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IB Physics Standard Level · Topic E.3

E.3 Radioactive decay

163 free, high-quality past-paper-style questions on Radioactive decay for IB Physics SL, each with a full mark scheme. Covers: Isotopes; Changes in the nucleus; The strong nuclear force; and more. For when you have finished the past papers, or want to practise one topic at a time.E.3 at Higher Level →Read the E.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.810 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-08Isotopes Paper 1AEasy1 mark

Two nuclides are isotopes of the same element.

Which quantities must be the same for both?

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Marking pointMarkNotes
Step 1Isotopes of an element have the same proton number Z but different neutron numbers, and therefore different nucleon numbers A.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: isotopes have the same number of protons, which fixes the element, but different numbers of neutrons.
  • BIsotopes have different nucleon numbers — that is what distinguishes them.
  • CIf both were the same the two nuclides would be identical, not isotopes.
  • DNuclides with the same neutron number but different proton numbers are different elements.
2E-1A-09Changes in the nucleus Paper 1AEasy1 mark

Which statement about a change in a nucleus during radioactive decay is correct?

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Step 1A gamma photon has no charge and no mass, so emitting it lowers the energy of the nucleus (it moves from an excited state to a lower one) without changing A or Z.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThe alpha particle takes away 2 protons and 2 neutrons: the nucleon number falls by 4 but the neutron number falls by only 2.
  • BIn beta-minus decay a neutron becomes a proton: the proton number rises by 1 and the nucleon number is unchanged.
  • CThat describes beta-minus decay; in beta-plus decay a proton changes into a neutron.
  • DCorrect: a gamma photon carries away energy but no charge and no nucleons.
3E-1A-16The strong nuclear force Paper 1AEasy1 mark

A nucleus contains several protons packed within a distance of about 10−15 m.

Why does the nucleus not fly apart?

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Step 1The protons repel electrostatically and gravity is far too weak to balance this, so a third force — the strong nuclear force, attractive between all nucleons over a range of about 10−15 m — must hold the nucleus together.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AGravity between two protons is about 1036 times weaker than their electrostatic repulsion.
  • BThe electrons are far outside the nucleus and do not cancel the repulsion between protons inside it.
  • CThe electrostatic force between two protons is always repulsive; a different, attractive force is needed.
  • DCorrect: at separations of about 10−15 m the strong nuclear force is attractive and larger than the electrostatic repulsion, and it acts on neutrons as well as protons.
4E-1A-36Mass–energy equivalence Paper 1AEasy1 mark

A stationary nucleus of mass M decays into a nucleus of mass m1 and an alpha particle of mass m2.

What is the energy released in the decay? (c is the speed of light.)

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Step 1The products have less mass than the original nucleus; the missing mass M − m1 − m2 is released as energy, E = (M − m1 − m2)c2.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AThis is the negative of the energy released: for a decay that happens, the products have less mass than the parent, so this expression is negative.
  • BCorrect: the decrease in mass is M − m1 − m2, and E = Δmc2.
  • CThe speed of light has not been squared; E = mc2.
  • DMc2 is the energy equivalent of the whole nucleus; only the decrease in mass is released.
5E-1A-51Half-life & activity Paper 1AEasy1 mark

A sealed sample contains only nuclei of a radioactive isotope X at time t = 0. X decays directly into a stable nuclide Y.

Which graph shows how the number N of nuclei of Y in the sample varies with time t?

Four sketch graphs A to D of the number N of nuclei of Y against time t. A: falls from a maximum with decreasing gradient. B: rises from zero with decreasing gradient towards a constant value. C: a straight line rising from zero. D: rises from zero to a maximum and then falls.A.Nt0B.Nt0C.Nt0D.Nt0
Sketch graphs, NOT to scale
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Step 1Every X nucleus that decays becomes a Y nucleus, so N = (initial number of X) − (number of X remaining).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2N starts at zero, rises to half the initial number of X after one half-life and three-quarters after two, and approaches the initial number of X: a rising curve whose gradient decreases.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis curve shows the number of X nuclei remaining; the number of Y nuclei grows as X decays.
  • BCorrect: Y increases quickly at first, when the activity of X is high, then more and more slowly as X runs out.
  • CA straight line would need a constant activity; the activity of X halves every half-life, so Y is produced more and more slowly.
  • DY is stable, so its number never falls; a rise followed by a fall would happen only if Y were itself radioactive.
6E-1A-52Penetration & ionization Paper 1AEasy1 mark

In a cloud chamber, alpha particles leave short, thick, straight tracks of droplets, while beta particles leave longer, thinner tracks. The droplets form on ions that the particles produce in the air.

Which statement explains why the alpha-particle tracks are thicker?

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Step 1A thick track contains many droplets per unit length, and each droplet forms on an ion.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Alpha particles are much more strongly ionizing than beta particles, so they produce far more ions per millimetre and lose their energy in a short distance.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: alpha particles are strongly ionizing, so they leave a dense trail of ions (and droplets) and stop after a short distance.
  • BThe droplets form on ions, not because the particle pushes the vapour; the size of the particle is not what makes the track.
  • CAlpha particles from radioactive sources are slower than beta particles; in any case the thickness depends on how strongly the particle ionizes.
  • DAlpha particles are positively charged; the track shows where ions were produced, not an attraction of droplets.
7E-1A-53Mass–energy equivalence Paper 1AEasy1 mark

The binding energy of a lithium-7 nucleus is 39.2 MeV.

What is the mass defect of this nucleus? (1 MeV = 1.60 × 10−13 J)

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Step 1E = 39.2 × 1.60 × 10−13 = 6.27 × 10−12 J.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Δm = E/c2 = 6.27 × 10−12/(3.00 × 108)2 = 7.0 × 10−29 kg.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis converts 39.2 eV, not 39.2 MeV, into joules: the factor of 106 has been lost.
  • BThis is the mass defect divided by 7, i.e. per nucleon; the question asks for the mass defect of the whole nucleus.
  • CCorrect: Δm = E/c2 = 6.27 × 10−12 J/(9.00 × 1016 m2 s−2) = 7.0 × 10−29 kg.
  • DThe energy has been divided by c instead of by c2.
8E-1A-75Isotopes Paper 1AEasy1 mark

Uranium-235 and uranium-238 are isotopes. To make reactor fuel, the fraction of uranium-235 is increased in gas centrifuges, which separate molecules of a uranium gas compound by their mass, because the two isotopes cannot be separated by chemical reactions.

Which statement explains why the two isotopes behave identically in chemical reactions but differ in mass?

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Step 1Isotopes have the same proton number (92), so neutral atoms have the same number of electrons (92) and the same chemistry; uranium-238 has 146 neutrons and uranium-235 has 143, so the atoms differ in mass.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: the equal numbers of protons and electrons fix the element and its chemistry, and the three extra neutrons in uranium-238 give it the larger mass.
  • BProtons and neutrons have been swapped: a different number of protons would make a different element, with different chemistry.
  • CThe nucleon numbers 235 and 238 are different, and the neutral atoms of one element have the same number of electrons; it is the electrons that set the chemical behaviour.
  • DThe nuclei are not identical: they contain different numbers of neutrons, and it is mainly this, not a difference in binding, that makes the masses differ.
9E-1A-77Background radiation Paper 1AEasy1 mark

Background radiation comes from both natural and artificial sources.

Which row gives one natural source and one artificial source of background radiation?

Natural sourceArtificial source
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Step 1Radon from uranium in rocks and soil is natural; fallout was produced by nuclear explosions, so it is artificial.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • ACarbon-14 is formed naturally in the upper atmosphere when cosmic-ray neutrons strike nitrogen nuclei, so it is a natural source even though a nuclear reaction produces it.
  • BThe two sources are the wrong way round: X-ray machines are made by people, while cosmic rays arrive naturally from space.
  • CBricks and concrete are made by people, but their radioactivity comes from natural uranium, thorium and potassium-40 in the rock they are made from, so this is a natural source.
  • DCorrect: radon is released naturally by the decay of uranium in the ground; fallout comes from nuclear weapons tests carried out by people.
10E-1A-78The strong nuclear force Paper 1AEasy1 mark

A helium-4 nucleus contains two protons and two neutrons.

Which row describes the electrostatic force and the strong nuclear force between the two neutrons?

Electrostatic forceStrong nuclear force
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Step 1Neutrons are uncharged, so there is no electrostatic force between them; the strong nuclear force acts between all nucleons at a separation of about 10−15 m and is attractive.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AOnly the protons repel each other; neutrons carry no charge, so there is no electrostatic force between them.
  • BThe strong nuclear force acts between any two nucleons — neutron–neutron as well as proton–proton and proton–neutron — not only where it must overcome a repulsion.
  • CCorrect: with no charge there is no electrostatic force, and the strong nuclear force attracts the two neutrons because they are within its short range.
  • DNeutrons are not held by an electric attraction, as electrons are held in an atom; they are uncharged, and it is the strong nuclear force that binds them.
11E-1A-85Background radiation Paper 1AEasy1 mark

A marble made of uranium glass is placed in front of a Geiger–Müller tube, and 4500 counts are recorded in 10.0 minutes. With the marble removed, 600 counts are recorded in 20.0 minutes.

What is the count rate due to the marble?

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Step 1Total count rate = 4500/600 s = 7.50 s−1; background count rate = 600/1200 s = 0.50 s−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Count rate due to the marble = 7.50 − 0.50 = 7.0 s−1.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe two counts were subtracted and divided by the total time of 30.0 minutes; the counts must first be turned into rates, each over its own counting time.
  • BThe background count of 600 was subtracted from the count of 4500 as if both had been recorded in 10.0 minutes; in 10.0 minutes the background gives only 300 counts.
  • CCorrect: each count is divided by its own time to give a rate, and the background rate is then subtracted.
  • DThis is the total count rate; the background has not been subtracted.
12E-1A-88Mass–energy equivalence Paper 1AEasy1 mark

Cobalt-60 decays by β− emission to a nickel-60 nucleus in an excited state. The nickel-60 nucleus then emits two gamma-ray photons, of energies 1.17 MeV and 1.33 MeV, as it reaches its ground state.

What is the decrease in the mass of the nickel-60 nucleus as it emits the two photons? (1 u = 931.5 MeV c−2)

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Step 1The nucleus loses 1.17 + 1.33 = 2.50 MeV of energy, so its mass falls by Δm = E/c2 = 2.50/931.5 = 2.68 × 10−3 u.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThe nucleon and proton numbers do not change, but the mass does: any loss of energy by the nucleus is a loss of mass, Δm = E/c2.
  • BThis uses only the 1.33 MeV photon; the nucleus emits both photons, so it loses 2.50 MeV.
  • CCorrect: 2.50 MeV ÷ 931.5 MeV u−1 = 2.68 × 10−3 u.
  • DThe energy has been multiplied by 931.5 instead of divided by it; a loss of a few MeV must be a tiny fraction of 1 u.
13IBP2S-038-04-01Random & spontaneous decay Paper 1AEasy1 mark

A household smoke detector contains a small sample of americium-241. During cooking, the detector warms from 20 °C to 60 °C. What happens to the half-life of the americium-241?

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Radioactive decay is spontaneous. It depends only on the instability of the nucleus and is not changed by external conditions such as temperature, pressure or chemical state. The half-life therefore stays the same. Option D is wrong because no realistic temperature change alters the half-life.✓ 1Answer C

Answer: C

Common error: Do not apply ideas from chemical reaction rates to nuclear decay. Heating speeds up chemical reactions but has no effect on the decay probability of a nucleus.

14IBP2S-038-04-02Isotopes Paper 1AEasy1 mark

Iodine-123 and iodine-131 are both used in thyroid medicine. The thyroid absorbs both of them in the same way. What is the reason for this?

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Isotopes have the same proton number. A neutral atom therefore has the same number and arrangement of electrons, and the electrons determine chemical behaviour. The body treats both isotopes as iodine. They differ in neutron number (option C is false) and in nucleon number (option A is false), and their half-lives are very different (option D is false).✓ 1Answer B

Answer: B

Exam tip: Link 'same chemistry' to 'same number of electrons', which follows from 'same number of protons'. Both steps are needed for a full explanation.

15IBP2S-038-04-04Random & spontaneous decay Paper 1AEasy1 mark

A Geiger counter is held over a granite kitchen worktop. The counts recorded in three successive one-minute intervals are 112, 98 and 105. What is the best explanation for the variation?

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Each nucleus decays randomly, so the number of decays in a fixed interval varies about a mean value. A short half-life (option B) would give a steady fall, not a rise and fall. Temperature does not affect decay (option D). Fluctuations of about ±10 in a count of about 100 are normal, so the counter need not be faulty (option C).✓ 1Answer A

Answer: A

Exam tip: Fluctuating counts are the experimental evidence for random decay. Random variation goes both up and down, whereas a short half-life gives a steady downward trend.

16IBP2S-038-04-09Changes in the nucleus Paper 1AEasy1 mark

A food-irradiation plant uses a cobalt-60 source. Which change would increase the activity of the source?

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Activity is proportional to the number of undecayed radioactive nuclei. Adding more cobalt-60 increases this number and so increases the activity. Decay is spontaneous, so chemical form (A), pressure (B) and temperature (C) have no effect.✓ 1Answer D

Answer: D

Common error: Changing the chemical compound does not change the nuclei. The cobalt-60 nuclei in cobalt chloride decay exactly as they do in the metal.

17IBP2S-038-02-01The strong nuclear force Paper 1AEasy1 mark

Which statement about the strong nuclear force is correct?

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The strong nuclear force acts between nucleons (proton–proton, neutron–neutron and proton–neutron) and is essentially independent of charge. Its range is only about 1–3 fm (10−15 m). At about 1 fm it is much stronger than the electrostatic repulsion between protons. This is why nuclei are stable. Its range is fixed and does not grow with nucleon number, so C is wrong.✓ 1Answer B

Answer: B

Common error: Neutrons feel the strong force too. In fact neutrons add attraction to a nucleus without adding any electrostatic repulsion.

18IBP2S-038-02-02Background radiation Paper 1AEasy1 mark

For a typical person living in a region of granite bedrock, which is most likely to be the largest single source of their annual background radiation dose?

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Granite contains uranium. Uranium decays through a chain that produces radon-222, a gas that seeps into buildings and is inhaled. In granite regions radon is by far the largest contribution to the background dose. Cosmic and medical sources contribute less. Nuclear power stations contribute a negligible fraction.✓ 1Answer C

Answer: C

Exam tip: Learn to sort background sources into natural ones (radon, rocks, cosmic rays, food) and artificial ones (medical, fallout, industry). Radon is usually the dominant natural source.

19IBP2S-038-02-07The strong nuclear force Paper 1AEasy1 mark

Two protons are held at a centre-to-centre separation of first 1.0 fm and then 4.0 fm. Which row gives the direction of the net force between them at each separation?

Separation 1.0 fm / Separation 4.0 fm

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At 1.0 fm the strong nuclear force is attractive and much larger than the electrostatic repulsion, so the net force is attractive. At 4.0 fm the separation is beyond the range of the strong force, which is now negligible. Only the electrostatic repulsion remains, so the net force is repulsive.✓ 1Answer C

Answer: C

Exam tip: Remember the shape of the strong force: repulsive below about 0.5 fm, strongly attractive around 1 fm, and negligible beyond about 3 fm.

20IBP2S-038-05-01Penetration & ionization Paper 1AEasy1 mark

A domestic smoke detector contains a small source of americium-241, an alpha emitter. The alpha particles ionize the air between two plates so that a small current flows. Smoke entering the chamber reduces this current and triggers the alarm.
Which property of alpha radiation makes it suitable for this use?

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The source must ionize the small volume of air between the plates strongly, so that a measurable current flows. Alpha particles are the most strongly ionizing radiation. Their range in air is only a few cm, and they cannot pass through the casing, so the device is safe to keep in a home. Alpha particles carry charge, so they are deflected by electric fields. They are emitted with discrete energies, not a continuous spectrum. Penetrating the casing would be a hazard, not an advantage.✓ 1Answer A

Answer: A

Exam tip: When you justify the choice of a source, give two links: the property of the radiation (ionizing ability or range), and why that property matters for the device.

21IBP2S-038-05-02Penetration & ionization Paper 1AEasy1 mark

A factory makes thin cardboard. A radioactive source sits on one side of the moving sheet and a detector on the other. The count rate is used to keep the thickness constant over many years of production.
Which source is most suitable?

sourcemoving sheetdetector
Diagram not to scale
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Beta particles are partly absorbed by thin cardboard. A small change in thickness therefore gives a noticeable change in count rate. A half-life of decades keeps the count rate almost constant during production, so any change in reading is caused by thickness and not by decay. Alpha particles would be almost completely stopped by the card. Gamma rays would pass through with almost no change in count rate. A 2-hour half-life would make the count rate fall continually.✓ 1Answer B

Answer: B

Common error: Choosing a source needs two checks: the type of radiation and the half-life. An answer with the right radiation but an unsuitable half-life is still wrong.

22IBP2S-038-05-03Isotopes Paper 1AEasy1 mark

In positron emission tomography (PET), carbon-11 atoms can replace carbon-12 atoms in a glucose molecule. The body then processes the molecule normally.
This is possible because carbon-11 and carbon-12 have the same

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Isotopes of an element have the same proton number. A neutral atom therefore has the same number of electrons, so the chemistry is the same. Carbon-11 has 5 neutrons and carbon-12 has 6, so their neutron numbers and nucleon numbers differ. Their binding energies also differ, which is why carbon-11 is unstable.✓ 1Answer A

Answer: A

Common error: Isotopes differ in neutron number, not proton number. The chemistry depends on the electrons, and the number of electrons is fixed by the proton number.

23IBP2S-038-03-01Beta decay & the neutrino Paper 1AEasy1 mark

Bananas contain a small amount of potassium-40. Most potassium-40 nuclei decay by beta-minus (β−) decay. Which particles are emitted from the nucleus in this decay?

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In β− decay a neutron in the nucleus changes into a proton, and an electron and an electron antineutrino are emitted. B pairs the electron with a neutrino, which is the wrong particle. C describes β+ decay. D leaves out the antineutrino, which carries away part of the energy and momentum.✓ 1Answer A

Answer: A

Common error: Match the pairs: β⁻ gives an electron + antineutrino, and β⁺ gives a positron + neutrino. Each decay has one matter particle and one antimatter particle.

24IBP2S-038-03-02Beta decay & the neutrino Paper 1AEasy1 mark

Which observation first suggested that an undetected particle is emitted in beta decay?

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A fixed amount of energy is released in each decay. If only the electron and the daughter nucleus were produced, every electron would have almost the same kinetic energy. The observed spread of energies, up to a maximum, shows that a third particle (the antineutrino) shares the energy. A is true but concerns alpha decay. B only shows that beta particles are charged. D is about the de-excitation of the daughter nucleus.✓ 1Answer C

Answer: C

Exam tip: Link 'continuous energy spectrum' to 'energy shared between electron and antineutrino'. This is the key idea that earns the mark.

25IBP2S-038-03-08Mass–energy equivalence Paper 1AEasy1 mark

Nuclear masses are often quoted in units based on the electronvolt. Which of these is a unit of mass?

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From E = mc2, m = E/c2, so an energy unit divided by c2 is a mass unit. MeV is a unit of energy. MeV c−1 is a unit of momentum. MeV c2 has no physical meaning here.✓ 1Answer C

Answer: C

Common error: Remember that 1 u ≈ 931.5 MeV c⁻². Writing '1 u = 931.5 MeV' mixes mass and energy units.

26IBP2S-038-01-01Mixtures of isotopes Paper 1AEasy1 mark

A hospital prepares a dose of technetium-99m (half-life 6.0 h) with an activity of 800 MBq at 08:00. What is the activity of the dose at 02:00 the following morning?

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From 08:00 to 02:00 is 18 h, which is 18/6.0 = 3 half-lives. The activity halves three times: 800 → 400 → 200 → 100 MBq. 267 MBq comes from dividing by 3 instead of halving three times. 200 MBq stops after two half-lives. 50 MBq uses four half-lives.✓ 1Answer B

Answer: B

Exam tip: Turn the elapsed time into a number of half-lives first. Then halve that many times. Do not divide by the number of half-lives.

27IBP2S-038-01-02The binding energy curve Paper 1AEasy1 mark

A uranium-235 nucleus undergoes fission into two medium-mass nuclei and some neutrons, and energy is released. Which statement explains the release of energy?

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Medium-mass nuclei lie near the peak of the binding energy curve. Their binding energy per nucleon is about 8.5 MeV, compared with about 7.6 MeV for uranium-235. The products are more tightly bound, so the total mass after fission is less than before. The mass difference is released as energy. Breaking nucleons apart always requires energy; it never releases it.✓ 1Answer A

Answer: A

Common error: Binding energy is the energy needed to pull a nucleus apart. It is not energy stored inside the nucleus. Energy is released when the products have a higher binding energy per nucleon.

28IBP2S-038-01-08Half-life & activity Paper 1AEasy1 mark

Two sealed sources P and Q contain the same radioactive isotope. Source P contains twice the mass of the isotope that source Q contains. Which comparison of P with Q is correct?

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Half-life is a property of the nuclide, so it does not depend on the amount present. P contains twice as many nuclei. Since each nucleus has the same probability of decaying per second, P has twice the activity (A = λN).✓ 1Answer A

Answer: A

Common error: A larger sample does not take longer to halve. Half-life is independent of the sample size, while activity is proportional to the number of nuclei.

29IBP2S-038-00-01Penetration & ionization Paper 1AEasy1 mark

A household smoke detector contains a small source of americium-241, which emits alpha particles across a short air gap between two metal plates. Why is an alpha emitter chosen for this device?

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Alpha particles cause intense ionization of the air, creating many ion pairs and a measurable current between the plates. Smoke particles absorb the ions and the current falls, which triggers the alarm. Their short range in air, and the fact that thin plastic stops them, means no radiation escapes the casing. Alpha particles have low penetration, carry a charge of +2e and travel well below the speed of light, so the other options are wrong.✓ 1Answer A

Answer: A

Common error: High ionizing ability and high penetrating ability do not go together. Alpha particles lose their energy quickly because they ionize strongly, so their range is short.

30IBP2S-038-00-02Changes in the nucleus Paper 1AEasy1 mark

Single-use surgical gloves are sealed in thick plastic packs and packed in cardboard boxes. The gloves are then sterilized by exposing the closed boxes to radiation. Which radiation is most suitable, and why?

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The radiation has to reach bacteria inside sealed, boxed packs. Alpha particles are stopped by the cardboard, and beta particles are absorbed by a few millimetres of material, so neither reaches the inner gloves. Gamma photons penetrate the whole box and ionize the material throughout it, which kills microorganisms everywhere. The gloves do not become radioactive.✓ 1Answer C

Answer: C

Exam tip: When you choose a radiation for an application, first ask whether it can actually reach the target. Penetration usually decides the choice before ionizing strength is considered.

31IBP2S-038-00-09Beta decay & the neutrino Paper 1AEasy1 mark

Which observation about beta-minus decay led physicists to propose that a third, almost undetectable particle (the antineutrino) is also emitted?

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If only the electron and the daughter nucleus were produced, conservation of energy and momentum would give every electron the same kinetic energy. The observed spread of energies up to a maximum shows that the released energy is shared with another particle, the antineutrino.✓ 1Answer B

Answer: B

Common error: Option D describes what would happen without the antineutrino. Alpha particles from one transition are monoenergetic, but beta particles are not.

32E-1A-10Beta decay & the neutrino Paper 1AMedium1 mark

Carbon-11 (116C), used in medical imaging, decays by β+ emission.

Which equation represents this decay?

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Step 1In β+ decay a proton changes into a neutron, so the nucleon number stays 11 and the proton number falls from 6 to 5 (boron).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The positron is always emitted together with a neutrino, ν; an antineutrino, ν̄, accompanies β− decay.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe proton number has been increased by 1, as happens in β− decay; in β+ decay it falls by 1.
  • BThe nuclide is right, but an antineutrino accompanies β− decay; a positron is emitted with a neutrino.
  • CThe positron is not a nucleon: the nucleon number does not change in β+ decay.
  • DCorrect: A is unchanged, Z falls by 1 (charge 6 = 5 + 1), and a neutrino is emitted with the positron.
33E-1A-11Penetration & ionization Paper 1AMedium1 mark

A factory makes paper in a continuous sheet about 0.1 mm thick. A radioactive source is placed on one side of the sheet and a detector on the other; the count rate is used to control the thickness of the sheet.

Which row gives the most suitable radiation and the reason for choosing it?

A paper sheet moves between rollers; a radioactive source below the sheet and a detector above it are connected to a control unit that adjusts the thickness.direction of motion of the sheetradioactive sourcedetectorcontrol unitpaper sheet
Diagram NOT accurately drawn
RadiationReason
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Step 1The radiation must be partly absorbed by the sheet: alpha particles are stopped completely by paper, and gamma rays pass through almost unaffected.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Beta particles are partly absorbed by paper, so a small change in thickness gives a measurable change in the count rate.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AAlpha particles are stopped completely by a sheet of paper, so the count rate would be almost zero whatever the thickness.
  • BCorrect: the fraction of beta particles absorbed depends on the thickness of the paper, so the count rate responds to small changes.
  • CGamma rays pass through paper almost unaffected, so the count rate would hardly change when the thickness changes.
  • DBeta is the right choice, but the reason is wrong: beta particles are less strongly ionizing than alpha particles.
34E-1A-12Choosing an isotope Paper 1AMedium1 mark

A radioactive tracer is injected into a patient so that the flow of blood through an organ can be followed by a detector outside the body. The measurement takes about one hour.

Which row describes the most suitable tracer?

Radiation emittedHalf-life
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Marking pointMarkNotes
Step 1The radiation must pass out of the body to the detector while causing little ionization inside it, so it must be gamma.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The half-life must be long enough to last through the one-hour measurement but short enough for the activity to become negligible soon afterwards: about 6 hours.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AAlpha particles cannot escape from the body to reach the detector, and they are strongly ionizing inside it.
  • BAfter one hour (30 half-lives) the activity would have fallen to almost nothing, long before the measurement is complete.
  • CCorrect: gamma rays leave the body and ionize weakly, and a 6-hour half-life lasts the measurement but decays away within a few days.
  • DThe patient would remain radioactive for decades after a one-hour test.
35E-1A-15The binding energy curve Paper 1AMedium1 mark

The graph shows the variation of binding energy per nucleon with nucleon number.

Which statement is correct?

Binding energy per nucleon against nucleon number: a steep rise to a maximum near iron-56 and a slow fall beyond it.050100150200250Nucleon number A0246810Binding energy per nucleon (MeV)⁴He⁵⁶Fe²³⁵U
Graph drawn to scale
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Marking pointMarkNotes
Step 1The curve rises steeply to a maximum near A = 56 and falls slowly beyond it, so nuclei near the peak are the most tightly bound.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Energy is released when the products have a greater total binding energy than the reactants, i.e. move towards the peak: fusion of very light nuclei and fission of very heavy ones both do this.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: both processes move the products towards the peak near iron, raising the binding energy per nucleon and releasing the difference.
  • BThe peak of the curve marks the most tightly bound, and therefore the most stable, nuclei.
  • CFusing nuclei heavier than iron would move down the curve and would require energy rather than release it.
  • DUranium lies well below the peak, at about 7.6 MeV per nucleon against about 8.8 MeV for iron.
36E-1A-31Decay chains Paper 1AMedium1 mark

A nucleus AZX emits one alpha particle and then two β− particles.

Which row gives the nucleon number and the proton number of the final nucleus?

Nucleon numberProton number
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Marking pointMarkNotes
Step 1The alpha particle (42He) lowers the nucleon number by 4 and the proton number by 2, giving A − 4 and Z − 2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each β− decay changes a neutron into a proton, leaving the nucleon number unchanged and raising the proton number by 1: after two, A − 4 and Z.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe β− decays have been treated as lowering the proton number; each one converts a neutron into a proton, so Z rises.
  • BCorrect: −2 from the alpha particle and +1 from each β− particle leave the proton number at Z — the final nucleus is an isotope of X.
  • CThe two β− decays have been left out of the proton-number count.
  • DThe alpha particle removes 4 nucleons, not 2; its charge number 2 has been used for the nucleon number.
37E-1A-32Mass defect & binding energy Paper 1AMedium1 mark

A nucleus AZX has mass M. The mass of a proton is mp and the mass of a neutron is mn.

What is the binding energy per nucleon of this nucleus? (c is the speed of light.)

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Marking pointMarkNotes
Step 1Mass defect Δm = Zmp + (A − Z)mn − M (separate nucleons minus nucleus), so the binding energy is Δmc2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Per nucleon: divide by the number of nucleons, A.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis is the total binding energy of the nucleus; it has not been divided by the number of nucleons.
  • BCorrect: the mass defect is Zmp + (A − Z)mn − M; multiplying by c2 gives the binding energy, and dividing by A gives the value per nucleon.
  • CA has been used as the number of neutrons; the neutron number is A − Z.
  • DThis divides by the number of protons Z instead of by the number of nucleons A.
38E-1A-33Half-life & activity Paper 1AMedium1 mark

Two radioactive sources X and Y have the same activity at time t = 0. The half-life of X is 5.0 days and the half-life of Y is 10.0 days.

What is (activity of Y)/(activity of X) at t = 20 days?

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Marking pointMarkNotes
Step 1After 20 days X has passed through 20/5.0 = 4 half-lives (activity × 1/16) and Y through 20/10.0 = 2 half-lives (activity × 1/4).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = (1/4)/(1/16) = 4.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis is the activity of X divided by that of Y — the ratio has been inverted.
  • BThis is the ratio of the half-lives, 10.0/5.0; the activities fall by factors of 24 and 22, so they differ by 22.
  • CCorrect: X falls to 1/16 and Y to 1/4 of the starting activity, so Y's activity is 4 times X's.
  • D16 is the factor by which X's activity falls; Y's activity has also fallen, by a factor of 4.
39E-1A-34Half-life & activity Paper 1AMedium1 mark

Sample P is a pure radioactive isotope. Sample Q is the same isotope but has twice the mass of P.

Which row gives the ratios for Q compared with P?

Half-life of Q ÷ half-life of PActivity of Q ÷ activity of P
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Marking pointMarkNotes
Step 1Half-life is a property of the isotope: every nucleus decays randomly and spontaneously, independently of how many others are present, so the ratio of half-lives is 1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Activity is the number of decays per second, which is proportional to the number of undecayed nuclei: twice the mass gives twice the activity.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe half-lives are equal, but the larger sample contains twice as many nuclei, so twice as many decay each second.
  • BCorrect: the half-life does not depend on the amount of the isotope, while the activity is proportional to the number of nuclei present.
  • CThis treats the half-life as if it depended on the amount present; each nucleus decays independently, so it does not.
  • DMore nuclei do not make each nucleus decay sooner, so the half-life is not shortened.
40E-1A-49Mass defect & binding energy Paper 1AMedium1 mark

The mass of a helium-4 nucleus is about 0.75 % less than the total mass of two free protons and two free neutrons.

Which statement explains this?

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Marking pointMarkNotes
Step 1When free nucleons bind to form a nucleus, energy (the binding energy) is released.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The nucleus therefore has less energy than the separated nucleons, and by E = mc2 it has less mass: the mass defect is the binding energy divided by c2.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe nucleus still contains two protons and two neutrons: nucleons are not destroyed. The mass of the whole system decreases slightly.
  • BCorrect: the binding energy is released when the nucleus forms, and the mass defect Δm equals this energy divided by c2.
  • CThe sign is reversed: energy is released, not supplied, when the nucleons bind. Supplying energy would increase the mass of the system.
  • DThe comparison is between the nucleus and free nucleons; neither contains electrons, so electrons cannot explain the difference.
41E-1A-50Changes in the nucleus Paper 1AMedium1 mark

Fluorine-18, 189F, is used in medical imaging. It decays by β+ emission.

The chart shows neutron number N against proton number Z. Each arrow starts at fluorine-18. Which arrow represents the β+ decay of fluorine-18?

Chart of neutron number N (vertical, 7 to 11) against proton number Z (horizontal, 7 to 11). Fluorine-18 is at Z = 9, N = 9. Arrow A goes to Z = 10, N = 8; arrow B to Z = 8, N = 9; arrow C to Z = 9, N = 10; arrow D to Z = 8, N = 10.¹⁸F78910117891011proton number Zneutron number NABCD
Diagram NOT accurately drawn
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Marking pointMarkNotes
Step 1In β+ decay a proton in the nucleus changes into a neutron (a positron and a neutrino are emitted).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2So Z falls by 1 (9 → 8) and N rises by 1 (9 → 10): the arrow goes one square left and one square up, to oxygen-18.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis arrow (Z + 1, N − 1) represents β− decay, in which a neutron changes into a proton.
  • BThis arrow lowers Z but leaves N unchanged, as if a proton were removed from the nucleus; the nucleon number would then fall, but in β+ decay it stays at 18.
  • CThis arrow adds a neutron without removing a proton, so the nucleon number would rise to 19; the proton that becomes the neutron must be taken away.
  • DCorrect: a proton becomes a neutron, so Z decreases by 1 and N increases by 1, and the nucleon number stays at 18.
42E-1A-76Half-life & activity Paper 1AMedium1 mark

A sealed capsule contains 2.0 g of pure polonium-210, which decays by alpha emission to stable lead-206 with a half-life of 138 days. The alpha particles are stopped inside the capsule.

Which row gives the mass of polonium-210 in the capsule and the total mass of its contents 276 days later?

Mass of polonium-210Total mass of contents
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Marking pointMarkNotes
Step 1276 days = 2 half-lives, so the polonium-210 falls to 2.0/22 = 0.50 g.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each decayed nucleus becomes a lead-206 nucleus and an alpha particle (helium), and both stay in the capsule; only about 3 × 10−5 of the decayed mass is carried away as energy, so the contents still have a mass of 2.0 g.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe polonium is right, but the decayed polonium does not vanish: its nucleons are now in lead-206 and helium, which stay in the sealed capsule.
  • BCorrect: two halvings leave 0.50 g of polonium-210, while the lead and helium produced keep the total mass at 2.0 g (to 2 s.f.).
  • CThe mass has been divided by the number of half-lives (2) instead of being halved twice (÷ 22 = 4).
  • D1.5 g is the mass of polonium-210 that has decayed, not the mass that remains.
43E-1A-80Changes in the nucleus Paper 1AMedium1 mark

Copper-64, 6429Cu, is used in medical imaging. A nucleus of copper-64 can decay either to zinc-64, 6430Zn, or to nickel-64, 6428Ni.

Which row identifies the decay that produces each nucleus?

