dwarf star releases energy by joining hydrogen nuclei together in its core.
State what is meant by nuclear fusion.
Explain why fusion only happens at very high temperatures and pressures.
Suggest why it is difficult to build a fusion power station on Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Two light (small) nuclei join together | ✓ 1 | |
| to form a heavier nucleus, releasing energy | ✓ 1 | |
| Part (b) | ||
| Nuclei are positively charged, so they repel each other electrostatically | ✓ 1 | |
| A high temperature gives the nuclei very high kinetic energy / speed, so they can get close enough to fuse | ✓ 1 | |
| A high pressure / density makes collisions between nuclei much more frequent | ✓ 1 | |
| Part (c) | ||
| It is very hard to produce and contain the extremely high temperature and pressure (no material can hold the hot plasma) | ✓ 1 | |
Exam tip: Link the temperature to overcoming the repulsion between positive nuclei. Just writing 'it needs to be hot' gains nothing.
A pool-type research reactor produces a steady power output over several hours. The fuel is uranium-235, the pool water acts as the moderator and the reactor has control rods. Describe how the chain reaction in the reactor is started and kept steady. You should refer to the moderator and the control rods in your answer.
Describe how the chain reaction is started and kept steady.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 a neutron is absorbed by a uranium-235 nucleus, which becomes unstable and splits (fission) | ✓ 1 | |
| B1 energy is released and two or three more neutrons are released in each fission | ✓ 1 | |
| B1 these neutrons can be absorbed by other uranium-235 nuclei, causing further fission, so a chain reaction occurs | ✓ 1 | |
| B1 the neutrons released are fast, so the moderator (water) slows them down | ✓ 1 | |
| B1 slow neutrons are more likely to be absorbed by uranium-235 nuclei | ✓ 1 | |
| B1 control rods absorb some neutrons, adjusted so that on average one neutron from each fission causes another fission (the rate stays constant) | ✓ 1 | |
Exam tip: In a six-mark description, aim for a clear sequence: fission, neutrons released, further fission, then the roles of the moderator and the rods.
A volcanologist collects a sample of gas containing a radioactive isotope. The count rate is measured every 3 hours. The background count rate is 20 counts per minute.
Time / h: 0, 3, 6, 9, 12
Measured count rate / counts per minute: 980, 500, 260, 140, 80
Calculate the corrected count rate at 6 hours.
Determine the half-life of the isotope.
Calculate the fraction of the radioactive nuclei in the sample that have decayed by 12 hours. Give your answer as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 260 − 20 = 240 counts per minute | ✓ 1 | |
| Part (b) | ||
| M1 corrected count rate halves from 960 to 480 between 0 and 3 hours | ✓ 1 | |
| A1 3 hours (3.0 hours) | ✓ 1 | |
| Part (c) | ||
| M1 subtract the background: 80 − 20 = 60 counts per minute at 12 hours | ✓ 1 | |
| M1 fraction remaining = 60 ÷ 960 = 1/16 | ✓ 1 | |
| A1 15/16 (0.9375) have decayed | ✓ 1 | |
Common error: Subtract the background count rate from every reading before comparing them.
In a nuclear power station, uranium-235 undergoes fission. Each fission releases 3.2×10−11 J. The reactor releases energy at a rate of 1.6×109 J/s.
Describe how a chain reaction occurs in the reactor.
Explain the purpose of the control rods.
Calculate the number of fissions each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a neutron is absorbed by a uranium-235 nucleus, which splits | ✓ 1 | |
| this releases further neutrons | ✓ 1 | |
| these cause more nuclei to split, so the reaction continues | ✓ 1 | |
| Part (b) | ||
| they absorb neutrons | ✓ 1 | |
| so that on average only one neutron per fission causes another fission and the reaction is controlled | ✓ 1 | |
| Part (c) | ||
| 1.6×109 ÷ 3.2×10−11 | ✓ 1 | |
| 5.0×1019 | ✓ 1 | |
Exam tip: Divide the total energy rate by the energy per event to get events per second.
A nuclear power station uses uranium-235. In one fission reaction a neutron is absorbed by a nucleus of uranium-235, which splits into a nucleus of xenon-140, a nucleus of strontium-94 and some neutrons. Xenon-140 has 54 protons.
23592U + 10n → 14054Xe + 94…Sr + … 10n
Determine the proton number of strontium and the number of neutrons released in this reaction.
Explain how a chain reaction can occur in the reactor.
State the purpose of the control rods in the reactor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| proton number of strontium = 92 − 54 = 38 | ✓ 1 | |
| mass number: 235 + 1 = 236 = 140 + 94 + x | ✓ 1 | |
| 2 neutrons released | ✓ 1 | |
| Part (b) | ||
| neutrons released in fission are absorbed by other uranium-235 nuclei | ✓ 1 | |
| which then undergo fission, releasing more neutrons | ✓ 1 | |
| this repeats, so the number of fissions increases (if not controlled) | ✓ 1 | |
| Part (c) | ||
| to absorb neutrons | ✓ 1 | |
| so that the rate of fission (power output) is kept constant or the reaction is stopped | ✓ 1 | |
Common error: Count both the neutron that starts the fission and the neutrons released when balancing mass numbers: 235 + 1 on the left.
Nuclear fission is used in power stations. Nuclear fusion takes place in stars.
