Both levels cover mass defect and binding energy, the binding energy per nucleon curve, the strong force, the properties of alpha, beta and gamma radiation, half-life and background radiation.
HL adds the decay constant and the exponential decay law, N = N₀e^(−λt) and A = λN, the link T½ = ln 2/λ, methods for measuring very long and very short half-lives, and the evidence that alpha and gamma spectra give for nuclear energy levels. Dating and medical tracers provide the contexts.
57 questions
298 marks
Paper 1A: 31
Paper 1B: 10
Paper 2: 16
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29 practice questions on E.3 Radioactive decay
1E-1A-07
Activity and decay law·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A sample contains a mass m of a pure radioactive nuclide. The molar mass of the nuclide is M and its decay constant is λ. The Avogadro constant is NA.
What is the activity of the sample?
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Notes
Step 1Number of moles = m/M, so the number of nuclei is N = mNA/M.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Activity A = λN = λmNA/M.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis divides by the Avogadro constant instead of multiplying: it gives a number of moles per second, not decays per second.
BThis divides by λ instead of multiplying (N/λ has the unit of a number × time, not s⁻¹).
CThe ratio m/M has been inverted, so a larger sample would give a smaller activity.
DCorrect: A = λN with N = (m/M)NA.
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN; B.3 — the amount of substance n = N/NACommand term: Determine
2E-1A-11
Activity and decay law·E.3 Radioactive decay
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two radioactive samples X and Y have equal activities at time t = 0. The half-life of X is 2.0 h and the half-life of Y is 6.0 h.
What is the ratio (number of undecayed nuclei in X) / (number of undecayed nuclei in Y) at t = 6.0 h?
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Notes
Step 1At t = 0, N = A/λ ∝ T½, so NX/NY = 2.0/6.0 = 1/3.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In 6.0 h, X passes through 3 half-lives (factor 1/8) and Y through 1 half-life (factor 1/2), so the ratio is multiplied by (1/8)/(1/2) = 1/4.
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Step 3Ratio = 1/3 × 1/4 = 1/12.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: (1/3) × (1/4).
BThis assumes the samples start with equal numbers of nuclei; equal activities mean the longer-lived Y has three times as many.
CThis is the ratio at t = 0; the decay over 6.0 h has been ignored.
DThis takes N ∝ λ (the starting ratio inverted to 3) before applying the factor 1/4.
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN; the relationship between half-life and the decay constant; the changes in activity using integer values of half-life Command term: Deduce
3E-1A-12
Mass defect and binding energy·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A nucleus contains Z protons and N neutrons. The mass of the nucleus is M, the mass of a proton is mp and the mass of a neutron is mn.
What is the binding energy of the nucleus?
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Notes
Step 1The mass defect is the separate nucleons' mass minus the nucleus's mass: Δm = Zmp + Nmn − M (positive).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Binding energy = Δmc² = (Zmp + Nmn − M)c².
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe subtraction is reversed: the nucleus has less mass than its separate nucleons, so this gives a negative energy.
BCorrect: binding energy = mass defect × c².
CThis divides by c² instead of multiplying (E = mc²).
DThis is the binding energy per nucleon, not the binding energy of the whole nucleus.
Syllabus understandingE.3 — nuclear binding energy and mass defect; the mass–energy equivalence as given by E = mc² in nuclear reactions Command term: Determine
4E-1A-13
Activity and decay law·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The activity of a radioactive sample falls from A0 to A in a time t. The time t is not a whole number of half-lives.
What is the half-life of the nuclide?
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Notes
Step 1A = A0e−λt, so λt = ln (A0/A) and λ = ln (A0/A)/t.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2T½ = ln 2/λ = t ln 2 / ln (A0/A).
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is t × λt/ln 2: the ratio of the logarithms has been inverted, so a faster decay would give a longer half-life.
BThe ratio inside the logarithm is inverted: ln (A/A0) is negative, giving a negative half-life.
CCorrect: λ = ln (A0/A)/t and T½ = ln 2/λ.
DThe logarithm of the ratio has been omitted; this treats the activity as falling in proportion to time rather than exponentially.
Syllabus understandingE.3 (HL) — the decay constant λ and the radioactive decay law N = N0e−λt; the activity A = λN0e−λt; T½ = ln 2/λ; application of the decay equations for arbitrary time intervals Command term: Determine
5E-1A-15
Binding energy per nucleon·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The graph shows how the binding energy per nucleon varies with nucleon number. Four nuclides are marked.
Which nucleus has the greatest total binding energy?
Binding energy per nucleon against nucleon number, with four nuclides marked.Show mark scheme
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Notes
Step 1Total binding energy = (binding energy per nucleon) × A.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3The binding energy per nucleon varies by only about 20 % between these nuclei, while A varies by a factor of 60, so U-238 has the largest total.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AHelium-4 is unusually tightly bound for a light nucleus (a peak on the curve), but it has only 4 nucleons: total ≈ 28 MeV.
BIron-56 has the greatest binding energy per nucleon (the most stable nucleus), not the greatest total: ≈ 493 MeV.
CTin-120 has a high value per nucleon and many nucleons, but only half as many as uranium-238: ≈ 1020 MeV.
DCorrect: the lowest value per nucleon of the heavy nuclei, but 238 nucleons give the largest total, ≈ 1809 MeV.
Syllabus understandingE.3 — nuclear binding energy and mass defect; the variation of the binding energy per nucleon with nucleon number (an interpretation of binding energy curves is required) Command term: Deduce
6E-1A-19
Absorption of radiation·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Sheets of aluminium of increasing total thickness are placed between a radioactive source and a detector, and the count rate is recorded. The graph shows the results; the dashed line is the background count rate.
When a single sheet of paper is placed between the source and the detector, the count rate does not change.
Which radiations does the source emit?
Count rate against thickness of aluminium between source and detector (dashed line: background).Show mark scheme
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Notes
Step 1Paper stops alpha particles; since paper causes no change, the source emits no alpha radiation.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The steep fall over the first 3 mm shows radiation stopped by a few millimetres of aluminium: beta.
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Step 3Beyond 3 mm the count rate stays between about 150 and 180 min⁻¹, far above the background of 25 min⁻¹, and falls only slowly: this is penetrating gamma radiation.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the level part of the graph as background; it is about six times the background, so another radiation must reach the detector.
BAlpha particles would be stopped by the paper and the count rate would fall; and the plateau above background shows gamma radiation.
CCorrect: beta is stopped by about 3 mm of aluminium; the remaining count above background is gamma.
DThe paper would have absorbed any alpha particles and reduced the count rate, but it did not.
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the effect of background radiation on count rate (thickness of materials as a real-life context) Command term: Deduce
7E-1A-24
Decay constant and half-life·E.3 Radioactive decay
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows how ln (A/Bq) varies with time t for a pure radioactive sample, where A is its activity. The line meets the vertical axis at c and the time axis at t1.
What was the number of undecayed nuclei in the sample at t = 0?
The natural logarithm of the activity A against time t (sketch).Show mark scheme
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Notes
Step 1ln A = ln A0 − λt: the intercept gives A0 = ec Bq.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The gradient is −c/t1 = −λ, so λ = c/t1.
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Step 3N0 = A0/λ = ect1/c.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses the intercept c as the initial activity itself instead of ln A0 (c ÷ c/t1).