Decay to zinc-64Decay to nickel-64
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Marking pointMarkNotes
Step 1To zinc: A is unchanged and Z rises by 1, so a neutron has changed into a proton — β− decay.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2To nickel: A is unchanged and Z falls by 1, so a proton has changed into a neutron — β+ decay.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe two decays are reversed. Emitting a positive particle lowers the charge of the nucleus, so β+ decay lowers Z; it is β− decay that raises Z.
  • BZ falls by 1 in the decay to nickel, but alpha emission lowers Z by 2 and A by 4, giving A = 60, not 64.
  • CThe unchanged nucleon number does not mean gamma emission: A is also unchanged in beta decay, and gamma emission cannot change Z from 29 to 30.
  • DCorrect: β− decay (n → p) raises Z by 1 to give zinc-64, and β+ decay (p → n) lowers Z by 1 to give nickel-64; A stays 64 in both.
44E-1A-81Beta decay & the neutrino Paper 1AMedium1 mark

Uranium-238, 23892U, decays through a series of alpha and β− decays to stable lead-206, 20682Pb. No other types of decay occur in the series.

How many antineutrinos are emitted in the series, for each uranium-238 nucleus that becomes lead-206?

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Marking pointMarkNotes
Step 1Only alpha decay changes A: (238 − 206)/4 = 8 alpha decays, which lower Z by 16 to 76.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Z must rise from 76 to 82, so there are 6 β− decays, and each β− particle is emitted with one antineutrino: 6 antineutrinos.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: 8 alpha decays take A from 238 to 206 and Z down to 76; 6 β− decays raise Z to 82, and each emits one antineutrino.
  • B8 is the number of alpha decays. Alpha decay emits no antineutrino; antineutrinos come only with β− particles.
  • C10 is the overall fall in Z (92 − 82), used as if it were the number of beta decays; the alpha decays lower Z by 16, so 6 β− decays are needed to bring it back up.
  • D14 counts one antineutrino for every decay in the series, alpha decays included.
45E-1A-83Activity & count rate Paper 1AMedium1 mark

A small gamma source emits one photon in each decay, equally in all directions. A detector with a window of area a faces the source at a distance d, where a is much smaller than d2. Every photon that reaches the window is counted. Absorption in the air and the background are negligible.

The count rate is R. What is the activity of the source?

sourcewindow ofarea adetectord
Diagram not to scale
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Marking pointMarkNotes
Step 1The photons spread over a sphere of area 4πd2, so the fraction that reaches the window is a/4πd2 and R = A × a/4πd2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Rearranging: A = 4πd2R/a.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe fraction a/4πd2 has been multiplied by R instead of divided into it; this expression is smaller than R, but the activity must be larger than the count rate.
  • Bπd2 is the area of a flat circle of radius d; the photons spread over a whole sphere of area 4πd2.
  • CCorrect: only the fraction a/4πd2 of the decays is counted, so the activity is the count rate divided by this fraction.
  • DThe distance has not been squared: the area of the sphere over which the photons spread is 4πd2.
46E-1A-86Mass defect & binding energy Paper 1AMedium1 mark

The binding energy of a carbon-12 nucleus is 92.2 MeV. The binding energy of a carbon-13 nucleus is 97.1 MeV.

What is the minimum energy needed to remove one neutron from a carbon-13 nucleus?

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Marking pointMarkNotes
Step 1The binding energy is the energy needed to separate a nucleus completely into its nucleons. Removing one neutron from carbon-13 leaves carbon-12 and a free neutron, which has no binding energy.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Energy needed = 97.1 − 92.2 = 4.9 MeV.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • A0.21 MeV is the difference between the binding energies per nucleon (7.68 − 7.47 MeV); the energy needed depends on the total binding energies.
  • BCorrect: separating carbon-13 completely needs 97.1 MeV and separating carbon-12 needs 92.2 MeV, so taking out the one extra neutron needs the difference, 4.9 MeV.
  • C7.5 MeV is the binding energy per nucleon of carbon-13, an average over all 13 nucleons; the last neutron need not be bound by the average amount.
  • DThe two binding energies have been added instead of subtracted.
47E-1A-87The binding energy curve Paper 1AMedium1 mark

The binding energy per nucleon of stable nuclei is between about 7.5 MeV and 8.8 MeV for all but the lightest nuclei.

Which graph shows how the total binding energy E of a stable nucleus varies with its nucleon number A?

Four sketch graphs, A to D, of binding energy against nucleon numberA.EA0B.EA0C.EA0D.EA0
Sketch graphs, NOT to scale
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Marking pointMarkNotes
Step 1E = A × (binding energy per nucleon), and the binding energy per nucleon is nearly constant at about 8 MeV, so E ≈ 8A: a rising, almost straight line through the origin (slightly less steep for the lightest nuclei).✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThis is the shape of the binding energy per nucleon against A, with its peak near A ≈ 60; the total binding energy keeps increasing as nucleons are added.
  • BThis assumes that, because the strong force is short-range, extra nucleons add no binding energy beyond a certain size. Each extra nucleon adds about 8 MeV, so the graph keeps rising.
  • CThis assumes that every nucleon is bound to every other nucleon, so that E ∝ A2. The strong force is short-range, so each nucleon is bound only to its neighbours and E ∝ A roughly.
  • DCorrect: with about 8 MeV per nucleon, the total binding energy is roughly proportional to A.
48IBP2S-038-04-03Random & spontaneous decay Paper 1AMedium1 mark

A technetium-99m nucleus has a half-life of 6.0 h. A single nucleus has already remained undecayed for 12 h. What is the probability that it decays during the next 6.0 h?

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Marking pointMarkNotes
Radioactive decay is random and memoryless: the probability that a nucleus decays does not depend on how long it has already existed. The probability of decaying within any one half-life is 1 − (1/2) = 0.50, whatever has happened before, so C is correct. Option B (0.25) is the probability of surviving the first 12 h, which wrongly brings past survival into the future probability. Option A (0.125) is the probability of surviving all 18 h from the start, the same mistake. Option D assumes the nucleus is 'due' to decay after surviving so long.✓ 1Answer C

Answer: C

Common error: Nuclei do not age. A nucleus that has survived for many half-lives has exactly the same chance of decaying in the next half-life as a newly formed one.

49IBP2S-038-04-05Isotopes Paper 1AMedium1 mark

A nuclide of iron has proton number 26 and nucleon number 56. Which of the following nuclides is an isotope of this iron nuclide?

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Marking pointMarkNotes
Isotopes have the same proton number but different neutron numbers. Proton number 26 with nucleon number 58 is iron with 32 neutrons instead of 30. Option A has the same nucleon number but is a different element. Option B has the same neutron number (30) but is a different element. Option D is a different element.✓ 1Answer C

Answer: C

Common error: Matching the nucleon number or the neutron number does not make two nuclides isotopes. Only the proton number must match.

50IBP2S-038-04-06Mass–energy equivalence Paper 1AMedium1 mark

A space probe is powered by plutonium-238, which has a half-life of 88 years. Its thermal power is proportional to the activity of the plutonium. What fraction of the original thermal power remains after 264 years?

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Marking pointMarkNotes
264 years is 264/88 = 3 half-lives. The activity after 3 half-lives is \((\frac{1}{2})^3 = \frac{1}{8}\) of its original value. The power is proportional to the activity, so the power also falls to \(\frac{1}{8}\) (option C).
Option A divides 1 by the number of half-lives. Option B divides 1 by 2 × 3 instead of by 23. Option D is the fraction that has decayed, not the fraction that remains.
✓ 1Answer C

Answer: C

Exam tip: After n half-lives the fraction remaining is (1/2)ⁿ. Check whether the question asks for the fraction remaining or the fraction decayed.

51IBP2S-038-04-08Beta decay & the neutrino Paper 1AMedium1 mark

Carbon-14 in a sample of ancient wood decays by β− emission to form nitrogen-14. Which row describes the change from the carbon-14 nucleus to the nitrogen-14 nucleus?

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Marking pointMarkNotes
In β− decay a neutron changes into a proton, and an electron and an antineutrino are emitted. Proton number rises by 1 (6 to 7), neutron number falls by 1 (8 to 7), and nucleon number stays 14. The product is a new element with the same nucleon number, so it is not an isotope of carbon. Option B describes β+ decay. Options C and D wrongly change the nucleon number.✓ 1Answer A

Answer: A

Exam tip: Beta decay never changes the nucleon number. It always changes the element, so the product is never an isotope of the parent.

52IBP2S-038-02-03Background radiation Paper 1AMedium1 mark

A hospital technician measures a sample containing a radioactive tracer. The measured count rate is 3.0 counts s−1, and this includes a background count rate of 0.50 counts s−1. What measured count rate is expected after two half-lives of the tracer?

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Marking pointMarkNotes
Only the count rate due to the source decays.
Corrected rate = 3.0 − 0.50 = 2.5 counts s−1.
After two half-lives the corrected rate is 2.5/4 = 0.625 counts s−1.
The detector still records the background, so the measured rate = 0.625 + 0.50 ≈ 1.1 counts s−1.
The distractors come from three mistakes: quartering the total gives A; forgetting to add the background back gives B; using only one half-life gives C.
✓ 1Answer D

Answer: D

Common error: Background does not decay. Subtract it before applying half-life, then add it back if the question asks for the reading on the detector.

53IBP2S-038-02-04The strong nuclear force Paper 1AMedium1 mark

Stable heavy nuclei, such as lead-208, contain significantly more neutrons than protons. What is the best explanation?

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Marking pointMarkNotes
The strong force is short-range, so each nucleon is attracted only by its nearest neighbours. The electrostatic force is long-range, so each proton is repelled by every other proton in the nucleus. As the proton number increases, the total repulsion grows faster than the total attraction. Extra neutrons add attraction but no repulsion. Neutrons are uncharged, so B is wrong. The strong force acts between all nucleons, so C is wrong.✓ 1Answer A

Answer: A

Exam tip: Explanations of nuclear stability score well when they contrast the two ranges: strong force is short-range (nearest neighbours), electrostatic force is long-range (all protons).

54IBP2S-038-02-05Background radiation Paper 1AMedium1 mark

Long-haul airline pilots receive a larger annual dose of background radiation than people who work at ground level. Which statement explains this?

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Marking pointMarkNotes
Cosmic radiation and the secondary particles it produces are partly absorbed by the atmosphere. At about 11 km altitude much less air lies above the aircraft, so the cosmic-ray dose rate is many times higher than at sea level. Engines are not radioactive sources, so A is wrong. Radon comes from the ground, so C is wrong. Decay rates do not depend on temperature, so D is wrong.✓ 1Answer B

Answer: B

Common error: Radioactive decay rates are not affected by temperature, pressure or chemical state.

55IBP2S-038-02-06Background radiation Paper 1AMedium1 mark

A student encloses a Geiger–Müller tube in a thick lead box, with no source present. The count rate falls but does not become zero. What is the most likely reason?

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Marking pointMarkNotes
Some components of background radiation, particularly high-energy cosmic-ray muons, are penetrating enough to pass through considerable thicknesses of lead. Trace radioactivity in the materials themselves can also contribute. Absorbing gamma radiation does not make lead radioactive, so A is wrong. Alpha particles are the easiest radiation to stop, so B is wrong. Randomness causes fluctuations about a mean, but it does not explain why the mean is non-zero, so D is wrong.✓ 1Answer C

Answer: C

Common error: Being irradiated by alpha, beta or gamma radiation does not make a material radioactive.

56IBP2S-038-02-10Decay chains Paper 1AMedium1 mark

Radon-222 (22286Rn), a major source of background radiation, decays by α emission. The product then decays by a second α emission, followed by a β− emission. What is the nuclide produced after these three decays?

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Marking pointMarkNotes
Each α decay reduces A by 4 and Z by 2. The β− decay leaves A unchanged and increases Z by 1.
A = 222 − 8 = 214
Z = 86 − 4 + 1 = 83, which is bismuth.
The distractors come from three mistakes: A ignores the β− decay; B treats β− as decreasing Z; C reduces A by only 2 per α decay.
✓ 1Answer D

Answer: D

Common error: In β⁻ decay a neutron becomes a proton, so Z goes up by 1 and A stays the same.

57IBP2S-038-05-04Mass defect & binding energy Paper 1AMedium1 mark

Nucleus X has a binding energy per nucleon of 8.8 MeV. Nucleus Y has a binding energy per nucleon of 7.6 MeV.
Which statement must be correct?

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Marking pointMarkNotes
Binding energy per nucleon is the energy needed, on average, to remove each nucleon when a nucleus is separated completely into free nucleons. A larger value therefore means more energy per nucleon is required for X. The total binding energy and the mass defect both depend on the number of nucleons as well. A very heavy nucleus such as uranium has a lower value per nucleon than iron but a much larger total. The value per nucleon alone tells us nothing certain about the number of nucleons.✓ 1Answer D

Answer: D

Common error: Keep total binding energy and binding energy per nucleon separate. Only the value per nucleon measures stability.

58IBP2S-038-05-05Mass defect & binding energy Paper 1AMedium1 mark

The mass defect of a helium-4 nucleus is 0.0304 u. Take 1 u = 931.5 MeV c−2.
What is the binding energy per nucleon of helium-4?

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Marking pointMarkNotes
Total binding energy = 0.0304 × 931.5 ≈ 28.3 MeV.
Helium-4 has 4 nucleons, so the binding energy per nucleon is 28.3 / 4 ≈ 7.1 MeV.
28 MeV is the total binding energy. 14 MeV comes from dividing by the number of protons only. 0.0076 MeV comes from dividing the mass defect by 4 without converting it to energy.
✓ 1Answer B

Answer: B

Exam tip: Divide by the nucleon number A, not by the proton number Z.

59IBP2S-038-05-06Choosing an isotope Paper 1AMedium1 mark

A hospital needs a radioactive tracer to study blood flow through a patient's lungs. The radiation will be detected by a camera outside the body. The isotopes available are:
P: alpha emitter, half-life 6 hours
Q: gamma emitter, half-life 6 hours
R: gamma emitter, half-life 30 years
S: beta emitter, half-life 8 days
Which isotope is most suitable?

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Marking pointMarkNotes
Q is the best choice. Gamma rays pass out of the body and reach the external camera, and they ionize weakly, so they cause little tissue damage. A half-life of a few hours lasts long enough for the scan, but the activity then falls quickly, so the total dose is small. R would stay in the body and keep irradiating the patient for years. Alpha radiation (P) and beta radiation (S) are absorbed inside the body, so they would not reach the camera and would deposit a large dose.✓ 1Answer B

Answer: B

Common error: A longer half-life is not 'better' for a medical tracer. The isotope only needs to last as long as the procedure.

60IBP2S-038-05-07Alpha decay & recoil Paper 1AMedium1 mark

Radioisotope power sources on deep-space probes use heat from alpha decay. Plutonium-238 (half-life 88 years) is used rather than plutonium-239 (half-life 24 000 years). Both release a similar energy per decay.
Why is plutonium-238 preferred?

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Marking pointMarkNotes
Activity is proportional to the number of nuclei divided by the half-life. A sample of plutonium-238 with about the same mass as a sample of plutonium-239 contains a similar number of nuclei. Its half-life is about 270 times shorter, so its activity, and therefore its power output, is about 270 times greater. A half-life of 88 years is still long compared with a mission lasting a decade or two, so the power falls only slowly. The energy per decay is similar for the two isotopes, and both are alpha emitters.✓ 1Answer D

Answer: D

Exam tip: For a power source, the half-life must be long compared with the mission but short enough to give a useful activity per kilogram.

61IBP2S-038-05-08Mass defect & binding energy Paper 1AMedium1 mark

Radium-226 decays by alpha emission to radon-222, and energy is released.
Which statement is correct?

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Marking pointMarkNotes
Energy is released because the products are more tightly bound overall. Their total binding energy is greater than that of the parent nucleus, and the difference is the energy released. Because energy is released, the total mass of the products is less than the mass of the parent, not greater. Helium-4 has about 7.1 MeV per nucleon, which is less than the value for radon (about 7.7 MeV). The energy released is only a few MeV, which is much smaller than helium's total binding energy of about 28 MeV.✓ 1Answer B

Answer: B

Common error: Binding energy is not stored energy that is 'let out'. It is the energy that would be needed to pull the nucleus apart. Energy is released when the total binding energy increases.

62IBP2S-038-05-10Mass–energy equivalence Paper 1AMedium1 mark

The mass defect of an oxygen-16 nucleus is 0.137 u. Take 1 u = 1.66 × 10−27 kg and c = 3.00 × 108 m s−1.
What is the binding energy of the nucleus?

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Marking pointMarkNotes
Δm = 0.137 × 1.66 × 10−27 = 2.27 × 10−28 kg.
E = Δmc2 = 2.27 × 10−28 × 9.00 × 1016 ≈ 2.0 × 10−11 J, which is about 128 MeV.
6.8 × 10−20 J comes from multiplying by c instead of c2. 2.0 × 10−17 J treats 128 MeV as 128 eV. 1.3 × 102 J writes the MeV value as if it were in joules.
✓ 1Answer C

Answer: C

Common error: 1 MeV = 1.60 × 10⁻¹³ J, not 1.60 × 10⁻¹⁹ J. Do not drop the 'mega'.

63IBP2S-038-03-03Changes in the nucleus Paper 1AMedium1 mark

Fluorine-18, 189F, is used in PET medical scanners. It decays by beta-plus (β+) decay. What is the daughter nucleus?

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Marking pointMarkNotes
In β+ decay a proton changes into a neutron. The proton number falls by 1 (9 → 8) and the nucleon number stays at 18, so the daughter is 188O. A applies the β− rule. C wrongly assumes a nucleon is lost. D corresponds to alpha emission.✓ 1Answer B

Answer: B

Common error: Beta decay never changes the nucleon number. Only the proton number changes, by ±1.

64IBP2S-038-03-04Mass–energy equivalence Paper 1AMedium1 mark

Nuclear fusion in the Sun converts about 4.0×109 kg of mass into energy every second. What is the approximate power output of the Sun?
(c = 3.0×108 m s−1)

Show mark scheme
Marking pointMarkNotes
P = (Δm/Δt)c2 = 4.0×109 × (3.0×108)2 = 3.6×1026 W. A uses c instead of c2. B wrongly includes a factor of ½, as if this were kinetic energy. D divides by c2 instead of multiplying.✓ 1Answer C

Answer: C

Exam tip: E = mc² has no factor of ½. A rate of mass loss in kg s⁻¹ multiplied by c² gives a power in W.

65IBP2S-038-03-05Mass defect & binding energy Paper 1AMedium1 mark

Which statement about a stable nucleus is correct?

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Marking pointMarkNotes
When nucleons bind together, energy is released, so the nucleus has less mass than its separated parts (the mass defect). To pull the nucleons apart again, an energy equal to the binding energy (Δm c2) must be supplied. A reverses the sign of the mass defect. C reverses the direction of energy flow: energy is released, not supplied, when the nucleus forms. D confuses the mass defect with the electron mass.✓ 1Answer B

Answer: B

Common error: Binding energy is not energy 'stored in' the nucleus. It is the energy needed to take the nucleus apart.

66IBP2S-038-03-06Mass defect & binding energy Paper 1AMedium1 mark

Self-luminous exit signs contain tritium, which decays by β− decay to helium-3. The energy released in each decay is 18.6 keV. Which statement about the emitted electrons is correct?

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Marking pointMarkNotes
The 18.6 keV is shared between the electron and the antineutrino in different proportions in each decay. The recoiling nucleus takes only a negligible share. Electron energies therefore range from near zero up to almost 18.6 keV. A ignores the antineutrino. C violates energy conservation. D wrongly gives the heavy nucleus a large share of the energy.✓ 1Answer B

Answer: B

Exam tip: The decay energy sets the maximum kinetic energy of the beta particle, not its typical value.

67IBP2S-038-03-09Decay chains Paper 1AMedium1 mark

In the thorium decay series, lead-212 (21282Pb) undergoes two successive β− decays. What is the nuclide produced after the second decay?

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Marking pointMarkNotes
Each β− decay raises the proton number by 1 and leaves the nucleon number unchanged: Z = 82 → 83 → 84, A = 212. The product is 21284Po. B lowers Z, which would be β+ decay. C wrongly removes one nucleon per decay. D applies the alpha-decay rule twice.✓ 1Answer A

Answer: A

Exam tip: Track Z and A separately, one decay at a time. In β⁻ decay, A stays the same and Z goes up by 1.

68IBP2S-038-03-10Mass–energy equivalence Paper 1AMedium1 mark

A mobile phone battery stores an extra 40 kJ of chemical energy when it is fully charged. Which statement about the mass of the battery is correct?
(c = 3.0×108 m s−1)

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Marking pointMarkNotes
Any increase in the energy of a system increases its mass by Δm = ΔE/c2 = 4.0×104/(9.0×1016) ≈ 4.4×10−13 kg. This is far too small to measure. A wrongly restricts E = mc2 to nuclear physics. B has the direction of the change wrong. C divides by c instead of c2.✓ 1Answer D

Answer: D

Common error: E = mc² applies to every energy change. In chemical changes the mass change is simply negligible.

69IBP2S-038-01-04Mass–energy equivalence Paper 1AMedium1 mark

A deep-space probe is powered by plutonium-238, which has a half-life of 88 years. Assume that the electrical power output is proportional to the activity of the plutonium. The probe can operate only while the output is at least 25% of its initial value. What is the maximum operating time?

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Marking pointMarkNotes
25% is one quarter of the initial output. Going from 100% to 50% to 25% takes two half-lives, so the time is 2 × 88 = 176 years. 66 years assumes a linear decrease, with 75% lost in 0.75 of a half-life. 44 years comes from halving the half-life. 264 years uses three half-lives, which would give 12.5%.✓ 1Answer C

Answer: C

Common error: Radioactive decay is exponential, not linear. Losing 75% of the activity takes two half-lives, not three-quarters of one.

70IBP2S-038-01-05Mixtures of isotopes Paper 1AMedium1 mark

A detector placed near a radioactive source records 260 counts per minute. The background count rate is 20 counts per minute. Two hours later the detector records 80 counts per minute. What is the half-life of the source?

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Marking pointMarkNotes
First subtract the background from both readings. The corrected count rates are 260 − 20 = 240 and 80 − 20 = 60 counts per minute. 240 → 60 is a factor of 4, which is two half-lives. So the half-life is 120 min / 2 = 60 min. 71 min comes from ignoring the background (260/80 = 3.25). 30 min divides 120 min by 4. 120 min assumes only one halving.✓ 1Answer B

Answer: B

Exam tip: Subtract the background from every reading before comparing count rates. The background count does not decay.

71IBP2S-038-01-06Mass defect & binding energy Paper 1AMedium1 mark

Which statement about the nuclear binding energy of a stable nucleus is correct?

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Marking pointMarkNotes
When nucleons bind together, energy is released, so the nucleus has less mass than its separate nucleons. This mass defect Δm corresponds to the binding energy, E = Δmc2. The binding energy is the work needed to separate the nucleus completely into its nucleons. The other options reverse the direction of the mass change or the energy flow, or confuse binding energy with decay energy.✓ 1Answer C

Answer: C

Common error: A nucleus has LESS mass than its separated parts. To pull it apart, energy must be supplied, and that energy reappears as the extra mass of the separated nucleons.

72IBP2S-038-01-09The binding energy curve Paper 1AMedium1 mark

In a proposed fusion reactor, two deuterium nuclei (\(^2_1\)H) fuse to form a helium-4 nucleus. The binding energy per nucleon is 1.1 MeV for \(^2_1\)H and 7.1 MeV for \(^4_2\)He. What is the energy released in one fusion event?

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Marking pointMarkNotes
Total binding energy of He-4 = 4 × 7.1 = 28.4 MeV. Total binding energy of two deuterons = 2 × (2 × 1.1) = 4.4 MeV. Energy released = 28.4 − 4.4 = 24 MeV. 6.0 MeV is only the difference per nucleon. 26 MeV counts 1.1 MeV per deuteron instead of per nucleon. 12 MeV is the energy per deuteron.✓ 1Answer C

Answer: C

Exam tip: Multiply the binding energy per nucleon by the nucleon number to get each total binding energy. The energy released is (total BE of products) − (total BE of reactants).

73IBP2S-038-00-03Decay chains Paper 1AMedium1 mark

A nucleus X undergoes one alpha decay. The daughter then undergoes two successive beta-minus decays. Which describes the final nucleus compared with X?

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Marking pointMarkNotes
The alpha decay changes Z by −2 and A by −4. Each beta-minus decay changes Z by +1 and leaves A unchanged, so the two together change Z by +2. The net change is ΔZ = 0 and ΔA = −4. The final nucleus is therefore the same element as X with four fewer nucleons, which makes it an isotope of X.✓ 1Answer B

Answer: B

Common error: Beta-minus decay increases the proton number, because a neutron becomes a proton. Do not treat it as a loss of charge from the nucleus.

74IBP2S-038-00-04Changes in the nucleus Paper 1AMedium1 mark

Fluorine-18 (proton number 9) is used as a tracer in positron emission tomography (PET) scans. It decays by β+ emission. Which statement about this decay is correct?

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Marking pointMarkNotes
In β+ decay a proton changes into a neutron, so Z falls from 9 to 8 (oxygen) and A stays at 18. The positron is an antilepton (lepton number −1), so a neutrino (lepton number +1) must also be emitted to conserve lepton number. An antineutrino goes with β− decay, not β+ decay.✓ 1Answer C

Answer: C

Exam tip: Pairing rule: β⁻ decay emits an electron with an antineutrino, and β⁺ decay emits a positron with a neutrino. Lepton number must total zero on the right-hand side.

75IBP2S-038-00-05Changes in the nucleus Paper 1AMedium1 mark

Technetium-99m, used in medical imaging, is an excited state of technetium-99. It decays to the ground state of technetium-99 by emitting a gamma photon. Which statement about this change is correct?

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Marking pointMarkNotes
A gamma photon has no charge and no rest mass and is not a nucleon, so neither Z nor A changes. The nucleus loses energy as it drops from an excited state to a lower one. Because it loses energy E, its mass decreases slightly, by E/c2. For this reason the option claiming unchanged mass is wrong, even though A is unchanged.✓ 1Answer B

Answer: B

Common error: An unchanged nucleon number does not mean an unchanged mass. Any release of energy reduces the mass of the nucleus by ΔE/c².

76IBP2S-038-00-06Choosing an isotope Paper 1AMedium1 mark

During maintenance work on a space station, an astronaut accidentally inhales dust containing an alpha-emitting nuclide. Why is the source more hazardous inside the body than the same source held a short distance outside it?

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Marking pointMarkNotes
Alpha particles have a range of only a few centimetres in air and are stopped by the dead surface layer of skin, so an external source does little harm. Once the source is inside the body, the strongly ionizing alpha particles transfer all their energy to living cells within a fraction of a millimetre. This concentrates the damage, for example to cell DNA. Activity is a property of the nuclide and is not affected by its surroundings.✓ 1Answer D

Answer: D

Exam tip: When you compare hazards, separate external sources (where penetration matters most) from internal sources (where ionization density matters most). Alpha emitters are the least hazardous outside the body and the most hazardous inside it.

77IBP2S-038-00-10Beta decay & the neutrino Paper 1AMedium1 mark

Carbon-11 (6 protons, 5 neutrons) is produced in a hospital cyclotron. The stable isotopes of carbon are carbon-12 and carbon-13. By which process is carbon-11 most likely to decay?

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Marking pointMarkNotes
Carbon-11 has fewer neutrons than the stable carbon isotopes, so it is proton-rich. β+ decay turns a proton into a neutron, giving boron-11 with 5 protons and 6 neutrons, which is closer to stability. β− decay would make the nucleus even more proton-rich. Alpha decay is typical of very heavy nuclei. Gamma emission alone cannot change the neutron–proton ratio.✓ 1Answer D

Answer: D

Exam tip: Neutron-rich nuclei tend to decay by β⁻ and proton-rich nuclei by β⁺. Compare the isotope with the stable isotopes of the same element to decide which applies.

78E-1A-13Background radiation Paper 1AHard1 mark

A detector placed near a radioactive source records a count rate C. With the source removed, the background count rate is B.

What count rate does the detector record, with the source in the same place, after three half-lives of the source?

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Marking pointMarkNotes
Step 1Only the part of the reading due to the source decays: C − B falls to (C − B)/23 = (C − B)/8 after three half-lives.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The background does not decay and is still recorded, so the reading is (C − B)/8 + B.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe source contribution has been divided by 3, the number of half-lives, instead of by 23 = 8.
  • BThe whole reading has been divided by 8, but the background does not decay.
  • CThis is the count rate due to the source alone; the detector still records the background as well.
  • DCorrect: the source part (C − B) halves three times to (C − B)/8, and the unchanged background B is added back.
79E-1A-14Random & spontaneous decay Paper 1AHard1 mark

Radioactive decay is described as a random and spontaneous process.

Which statement describes what is meant by spontaneous?

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Marking pointMarkNotes
Step 1'Random' means that the moment at which a particular nucleus decays cannot be predicted; 'spontaneous' means that decay needs no external cause and cannot be changed by external conditions such as temperature, pressure or chemical combination.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AThis is true of radioactive decay, but it describes its random nature, not what spontaneous means.
  • BCorrect: a spontaneous process happens without any external trigger, and its rate cannot be altered by changing the physical or chemical conditions.
  • CHalf-life is a statistical property of a large number of nuclei; individual nuclei decay at unpredictable times, some long before and some long after one half-life.
  • DDecay needs no trigger; a nucleus that splits only when struck by a neutron is undergoing induced fission, not spontaneous decay.
80E-1A-35Radioactive dating Paper 1AHard1 mark

When a rock solidified it contained atoms of a radioactive isotope X but none of Y, the stable nuclide into which X decays directly. The age of the rock is now three half-lives of X.

What is now (number of Y atoms)/(number of X atoms)?

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Marking pointMarkNotes
Step 1After three half-lives the fraction of X remaining is (1/2)3 = 1/8, so 7/8 of the original X atoms have become Y.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2NY/NX = (7/8)/(1/8) = 7.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A1/8 is the fraction of the original X that remains, not the ratio of Y to X.
  • BThis is NX/NY, the ratio the wrong way round.
  • CCorrect: 1/8 of X remains and 7/8 has become Y, so there are 7 atoms of Y for every atom of X.
  • D8 compares the original number of X atoms with the number remaining; the atoms that have decayed number 7/8, not 8/8, of the original.
81E-1A-79Random & spontaneous decay Paper 1AHard1 mark

Iodine-131 has a half-life of 8.0 days. One particular nucleus in a sample of iodine-131 has not decayed after 16 days.

What is the probability that this nucleus decays during the next 8.0 days?

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Marking pointMarkNotes
Step 1Decay is random: every undecayed nucleus has the same, constant probability of decaying in a given time, whatever its age.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Half of the nuclei present at any moment decay in the next half-life, so the probability is 1/2.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • A1/4 = (1/2)2 is the probability, judged at t = 0, of surviving the first 16 days. That has already happened; it does not change the chance of decaying from now on.
  • BCorrect: a nucleus does not 'age', so its probability of decaying in any interval of one half-life is always 1/2.
  • C3/4 is the fraction of the original sample that has decayed after 16 days; the nucleus is not more likely to decay because others have done so.
  • D7/8 is the fraction of the original sample that will have decayed after 24 days; this treats the nucleus as 'overdue' to decay.
82E-1A-82Penetration & ionization Paper 1AHard1 mark

An alpha particle and a β− particle have the same initial kinetic energy. Each is brought to rest in air. The average energy needed to produce one ion pair in air is the same for both particles.

Which row compares the alpha particle with the β− particle?

Range in air of alpha particleTotal number of ion pairs produced by alpha particle
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Marking pointMarkNotes
Step 1The alpha particle ionizes far more strongly — many more ion pairs per millimetre — so it loses its energy in a much shorter distance.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each particle loses all of its kinetic energy in producing ion pairs, so the total number is Ek ÷ (energy per ion pair): the same for both.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the alpha particle produces many more ion pairs per millimetre and stops sooner, but the total number of ion pairs is set by the energy it has to give away, which is the same.
  • BThis confuses ionization per unit length with total ionization. The alpha particle ionizes more strongly along its path, but its path is correspondingly shorter.
  • CThe alpha particle is the more massive, but it is the more strongly ionizing particle, so it loses its energy faster and stops in a much shorter distance.
  • DA short track does not mean fewer ions in total: the ion pairs are packed more densely along the alpha track, and the energy available for ionization is the same.
83E-1A-84Half-life & activity Paper 1AHard1 mark

A radioactive sample is placed near a detector. The graph shows how the count rate C recorded by the detector varies with time t. The values of C include a background count rate of 40 min−1.

What is the half-life of the radioactive sample?

Total count rate C, including background, against time t020406080100t / min04080120160200240C / min⁻¹
Graph drawn to scale
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Marking pointMarkNotes
Step 1At t = 0 the count rate due to the sample is 200 − 40 = 160 min−1; it has halved when it is 80 min−1, i.e. when the graph reads 80 + 40 = 120 min−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The graph reads 120 min−1 at t = 20 min.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: subtract the background first (160 min−1), halve it (80 min−1), add the background back (120 min−1) and read the time from the graph.
  • BThe background has not been subtracted: the total reading halves (200 → 100 min−1) at about 28 min, but the background does not decay.
  • CThe sample's count rate was halved to 80 min−1, but the graph was then read at 80 min−1, forgetting that the plotted values include the background; this is two half-lives.
  • DThe background was subtracted from half the initial total (100 − 40 = 60 min−1) and the graph read at 60 min−1; this is three half-lives.
84IBP2S-038-04-10Random & spontaneous decay Paper 1AHard1 mark

A detector in a research lab contains exactly 4 undecayed nuclei of the same radioactive isotope. What is the probability that exactly 2 nuclei are still undecayed after one half-life?

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Marking pointMarkNotes
Each nucleus independently has a probability of 1/2 of surviving one half-life. Any one particular pattern of survivors and decays among the 4 nuclei therefore has probability (1/2)4 = 1/16. There are 6 different ways to choose which 2 nuclei survive (AB, AC, AD, BC, BD, CD). The probability is 6 × 1/16 = 6/16 = 3/8, so B is correct. Half-life predicts exact halving only for very large numbers of nuclei. Option D treats decay as deterministic. Option C confuses the survival probability of one nucleus with the probability of this outcome. Option A is the probability of one particular arrangement only.✓ 1Answer B

Answer: B

Exam tip: Half-life is a statistical quantity. For a handful of nuclei, you can state probabilities but you cannot predict exact numbers.