Describe how a chain reaction takes place in uranium-235.
State what is meant by nuclear fusion.
Explain why fusion needs very high temperatures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A neutron is absorbed by a uranium-235 nucleus, which splits into two smaller nuclei | ✓ 1 | |
| More neutrons are released | ✓ 1 | |
| These neutrons go on to cause further nuclei to split | ✓ 1 | |
| Part (b) | ||
| Two light (small) nuclei join to form a larger nucleus | ✓ 1 | |
| Part (c) | ||
| Nuclei are positively charged and repel each other | ✓ 1 | |
| High temperature gives nuclei enough kinetic energy to get close enough to join (overcoming repulsion) | ✓ 1 | |
Exam tip: For fusion, always mention electrostatic repulsion between nuclei when explaining the need for high temperature or pressure.
Caesium-137 is a contaminant that can reach fisheries after a nuclear accident. A nucleus of caesium-137 (mass number 137, proton number 55) decays by beta-minus emission to barium (proton number 56).
State the mass number of the barium nucleus formed.
Explain why the proton number increases by one in beta-minus decay.
Calculate the number of neutrons in the barium nucleus.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 137 | ✓ 1 | |
| Part (b) | ||
| A neutron changes into a proton | ✓ 1 | |
| An electron (beta particle) is emitted | ✓ 1 | |
| Part (c) | ||
| 137 − 56 | ✓ 1 | |
| 81 | ✓ 1 | |
Exam tip: In beta-minus decay the mass number stays the same while the proton number goes up by one.
An experimental fusion reactor heats a gas of deuterium and tritium until the nuclei fuse. One reaction can be written as
21H + 31H → 42He + X
Each reaction releases 2.8 × 10−12 J of energy.
Give the mass number and the atomic number of particle X, and name it.
Explain why the gas must be heated to a very high temperature for fusion to happen.
State one advantage of nuclear fusion over nuclear fission as a source of energy.
The reactor is designed to have a thermal power output of 500 MW (5.0 × 108 W). Calculate the number of fusion reactions needed each second. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Mass number 1 and atomic number 0 (from 2 + 3 − 4 = 1 and 1 + 1 − 2 = 0) | ✓ 1 | |
| X is a neutron | ✓ 1 | |
| Part (b) | ||
| Nuclei are positively charged so they repel each other | ✓ 1 | |
| High temperature gives the nuclei enough kinetic energy (speed) to get close enough together for fusion to happen | ✓ 1 | |
| Part (c) | ||
| Any one from: fusion products are not (as) long-lived radioactive waste, fuel (hydrogen isotopes) is plentiful, no risk of an uncontrolled chain reaction | ✓ 1 | |
| Part (d) | ||
| Number per second = power ÷ energy per reaction | ✓ 1 | |
| 5.0 × 108 ÷ 2.8 × 10−12 = 1.7857 × 1020 | ✓ 1 | |
| 1.8 × 1020 reactions per second | ✓ 1 | |
Exam tip: In a nuclear equation, the mass numbers on each side must add up, and so must the atomic numbers.
A university has a pool-type research reactor. The fuel is held in a core at the bottom of a deep pool of water. Control rods can be raised or lowered between the fuel elements. The power of the reactor is 2.4 MW. Each fission of a uranium-235 nucleus releases 3.2 × 10−11 J.
State the job of the moderator in a nuclear reactor.
Explain how the control rods are used to keep the power of the reactor constant.
Several metres of water lie above the core. Explain how this water helps to protect people working at the edge of the pool.
Calculate the number of fission events in the reactor each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| To slow down the neutrons so that they are more likely to cause further fission | ✓ 1 | |
| Part (b) | ||
| Control rods absorb neutrons | ✓ 1 | |
| They are lowered further to absorb more neutrons (or raised to absorb fewer) so that, on average, one neutron from each fission goes on to cause another fission | ✓ 1 | |
| Part (c) | ||
| The water absorbs (stops) neutrons and gamma radiation coming from the core | ✓ 1 | |
| This reduces the amount of radiation reaching the people, so their exposure is lower | ✓ 1 | |
| Part (d) | ||
| Number per second = power ÷ energy per fission = 2.4 × 106 ÷ 3.2 × 10−11 | ✓ 1 | |
| 7.5 × 1016 fissions per second | ✓ 1 | |
Exam tip: Write the power in watts (1 MW = 10⁶ W) before dividing by the energy of each event in joules.
A nuclear reactor has been operating at constant power. The control rods are then pushed further into the core. What is the effect?
| Marking point | Mark | Notes |
|---|---|---|
| Control rods absorb neutrons. With more absorbed, fewer than one neutron per fission goes on to cause a further fission, so the chain reaction dies away and the power falls. The number of neutrons released per fission is a property of the fission and is not changed by the rods. Exactly one neutron per fission causing another fission is the condition for constant power, which no longer holds once more neutrons are absorbed. Rods absorb neutrons; they do not cause fission. | ✓ 1 | Answer A |
Answer: A
Common error: Control rods change how many neutrons are absorbed, not how many are released per fission.
One fission of a uranium-235 nucleus begins when it absorbs a neutron. The nucleus splits into a barium-141 nucleus, a krypton nucleus and three neutrons. The barium nucleus has 56 protons and uranium has 92 protons.
Determine the nucleon number and the proton number of the krypton nucleus.