BCorrect: N0 = A0/λ with A0 = ec and λ = c/t1.
CThe gradient has been inverted (λ taken as t1/c).
DThis is the initial activity A0, not the number of nuclei: it has not been divided by λ.
Syllabus understandingE.3 (HL) — the decay constant λ and the radioactive decay law N = N0e−λt; the activity as the rate of decay A = λN; Tool 3 — linearize graphs and interpret gradient and intercept Command term: Determine
8E-1A-29
Decays on the N–Z chart·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The chart shows nuclide X (Z = 88, N = 130) and four other positions.
Which position represents the nucleus formed when X undergoes an alpha decay followed by a beta-minus decay?
Part of the chart of nuclides: X is the parent nucleus; P, Q, R and S are possible positions of a daughter nucleus.Show mark scheme
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Notes
Step 1Alpha decay removes 2 protons and 2 neutrons: (Z, N) = (88, 130) → (86, 128), which is position P.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Beta-minus decay converts a neutron into a proton: Z → Z + 1, N → N − 1: (86, 128) → (87, 127).
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Step 3Position R is at (87, 127) — one step right and one step down from P.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AP is the nucleus after the alpha decay alone.
BQ (86, 130) would need the alpha decay to remove only protons.
CCorrect: (88 − 2 + 1, 130 − 2 − 1) = (87, 127).
DS (84, 126) is the result of two alpha decays.
Syllabus understandingE.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving α, β−, β+, γ Command term: Deduce
9E-1A-40
Beta decay and the neutrino·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In every β⁻ decay of a particular nuclide, the parent nucleus changes to the daughter nucleus in its ground state, and the total energy released is 1.20 MeV. The graph shows the measured kinetic energies of the β⁻ particles emitted from a thin source in a vacuum.
Which conclusion is supported by the graph?
Distribution of the kinetic energies of the β⁻ particles from a thin source in a vacuum.Show mark scheme
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Notes
Step 1If only two bodies (β⁻ particle and daughter nucleus) shared a fixed 1.20 MeV, momentum and energy conservation would give every β⁻ particle the same energy: a single line.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The spectrum is continuous from 0 up to a maximum of 1.20 MeV.
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Step 3So a third, undetected particle — the antineutrino — shares the energy, and in each decay the total is 1.20 MeV.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ADiscrete excited states would give a few discrete β energies (a line spectrum), and the stem says the daughter is formed in its ground state.
BThis was once proposed, but the maximum energy equals the full 1.20 MeV in every case; conservation holds in each decay when the antineutrino is included.
CThe source is thin and in a vacuum, so energy losses on the way out are negligible; they could not produce a sharp upper limit at exactly 1.20 MeV.
DCorrect: the continuous spectrum with an end-point at the decay energy is evidence for the antineutrino.
Syllabus understandingE.3 (HL) — the continuous spectrum of beta decay as evidence for the neutrino; E.3 — the existence of neutrinos ν and antineutrinos ν̄ Command term: Deduce
10E-1A-41
Uses of radioisotopes·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
A water company needs to find a leak in a pipe buried about 1 m below the ground. A small quantity of a radioisotope is dissolved in the water in the pipe and a detector is moved along the ground surface above the pipe. Four isotopes are available:
Step 1The radiation must pass through about 1 m of soil to reach the detector, so only gamma radiation is penetrating enough.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The half-life must be long enough for the search (hours) but short enough that the water supply is not contaminated for long: 12 hours rather than 30 years.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AAlpha particles are the most ionising but are stopped by a few centimetres of air or a sheet of paper; none would reach the surface through 1 m of soil.
BCorrect: gamma rays penetrate the soil, and after a few days (several half-lives of 12 h) the activity in the water is negligible.
CThe radiation is suitable, but a 30-year half-life would leave the drinking water radioactive for decades.
DThe half-life is suitable, but beta particles are absorbed by a few millimetres of aluminium, and so by far less than 1 m of soil.
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; guidance: the choice of isotope for leaks in underground pipes Command term: Identify
11E-1A-42
Beta-plus decay·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
Sodium-22 (2211Na), used as a positron source in teaching laboratories, decays by β+ emission.
Which row gives the daughter nucleus and the particle emitted together with the positron?
Daughter nucleusOther particle emitted
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Notes
Step 1In β+ decay a proton in the nucleus becomes a neutron: Z falls by 1 and A is unchanged, giving 2210Ne.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The positron is emitted with a neutrino: 2211Na → 2210Ne + 0+1e + ν (charge: 11 = 10 + 1).
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: 2211Na → 2210Ne + 0+1e + ν.
BThe antineutrino accompanies an electron in β− decay; the positron in β+ decay is accompanied by a neutrino.
CThis is the β− result: proton number increased by 1 and an antineutrino emitted.
DThe proton number has been increased, as in β− decay; charge would not be conserved with a positron emitted (11 ≠ 12 + 1).
Syllabus understandingE.3 — the radioactive decay equations involving α, β−, β+, γ; the existence of neutrinos ν and antineutrinos ν̄ Command term: Identify
12E-1A-43
Nuclear energy levels·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
After an alpha decay the daughter nucleus may be left in either of two excited states, 0.24 MeV and 0.61 MeV above its ground state, as shown in the diagram. It returns to the ground state by emitting one or more gamma-ray photons.
Which list gives all the possible gamma-ray photon energies?
Energy levels of the daughter nucleus (to scale).Show mark scheme
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Notes
Step 1Each photon energy equals the difference between two nuclear energy levels.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3Three discrete energies: this line spectrum is evidence for discrete nuclear energy levels.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AOmits the transition between the two excited states (0.61 → 0.24 MeV), which gives 0.37 MeV.
B0.85 MeV is the sum 0.24 + 0.61, which is not the difference between any two levels.
CUses the difference and the sum of the two excitation energies and ignores the direct transitions to the ground state.
DCorrect: the three possible level differences.
Syllabus understandingE.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; the changes in the state of the nucleus following gamma decay Command term: Deduce
13E-1A-45
Binding energy per nucleon·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksExplain
For nuclei with nucleon number greater than about 60, the binding energy per nucleon is approximately constant at about 8 MeV.
Which statement best explains this?
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Notes
Step 1The strong nuclear force has a range of about 10−15 m, similar to the separation of neighbouring nucleons.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In a large nucleus each nucleon is surrounded by roughly the same number of neighbours whatever the size of the nucleus.
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Step 3So each added nucleon adds roughly the same binding energy, and binding energy per nucleon stays nearly constant (total binding energy ∝ A).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the force saturates — a nucleon interacts only with its neighbours.
BAn inverse-square force is long-range (like the electric force); every nucleon would then attract all others and the binding energy per nucleon would keep rising with A.
CThe repulsion is not negligible: it grows roughly as Z² and causes the slow fall of binding energy per nucleon from about 8.8 MeV to 7.6 MeV for the heaviest nuclei.
DConfuses total with per-nucleon binding energy; the total rises almost in proportion to A (about 490 MeV for A = 56, about 1800 MeV for A = 238).
Syllabus understandingE.3 (HL) — the approximate constancy of binding energy curve above a nucleon number of 60; the existence of the strong nuclear force, a short-range, attractive force between nucleons Command term: Explain
14E-1A-47
Background radiation·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A detector placed near a radioactive sample records 680 counts per minute. The background count rate is 40 counts per minute. The half-life of the sample is 3.0 hours.