85IBP2S-038-02-09Half-life & activity Paper 1AHard1 mark

A radioactive sample gives an initial measured count rate of 165 counts min−1. The background count rate is 5 counts min−1. After how many half-lives will the measured count rate be twice the background count rate?

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Marking pointMarkNotes
Corrected initial rate = 165 − 5 = 160 counts min−1.
Twice the background is a measured rate of 10 counts min−1, so the corrected rate is 10 − 5 = 5 counts min−1.
Reduction factor = 160/5 = 32 = 25, so 5 half-lives.
The distractors come from three mistakes: ignoring background (165 → about 10) gives about 4 half-lives; 16 and 32 are reduction factors, not numbers of half-lives.
✓ 1Answer B

Answer: B

Exam tip: Convert both the initial and the final measured rates into corrected rates before counting half-lives.

86IBP2S-038-05-09Mass defect & binding energy Paper 1AHard1 mark

Polonium-210 decays by alpha emission to lead-206. The binding energies per nucleon are:
polonium-210: 7.834 MeV
lead-206: 7.875 MeV
helium-4: 7.074 MeV
What is the energy released in one decay?

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Marking pointMarkNotes
Energy released = (total binding energy of products) − (total binding energy of parent).
= (206 × 7.875 + 4 × 7.074) − 210 × 7.834
= (1622.25 + 28.30) − 1645.14
≈ 5.4 MeV.
0.041 MeV subtracts the values per nucleon without multiplying by the nucleon numbers. 7.1 MeV adds and subtracts the values per nucleon directly. 23 MeV leaves out the alpha particle's binding energy.
✓ 1Answer B

Answer: B

Exam tip: Always convert to total binding energies (multiply by A) before comparing the products with the parent.

87IBP2S-038-03-07Alpha decay & recoil Paper 1AHard1 mark

A nucleus at rest undergoes β− decay inside a particle detector. The tracks of the emitted electron and the recoiling daughter nucleus are both recorded. Which observation provides evidence that a third, undetected particle is also emitted?

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Marking pointMarkNotes
The initial momentum is zero. If only two particles were produced, their momenta would have to be equal and opposite, so the tracks would lie along one line. Tracks that are not collinear mean that a third particle carries momentum. B is explained by the electron being lighter and faster. C only shows that the particles have opposite charges. D shows charge conservation, which a neutral third particle would not change.✓ 1Answer A

Answer: A

Exam tip: Momentum conservation, as well as energy conservation, can be used to argue for the neutrino. Two products from a decay at rest must recoil back-to-back.

88IBP2S-038-01-10The binding energy curve Paper 1AHard1 mark

The binding energy per nucleon is about 8.8 MeV for \(^{56}\)Fe and about 7.6 MeV for \(^{238}\)U. Which statement is correct?

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Marking pointMarkNotes
Total binding energy of U-238 ≈ 238 × 7.6 ≈ 1810 MeV. Total binding energy of Fe-56 ≈ 56 × 8.8 ≈ 490 MeV. U-238 therefore has the larger total binding energy, and so also the larger mass defect. Stability is judged by binding energy per nucleon, which is higher for Fe. Fe-56 is near the peak of the curve, so splitting it would lower the binding energy per nucleon and would absorb energy.✓ 1Answer B

Answer: B

Common error: Stability is indicated by binding energy PER NUCLEON, not by total binding energy. Heavy nuclei have large total binding energies simply because they contain many nucleons.

89IBP2S-038-00-07Penetration & ionization Paper 1AHard1 mark

Narrow beams of alpha particles, beta-minus particles and gamma radiation enter the same region of uniform magnetic field, all moving at right angles to the field. Each alpha particle has a kinetic energy of about 5 MeV and each beta particle about 1 MeV. Which describes the paths in the field?

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Marking pointMarkNotes
The charges have opposite signs, so the magnetic forces act in opposite directions. Gamma photons are uncharged and are not deflected. The radius of the path is r = p/(qB). An alpha particle has a momentum of about 1×10−19 kg m s−1, more than 100 times that of the beta particle. Its charge is only twice as large, so its radius is tens of times larger. The beta path is therefore far more curved.✓ 1Answer A

Answer: A

Common error: Curvature depends on charge and momentum. The very small mass of the electron gives it a much smaller momentum, so the beta path curves far more tightly even though its charge is smaller.

90IBP2S-038-00-08Changes in the nucleus Paper 1AHard1 mark

Neptunium-237 (Z = 93) decays through a chain of alpha and beta-minus decays to the stable nuclide bismuth-209 (Z = 83). How many alpha decays and beta-minus decays occur in the chain?

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Marking pointMarkNotes
Only alpha decay changes A. The fall in A is 237 − 209 = 28, so there are 28/4 = 7 alpha decays. These seven alpha decays reduce Z by 14, from 93 to 79. Bismuth has Z = 83, so Z must then rise by 4. Each beta-minus decay raises Z by 1, so there are 4 beta-minus decays.✓ 1Answer C

Answer: C

Exam tip: Always use the nucleon number first to fix the number of alpha decays. Then use the proton number to find how many beta decays are needed and of which type.

91E-1B-03Half-life & activity Paper 1BEasy9 marks

In a hospital radiopharmacy, the activity A of a vial of a radioactive tracer is measured in a dose calibrator every 2.0 h. The label naming the isotope has been lost. The manufacturer states that each reading of the calibrator has an uncertainty of ±3 %. The readings are shown in the graph with their error bars.

The table gives the half-lives of five isotopes used in hospitals, taken from a nuclear data library.

IsotopeHalf-life
fluorine-181.83 h
technetium-99m6.01 h
iodine-12313.2 h
indium-11167.3 h
gallium-6778.3 h
Activity A of the tracer against time t, points with ±3 % error bars02468101214161820t / h0100200300400500600700800900A / MBq
Graph drawn to scale
(a)

The reading at t = 0 is 825 MBq. Calculate its absolute uncertainty.

(1)
(b)

Draw the curve of best fit for the data.

(1)
(c)

Determine the half-life of the tracer, using your curve.

(2)
(d)

Identify the isotope, using the table.

(1)
(e)

Using a whole number of half-lives, predict the activity of the vial at t = 30 h.

(2)
(f)

With no vial inside it, the calibrator reads 4 MBq. State the type of error this causes, and explain its effect on your answer to (c).

(2)
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Marking pointMarkNotes
Part (a)
0.03 × 825 = ±25 MBq✓ 1Accept ±24.8 MBq, or ±20 MBq (1 s.f.).
Part (b)
Smooth curve of decreasing gradient passing through all the error bars✓ 1Not dot-to-dot.
Part (c)
Reads a halving from the curve, e.g. 825 MBq → 412 MBq at t ≈ 6.1 h, and a second halving to 206 MBq at t ≈ 12 h✓ 1
Half-life ≈ 6.1 h✓ 1Accept 5.6–6.4 h. Award [1] max for a single reading taken from the table rather than the curve.
Part (d)
Technetium-99m, whose half-life (6.01 h) matches the result of (c)✓ 1Allow ECF from (c).
Part (e)
30 h is 5 half-lives, so A falls by 25 = 32✓ 1Allow ECF from (c) if a whole number of half-lives results.
A = 825/32 = 26 MBq✓ 1Accept 25.8 MBq.
Part (f)
Systematic (zero-offset) error: every reading is 4 MBq too high✓ 1
The late, small readings are affected proportionally more, so the activity appears to fall more slowly and the half-life found is slightly too large✓ 1Accept "half-life overestimated" with this reasoning. OWTTE

Answers: (a) ±25 MBq  ·  (c) ≈ 6.1 h  ·  (d) technetium-99m  ·  (e) 26 MBq (the remaining parts are explanations — see the table above)

92E-1B-10Half-life & activity Paper 1BEasy8 marks

A class models radioactive decay with small cubes, each with one face painted red. The cubes are shaken and thrown; every cube that lands red face up is said to have "decayed" and is removed, and the rest are counted and thrown again. Five groups each start with 500 cubes. The table gives, after selected throws, the mean number of cubes remaining and its uncertainty (half the range of the five groups' results). The graph shows these data, except for throw 4.

For this model the expected half-life is 3.8 throws.

Throw number nMean number remainingUncertainty
05000
23449
4
617010
81148
10827
12556
14405
Mean number of cubes remaining against throw number, points with error bars02468101214throw number n0100200300400500mean number remaining
Graph drawn to scale
(a)

State how the model shows that radioactive decay is random.

(1)
(b)

After throw 4 the five groups had 243, 229, 251, 236, 247 cubes remaining. Calculate the mean number remaining and its uncertainty.

(2)
(c)

Plot your result from (b) on the graph and draw the curve of best fit.

(1)
(d)

Determine the half-life of the model, in throws, using two successive halvings.

(2)
(e)

Comment on your answer to (d).

(1)
(f)

Explain why the uncertainty is a larger fraction of the mean after 12 throws than after 2 throws.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
It is impossible to predict which cube will land red face up on a throw / the groups' results differ even under identical conditions✓ 1OWTTE
Part (b)
Mean = 1206/5 = 241.2 ≈ 241✓ 1
Uncertainty = (251 − 229)/2 = ±11✓ 1Accept (241 ± 11).
Part (c)
Point plotted correctly and a smooth curve of decreasing gradient through all the error bars✓ 1Do not accept a dot-to-dot line.
Part (d)
500 → 250 at n ≈ 3.8, and 250 → 125 at n ≈ 7.7✓ 1
Half-life ≈ 3.8 throws✓ 1Accept 3.5–4.1 throws.
Part (e)
The result agrees with the expected 3.8 throws (within the precision of reading the graph)✓ 1Allow ECF from (d).
Part (f)
Fewer cubes remain after 12 throws, so the random variation in the number decaying is a larger fraction of the number remaining✓ 1Accept: fractional fluctuation is larger for small numbers (∝ 1/√N). OWTTE

Answers: (b) 241 ± 11  ·  (d) ≈ 3.8 throws (the remaining parts are explanations — see the table above)

93E-1B-11Background radiation Paper 1BEasy10 marks

Two students test the hypothesis that the background count rate is higher in room A, which has granite walls, than in room B, which has brick walls. A GM tube and counter, with no source nearby, are used to record ten separate one-minute counts in each room.

For room B the result is (22.5 ± 6.0) counts per minute. The uncertainty in a single count N is ±√N.

Room12345678910
A (granite walls)28353140263337293438
B (brick walls)24192721162522282023
(a)

State one reason why the background in a granite building may be higher.

(1)
(b)

Calculate the mean count per minute in room A and its absolute uncertainty.

(2)
(c)

Discuss whether the data support the hypothesis.

(2)
(d)

Suggest how the procedure could be improved to test the hypothesis more convincingly.

(1)
(e)

The students then record a single count of 318 in 10.0 minutes in room A. Calculate the background count rate, in counts per minute, and its percentage uncertainty.

(2)
(f)

A source placed near the GM tube in room A gives 1452 counts in one minute. Using your answer to (e), determine the count rate due to the source and its absolute uncertainty.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Granite contains radioactive isotopes (e.g. of uranium, thorium or potassium) / releases radon gas✓ 1OWTTE
Part (b)
Mean = 331/10 = 33.1 ≈ 33 counts per minute✓ 1
Uncertainty = (40 − 26)/2 = ±7, so (33 ± 7) counts per minute✓ 1Accept (33.1 ± 7.0).
Part (c)
The mean for A (33.1) is higher than for B (22.5)✓ 1
but the ranges overlap (26.1–40.1 and 16.5–28.5), so the hypothesis is not conclusively supported✓ 1MP2 requires reference to the overlap of the uncertainties. Allow ECF from (b).
Part (d)
Count for much longer in each room (e.g. several 10-minute counts) so that the percentage uncertainty is smaller✓ 1Accept: take many more one-minute counts. Do not accept "repeat" alone.
Part (e)
Rate = 318/10.0 = 31.8 counts per minute✓ 1
Percentage uncertainty = √318/318 × 100 = 5.6 %✓ 1Accept 5.6 % or 6 %; compare ≈ 17 % for a single one-minute count.
Part (f)
1452 − 31.8 = 1420 counts per minute✓ 1
Uncertainty = √1452 + √318/10 = 38.1 + 1.8 ≈ ±40, so (1420 ± 40) counts per minute✓ 1Absolute uncertainties add when subtracting. Accept ±38 (background uncertainty ignored).

Answers: (b) (33 ± 7) counts per minute  ·  (e) 31.8 counts per minute, 5.6 %  ·  (f) (1420 ± 40) counts per minute (the remaining parts are explanations — see the table above)

94E-1B-28Background radiation Paper 1BEasy10 marks

Students launch a weather balloon carrying a Geiger–Müller (GM) counter and a data logger. The logger records the number of counts in each one-minute interval and the altitude h of the balloon. No radioactive source is carried. The table and the graph show selected readings; the uncertainty in a number of counts N is ±√N.

h / kmCounts per minute
027
245
479
6174
8329
10690
12941
151196
181218
211207
241072
271009
30893
Graph of counts per minute against altitude h / km from 0 to 30 km, with error bars; points only. Vertical axis 0 to 1400.048121620242832h / km0200400600800100012001400counts per minute
Graph drawn to scale
(a)

State one source, other than cosmic rays, of the counts recorded at ground level.

(1)
(b)

Calculate the percentage uncertainty in the count recorded at ground level.

(1)
(c)

Draw the curve of best fit for the data.

(1)
(d)

Determine, using your curve, the altitude at which the count rate is greatest, and its uncertainty.

(2)
(e)

A student claims that, between h = 2 km and h = 10 km, the count rate doubles for every 2 km increase in altitude. Evaluate this claim, referring to the uncertainties.

(2)
(f)

After each count, a GM tube cannot record another count for a short time (its dead time). Explain the effect of this on the maximum count rate measured.

(2)
(g)

Suggest one improvement to the experiment that would reduce the uncertainty in the altitude found in (d).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any one of: radioactive isotopes in rocks, soil or building materials / radon gas in the air / potassium-40 in the ground «or in the apparatus»✓ 1Do not accept 'the Sun' or 'cosmic rays'.
Part (b)
√27/27 × 100 = 19 %✓ 1Accept 19 %, 19.2 % or 20 %.
Part (c)
Smooth curve rising with increasing gradient at low altitude, reaching a single broad maximum and then falling slowly, passing through all the error bars✓ 1Do not accept a dot-to-dot line.
Part (d)
Maximum of the curve at h ≈ 18–19 km✓ 1Accept 16–21 km from the candidate's curve.
Uncertainty about ±2 km «the curve is nearly flat between about 15 km and 21 km»✓ 1Accept ±1 km to ±4 km with a sensible method.
Part (e)
Four doublings from 2 km to 10 km predict 16 × 45 = 720 ± 107 counts per minute «uncertainty 16 × √45»; the measured value is 690 ± 26✓ 1Accept successive ratios 1.76, 2.20, 1.89, 2.10, all close to 2 within their uncertainties.
The ranges overlap, so the data support the claim «within the uncertainties»✓ 1MP2 needs a comparison with the uncertainties. Allow a consistent conclusion.
Part (f)
Some particles arrive during the dead time and are not counted, so the measured count rate is lower than the true rate «a systematic error»✓ 1
The loss is greatest at the highest count rates, so the measured maximum count rate is too low «and the peak appears flatter»✓ 1OWTTE
Part (g)
Use longer counting intervals near the peak «e.g. a slower ascent or several 1-minute counts at each altitude» / fly several GM tubes and combine their counts, so that the percentage uncertainty in each count is smaller✓ 1Accept: use a detector with a shorter dead time. Do not accept 'repeat' alone.

Answers: (b) 19 %  ·  (d) ≈ 18 km ± 2 km  ·  (e) prediction 720 ± 107; measured 690 ± 26 (the remaining parts are explanations — see the table above)

95E-1B-38Half-life & activity Paper 1BEasy8 marks

After a leak from a nuclear power station, iodine-131 was deposited on the grass of a dairy farm. A food laboratory measured the activity A of iodine-131 per litre of the farm's milk every two days. Each measurement has an uncertainty of ±5 %. Time t is measured from the first sample.

The graph shows the data, with error bars. Milk may be sold only when A is less than 500 Bq L−1.

t / daysA / Bq L−1
03960
23100
42240
61780
81290
101020
12740
14590
16430
18340
20250
Activity of iodine-131 per litre of milk against time02468101214161820t / days050010001500200025003000350040004500A / Bq L⁻¹
Graph drawn to scale
(a)

Calculate the absolute uncertainty in the measurement at t = 0.

(1)
(b)

Draw the curve of best fit for the data.

(1)
(c)

Determine, using your curve and two successive halvings, the time in which the activity of iodine-131 in the milk halves.

(2)
(d)

The half-life of iodine-131 is 8.0 days. Compare your answer to (c) with this value and suggest a reason for the difference.

(2)
(e)

Using your curve at t = 0 and a whole number of halving times, predict the day on which the milk may first be sold.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.05 × 3960 = ±198 Bq L−1 «≈ ±200 Bq L−1»✓ 1
Part (b)
Smooth curve of decreasing gradient passing through all the error bars✓ 1Not dot-to-dot.
Part (c)
Two successive halvings read from the curve, e.g. 4000 → 2000 Bq L−1 in about 5 days and 2000 → 1000 Bq L−1 in about another 5 days✓ 1Evidence of use of the curve needed.
Halving time = 5.0 days✓ 1Accept 4.5–5.5 days.
Part (d)
The activity in the milk halves faster than the 8.0-day half-life of iodine-131✓ 1Allow ECF from (c).
Iodine-131 is also removed from the milk by processes other than radioactive decay, e.g. rain washes it off the grass, the grass grows or is replaced, or the cows are given uncontaminated feed✓ 1OWTTE. Accept any sensible non-radioactive removal process.
Part (e)
4000/500 = 8 = 23: three halving times✓ 1Allow ECF from (c).
3 × 5.0 = 15 days after the first sample✓ 1Award [2] for CNA. Accept 14–16 days.

Answers: (a) ±198 Bq L−1  ·  (c) 5.0 days  ·  (e) day 15 (the remaining parts are explanations — see the table above)

96E-1B-01Half-life & activity Paper 1BMedium11 marks

A student investigates the decay of barium-137m. A solution containing barium-137m is washed out of a sealed caesium-137 generator and a few drops are placed in a dish just below the window of a Geiger–Müller (GM) tube connected to a data-logger. Starting at time t = 0, the logger records the number of counts N in a 10.0 s interval that begins every 60 s. Before preparing the solution, the student recorded 111 counts in 300 s with no source present.

The corrected count rate C is the count rate due to the barium-137m alone. The uncertainty in a number of counts N is ±√N. The graph shows C against t, with error bars, for all but two of the readings.

t / sN (counts in 10.0 s)C / s−1
059158.7
6043342.9
12035535.1
180257
24020620.2
30014213.8
36012211.8
42087
480757.1
540504.6
600433.9
Corrected count rate C against time t for barium-137m, points with error bars0100200300400500600t / s0102030405060C / s⁻¹
Graph drawn to scale
(a)

State one reason for measuring the background over 300 s rather than over 10 s.

(1)
(b)

Calculate C at t = 180 s and at t = 420 s.

(2)
(c)

Show that the percentage uncertainty in N at t = 600 s is about 15 %.

(1)
(d)

Plot your two values from (b) on the graph and draw the curve of best fit.

(2)
(e)

Using your curve, determine the half-life of barium-137m. Use at least two successive halvings of C and give your answer with its absolute uncertainty.

(3)
(f)

The accepted value of the half-life of barium-137m is 153 s. Comment on your answer to (e).

(1)
(g)

Suggest one change to the method that would reduce the random uncertainty in the half-life.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A larger number of counts gives a smaller percentage (fractional) uncertainty in the background count rate✓ 1Accept: decay is random, so a longer count gives a more reliable mean rate. OWTTE
Part (b)
t = 180 s: C = 257/10.0 − 0.37 = 25.3 s−1✓ 1Background rate = 111/300 = 0.37 s−1 must be used. Accept 25.33.
t = 420 s: C = 87/10.0 − 0.37 = 8.3 s−1✓ 1Accept 8.33. Award [1] max if the background is not subtracted in both.
Part (c)
√43/43 × 100 = 15.2 %✓ 1Must see √N/N with N = 43 substituted OR an answer to at least 3 s.f.
Part (d)
Both points plotted correctly (±1 s−1, i.e. within half a small square) at (180, 25.3) and (420, 8.3)✓ 1
A single smooth curve of decreasing gradient that passes through all the error bars✓ 1Do not accept a dot-to-dot line or straight segments.
Part (e)
Reads a first halving time from the curve, e.g. C falls from 58 to 29 s−1 in about 153 s✓ 1Values read must come from the candidate's curve, not from the table.
Reads at least one further halving, e.g. 29 to 15 s−1, and averages the half-lives✓ 1Accept a half-life in the range 140–165 s.
Uncertainty = half the range of the values found, e.g. (153 ± 4) s; value and uncertainty quoted to the same precision✓ 1MP3 is for a sensible uncertainty (±2 to ±10 s) with matching precision. Allow ECF from their readings.
Part (f)
153 s lies within the range of (e), so the result agrees with the accepted value within its uncertainty✓ 1Allow ECF: if 153 s is outside their range, a statement that it does not agree (and so there is a systematic error) scores.
Part (g)
Use a more active sample / place the dish closer to the window, so that more counts are recorded in each interval✓ 1Accept: repeat with several fresh samples and average the half-lives; record counts over consecutive intervals with no gaps. Do not accept "repeat the readings" alone.

Answers: (b) 25.3 s−1 and 8.3 s−1  ·  (c) 15.2 %  ·  (e) about 153 s (accept 140–165 s), e.g. (153 ± 4) s (the remaining parts are explanations — see the table above)

97E-1B-04Penetration & ionization Paper 1BMedium9 marks

A student is asked to find which types of radiation are emitted by an unlabelled sealed source. The source is placed 2.0 cm from the window of a GM tube, and the number of counts in 100 s is recorded with different absorbers between the source and the tube. The source–tube distance is not changed.

The uncertainty in a number of counts N is ±√N.

ArrangementCounts in 100 s
no source (background)42
source, no absorber3960
source, paper 0.1 mm3885
source, aluminium 5 mm468
source, aluminium 10 mm451
source, lead 20 mm176
sourceabsorber(varied)GM tube2.0 cmcounter
Diagram not to scale
(a)
(i)

State the absolute uncertainty in the count of 3960.

(1)
(ii)

Determine the count rate due to the source alone, with no absorber, and its absolute uncertainty.

(2)
(b)

Deduce, with reference to the uncertainties, whether there is evidence that the source emits alpha particles.

(2)
(c)

Deduce which other type or types of radiation the source emits. Refer to the data.

(2)
(d)

Suggest why the source was placed only 2.0 cm from the tube.

(1)
(e)

Suggest one change to the procedure that would make the decision in (b) more reliable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
√3960 = ±63 counts✓ 1Accept ±62.9.
Part (a)(ii)
(3960 − 42)/100 = 39.2 s−1✓ 1
Uncertainty = (√3960 + √42)/100 = ±0.7 s−1, so (39.2 ± 0.7) s−1✓ 1Absolute uncertainties add when subtracting. MP2 is for value and uncertainty to matching precision; accept ±0.6 s−1 (background uncertainty ignored).
Part (b)
The paper reduces the count by only 3960 − 3885 = 75✓ 1
This is smaller than the combined uncertainty of about ±125 counts, so there is no significant evidence of alpha particles✓ 1Award MP2 only with a comparison to the uncertainty. Accept "no alpha detected".
Part (c)
Beta: 5 mm of aluminium reduces the count greatly (from 3885 to 468), far more than the uncertainty✓ 1
Gamma: doubling the aluminium to 10 mm makes no significant change (468 → 451, within ±43) and the count stays far above background; 20 mm of lead reduces it but not to background✓ 1Both types needed for [2] with a reason each from the data.
Part (d)
Alpha particles have a range of only a few centimetres in air, so any alpha particles would otherwise not reach the tube✓ 1OWTTE
Part (e)
Count for much longer (e.g. 1000 s) so that the fractional uncertainty (∝ 1/√N) is smaller and a small difference could be detected✓ 1Accept: repeat each count several times and average. Do not accept "repeat" without a reason.

Answers: (a)(i) ±63 counts  ·  (a)(ii) (39.2 ± 0.7) s−1 (the remaining parts are explanations — see the table above)

98E-1B-12The binding energy curve Paper 1BMedium10 marks

A student uses an online database of atomic masses M to investigate how the binding energy per nucleon varies with nucleon number A. For each nuclide the mass defect is calculated from

Δm = ZmH + (A − Z)mn − M

where mH = 1.007825 u is the mass of a hydrogen-1 atom and mn = 1.008665 u is the mass of a neutron. (1 u = 931.5 MeV c−2) The graph shows the results for ten of the nuclides.

NuclideZAM / uBinding energy per nucleon / MeV
4He244.0026027.07
7Li377.0160035.61
12C61212.0000007.68
16O81615.9949157.98
24Mg122423.9850428.26
40Ca204039.9625918.55
56Fe265655.9349368.79
90Zr409089.9046988.71
120Sn50120119.902199
158Gd64158157.9241048.20
208Pb82208207.976652
238U92238238.0507887.57
Binding energy per nucleon against nucleon number from database masses04080120160200240nucleon number A5.05.56.06.57.07.58.08.59.0binding energy per nucleon / MeV
Graph drawn to scale
(a)

Explain why the mass of a hydrogen-1 atom, rather than the mass of a proton, is used with atomic masses in this equation.

(1)
(b)

Show that the binding energy per nucleon of iron-56 is about 8.8 MeV.

(1)
(c)

Complete the table for tin-120 and lead-208.

(2)
(d)

Plot your two values from (c) on the graph and draw the curve of best fit.

(2)
(e)

Using the graph, explain why energy is released when a nucleus with A ≈ 240 splits into two nuclei of roughly equal mass.

(2)
(f)

State the nucleon number at which the binding energy per nucleon is greatest.

(1)
(g)

The helium-4 point lies well above a smooth curve through the other light nuclides. A student suggests removing it as an outlier. Comment on this suggestion.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The atomic mass M includes Z electrons; Z hydrogen atoms contain the same Z electrons, so the electron masses cancel✓ 1OWTTE
Part (b)
Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934936 = 0.528464 u; × 931.5/56 = 8.790 MeV✓ 1Must see the substitution OR an answer to at least 3 s.f.
Part (c)
Tin-120: 8.50 MeV✓ 1Accept 8.5 MeV.
Lead-208: 7.87 MeV✓ 1Accept 7.9 MeV.
Part (d)
Both points plotted correctly to within half a small square✓ 1Allow ECF from (c).
Smooth curve rising steeply at small A, peaking near A ≈ 60 and falling gently to large A✓ 1The curve need not pass through the helium-4 point.
Part (e)
Nuclei of A ≈ 120 have a higher binding energy per nucleon (≈ 8.5 MeV) than those of A ≈ 240 (≈ 7.6 MeV)✓ 1
So the products have a greater total binding energy / are more tightly bound; the increase in binding energy (≈ 0.9 MeV × 240 ≈ 200 MeV) is released✓ 1OWTTE
Part (f)
A ≈ 56–60 (iron-56 in these data)✓ 1Accept 50–65.
Part (g)
It should not be removed: database masses are very precise, so the point is a genuine result (helium-4 is unusually tightly bound), not a measurement error✓ 1OWTTE

Answers: (b) 8.790 MeV  ·  (c) 8.50 MeV and 7.87 MeV  ·  (f) A ≈ 56–60 (the remaining parts are explanations — see the table above)

99E-1B-17Penetration & ionization Paper 1BMedium9 marks

A factory monitors the thickness of aluminium foil with a gauge. A krypton-85 beta source is placed on one side of the foil and a detector on the other. The gauge is calibrated by recording the counts N in 10.0 s for standard foils whose thicknesses are known precisely. The background is negligible. The uncertainty in a number of counts N is ±√N.

The graph shows the count rate R against thickness, with error bars.

Thickness / μmNR / s−1
08537853.7
505943594.3
1004266426.6
1502958295.8
2002119211.9
2501446144.6
3001051105.1
Count rate R against foil thickness for the calibration standards, with error bars050100150200250300thickness / μm0100200300400500600700800900R / s⁻¹
Graph drawn to scale
(a)

Explain why a beta source, rather than an alpha or a gamma source, is used in this gauge.

(2)
(b)

Draw the curve of best fit for the data.

(1)
(c)

A sample of foil gives 3510 counts in 10.0 s. Calculate R for this sample and its absolute uncertainty.

(2)
(d)

Using the graph, determine the thickness of the sample and its absolute uncertainty.

(2)
(e)

The foil is specified to be (125 ± 5) μm thick. Deduce whether this sample meets the specification.

(1)
(f)

Suggest one change to the gauge that would reduce the uncertainty in the thickness.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Alpha particles would be completely absorbed by the foil, so there would be no reading✓ 1
Gamma rays would pass through almost unaffected, so the count rate would hardly change with thickness; beta particles are partly absorbed, so R depends strongly on thickness✓ 1OWTTE
Part (b)
Smooth curve of decreasing gradient passing through all the error bars✓ 1Not dot-to-dot.
Part (c)
R = 3510/10.0 = 351 s−1✓ 1
Uncertainty = √3510/10.0 = ±6 s−1, so (351 ± 6) s−1✓ 1Accept ±5.9 s−1.
Part (d)
Reads the thickness at R = 351 s−1: ≈ 126 μm✓ 1Accept 122–130 μm.
Reads the thicknesses at 357 and 345 s−1 (≈ 124 and 129 μm), giving ±2 μm✓ 1Accept ±2 μm to ±4 μm. Allow ECF from (c).
Part (e)
The range 124–129 μm lies entirely within 120–130 μm, so the sample meets the specification✓ 1Allow ECF from (d).
Part (f)
Count for longer / use a more active source, so that N is larger and the fractional uncertainty √N/N is smaller✓ 1OWTTE

Answers: (c) (351 ± 6) s−1  ·  (d) (126 ± 2) μm (the remaining parts are explanations — see the table above)

100E-1B-23Activity & count rate Paper 1BMedium12 marks

Potassium chloride (KCl) is sold as a low-sodium salt substitute. Natural potassium contains a small fraction of radioactive potassium-40, which mainly emits β− particles and has a very long half-life. A student spreads a mass m of KCl, measured on a balance with a resolution of 0.01 g, evenly over a large tray, and places the window of a large-area radiation monitor just above it. For each mass the number of counts N in 600 s is recorded. With an empty tray, 5400 counts were recorded in 1800 s.

C is the count rate corrected for background. The uncertainty in a number of counts N is ±√N. The graph shows C against m, with error bars, for five of the masses.

m / gN (counts in 600 s)C / s−1Uncertainty in C / s−1
0.2022540.76±0.12
0.4026401.40±0.13
0.6031082.18±0.13
0.803492
1.0039893.65±0.15
1.2043624.27±0.15
Graph of corrected count rate C / s^−1 against mass m / g with error bars for m = 0.20, 0.40, 0.60, 1.00 and 1.20 g; points only. C axis 0 to 5 s^−1, m axis 0 to 1.4 g.00.20.40.60.811.21.4m / g012345C / s⁻¹
Graph drawn to scale
(a)

Calculate the background count rate and its absolute uncertainty.

(2)
(b)

Calculate C for m = 0.80 g and its absolute uncertainty.

(2)
(c)

Plot your result from (b) on the graph, with its error bar, and draw the line of best fit for the data.

(2)
(d)

Discuss whether the data support the hypothesis that C is proportional to m.

(2)
(e)

The monitor records 22 % of the decays that occur in a thin layer of KCl. The accepted activity of KCl is 16.3 Bq g−1. Determine the activity per gram of KCl from your graph and compare it with the accepted value.

(3)
(f)

When 50 g of KCl is spread over the same tray, C is much smaller than the value predicted by extending the line. Suggest why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
5400/1800 = 3.00 s−1✓ 1
√5400/1800 = ±0.041 s−1, so (3.00 ± 0.04) s−1✓ 1Accept ±0.041 s−1.
Part (b)
C = 3492/600 − 3.00 = 2.82 s−1✓ 1
Uncertainty = √3492/600 + 0.04 = ±0.14 s−1, so (2.82 ± 0.14) s−1✓ 1Absolute uncertainties add when the background is subtracted. Accept ±0.10 s−1 (background uncertainty ignored). MP2 is for matching precision.
Part (c)
Point plotted at (0.80, 2.82) to within half a small square, with an error bar of about ±0.14 s−1✓ 1Allow ECF from (b).
Single straight line passing through all the error bars✓ 1
Part (d)
A straight line can be drawn through all the error bars, and it passes through (or within the error bars very close to) the origin✓ 1
So, within the uncertainties, C ∝ m and the data support the hypothesis✓ 1Award MP2 only with reference to the origin and the error bars. OWTTE
Part (e)
Gradient ≈ 3.58 s−1 g−1✓ 1Accept 3.4–3.8 s−1 g−1.
Activity per gram = gradient/0.22 = 16.3 Bq g−1✓ 1Allow ECF. Accept 15.5–17.3 Bq g−1.
This is within about 1 % of 16.3 Bq g−1, which is smaller than the uncertainty in the gradient, so the result agrees with the accepted value✓ 1Allow a consistent comparison with an ECF value; a comparison must be quantitative or refer to the uncertainty.
Part (f)
A thick layer absorbs beta particles: those emitted by the KCl near the bottom are absorbed by the KCl above before reaching the monitor «so only the top layer contributes»✓ 1OWTTE

Answers: (a) (3.00 ± 0.04) s−1  ·  (b) (2.82 ± 0.14) s−1  ·  (e) ≈ 3.58 s−1 g−1; ≈ 16 Bq g−1 (the remaining parts are explanations — see the table above)

101E-1B-34Penetration & ionization Paper 1BMedium12 marks

A student measures the range in air of the alpha particles emitted by a sealed americium-241 source. The source and a thin-window Geiger–Müller (GM) tube are mounted on an optical bench, and the distance d from the face of the source to the window is read from the bench scale with an uncertainty of ±0.05 cm. At each distance the number of counts N in 60 s is recorded. With the source removed, 135 counts were recorded in 300 s.

C is the count rate corrected for background. The uncertainty in a number of counts N is ±√N. The graph shows C against d, with error bars. The alpha particles leave the source with almost the same kinetic energy; the thin cover of the source may be ignored.

d / cmN (counts in 60 s)C / s−1
1.0419869.5
1.5302349.9
2.02369
2.5183530.1
2.8163526.8
2.9139222.8
3.0104717.0
3.15288.4
3.22213.2
3.3881.0
3.5720.8
4.0560.5
5.0540.5
6.0440.3
Corrected count rate C against distance d from the source0123456d / cm010203040506070C / s⁻¹
Graph drawn to scale
(a)

State the absolute uncertainty in the count recorded at d = 2.0 cm.