Explain how a chain reaction can occur in uranium-235.
Explain why used fuel rods are kept under water for several years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number 235 + 1 − 141 − 3 = 92 | ✓ 1 | |
| proton number 92 − 56 = 36 | ✓ 1 | |
| Part (b) | ||
| the neutrons released can be absorbed by other uranium-235 nuclei and cause them to fission | ✓ 1 | |
| each fission releases more neutrons, so the reaction can continue by itself | ✓ 1 | |
| Part (c) | ||
| the fission products are radioactive and keep producing thermal energy, and the water removes it | ✓ 1 | |
| the water absorbs radiation from the fuel and shields workers | ✓ 1 | |
Common error: Check that nucleon numbers balance, including the neutrons on both sides.
Fusion of hydrogen nuclei in the core of a star needs a temperature of millions of degrees. Why is such a high temperature needed?
| Marking point | Mark | Notes |
|---|---|---|
| Nuclei are all positively charged and repel each other. At very high temperature they move fast enough to approach closely, where fusion can happen, so this is correct. Fusion does not require nuclei to be split first. Gravity between single nuclei is far too weak to matter. Nuclei do not become negative, they stay positive. | ✓ 1 | Answer D |
Answer: D
Common error: Fusion is about overcoming repulsion between positive nuclei, not about melting or splitting them.
In the core of some stars, two helium-3 nuclei fuse to form a helium-4 nucleus and two protons:
32He + 32He → 42He + 211H
State the number of neutrons in a helium-3 nucleus.
Show that the nucleon number and the proton number are both conserved in this reaction.
Explain why the temperature in the core must be very high for helium-3 nuclei to fuse.
Explain why a star in which fusion is taking place does not collapse under its own gravity.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1 neutron | ✓ 1 | |
| Part (b) | ||
| nucleon number: 3 + 3 = 6 and 4 + 1 + 1 = 6 | ✓ 1 | |
| proton number: 2 + 2 = 4 and 2 + 1 + 1 = 4 | ✓ 1 | |
| Part (c) | ||
| nuclei are positively charged and repel each other | ✓ 1 | |
| high temperature gives the nuclei enough kinetic energy to get close enough to fuse | ✓ 1 | |
| Part (d) | ||
| gravity pulls the material of the star inwards | ✓ 1 | |
| the outward pressure of the very hot gas and radiation from fusion balances the inward pull | ✓ 1 | |
Common error: Fusion joins small nuclei into a larger one; fission splits a large nucleus into smaller ones.
A student writes this description of nuclear fission in a power-station reactor.
“A uranium-235 nucleus absorbs a proton and splits into two identical smaller nuclei. The energy released comes from the nucleons being destroyed. Control rods are used to speed up the reaction.”
Identify three errors in the student’s description and correct each one.
Describe how a chain reaction occurs in the reactor.
Explain why control rods are needed when the reactor is operating at constant power.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| first valid correction given | ✓ 1 | |
| second valid correction given | ✓ 1 | |
| third valid correction given | ✓ 1 | |
| Accept any three of: the nucleus absorbs a neutron, not a proton; the nucleus splits into two smaller nuclei that are usually of different sizes, not identical; nucleons are not destroyed, the energy is released as kinetic energy of the fission fragments (and neutrons); control rods absorb neutrons to slow the reaction, they do not speed it up | — | |
| Part (b) | ||
| each fission releases (two or three) neutrons | ✓ 1 | |
| these neutrons are absorbed by other uranium-235 nuclei, which then undergo fission and release more neutrons | ✓ 1 | |
| Part (c) | ||
| at constant power, on average only one neutron from each fission must cause another fission | ✓ 1 | |
| control rods absorb the surplus neutrons, so the number of fissions per second stays constant | ✓ 1 | |
Exam tip: In fission answers name the particle that triggers the splitting (a neutron) and say where the released energy goes (kinetic energy of the products).
A nuclear power station uses uranium-235 as fuel. A neutron is absorbed by a uranium-235 nucleus, which then splits into two smaller nuclei and releases several more neutrons.
Describe how a chain reaction is sustained in the reactor.
State the purpose of the control rods in the reactor.
In stars, energy is released by nuclear fusion. Explain why very high temperatures are needed for fusion to happen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The neutrons released by one fission can be absorbed by other uranium-235 nuclei | ✓ 1 | |
| These nuclei then split and release further neutrons | ✓ 1 | |
| So the process continues by itself, with each fission causing more fissions | ✓ 1 | |
| Part (b) | ||
| They absorb neutrons to control the rate of the reaction | ✓ 1 | |
| Part (c) | ||
| Nuclei are positively charged so they repel each other | ✓ 1 | |
| High temperature means high kinetic energy, so nuclei can get close enough to join | ✓ 1 | |
Common error: Fission is the splitting of a large nucleus and fusion is the joining of small nuclei; do not swap the two terms.
A hospital store holds old radium needles. Radium-226, 22688Ra, decays by alpha emission to an isotope of radon (Rn). Later in the decay series, lead-214, 21482Pb, decays by beta-minus emission to an isotope of bismuth (Bi).
Determine the nucleon number and the proton number of the radon nucleus.
Determine the nucleon number and the proton number of the bismuth nucleus.