After what time will the detector record 80 counts per minute?
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Notes
Step 1Count rate due to the sample at the start = 680 − 40 = 640 counts min−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2When the detector records 80 counts min−1, the sample contributes only 80 − 40 = 40 counts min−1, because the background does not decay.
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Step 3640/40 = 16 = 24: four half-lives, so the time is 4 × 3.0 h = 12.0 h.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis subtracts the background from the first reading only: 640/80 = 8 = 23, three half-lives.
BNo background correction: 680/80 = 8.5, which is 23.09, so 3.09 half-lives; the background is wrongly allowed to decay.
CCorrect: (680 − 40)/(80 − 40) = 16 = 24, so four half-lives.
DThis subtracts the background from the final reading only: 680/40 = 17 = 24.09, so 4.09 half-lives.
Syllabus understandingE.3 — the effect of background radiation on count rate; the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Determine
15E-1A-49
Decay constant as a probability·E.3 Radioactive decay
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A radioactive nuclide has a decay constant λ = 0.10 s−1.
What is the probability that a particular nucleus of this nuclide decays within the next 5.0 s?
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Notes
Step 1The probability of the nucleus surviving for time t is N/N0 = e−λt = e−0.50 = 0.61.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Probability of decaying = 1 − 0.61 = 0.39.
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Step 3λt = 0.50 is not small, so λt is not a valid estimate here; it works only for small λt, e.g. for 0.10 s: 1 − e−0.010 = 0.00995 ≈ λt = 0.010.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is λ itself, the approximate probability of decay in 1 s, not in 5.0 s.
BCorrect: 1 − e−0.50 = 0.39.
CTreats λ as the probability per unit time over a long interval: λt = 0.10 × 5.0 overestimates because λt is not small.
DThis is the probability that the nucleus does not decay in 5.0 s.
Syllabus understandingE.3 (HL) — that the decay constant approximates the probability of decay in unit time only in the limit of sufficiently small λt; the radioactive decay law N = N0e−λtCommand term: Determine
16E-1A-65
The random nature of decay·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Three statements about a sample of a radioactive nuclide are:
I. Heating the sample to a much higher temperature does not change its activity. II. It is not possible to predict which nucleus in the sample will decay next. III. A nucleus that has not decayed during one half-life has a probability of one half of decaying during the next half-life.
Which statements are correct?
Show mark scheme
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Notes
Step 1Decay is spontaneous: the decay constant is a property of the nucleus and is not affected by temperature or other physical or chemical conditions, so I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Decay is random: every undecayed nucleus has the same probability of decaying in the next second, so which one decays next cannot be predicted; II is correct.
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Step 3A nucleus does not "age": its probability of decaying in any interval of one half-life is ½, whatever its history, so III is correct.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis rejects III, treating a nucleus that has survived for one half-life as being more likely ("due") to decay soon; the probability of decay does not depend on how long the nucleus has existed.
BThis rejects II, assuming that the oldest nuclei decay first; but every undecayed nucleus has the same chance of decaying next.
CThis rejects I by analogy with chemical reactions, whose rates rise with temperature; nuclear decay is unaffected by the temperatures reached in a laboratory.
DCorrect: decay is spontaneous (I) and random (II), and the probability of decay per unit time is constant (III).
Syllabus understandingE.3 — the random and spontaneous nature of radioactive decay; the activity, count rate and half-life in radioactive decay Command term: Deduce
17E-1A-66
Activity and decay law·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows how the activity A of a pure sample of a radioactive nuclide varies with time t.
What fraction of the nuclei originally present has decayed at t = 12 h?
Activity against time (graph drawn to scale)Show mark scheme
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Notes
Step 1From the graph the activity falls from 800 Bq to 400 Bq in 3.0 h, so the half-life is 3.0 h.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 212 h is four half-lives, so the fraction of nuclei remaining is (½)⁴ = 1/16 (the graph confirms A = 50 Bq = 800/16 Bq, and A ∝ N).
—
Step 3The fraction that has decayed is 1 − 1/16 = 15/16.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is the fraction remaining, not the fraction that has decayed.
BThis is 1 − ¼: it divides by the number of half-lives (4) instead of by 2⁴.
CThis is 1 − (½)³: it counts only three halvings between 3 h and 12 h, forgetting the first half-life from 0 to 3 h.
DCorrect: four half-lives leave 1/16, so 15/16 has decayed.
Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Determine
18E-1A-67
Decay equations·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Thorium-232 (Z = 90) decays through a series of α and β− decays to the stable nuclide lead-208 (Z = 82). Any γ emissions are ignored.
How many α decays and how many β− decays occur in the series?
Show mark scheme
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Notes
Step 1Only α decay changes the nucleon number, by 4 each time: (232 − 208)/4 = 6 α decays.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Six α decays lower Z by 12, from 90 to 78.
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Step 3Each β− decay raises Z by 1 (a neutron becomes a proton), so 82 − 78 = 4 β− decays are needed.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis swaps the two numbers.
BCorrect: 6 α decays account for the change in nucleon number; 4 β− decays then restore Z from 78 to 82.
CThis takes the number of β− decays as the overall fall in proton number (90 − 82 = 8), forgetting that each α decay also lowers Z by 2 and that β− decay raises Z.
DThis treats each α decay as lowering the nucleon number by 2 (giving 12 α decays, which lower Z by 24, so 16 β− decays would be needed).
Syllabus understandingE.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving α, β−, β+, γ Command term: Deduce
19E-1A-98
Isotopes·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Uranium-235 and uranium-238 are isotopes of uranium.
Which statements about these two isotopes are correct?
I. Their nuclei have the same charge. II. Their nuclei have the same binding energy. III. Their neutral atoms contain the same number of electrons.
Show mark scheme
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Notes
Step 1Isotopes have the same proton number Z (92) but different neutron numbers (143 and 146).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The nuclear charge is +Ze for both (I is correct), and a neutral atom has Z electrons, so both have 92 electrons (III is correct).
—
Step 3The binding energy is roughly (binding energy per nucleon) × (number of nucleons); uranium-238 has three more nucleons and a binding energy of about 1802 MeV, compared with about 1784 MeV for uranium-235 (II is wrong).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis accepts II, treating the binding energy as a property of the element (set by Z) rather than of the whole nucleus, and rejects III by matching the number of electrons to the number of nucleons.
BCorrect: the same proton number gives the same nuclear charge and the same number of electrons in the neutral atom; the different numbers of neutrons give different binding energies.
CThis rejects I, as if the extra neutrons changed the charge of the nucleus; neutrons are uncharged.
DThis treats isotopes as identical apart from their mass; the three extra nucleons in uranium-238 add about 18 MeV of binding energy.
Syllabus understandingE.3 — isotopes; nuclear binding energy and mass defect; E.1 — nuclear notation AZX Command term: Deduce
20E-1A-99
Stability and the N/Z ratio·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The chart shows the band in which stable nuclides lie on a graph of neutron number N against proton number Z. Nuclides P and Q are both unstable.
Which row gives the most likely mode of decay of P and of Q?
Neutron number against proton number, showing the band of stable nuclides (drawn to scale).