(1)
(b)

Show that C at d = 2.0 cm is about 39 s−1.

(1)
(c)

Draw a straight line through the points on the steeply falling part of the graph and extend it to the d-axis. Hence determine the range of the alpha particles in air, with its absolute uncertainty.

(3)
(d)

The window of the GM tube absorbs alpha particles as much as 0.9 cm of air does. State the type of error that the window introduces into the range found in (c), and determine the total range in air of the alpha particles.

(2)
(e)

The range R in air, in cm, of alpha particles of initial kinetic energy E, in MeV, is given by R = 0.318E3/2. Determine E and its absolute uncertainty, and compare your result with the accepted value of 5.49 MeV.

(3)
(f)

Explain why C does not fall to zero for d > 3.5 cm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
±√2369 = ±49 counts✓ 1Accept ±48.7.
Part (b)
C = 2369/60 − 135/300 = 39.48 − 0.45 = 39.0 s−1✓ 1Must see the background subtracted OR an answer to at least 3 s.f.
Part (c)
Single straight line through the points between d ≈ 2.9 cm and 3.2 cm, extended to C = 0✓ 1Ignore the points of the gradual decrease and of the tail.
Range read where the line meets the d-axis: 3.2 cm✓ 1Accept 3.1–3.3 cm (least-squares value 3.24 cm).
Uncertainty ±0.1 cm, e.g. from the spacing of the readings near the range, or from steepest and least steep lines through the error bars of the steep points✓ 1Accept ±0.05 cm to ±0.2 cm with a reason. Value and uncertainty must have matching precision.
Part (d)
Systematic error: every range found is too small by the same amount «0.9 cm»✓ 1
Total range = 3.2 + 0.9 = 4.1 cm✓ 1Allow ECF from (c).
Part (e)
E = (4.1/0.318)2/3 = 5.50 MeV✓ 1Allow ECF from (d).
ΔE/E = (2/3) × ΔR/R = (2/3) × 0.1/4.1 = 1.6 %, so E = (5.5 ± 0.1) MeV✓ 1Accept the range method: 5.41–5.59 MeV. MP2 is for matching precision of value and uncertainty.
5.49 MeV lies within the range of the result, so the experiment agrees with the accepted value✓ 1Allow a consistent conclusion from an ECF value.
Part (f)
Americium-241 also emits gamma rays «as well as alpha particles»✓ 1Do not accept 'background' — it has been subtracted.
Gamma rays are weakly ionizing and very penetrating, so they are not stopped by a few cm of air and a small fraction of them is detected by the GM tube «at all distances»✓ 1OWTTE

Answers: (a) ±49 counts  ·  (c) (3.2 ± 0.1) cm  ·  (d) 4.1 cm  ·  (e) (5.5 ± 0.1) MeV (the remaining parts are explanations — see the table above)

102E-1B-37Penetration & ionization Paper 1BMedium10 marks

The half-value thickness x½ of a material is the thickness that halves the intensity of a beam of gamma rays. The table gives published values of x½ for the 662 keV gamma rays emitted after the decay of caesium-137, for six materials of density ρ. Each value of x½ has an uncertainty of ±5 %; the density of concrete varies from one mix to another.

A student suggests that x½ = k/ρ, where k is the same constant for every material. The graph shows x½ against 1/ρ, with error bars, for four of the materials.

Materialρ / g cm−3x½ / cm(1/ρ) / cm3 g−1
concrete2.30 ± 0.103.88
aluminium2.703.430.370
titanium4.512.110.222
iron7.871.200.127
copper8.961.060.112
lead11.350.555
Half-value thickness against 1/density00.10.20.30.40.5(1/ρ) / cm³ g⁻¹012345x½ / cm
Graph drawn to scale
(a)

Calculate 1/ρ for concrete and its absolute uncertainty.

(2)
(b)

Plot the points for concrete and for lead on the graph, with the error bar in 1/ρ for concrete.

(1)
(c)

Draw the line of best fit for the data that support the student's suggestion, and determine its gradient. Include the unit.

(3)
(d)

Outline what the gradient represents physically.

(1)
(e)

Comment on the point for lead.

(1)
(f)

A wall is to reduce the intensity of the gamma rays from a caesium-137 store to one-eighth. Determine the thickness of concrete needed, and the thickness of lead that would do the same.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
1/2.30 = 0.435 cm3 g−1✓ 1
Δ(1/ρ) = (Δρ/ρ) × (1/ρ) = (0.10/2.30) × 0.435 = ±0.019 cm3 g−1, so (0.435 ± 0.019) cm3 g−1✓ 1Accept ±0.02. Fractional uncertainty of 1/ρ equals that of ρ «4.3 %».
Part (b)
Concrete at (0.435, 3.88) with a horizontal error bar of ±0.02, and lead at (0.088, 0.555), both to within half a small square✓ 1Allow ECF from (a). Both points needed.
Part (c)
Single straight line through the error bars of the five materials other than lead «it passes through, or very close to, the origin»✓ 1The line must not be forced through the lead point.
Gradient from a large triangle «at least half the line»✓ 1
k = 9.1 g cm−2✓ 1Accept 8.7–9.6 g cm−2. Unit needed.
Part (d)
The mass per unit area «ρx½» of material that halves the intensity: equal masses per unit area of these materials absorb the gamma rays equally✓ 1OWTTE
Part (e)
Lead lies well below the line «0.555 cm against a predicted 0.81 cm»: it halves the intensity with a smaller mass per unit area than the other materials, so the suggestion does not hold for lead «lead is a better absorber per kilogram»✓ 1OWTTE. Must refer to the data.
Part (f)
One-eighth = (1/2)3: three half-value thicknesses; concrete: 3 × 3.88 = 11.6 cm✓ 1
Lead: 3 × 0.555 = 1.67 cm «≈ 1.7 cm»✓ 1Award [1 max] for 8 × x½.

Answers: (a) (0.435 ± 0.019) cm3 g−1  ·  (c) 9.1 g cm−2  ·  (f) 11.6 cm; 1.67 cm (the remaining parts are explanations — see the table above)

103E-1B-39Background radiation Paper 1BMedium11 marks

On a geology field trip, students hold a Geiger–Müller survey meter 1.0 m above six flat rock outcrops and record five separate two-minute counts above each. The uranium concentration u of each rock, in parts per million (ppm), is known from laboratory analysis. The students assume that the rocks contain similar amounts of the other natural radioactive elements, thorium and potassium-40.

The table gives the counts, the mean count rate C and its uncertainty (half the range of the five count rates). The graph shows C against u, with error bars, for five of the outcrops.

u / ppmCounts in five two-minute intervalsC / min−1Uncertainty / min−1
1.255, 45, 51, 41, 5324.53.5
2.557, 59, 69, 61, 7131.73.5
3.685, 73, 65, 81, 7638.05.0
5.194, 74, 83, 81, 96
7.8119, 104, 99, 116, 10654.45.0
11.4136, 138, 143, 143, 12368.35.0
Mean count rate against uranium concentration of the rock024681012u / ppm01020304050607080C / min⁻¹
Graph drawn to scale
(a)

Calculate C and its uncertainty for the outcrop with u = 5.1 ppm.

(2)
(b)

Plot your result from (a) on the graph, with its error bar, and draw the line of best fit.

(2)
(c)

Determine the gradient of your line. Include the unit.

(2)
(d)

By drawing lines of maximum and minimum gradient, determine the intercept on the C-axis and its absolute uncertainty.

(2)
(e)

Over the middle of a deep lake, where the water absorbs the radiation from the ground, the meter reads (10 ± 2) min−1. Discuss what the intercept represents, by comparing it with this reading.

(2)
(f)

Suggest one change to the method that would reduce the uncertainty in each value of C.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
C = 428/(5 × 2.0 min) = 42.8 min−1✓ 1Award [0] for 85.6 min−1 «counts not divided by the 2.0 min counting time».
Uncertainty = (96 − 74)/(2 × 2.0) = ±5.5 min−1✓ 1Accept ±5 or ±6 min−1. Allow ECF for the counting time.
Part (b)
Point plotted at (5.1, 42.8) with an error bar of ±5.5, to within half a small square✓ 1Allow ECF from (a).
Single straight line passing through all the error bars✓ 1
Part (c)
Gradient from a large triangle «at least half the line»✓ 1
4.2 min−1 ppm−1✓ 1Accept 3.9–4.5. Unit needed: «counts» min−1 ppm−1.
Part (d)
Steepest and least steep lines through all the error bars, extended to u = 0✓ 1
Intercept = (21 ± 4) min−1 «half the difference of the two intercepts»✓ 1Accept an intercept of 19–23 min−1 and an uncertainty of ±2 to ±6 min−1 from the candidate's lines; value and uncertainty to matching precision.
Part (e)
The intercept is the count rate above a rock with no uranium «≈ 21 min−1», which is larger than the lake reading even allowing for the uncertainties «about 17–25 against 8–12 min−1»✓ 1Allow ECF from (d).
The lake reading is due only to cosmic rays «and radon in the air / the meter itself»; the intercept also includes radiation from the thorium and potassium-40 in the rocks, so it is not just the non-ground background✓ 1OWTTE
Part (f)
Count for longer above each outcrop «e.g. 10 minutes» so that the number of counts is larger and its percentage uncertainty «∝ 1/√N» is smaller✓ 1Do not accept 'repeat' alone.

Answers: (a) (42.8 ± 5.5) min−1  ·  (c) 4.2 min−1 ppm−1  ·  (d) (21 ± 4) min−1 (the remaining parts are explanations — see the table above)

104E-1B-09Penetration & ionization Paper 1BHard12 marks

A student tests whether the count rate from a sealed gamma source obeys an inverse-square law. The distance x is measured with a ruler from the front face of the source holder to the window of a GM tube (uncertainty ±0.1 cm). At each distance the counts N in 100 s are recorded. With the source removed, 90 counts were recorded in 300 s. C is the corrected count rate.

The source is set back inside its holder, so the true source–detector distance is x + x0, where x0 is unknown. The student's model is C = k/(x + x0)2, where k is a constant.

The graph shows C−½ against x for all seven readings, with error bars.

x / cmNC / s−1C−½ / s½
2.0807780.470.111
4.0316331.330.179
6.0173217.02
8.0110310.730.305
10.07317.010.378
12.05595.29
14.04103.800.513
C to the power minus one half against x, points with error bars-4-20246810121416x / cm0.00.10.20.30.40.50.6C^(−1/2) / s^(1/2)
Graph drawn to scale
(a)

Show that the corrected count rate at x = 6.0 cm is about 17 s−1.

(1)
(b)

Complete the table for x = 6.0 cm and x = 12.0 cm.

(2)
(c)

Explain why, if the model is correct, the graph of C−½ against x is a straight line that meets the x-axis at x = −x0.

(2)
(d)

Draw the line of best fit and hence determine x0.

(3)
(e)

By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in x0.

(2)
(f)

State what type of error x0 represents in the measurement of distance, and outline whether the data support the inverse-square model.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
C = 1732/100 − 90/300 = 17.32 − 0.30 = 17.02 s−1✓ 1Must see the background subtracted OR an answer to at least 3 s.f.
Part (b)
x = 6.0 cm: C−½ = 17.02−½ = 0.242 s½✓ 1Accept 0.24.
x = 12.0 cm: C−½ = 5.29−½ = 0.435 s½✓ 1Accept 0.43.
Part (c)
Rearranging gives C−½ = (x + x0)/√k = x/√k + x0/√k, a straight line with gradient 1/√k✓ 1
C−½ = 0 when x + x0 = 0, i.e. at x = −x0✓ 1
Part (d)
Straight line through all the error bars, extended to cross the x-axis✓ 1
x-intercept read, or calculated from the gradient and y-intercept (x0 = y-intercept/gradient)✓ 1e.g. gradient ≈ 0.0329 s½ cm−1, y-intercept ≈ 0.046 s½.
x0 = 1.4 cm✓ 1Accept 1.2–1.6 cm.
Part (e)
Steepest and shallowest lines through all the error bars, both extended to the x-axis✓ 1
Uncertainty = half the difference of their x-intercepts, e.g. ±0.2 cm✓ 1Accept ±0.2 cm to ±0.6 cm from the candidate's lines.
Part (f)
A systematic error: every measured distance is too small by the same amount x0✓ 1
The points lie on a straight line within their error bars, so the data support C ∝ 1/(x + x0)2, i.e. the inverse-square law once the offset is included✓ 1OWTTE

Answers: (a) 17.02 s−1  ·  (b) 0.242 and 0.435 s½  ·  (d) x0 ≈ 1.4 cm  ·  (e) ≈ ±0.2 cm (the remaining parts are explanations — see the table above)

105E-1B-20Mixtures of isotopes Paper 1BHard12 marks

A metal sample is irradiated with neutrons, which produces two radioactive isotopes with different half-lives. The sample is then placed near a GM tube, and the count rate C, corrected for background, is found by counting for 10.0 s at the times t shown. The uncertainty in a number of counts N is ±√N.

The graph shows the data, with error bars.

t / sC / s−1
0301.1
40171.0
80106.8
12069.1
16050.1
20037.9
30021.9
40015.0
50010.3
6007.4
7005.6
8003.8
Total corrected count rate against time for the irradiated sample, with error bars0100200300400500600700800t / s04080120160200240280320C / s⁻¹
Graph drawn to scale
(a)

Outline how the data show that the sample contains more than one radioactive isotope.

(1)
(b)

Draw the curve of best fit for the data.

(1)
(c)

Show that the half-life of the longer-lived isotope is about 200 s, using the data for t ≥ 400 s.

(1)
(d)

Determine the count rate due to the longer-lived isotope at t = 200 s and at t = 0.

(2)
(e)

Hence determine the half-life of the shorter-lived isotope.

(3)
(f)

Determine the percentage uncertainty in the count rate due to the shorter-lived isotope at t = 200 s, considering only the uncertainty in the count recorded at that time. Suggest how the experiment could be changed to reduce it.

(3)
(g)

Predict the count rate at t = 1000 s.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
For a single isotope C would fall by the same fraction in equal times, but it falls from 301 to 171 s−1 in the first 40 s, while later it only halves in 200 s✓ 1OWTTE. Accept: the curve does not have a constant half-life.
Part (b)
Smooth curve through all the error bars, steep at first and then much flatter✓ 1Not dot-to-dot.
Part (c)
C falls from 15.0 to 7.4 s−1 between 400 s and 600 s (and to 3.8 s−1 at 800 s): it halves every 200 s✓ 1Must see two pairs of values or a ratio of ≈ 2 over 200 s.
Part (d)
t = 200 s is one half-life before 400 s: 2 × 15.0 = 30.0 s−1✓ 1Accept 28–32 s−1.
t = 0 is two half-lives before 400 s: 4 × 15.0 = 60 s−1✓ 1Accept 57–63 s−1. Allow ECF from (c).
Part (e)
Short-lived contribution at t = 0: 301 − 60 = 241 s−1✓ 1
At t = 200 s: 37.9 − 30.0 = 7.9 s−1✓ 1
Ratio 241/7.9 ≈ 31 ≈ 25, so 5 half-lives in 200 s: half-life = 40 s✓ 1Allow ECF from (d). Accept 38–42 s.
Part (f)
Counts at 200 s ≈ 379, uncertainty √379 = ±19 counts, i.e. ±1.9 s−1✓ 1
Percentage uncertainty = 1.9/7.9 × 100 ≈ 25 %✓ 1Allow ECF from (e). Accept 20–35 %.
Count for longer at each time / use a more active sample / repeat with fresh samples and average✓ 1Accept: use an earlier time, where the short-lived contribution is larger. OWTTE
Part (g)
Only the long-lived isotope remains: 1000 s is 5 half-lives, so C ≈ 60/32 = 1.9 s−1✓ 1Accept 1.8–2.0 s−1. Allow ECF.

Answers: (d) 30.0 s−1 and 60 s−1  ·  (e) 40 s  ·  (f) ≈ 25 %  ·  (g) 1.9 s−1 (the remaining parts are explanations — see the table above)

106E-1B-22Penetration & ionization Paper 1BHard12 marks

A company makes flexible radiation-shielding sheets, 0.100 mm thick, from tungsten powder set in a polymer, for aprons worn by hospital staff. It claims that 0.20 mm of the material halves the intensity of the gamma rays from a cobalt-57 source. A student tests the claim by stacking sheets between a sealed cobalt-57 source and a detector and recording the number of counts N in 100 s for each total thickness x. With the source removed, 312 counts were recorded in 600 s. C is the count rate corrected for background.

The student's model is C = C0 × 2−x/x½, where x½ is the thickness that halves the corrected count rate. The uncertainty in a number of counts N is ±√N. The graph shows log(C/s−1) against x, with error bars, for five of the thicknesses.

x / mmN (counts in 100 s)C / s−1log(C / s−1)
0.00848984.371.926
0.10629762.451.796
0.20478747.351.675
0.30361435.62
0.4026771.419
0.50206620.14
0.60152514.731.168
Graph of log(C / s^−1) against thickness x / mm for x = 0, 0.10, 0.20, 0.40 and 0.60 mm, with small error bars; points only. Vertical axis 1.0 to 2.0, horizontal axis 0 to 0.70 mm.0.00.10.20.30.40.50.60.7x / mm1.01.21.41.61.82.0log(C / s⁻¹)
Graph drawn to scale
(a)

Show that the corrected count rate at x = 0.40 mm is about 26 s−1.

(1)
(b)

Complete the table for x = 0.30 mm and x = 0.50 mm, and plot these two points on the graph.

(2)
(c)

Explain why, if the model is correct, the graph is a straight line of gradient −(log 2)/x½.

(2)
(d)

Draw the line of best fit and determine its gradient.

(3)
(e)

Determine x½.

(1)
(f)

The student estimates that the uncertainty in x½ is ±0.01 mm. Comment on the company's claim.

(1)
(g)

In a first trial the student forgot to subtract the background count rate. Explain the effect that this would have on the graph and on the value of x½.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
C = 2677/100 − 312/600 = 26.77 − 0.52 = 26.25 s−1✓ 1Must see the background subtracted OR an answer to at least 3 s.f.
Part (b)
x = 0.30 mm: log C = log 35.62 = 1.552; x = 0.50 mm: log C = log 20.14 = 1.304✓ 1Both needed. Accept 2 or more decimal places.
Both points plotted to within half a small square✓ 1Allow ECF from the logarithms.
Part (c)
Taking logarithms of the model: log C = log C0 − (x/x½) log 2✓ 1
This has the form y = c + mx: log C0 is constant and the gradient −(log 2)/x½ is constant, so the graph is a straight line✓ 1OWTTE
Part (d)
Single straight line passing through all the error bars✓ 1
Gradient calculated from a large triangle (at least half the line)✓ 1
Gradient = −1.25 mm−1✓ 1Accept −1.31 to −1.19 mm−1. Negative sign needed; unit not required.
Part (e)
x½ = log 2/|gradient| = 0.301/1.25 = 0.24 mm✓ 1Allow ECF from (d). Accept 0.23–0.25 mm.
Part (f)
0.20 mm lies outside the range (0.24 ± 0.01) mm, so the claim is not supported: a thicker sheet than claimed is needed to halve the count rate✓ 1Allow ECF from (e). Must refer to the uncertainty range.
Part (g)
Each C would be too large by 0.52 s−1, which is a larger fraction of C at large x (about 4 % at 0.60 mm compared with 0.6 % at 0), so the points at large x are raised and the graph curves upwards / its gradient is less steep✓ 1OWTTE
So |gradient| is smaller and x½ is overestimated (a systematic error)✓ 1MP2 only scores if MP1 scores.

Answers: (a) 26.25 s−1  ·  (b) 1.552 and 1.304  ·  (d) ≈ −1.25 mm−1  ·  (e) ≈ 0.24 mm (the remaining parts are explanations — see the table above)

107E-1B-35Random & spontaneous decay Paper 1BHard12 marks

A sealed cobalt-60 source (half-life 5.3 years) is fixed in front of a Geiger–Müller tube connected to a data-logger. For each counting interval Δt, the logger records the number of counts in 200 successive intervals and calculates the mean count N̄ and the spread s of the 200 counts (their standard deviation). The background is negligible compared with the counts from the source.

A student tests the hypothesis that s = N̄n with n = 0.5, i.e. that s = √N̄. The uncertainty in each value of s is ±5 %; the uncertainty in N̄ is negligible. The graph shows log s against log N̄, with error bars, for four of the six intervals.

Δt / sN̄slog N̄log s
0.526.15.31.4170.724
1.051.87.01.7140.845
2.0104.610.5
5.0259.315.62.4141.193
10.0521.723.32.7171.367
20.01038.931.5
log s against log N̄11.522.533.5log (N̄)0.50.7511.251.51.75log (s)
Graph drawn to scale
(a)

Outline why the counts recorded in successive intervals of the same length are different, although nothing in the apparatus is changed.

(1)
(b)

The complete experiment takes about 3 hours. Explain why the activity of the source can be treated as constant.

(1)
(c)

Complete the table for Δt = 2.0 s and Δt = 20.0 s.

(2)
(d)

Plot your two points from (c) on the graph, with their error bars, and draw the line of best fit.

(2)
(e)

Determine the gradient of your line.

(2)
(f)

Lines of maximum and minimum gradient through the error bars have gradients 0.51 and 0.48. Deduce whether the data support the hypothesis. Refer to both the gradient and the intercept of the graph.

(2)
(g)

The mean count rate is 52 s−1. Assuming that s = √N, determine the minimum counting time needed for the percentage uncertainty in a single count to be 1 %.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Radioactive decay is random: it is impossible to predict when a particular nucleus will decay, so the number of decays in equal intervals fluctuates✓ 1OWTTE
Part (b)
3 hours is very short compared with the half-life of 5.3 years, so the activity falls by a negligible fraction during the experiment✓ 1Must refer to the half-life.
Part (c)
Δt = 2.0 s: log N̄ = 2.020, log s = 1.021✓ 1Both values needed.
Δt = 20.0 s: log N̄ = 3.017, log s = 1.498✓ 1Both values needed. Accept values to 2 or more decimal places; MP2 also requires a consistent number of decimal places for log N̄ and log s.
Part (d)
Both points plotted to within half a small square, with error bars of about ±0.02✓ 1Allow ECF from (c).
Single straight line passing through all six error bars✓ 1
Part (e)
Gradient calculated from a large triangle «covering at least half the line»✓ 1
Gradient = 0.49✓ 1Accept 0.46–0.52. No unit.
Part (f)
0.5 lies between 0.48 and 0.51, so n = 0.5 is consistent with the data✓ 1
The line passes through «or very close to» the origin, so the constant is log 1 = 0 and s = 1 × N̄0.5: the data support s = √N̄✓ 1Accept reading the intercept at log N̄ = 0 by extrapolation or from y = mx + c; intercept ≈ 0.02. MP2 needs a reference to the intercept.
Part (g)
√N/N = 1/√N = 0.01, so N = 1.0 × 104 counts✓ 1
t = 1.0 × 104/52 = 192 s «≈ 3.2 min»✓ 1Award [2] for CNA. Accept 190 s to 200 s.

Answers: (c) 2.020, 1.021; 3.017, 1.498  ·  (e) 0.49  ·  (g) 192 s (the remaining parts are explanations — see the table above)

108E-1B-36Radioactive dating Paper 1BHard12 marks

Lead-210 (half-life 22.3 years) falls from the atmosphere onto a lake and is buried in the sediment that settles on the lake bed. In 2023 a core of sediment was taken from the lake bed and cut into slices 2 cm thick. The activity of lead-210 per kilogram of each dried slice was measured; z is the depth of the middle of the slice below the top of the core.

The sediment also contains radium-226, which has a very long half-life and continuously produces lead-210 in the sediment. This 'supported' lead-210 has an activity of (25 ± 3) Bq kg−1 at all depths. The unsupported activity AU is the measured activity minus the supported activity. The graph shows AU against z, with error bars, for nine of the slices.

z / cmMeasured activity / Bq kg−1AU / Bq kg−1
1384 ± 15359 ± 18
3270 ± 11245 ± 14
5206 ± 8
7146 ± 6121 ± 9
9116 ± 591 ± 8
1184 ± 459 ± 7
1371 ± 4
1558 ± 433 ± 7
1745 ± 420 ± 7
1942 ± 417 ± 7
2135 ± 410 ± 7
Unsupported lead-210 activity against depth0246810121416182022z / cm050100150200250300350400450unsupported activity / Bq kg⁻¹
Graph drawn to scale
(a)

Suggest why the supported activity must be subtracted before the data are used for dating.

(1)
(b)

Calculate AU, with its absolute uncertainty, for z = 5 cm and z = 13 cm.

(2)
(c)

Plot your two values from (b) on the graph and draw the curve of best fit.

(2)
(d)

Determine, using your curve and at least two successive halvings of AU, the increase in depth over which AU halves.

(2)
(e)

Assume that sediment has built up at a constant rate. Using your answer to (d), determine the rate at which sediment builds up, in cm per year. Take the uncertainty in your answer to (d) to be ±0.3 cm, and give your answer with its absolute uncertainty.

(2)
(f)

Caesium-137 from nuclear weapons tests reached the lake mainly in 1963. The activity of caesium-137 in the core is greatest at a depth of (10.5 ± 0.5) cm. Deduce whether this supports your answer to (e).

(2)
(g)

Suggest why this method cannot be used to date sediment that lies more than about 30 cm below the top of the core.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The supported lead-210 is replaced by the decay of radium-226 as fast as it decays, so its activity does not decrease with the age of the layer; only the unsupported lead-210 decays with a half-life of 22.3 years «it acts like a background»✓ 1OWTTE
Part (b)
z = 5 cm: 206 − 25 = 181 Bq kg−1; z = 13 cm: 71 − 25 = 46 Bq kg−1✓ 1Both values needed.
Uncertainties add when subtracting: ±(8 + 3) = ±11 and ±(4 + 3) = ±7 Bq kg−1✓ 1Both uncertainties needed.
Part (c)
Both points plotted to within half a small square✓ 1Allow ECF from (b).
Smooth curve of decreasing gradient passing through all the error bars✓ 1Not dot-to-dot.
Part (d)
Two successive halvings read from the curve, e.g. 400 → 200 Bq kg−1 between z ≈ 0.3 cm and 4.3 cm, and 200 → 100 Bq kg−1 between 4.3 cm and 8.3 cm✓ 1Evidence of use of the curve needed, e.g. construction lines.
Halving depth = 4.0 cm✓ 1Accept 3.6–4.4 cm.
Part (e)
Rate = 4.0 cm/22.3 y = 0.179 cm y−1✓ 1Allow ECF from (d).
Uncertainty = 0.179 × 0.3/4.0 = ±0.013, so (0.18 ± 0.01) cm y−1✓ 1MP2 is for matching precision of value and uncertainty. Accept the range method (3.7/22.3 to 4.3/22.3).
Part (f)
Predicted depth of the 1963 layer = 60 y × 0.179 cm y−1 = 10.8 cm «range 10.0–11.6 cm»✓ 1Allow ECF from (e).
This overlaps 10.0–11.0 cm, so the caesium-137 marker supports the rate found from lead-210✓ 1The conclusion must be consistent with the candidate's values and refer to the uncertainties.
Part (g)
Below about 30 cm «more than 7 halvings» AU is less than a few Bq kg−1, much smaller than the uncertainty in the supported activity «±3 Bq kg−1», so it cannot be measured reliably✓ 1OWTTE. Accept: the unsupported activity cannot be distinguished from the supported activity.

Answers: (b) (181 ± 11) Bq kg−1, (46 ± 7) Bq kg−1  ·  (d) 4.0 cm  ·  (e) (0.18 ± 0.01) cm y−1  ·  (f) 10.8 cm (the remaining parts are explanations — see the table above)

109E-2-05Beta decay & the neutrino Paper 2Easy11 marks

Sodium has several radioactive isotopes. Sodium-24 decays by β− emission to magnesium-24. Sodium-22 decays by β+ emission to neon-22; the neon-22 nucleus is formed in an excited state and then emits a gamma-ray photon of energy 1.27 MeV.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, 1 MeV = 1.60 × 10−13 J)

(a)

The two beta decays.

(i)

Complete the equation for the decay of sodium-24: 2411Na → ……Mg + ……e + …

(2)
(ii)

Complete the equation for the decay of sodium-22: 2211Na → ……Ne + ……e + …

(2)
(iii)

Outline the change that takes place inside the nucleus during β+ decay.

(1)
(b)

The gamma ray.

(i)

State the changes, if any, in the nucleon number and in the proton number of the neon-22 nucleus when it emits the gamma ray.

(1)
(ii)

Calculate the wavelength of the gamma-ray photon.

(2)
(c)

A detector close to a sodium-22 source records a count rate well above background. A sheet of aluminium 5 mm thick is then placed between the source and the detector. Predict, with a reason, the effect on the count rate.

(2)
(d)

State one reason why the neutrinos emitted by the source are not registered by the detector.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
2412Mg✓ 1Both numbers needed.
0−1e and an antineutrino, ν̄✓ 1Accept β− for the electron. Do not accept a neutrino.
Part (a)(ii)
2210Ne✓ 1Both numbers needed.
0+1e «a positron» and a neutrino, ν✓ 1Accept β+ for the positron. Do not accept an antineutrino.
Part (a)(iii)
A proton changes into a neutron «and a positron and a neutrino are emitted»✓ 1
Part (b)(i)
Neither changes: the gamma ray carries no charge and no nucleons «the nucleus only loses energy»✓ 1
Part (b)(ii)
E = 1.27 × 1.60 × 10−13 = 2.03 × 10−13 J✓ 1
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/2.03 × 10−13 = 9.8 × 10−13 m✓ 1Award [2] for CNA.
Part (c)
The count rate falls, but stays above background✓ 1
The positrons «β+ particles» are absorbed by a few millimetres of aluminium, but the gamma rays pass through it✓ 1MP2 is for the reason.
Part (d)
Neutrinos are uncharged and have a very small mass, so they interact with matter extremely rarely and pass straight through the detector✓ 1OWTTE

Answers: (b)(ii) 9.8 × 10−13 m (the remaining parts are explanations — see the table above)

110E-2-06Half-life & activity Paper 2Easy10 marks

A self-powered emergency exit sign contains tritium gas, 31H, sealed inside glass tubes that are coated on the inside with a phosphor. Beta particles emitted by the tritium strike the phosphor, which emits light. Tritium has a half-life of 12.3 years.

When the sign is new, the activity of the tritium is 7.4 × 1011 Bq. Assume that the brightness of the sign is proportional to this activity.

Blank axes: activity of the tritium in units of 10^11 Bq (0 to 8) against time in years (0 to 50); the initial activity 7.4 × 10^11 Bq is plotted at t = 0.01020304050time / years02468activity / 10¹¹ Bq
Graph axes drawn to scale
(a)

State what is meant by the half-life of a radioactive isotope.

(1)
(b)

Draw, on the axes, a graph to show how the activity of the tritium varies during the first 49.2 years.

(3)
(c)

The sign is acceptable for use while the activity is at least 2.0 × 1011 Bq. Determine whether a sign that is 24.6 years old is acceptable.

(2)
(d)

Explain why the beta particles from the tritium present no hazard to a person standing next to an undamaged sign.

(2)
(e)

Suggest why tritium released from a broken sign could be hazardous if it were inhaled.

(1)
(f)

A sign stored in an unheated warehouse at −20 °C has the same activity as an identical sign kept indoors at 20 °C. State the property of radioactive decay that this illustrates.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The time taken for the activity «or the number of undecayed nuclei» of a sample of the isotope to fall to half of its initial value✓ 1Accept: the time for half of the radioactive nuclei to decay.
Part (b)
Curve starts at 7.4 × 1011 Bq at t = 0 and passes through 3.7 × 1011 Bq at 12.3 years✓ 1
Passes through ≈ 1.85 × 1011 Bq at 24.6 years, ≈ 0.93 × 1011 Bq at 36.9 years and ≈ 0.46 × 1011 Bq at 49.2 years✓ 1Within half a small square.
A smooth curve whose gradient decreases in magnitude, not reaching zero activity✓ 1Do not accept a straight line or a curve that reaches zero.
Part (c)
24.6 years = 2 half-lives, so A = 7.4 × 1011/4 = 1.85 × 1011 Bq✓ 1
1.85 × 1011 Bq < 2.0 × 1011 Bq, so the sign is no longer acceptable✓ 1Allow ECF from a value read from the candidate's graph in (b).
Part (d)
Beta particles from tritium have a very short range: they are absorbed by the phosphor and the glass wall of the tube✓ 1Accept: beta particles are stopped by a few mm of glass/aluminium.
Tritium emits no gamma rays, so no ionizing radiation leaves the sign✓ 1OWTTE
Part (e)
Inside the body there is no shielding, so the beta particles ionize the surrounding tissue directly «damaging cells»✓ 1OWTTE
Part (f)
Radioactive decay is spontaneous: it is not affected by external conditions such as temperature✓ 1Do not accept 'random' alone.

Answers: (c) 1.85 × 1011 Bq — not acceptable (the remaining parts are explanations — see the table above)

111E-2-19Decay chains Paper 2Easy10 marks

Thorium-228, 22890Th, is part of a natural radioactive decay series. It decays by four successive alpha emissions to lead-212, 21282Pb.

The chart shows the neutron number N against the proton number Z for thorium-228 and lead-212.

Chart of neutron number N against proton number Z with thorium-228 at Z = 90, N = 138 and lead-212 at Z = 82, N = 130 marked; grid lines at every whole number.80828486889092proton number Z122124126128130132134136138140neutron number Nthorium-228lead-212
Chart drawn to scale
(a)
(i)

Thorium-228 decays by alpha emission to an isotope of radium (Ra). State the nuclear equation for this decay.

(2)
(ii)

Show that four alpha decays, and no beta decays, are needed to change 22890Th into 21282Pb.

(1)
(b)

Lead-212 decays by β− emission to bismuth-212, which then decays by β− emission to polonium-212. Polonium-212 decays by alpha emission to a nuclide F.

(i)

Draw arrows on the chart to show the two β− decays, starting from lead-212.

(2)
(ii)

Draw an arrow on the chart to show the alpha decay of polonium-212. Label the final nuclide F.

(1)
(iii)

State and explain the relationship between nuclide F and lead-212.