Explain why the nucleon number does not change when a nucleus emits a beta-minus particle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number 226 − 4 = 222 | ✓ 1 | |
| proton number 88 − 2 = 86 | ✓ 1 | |
| Part (b) | ||
| nucleon number 214 | ✓ 1 | |
| proton number 82 + 1 = 83 | ✓ 1 | |
| Part (c) | ||
| A neutron in the nucleus changes into a proton (and an electron is emitted) | ✓ 1 | |
| The total number of protons plus neutrons stays the same | ✓ 1 | |
Exam tip: For a nuclear equation, check that the nucleon numbers add up to the same total on both sides, and so do the proton numbers.
Which statement gives a reason why nuclear fusion is not yet used to generate electricity commercially?
| Marking point | Mark | Notes |
|---|---|---|
| Light nuclei repel each other, so they must be at extremely high temperature to fuse, and the hot plasma is very hard to contain. The waste from fusion is small in quantity and short-lived, so that option is wrong. Fusion fuels are hydrogen isotopes, not uranium. Fusion releases more energy per kilogram than fission, not less. | ✓ 1 | Answer A |
Answer: A
Common error: Do not mix up the two processes: fission splits large nuclei such as uranium, while fusion joins small nuclei such as hydrogen isotopes.
Engineers are comparing a fission power station with a fusion power station that might be built in the future.
Describe how a chain reaction is sustained in a fission reactor.
Give one advantage and one disadvantage of a fusion power station compared with a fission power station.
The fusion power station is designed to release energy at a rate of 500 MW. Each fusion reaction releases 2.8 × 10−12 J. Calculate the number of fusion reactions per second. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a neutron is absorbed by a uranium-235 nucleus, which splits | ✓ 1 | |
| releasing further neutrons that go on to split other nuclei | ✓ 1 | |
| Part (b) | ||
| advantage, any one from: fuel (hydrogen isotopes) is abundant; waste is much less radioactive or shorter-lived; no runaway chain reaction | ✓ 1 | |
| disadvantage, any one from: extremely high temperatures needed and difficult to contain; no working commercial plant yet; tritium fuel has to be produced | ✓ 1 | |
| Part (c) | ||
| number per second = power ÷ energy per reaction = 500 × 106 ÷ 2.8 × 10−12 | ✓ 1 | |
| = 1.8 × 1020 per second | ✓ 1 | |
Exam tip: Write 500 MW as 500 × 10⁶ W before dividing, so the power of ten is correct.
Potassium-40 decays by emitting a beta-minus particle and forms calcium-40.
State what a beta-minus particle is.
Describe how a beta-minus particle is produced in the nucleus.
Calculate the nucleon number and the proton number of the nucleus formed when potassium-40 (19 protons, nucleon number 40) emits a beta-minus particle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a fast-moving electron | ✓ 1 | |
| Part (b) | ||
| a neutron changes into a proton | ✓ 1 | |
| and an electron, which is emitted from the nucleus | ✓ 1 | |
| Part (c) | ||
| nucleon number stays 40 | ✓ 1 | |
| proton number 19 + 1 = 20 | ✓ 1 | |
Exam tip: In nuclear equations the top numbers and the bottom numbers must each add up to the same total on both sides.
A small nuclear reactor supplies heat to a data centre. The reactor uses fission of uranium-235. Each fission releases 3.2 × 10−11 J of energy.
Describe how a chain reaction is sustained in the reactor.
State the purpose of the control rods.
The reactor releases 2.4 × 107 J of energy each second. Calculate the number of fissions each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a neutron is absorbed by a uranium-235 nucleus, which splits and releases further neutrons | ✓ 1 | |
| these neutrons go on to cause further fissions | ✓ 1 | |
| Part (b) | ||
| absorb neutrons to control (slow or stop) the rate of fission | ✓ 1 | |
| Part (c) | ||
| 2.4 × 107 ÷ 3.2 × 10−11 | ✓ 1 | |
| 7.5 × 1017 | ✓ 1 | |
Exam tip: Divide the total energy by the energy released in one event to find how many events took place.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated statements about dangers OR about storage, with little linkage to the type of waste. |
| Level 2 (3–4) | Dangers and some risk-reduction methods are described; some linkage to the properties of at least one kind of waste (activity, half-life or heat). |
| Level 3 (5–6) | Dangers explained and risk-reduction methods for BOTH kinds of waste discussed, clearly linked to their activity, half-life and heat; logically structured. |
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated statements about fission or fusion, for example a simple description of one process or one advantage, with little comparison. |
| Level 2 (3–4) | Describes both processes and gives some advantages or disadvantages, with some linked comparison; reasons may be incomplete (e.g. high temperature stated without the repulsion of nuclei). |
| Level 3 (5–6) | Clear, balanced comparison of both processes covering how energy is released and at least two linked advantages and disadvantages of each, with physics reasons (e.g. radioactive daughter nuclei, electrostatic repulsion requiring high temperature and pressure). |
A uranium-236 nucleus, initially at rest, splits into two fragments. Ignore the neutrons released.
Fragment X has a mass of 1.6×10−25 kg and moves at 1.4×107 m/s.
Fragment Y has a mass of 2.3×10−25 kg.
Explain why the two fragments must move in opposite directions.
Calculate the speed of fragment Y.
Calculate the total kinetic energy of the two fragments.