PQ
Show mark scheme
Marking point
Mark
Notes
Step 1P (Z = 40, N = 62) lies above the band: it has too many neutrons for its proton number (N/Z = 1.55, whereas stable nuclides with Z = 40 have N/Z ≈ 1.25).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2β− decay turns a neutron into a proton (N − 1, Z + 1), moving P down and to the right, towards the band.
—
Step 3Q (Z = 40, N = 42) lies below the band: it has too few neutrons. β+ decay turns a proton into a neutron (N + 1, Z − 1), moving Q towards the band.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis reverses the two decays: β+ decay would raise N/Z for P, moving it further from the band.
BThis treats P as a nucleus that is too large; alpha decay lowers N and Z equally, which hardly changes the large excess of neutrons in P, and nuclides with Z = 40 are far too light for alpha decay.
CThis treats Q as a nucleus in an excited state; γ emission changes neither N nor Z, so Q would stay below the band.
DCorrect: the neutron-rich nuclide P decays by β− emission and the proton-rich nuclide Q by β+ emission, each moving towards the band of stability.
Syllabus understandingE.3 (HL) — the role of the ratio of neutrons to protons for the stability of nuclides; E.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving α, β−, β+, γ Command term: Deduce
21E-1A-100
Total binding energy of heavy nuclei·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
For nuclei with nucleon number A greater than about 60, the binding energy per nucleon is approximately constant.
Which sketch graph best shows how the total binding energy B of a nucleus varies with A in this range?
Four sketch graphs of total binding energy B against nucleon number A, for A from 60 to 240.Show mark scheme
Marking point
Mark
Notes
Step 1Total binding energy B = (binding energy per nucleon) × A.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2If the binding energy per nucleon is approximately constant (it falls only from about 8.8 MeV to 7.6 MeV between A = 60 and A = 240), B is approximately proportional to A: from about 527 MeV to about 1802 MeV.
—
Step 3So the graph is a nearly straight line through (close to) the origin, bending very slightly downward because the binding energy per nucleon falls slowly.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis copies the shape of the graph of binding energy per nucleon against A; the total binding energy increases about 3.4 times between A = 60 and A = 238.
BThis confuses the total binding energy with the binding energy per nucleon, which is the quantity that is approximately constant.
CThis would be true if every nucleon attracted every other nucleon (a long-range force, with about A2/2 interacting pairs); the short range of the strong force means that each nucleon is bound only to its neighbours.
DCorrect: an approximately constant binding energy per nucleon makes the total binding energy approximately proportional to A.
Syllabus understandingE.3 (HL) — the approximate constancy of binding energy curve above a nucleon number of 60; E.3 — the variation of the binding energy per nucleon with nucleon number; the existence of the strong nuclear force, a short-range, attractive force between nucleons Command term: Deduce
22E-1A-101
Mass–energy equivalence·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A sealed heat source containing an alpha-emitting nuclide keeps the instruments of a remote weather station warm. All the alpha particles and recoiling nuclei stop inside the source, which releases thermal energy to its surroundings at a constant rate of 240 W.
What is the decrease in the mass of the source in one year? (1 year = 3.16 × 107 s)
Show mark scheme
Marking point
Mark
Notes
Step 1The alpha particles become helium atoms that stay inside the source, so the only mass that leaves is the mass equivalent of the energy released to the surroundings.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy released in one year: E = 240 × 3.16 × 107 = 7.58 × 109 J.
CThis uses E = ½mc2, a kinetic-energy expression, which doubles the mass.
DThis divides by c instead of c2.
Syllabus understandingE.3 — the mass–energy equivalence as given by E = mc2 in nuclear reactions; A.3 — power as the rate of energy transfer Command term: Determine
23E-1A-102
Activity and the decay constant·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The graph shows how the activity A of a pure sample varies with the number N of undecayed nuclei in the sample, for two different nuclides X and Y.
What is (half-life of X)/(half-life of Y)?
Activity A against number of undecayed nuclei N for nuclides X and Y (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1A = λN, so the gradient of each line is the decay constant.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3T½ = ln 2/λ, so the half-life ratio is the inverse of the gradient ratio: 1/6.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: X has the steeper line, six times the decay constant of Y, and so one-sixth of its half-life.
BThis compares the activities read at two points, 2.0 × 108 Bq and 6.0 × 108 Bq, as if half-life were inversely proportional to activity, ignoring the different numbers of nuclei at those points.
CThis compares the same two activities, 6.0 and 2.0 × 108 Bq, and takes a larger activity to mean a longer half-life; the numbers of nuclei have been ignored.
DThis finds the ratio of the gradients correctly (6) but treats the half-life as proportional to the decay constant.
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λCommand term: Deduce
24E-1A-103
Evidence for nuclear energy levels·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Three observations about the radiations emitted by radioactive nuclides are:
I. The alpha particles emitted by one nuclide have only a few discrete kinetic energies. II. The gamma-ray photons emitted by one nuclide have only a few discrete energies. III. The β− particles emitted by one nuclide have a range of kinetic energies up to a definite maximum value.
Which observations provide evidence that nuclei have discrete energy levels?
Show mark scheme
Marking point
Mark
Notes
Step 1In alpha decay the energy released is shared between only two bodies, so each alpha energy corresponds to a definite final state of the daughter nucleus: a few discrete energies imply a few discrete levels (I).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each gamma-ray photon energy equals the difference between two nuclear energy levels, so a line spectrum of photons implies discrete levels (II).
—
Step 3The continuous β− spectrum shows that the energy is shared with a third particle, the antineutrino; it is evidence for the antineutrino, not for discrete levels (III).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the discrete alpha and gamma energies reveal discrete nuclear energy levels; the continuous beta spectrum is evidence for the (anti)neutrino.
BThis rejects II and accepts III, taking the definite maximum β− energy as the evidence; the continuous spread below the maximum is explained by the antineutrino, and gamma photons of discrete energies are the most direct evidence of all.
CThis rejects I, treating alpha energies as continuous like beta energies; with only two bodies sharing the energy, alpha particles from one transition all have the same energy.
DThis treats any spectrum with a sharp limit as evidence of levels; the continuous β− spectrum is evidence for the antineutrino instead.
Syllabus understandingE.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; the continuous spectrum of beta decay as evidence for the neutrino Command term: Deduce
25E-1A-104
Comparing activities of two nuclides·E.3 Radioactive decay
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
At time t = 0, two pure radioactive samples X and Y contain equal numbers of undecayed nuclei. The half-life of X is T and the half-life of Y is 2T.
When are the activities of X and Y equal?
Show mark scheme
Marking point
Mark
Notes
Step 1A = λN and λ ∝ 1/T½, so at t = 0 the activity of X is twice that of Y.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2In a time 2T, X passes through two half-lives (N falls to N0/4) and Y through one (N falls to N0/2).
—
Step 3Then AX = (2λY)(N0/4) = λY(N0/2) = AY: the activities are equal at t = 2T (after which Y has the greater activity).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes the activity to be proportional to the number of nuclei alone, ignoring the decay constant: equal numbers would then mean equal activities only at the start.
BThis finds when the activity of X has fallen to the initial activity of Y (half its own), forgetting that the activity of Y is also falling.
CCorrect: AX/AY = 2 × (½)t/2T, which equals 1 when t = 2T.
DThis assumes that the longer-lived sample Y starts with the greater activity (confusing a long half-life with a high activity); since the nuclei of Y also decay more slowly, X could then never catch up. In fact X starts with twice the activity of Y.