(2)
(c)

Some nuclides decay by β+ emission. State the change in the proton number and in the neutron number of a nucleus that emits a β+ particle, and name the other particle that is emitted in this decay.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
22488Ra as the product nucleus✓ 1Do not accept a product with A = 228 or Z = 90.
42α OR 42He as the second product «equation balanced in A and in Z»✓ 1
Part (a)(ii)
the nucleon number falls by 228 − 212 = 16 = 4 × 4, so four alpha decays, AND these lower the proton number by 4 × 2 = 8 = 90 − 82, so no beta decay is needed✓ 1Must see both the change in A and the change in Z.
Part (b)(i)
each arrow goes one square to the right «Z increases by 1» and one square down «N decreases by 1»✓ 1OWTTE
two such arrows, ending at Z = 84, N = 128 «polonium-212»✓ 1Award MP2 only if both arrows keep N + Z constant.
Part (b)(ii)
arrow two squares to the left and two squares down, ending at Z = 82, N = 126, labelled F✓ 1Allow ECF from (b)(i): an arrow of −2 in Z and −2 in N from the candidate's end point.
Part (b)(iii)
F is lead-208 «Z = 82», so F and lead-212 are isotopes «of lead»✓ 1
isotopes have the same number of protons but different numbers of neutrons «126 and 130» / different nucleon numbers «208 and 212»✓ 1
Part (c)
the proton number decreases by 1 AND the neutron number increases by 1 «a proton changes into a neutron»✓ 1
a neutrino «ν» is also emitted✓ 1Do not accept antineutrino.

Answers: (a)(i) 22890Th → 22488Ra + 42α  ·  (b)(ii) F: Z = 82, N = 126 (lead-208) (the remaining parts are explanations — see the table above)

112E-2-23Radioactive dating Paper 2Easy9 marks

Living wood contains radioactive carbon-14, so that every gram of carbon in a living tree has an activity of 0.25 Bq. When the tree dies no further carbon-14 is taken in. Carbon-14 decays by β− emission with a half-life of 5700 years.

The graph shows how the activity per gram of carbon varies with the time since the tree died.

Graph of activity per gram of carbon, falling from 0.25 Bq at time 0 and halving every 5700 years, for times since death from 0 to 25 thousand years.0510152025time since death / thousands of years00.050.10.150.20.25activity per gram / Bq
Graph drawn to scale
(a)

State the nuclear equation for the β− decay of carbon-14, 146C, to nitrogen (N).

(2)
(b)
(i)

A wooden bowl is found at an ancient site. One gram of carbon from the bowl has an activity of 0.031 Bq. Show that the bowl is about 17 000 years old.

(1)
(ii)

Using the graph, determine the age of a piece of wood for which the activity per gram of carbon is 0.10 Bq.

(2)
(c)

Carbon-14 dating cannot be used for every object.

(i)

Explain why it cannot be used to find the age of a wooden chair made 50 years ago.

(1)
(ii)

Explain why it cannot be used to find the age of a fossil that is about 1 million years old.

(2)
(d)

Suggest why the age found for a very small sample of wood has a large uncertainty.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
146C → 147N + …✓ 1
… + 0−1e «β−» + ν̄ «antineutrino»✓ 1Do not accept a neutrino for MP2.
Part (b)(i)
0.25/0.031 ≈ 8 = 23, so three half-lives: 3 × 5700 = 17 100 years✓ 1Must see the factor 8 OR three half-lives.
Part (b)(ii)
reading taken from the graph at 0.10 Bq✓ 1Evidence of use of the graph, e.g. lines drawn on it.
7.5 × 103 years✓ 1Accept 7.2–7.8 × 103 years.
Part (c)(i)
50 years is very short compared with the half-life, so the activity has fallen by less than 1 %, too little to measure against the random variation in the count✓ 1OWTTE
Part (c)(ii)
1 million years is about 175 half-lives, so the fraction of carbon-14 remaining is negligible✓ 1
the count rate cannot be distinguished from the background count rate✓ 1
Part (d)
few decays are recorded, so the random variation in the count is a large fraction of it / the count is comparable with the background✓ 1OWTTE

Answers: (b)(i) 1.71 × 104 years  ·  (b)(ii) 7.5 × 103 years (the remaining parts are explanations — see the table above)

113E-2-34Random & spontaneous decay Paper 2Easy11 marks

A student places a sealed strontium-90 source in front of a Geiger–Müller (GM) tube and records the number of counts in each of twelve successive 10 s intervals. Strontium-90 has a half-life of 29 years. With the source removed, the background count is 4 counts per 10 s on average.

The graph shows the number of counts recorded in each interval.

Interval123456789101112
Counts in 10 s318296341305287329312300334291322309
Graph of the number of counts recorded in each 10 s interval against interval number from 1 to 12: points at 318, 296, 341, 305, 287, 329, 312, 300, 334, 291, 322, 309.024681012interval number260280300320340360counts in 10 s
Graph drawn to scale
(a)

Explain what is meant by the random nature of radioactive decay, referring to the data.

(2)
(b)

Calculate the mean count rate due to the source, in s−1.

(2)
(c)

Draw, on the graph, a line to show the mean number of counts recorded in 10 s.

(1)
(d)

The student claims that the activity of the source must be changing, because the counts vary from 287 to 341. Explain why this claim is incorrect.

(2)
(e)

A second source contains the same number of strontium-90 nuclei, but as strontium metal instead of a compound. State, with a reason, whether its activity is different.

(1)
(f)

Predict the mean number of counts in 10 s if the experiment is repeated with the same source and apparatus 58 years later.

(2)
(g)

Suggest one way of reducing the effect of the random variation on the measured count rate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
it is not possible to predict when a particular nucleus will decay; each nucleus has the same constant chance of decaying in a given time✓ 1
so the number of decays in equal time intervals varies, as the counts do «from 287 to 341»✓ 1MP2 must refer to the data.
Part (b)
mean count = 3744/12 = 312 counts in 10 s✓ 1
(312 − 4)/10 = 30.8 s−1✓ 1Award [2] for CNA. Award [1 max] for 31.2 s−1 «background not subtracted».
Part (c)
horizontal straight line at 312 counts across the range of the data✓ 1Accept 310–314. Allow ECF from (b) for the candidate's mean count.
Part (d)
the half-life «29 years» is very long compared with the 2 minutes of the experiment, so the activity is effectively constant✓ 1
the variation is due to the random nature of the decay «random fluctuation about the mean»✓ 1
Part (e)
no difference: decay is spontaneous, so it is not affected by chemical combination✓ 1Both the answer and the reason are needed.
Part (f)
58 years = 2 half-lives, so the count due to the source = (312 − 4)/4 = 77✓ 1
77 + 4 = 81 counts✓ 1Award [2] for CNA. Award [1 max] for 78 «background not removed before halving». Allow ECF from (b).
Part (g)
count for a longer time / take the mean of many more intervals✓ 1Accept any one. Do not accept 'repeat the experiment' without more counts.

Answers: (b) 30.8 s−1  ·  (f) 81 counts in 10 s (the remaining parts are explanations — see the table above)

114E-2-41Half-life & activity Paper 2Easy12 marks

Phosphorus-32, 3215P, is used in biology laboratories to label DNA. It decays by β− emission to sulfur-32, which is stable. Liquid waste containing phosphorus-32 is kept in a store until its activity has fallen to less than 5 % of its initial value; it may then be disposed of.

A container of waste initially holds 1.2 × 1015 phosphorus-32 nuclei and no sulfur-32. The graph shows how the number of phosphorus-32 nuclei in the container varies with time.

Graph of the number of phosphorus-32 nuclei, in units of 10^14, against time from 0 to 80 days: a decay curve starting at 12 × 10^14 and halving every 14.3 days.01020304050607080time / days02468101214number of nuclei / 10¹⁴phosphorus-32
Graph drawn to scale
(a)

Complete the equation for the decay of phosphorus-32: 3215P → … S + … + …

(2)
(b)

Use the graph to show that the half-life of phosphorus-32 is about 14 days.

(1)
(c)

Draw, on the graph, a line to show how the number of sulfur-32 nuclei in the container varies with time.

(3)
(d)

The laboratory manager plans to store the waste for 8 weeks before disposal. Determine whether this is long enough.

(3)
(e)

The container is kept inside a box made of perspex (acrylic) 10 mm thick, rather than inside a lead box. Explain why the perspex is sufficient to protect the laboratory staff.

(2)
(f)

Outline why the sulfur-32 formed in the container does not contribute to the activity of the waste.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
3216S✓ 1Both numbers needed.
0−1e «β−» AND ν̄ «antineutrino»✓ 1Do not accept a neutrino. Accept 0−1β.
Part (b)
the number falls from 12 × 1014 to 6 × 1014 «or 6 to 3 × 1014» in about 14 days✓ 1Must see a pair of values read from the graph «or lines drawn on it».
Part (c)
line starts at zero at t = 0 and rises with a gradient that decreases with time✓ 1
crosses the phosphorus-32 curve at 6 × 1014 at t ≈ 14 days✓ 1
at every time the two values add up to 12 × 1014 «e.g. 9 × 1014 at 28.6 days, 10.5 × 1014 at 42.9 days», approaching but never exceeding 12 × 1014✓ 1Within half a large square.
Part (d)
8 weeks = 56 days ≈ 4 half-lives «56/14 = 4.0, or 3.9 using 14.3 days»✓ 1
after 4 half-lives the activity is 1/16 = 6.25 % of its initial value, «slightly more after 3.9 half-lives»✓ 1Allow a value read from the graph: ≈ 0.79 × 1014 nuclei ≈ 6.6 %.
6.25 % > 5 %, so 8 weeks is not long enough «5 half-lives, about 72 days, are needed»✓ 1The conclusion must be consistent with the working.
Part (e)
phosphorus-32 emits only β− particles «no gamma rays»✓ 1
beta particles have a range of only a few millimetres in perspex, so they are all absorbed by the 10 mm wall✓ 1OWTTE
Part (f)
sulfur-32 is stable: its nuclei do not decay✓ 1

Answers: (b) ≈ 14 days  ·  (d) 56 days ≈ 4 half-lives, about 6.25 % remains — not long enough (≈ 72 days needed) (the remaining parts are explanations — see the table above)

115E-2-42Penetration & ionization Paper 2Easy13 marks

Welds in steel pipelines are checked by gamma radiography. A small sealed source of iridium-192, 19277Ir, is placed on one side of the weld and a detector plate on the other; flaws in the weld show up as regions that receive more radiation. Iridium-192 decays mainly by β− emission to platinum (Pt) and emits gamma rays. Its half-life is 74 days.

The time for which the detector must be exposed is inversely proportional to the intensity of the gamma radiation at the detector. With a new source at a distance of 0.50 m from the detector, the exposure time is 40 s. Treat the source as a point source and ignore absorption in air.

Axes for the exposure time needed at 0.50 m, from 0 to 350 s, against the age of the source from 0 to 250 days; the point for a new source, 40 s at 0 days, is plotted.050100150200250age of source / days050100150200250300350exposure time / s
Graph axes drawn to scale
(a)

State the number of protons and the number of neutrons in the platinum nucleus formed in the decay.

(1)
(b)

The steel wall of the pipe is 25 mm thick. Explain why a gamma source, rather than an alpha or a beta source, is used.

(3)
(c)

The same source is used for several months.

(i)

Calculate the exposure time needed at 0.50 m when the source is 222 days old.

(2)
(ii)

The crew must keep each exposure below 120 s. Determine whether this is possible with the 222-day-old source if the distance from the source to the detector is reduced to 0.30 m.

(2)
(iii)

Sketch, on the axes, a graph to show how the exposure time needed at 0.50 m varies with the age of the source over the first 222 days.

(2)
(d)

During an exposure the radiographers stand 4.0 m from the source. Calculate the ratio intensity of the gamma radiation at 4.0 m / intensity at 0.50 m.

(1)
(e)

Suggest one advantage and one disadvantage of using a gamma source with a much longer half-life, such as caesium-137 (30 years).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
78 protons AND 192 − 78 = 114 neutrons✓ 1Both needed.
Part (b)
alpha particles are stopped by a sheet of paper / a few cm of air, so none would pass through the steel✓ 1
beta particles are stopped by a few mm of metal, so none would pass through 25 mm of steel✓ 1
gamma rays are the most penetrating: a measurable fraction passes through 25 mm of steel, and the fraction transmitted depends on the thickness, so flaws «thinner regions» show up✓ 1OWTTE
Part (c)(i)
222 days = 3 half-lives, so the intensity at 0.50 m falls to 1/8 of its initial value✓ 1
exposure time = 8 × 40 = 320 s✓ 1Award [2] for CNA. Award [0] for 40/8 = 5 s.
Part (c)(ii)
intensity ∝ 1/d2, so the exposure time ∝ d2: 320 × (0.30/0.50)2 = 115 s✓ 1Allow ECF from (c)(i).
115 s < 120 s, so the exposure limit can be met✓ 1The conclusion must be consistent with the calculation. Award [0] for a calculation using (0.30/0.50) not squared.
Part (c)(iii)
curve through «0, 40 s», «74 days, 80 s», «148 days, 160 s» and «222 days, 320 s»✓ 1Within half a large square. Allow ECF from (c)(i).
smooth curve whose gradient increases with time «doubling every 74 days»✓ 1Do not accept a straight line.
Part (d)
(0.50/4.0)2 = 1/64 «= 0.016»✓ 1
Part (e)
advantage: the activity «and the exposure time» stays almost constant, so the source need not be replaced for many years✓ 1
disadvantage: a lost or discarded source remains hazardous for a very long time «centuries», so storage and disposal are more difficult✓ 1Accept any one valid advantage and one valid disadvantage.

Answers: (a) 78 protons, 114 neutrons  ·  (c)(i) 320 s  ·  (c)(ii) 115 s — yes, below 120 s  ·  (d) 1/64 ≈ 0.016 (the remaining parts are explanations — see the table above)

116E-2-64Beta decay & the neutrino Paper 2Easy11 marks

Nitrogen-13, 137N, is made in a small cyclotron inside a hospital and is used, in the form of ammonia, to image the blood flow in the heart muscle of a patient. Nitrogen-13 decays by β+ emission to an isotope of carbon (C) with a half-life of 10.0 minutes. The nitrogen-13 is made by bombarding oxygen-16 nuclei with protons from the cyclotron.

Each positron travels about 1 mm through the tissue and then meets an electron. The positron and the electron, both almost at rest, are annihilated and two gamma-ray photons of equal energy are produced. The photons are detected by a ring of detectors around the patient.

(1 u = 931.5 MeV c−2, mass of an electron = 9.11 × 10−31 kg, c = 3.00 × 108 m s−1)

ParticleMass / u
nitrogen-13 nucleus13.001899
carbon-13 nucleus13.000063
positron (or electron)0.000549
(a)

The nuclear equation for the production of nitrogen-13 is 168O + 11p → 137N + X. Identify X.

(1)
(b)

Complete the equation for the decay of nitrogen-13. 137N → ……C + …… + ……

(2)
(c)
(i)

Show that the energy released in the decay of one nitrogen-13 nucleus is about 1.2 MeV.

(1)
(ii)

Calculate, in J, the energy of each photon produced when a positron and an electron are annihilated.

(2)
(d)

Explain why the photons, but not the positrons, can be detected by the ring of detectors outside the body.

(2)
(e)

A typical injection has an activity of 740 MBq. A supplier whose cyclotron is 1.5 hours' drive away offers to deliver nitrogen-13 to the hospital. Suggest, with a calculation, why the hospital makes its own nitrogen-13 instead.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
X is an alpha particle, 42He «nucleon number 16 + 1 − 13 = 4, proton number 8 + 1 − 7 = 2»✓ 1Accept 42α or 'helium-4 nucleus'. Both numbers needed if the symbol is given.
Part (b)
136C✓ 1Both numbers needed.
0+1e «β+» AND ν «neutrino»✓ 1Accept 0+1β or e+. Do not accept an antineutrino.
Part (c)(i)
Δm = 13.001899 − 13.000063 − 0.000549 = 0.001287 u; E = 0.001287 × 931.5 = 1.199 MeV✓ 1Must see full substitution OR answer to at least 3 s.f. The positron mass must be subtracted because nuclear masses are used.
Part (c)(ii)
the energy of the two photons equals the rest energy of the positron and the electron, 2mec2, shared equally, so each photon has E = mec2✓ 1Accept direct use of E = mc2 with the mass of one electron.
E = 9.11 × 10−31 × (3.00 × 108)2 = 8.2 × 10−14 J✓ 1Award [2] for CNA. Award [1 max] for 1.6 × 10−13 J «total energy not shared». Accept 0.51 MeV converted to J.
Part (d)
positrons are charged and ionize the tissue continually along their path, so they lose their energy and are annihilated within about 1 mm «they never leave the body»✓ 1
gamma photons are uncharged and only weakly ionizing / the most penetrating radiation, so most pass through the tissue to the detectors✓ 1OWTTE
Part (e)
1.5 h = 90 min = 9 half-lives✓ 1
the activity falls by a factor of 29 = 512, so about 379 GBq would have to be dispatched to give 740 MBq on arrival✓ 1Accept 3.8 × 105 MBq. Allow ECF from an incorrect number of half-lives.
almost all of the nitrogen-13 decays on the journey / far too large an activity to produce and transport safely, so it must be made on site and used at once✓ 1MP3 needs a judgement linked to the calculation. OWTTE

Answers: (c)(i) 1.20 MeV  ·  (c)(ii) 8.2 × 10−14 J  ·  (e) ≈ 379 GBq at dispatch (the remaining parts are explanations — see the table above)

117E-2-65Random & spontaneous decay Paper 2Easy13 marks

A hardware random-number generator, used to make encryption keys, contains a sealed source of promethium-147, 14761Pm. Promethium-147 decays by β− emission to samarium (Sm) with a half-life of 2.6 years; it emits almost no gamma rays. The source and a detector are sealed together inside a metal case 1 mm thick. The detector registers the beta particles, and the electronics measure the time interval between each count and the next.

The intervals are compared in pairs. If the first interval of a pair is shorter than the second, the device outputs the binary digit 0; otherwise it outputs 1. Each digit therefore uses two intervals.

When the device is new the mean count rate is 2400 s−1. The background count rate is 0.5 s−1.

(a)

Complete the equation for the decay of promethium-147. 14761Pm → ……Sm + …… + ……

(2)
(b)

Explain, with reference to the nature of radioactive decay, why the next digit produced by the device cannot be predicted.

(2)
(c)

Another type of generator uses the electrical noise in a resistor, which depends on the temperature of the resistor. Outline one advantage of the radioactive generator.

(1)
(d)

The beta particles emitted by promethium-147 have kinetic energies of less than 0.23 MeV. Explain why the sealed device is safe to handle.

(2)
(e)
(i)

Show that the device produces about 1200 digits per second when it is new.

(1)
(ii)

The specification requires at least 250 digits per second. Determine whether a device that is 7.8 years old meets the specification.

(3)
(f)

A student argues that background radiation will spoil the randomness of the output. Discuss this argument.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
14762Sm✓ 1Both numbers needed.
0−1e «β−» AND ν̄ «antineutrino»✓ 1Accept 0−1β. Do not accept a neutrino.
Part (b)
radioactive decay is random: it is impossible to predict when any particular nucleus will decay «each nucleus has the same constant chance of decaying in a given time»✓ 1
so the length of each interval between counts, and therefore which interval of a pair is shorter, cannot be predicted✓ 1MP2 must link the randomness to the intervals or digits.
Part (c)
decay is spontaneous: its rate is not affected by temperature «pressure or other external conditions», so the output cannot be biased by heating or cooling the device✓ 1OWTTE. Do not accept 'it is random' alone.
Part (d)
beta particles of such low energy have a very short range: they are all absorbed by the detector and the 1 mm metal case✓ 1
almost no gamma rays, the only radiation that could penetrate the case, are emitted, so almost no radiation leaves the device✓ 1OWTTE
Part (e)(i)
2400 counts per second give 2400 intervals per second; two intervals per digit gives 2400/2 = 1200 digits per second✓ 1Must see the division by 2 with a reason.
Part (e)(ii)
7.8 years = 3 half-lives, so the count rate falls by a factor of 23 = 8✓ 1
count rate = 2400/8 = 300 s−1, so the device produces 150 digits per second✓ 1Allow ECF from (e)(i).
150 < 250, so the device does not meet the specification✓ 1MP3 needs a comparison with 250 and a conclusion consistent with the candidate's value.
Part (f)
background radiation also arrives at random «it comes from random decays and cosmic rays», so the extra counts are just as unpredictable and the output is still random✓ 1
the background adds only 0.5 s−1 to 2400 s−1 «about 0.02 %», so its effect on the digit rate is negligible: the argument is not valid✓ 1A judgement is needed for MP2. OWTTE

Answers: (e)(ii) 150 digits per second < 250 (the remaining parts are explanations — see the table above)

118E-2-66Isotopes Paper 2Easy10 marks

Lutetium-177, 17771Lu, is used to treat some cancers. It is attached to a drug that collects on tumour cells. Lutetium-177 decays by β− emission to hafnium (Hf) with a half-life of 6.6 days. The beta particles have a range of up to about 2 mm in tissue.

Lutetium-177 is made in a nuclear reactor, where nuclei of lutetium-176 each absorb one neutron.

(a)
(i)

State the number of protons and the number of neutrons in a nucleus of lutetium-177.

(1)
(ii)

Explain why the nucleus formed when a lutetium-176 nucleus absorbs a neutron is an isotope of lutetium.

(1)
(b)

Complete the equation for the decay of lutetium-177. 17771Lu → ……Hf + …… + ……

(2)
(c)

Outline the change that takes place inside the nucleus during β− decay.

(1)
(d)

Alpha emitters, whose alpha particles have a range of less than 0.1 mm in tissue, are also used to treat tumours. A tumour is several millimetres across and the drug does not reach every cell in it. Suggest why lutetium-177 may be more effective than an alpha emitter for this tumour.

(2)
(e)

A patient is given an injection of lutetium-177 with an activity of 7.4 GBq. A second injection of the same activity is given 39.6 days later.

(i)

Determine the total activity of lutetium-177 in the patient just after the second injection. Ignore the removal of lutetium from the body.

(2)
(ii)

Suggest why the actual activity is smaller than your answer to (e)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
71 protons AND 106 neutrons✓ 1Both needed.
Part (a)(ii)
the neutron has no charge, so the proton number stays 71 «same element» while the neutron «nucleon» number increases by 1: same proton number, different number of neutrons✓ 1OWTTE
Part (b)
17772Hf✓ 1Both numbers needed.
0−1e «β−» AND ν̄ «antineutrino»✓ 1Accept 0−1β. Do not accept a neutrino.
Part (c)
a neutron changes into a proton «and an electron and an antineutrino are created and emitted»✓ 1Do not accept 'an electron leaves the nucleus' alone.
Part (d)
beta particles travel up to about 2 mm, so they ionize «and damage» cells up to about 2 mm from the cells that have taken up the drug✓ 1
alpha particles travel less than 0.1 mm «a few cells», so cells that the drug has not reached would not be irradiated and could survive✓ 1OWTTE. MP2 needs the comparison with the alpha range.
Part (e)(i)
39.6 days = 6 half-lives, so the first dose has fallen to 7.4/26 = 0.12 GBq✓ 1
total = 7.4 + 0.12 = 7.5 GBq✓ 1Award [2] for CNA. Accept 7.5–7.6 GBq. Award [1 max] for 8.6 GBq «divided by the number of half-lives».
Part (e)(ii)
some of the lutetium «drug» is removed from the body «e.g. by the kidneys in the urine» between the injections✓ 1OWTTE

Answers: (a)(i) 71 and 106  ·  (e)(i) 7.5 GBq (the remaining parts are explanations — see the table above)

119E-2-74Penetration & ionization Paper 2Easy10 marks

Some electrodes used for welding are made of tungsten containing 2 % of thorium oxide. The thorium is thorium-232, 23290Th, which decays by alpha emission to radium-228 and has a half-life of 1.4 × 1010 years. Radium-228 and other decay products, which build up in the electrode, emit beta particles and gamma rays.

Before use, the tip of each electrode is sharpened on a grinding wheel. This produces a fine dust that contains thorium.

Blank axes: number of thorium-232 nuclei in an electrode, from 0 to 1.2 N-nought, against time from 0 to 50 years01020304050time / years00.5 N₀N₀number of Th-232 nuclei
Axes for (c). N0 is the number of thorium-232 nuclei in a new electrode.
(a)
(i)

State the change in the number of protons and the change in the number of neutrons in a nucleus of thorium-232 when it emits an alpha particle.

(1)
(ii)

Radium-228 decays by β− emission to actinium (Ac). State the proton number and the nucleon number of this actinium nucleus.

(1)
(b)

A welder holds a new electrode in a gloved hand. Explain why the alpha particles from the electrode present little hazard to the welder, but the dust from grinding is a serious hazard if it is breathed in.

(3)
(c)

Sketch, on the axes, a graph to show how the number of thorium-232 nuclei in an electrode varies during the first 50 years after it is made. Explain the shape of your graph.

(2)
(d)

A Geiger–Müller tube held close to a new electrode records 412 counts in 5.0 minutes. With the electrode removed, 145 counts are recorded in 5.0 minutes. Calculate the count rate due to the electrode, in s−1.

(2)
(e)

Suggest one precaution, other than wearing gloves, that the welder should take while grinding an electrode.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
number of protons decreases by 2 AND number of neutrons decreases by 2✓ 1Both needed.
Part (a)(ii)
proton number 89 AND nucleon number 228✓ 1Both needed.
Part (b)
alpha particles have a very short range «a few cm in air» and are stopped by the glove «or the dead outer layer of skin», so outside the body they do not reach living cells✓ 1
dust that is breathed in stays in the lungs, so the alpha particles are emitted inside the body, next to living tissue, with nothing to absorb them✓ 1
alpha particles are strongly ionizing, so they deposit all their energy in a small volume of tissue and damage «kill or mutate» the cells there✓ 1OWTTE
Part (c)
horizontal straight line at N0 across the 50 years✓ 1
50 years is a negligible fraction of the half-life «1.4 × 1010 years», so the number of nuclei decreases by a negligible amount✓ 1MP2 needs the comparison with the half-life.
Part (d)
net count = 412 − 145 = 267✓ 1
count rate = 267/300 = 0.89 s−1✓ 1Award [2] for CNA. Award [1 max] for 1.37 s−1 «background not subtracted» or for 267 counts per 5 minutes.
Part (e)
grind with local extraction «ventilation» that removes the dust / wear a dust mask «respirator» / grind with a dedicated wheel and dispose of the dust safely✓ 1Accept any one sensible precaution against breathing in or spreading the dust.

Answers: (a)(ii) 89 and 228  ·  (d) 0.89 s−1 (the remaining parts are explanations — see the table above)

120IBP2S-038-04-11Isotopes Paper 2Easy7 marks

Experimental fusion reactors use deuterium (\(^{2}_{1}\)H) and tritium (\(^{3}_{1}\)H) as fuel. Tritium is radioactive. It decays by β− emission to helium-3 with a half-life of 12.3 years.

(a)

Deuterium and tritium are isotopes of hydrogen. Outline what is meant by isotopes.

(2)
(b)

State the number of neutrons in a tritium nucleus.

(1)
(c)

Write down the nuclear equation for the β− decay of tritium.

(2)
(d)

A sealed container of tritium is stored for 36.9 years. Calculate the percentage of the original tritium that remains undecayed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Nuclei with the same number of protons / same proton number / of the same element . Have different numbers of neutrons / different nucleon numbers .✓ 2
Part (b)
2✓ 1
Part (c)
\(^{3}_{1}\)H → \(^{3}_{2}\)He with correct nucleon and proton numbers . + \(^{0}_{-1}\)e (accept \(^{0}_{-1}β\) or β−) + electron antineutrino \(\bar{\nu}_e\) (both required; a neutrino ν without the bar does not earn this mark) .✓ 2
Part (d)
36.9 / 12.3 = 3 half-lives . (1/2)3 = 0.125, so 12.5 % remains .✓ 2

Common error: In β⁻ decay the proton number rises by one, so the product is a different element (helium), not another isotope of hydrogen. The particle emitted with the electron is an antineutrino, not a neutrino.

121IBP2S-038-02-11Background radiation Paper 2Easy6 marks

A sports physiotherapy clinic is checking a Geiger–Müller counter before using it with a small calibration source. With no source present, the counter records the following counts in six consecutive one-minute intervals:

Interval: 1, 2, 3, 4, 5, 6
Counts: 18, 22, 19, 25, 21, 21

(a)

State two natural sources of background radiation.

(2)
(b)

Calculate the mean background count rate in counts min−1.

(1)
(c)

The calibration source is placed near the tube, and 386 counts are recorded in 2.0 minutes. Determine the count rate due to the source alone.

(2)
(d)

Outline why the background count was measured over several intervals rather than just one.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any two from: radon gas from the ground or buildings; cosmic radiation; radioactive rocks or soil (e.g. granite); radioactive isotopes in food or the body (e.g. K-40, C-14).—
First valid source✓ 1
Second valid source✓ 1
Part (b)
Mean = (18 + 22 + 19 + 25 + 21 + 21)/6 = 126/6 = 21 counts min−1✓ 1
Part (c)
Measured rate = 386/2.0 = 193 counts min−1✓ 1
Corrected rate = 193 − 21 = 172 counts min−1. Allow an error carried forward from the mean background rate calculated earlier.✓ 1
Part (d)
Radioactive decay is random, so counts fluctuate from interval to interval. Averaging several intervals gives a more reliable value with a smaller relative uncertainty.✓ 1

Exam tip: Always check the time interval. Here the source count was recorded over 2 minutes, so convert it to a per-minute rate before subtracting the background.

122IBP2S-038-02-12The strong nuclear force Paper 2Easy6 marks

A helium-4 nucleus contains two protons and two neutrons. In this nucleus the centres of the two protons are about 1.6 fm apart.
Use e = 1.60×10−19 C and k = 8.99×109 N m2 C−2.

(a)

State two properties of the strong nuclear force.

(2)
(b)

Calculate the electrostatic force between the two protons.

(2)
(c)

Explain why the helium-4 nucleus does not fly apart, despite the electrostatic force between the protons that you calculated.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any two from: acts between all nucleons (charge-independent); very short range, about 10−15 m (a few fm); attractive at separations of about 1 fm; repulsive at separations below about 0.5 fm; much stronger than the electrostatic force at nuclear separations. One mark for each property.✓ 2
Part (b)
F = ke2/r2 = 8.99×109 × (1.60×10−19)2 / (1.6×10−15)2✓ 1
F ≈ 90 N (repulsive).✓ 1
Part (c)
The strong nuclear force between nucleons at about 1.6 fm is attractive and much larger than the electrostatic repulsion.✓ 1
The neutrons add strong-force attraction without adding any electrostatic repulsion, so the net force holds the nucleons together.✓ 1

Exam tip: A force of about 90 N on a single proton is enormous. Quote this value when you explain why the nucleus stays together: it shows that the strong force must be stronger still.

123IBP2S-038-05-11Mass defect & binding energy Paper 2Easy7 marks

Some designs of molten-salt reactor use a coolant rich in lithium-7, 73Li.
Data:
mass of a lithium-7 nucleus = 7.014 36 u
mass of a proton = 1.007 276 u
mass of a neutron = 1.008 665 u
1 u = 931.5 MeV c−2

(a)

State the number of protons and the number of neutrons in a lithium-7 nucleus.

(1)
(b)

Explain why the mass of the nucleus is less than the total mass of its separate nucleons.

(2)
(c)

Calculate the mass defect of lithium-7 in u.

(2)
(d)

Determine the binding energy per nucleon of lithium-7 in MeV.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
3 protons and 4 neutrons✓ 1
Part (b)
Energy must be supplied (work done against the strong nuclear force) to separate the nucleus into free nucleons, so the separated nucleons have more energy. Equivalently, energy is released when the nucleons bind.✓ 1
By E = mc2, this energy difference corresponds to a mass difference, so the bound nucleus has less mass than its separate nucleons.✓ 1
Part (c)
Total mass of nucleons = 3 × 1.007 276 + 4 × 1.008 665 = 7.056 488 u✓ 1
Mass defect = 7.056 488 − 7.014 36 = 0.0421 u✓ 1
Part (d)
Binding energy = 0.042 13 × 931.5 = 39.2 MeV✓ 1
Binding energy per nucleon = 39.2 / 7 = 5.61 MeV (accept 5.6 MeV)✓ 1
Allow error carried forward from the mass defect calculated in the previous part.—

Exam tip: Keep at least five decimal places when you subtract masses in u. The mass defect is a small difference between two large numbers, so rounding early ruins it.

124IBP2S-038-05-12Choosing an isotope Paper 2Easy7 marks

A hospital wants a tracer to image how well a patient's kidneys are working. The images will be made with a gamma camera outside the body. The isotopes available are shown in the table.

Isotope — main emission — half-life
technetium-99m — gamma — 6.0 h
strontium-90 — beta — 29 years
polonium-210 — alpha — 138 days
caesium-137 — beta and gamma — 30 years

(a)

State which isotope is most suitable and explain your choice.

(3)
(b)

Outline why polonium-210 would be unsuitable.

(2)
(c)

A dose of the chosen isotope is injected. Calculate the fraction of the original activity that remains 24 hours later.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Technetium-99m✓ 1
It emits only gamma rays, which pass out of the body to reach the camera and are weakly ionizing, so they cause little tissue damage.✓ 1
Its half-life of a few hours is long enough for the scan but short enough that the activity, and so the dose, falls quickly afterwards.✓ 1
Part (b)
Alpha particles have a very short range (well under 1 mm in tissue; only a few cm in air), so they would be absorbed inside the body and would not reach the external camera.✓ 1
Being strongly ionizing, they would cause severe damage to the tissue around the source.✓ 1
Part (c)
24 h ÷ 6.0 h = 4 half-lives✓ 1
Fraction = (1/2)4 = 1/16 (= 0.0625)✓ 1

Exam tip: A full justification links each property of the isotope (type of emission, half-life) to a specific need of the procedure.

125IBP2S-038-03-12Beta decay & the neutrino Paper 2Easy5 marks

In a PET scanner, a positron emitted by a tracer meets an electron in the patient's tissue. Both particles are effectively at rest when they annihilate, producing gamma-ray photons.
Mass of electron = mass of positron = 9.11×10−31 kg; c = 3.00×108 m s−1; h = 6.63×10−34 J s; e = 1.60×10−19 C.

(a)

Outline why two photons, rather than one, are produced.

(1)
(b)

Show that the energy of each photon is about 0.51 MeV.