Explain how this kinetic energy ends up heating the coolant in the reactor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Momentum is conserved, and the momentum before the split is zero because the nucleus is at rest . The total momentum afterwards must also be zero, so the fragments have equal and opposite momenta . | ✓ 2 | |
| Part (b) | ||
| mXvX = mYvY | ✓ 1 | |
| vY = (1.6×10−25 × 1.4×107) ÷ 2.3×10−25 | ✓ 1 | |
| = 9.7×106 m/s | ✓ 1 | |
| Part (c) | ||
| KE = ½mv2 used | ✓ 1 | |
| KE of X = 1.57×10−11 J and KE of Y = 1.08–1.09×10−11 J | ✓ 1 | |
| total = 2.7×10−11 J (accept 2.6×10−11 J if the rounded speed of Y is used) | ✓ 1 | |
| Part (d) | ||
| The fragments collide with the surrounding atoms in the fuel and slow down, increasing the thermal / internal energy of the fuel rods . Energy is then transferred to the coolant flowing past the fuel rods by heating / conduction . | ✓ 2 | |
Common error: The lighter fragment moves faster and carries more kinetic energy. The fragments share momentum equally, but not kinetic energy.
Which statement about nuclear fusion in a future power station is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Fusion joins small positive nuclei, which repel, so they need very high speeds and hence very high temperature (B). A describes fission. C is wrong because the main product is helium. D describes fission. | ✓ 1 | Answer B |
Answer: B
Exam tip: Fusion joins small nuclei; fission splits large ones.
Fission and fusion are both considered for future power stations.
State what happens to a nucleus in nuclear fission.
Explain why nuclear fusion needs a very high temperature.
Evaluate the use of fission and fusion for a future power station.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a large nucleus splits into two smaller nuclei, releasing neutrons | ✓ 1 | |
| Part (b) | ||
| nuclei are positively charged and repel one another | ✓ 1 | |
| high temperature gives them enough kinetic energy to get close enough to fuse | ✓ 1 | |
| Part (c) | ||
| fission splits large nuclei such as uranium; fusion joins small nuclei such as hydrogen isotopes | ✓ 1 | |
| fusion fuel (hydrogen isotopes from water) is very plentiful, whereas uranium is a limited mined resource | ✓ 1 | |
| fission waste is highly radioactive for thousands of years and must be stored; fusion makes helium, which is not radioactive | ✓ 1 | |
| a fission reactor could overheat if cooling failed; fusion stops if conditions are lost | ✓ 1 | |
| fission power stations work now, but sustained fusion power has not yet been achieved on a commercial scale | ✓ 1 | |
| a justified conclusion that weighs these points, e.g. fusion is preferable in the long term but fission is available now | ✓ 1 | |
Exam tip: In an 'evaluate' answer compare both options on several points and finish with a reasoned judgement.
A nuclear power station has a thermal power output of 2.4 GW (2.4 × 109 W) from uranium-235 fission. Each fission releases 3.2 × 10−11 J of energy. The mass of one uranium-235 nucleus is 3.9 × 10−25 kg. There are 86 400 seconds in one day.
Calculate the number of fissions that occur each second. Give your answer in standard form to 2 significant figures.
Calculate the mass of uranium-235 that undergoes fission in one day. Give your answer in kg to 2 significant figures. Use the unrounded answer to (a), or 7.5 × 1019; both give the same result.
A coal power station releases 3.0 × 107 J of energy for every kilogram of coal burned. Calculate the mass of coal needed to release the same amount of energy in one day as the nuclear station. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| fissions per second = power ÷ energy per fission = 2.4 × 109 ÷ 3.2 × 10−11 | ✓ 1 | |
| 7.5 × 1019 per second | ✓ 1 | |
| Part (b) | ||
| fissions in one day = 7.5 × 1019 × 86 400 (= 6.48 × 1024) | ✓ 1 | |
| mass = number of fissions × 3.9 × 10−25 kg | ✓ 1 | |
| 2.5 kg (2.527 kg) | ✓ 1 | |
| Part (c) | ||
| energy in one day = 2.4 × 109 × 86 400 = 2.07 × 1014 J | ✓ 1 | |
| mass = energy ÷ 3.0 × 107 (= 6.912 × 106 kg) | ✓ 1 | |
| 6.9 × 106 kg (6 900 000 kg) | ✓ 1 | |
Exam tip: For large and small numbers, enter standard form using the EXP key and keep unrounded values on the calculator between parts.
A source has a half-life of 4.0 hours. A detector reads 120 counts per minute with the source present. The background count rate is 30 counts per minute. What will the detector read after 8.0 hours?
| Marking point | Mark | Notes |
|---|---|---|
| Correct for background first: 120 − 30 = 90 counts per minute from the source. 8.0 hours is 2 half-lives, so this falls to 90 ÷ 4 = 22.5 counts per minute. The detector then also reads the background: 22.5 + 30 = 52.5 counts per minute, so C is correct. A forgets to add the background back. B quarters the total reading including background. D uses only one half-life (45 + 30). | ✓ 1 | Answer C |
Answer: C
Common error: Only the count rate from the source halves each half-life; the background stays constant.
Radium-226 (88 protons) decays through a series of alpha and beta-minus decays to form lead-210 (82 protons). One step in the series is the decay of lead-214 (82 protons) to bismuth-214 (83 protons).
Write a nuclear equation for the beta-minus decay of 21482Pb to bismuth-214.