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; the decay constant λ and the radioactive decay law as given by N = N0e−λt; E.3 — the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Deduce
26E-1A-105
Parent and daughter nuclei·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A radioactive parent nuclide decays into a stable daughter nuclide. The graph shows how the numbers of parent and daughter nuclei in a sample vary with time t. At t = 0 the sample contained no daughter nuclei.
What is (number of daughter nuclei)/(number of parent nuclei) at t = 6.0 h?
Numbers of parent and daughter nuclei against time (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The curves cross when half the parent nuclei have decayed: the half-life is 2.0 h.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 26.0 h is three half-lives, so the fraction of parent nuclei remaining is (½)3 = 1/8, and 7/8 have become daughter nuclei.
—
Step 3Ratio = (7/8)/(1/8) = 7 (the graph confirms 7.0 × 1020 and 1.0 × 1020).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the ratio: it is (parent)/(daughter).
BThis gives the number of half-lives that have passed, not a ratio of numbers of nuclei.
CCorrect: 7/8 of the original nuclei are daughter nuclei and 1/8 are parent nuclei.
DThis is (original number of parent nuclei)/(parent nuclei remaining); the daughter nuclei number N0 − N0/8, not N0.
Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life; the activity, count rate and half-life in radioactive decay Command term: Determine
27E-1A-106
Decay constant as a probability·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The probability that a particular nucleus of a radioactive nuclide decays during a short time interval Δt is p, where p ≪ 1.
What is the half-life of the nuclide?
Show mark scheme
Marking point
Mark
Notes
Step 1Because p ≪ 1, λΔt is small and the decay constant is the probability of decay per unit time: λ = p/Δt.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2T½ = ln 2/λ = Δt ln 2/p.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes that the probability keeps adding up linearly, so that half the nuclei have decayed after 1/(2p) intervals; the approximation p ≈ λΔt fails when the time is not short.
BThis inverts the decay constant (λ taken as Δt/p): the result has the unit of s−1, not s.
CThis divides by ln 2 instead of multiplying: it is 1/(λ ln 2).
DCorrect: λ = p/Δt for a short interval, and T½ = ln 2/λ.
Syllabus understandingE.3 (HL) — that the decay constant approximates the probability of decay in unit time only in the limit of sufficiently small λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λCommand term: Determine
28E-1A-107
Decays during one half-life·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
At t = 0 a pure sample contains N0 nuclei of a radioactive nuclide whose half-life is 1.0 h.
How many nuclei decay between t = 2.0 h and t = 3.0 h?
Show mark scheme
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Mark
Notes
Step 1At t = 2.0 h, after two half-lives, N0/4 nuclei remain.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2During the next half-life half of these decay: N0/8 nuclei decay and N0/8 remain at t = 3.0 h.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: N0/4 − N0/8 = N0/8.
BThis is the number remaining at t = 2.0 h; only half of these decay during the third hour.
CThis assumes that half of the original sample decays in every half-life, as if the decay were linear.
DThis is the total number decayed by t = 3.0 h, not the number decaying during the third hour.
Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life; the activity, count rate and half-life in radioactive decay Command term: Determine
29E-1A-108
The strong nuclear force·E.3 Radioactive decay
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
The deuteron, which consists of one proton and one neutron, is a stable nucleus. For nuclei with nucleon number greater than about 60, the binding energy per nucleon is approximately constant.
Which row describes the strong nuclear force in a way that is consistent with both observations?
Acts betweenRange
Show mark scheme
Marking point
Mark
Notes
Step 1The deuteron contains only one proton, so the attraction that binds it must act between a neutron and a proton: the force acts between all nucleons.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2A constant binding energy per nucleon means each nucleon is bound only to the few nucleons next to it, whatever the size of the nucleus: the range is about the separation of neighbouring nucleons, 10−15 m.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AA force acting only between protons could not bind the deuteron, which has a single proton.
B10−10 m is the size of an atom, far larger than any nucleus; every nucleon would attract every other one, and the binding energy per nucleon would keep rising with nucleon number.
CCorrect: an attractive force between all nucleons (n–n, n–p and p–p) with a range of about 10−15 m.
DAn inverse-square force has unlimited range, so each nucleon would attract all the others and the binding energy per nucleon would grow with the nucleon number.
Syllabus understandingE.3 — the existence of the strong nuclear force, a short-range, attractive force between nucleons; E.3 (HL) — the evidence for the strong nuclear force; the approximate constancy of binding energy curve above a nucleon number of 60 Command term: Deduce
30E-1A-109
Decay over a non-integer number of half-lives·E.3 Radioactive decay
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A dose of technetium-99m, a gamma emitter used in medical imaging, has an activity of 720 MBq when it is prepared at 07:30. The half-life of technetium-99m is 6.0 h.
What is the activity of the dose when it is injected into a patient at 11:00?
Show mark scheme
Marking point
Mark
Notes
Step 1The time elapsed is 3.5 h, which is 3.5/6.0 = 0.583 half-lives.
—
All 3 steps must be completed — there is no mark for a part-answer.
DThis assumes the activity falls linearly, by 360 MBq in 6.0 h; the decay is exponential, so the activity falls faster at first.
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; guidance: application of the decay equations for arbitrary time intervals; guidance: medical isotopes Command term: Determine
31E-1A-110
Half-life from the initial rate of decay·E.3 Radioactive decay
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The graph shows how the number N of undecayed nuclei in a pure radioactive sample varies with time t. The tangent to the curve at t = 0 meets the time axis at t = τ.
What is the half-life of the nuclide?
Number of undecayed nuclei N against time t, with the tangent to the curve at t = 0 (sketch).Show mark scheme
Marking point
Mark
Notes
Step 1The gradient of the tangent at t = 0 is −N0/τ; its magnitude is the initial rate of decay, the activity A0 = λN0.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2So λN0 = N0/τ, giving λ = 1/τ.
—
Step 3T½ = ln 2/λ = τ ln 2 ≈ 0.69τ: the curve reaches N0/2 before the tangent reaches the axis.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes the nuclei go on decaying at the initial rate until half have gone; the rate falls as N falls, so halving takes longer than τ/2.
BCorrect: λ = 1/τ from the initial gradient, and T½ = ln 2/λ = τ ln 2.
CThis takes the intercept of the tangent as the half-life; at t = τ the curve has fallen to N0/e ≈ 0.37N0, not to N0/2.
DThis finds λ = 1/τ correctly but divides by ln 2 instead of multiplying (T½ = 1/(λ ln 2)).
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the decay constant λ and the radioactive decay law as given by N = N0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λCommand term: Deduce
32E-1B-01
Half-life with background correction·E.3 Radioactive decay
Paper 1BMedium6 marks
Data-based question5 steps to full marksDetermine
A student investigates the decay of barium-137m, which is washed out of a sealed isotope generator into a small shallow dish. A Geiger–Müller tube, clamped 3 cm above the dish, is connected to a ratemeter. The student records the count rate R shown by the ratemeter every minute for 10 minutes and then every 5 minutes until t = 30 min, without moving the tube or the dish. No separate measurement of the background is made.
The graph shows all the readings with a smooth curve of best fit; the table lists five of the readings.
t / min
0
2
4
6
8
R / s−1
42.3
25.5
15.6
9.9
6.6
Count rate against time (graph drawn to scale)
(a)
(i)
Determine, using the graph, the background count rate.