(2)
(c)

Calculate the wavelength of each photon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The initial momentum is zero, so two photons with equal and opposite momenta (travelling in opposite directions) are needed for momentum to be conserved .✓ 1
Part (b)
The total energy released is the rest energy of both particles, 2mc2, shared equally between the two photons, so each photon has E = mc2 .✓ 1
E = 9.11×10−31 × (3.00×108)2 = 8.20×10−14 J, and 8.20×10−14 / 1.60×10−19 = 5.12×105 eV ≈ 0.51 MeV .✓ 1
Part (c)
λ = hc/E = (6.63×10−34 × 3.00×108) / 8.20×10−14 .✓ 1
λ = 2.43×10−12 m . Allow ECF from (b).✓ 1

Common error: Each photon gets the rest energy of one particle, not the combined rest energy of both. Using 2mc² for one photon doubles the energy and halves the wavelength.

126IBP2S-038-01-11Choosing an isotope Paper 2Easy5 marks

Iodine-131 is a beta-emitting isotope with a half-life of 8.0 days. It is used to treat an overactive thyroid gland. A patient is given a dose with an initial activity of 4.0 GBq. Ignore any iodine removed from the body by biological processes.

(a)

State what is meant by the half-life of a radioactive isotope.

(1)
(b)

Determine the time taken for the activity of the dose to fall to 0.50 GBq.

(2)
(c)

Outline why a beta emitter, rather than a gamma emitter, is suitable for destroying thyroid tissue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The time taken for the number of undecayed nuclei (or the activity) of a sample to fall to half its initial value✓ 1
Part (b)
4.0/0.50 = 8 = 23, so three half-lives✓ 1
t = 3 × 8.0 = 24 days✓ 1
Part (c)
Beta particles have a short range in tissue (a few mm), so their energy is deposited within the thyroid✓ 1
Beta particles are more strongly ionising than gamma rays, so they damage nearby cells; gamma rays would mostly pass out of the body and irradiate other tissue✓ 1

Exam tip: When the ratio of activities is a power of 2, count the halvings directly. This avoids logarithms and reduces arithmetic slips.

127IBP2S-038-01-12Mass defect & binding energy Paper 2Easy6 marks

Helium-3 (\(^3_2\)He) is a proposed fuel for future fusion reactors and may be mined from the lunar surface. Use the following data:
mass of \(^3_2\)He nucleus = 3.014932 u
mass of proton = 1.007276 u
mass of neutron = 1.008665 u
1 u = 931.5 MeV c−2

(a)

State what is meant by the mass defect of a nucleus.

(1)
(b)

Calculate the mass defect of a \(^3_2\)He nucleus.

(2)
(c)

Calculate the binding energy per nucleon of \(^3_2\)He, in MeV.

(2)
(d)

The binding energy per nucleon of \(^4_2\)He is 7.07 MeV. State, with a reason, which of the two helium isotopes is more stable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The difference between the total mass of the separated nucleons and the mass of the nucleus✓ 1
Part (b)
Total mass of nucleons = 2(1.007276) + 1.008665 = 3.023217 u✓ 1
Δm = 3.023217 − 3.014932 = 0.008285 u✓ 1
Part (c)
Binding energy = 0.008285 × 931.5 = 7.72 MeV✓ 1
Binding energy per nucleon = 7.72/3 = 2.57 MeV✓ 1
ECF from (b)—
Part (d)
\(^4_2\)He is more stable, because it has the greater binding energy per nucleon✓ 1

Common error: Use the mass of the NUCLEUS, not the atomic mass, when the nucleon masses are given for protons and neutrons only. Mixing the two gives a mass defect that is out by the electron masses.

128IBP2S-038-00-12Penetration & ionization Paper 2Easy5 marks

Engineers check welds in steel gas pipelines by industrial radiography. A sealed iridium-192 source, which emits gamma radiation, is placed on one side of a weld. A detector film is placed on the other side, beyond 2.5 cm of steel.

iridium-192 sourcetechnician at 2.0 mtechnician at 8.0 m2.0 m8.0 m
Diagram not to scale (the technician moves from 2.0 m to 8.0 m)
(a)

State why gamma radiation, rather than alpha or beta radiation, is used.

(1)
(b)

Explain why gamma radiation is much less ionizing than alpha radiation.

(2)
(c)

The source can be treated as a point source with negligible absorption in air. A technician moves from 2.0 m to 8.0 m away from the source. Calculate the factor by which the intensity of the gamma radiation received by the technician decreases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Only gamma radiation is penetrating enough to pass through 2.5 cm of steel and reach the film; alpha and beta would be absorbed by the steel✓ 1
Part (b)
Alpha particles are charged (+2e), massive and relatively slow, so they exert strong electric forces on atomic electrons all along their path and ionize densely✓ 1
Gamma photons are uncharged, so they interact with atoms only occasionally and produce far fewer ion pairs per unit length of path✓ 1
Part (c)
Uses the inverse-square law, I ∝ 1/r2✓ 1
Factor = (8.0/2.0)2 = 16✓ 1

Common error: Quadrupling the distance does not reduce the intensity by a factor of 4. The inverse-square law gives a factor of 4² = 16.

129E-2-07Background radiation Paper 2Medium11 marks

Radon-222, 22286Rn, is a radioactive gas produced in rocks and soil by the decay of radium-226, which has a half-life of 1600 years. Radon seeps out of the ground into buildings and is the largest natural source of background radiation for many people. Radon-222 decays by alpha emission to polonium-218 with a half-life of 3.8 days.

To test the air in the basement of a house, a canister is filled with basement air and sealed. The canister reaches a laboratory 7.6 days later. There it is placed next to a gamma-ray detector, which detects the gamma rays emitted by the short-lived decay products of the radon-222, so that the count rate is proportional to the activity of the radon-222 in the canister. 1440 counts are recorded in 60.0 minutes. With the canister removed, 900 counts are recorded in another 60.0 minutes.

(a)

Radon and background radiation.

(i)

State the number of protons and the number of neutrons in a nucleus of polonium-218.

(1)
(ii)

State one source of background radiation other than radon.

(1)
(b)

The laboratory measurement.

(i)

Show that the count rate due to the radon-222 in the canister, when it is measured, is about 0.15 s−1.

(1)
(ii)

Explain why the laboratory needs to know the date on which the canister was filled.

(2)
(iii)

Calculate the count rate due to the radon-222 that would have been recorded if the canister had been measured as soon as it was sealed.

(1)
(c)

Explain why the laboratory records counts for 60.0 minutes rather than for 1 minute.

(2)
(d)

The householder argues that, because the half-life of radon-222 is only 3.8 days, the radon in the basement will fall to a negligible level within a few weeks without any action. Evaluate this argument.

(2)
(e)

Suggest why the radon concentration in a basement is usually much higher than in the open air.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
84 protons and 218 − 84 = 134 neutrons✓ 1Both values needed.
Part (a)(ii)
Any one of: cosmic rays / radioactive isotopes in rocks, soil or building materials «other than radon» / potassium-40 or carbon-14 in food and in the body / medical X-rays / fallout from nuclear weapons tests✓ 1Do not accept 'the Sun' alone.
Part (b)(i)
Net count rate = (1440 − 900)/3600 = 0.150 s−1✓ 1Must see the background subtracted OR answer to 3 s.f. Accept 9.0 counts per minute.
Part (b)(ii)
The half-life of radon-222 «3.8 days» is comparable with the time between filling and measurement, so the activity in the canister falls significantly «by a factor of 4 in 7.6 days» before it is measured✓ 1
The measured value must be corrected back to the time of filling to find the level in the basement air✓ 1OWTTE
Part (b)(iii)
7.6 days = 2 half-lives, so 22 × 0.150 = 0.60 s−1✓ 1Allow ECF from (b)(i). Award [0] for 0.30 s−1 «one half-life» or 0.038 s−1 «decay applied the wrong way».
Part (c)
Radioactive decay is random, so the number of counts recorded in a given time fluctuates✓ 1
The net count «540» is smaller than the background count «900» and both fluctuate; recording more counts makes the fractional «percentage» uncertainty in the net count rate smaller✓ 1OWTTE
Part (d)
The radon in a sealed sample would fall to 1/16 of its activity in about 15 days, so the argument is true for radon that is already in the basement✓ 1
But radium-226 in the ground «half-life 1600 years» keeps producing radon at an almost constant rate, which keeps seeping in; the concentration stays about the same, so the argument is not valid «ventilation or sealing is needed»✓ 1MP2 is for the continuous supply and a judgement.
Part (e)
The basement is enclosed and in contact with the ground, so the radon builds up instead of mixing into a large volume of air as it does outdoors✓ 1OWTTE

Answers: (a)(i) 84 protons, 134 neutrons  ·  (b)(i) 0.150 s−1  ·  (b)(iii) 0.60 s−1 (the remaining parts are explanations — see the table above)

130E-2-08The binding energy curve Paper 2Medium11 marks

The graph shows the variation of the binding energy per nucleon with nucleon number A for the most stable nuclides.

Binding energies per nucleon: 32He 2.57 MeV, 42He 7.07 MeV.

Binding energy per nucleon in MeV (0 to 9) against nucleon number A (0 to 250) for the most stable nuclides: a steep rise through 2H, 4He (7.07) and 12C, a broad maximum of about 8.8 MeV near A = 56–62, then a slow fall to about 7.6 MeV at A = 235.050100150200250nucleon number A0123456789binding energy per nucleon / MeV⁴He⁵⁶Fe²³⁵U²H
Graph drawn to scale
(a)

The curve.

(i)

Identify, using the graph, the nucleon number of the most tightly bound nuclides.

(1)
(ii)

Explain why energy is released in a nuclear reaction in which the total binding energy of the products is greater than that of the reactants.

(2)
(b)

In one fission, a nucleus of uranium-236 splits into two fragments with nucleon numbers 96 and 138, and two neutrons.

(i)

Draw, on the graph, arrows to show how the binding energy per nucleon changes in this fission.

(1)
(ii)

Estimate, using the graph, the mass defect of the uranium-236 nucleus. Give your answer in u. (1 u = 931.5 MeV c−2)

(2)
(c)

In the core of the Sun, one of the reactions that forms helium-4 is 32He + 32He → 42He + 2 11H.

(i)

Show that about 13 MeV is released in this reaction.

(1)
(ii)

Explain, with reference to the graph, why energy can be released by the fusion of very light nuclei and by the fission of very heavy nuclei, but not by the fusion of two nuclei heavier than iron.

(3)
(iii)

The core of a very massive star eventually consists mainly of iron. Suggest why the core then collapses.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
A ≈ 56–62 «iron and nickel»✓ 1Accept any value from 55 to 65.
Part (a)(ii)
Binding energy is the energy released when a nucleus forms from separate nucleons, so a larger total binding energy means the products have less energy «and mass» than the reactants✓ 1
The difference is released «as kinetic energy of the products and photons»; by E = mc2 the total mass of the products is less than that of the reactants✓ 1OWTTE
Part (b)(i)
Arrows from the curve at A ≈ 236 to the curve at A ≈ 96 and at A ≈ 138, both pointing upwards «to a higher binding energy per nucleon»✓ 1Both arrows needed.
Part (b)(ii)
Binding energy of uranium-236 ≈ 236 × 7.6 ≈ 1.79 × 103 MeV✓ 1Accept 7.5–7.7 MeV per nucleon read from the graph.
Δm = 1.79 × 103/931.5 ≈ 1.9 u✓ 1Accept 1.90–1.95 u. Allow ECF from the value read from the graph. Award [1 max] for 1.79 × 103 u «no conversion» or 7.6/931.5 = 8.2 × 10−3 u «binding energy per nucleon used».
Part (c)(i)
4 × 7.07 − 2 × 3 × 2.57 = 28.28 − 15.42 = 12.86 MeV✓ 1Must see full substitution OR answer to 3 s.f. The 11H nuclei «single protons» have no binding energy.
Part (c)(ii)
Light nuclei: the binding energy per nucleon rises steeply with A at small A, so the product of fusion has a greater binding energy per nucleon than the reactants✓ 1
Heavy nuclei: the fragments of fission lie nearer the peak «A ≈ 56–62» than the original nucleus, so they also have a greater binding energy per nucleon✓ 1
Beyond the peak the binding energy per nucleon decreases as A increases, so fusing two nuclei heavier than iron gives a less tightly bound product: energy would have to be supplied rather than released✓ 1OWTTE. Award [1 max] for 'energy is released by moving towards iron' with no reference to the graph.
Part (c)(iii)
Fusion of iron releases no energy, so the core can no longer maintain the pressure that balances the inward gravitational force✓ 1OWTTE

Answers: (a)(i) A ≈ 56–62  ·  (b)(ii) ≈ 1.9 u  ·  (c)(i) 12.86 MeV (the remaining parts are explanations — see the table above)

131E-2-21The strong nuclear force Paper 2Medium12 marks

The nucleus of deuterium, 21H, consists of one proton and one neutron. The table gives some masses.

(1 u = 931.5 MeV c−2 = 1.661 × 10−27 kg, 1 MeV = 1.60 × 10−13 J, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, G = 6.67 × 10−11 N m2 kg−2)

ParticleMass / u
proton1.007276
neutron1.008665
deuterium nucleus2.013553
(a)

Show that the mass defect of the deuterium nucleus is about 2.4 × 10−3 u.

(1)
(b)
(i)

Calculate, in MeV, the binding energy of the deuterium nucleus.

(1)
(ii)

The binding energy per nucleon of iron-56 is 8.8 MeV. Compare the binding energy per nucleon of deuterium with that of iron-56, and outline what the comparison shows.

(2)
(c)

A deuterium nucleus can be split into a free proton and a free neutron by absorbing a gamma-ray photon. Determine the maximum wavelength of a photon that can do this. Assume that the proton and the neutron are left with negligible kinetic energy.

(2)
(d)

In the deuterium nucleus the centres of the proton and the neutron are about 2 × 10−15 m apart.

(i)

Calculate the gravitational force between the proton and the neutron.

(1)
(ii)

Estimate the average force needed to pull the proton and the neutron apart, assuming that the binding energy is supplied over a distance of 2 × 10−15 m.

(1)
(iii)

Hence explain why a force other than gravity must hold the nucleons together, and state the nature of this force.

(2)
(e)

In deuterium gas at room temperature, the nuclei of neighbouring molecules are about 3 × 10−9 m apart. Explain why the force that holds the proton and the neutron together does not pull these nuclei together.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δm = 1.007276 + 1.008665 − 2.013553 = 0.002388 u✓ 1Must see full substitution OR answer to 3 s.f. «2.39 × 10−3 u».
Part (b)(i)
E = 0.002388 × 931.5 = 2.22 MeV✓ 1Allow ECF from (a). Accept 2.2 MeV.
Part (b)(ii)
2.22/2 = 1.11 MeV per nucleon, about one eighth of 8.8 MeV✓ 1Allow ECF from (b)(i).
Far less energy is needed per nucleon to separate deuterium into its nucleons: its nucleons are much less tightly bound than those in iron-56✓ 1OWTTE. Do not accept 'deuterium is unstable'.
Part (c)
Photon energy = 2.22 × 1.60 × 10−13 = 3.56 × 10−13 J✓ 1Allow ECF from (b)(i).
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/3.56 × 10−13 = 5.6 × 10−13 m✓ 1Award [2] for CNA. Accept 5.5–5.6 × 10−13 m.
Part (d)(i)
F = 6.67 × 10−11 × 1.67 × 10−27 × 1.68 × 10−27/(2 × 10−15)2 = 4.7 × 10−35 N✓ 1Accept 4.6–4.7 × 10−35 N «1.67 × 10−27 kg used for both».
Part (d)(ii)
F = W/s = 3.56 × 10−13/2 × 10−15 ≈ 178 N «≈ 2 × 102 N»✓ 1Allow ECF from (c). Accept 1.8 × 102 N.
Part (d)(iii)
The gravitational force is about 4 × 1036 times smaller than the force needed, and the neutron is uncharged, so there is no electric attraction: another force must act✓ 1Allow ECF from (d)(i) and (d)(ii). Must see a comparison of the two forces.
The strong nuclear force: an attractive force that acts between nucleons «proton–neutron as well as proton–proton and neutron–neutron»✓ 1Do not accept 'nuclear force' without 'strong'.
Part (e)
The strong nuclear force has a very short range «about 10−15 m», so it is negligible between nuclei 3 × 10−9 m apart✓ 1
The nuclei are both positive and repel electrically; at room temperature they have far too little kinetic energy to come within range of the strong force✓ 1OWTTE

Answers: (a) 0.002388 u  ·  (b)(i) 2.22 MeV  ·  (b)(ii) 1.11 MeV per nucleon  ·  (c) 5.6 × 10−13 m  ·  (d)(i) 4.7 × 10−35 N  ·  (d)(ii) ≈ 180 N (the remaining parts are explanations — see the table above)

132E-2-22Choosing an isotope Paper 2Medium13 marks

A hospital needs a radioactive tracer that is injected into a patient so that the kidneys can be imaged by a gamma camera placed outside the body. Each scan takes about 30 minutes. The table shows four isotopes that could be used.

IsotopeRadiation emittedHalf-life
Palpha6.0 hours
Qgamma6.0 hours
Rgamma2.0 minutes
Sgamma30 years
(a)

Discuss which isotope should be chosen, giving a reason for rejecting each of the other three.

(4)
(b)

The gamma camera records a count rate that is much smaller than the activity of the tracer in the patient.

(i)

State what is meant by the activity of a radioactive sample.

(1)
(ii)

Suggest two reasons why the count rate is much smaller than the activity.

(2)
(c)

A dose of isotope Q has an activity of 1.6 GBq when it is prepared at 06:00. It must be injected while its activity is at least 400 MBq. Determine the latest time at which it can be injected.

(2)
(d)

Isotope Q is also removed from the body by the kidneys, so that, ignoring radioactive decay, the amount in the body would halve every 12 hours. Deduce the fraction of the injected activity that remains in the patient's body 24 hours after the injection.

(3)
(e)

Suggest one precaution, other than shielding, that staff preparing the injections can take to reduce the dose they receive.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q: gamma rays leave the body and reach the camera, and they are only weakly ionizing✓ 1
P: alpha particles are absorbed within the body, so cannot be detected outside it, AND are strongly ionizing, damaging tissue✓ 1
R: its activity falls to a tiny fraction during the 30-minute scan «15 half-lives» / it decays before it can be prepared and injected✓ 1
S: its activity hardly falls, so the patient would remain radioactive «and receive a dose» for many years✓ 1Award [4] only if Q is chosen; MP2–MP4 can be awarded without MP1.
Part (b)(i)
The number of nuclei that decay per unit time «per second»✓ 1Accept 'the rate of decay'. Do not accept 'the number of particles detected per second'.
Part (b)(ii)
Gamma rays are emitted in all directions, so only a small fraction travel towards the camera✓ 1
Some gamma rays are absorbed or scattered by the tissue between the kidneys and the camera OR the camera does not register every gamma ray that reaches it✓ 1Accept any two distinct reasons, one per mark.
Part (c)
1.6 GBq/400 MBq = 4 = 22, so 2 half-lives «= 12 hours»✓ 1
18:00✓ 1Award [2] for CNA.
Part (d)
radioactive decay alone: 24 h = 4 half-lives, factor 1/16✓ 1
removal by the kidneys alone: 2 halvings, factor 1/4✓ 1
both act together, so fraction = 1/16 × 1/4 = 1/64 «≈ 0.016»✓ 1Award [3] for CNA. Do not award MP3 for adding the factors or for 1/32.
Part (e)
minimise the time spent close to the source / increase the distance, e.g. handle it with tongs✓ 1Accept any one.

Answers: (c) 18:00  ·  (d) 1/64 ≈ 0.016 (the remaining parts are explanations — see the table above)

133E-2-24Penetration & ionization Paper 2Medium12 marks

A smoke detector contains a source of americium-241 that emits alpha particles, each with a kinetic energy of 5.5 MeV. The activity of the source is 3.7 × 104 Bq. The alpha particles enter a small chamber of air between two metal plates that are connected to a battery and a sensitive ammeter.

On average, 34 eV of energy is needed to produce one ion pair (a positive ion and a free electron) in air. (e = 1.60 × 10−19 C)

Smoke-detector chamber: an americium-241 source sends alpha particles into the air between two parallel metal plates; the plates are connected in series with a battery and an ammeter.metal platesairamericium-241sourceαAbattery
Diagram NOT accurately drawn
(a)

Americium-241, 24195Am, decays by alpha emission to an isotope of neptunium (Np). State the nucleon number and the proton number of this neptunium nucleus.

(1)
(b)

Show that one alpha particle can produce about 1.6 × 105 ion pairs.

(1)
(c)

Half of the alpha particles emitted by the source enter the chamber, and each gives all of its kinetic energy to the air in the chamber. Determine the current in the ammeter.

(3)
(d)

Explain why the current falls when smoke enters the chamber.

(2)
(e)

Explain why an alpha source is used rather than a beta source of the same activity.

(3)
(f)

In a real smoke detector the current is much smaller than your answer to (c). Suggest one reason for this.

(1)
(g)

The half-life of americium-241 is about 430 years. Suggest why a source with a long half-life is chosen.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
nucleon number 237 AND proton number 93✓ 1
Part (b)
5.5 × 106/34 «= 1.62 × 105»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
alpha particles entering per second = 0.5 × 3.7 × 104 = 1.85 × 104 s−1✓ 1
ion pairs produced per second = 1.85 × 104 × 1.62 × 105 = 3.0 × 109 s−1✓ 1Allow ECF from (b).
I = 3.0 × 109 × 1.60 × 10−19 = 4.8 × 10−10 A✓ 1Each ion pair transfers a charge e round the circuit. Award [3] for CNA. Accept 4.7–4.8 × 10−10 A.
Part (d)
smoke particles absorb alpha particles / ions attach to the «much more massive» smoke particles✓ 1
so fewer ions are produced or fewer ions reach the plates per second, and the current falls✓ 1
Part (e)
alpha particles are much more strongly ionizing, so each one produces a large number of ion pairs within the small chamber✓ 1
beta particles are weakly ionizing and travel much further in air, so they would produce few ion pairs in the chamber and the current would be far too small «and hardly affected by smoke»✓ 1
alpha particles are absorbed by a few centimetres of air or by the casing, so no radiation escapes from the detector✓ 1
Part (f)
the alpha particles have a range of a few cm in air, so they reach the plates/walls before giving all their energy to the air in the small chamber / some ions recombine before reaching the plates✓ 1Accept any one.
Part (g)
the activity, and so the current, stays almost constant over the working life of the detector✓ 1

Answers: (a) A = 237, Z = 93  ·  (b) 1.62 × 105  ·  (c) 4.8 × 10−10 A (the remaining parts are explanations — see the table above)

134E-2-43Choosing an isotope Paper 2Medium14 marks

Radium-223, 22388Ra, is injected into patients whose cancer has spread to their bones. Radium is taken up by bone, mostly where tumours are growing. Radium-223 decays by alpha emission with a half-life of 11.4 days to radon-219, which has a half-life of 4 s; radon-219 and the nuclides that follow it in the decay series decay quickly. The alpha particles have a range of less than 0.1 mm in tissue.

In one treatment 2.4 × 1012 radium-223 nuclei decay in a region of bone and tumour tissue of mass 0.20 kg. Each decay of a radium-223 nucleus, together with the decays of its short-lived daughter nuclides, deposits 27 MeV in this region. (specific heat capacity of the tissue = 3.5 × 103 J kg−1 K−1, 1 MeV = 1.60 × 10−13 J)

(a)

Complete the equation for the alpha decay of radium-223: 22388Ra → … Rn + …

(2)
(b)

Explain why an alpha emitter taken up by bone can destroy tumour cells there while doing very little damage to healthy tissue a few millimetres away.

(3)
(c)

Energy deposited in the tissue.

(i)

Show that the energy deposited in the 0.20 kg region is about 10 J.

(1)
(ii)

Calculate the rise in temperature of the region if all of this energy became internal energy of the tissue.

(2)
(iii)

Explain why the treatment destroys many tumour cells although your answer to (c)(ii) is very small.

(2)
(d)

Calculate the percentage of the injected radium-223 that has not decayed 57 days after the injection, ignoring any that leaves the body.

(1)
(e)

Radon is a gas. Suggest why it is important that radon-219 has a very short half-life.

(1)
(f)

Some decays in the series also emit gamma-ray photons. Suggest one use of these photons and one precaution that their emission makes necessary.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
21986Rn✓ 1Both numbers needed.
42α OR 42He✓ 1
Part (b)
alpha particles are strongly ionizing: they produce many ion pairs per mm of their path «and so damage cells, e.g. their DNA»✓ 1
they have a very short range «< 0.1 mm, a few cell diameters», so all of their energy is deposited close to the bone surface where the radium is taken up✓ 1
tissue a few millimetres away is far beyond their range and receives almost no radiation✓ 1OWTTE
Part (c)(i)
E = 2.4 × 1012 × 27 × 1.60 × 10−13 «= 10.4 J»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
ΔT = Q/mc = 10.4/(0.20 × 3.5 × 103)✓ 1Allow 10 J from (c)(i).
= 0.015 K «about 0.01 K»✓ 1Award [2] for CNA. Accept 0.014–0.015 K.
Part (c)(iii)
the energy is not spread evenly as random kinetic energy: it is delivered in large amounts «tens of eV per ionization» along the short tracks of the alpha particles✓ 1
each ionization can break chemical bonds, e.g. in the DNA of the cells, which kills the cells even though the total energy and the temperature rise are tiny✓ 1OWTTE
Part (d)
57 days = 5 half-lives, so 1/32 = 3.1 % remains✓ 1Accept 3 %.
Part (e)
the radon-219 decays within seconds, before it can diffuse away from the bone, so its alpha particles «and those of its daughters» are also emitted at the tumour site✓ 1OWTTE
Part (f)
use: the gamma rays leave the body, so a gamma camera can image where the radium has been taken up✓ 1
precaution: the patient irradiates people nearby, so staff limit time near the patient / keep their distance / use shielding✓ 1Accept any sensible precaution linked to gamma rays leaving the body.

Answers: (c)(i) 10.4 J  ·  (c)(ii) 0.015 K  ·  (d) 3.1 % (the remaining parts are explanations — see the table above)

135E-2-44Beta decay & the neutrino Paper 2Medium14 marks

A nuclear reactor has a thermal power of 3.0 GW. In a typical fission, 23592U + 10n → 14054Xe + 9438Sr + 210n, about 200 MeV is released. The fission products are radioactive and decay by a series of β− decays until stable nuclides are reached; xenon-140 ends as cerium-140, 14058Ce, and strontium-94 ends as zirconium-94, 9440Zr.

A detector of cross-sectional area 10 m2 is placed 1.5 km from the reactor, deep underground. On average only one antineutrino in every 5.0 × 1017 that pass through the detector is detected. (1 MeV = 1.60 × 10−13 J)

(a)

The fission products.

(i)

Complete the equation for the first decay of xenon-140: 14054Xe → … Cs + … + …

(2)
(ii)

Show that each fission is followed by a total of six β− decays of the two fission products.

(1)
(b)

Calculate the number of antineutrinos emitted by the reactor per second. Assume that every fission is of the type shown.

(2)
(c)

Calculate the number of antineutrinos per second that pass through each square metre at the detector. Assume that they are emitted equally in all directions.

(2)
(d)

Estimate the number of antineutrinos detected per day.

(3)
(e)

The beta particles and gamma rays from the fission products are absorbed in the reactor and its shielding, but almost all the antineutrinos pass through the Earth. Explain this difference.

(2)
(f)

Suggest why the detector is placed deep underground.

(1)
(g)

Suggest one advantage of using an antineutrino detector to monitor the power of a reactor from outside the site.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
14055Cs✓ 1Both numbers needed.
0−1e «β−» AND ν̄ «antineutrino»✓ 1Do not accept a neutrino.
Part (a)(ii)
each β− decay raises the proton number by 1 «at constant nucleon number»: xenon 54 → 58 gives 4 decays and strontium 38 → 40 gives 2, a total of 6✓ 1Must see both changes in proton number.
Part (b)
fissions per second = 3.0 × 109/(200 × 1.60 × 10−13) = 9.4 × 1019 s−1✓ 1
antineutrinos per second = 6 × 9.4 × 1019 = 5.6 × 1020 s−1✓ 1Award [2] for CNA. Award [1 max] for 9.4 × 1019 s−1 «one per fission».
Part (c)
number per unit area = rate/4πd2✓ 1Inverse-square spreading from a point source.
= 5.6 × 1020/(4π × 15002) = 2.0 × 1013 m−2 s−1✓ 1Allow ECF from (b). Award [2] for CNA.
Part (d)
number passing through the detector per second = 2.0 × 1013 × 10 = 2.0 × 1014 s−1✓ 1Allow ECF from (c).
number detected per second = 2.0 × 1014/5.0 × 1017 = 4.0 × 10−4 s−1✓ 1
per day: × 8.64 × 104 s ≈ 34✓ 1Accept 30–40 per day. Award [3] for CNA.
Part (e)
antineutrinos have no charge and almost no mass, so they interact with matter extremely rarely «they cause almost no ionization»✓ 1
beta particles are charged and gamma photons interact with electrons, so both lose energy by ionizing atoms and are absorbed within a few metres of concrete and steel✓ 1OWTTE
Part (f)
the rock above absorbs cosmic rays «and other background radiation», which would otherwise give many more counts than the few antineutrino events✓ 1OWTTE
Part (g)
the detection rate is proportional to the fission rate and so to the power, and the antineutrinos cannot be shielded or hidden, so the power can be checked without access to the reactor✓ 1OWTTE

Answers: (b) 5.6 × 1020 s−1  ·  (c) 2.0 × 1013 m−2 s−1  ·  (d) ≈ 34 per day (the remaining parts are explanations — see the table above)

136E-2-67Background radiation Paper 2Medium12 marks

At the exit of a hospital radiopharmacy, every member of staff stands for 5.0 s on a hand-and-foot monitor. Large detectors face the hands and the soles of the shoes. The radiopharmacy prepares doses of technetium-99m, a gamma emitter with a half-life of 6.0 hours.

At the start of each day, with nobody on the monitor, 2500 counts are recorded in 100 s. The monitor sounds an alarm if the number of counts recorded in 5.0 s is greater than B + 3√B, where B is the expected number of background counts in 5.0 s.

(a)

State one source of the background radiation recorded by the monitor.

(1)
(b)
(i)

Calculate B.

(1)
(ii)

Show that the alarm sounds if more than about 160 counts are recorded.

(1)
(c)

A person with no contamination does not always record exactly B counts. Explain why, and why the alarm level is set well above B.

(2)
(d)

The monitor records 12 % of the decays that take place in contamination on a hand. Determine whether contamination with an activity of 40 Bq on a hand will normally set off the alarm.

(3)
(e)

The same design of monitor is used in a laboratory that handles alpha emitters. State why its detectors must have very thin windows.

(1)
(f)

A supervisor proposes increasing the counting time from 5.0 s to 20 s. Discuss one advantage and one disadvantage of this change.

(2)
(g)

A laboratory coat is found to be contaminated with technetium-99m. Suggest why it can be kept in a shielded bin for two days and then washed in the normal way.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
any one of: cosmic rays / radioactive isotopes in rocks, soil or building materials / radon gas in the air / potassium-40 in the body or in food / radioactive sources used in the building✓ 1Do not accept 'the Sun' alone.
Part (b)(i)
B = 2500/100 × 5.0 = 125 «counts»✓ 1
Part (b)(ii)
125 + 3√125 = 125 + 33.5 = 158.5 «≈ 160»✓ 1Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (c)
radioactive decay «and the arrival of cosmic rays» is random, so the number of background counts in 5.0 s fluctuates «typically by about ±√B ≈ ±11»✓ 1
setting the alarm level 3√B above B makes it very unlikely that a random fluctuation of the background alone sets off the alarm «few false alarms»✓ 1OWTTE
Part (d)
count rate due to the contamination = 0.12 × 40 = 4.8 s−1✓ 1
expected number of counts in 5.0 s = 125 + 4.8 × 5.0 = 149✓ 1Allow ECF from (b)(i).
149 < 158.5 «≈ 159», so the alarm will not normally sound✓ 1MP3 needs a comparison with the alarm level and a consistent conclusion. ALTERNATIVE: the smallest activity detected is (158.5 − 125)/(5.0 × 0.12) ≈ 56 Bq > 40 Bq.
Part (e)
alpha particles have a very short range «are stopped by a sheet of paper», so a thick window would absorb them before they enter the detector✓ 1OWTTE
Part (f)
advantage: more counts are recorded, so the fractional fluctuation (√N/N) is smaller and a smaller contamination can be detected «about 28 Bq instead of 56 Bq»✓ 1The numerical estimate is not required.
disadvantage: each person must stand on the monitor four times as long «queues at the end of a shift»✓ 1Accept any sensible disadvantage.
Part (g)
48 h = 8 half-lives, so the activity falls to 1/256 «0.4 %» of its value✓ 1Accept 'about 8 half-lives' with a statement that the activity becomes negligible.

Answers: (b)(i) B = 125  ·  (b)(ii) 158.5  ·  (d) 149 counts expected — no alarm (the remaining parts are explanations — see the table above)

137E-2-68Mass defect & binding energy Paper 2Medium13 marks

The luminous paint on some aircraft instrument dials made before 1960 contains radium-226, 22688Ra. Radium-226 decays by alpha emission to radon-222 with a half-life of 1600 years. A museum displays a collection of these dials.

The table gives some masses. The atomic masses include the masses of the electrons; in each decay considered here the numbers of electrons on the two sides are equal. (1 u = 931.5 MeV c−2, k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C)

ParticleMass / u
radium-226 atom226.025410
radon-222 atom222.017578
helium-4 atom4.002603
francium-225 atom225.025573
hydrogen-1 atom1.007825
proton1.007276
neutron1.008665
helium-4 nucleus4.001506
(a)

State the nuclear equation for the alpha decay of radium-226.

(1)
(b)

Show that the energy released in this decay is about 4.9 MeV.

(1)
(c)

Calculate the binding energy per nucleon of the helium-4 nucleus.

(2)
(d)

A student suggests that radium-226 could instead decay by emitting a single proton, becoming francium-225. Deduce, using the data, that this decay cannot happen.

(2)
(e)

The binding energy per nucleon of radium-226 is 7.6620 MeV and that of radon-222 is 7.6946 MeV.

(i)

Outline, with reference to the variation of the binding energy per nucleon with nucleon number, why radon-222 has the larger binding energy per nucleon.

(1)
(ii)

The binding energy per nucleon of helium-4 is smaller than that of radium-226. Explain why energy is nevertheless released when radium-226 emits an alpha particle.