Determine the number of alpha particles and the number of beta-minus particles emitted altogether when a radium-226 nucleus decays to lead-210.
Explain why beta-minus decay changes the proton number of a nucleus but not its nucleon number.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 21483Bi as the product | ✓ 1 | |
| B1 an electron (beta particle) 0−1e as the other product, with the equation balanced | ✓ 1 | |
| Part (b) | ||
| M1 for nucleon number change 226 − 210 = 16 | ✓ 1 | |
| A1 for 4 alpha particles | ✓ 1 | |
| M1 for 4 alpha particles lowering the proton number by 8, to 80, so 2 more protons are needed to reach 82 | ✓ 1 | |
| A1 for 2 beta-minus particles | ✓ 1 | |
| Part (c) | ||
| B1 a neutron changes into a proton and an electron | ✓ 1 | |
| B1 the electron is emitted, so the total number of protons plus neutrons stays the same while the number of protons increases by one | ✓ 1 | |
Common error: Count the total change in nucleon number to find the alpha particles first, then use the proton number to find the beta particles.
A volcanologist collects a sample of ash that contains a radioactive isotope. The detector reading is taken at different times. The background count rate is 20 counts per minute.
| Time / h | 0 | 2 | 4 | 6 | 8 | 10 |
| Count rate / counts per minute | 380 | 275 | 200 | 147 | 110 | 84 |
Calculate the count rate due to the ash at 4 hours.
Determine the half-life of the isotope.
Predict the count rate that the detector will read at 12 hours.
Explain why the background count rate must be subtracted before the half-life is found.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 180 (counts per minute) | ✓ 1 | |
| Part (b) | ||
| M1 for corrected count rate falling from 360 to 180 (counts per minute) | ✓ 1 | |
| A1 for 4.0 h | ✓ 1 | |
| Part (c) | ||
| M1 for 12 h = 3 half-lives, e.g. \(360\div2^3\) | ✓ 1 | |
| A1 for 45 (counts per minute) from the source | ✓ 1 | |
| A1 for 65 (counts per minute) after adding the background (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| B1 the background does not come from the sample and does not decay (it stays constant) | ✓ 1 | |
| B1 so the total reading would not halve in the same time, and the half-life found would be wrong (too long) | ✓ 1 | |
Exam tip: Correct every reading for background first, then look for the time taken for that corrected count rate to halve.
Archaeologists date ancient dyed flax fibres from a cave. Carbon-14 has a half-life of 5700 years. Living flax gives a corrected count rate of 12.0 counts/min per gram. A fibre sample gives a measured count rate of 4.0 counts/min per gram; the background rate is 1.0 counts/min per gram.
State one difference between atoms of carbon-12 and carbon-14.
Calculate the age of the fibres, in years.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Carbon-14 has 8 neutrons and carbon-12 has 6 neutrons (carbon-14 has two more neutrons) | ✓ 1 | |
| Part (b) | ||
| Corrected count rate = 4.0 − 1.0 = 3.0 counts/min per gram | ✓ 1 | |
| 3.0 ÷ 12.0 = 1/4 of the original | ✓ 1 | |
| This is 2 half-lives | ✓ 1 | |
| 2 × 5700 = 11400 years | ✓ 1 | |
Common error: Subtract the background count before comparing the sample with living material.
Scientists are comparing nuclear fission and nuclear fusion as energy sources for future power stations.
Explain why fusion of nuclei requires a temperature of millions of degrees.
Give two advantages of fusion over fission as the energy source in a power station.
In one fusion reaction a deuterium nucleus and a tritium nucleus release 2.8 × 10−12 J. The total mass of these two nuclei is 8.3 × 10−27 kg. Calculate the energy released per kilogram of this fuel. Give your answer to 2 significant figures.
Fission of uranium-235 releases 8.2 × 1013 J/kg. Use the value 3.4 × 1014 J/kg from (c) to calculate how many times more energy per kilogram is released by this fusion fuel. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nuclei are positively charged so they repel each other | ✓ 1 | |
| the nuclei must have very high kinetic energy to get close enough for fusion to occur | ✓ 1 | |
| Part (b) | ||
| any valid advantage, for example: the fuel (hydrogen isotopes) is plentiful | ✓ 1 | |
| a second different valid advantage, for example: no long-lived radioactive waste, or no chain reaction that can run out of control | ✓ 1 | |
| Part (c) | ||
| energy per kg = 2.8 × 10−12 / 8.3 × 10−27 | ✓ 1 | |
| 3.4 × 1014 J/kg | ✓ 1 | |
| Part (d) | ||
| 3.4 × 1014 / 8.2 × 1013 | ✓ 1 | |
| 4.1 (times) | ✓ 1 | |
Exam tip: Give two advantages that are clearly different from each other; repeating the same idea in other words earns only one mark.
One fission reaction in a nuclear power station is shown below.
23592U + 10n → 14156Ba + 9236Kr + x 10n
Determine the value of x.
Explain how a chain reaction occurs.
The power station has an electrical output of 900 MW and an efficiency of 36%. Calculate the thermal power produced by the reactor.