(1)
(b)
(i)
The student claims that the barium-137m decays with a constant half-life. Test this claim using at least three readings from the table.
(2)
(c)
(i)
Determine the half-life of barium-137m, using the full range of the data in the table.
(2)
(d)
(i)
Predict the count rate shown by the ratemeter at t = 12 min.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The graph levels off for t ≥ 20 min, where the source contributes almost nothing: background = 2.0 s−1
✓ 1
Accept 1.8–2.2 s−1.
Part (b)(i)
Background subtracted from each reading: 40.3, 23.5, 13.6, 7.9, 4.6 s−1
✓ 1
Allow ECF from (a). Ratios of the uncorrected readings (0.60, 0.61, 0.63, 0.67) rise steadily: 0 for this mark, but the second mark may be awarded for a correct comparison.
Ratio over each equal 2 min interval: 0.583, 0.579, 0.581, 0.582 — constant (within about 1 %), so the claim is supported
✓ 1
At least three ratios (or three successive equal-ratio / halving times) and a conclusion are needed.
Part (c)(i)
λ = ln(40.3/4.6)/8 = 0.271 min−1
✓ 1
Allow ECF from (a) and (b). ALT: mean ratio r over 2 min, T½ = 2 ln 2/ln(1/r).
T½ = ln 2/λ = 2.56 ≈ 2.6 min
✓ 1
Accept 2.5–2.6 min. A value from a single 2 min interval only: [1 max].
Part (d)(i)
40.3 e−0.271 × 12 + 2.0 = 1.55 + 2.0 = 3.6 s−1
✓ 1
Allow ECF from (a) and (c). Accept 3.4–3.7 s−1. The source alone (1.6 s−1) scores 0: the ratemeter also records the background.
Answers: (a)(i) 2.0 s−1 · (c)(i) 2.6 min · (d)(i) 3.6 s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the effect of background radiation on count rate; the activity, count rate and half-life in radioactive decay; (HL) the radioactive decay law N = N0e−λt and T½ = ln 2/λ; guidance: the determination of the half-life of a nuclide; Tools 3 — testing a relationship, using the full data range Command term: Determine
33E-1B-08
Range of alpha particles in air·E.3 Radioactive decay
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
An americium-241 alpha source and a solid-state detector face each other inside a sealed chamber connected to a vacuum pump and a pressure gauge. The source is mounted on a sliding rod, and its distance from the detector is read from a scale engraved on the rod. At each air pressure p, with the air at 20 °C, the student moves the source away until the count rate just falls to the background value, and records the scale reading R as the range of the alpha particles.
The graph shows R against 1/p with the line of best fit.
p / kPa
101
85
70
55
45
35
1/p / 10−3 kPa−1
9.9
11.8
14.3
18.2
22.2
28.6
R / cm
3.88
4.59
5.71
7.29
9.05
11.66
Scale reading R against 1/p (graph drawn to scale)
(a)
(i)
Explain, using the ideal gas model, why the range is expected to be inversely proportional to p at constant temperature.
(2)
(b)
(i)
Determine the gradient of the line, giving its unit.
(2)
(c)
(i)
The line does not pass through the origin. Determine the intercept on the R axis and hence the true range of the alpha particles in air at 101 kPa and 20 °C.
(2)
(d)
(i)
Predict the range of these alpha particles in air at 101 kPa and 60 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
pV = NkBT, so the number of molecules per unit volume N/V = p/kBT is proportional to p
✓ 1
An alpha particle loses about the same energy in each ionising collision, so it stops after meeting a fixed number of molecules; the distance needed is proportional to 1/(N/V), i.e. to 1/p
✓ 1
Part (b)(i)
Gradient from two well-separated points on the line = 0.419 cm per 10−3 kPa−1 = 4.2 × 102
✓ 1
Accept 4.0–4.4 × 102.
Unit: cm kPa
✓ 1
Or equivalent, e.g. 4.2 × 103 Pa m.
Part (c)(i)
Extrapolated intercept = −0.3 cm: each scale reading is 0.3 cm less than the true source–detector distance (zero error on the rod scale)
✓ 1
Accept −0.2 to −0.4 cm.
True range = gradient/101 = 419/101 = 4.15 cm
✓ 1
Allow ECF from (b). ALT: scale reading at 101 kPa (3.85 cm) + 0.3 cm.
Part (d)(i)
N/V ∝ p/T, so range ∝ T/p: 4.15 × 333/293 = 4.7 cm
✓ 1
Allow ECF from (c). Kelvin temperatures required; using 60/20 scores 0.
Answers: (b)(i) 4.2 × 102 cm kPa · (c)(i) 4.15 cm · (d)(i) 4.7 cm (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; B.3 — the ideal gas equation pV = NkBT and the number of molecules per unit volume; Tools 3 — gradient with its unit, extrapolation to an intercept, systematic (zero) error Command term: Determine
34E-1B-10
Inverse-square law for gamma radiation·E.3 Radioactive decay
Paper 1BMedium6 marks
Data-based question5 steps to full marksDetermine
A student investigates how the corrected count rate C from a small gamma source depends on the distance d between the source and the front face of a Geiger–Müller tube. Each reading is the number of counts N recorded in 100 s; the background count rate, measured beforehand over 1000 s, is 0.40 s−1. The gamma photons are detected at an unknown distance x behind the front face of the tube, so that C = k/(d + x)², where k is a constant.
The graph shows d against C−1/2 with a line of best fit.
d / cm
N
C / s−1
C−1/2 / s1/2
5.0
26954
269.14
0.0610
10.0
7687
76.47
0.1144
15.0
3674
36.34
0.1659
20.0
2099
20.59
0.2204
30.0
1000
9.60
0.3227
40.0
574
5.34
0.4327
d against C−1/2 (graph drawn to scale)
(a)
(i)
Show that the intercept on the d axis is equal to −x, and determine x.
(2)
(b)
(i)
Predict the corrected count rate when d = 50.0 cm.
(2)
(c)
(i)
Estimate the percentage uncertainty in a 100 s count at d = 50.0 cm. Suggest how, with the same apparatus, the student could reduce it to below 3 %.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
d + x = √k × C−1/2, so d = √kC−1/2 − x: straight line with intercept −x
✓ 1
Line extrapolated to C−1/2 = 0 gives −0.75 cm, so x = 0.75 cm
✓ 1
Accept 0.5–1.0 cm.
Part (b)(i)
Gradient = √k = 94.5 cm s−1/2, so k = 8.93 × 103 s−1 cm²
✓ 1
Accept a gradient of 92–97 cm s−1/2.
C = 8.93 × 103/(50.0 + 0.75)² = 3.47 s−1
✓ 1
Allow ECF from (a). Accept 3.3–3.6 s−1. Omitting x gives 3.57 s−1: [1 max].
Allow ECF from (b). The counter records the background too; omitting it (5.4 %) scores 0.
Need N > (100/3)² ≈ 1.1 × 103 counts, so count for at least ≈ 290 s (e.g. 300 s) at this distance
✓ 1
Allow ECF. Accept "repeat the count several times and add the counts". Do not accept "use a stronger source" or "use a different detector".