(3)
(f)
(i)

In the radium nucleus the centres of two neighbouring protons are about 1.5 × 10−15 m apart. Calculate the electric force between them.

(1)
(ii)

Explain why the nucleus does not fly apart, even though each proton is repelled by 87 other protons.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
22688Ra → 22286Rn + 42α✓ 1All numbers needed. Accept 42He.
Part (b)
Δm = 226.025410 − 222.017578 − 4.002603 = 0.005229 u; E = 0.005229 × 931.5 = 4.87 MeV✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (c)
Δm = 2 × 1.007276 + 2 × 1.008665 − 4.001506 = 0.030376 u, so the binding energy = 28.3 MeV✓ 1
binding energy per nucleon = 28.3/4 = 7.07 MeV✓ 1Award [2] for CNA. Award [1 max] for 28.3 MeV «not divided by 4».
Part (d)
Δm = 226.025410 − 225.025573 − 1.007825 = −0.007988 u «negative»✓ 1
the products would have more mass «energy» than the radium-226 atom, so about 7.4 MeV would have to be supplied: the decay cannot happen spontaneously✓ 1MP2 needs the sign of Δm interpreted. Accept the comparison of the two total masses.
Part (e)(i)
beyond the peak of the curve «A ≈ 60», the binding energy per nucleon decreases as the nucleon number increases, so radon-222, with fewer nucleons, has the larger value✓ 1OWTTE
Part (e)(ii)
total binding energy of the products = 222 × 7.6946 + 28.3 = 1736.5 MeV✓ 1Allow ECF from (c): 4 × «binding energy per nucleon of helium-4».
total binding energy of radium-226 = 226 × 7.6620 = 1731.6 MeV✓ 1
the products have the greater total binding energy «are more tightly bound overall», so the difference, about 4.9 MeV, is released, in agreement with (b)✓ 1MP3 needs the comparison of totals. ALTERNATIVE: the 222 nucleons that stay gain 222 × 0.0326 ≈ 7.2 MeV, more than the ≈ 2.4 MeV lost by the four nucleons of the alpha particle. OWTTE
Part (f)(i)
F = 8.99 × 109 × (1.60 × 10−19)2/(1.5 × 10−15)2 = 102 N «≈ 1 × 102 N»✓ 1Accept 1.0 × 102 N.
Part (f)(ii)
the strong nuclear force acts between all nucleons «p–p, p–n and n–n» and is attractive✓ 1
at separations of about 10−15 m it is larger than the electric repulsion «of about 100 N between neighbouring protons», and the neutrons add attraction without adding repulsion✓ 1Do not accept 'nuclear force' without 'strong'. OWTTE

Answers: (b) 4.87 MeV  ·  (c) 7.07 MeV  ·  (d) Δm = −0.007988 u «−7.4 MeV»  ·  (e)(ii) 1736.5 MeV > 1731.6 MeV; ≈ 4.9 MeV released  ·  (f)(i) 102 N (the remaining parts are explanations — see the table above)

138E-2-71Penetration & ionization Paper 2Medium12 marks

A science centre displays a large cloud chamber. The air in the chamber contains a vapour that condenses into tiny droplets when ions are present, so the paths of ionizing particles show up as lines of droplets (tracks). A uniform magnetic field of flux density 0.50 T is directed into the plane of the floor of the chamber.

A source S placed at the edge of the chamber emits alpha particles and beta-minus particles. The diagram is a view from above. Consider particles that start moving to the right, along the initial direction shown, in the plane of the diagram.

(mass of an alpha particle = 6.64 × 10−27 kg, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)

Top view of the floor of a cloud chamber in a uniform magnetic field directed into the page, shown by crosses. A source S on the left edge emits particles that start moving to the right.×××××××××××××××××××××××××××××××××××××××××××Sinitial directionmagnetic field into the page (×)
Diagram NOT accurately drawn
(a)

Outline how an alpha particle produces a track in the chamber.

(2)
(b)

The alpha particles have a kinetic energy of 5.5 MeV.

(i)

Show that the radius of the path of an alpha particle is about 0.7 m.

(2)
(ii)

The alpha-particle tracks are about 4.0 cm long. Explain why they appear almost straight.

(1)
(c)

The tracks of the beta-minus particles are curved much more strongly than those of the alpha particles, and they form spirals whose radius decreases along the track. Explain these two observations.

(2)
(d)

Draw, on the diagram, the track of an alpha particle and the track of a beta-minus particle that both start from S in the initial direction shown. Label each track.

(2)
(e)

In another demonstration a different source is used. Its tracks have the same shape as those of the beta-minus particles but curve the opposite way. Identify the particles and the decay that produces them.

(2)
(f)

The source in (e) also emits gamma rays. State why the gamma rays do not leave continuous tracks.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the alpha particle ionizes the air molecules along its path «knocking electrons out of them»✓ 1
the vapour condenses on these ions, forming a line of droplets along the path✓ 1OWTTE
Part (b)(i)
v = √(2Ek/m) = √(2 × 5.5 × 1.60 × 10−13/6.64 × 10−27) = 1.63 × 107 m s−1✓ 1
qvB = mv2/r, so r = mv/qB = 6.64 × 10−27 × 1.63 × 107/(3.20 × 10−19 × 0.50) = 0.676 m✓ 1Must see full substitution OR answer to at least 3 s.f. Charge 2e needed: 1.4 m with e scores MP1 only.
Part (b)(ii)
the track length (4.0 cm) is much smaller than the radius «≈ 6 % of it», so the particle turns through only a few degrees «≈ 3°»✓ 1OWTTE. Allow ECF from (b)(i).
Part (c)
a beta particle has a much smaller mass and so a much smaller momentum than an alpha particle; r = mv/qB, so its radius is much smaller✓ 1
the beta particle loses kinetic energy by ionizing the air, so its speed and hence the radius of its path decrease along the track✓ 1OWTTE
Part (d)
alpha track curving towards the top of the page AND beta-minus track curving towards the bottom of the page✓ 1Directions from F = qvB with the field into the page. Both needed.
beta-minus track much more strongly curved «a spiral of decreasing radius», alpha track short and almost straight, each labelled✓ 1MP2 only scores if MP1 scores. Allow ECF from (b) and (c).
Part (e)
positrons «β+ particles»: same mass as electrons «same curvature» but positive charge «opposite curvature»✓ 1
β+ decay, in which a proton in the nucleus changes into a neutron and a positron and a neutrino are emitted✓ 1Do not accept antineutrino.
Part (f)
gamma rays are uncharged and only weakly ionizing, so they produce very few ions along their path «too few to form a line of droplets»✓ 1OWTTE

Answers: (b)(i) r = 0.68 m (the remaining parts are explanations — see the table above)

139E-2-72Changes in the nucleus Paper 2Medium12 marks

Single-use medical items, such as syringes and gloves, are sterilised after they have been sealed in cardboard boxes about 0.5 m thick. The boxes are carried on a conveyor around a rack of sealed cobalt-60 sources. When the plant is not in use, the rack is lowered into a deep pool of water.

Cobalt-60 decays by β− emission with a half-life of 5.3 years. The nickel-60 nucleus that is formed is in an excited state; it then emits two gamma-ray photons, of energies 1.17 MeV and 1.33 MeV. The beta particles are absorbed by the casings of the sources. The activity of the cobalt-60 is 1.10 × 1017 Bq. (1 MeV = 1.60 × 10−13 J)

Blank axes: activity of the cobalt-60 source, in units of 10 to the 16 becquerel, from 0 to 12, against time from 0 to 22 years0246810121416182022time / years024681012activity / 10¹⁶ Bq
Axes for (e)(ii)
(a)

Outline the change in the nickel-60 nucleus when it emits the gamma-ray photons.

(1)
(b)
(i)

Show that the power emitted by the sources as gamma radiation is about 44 kW.

(1)
(ii)

30 % of this power is absorbed in the boxes. To be sterilised, each kilogram of product must absorb 25 kJ. Calculate the greatest mass of product that can be sterilised in one hour.

(2)
(c)

Explain why gamma rays, rather than a beam of fast electrons, are used to sterilise the sealed boxes.

(2)
(d)

Explain why the sterilised items do not become radioactive.

(2)
(e)

The operator adds new cobalt-60 to the rack every 10.6 years, so that the activity returns to 1.10 × 1017 Bq.

(i)

Determine the activity of the new cobalt-60 that must be added each time.

(2)
(ii)

Sketch, on the axes, a graph to show how the activity of the sources on the rack varies over the first 21.2 years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the nucleus loses energy «falls from an excited state to its ground state» while its proton number and nucleon number do not change✓ 1Both ideas needed.
Part (b)(i)
P = 1.10 × 1017 × (1.17 + 1.33) × 1.60 × 10−13 = 4.4 × 104 W✓ 1Must see full substitution OR answer to at least 3 s.f. Both photon energies must be used.
Part (b)(ii)
power absorbed = 0.3 × 4.4 × 104 = 1.32 × 104 W, so energy absorbed in one hour = 4.75 × 107 J✓ 1
mass = 4.75 × 107/2.5 × 104 = 1.9 × 103 kg✓ 1Award [2] for CNA. Allow ECF from (b)(i). Award [1 max] for 6.3 × 103 kg «30 % not used».
Part (c)
fast electrons «beta particles» are charged and ionize continually along their path, so they are absorbed within a few centimetres and would sterilise only the outer layer of each box✓ 1
gamma rays are uncharged and only weakly absorbed «the most penetrating», so a large fraction passes right through 0.5 m of boxes and every item receives energy✓ 1OWTTE
Part (d)
the gamma photons interact with the electrons of the atoms «ionization»; they do not add nucleons to, or remove nucleons from, the nuclei of the items✓ 1
the photon energies «≤ 1.33 MeV» are far less than the energy needed to remove a nucleon from a nucleus «binding energy per nucleon ≈ 8 MeV», so no unstable nuclei are formed✓ 1Accept: the cobalt-60 is sealed, so none of it reaches the items, for MP2 only if MP1 scores. OWTTE
Part (e)(i)
10.6 years = 2 half-lives, so the activity has fallen to 1.10 × 1017/4 = 2.75 × 1016 Bq✓ 1
activity to add = 1.10 × 1017 − 2.75 × 1016 = 8.25 × 1016 Bq✓ 1Award [2] for CNA. Accept 8.2–8.3 × 1016 Bq.
Part (e)(ii)
curve falling from 11 × 1016 Bq at t = 0 through 5.5 × 1016 Bq at 5.3 years to 2.75 × 1016 Bq at 10.6 years, with decreasing gradient✓ 1Within half a large square. Do not accept straight lines.
vertical rise to 11 × 1016 Bq at 10.6 years, followed by the same curve repeated, reaching 2.75 × 1016 Bq at 21.2 years✓ 1Allow ECF from (e)(i).

Answers: (b)(i) 4.4 × 104 W  ·  (b)(ii) 1.9 × 103 kg  ·  (e)(i) 8.25 × 1016 Bq (the remaining parts are explanations — see the table above)

140E-2-73Half-life & activity Paper 2Medium12 marks

Potassium-40, 4019K, is a naturally occurring radioactive isotope with a half-life of 1.25 × 109 years. In this question, assume that every potassium-40 nucleus decays by β− emission to calcium (Ca).

The body of an adult of mass 70 kg contains potassium-40 with an activity of 4.4 × 103 Bq. The energy released in each decay is 1.31 MeV, but on average only 0.50 MeV of it is absorbed in the body. (1 MeV = 1.60 × 10−13 J, 1 year = 3.16 × 107 s)

Blank axes: power released by potassium-40 decay in the Earth, in terawatts from 0 to 35, against time from 3.75 billion years ago to now−3.75−2.50−1.250time / 10⁹ years (0 = now)05101520253035power / TW
Axes for (e)(ii)
(a)

State the nucleon number and the proton number of the calcium nucleus formed in the decay.

(1)
(b)
(i)

Show that the energy absorbed in the body in one year from the decay of potassium-40 is about 1.1 × 10−2 J.

(1)
(ii)

Explain why, on average, much less than 1.31 MeV is absorbed in the body in each decay.

(2)
(c)

Explain why the activity of the potassium-40 in an adult's body hardly changes during their lifetime, even though potassium-40 is continually decaying.

(1)
(d)

On average, a person absorbs about 2.0 × 10−3 J per kilogram of body mass each year from all natural sources of radiation. Determine whether potassium-40 in the body provides more than 10 % of this.

(3)
(e)

The decay of potassium-40 inside the Earth now releases thermal energy at a rate of about 4.0 TW.

(i)

Determine the rate at which potassium-40 released thermal energy inside the Earth 2.5 × 109 years ago.

(2)
(ii)

Sketch, on the axes, a graph to show how this rate has varied from 3.75 × 109 years ago to now.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
nucleon number 40 AND proton number 20✓ 1Both needed.
Part (b)(i)
E = 4.4 × 103 × 0.50 × 1.60 × 10−13 × 3.16 × 107 = 1.11 × 10−2 J✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
the energy released is shared between the beta particle and the antineutrino «in different proportions in each decay»✓ 1
antineutrinos interact with matter extremely rarely, so they leave the body and their energy is not absorbed✓ 1OWTTE
Part (c)
the half-life «1.25 × 109 years» is enormous compared with a human lifetime, so only a negligible fraction of the potassium-40 nuclei decays during a life✓ 1Accept also: the potassium in the body is continually replaced from food. OWTTE
Part (d)
energy absorbed per kilogram per year = 1.1 × 10−2/70 = 1.6 × 10−4 J kg−1✓ 1Allow ECF from (b)(i).
fraction = 1.6 × 10−4/2.0 × 10−3 × 100 = 8 %✓ 1
8 % < 10 %, so potassium-40 in the body does not provide more than 10 %✓ 1MP3 needs a comparison and a conclusion consistent with the candidate's value.
Part (e)(i)
2.5 × 109 years is 2 half-lives, and there were then 22 = 4 times as many potassium-40 nuclei, so the activity was 4 times larger✓ 1
rate = 4 × 4.0 = 16 TW✓ 1Award [2] for CNA. Award [1 max] for 1.0 TW «decay applied forwards in time».
Part (e)(ii)
curve through «−3.75, 32 TW», «−2.5, 16 TW», «−1.25, 8 TW» and «0, 4 TW»✓ 1At least three of the four points within half a small square. Allow ECF from (e)(i).
smooth curve that decreases towards the present with a gradient whose magnitude decreases✓ 1Do not accept straight-line segments.

Answers: (a) 40 and 20  ·  (b)(i) 1.11 × 10−2 J  ·  (d) 8 % — less than 10 %  ·  (e)(i) 16 TW (the remaining parts are explanations — see the table above)

141IBP2S-038-04-12Half-life & activity Paper 2Medium6 marks

A class models radioactive decay with 600 small wooden cubes. One face of each cube is painted red. The cubes are thrown onto a table together. Every cube that lands red face up is removed, as if it has 'decayed'. The remaining cubes are collected and thrown again, and the process is repeated.

(a)

Outline two ways in which this model represents radioactive decay.

(2)
(b)

Show that the expected number of cubes remaining after the first throw is 500.

(1)
(c)

In the first throw, the class removes 87 cubes instead of the expected 100. Explain why this result does not show that the model is wrong.

(1)
(d)

Determine the 'half-life' of the cubes, measured in throws.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Award one mark for each of any two valid points. First point: it is impossible to predict which particular cube will be removed on a throw, as the decay of an individual nucleus is random . Second point: each cube has the same constant probability (1/6) of removal per throw, independent of how many throws it has survived, like a constant decay probability per unit time; OR the number removed per throw is proportional to the number remaining, like activity being proportional to the number of undecayed nuclei; OR removed cubes take no further part, like decayed nuclei .✓ 2
Part (b)
The probability of remaining is 5/6, so 600 × 5/6 = 500 .✓ 1
Part (c)
Removal is random, so the actual number fluctuates about the expected value. The difference of 13 is comparable to the expected statistical spread of about √100 ≈ 10 (σ ≈ 9), so it is a normal random variation .✓ 1
Part (d)
(5/6)n = 0.5 . n = ln 0.5 / ln(5/6) = 3.8 throws .✓ 2

Exam tip: A half-life does not have to be a whole number of throws or time units. Set the surviving fraction equal to 0.5 and solve using logarithms.

142IBP2S-038-04-13Mixtures of isotopes Paper 2Medium8 marks

Engineers inject a solution containing sodium-24 into an underground water pipe to find a leak. A detector placed near the injection point records the count rate shown below. The background count rate is measured separately as 30 counts min−1.

Time / h: 0, 5, 10, 15, 20, 25
Measured count rate / counts min−1: 830, 665, 534, 430, 347, 282

05101520250200400600800Time / hCount rate / counts min⁻¹
Measured count rate (background not subtracted)
(a)

Explain why the background count rate must be subtracted from each reading.

(2)
(b)

Determine the half-life of sodium-24 using the data.

(2)
(c)

Show that the reading at 25 h is consistent with your answer to (b).

(2)
(d)

Estimate the time after injection at which the corrected count rate will have fallen to 50 counts min−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Background radiation (from rocks, cosmic rays, radon) is detected whether or not the tracer is present . Subtracting it gives the count rate due to the sodium-24 alone; otherwise the half-life found would be too long .✓ 2
Part (b)
Corrected count rates: 800 at 0 h and 400 at 15 h . Half-life = 15 h .✓ 2
Part (c)
Predicted corrected rate = 800 × (1/2)25/15 = 252 counts min−1 . Measured corrected rate = 282 − 30 = 252 counts min−1, which agrees . ECF from (b).✓ 2
Part (d)
800 / 50 = 16 = 24, so 4 half-lives . Time = 4 × 15 = 60 h . ECF from (b).✓ 2

Common error: If you halve the raw readings without first subtracting background, you get the wrong half-life. Correct every reading before using it.

143IBP2S-038-04-15Isotopes Paper 2Medium7 marks

Nuclear fuel is made from uranium in which the proportion of uranium-235 has been increased. Natural uranium is mainly uranium-238 (\(^{238}_{92}\)U), with 0.72 % of its atoms being uranium-235 (\(^{235}_{92}\)U). Enrichment plants separate the isotopes using gas centrifuges, which spin uranium hexafluoride gas at high speed.

(a)

State the number of protons and the number of neutrons in a nucleus of (i) uranium-235 and (ii) uranium-238.

(2)
(b)

Explain why the two isotopes cannot be separated by a chemical reaction.

(2)
(c)

Outline why a centrifuge is able to separate the isotopes.

(1)
(d)

Calculate the number of uranium-235 nuclei in 1.0 kg of natural uranium. Take the molar mass of natural uranium as 238 g mol−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
(i) 92 protons, 143 neutrons . (ii) 92 protons, 146 neutrons .✓ 2
Part (b)
Both have 92 protons, so their atoms have the same number and arrangement of electrons . Chemical behaviour depends on the electrons, so the isotopes react identically .✓ 2
Part (c)
The molecules containing U-238 are more massive (3 extra neutrons), so they move preferentially towards the outside of the spinning centrifuge .✓ 1
Part (d)
Total atoms = (1000 / 238) × 6.02×1023 = 2.53×1024 . U-235 nuclei = 0.0072 × 2.53×1024 = 1.8×1022 .✓ 2

Common error: 0.72 % must be converted to 0.0072, not 0.72, before multiplying. Also check that mass and molar mass are in the same units (grams with g mol⁻¹).

144IBP2S-038-02-13Background radiation Paper 2Medium7 marks

A home radiation survey measures a radon-222 activity concentration of 400 Bq m−3 in the air of a poorly ventilated basement. The volume of the basement is 60 m3. Radon-222 is an alpha emitter with a half-life of 3.8 days.

(a)

Calculate the total activity of radon in the basement air.

(1)
(b)

Assume that no further radon enters the basement and none escapes. Determine the time for the activity concentration to fall to 50 Bq m−3.

(2)
(c)

In practice the measured concentration stays roughly constant over many weeks. Explain why.

(2)
(d)

Outline why radon is a significant health hazard even though alpha particles cannot penetrate skin.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A = 400 × 60 = 2.4×104 Bq✓ 1
Part (b)
400 → 50 is a reduction by a factor of 8 = 23, so 3 half-lives✓ 1
t = 3 × 3.8 = 11.4 days ≈ 11 days✓ 1
Part (c)
Radon is continuously produced by the decay of uranium/radium in the rock and soil beneath the house and seeps into the basement✓ 1
The rate at which new radon is supplied equals the rate at which it is lost (by decay and slow leakage), so the amount present stays approximately steady✓ 1
Part (d)
Radon is a gas, so it is inhaled and decays inside the lungs, where its solid radioactive decay products can also lodge✓ 1
Alpha particles are highly ionizing, so they deposit their energy in a small volume of unprotected lung tissue, damaging cells/DNA✓ 1

Common error: Low penetration does not mean low hazard. An alpha emitter inside the body is very dangerous because all of its energy is deposited in a small volume of tissue.

145IBP2S-038-05-14The binding energy curve Paper 2Medium7 marks

An experimental fusion reactor (tokamak) uses the reaction
21H + 31H → 42He + 10n
Binding energies per nucleon: hydrogen-2 = 1.112 MeV; hydrogen-3 = 2.827 MeV; helium-4 = 7.074 MeV.

(a)

Outline, in terms of binding energy, why energy is released in this reaction.

(2)
(b)

State why the neutron does not contribute to the binding energy calculation.

(1)
(c)

Calculate the energy released in one reaction.

(2)
(d)

One fission of uranium-235, triggered by one neutron, releases about 200 MeV. The reactants are the uranium-235 nucleus and the neutron. Determine how many times greater the energy released per nucleon of reactants is for the fusion reaction than for this fission.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The total binding energy of the helium-4 nucleus is greater than the combined total binding energy of the hydrogen-2 and hydrogen-3 nuclei.✓ 1
The products are more tightly bound and have less total mass. The mass difference is released as kinetic energy of the products.✓ 1
Part (b)
A free neutron is not bound to any other nucleons, so its binding energy is zero.✓ 1
Part (c)
Binding energy of reactants = 2 × 1.112 + 3 × 2.827 = 10.705 MeV and binding energy of products = 4 × 7.074 = 28.296 MeV✓ 1
Energy released = 28.296 − 10.705 = 17.6 MeV✓ 1
Part (d)
Fusion: 17.6 / 5 = 3.52 MeV per nucleon. Fission: 200 / 236 = 0.847 MeV per nucleon✓ 1
Ratio = 3.52 / 0.847 ≈ 4.2 (accept 4.1 if 235 nucleons are used; allow error carried forward from the energy calculated in the previous part)✓ 1

Exam tip: Per nucleon of fuel, fusion releases several times more energy than fission. This is why fusion fuel is so attractive, even though a single fission releases more energy than a single fusion.

146IBP2S-038-03-13Changes in the nucleus Paper 2Medium7 marks

Cobalt-60 is used to sterilise surgical instruments. It decays by β− decay to nickel-60. The nickel-60 nucleus is formed in an excited state and then emits two gamma photons in succession, of energies 1.17 MeV and 1.33 MeV.

Data

NuclideAtomic mass / u
cobalt-6059.933816
nickel-6059.930786
1 u = 931.5 MeV c−2. When atomic masses are used for β− decay, the electron masses cancel, so the energy released equals the difference in atomic masses multiplied by c2.

(a)

State the change that occurs inside the cobalt-60 nucleus during β− decay.

(1)
(b)

Show that the energy released in the decay is about 2.8 MeV.

(2)
(c)

Determine the maximum kinetic energy of a beta particle emitted in this decay.

(2)
(d)

Explain why most of the beta particles have kinetic energies less than the value in (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A neutron changes into a proton (and an electron and an antineutrino are emitted) .✓ 1
Part (b)
Δm = 59.933816 − 59.930786 = 0.003030 u .✓ 1
E = 0.003030 × 931.5 = 2.82 MeV .✓ 1
Part (c)
The energy carried by the gamma photons is 1.17 + 1.33 = 2.50 MeV .✓ 1
Ek,max = 2.82 − 2.50 = 0.32 MeV . ECF from (b).✓ 1
Part (d)
The energy available is shared between the electron and the antineutrino .✓ 1
The maximum occurs only when the antineutrino receives (almost) no energy, so typical electron energies are lower .✓ 1

Exam tip: Keep all six decimal places when subtracting atomic masses. The difference is small, and rounding first can destroy the answer.

147IBP2S-038-03-14The binding energy curve Paper 2Medium7 marks

Iron-56 (5626Fe) builds up in the cores of massive stars near the end of their lives.
Nuclear mass of iron-56 = 55.920662 u; mass of proton = 1.007276 u; mass of neutron = 1.008665 u; 1 u = 931.5 MeV c−2.

(a)

State what is meant by the mass defect of a nucleus.

(1)
(b)

Calculate the mass defect of an iron-56 nucleus, in u.

(2)
(c)

Determine the binding energy per nucleon of iron-56, in MeV.

(2)
(d)

Iron-56 has one of the highest binding energies per nucleon of any nuclide. Explain why fusion reactions that produce elements heavier than iron cannot supply energy to the star.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The difference between the total mass of the separated nucleons and the mass of the nucleus .✓ 1
Part (b)
Total mass of nucleons = 26 × 1.007276 + 30 × 1.008665 = 56.449126 u .✓ 1
Δm = 56.449126 − 55.920662 = 0.528464 u ≈ 0.5285 u .✓ 1
Part (c)
Binding energy = 0.528464 × 931.5 = 492 MeV .✓ 1
Per nucleon = 492/56 = 8.79 MeV . ECF from (b).✓ 1
Part (d)
Beyond iron, binding energy per nucleon decreases as nucleon number increases .✓ 1
So fusion products would be less tightly bound than the reactants: their total mass would be greater, and energy would have to be supplied rather than released .✓ 1

Common error: Use the number of neutrons (A − Z = 30), not A, when adding up the neutron masses.

148IBP2S-038-01-14Mixtures of isotopes Paper 2Medium6 marks

A water company injects a short-lived gamma-emitting tracer into an underground pipe to locate a leak. A detector held in a fixed position monitors a sample of the tracer. The background count rate at the site is 30 counts per minute.

Time / h: 0, 2, 4, 6, 8, 10
Measured count rate / counts min−1: 1650, 1176, 840, 603, 435, 316

02468100300600900120015001800Time / hCount rate / counts min⁻¹
Measured count rate (background not subtracted)
(a)

State why the background count rate must be subtracted from each reading before the half-life is found.

(1)
(b)

Determine the half-life of the tracer, using the data.

(2)
(c)

Calculate the corrected count rate expected at t = 14 h.

(2)
(d)

Outline one reason why a tracer with a half-life of a few hours is suitable for this application.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The background radiation does not come from the tracer and does not decay, so leaving it in would distort the ratio of count rates and give a wrong half-life✓ 1
Part (b)
Corrected count rate at t = 0 is 1620; at t = 4 h it is 810, which is half✓ 1
Half-life = 4.0 h, confirmed by the corrected rate of 405 at 8 h✓ 1
Part (c)
14 h = 3.5 half-lives, or use λ = ln2 / 4.0 h✓ 1
Corrected rate = 1620 × 2−3.5 = 143 counts min−1✓ 1
Allow error carried forward from the half-life found in the previous part.—
Part (d)
It stays active long enough to complete the survey, but decays to negligible activity soon afterwards, so the water supply is not contaminated for long✓ 1

Common error: Do not use the raw count rates. Halving 1650 gives 825, which does not appear at a regular time interval and leads to an inconsistent half-life.

149IBP2S-038-01-16The binding energy curve Paper 2Medium8 marks

In a nuclear power station, a typical fission reaction is
\(^1_0\)n + \(^{235}_{92}\)U → \(^{141}_{56}\)Ba + \(^{92}_{36}\)Kr + 3 \(^1_0\)n
Approximate binding energies per nucleon are: \(^{235}\)U 7.6 MeV, \(^{141}\)Ba 8.3 MeV, \(^{92}\)Kr 8.5 MeV. Take 1 MeV = 1.60 × 10−13 J.

(a)

Outline the general shape of a graph of binding energy per nucleon against nucleon number for nuclei from hydrogen to uranium.

(2)
(b)

Explain, with reference to the graph, why the fission of a heavy nucleus releases energy.

(2)
(c)

Estimate the energy released in one fission event, in MeV.

(2)
(d)

The reactor produces a thermal power of 3.0 GW. Calculate the number of fissions occurring per second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Rises steeply from zero (hydrogen-1) for light nuclei, with He-4 lying above the general trend✓ 1
Reaches a broad maximum of about 8.8 MeV near A ≈ 56–62 (Fe/Ni), then decreases slowly to about 7.6 MeV for uranium✓ 1
Part (b)
The fission products are closer to the peak of the curve, so they have a greater binding energy per nucleon than uranium✓ 1
The total binding energy increases, so the total mass decreases, and the mass difference is released as energy (E = Δmc2)✓ 1
Part (c)
Binding energy of products = 141 × 8.3 + 92 × 8.5 = 1170 + 782 = 1952 MeV, compared with 235 × 7.6 = 1786 MeV for U-235✓ 1
Energy released = 1952 − 1786 ≈ 166 MeV (accept 170 MeV)✓ 1
Part (d)
Energy per fission = 166 × 1.60 × 10−13 = 2.66 × 10−11 J✓ 1
Rate = 3.0 × 109 / 2.66 × 10−11 = 1.13 × 1020 ≈ 1.1 × 1020 s−1✓ 1
Allow error carried forward from the previous part; using 170 MeV gives 1.1 × 1020 s−1.—

Common error: Free neutrons have zero binding energy, so they add nothing to either side. Do not include the neutrons when you total the binding energies.

150IBP2S-038-00-13Penetration & ionization Paper 2Medium9 marks

A factory that rolls thin aluminium sheet tests a radioactive source for use in a thickness gauge. A Geiger–Müller tube is fixed 3 cm from the source, and absorbers are placed between them. The background count rate is 30 counts min−1.

Absorber — count rate / counts min−1
none — 1250
one sheet of paper — 1248
0.5 mm aluminium — 820
1.0 mm aluminium — 530
1.5 mm aluminium — 345
2.0 mm aluminium — 225
5.0 mm aluminium — 31

(a)

Deduce, using the data, that the source does not emit alpha particles.

(2)
(b)

Explain whether the data show that the source emits a significant amount of gamma radiation.

(2)
(c)

Estimate the thickness of aluminium that halves the corrected count rate.

(3)
(d)

Outline why this source is suitable for monitoring aluminium sheet about 0.8 mm thick.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The count rate with paper (1248) is essentially the same as with no absorber (1250), so the difference is negligible✓ 1
Any alpha particles reaching the tube would be stopped by the paper, so the count rate would fall noticeably if alpha were present✓ 1
Part (b)
With 5.0 mm of aluminium the count rate (31) is equal to background (30) within uncertainty✓ 1
Gamma radiation would pass through 5 mm of aluminium largely unabsorbed, so the count would stay well above background. There is therefore no significant gamma emission, and the source is a beta emitter✓ 1
Part (c)
Corrected count rates are found by subtracting 30 (no absorber: 1220; 0.5 mm: 790; 1.0 mm: 500)✓ 1
Half the corrected rate = 610 counts min−1✓ 1
Interpolation between 0.5 mm and 1.0 mm gives ≈ 0.8 mm (accept 0.75–0.85 mm)✓ 1
Allow error carried forward for an incorrect corrected count rate.—
Part (d)
At this thickness the corrected count rate (about 600 counts min−1) changes significantly with small changes in thickness, so variations are easily detected✓ 1
Alpha would be completely stopped, and gamma would hardly be affected by changes in thickness✓ 1

Exam tip: Subtract the background before halving any count rate. Halving the raw reading of 1250 gives the wrong target value.

151IBP2S-038-00-14Choosing an isotope Paper 2Medium7 marks

Radium-223 (proton number 88) is injected into patients to treat cancer that has spread to the bones. It collects in bone tissue and decays by alpha emission to radon (Rn).

(a)

Write the nuclear equation for this decay.

(2)
(b)

Explain why an alpha emitter is suitable for destroying tumour cells in bone while limiting damage to nearby healthy tissue.

(2)
(c)

Each alpha particle is emitted with a kinetic energy of 5.7 MeV. On average, 34 eV of energy is needed to produce one ion pair in tissue. Calculate the number of ion pairs produced by one alpha particle.

(2)
(d)

The range of these alpha particles in tissue is 50 µm. Determine the average number of ion pairs produced per micrometre of path.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\({}^{223}_{\;88}\text{Ra} \rightarrow {}^{219}_{\;86}\text{Rn} + {}^{4}_{2}\alpha\)—
Radon with nucleon number 219 and proton number 86✓ 1
Alpha particle shown as \({}^{4}_{2}\alpha\) or \({}^{4}_{2}\text{He}\)✓ 1
Part (b)
Alpha particles are very strongly ionizing, so they deposit a large amount of energy in the tumour cells and damage their DNA✓ 1
Their range in tissue is very short (less than about 0.1 mm), so tissue more than a few cells away receives little radiation✓ 1
Part (c)
N = 5.7×106 eV / 34 eV✓ 1
N = 1.7×105✓ 1
Part (d)
1.7×105 / 50 = 3.4×103 µm−1✓ 1
ECF from (c)—

Common error: Keep the energies in eV when you divide. Converting only one of the two energies to joules gives an answer that is wrong by a factor of about 10¹⁹.

152IBP2S-038-00-16Half-life & activity Paper 2Medium8 marks

Some emergency exit signs need no electrical supply. They contain sealed glass tubes filled with tritium gas (hydrogen-3), coated inside with a phosphor. Tritium is a beta-minus emitter. The beta particles cause the phosphor to glow.

(a)

Write the nuclear equation for the decay of tritium.

(2)
(b)

Explain why an intact sign presents no significant radiation hazard to people nearby.

(2)
(c)

Explain why the beta particles emitted by tritium have a range of kinetic energies from almost zero up to 18.6 keV.

(2)
(d)

A sign contains tritium with an activity of 9.0×1010 Bq. The mean kinetic energy of the beta particles is 5.7 keV. Calculate the rate at which energy is delivered by the beta particles.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\({}^{3}_{1}\text{H} \rightarrow {}^{3}_{2}\text{He} + {}^{\;0}_{-1}\text{e} + \bar{\nu}_e\)—
Helium-3 with nucleon number 3 and proton number 2✓ 1
Electron with correct numbers and an electron antineutrino✓ 1
Part (b)
Tritium beta particles have very low energy (maximum 18.6 keV), so their range is tiny and they are absorbed by the phosphor or the glass wall✓ 1
Tritium emits no gamma radiation, so no penetrating radiation leaves the tube✓ 1
Part (c)
The energy released in each decay is shared between the electron and the antineutrino (the recoil of the nucleus is negligible)✓ 1
The sharing varies from decay to decay; the maximum of 18.6 keV occurs when the antineutrino receives almost no energy✓ 1
Part (d)
P = activity × mean energy = 9.0×1010 × 5.7×103 × 1.6×10−19✓ 1
P = 8.2×10−5 W✓ 1

Common error: Use the mean beta energy, not the maximum, when you find a total power. Using 18.6 keV overestimates the power by a factor of about 3.