Each fission releases 3.2 × 10−11 J of energy. Use the value 2.5 × 109 W to calculate the number of fission reactions each second. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number: 235 + 1 = 236 = 141 + 92 + x | ✓ 1 | |
| x = 3 | ✓ 1 | |
| Part (b) | ||
| neutrons released in one fission are absorbed by other uranium-235 nuclei and cause them to split | ✓ 1 | |
| these fissions release more neutrons, so the process continues by itself | ✓ 1 | |
| Part (c) | ||
| thermal power = 900 × 106 / 0.36 | ✓ 1 | |
| 2.5 × 109 W | ✓ 1 | |
| Part (d) | ||
| number per second = 2.5 × 109 / 3.2 × 10−11 | ✓ 1 | |
| 7.8 × 1019 per second | ✓ 1 | |
Exam tip: For a nuclear equation, check the nucleon numbers first; then use the proton numbers as a second check.
In a research reactor each fission of a uranium-235 nucleus releases on average 2.5 neutrons. Some of these neutrons are absorbed by the control rods, some are absorbed without causing fission and some escape. Only the remainder cause another fission.
Calculate the fraction of the released neutrons that must cause another fission when the reactor runs at constant power.
An operator sets the control rods so that only 36% of the released neutrons cause another fission. The first generation has 1000 fissions. Calculate the number of fissions in the third generation.
Explain what happens to the power of the reactor in the situation described in (b), and state what the operator must do to restore constant power.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 on average one neutron from each fission must cause one further fission | ✓ 1 | |
| A1 1 ÷ 2.5 = 0.40 | ✓ 1 | |
| Part (b) | ||
| M1 each fission leads on average to 2.5 × 0.36 = 0.90 fissions in the next generation | ✓ 1 | |
| M1 apply this factor twice: 1000 × 0.90 × 0.90 | ✓ 1 | |
| A1 810 fissions | ✓ 1 | |
| Part (c) | ||
| fewer fissions in each generation, so the power falls | ✓ 1 | |
| withdraw the control rods slightly so that fewer neutrons are absorbed | ✓ 1 | |
Common error: Constant power needs one further fission per fission, not one neutron released per fission.
In a fusion experiment, deuterium nuclei (hydrogen-2) fuse with tritium nuclei (hydrogen-3) to form helium-4 and a particle X: hydrogen-2 + hydrogen-3 → helium-4 + X. Hydrogen has 1 proton and helium has 2 protons.
Determine the nucleon number and proton number of X and name the particle.
Explain why fusion requires a very high temperature.
Explain why the hot plasma cannot be held in a solid container, and state how it is contained.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number 2 + 3 − 4 = 1 and proton number 1 + 1 − 2 = 0 | ✓ 1 | |
| X is a neutron | ✓ 1 | |
| Part (b) | ||
| nuclei are positively charged and repel each other | ✓ 1 | |
| at high temperature the nuclei have enough kinetic energy to get close enough to fuse | ✓ 1 | |
| Part (c) | ||
| the plasma would melt the container or be cooled by contact with it | ✓ 1 | |
| it is held in place by strong magnetic fields | ✓ 1 | |
Exam tip: For nuclear equations, check that nucleon numbers and proton numbers each balance on both sides.
A narrow beam containing alpha particles, beta particles and gamma rays travels horizontally in a vacuum between two parallel horizontal metal plates. The upper plate is positive and the lower plate is negative.
Describe the path of each type of radiation between the plates.
Explain why the beta particles are deflected much more than the alpha particles, even though an alpha particle has twice the charge.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| beta particles curve upwards towards the positive plate | ✓ 1 | |
| alpha particles curve downwards towards the negative plate | ✓ 1 | |
| gamma rays continue in a straight line | ✓ 1 | |
| Part (b) | ||
| the force on an alpha particle is larger because of its larger charge | ✓ 1 | |
| an alpha particle has thousands of times the mass, so its acceleration and deflection are much smaller | ✓ 1 | |
Exam tip: Deflection depends on the force and on the mass: compare both.
A shipment contains two radioactive tracers, A and B, with equal numbers of nuclei at the start. The half-life of A is 2.0 years and the half-life of B is 8.0 years. The shipment is held in a warehouse for 8.0 years.
Calculate the fraction of the nuclei of A that remain after 8.0 years.
State the fraction of the nuclei of B that remain after 8.0 years.
Calculate the percentage of all the nuclei in the shipment that have decayed after 8.0 years. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 8.0 years is 4 half-lives of A | ✓ 1 | |
| A1 (1/2)4 = 1/16 | ✓ 1 | |
| Part (b) | ||
| 1/2 | ✓ 1 | |
| Part (c) | ||
| M1 fraction of all nuclei remaining = (1/16 + 1/2) ÷ 2 = 0.28125 | ✓ 1 | |
| M1 fraction decayed = 1 − 0.28125 | ✓ 1 | |
| A1 71.9 % | ✓ 1 | |
Common error: With equal starting numbers, add the remaining fractions and divide by two; do not add the half-lives.
In the core of a massive star near the end of its life, two carbon-12 nuclei can fuse to form a neon nucleus and a helium-4 nucleus:
126C + 126C → abNe + 42He
State the values of a and b.
Explain why carbon nuclei need a higher temperature to fuse than hydrogen nuclei.
Each fusion reaction releases 7.0 × 10−13 J. The carbon-burning region of the star releases energy at a rate of 2.1 × 1030 W. Calculate the number of reactions that occur each second.