Answers: (a)(i) 0.75 cm · (b)(i) 3.5 s−1 · (c)(i) 5.1 % (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of gamma rays; the effect of background radiation on count rate; Tools 3 — linearising a relationship, extrapolation to an intercept, systematic error, random uncertainty in a count Command term: Determine
35E-1B-15
Absorption of beta radiation·E.3 Radioactive decay
Paper 1BEasy6 marks
Data-based question4 steps to full marksDetermine
A factory monitors the thickness of aluminium sheet using a beta source below the sheet and a Geiger–Müller tube above it, connected to a ratemeter. To calibrate the gauge, a technician places aluminium of known total thickness x between the source and the tube and records the number of counts in 100 s. The background count rate, measured with the source removed, is 0.40 s−1. Over this range of thickness the corrected count rate C is modelled by C = C0e−μx, where μ is a constant.
The graph shows ln(C / s−1) against x with a line of best fit.
x / mm
Counts in 100 s
C / s−1
ln(C / s−1)
0.00
24328
242.9
5.49
0.50
17642
176.0
5.17
1.00
13172
131.3
4.88
1.50
9846
98.1
4.59
2.00
7240
72.0
4.28
2.50
5427
53.9
3.99
3.00
3987
39.5
3.68
ln(corrected count rate) against thickness (graph drawn to scale)
(a)
(i)
Determine μ, giving its unit.
(2)
(b)
(i)
Determine the thickness of aluminium that halves the corrected count rate.
(1)
(c)
(i)
During production the reading on the ratemeter rises from 72.4 s−1 to 80.4 s−1. Determine the change in the thickness of the sheet, stating whether it has become thicker or thinner.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from points at opposite ends of the line, e.g. (3.68 − 5.48)/(3.00 − 0) = −0.60 mm−1
✓ 1
Using only the first two readings gives 0.64 mm−1: [1 max].
μ = 0.60 mm−1
✓ 1
Unit required. Accept 0.58–0.62 mm−1 (5.8–6.2 × 102 m−1).
Part (b)(i)
x½ = ln 2/μ = 0.693/0.60 = 1.2 mm
✓ 1
Allow ECF from (a). Accept 1.1–1.2 mm.
Part (c)(i)
Corrected rates: 72.0 s−1 and 80.0 s−1
✓ 1
Background subtracted.
Δx = ln(80.0/72.0)/μ = 0.105/0.60 = 0.18 mm
✓ 1
Allow ECF from (a). Accept 0.17–0.18 mm. Without the background correction (0.17 mm) the first mark is lost but the second may be awarded.
The sheet has become thinner (more beta particles get through)
✓ 1
Answers: (a)(i) 0.60 mm−1 · (b)(i) 1.2 mm · (c)(i) 0.18 mm thinner (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the effect of background radiation on count rate; guidance: real-life contexts include the thickness of materials; Tools 3 — linearising with a logarithm, gradient with its unit, using the full range of data Command term: Determine
36E-1B-16
Determining a long half-life·E.3 Radioactive decay
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
Potassium chloride, KCl (molar mass 74.55 g mol−1), is sold as a low-sodium salt. A fraction 1.17 × 10−4 of all potassium atoms are the radioactive isotope potassium-40. A student spreads samples of KCl of mass m evenly over a tray of fixed area beneath a detector that records 4.0 % of all the potassium-40 decays in the sample. Each net count rate C (background subtracted) is the result of a long count and is uncertain by ±4 min−1.
The graph shows C against m, with a line of best fit through the origin. (NA = 6.02 × 1023 mol−1; 1 year = 3.16 × 107 s)
m / g
2.0
4.0
6.0
8.0
10.0
12.0
14.0
C / min−1
83
155
241
323
395
448
496
Net count rate against mass of KCl (graph drawn to scale)
(a)
(i)
Determine the gradient of the line of best fit, giving its unit and explaining which data you have used.
(2)
(b)
(i)
Determine the half-life of potassium-40, in years.
(3)
(c)
(i)
The detection efficiency is (4.0 ± 0.2) % and the gradient is uncertain by ±2 %. Determine the absolute uncertainty in your answer to (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Only the points for m ≤ 10 g are used: at 12 g and 14 g the thicker layer absorbs some of its own beta particles, so these points fall below the line
✓ 1
Using all seven points (about 38 min−1 g−1) scores 0 for the second mark.
Gradient = 40 min−1 g−1
✓ 1
Accept 39–41; unit required.
Part (b)(i)
Number of K-40 nuclei in 1.00 g = (6.02 × 1023/74.55) × 1.17 × 10−4 = 9.45 × 1017
✓ 1
One potassium atom per formula unit of KCl.
Activity of 1.00 g: A = gradient/(60 × 0.040) = 16.6 Bq
Allow ECF from (b). 1 s.f. required: (1.25 ± 0.09) × 109 years.
Answers: (a)(i) 40 min−1 g−1 · (b)(i) 1.25 × 109 years · (c)(i) ±9 × 107 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — the activity as the rate of decay as given by A = λN; the relationship T½ = ln 2/λ; guidance: the determination of the half-life of a nuclide; B.3 — amount of substance n = N/NA and molar mass; Tools 3 — choosing the linear range, gradient with units, propagation of uncertainties Command term: Determine
37E-1B-21
Modelling decay with dice·E.3 Radioactive decay
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
To model radioactive decay, a student throws 1000 identical fair cubes, each with one face painted red. After each throw, every cube that lands red face up is removed and counted as "decayed"; the remaining cubes are thrown again. The table shows the number N of cubes remaining after throw number n. The probability that any one cube lands red face up in a single throw is 1/6.
The graph shows ln N against n, with the line of best fit.
n
0
1
2
3
4
5
6
7
8
9
10
11
12
N
1000
813
687
579
484
394
325
279
235
206
166
139
114
ln N against throw number (graph drawn to scale)
(a)
(i)
Determine, using the graph, the decay constant λ of the cubes in throw−1.
(2)
(b)
(i)
The decay constant is often described as the probability of decay per unit time. Compare your answer to (a) with the probability 1/6 and explain the difference.
(2)
(c)
(i)
Determine the half-life of the cubes, in throws.
(1)
(d)
(i)
The points for large n lie further from the line than those for small n. Explain why, and suggest how the student could reduce this scatter using the same cubes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (0, 6.89) and (12, 4.75): (4.75 − 6.89)/12 = −0.178
✓ 1
Accept −0.17 to −0.19.
λ = −gradient = 0.18 throw−1
✓ 1
Accept 0.17–0.19 throw−1. A value from a single pair of table readings, e.g. ln(1000/813): [1 max].
Part (b)(i)
λ (0.18 throw−1) is greater than 1/6 = 0.167, by about 7 %
✓ 1
Allow ECF from (a).
After each throw a fraction 5/6 remains, so N = 1000(5/6)n and λ = ln(6/5) = 0.182; λ equals the probability per time step only when that probability is small (λΔt ≪ 1), which 1/6 is not
✓ 1
OWTTE. Do not accept "random error" or "too few throws" as the explanation for this mark.
Part (c)(i)
T½ = ln 2/λ = 0.693/0.18 = 3.9 throws
✓ 1
Allow ECF from (a). Accept 3.6–4.1 throws.
Part (d)(i)
The number removed in each throw is random, with a fluctuation of about √N; as N falls the fractional fluctuation (≈ 1/√N) grows, so ln N scatters more
✓ 1
Accept any answer linking small N to a larger relative random fluctuation.