153E-2-04Mass–energy equivalence Paper 2Hard13 marks

A space probe that will orbit Saturn is powered by a radioisotope generator. The fuel is plutonium-238, which decays by alpha emission to uranium-234 with a half-life of 88 years. The kinetic energy of the alpha particles and of the recoiling uranium nuclei is absorbed inside the fuel, so the decays heat it. Thermocouples convert 6.0 % of the thermal power of the fuel into electrical power. At launch the thermal power of the fuel is 4.40 kW.

(1 u = 931.5 MeV c−2, 1 MeV = 1.60 × 10−13 J, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s)

ParticleMass / u
plutonium-238 atom238.049558
uranium-234 atom234.040950
helium-4 atom4.002603
(a)

The decay of plutonium-238.

(i)

State the nuclear equation for the alpha decay of 23894Pu.

(1)
(ii)

Show that the energy released in one decay is about 5.6 MeV.

(1)
(iii)

Determine the activity of the plutonium-238 at launch.

(2)
(b)

The energy released in the fuel leaves the generator as thermal radiation and as electrical energy. Assuming that the thermal power stays at 4.40 kW, calculate the decrease in the mass of the generator during the first year after launch.

(2)
(c)

The instruments on the probe need an electrical power of at least 120 W. Determine whether the generator can still supply this power 88 years after launch. Assume that the efficiency of the thermocouples does not change.

(3)
(d)

Explain why an alpha emitter is a good choice of fuel for this generator.

(2)
(e)

Saturn is 9.5 AU from the Sun. At the Earth, 1.0 AU from the Sun, the intensity of sunlight is 1360 W m−2. Suggest, with a calculation, why solar panels are not used to power this probe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
23894Pu → 23492U + 42He «42α»✓ 1All four numbers on the right-hand side needed. Accept 42α.
Part (a)(ii)
Δm = 238.049558 − 234.040950 − 4.002603 = 0.006005 u; E = 0.006005 × 931.5 = 5.59 MeV✓ 1Must see full substitution OR answer to 3 s.f. Atomic masses may be used because the 94 electrons on the left balance 92 + 2 on the right.
Part (a)(iii)
Energy per decay = 5.594 × 1.60 × 10−13 = 8.95 × 10−13 J✓ 1Accept 8.96 × 10−13 J from 5.6 MeV.
A = 4400/8.95 × 10−13 = 4.9 × 1015 Bq✓ 1Award [2] for CNA. Award [1 max] for 4.9 × 1012 Bq «4.40 W used».
Part (b)
Energy transferred in one year = 4400 × 3.16 × 107 = 1.39 × 1011 J✓ 1
Δm = E/c2 = 1.39 × 1011/(3.00 × 108)2 = 1.5 × 10−6 kg✓ 1Accept 1.5–1.6 × 10−6 kg. Award [2] for CNA. Allow ECF for the energy.
Part (c)
Electrical power at launch = 0.060 × 4400 = 264 W✓ 1
88 years is one half-life, so the activity and the thermal power halve «to 2200 W»✓ 1
Electrical power = 132 W, which is greater than 120 W, so the generator can still supply the instruments✓ 1MP3 needs a comparison with 120 W. Allow a consistent conclusion from an ECF value.
Part (d)
Alpha particles have a very short range «a few μm in the solid fuel», so they are absorbed within the fuel and almost all of the decay energy becomes thermal energy of the fuel✓ 1
Very little penetrating radiation leaves the fuel, so only light shielding is needed to protect the instruments «keeping the mass of the probe low»✓ 1OWTTE
Part (e)
Intensity ∝ 1/d2: at Saturn it is 1360/9.52 = 15 W m−2✓ 1Accept 'about 1/90 of the intensity at the Earth'.
So the panels would need an area about 90 times larger than near the Earth to give the same power — far too large and heavy for the probe✓ 1OWTTE. Do not accept answers based on the probe being in the shadow of Saturn alone.

Answers: (a)(ii) 5.59 MeV  ·  (a)(iii) 4.9 × 1015 Bq  ·  (b) 1.5 × 10−6 kg  ·  (c) 132 W — yes, it can  ·  (e) 15 W m−2 (the remaining parts are explanations — see the table above)

154E-2-20Background radiation Paper 2Hard12 marks

A food-safety laboratory tests dried mushrooms for contamination by caesium-137. Caesium-137 is a β− emitter with a half-life of 30 years; most of its decays are followed by the emission of a gamma-ray photon.

A 0.40 kg sample of mushrooms is sealed in a plastic pot and placed on a gamma-ray detector inside a thick lead shield. On average the detector records one count for every 25 decays of caesium-137 in the sample. With an empty pot in place, five background counts were taken, each over 600 s. With the sample in place, 6450 counts were recorded in 600 s.

The legal limit for caesium-137 in food is 600 Bq per kilogram.

Background count over 600 s12345
Number of counts198221204230197
(a)

Outline why the detector is surrounded by a thick lead shield.

(1)
(b)

Explain why the detector records gamma rays from the sample but no β− particles.

(2)
(c)
(i)

Show that the count rate due to the caesium-137 in the sample is about 10 s−1.

(1)
(ii)

Determine whether these mushrooms may be sold.

(3)
(d)

The mushrooms were picked in a forest that was contaminated by caesium-137.

(i)

Predict the activity per kilogram of mushrooms picked in the same forest 60 years later, assuming that they take up caesium-137 from the soil in the same way.

(1)
(ii)

Suggest one reason why the actual activity could be lower than your prediction.

(1)
(e)

A second 0.40 kg sample gives 226 counts in 600 s. A technician concludes that this sample contains no caesium-137. Discuss whether the technician's conclusion is justified.

(2)
(f)

Suggest one change to the method that would allow smaller activities to be detected.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
to absorb gamma radiation from the surroundings «rocks, building materials», so that the background count is reduced✓ 1Accept 'to reduce the background count'. Do not accept 'to protect the operator'.
Part (b)
β− particles have a short range in solids and are absorbed within the mushrooms or the wall of the pot «a few mm»✓ 1
gamma rays are only weakly absorbed «most penetrating», so many pass through the sample and the pot to the detector✓ 1
Part (c)(i)
background = 1050/5 = 210 counts, so rate = (6450 − 210)/600 «= 10.40 s−1»✓ 1Must see full substitution OR answer to 3 s.f. (10.4 s−1).
Part (c)(ii)
activity of the sample = 10.4 × 25 = 260 Bq✓ 1Allow ECF from (c)(i): 10 × 25 = 250 Bq.
activity per kilogram = 260/0.40 = 650 Bq kg−1✓ 1Accept 625 Bq kg−1 from 10 s−1.
650 Bq kg−1 > 600 Bq kg−1, so the mushrooms may not be sold✓ 1The conclusion must be consistent with the candidate's value. Do not award MP3 for comparing 260 Bq «not divided by the mass» with the limit.
Part (d)(i)
60 years = 2 half-lives, so 650/4 ≈ 160 Bq kg−1✓ 1Allow ECF from (c)(ii). Accept 160–163 Bq kg−1.
Part (d)(ii)
caesium is washed deeper into the soil «out of reach of the mushrooms» / is carried away by water / is taken up less over time✓ 1Accept any sensible physical reason. OWTTE
Part (e)
226 counts lies within the range of the background counts «197 to 230», so the sample cannot be distinguished from background «random variation»✓ 1
so the conclusion is not justified: the sample may contain caesium-137 with an activity too small to detect «although any activity is far below the legal limit»✓ 1MP2 only scores if MP1 scores. OWTTE
Part (f)
count for a longer time / use a larger sample / take more background counts / use a more efficient detector✓ 1Accept any one.

Answers: (c)(i) 10.4 s−1  ·  (c)(ii) 260 Bq; 650 Bq kg−1 — above the limit, may not be sold  ·  (d)(i) ≈ 160 Bq kg−1 (the remaining parts are explanations — see the table above)

155E-2-45Mass defect & binding energy Paper 2Hard13 marks

Tritium, 31H, and helium-3, 32He, both have three nucleons. Tritium is radioactive and decays by β− emission to helium-3; helium-3 is stable. The table gives some masses. (1 u = 931.5 MeV c−2)

ParticleMass / u
proton1.0072765
neutron1.0086649
electron0.0005486
tritium nucleus3.0155007
helium-3 nucleus3.0149322
(a)

State, with a reason, whether tritium and helium-3 are isotopes.

(1)
(b)

Binding energies.

(i)

Show that the binding energy of a tritium nucleus is about 8.5 MeV.

(1)
(ii)

Calculate the binding energy per nucleon of helium-3.

(2)
(c)

Explain, with reference to the forces between the nucleons, why the binding energy of helium-3 is smaller than that of tritium.

(2)
(d)

The decay of tritium.

(i)

Complete the equation for the decay: 31H → 32He + … + …

(1)
(ii)

Determine, in keV, the energy released in the decay of one tritium nucleus.

(2)
(iii)

Discuss, using the data, the following argument made by a student: “Tritium has the larger binding energy, so it is more tightly bound than helium-3. It cannot release energy by changing into helium-3.”

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
no: they have different proton numbers «1 and 2», so they are different elements «isotopes have the same proton number»✓ 1The reason is needed for the mark.
Part (b)(i)
Δm = 1.0072765 + 2 × 1.0086649 − 3.0155007 = 0.0091056 u; E = 0.0091056 × 931.5 = 8.48 MeV✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
Δm = 2 × 1.0072765 + 1.0086649 − 3.0149322 = 0.0082857 u, so the binding energy is 7.72 MeV✓ 1
per nucleon: 7.72/3 = 2.57 MeV✓ 1Award [2] for CNA. Award [1 max] for 7.72 MeV «not divided by 3».
Part (c)
the strong nuclear force is attractive and acts in the same way between any two nucleons «p–p, n–n, n–p»; both nuclei have three nucleons✓ 1
helium-3 has two protons, which repel each other electrically, whereas tritium has only one; the repulsion means that less energy is needed to separate helium-3 into its nucleons✓ 1OWTTE
Part (d)(i)
0−1e AND ν̄✓ 1Both needed. Do not accept a neutrino.
Part (d)(ii)
Δm = 3.0155007 − 3.0149322 − 0.0005486 = 0.0000199 u✓ 1The electron mass must be subtracted when nuclear masses are used.
E = 0.0000199 × 931.5 = 0.0185 MeV ≈ 18.5 keV✓ 1Award [2] for CNA. Accept 18–19 keV. Award [1 max] for 530 keV «electron mass not subtracted».
Part (d)(iii)
whether a decay can happen depends on whether the total mass «energy» of the products is less than that of the parent, not on binding energy alone✓ 1
each binding energy compares a nucleus with its own separate nucleons, and the two nuclei are made of different nucleons «2n + 1p and 1n + 2p», so the binding energies cannot be compared directly✓ 1
a neutron has more mass than a proton and an electron together: (1.0086649 − 1.0072765 − 0.0005486) × 931.5 = 0.78 MeV✓ 1
this is larger than the difference in binding energy, 8.48 − 7.72 = 0.76 MeV, so the mass of tritium is greater than that of helium-3 plus an electron and 18.5 keV is released, as in (d)(ii): the argument is wrong✓ 1MP4 needs a comparison of the two energies and a conclusion. Allow ECF from (b) and (d)(ii).

Answers: (b)(i) 8.48 MeV  ·  (b)(ii) 2.57 MeV  ·  (d)(ii) 18.5 keV (the remaining parts are explanations — see the table above)

156E-2-69Alpha decay & recoil Paper 2Hard13 marks

An instrument package left on the Moon must be kept warm during the 14-day lunar night. Its heater is a sealed capsule containing 1.5 g of curium-242, 24296Cm, which decays by alpha emission to plutonium-238 (Pu) with a half-life of 163 days. Each decay releases 6.22 MeV, which is shared as kinetic energy between the alpha particle and the plutonium nucleus.

Each alpha particle stops inside the capsule and becomes an atom of helium. The helium collects in a free volume of 2.5 cm3 inside the capsule. Plutonium-238 decays so slowly that its decays may be ignored.

(molar mass of curium-242 = 242 g mol−1, R = 8.31 J K−1 mol−1)

Left: a curium-242 nucleus at rest. Right: just after the decay, a small alpha particle on the left and a larger plutonium-238 nucleus on the right, a short distance apart, with space around them for arrows.before the decayCmnucleus at restjust after the decayαPuPu-238
Diagram NOT accurately drawn
(a)

State the nuclear equation for this decay.

(1)
(b)

The diagram shows the curium-242 nucleus at rest before the decay, and the alpha particle and the plutonium nucleus just after it.

(i)

Draw, on the diagram, arrows to represent the momentum of the alpha particle and the momentum of the plutonium nucleus just after the decay.

(2)
(ii)

Show that (kinetic energy of the alpha particle)/(kinetic energy of the plutonium nucleus) is about 60.

(2)
(iii)

Calculate the kinetic energy of the alpha particle.

(1)
(c)

The planned mission lasts 326 days. The temperature of the capsule is 420 K.

(i)

Show that the amount of helium in the capsule at the end of the mission is about 4.6 × 10−3 mol.

(2)
(ii)

Calculate the pressure of the helium at the end of the mission.

(1)
(iii)

The capsule is designed to withstand a pressure of 7.0 × 106 Pa. Determine whether the capsule could safely be kept sealed at 420 K for 489 days.

(3)
(d)

Suggest why curium-242 would not be a suitable heat source for a mission lasting 10 years.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
24296Cm → 23894Pu + 42α✓ 1All numbers needed. Accept 42He.
Part (b)(i)
two arrows starting at the particles, pointing in opposite directions along the same line «α to the left, Pu to the right»✓ 1
arrows of equal length, each labelled✓ 1MP2 only scores if MP1 scores.
Part (b)(ii)
the total momentum is zero before the decay, so the two momenta are equal in magnitude «and opposite» after it✓ 1
Ek = p2/2m, so Eα/EPu = mPu/mα ≈ 238/4 = 59.5✓ 1Must see the use of equal momenta with Ek = p2/2m or ½mv2.
Part (b)(iii)
6.22 × 238/242 = 6.12 MeV✓ 1Accept 6.22 × 59.5/60.5. Allow ECF from (b)(ii). Do not accept 6.22 MeV.
Part (c)(i)
326 days = 2 half-lives, so 3/4 of the curium nuclei have decayed «one helium atom per decay»✓ 1
n = 0.75 × 1.5/242 = 4.65 × 10−3 mol✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (c)(ii)
p = nRT/V = 4.65 × 10−3 × 8.31 × 420/2.5 × 10−6 = 6.5 × 106 Pa✓ 1Volume must be in m3. Allow ECF from (c)(i).
Part (c)(iii)
489 days = 3 half-lives, so 7/8 of the curium has decayed: n = 0.875 × 1.5/242 = 5.42 × 10−3 mol✓ 1
p = 7.6 × 106 Pa✓ 1ALTERNATIVE: p ∝ n, so p = 6.5 × 106 × (7/8)/(3/4). Allow ECF from (c)(ii).
7.6 × 106 Pa > 7.0 × 106 Pa, so the capsule could not safely be kept sealed «it must be vented or made stronger»✓ 1MP3 needs a comparison and a consistent conclusion.
Part (d)
10 years is more than 20 half-lives «≈ 22», so almost all of the curium-242 will have decayed and the heating power becomes negligible✓ 1Accept: the power halves every 163 days, so it falls to a tiny fraction long before 10 years. OWTTE

Answers: (b)(ii) 59.5  ·  (b)(iii) 6.12 MeV  ·  (c)(i) 4.65 × 10−3 mol  ·  (c)(ii) 6.5 × 106 Pa  ·  (c)(iii) 7.6 × 106 Pa — not safe (the remaining parts are explanations — see the table above)

157E-2-70Activity & count rate Paper 2Hard14 marks

A district-heating network is a closed system of underground pipes that carries hot water to the buildings of a town. To find the volume of water in the network, an engineer uses bromine-82, which emits beta particles and gamma rays and has a half-life of 35 hours. Bromine-82 dissolves in water and becomes spread uniformly through it.

A solution containing bromine-82 is prepared. 100 cm3 of this solution is injected into the network. At the same time, 1.00 cm3 of the same solution is mixed with 20.0 L of water to make a standard.

70 hours later, when the bromine-82 has spread uniformly through the network, a 1.00 L sample of network water is taken. This sample and a 1.00 L sample of the standard are then counted, one after the other, in the same plastic beaker in front of the same gamma-ray detector, for 1000 s each. Before the injection, a 1.00 L sample of network water was counted in the same way. The uncertainty in a number of counts N is ±√N.

Sample in the beakerCounts in 1000 s
network water before the injection2000
standard152 000
network water 70 h after the injection3000
(a)

Outline why the background count is measured with a sample of network water in the beaker, rather than with the beaker empty.

(1)
(b)
(i)

Show that the count rate due to the bromine-82 in the sample of network water is about 1.0 s−1.

(1)
(ii)

Determine the volume of water in the network.

(3)
(c)

Explain why the engineer does not need to correct the count rates for the decay of the bromine-82 during the 70 hours.

(2)
(d)

The detector records 4.0 % of the decays that take place in a 1.00 L sample in the beaker. Determine the activity of the bromine-82 that was injected into the network.

(3)
(e)

Suggest why bromine-82 is more suitable for this measurement than a gamma emitter with a half-life of 15 minutes and than a gamma emitter with a half-life of 30 years.

(2)
(f)

Estimate the percentage uncertainty in the count due to the bromine-82 in the sample of network water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the water itself «and the beaker» contains naturally radioactive nuclides «e.g. potassium-40, radon» and the sample absorbs some of the background radiation, so the background must be measured in exactly the arrangement used for the samples✓ 1OWTTE
Part (b)(i)
(3000 − 2000)/1000 = 1.00 s−1✓ 1Must see full substitution OR answer to at least 3 s.f. The background must be subtracted.
Part (b)(ii)
count rate due to the standard = (152000 − 2000)/1000 = 150 s−1, which corresponds to 1.00 cm3 of solution in 20.0 L «0.050 cm3 per litre»✓ 1
each litre of network water contains 0.050 × 1.00/150 = 3.3 × 10−4 cm3 of the solution✓ 1Allow ECF from (b)(i).
V = 100/3.3 × 10−4 = 3.0 × 105 L «300 m3»✓ 1Award [3] for CNA. Award [2 max] for 1.0 × 105 L «background not subtracted from the network count».
Part (c)
the standard and the network sample come from the same solution, and their bromine-82 decays with the same half-life, so both activities fall by the same factor «22» before counting✓ 1
the volume depends only on the ratio of the two count rates «the dilution», which decay does not change, because the samples are counted within a short time of each other✓ 1OWTTE
Part (d)
activity of a 1.00 L sample = 1.00/0.04 = 25 Bq, so the activity in the network when counted = 25 × 3.0 × 105 = 7.5 × 106 Bq✓ 1Allow ECF from (b).
70 h = 2 half-lives, so the activity was 22 = 4 times larger at the injection✓ 1
injected activity = 3.0 × 107 Bq✓ 1Award [3] for CNA. Award [2 max] for 1.9 × 106 Bq «decay applied the wrong way».
Part (e)
15 minutes: the bromine needs about 70 h to spread through the network, which is 280 half-lives of that isotope, so its activity would be far too small to distinguish from background✓ 1OWTTE
30 years: the water in the network «and any water drained or leaked from it» would stay radioactive for decades after the measurement, whereas bromine-82 has decayed to a negligible activity within a few weeks✓ 1OWTTE
Part (f)
net count = 3000 − 2000 = 1000; uncertainty = √3000 + √2000 = 54.8 + 44.7 = 99.5 counts✓ 1Absolute uncertainties add when the background is subtracted.
percentage uncertainty = 99.5/1000 × 100 ≈ 10 %✓ 1Accept 9–10 %. Award [1 max] for 5.5 % «uncertainty in the background ignored».

Answers: (b)(i) 1.00 s−1  ·  (b)(ii) 3.0 × 105 L «300 m3»  ·  (d) 3.0 × 107 Bq  ·  (f) ≈ 10 % (the remaining parts are explanations — see the table above)

158E-2-75The strong nuclear force Paper 2Hard13 marks

Tennessine, element 117, does not occur naturally. Its nuclei are made by firing a beam of calcium-48 nuclei, 4820Ca, at a target containing berkelium-249, 24997Bk. Very rarely a calcium nucleus fuses with a berkelium nucleus to form tennessine-294, 294117Ts, and some free neutrons.

Each tennessine nucleus is carried into a silicon detector, which is divided into small pixels. The pixel that it enters records its arrival and then the alpha particles emitted by it and by its decay products.

(1 u = 931.5 MeV c−2)

Atom12345
Time from arrival to alpha decay / ms1852913144
(a)
(i)

Deduce the number of neutrons released when one nucleus of tennessine-294 is formed.

(1)
(ii)

Tennessine-294 and its decay products undergo six successive alpha decays. Determine the proton number and the nucleon number of the nucleus that is formed.

(1)
(b)

Explain, with reference to the forces between nuclei, why the calcium-48 nuclei must have a very large kinetic energy for fusion to take place.

(2)
(c)

The table shows the time between the arrival and the alpha decay for each of five tennessine-294 nuclei.

(i)

Explain why identical nuclei decay after such different times.

(2)
(ii)

Suggest why the half-life of tennessine-294 found from these data has a large uncertainty.

(1)
(d)

In a pixel, events that look like alpha decays are caused by background radiation at an average rate of one every 2.0 hours. In one event, the arrival of a nucleus and three alpha decays were all recorded in the same pixel within 12 s. Explain why the experimenters are confident that these decays were not caused by background radiation.

(2)
(e)

The alpha particle emitted by tennessine-294 has a kinetic energy of 10.8 MeV and the recoiling nucleus has a kinetic energy of 0.15 MeV. Calculate, in u, the decrease in mass in this decay.

(2)
(f)

The alpha particles are emitted in random directions, yet almost all of them are recorded in the same pixel as the tennessine nucleus that emitted them. Each pixel is about 1 mm across. Suggest why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
48 + 249 − 294 = 3 neutrons✓ 1Proton numbers balance: 20 + 97 = 117.
Part (a)(ii)
proton number 117 − 6 × 2 = 105 AND nucleon number 294 − 6 × 4 = 270✓ 1Both needed.
Part (b)
both nuclei are positively charged, so they repel electrically, and the repulsion increases as they approach✓ 1
the strong nuclear force that would bind them is attractive but has a very short range «≈ 10−15 m», so the calcium nucleus must have enough kinetic energy to get close enough «almost touching» for it to act✓ 1OWTTE
Part (c)(i)
radioactive decay is random: the moment at which a particular nucleus decays cannot be predicted✓ 1
each nucleus has the same probability of decaying in a given time, so identical nuclei decay at times that vary widely «from 9 ms to 131 ms»✓ 1MP2 should refer to the data. OWTTE
Part (c)(ii)
there are very few nuclei, so the random variation in the decay times is large compared with their mean «half-life is a statistical quantity for large numbers of nuclei»✓ 1OWTTE
Part (d)
expected number of background events in 12 s = 12/7200 ≈ 1.7 × 10−3✓ 1Accept a comparison of 12 s with 2 hours.
so it is extremely unlikely that even one, let alone three, of the alpha events within 12 s in the same pixel was background✓ 1MP2 needs a conclusion linked to the small expected number. OWTTE
Part (e)
energy released = 10.8 + 0.15 = 10.95 MeV✓ 1
Δm = 10.95/931.5 = 0.0118 u✓ 1Award [2] for CNA. Award [1 max] for 0.0116 u «recoil energy omitted».
Part (f)
alpha particles are strongly ionizing «they produce many ion pairs per unit length of path», so they lose their kinetic energy over a very short distance in the silicon✓ 1
their range «a small fraction of a millimetre» is therefore much smaller than the width of a pixel, and each alpha particle stops, depositing its energy, close to the point where the nucleus came to rest✓ 1OWTTE

Answers: (a)(i) 3  ·  (a)(ii) Z = 105, A = 270  ·  (d) 1.7 × 10−3  ·  (e) 0.0118 u (the remaining parts are explanations — see the table above)

159IBP2S-038-02-15The binding energy curve Paper 2Hard8 marks

The table gives the binding energy per nucleon for some nuclides being considered by an engineering team assessing fusion and fission fuels.

Nuclide: 2H, 4He, 12C, 56Fe, 238U
Binding energy per nucleon / MeV: 1.11, 7.07, 7.68, 8.79, 7.57

Use 1 u = 931.5 MeV c−2.

0501001502002500246810Nucleon number ABinding energy per nucleon / MeV²H⁴He¹²C⁵⁶Fe²³⁸U
Plotted from the table in the question
(a)

Calculate the mass defect of a helium-4 nucleus, in u.

(2)
(b)

The binding energy per nucleon changes far less between 12C and 238U than it does between 2H and 4He. Outline, with reference to the strong nuclear force, why this is so.

(2)
(c)

Explain why the binding energy per nucleon of uranium-238 is lower than that of iron-56.

(2)
(d)

Determine the energy released when two 2H nuclei fuse to form one 4He nucleus.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Total binding energy = 4 × 7.07 = 28.28 MeV.✓ 1
Mass defect = 28.28 / 931.5 = 0.0304 u.✓ 1
Part (b)
The strong force is short-range, so each nucleon attracts only its nearest neighbours.✓ 1
In larger nuclei most nucleons have a full set of neighbours, so each contributes a similar binding energy, whereas in very light nuclei most nucleons are at the surface with few neighbours.✓ 1
Part (c)
The electrostatic repulsion is long-range, so each proton repels every other proton in the nucleus.✓ 1
In a large nucleus the total repulsion grows faster than the short-range strong attraction, which reduces the binding energy per nucleon.✓ 1
Part (d)
Binding energy of 4He = 28.28 MeV, and the binding energy of two 2H nuclei = 2 × 2 × 1.11 = 4.44 MeV.✓ 1
Energy released = 28.28 − 4.44 = 23.8 MeV.✓ 1
Allow error carried forward from the 4He binding energy calculated in the first part.—

Common error: Multiply the binding energy per nucleon by the nucleon number before comparing reactants and products. Subtracting the per-nucleon values directly gives a meaningless answer.

160IBP2S-038-05-15Beta decay & the neutrino Paper 2Hard10 marks

Fluorine-18 is widely used in positron emission tomography (PET). It decays by β+ emission to oxygen-18 with a half-life of 110 minutes.
Atomic masses: fluorine-18 = 18.000 938 u; oxygen-18 = 17.999 160 u.
Electron mass = 0.000 549 u; 1 u = 931.5 MeV c−2.

(a)

Outline two reasons why fluorine-18 is suitable for imaging a patient.

(2)
(b)

Write down the full nuclear equation for the decay, including proton and nucleon numbers.

(2)
(c)

A dose with an activity of 400 MBq is prepared at 08:00. Calculate its activity when the scan starts at 11:40.

(2)
(d)

Explain why, when atomic masses are used, two electron masses must be subtracted from the difference in atomic masses to find the energy released.

(2)
(e)

Determine the maximum kinetic energy shared by the positron and neutrino.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
It emits positrons. Each positron annihilates with an electron to give two gamma photons, which are penetrating enough to leave the body and be detected outside it.✓ 1
Its half-life (110 min, about 2 h) is long enough to prepare the tracer and complete a scan, but short enough that the activity, and so the patient's dose, falls quickly afterwards.✓ 1
Part (b)
189F → 188O + 0+1e, with all nucleon and proton numbers correct. Accept 0+1β or e+.✓ 1
+ νe, an electron neutrino (not an antineutrino); accept 00ν.✓ 1
Part (c)
Elapsed time = 220 min = 2 half-lives.✓ 1
A = 400 / 22 = 100 MBq.✓ 1
Part (d)
The neutral fluorine atom contains 9 electrons but the neutral oxygen atom only 8. So the difference in atomic masses is one electron mass greater than the difference in nuclear masses.✓ 1
The decay also creates a positron, whose mass equals one electron mass, and this must be subtracted from the nuclear mass difference. In total, 2me is subtracted from the atomic mass difference.✓ 1
Part (e)
Δm = 18.000 938 − 17.999 160 − 2 × 0.000 549 = 0.000 680 u.✓ 1
E = 0.000 680 × 931.5 ≈ 0.63 MeV.✓ 1

Common error: Forgetting the two electron masses in β⁺ decay gives 1.66 MeV instead of 0.63 MeV. With nuclear masses, subtract one positron mass. With atomic masses, subtract two electron masses.

161IBP2S-038-05-16The binding energy curve Paper 2Hard9 marks

The table gives binding energies per nucleon for some nuclides.

Nuclide — binding energy per nucleon / MeV
hydrogen-2 — 1.11
helium-4 — 7.07
iron-56 — 8.79
krypton-92 — 8.51
barium-141 — 8.33
uranium-235 — 7.59

A submarine reactor uses the fission reaction
23592U + 10n → 14156Ba + 9236Kr + 3 10n
1 u = 1.66 × 10−27 kg; 1 MeV = 1.60 × 10−13 J.

(a)

State which nuclide in the table is the most stable.

(1)
(b)

Using the data, explain why both the fusion of light nuclei and the fission of heavy nuclei can release energy.

(2)
(c)

Determine the energy released in one fission.

(3)
(d)

The reactor's thermal power is 150 MW. Estimate the mass of uranium-235 that undergoes fission in one day.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Iron-56, which has the greatest binding energy per nucleon✓ 1
Part (b)
Binding energy per nucleon rises from light nuclei (e.g. hydrogen-2) to a maximum near iron-56, then falls for heavy nuclei (e.g. uranium-235).✓ 1
Fusion of light nuclei and fission of heavy nuclei both produce nuclei closer to iron, with higher binding energy per nucleon. The total binding energy therefore increases, and the difference is released as energy.✓ 1
Part (c)
Binding energy of products = 141 × 8.33 + 92 × 8.51 = 1174.5 + 782.9 = 1957.5 MeV✓ 1
Binding energy of uranium-235 = 235 × 7.59 = 1783.7 MeV✓ 1
Energy released = 1957.5 − 1783.7 ≈ 174 MeV (accept 173–174 MeV)✓ 1
Part (d)
Energy per fission = 174 × 1.60 × 10−13 = 2.78 × 10−11 J✓ 1
Fissions per day = (1.50 × 108 × 86 400) / 2.78 × 10−11 ≈ 4.7 × 1023✓ 1
Mass = 4.7 × 1023 × 235 × 1.66 × 10−27 ≈ 0.18 kg✓ 1
Allow error carried forward from the answer to (c).—

Exam tip: Free neutrons have zero binding energy. Only the uranium nucleus counts on the left-hand side and only the barium and krypton nuclei count on the right.

162IBP2S-038-03-15Beta decay & the neutrino Paper 2Hard8 marks

Fluorine-18 (189F) is a β+ emitter used in PET imaging. It decays to oxygen-18.
Atomic mass of fluorine-18 = 18.000938 u; atomic mass of oxygen-18 = 17.999159 u; 1 u = 931.5 MeV c−2; rest energy of electron = 0.511 MeV.

(a)

Write down the nuclear equation for this decay.

(2)
(b)

Explain why, when atomic masses are used, the energy released is (difference in atomic masses)c2 − 2mec2.

(2)
(c)

Calculate the maximum kinetic energy of the emitted positron.

(2)
(d)

The PET scanner detects the gamma photons from annihilation but not the neutrinos. Outline why the neutrinos are almost never detected.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
189F → 188O + 0+1e✓ 1
+ electron neutrino, νe, included✓ 1
Part (b)
The fluorine atom has 9 electrons, but the oxygen atom formed has only 8, so the atomic-mass difference includes one surplus electron mass .✓ 1
A positron of mass me is also created, so a further me must be subtracted, making 2me in total .✓ 1
Part (c)
(18.000938 − 17.999159) × 931.5 = 1.657 MeV .✓ 1
1.657 − 2 × 0.511 = 0.635 MeV (≈0.63 MeV) . ECF for arithmetic errors.✓ 1
Part (d)
Neutrinos are uncharged and have very small mass .✓ 1
They interact only very weakly with matter, so almost all pass straight through the patient and the detector .✓ 1

Common error: In β⁺ decay the electron masses do not cancel. Forgetting the 2mec² correction more than doubles the answer.

163IBP2S-038-00-15Mass–energy equivalence Paper 2Hard9 marks

Radioisotope power systems on deep-space probes are heated by the alpha decay of plutonium-238 (proton number 94) to uranium-234.
Data: mass of Pu-238 atom = 238.049560 u; mass of U-234 atom = 234.040952 u; mass of He-4 atom = 4.002603 u; 1 u = 931.5 MeV c−2.

(a)

Write the nuclear equation for this decay.

(2)
(b)

Explain why an alpha emitter is particularly suitable as a heat source on a spacecraft.

(2)
(c)

Show that the energy released in one decay is about 5.6 MeV.

(2)
(d)

The Pu-238 nucleus is initially at rest. Determine the kinetic energy of the emitted alpha particle.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\({}^{238}_{\;94}\text{Pu} \rightarrow {}^{234}_{\;92}\text{U} + {}^{4}_{2}\alpha\)—
Uranium with nucleon number 234 and proton number 92✓ 1
Alpha particle with correct numbers✓ 1
Part (b)
Alpha particles are absorbed within the fuel or its casing, so almost all of their kinetic energy is converted to internal energy (heat)✓ 1
Little penetrating radiation escapes, so only light shielding is needed to protect the instruments and electronics, which saves launch mass✓ 1
Part (c)
Mass defect Δm = 238.049560 − 234.040952 − 4.002603 = 0.006005 u✓ 1
E = 0.006005 × 931.5 = 5.59 MeV ≈ 5.6 MeV✓ 1
Part (d)
Momentum is conserved, so the alpha particle and the U-234 nucleus have equal and opposite momenta✓ 1
Ek = p2/2m, so the kinetic energy is shared in inverse proportion to mass: Eα = Q × 234/(234 + 4)✓ 1
Eα = 5.59 × 234/238 = 5.50 MeV✓ 1
ECF from (c)—

Exam tip: Atomic masses can be used directly for alpha decay because the electron masses cancel. Carry at least six decimal places in the mass difference, because it is a small difference between large numbers.