Each reaction uses two carbon-12 nuclei, each of mass 2.0 × 10−26 kg. Calculate the mass of carbon that is fused each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = 20 | ✓ 1 | |
| b = 10 | ✓ 1 | |
| Part (b) | ||
| a carbon nucleus has six protons, so it has a greater positive charge and the repulsion between two nuclei is greater | ✓ 1 | |
| so the nuclei need greater kinetic energy, which means a higher temperature | ✓ 1 | |
| Part (c) | ||
| M1 number = 2.1 × 1030 ÷ 7.0 × 10−13 | ✓ 1 | |
| A1 3.0 × 1042 per second | ✓ 1 | |
| Part (d) | ||
| M1 3.0 × 1042 × 2 × 2.0 × 10−26 | ✓ 1 | |
| A1 1.2 × 1017 kg (ECF from (c)) | ✓ 1 | |
Exam tip: Write the power in watts as joules per second; dividing by the energy per reaction then gives reactions per second.
A nuclear reactor operates at a constant thermal power of 120 MW. Each fission of a uranium-235 nucleus releases 3.2 × 10−11 J of energy and releases on average 2.4 neutrons. The mass of one uranium-235 nucleus is 3.9 × 10−25 kg.
Calculate the number of fissions occurring each second.
Calculate the number of neutrons released each second.
State how many of these neutrons must cause another fission each second, explain why, and calculate how many neutrons per second are lost without causing fission.
Calculate the mass of uranium-235 that undergoes fission in 24 hours. Use the unrounded value from part (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 120 × 106 ÷ 3.2 × 10−11 | ✓ 1 | |
| = 3.75 × 1018 fissions per second | ✓ 1 | |
| Part (b) | ||
| 2.4 × 3.75 × 1018 | ✓ 1 | |
| = 9.00 × 1018 neutrons per second | ✓ 1 | |
| (ECF from (a)) | — | |
| Part (c) | ||
| 3.75 × 1018 per second | ✓ 1 | |
| power is constant so the fission rate is constant, meaning on average one neutron per fission causes a further fission | ✓ 1 | |
| 9.00 × 1018 − 3.75 × 1018 = 5.25 × 1018 neutrons per second | ✓ 1 | |
| (ECF from (a) and (b)) | — | |
| Part (d) | ||
| 24 hours = 86 400 s, so number of fissions = 3.75 × 1018 × 86 400 = 3.24 × 1023 | ✓ 1 | |
| mass = 3.24 × 1023 × 3.9 × 10−25 | ✓ 1 | |
| = 0.126 kg (3 s.f.) | ✓ 1 | |
Exam tip: Convert hours to seconds before multiplying by a rate in “per second”, and keep unrounded values until the final line.
In the core of the Sun, hydrogen nuclei undergo fusion to form helium nuclei. The Sun radiates energy at a power of 3.6 × 1026 W. Assume that each fusion process that forms one helium nucleus releases 4.0 × 10−12 J, and that one helium nucleus has a mass of 6.6 × 10−27 kg.
Explain why fusion can happen only at very high temperatures.
Calculate the number of helium nuclei formed in the Sun each second.
Calculate the mass of helium formed in the Sun each second.
Assume the mass of hydrogen used each second equals the mass of helium formed. The core of the Sun contains 1.5 × 1029 kg of hydrogen. Calculate the time in seconds for all of this hydrogen to be used at this rate. Use the unrounded value from part (c).
State one difference between nuclear fusion and nuclear fission.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nuclei are positively charged so they repel each other | ✓ 1 | |
| a high temperature gives the nuclei a high speed (kinetic energy) | ✓ 1 | |
| so they can overcome the repulsion and get close enough to fuse | ✓ 1 | |
| Part (b) | ||
| 3.6 × 1026 ÷ 4.0 × 10−12 | ✓ 1 | |
| = 9.0 × 1037 per second | ✓ 1 | |
| Part (c) | ||
| 9.0 × 1037 × 6.6 × 10−27 | ✓ 1 | |
| = 5.94 × 1011 kg per second | ✓ 1 | |
| (ECF from (b)) | — | |
| Part (d) | ||
| time = mass ÷ rate | ✓ 1 | |
| 1.5 × 1029 ÷ 5.94 × 1011 | ✓ 1 | |
| = 2.53 × 1017 s | ✓ 1 | |
| (ECF from (c)) | — | |
| Part (e) | ||
| fusion joins small nuclei to form a larger nucleus, whereas fission splits a large nucleus into smaller nuclei | ✓ 1 | |
Exam tip: Check that the mass and energy figures use the same time basis, here “per second”, before dividing or multiplying.
A nucleus of uranium-238, 23892U, emits one alpha particle and then two beta-minus particles. What are the nucleon number and proton number of the nucleus that results?
| Marking point | Mark | Notes |
|---|---|---|
| The alpha particle removes 4 nucleons and 2 protons, giving 234 and 90. Each beta-minus emission leaves the nucleon number unchanged and raises the proton number by 1 (a neutron becomes a proton), so two emissions give 234 and 92, which is the second option. The first option wrongly lowers the proton number for each beta particle. The third option ignores the alpha particle. The fourth option ignores the beta particles and also leaves the nucleon number unchanged after the alpha emission. | ✓ 1 | Answer B |
Answer: B
Common error: Beta-minus emission raises the proton number by one; it does not lower it, and the nucleon number stays the same.
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