Repeat the whole experiment several times with the same cubes and plot the mean N for each n
✓ 1
Do not accept "use more cubes": that changes the apparatus.
Answers: (a)(i) 0.18 throw−1 · (c)(i) 3.9 throws (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — the decay constant λ and the radioactive decay law as given by N = N0e−λt; that the decay constant approximates the probability of decay in unit time only in the limit of sufficiently small λt; E.3 — the random and spontaneous nature of radioactive decay; Tools 3 — linearising a relationship, gradient of a line of best fit; Inquiry 2 — reducing random error by repeating Command term: Determine
38E-1B-23
Two nuclides decaying together·E.3 Radioactive decay
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
A silver foil is irradiated with neutrons and then placed next to a Geiger–Müller tube. The irradiation produces two radioactive isotopes of silver, which decay with different half-lives into stable nuclides. The count rate C, corrected for background, is determined every 20 s.
The graph shows ln(C / s−1) against time t. The solid straight line is the line of best fit to the points for t ≥ 240 s; it is extended back to t = 0 as a dashed line. The table gives the first three readings.
t / s
0
20
40
C / s−1
302.8
189.1
126.1
ln C against time (graph drawn to scale)
(a)
(i)
Explain why the graph is curved for small t but straight for large t.
(1)
(b)
(i)
Determine the half-life of the longer-lived isotope.
(2)
(c)
(i)
Determine, using the table and the dashed line, the half-life of the shorter-lived isotope.
(3)
(d)
(i)
Predict the corrected count rate at t = 800 s.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
C is the sum of two exponential terms, so ln C is not linear while both contribute; after about 200 s the shorter-lived isotope has decayed to a negligible activity, so only the longer-lived one contributes and ln C falls linearly with t
✓ 1
Both ideas (sum of two decays; the short-lived one becomes negligible) needed.
Part (b)(i)
Gradient of the straight line = (1.21 − 2.88)/(600 − 240) = −4.64 × 10−3 s−1
✓ 1
Accept −4.3 × 10−3 to −4.9 × 10−3 s−1.
T½ = ln 2/4.64 × 10−3 = 149 s
✓ 1
Accept 140–160 s.
Part (c)(i)
Longer-lived contribution from the line: at t = 0, ln C = 4.00 so CL = 54 s−1; at t = 40 s, ln C = 3.81 so CL = 45 s−1
✓ 1
Allow ECF from (b). Accept 50–58 s−1 and 42–48 s−1.
Subtraction of the longer-lived contribution at both times is the key step; using the raw readings 302.8 and 126.1 s−1 gives about 32 s: [1 max].
λ = ln(248/81)/40 = 2.8 × 10−2 s−1, so T½ = 25 s
✓ 1
Accept 22–27 s.
Part (d)(i)
C = exp(4.00 − 4.64 × 10−3 × 800) = 1.3 s−1
✓ 1
Allow ECF from (b). Accept 1.1–1.6 s−1 (the rounded gradient 4.6 × 10−3 s−1 gives 1.4 s−1).
Answers: (b)(i) 150 s · (c)(i) 25 s · (d)(i) 1.3 s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — the decay constant λ and the radioactive decay law as given by N = N0e−λt; the activity as the rate of decay A = λN; the relationship T½ = ln 2/λ; guidance: the determination of the half-life of a nuclide; Tools 3 — linearising with logarithms, gradient with its unit, extrapolation to an intercept Command term: Determine
39E-1B-30
Is a granite worktop radioactive?·E.3 Radioactive decay
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
A student investigates whether a granite kitchen worktop is radioactive. A Geiger–Müller tube connected to a counter is placed face down on the granite, and the number of counts in 200 s is recorded ten times. The tube is then taken to a room with no granite, and ten more 200 s counts of the background are recorded. The results are shown in the table.
The student takes the uncertainty in a mean count to be half the range of the readings.
Reading
1
2
3
4
5
6
7
8
9
10
Count in 200 s, tube on granite
137
132
140
168
131
156
133
163
142
128
Count in 200 s, background
78
78
66
95
77
86
90
76
75
104
(a)
(i)
Suggest why the ten counts recorded with the tube on the granite are not all the same.
(1)
(b)
(i)
Determine the mean count rate, in s−1, with the tube on the granite, and its absolute uncertainty.
(2)
(c)
(i)
Determine the count rate due to the granite, with its absolute uncertainty.
(2)
(d)
(i)
Deduce whether the data show that the granite is radioactive.
(1)
(e)
(i)
Suggest, with a reason, how the student could reduce the percentage uncertainty in each count without changing the apparatus.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Radioactive decay is random: the number of nuclei that decay (and are detected) in each 200 s interval varies by chance
✓ 1
Do not accept "human error", "the counter is inaccurate" or "reaction time".
Rate due to granite = 0.715 − 0.412 = 0.30 s−1; uncertainties add for a difference: ±(0.10 + 0.095) = ±0.20 s−1
✓ 1
Allow ECF from (b). Accept 0.30 ± 0.20 s−1 or 0.3 ± 0.2 s−1. Subtracting the uncertainties: [0] for this mark.
Part (d)(i)
0.30 − 0.20 = 0.10 s−1 is greater than zero: even at the lower limit the granite adds to the count rate, so it is (slightly) radioactive
✓ 1
Allow ECF from (c). The conclusion must be consistent with the candidate's own value and uncertainty; "the counts are higher" without reference to the uncertainty: [0].
Part (e)(i)
Count for a longer time (e.g. 2000 s) for each reading: the random fluctuation in a count N is about √N, so the percentage fluctuation (≈ 1/√N) falls as N increases
✓ 1
Reason needed. Do not accept "use a more sensitive detector", "use a stronger source" or "repeat the readings" (this does not change the spread of single counts).
Answers: (b)(i) 0.71 ± 0.10 s−1 · (c)(i) 0.30 ± 0.20 s−1 · (d)(i) yes: 0.30 − 0.20 > 0 (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the random and spontaneous nature of radioactive decay; the effect of background radiation on count rate; the activity, count rate and half-life in radioactive decay; Tools 3 — mean and half-range as uncertainty, propagating absolute uncertainties through a difference, deciding whether a difference is significant; Inquiry 2 — improving the precision of a measurement Command term: Determine
40E-1B-31
Range of alpha particles from different nuclides·E.3 Radioactive decay
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A student uses photographs of alpha-particle tracks in a cloud chamber containing air at atmospheric pressure. Each source is a single alpha-emitting nuclide, and all the tracks from one source are very nearly the same length. The student measures the mean track length (the range) R for each source and finds the kinetic energy E of its alpha particles in a data book.
It is suggested that R = kEn, where k and n are constants. The graph shows lg(R / cm) against lg(E / MeV) with the line of best fit.
Nuclide
E / MeV
R / cm
lg(E / MeV)
lg(R / cm)
thorium-232
4.01
2.55
0.603
0.407
uranium-238
4.20
2.75
0.623
0.439
radium-226
4.78
3.31
0.679
0.520
polonium-210
5.30
3.87
0.724
0.588
radon-222
5.49
4.04
0.740
0.606
polonium-218
6.00
4.66
0.778
0.668
polonium-214
7.69
6.87
0.886
0.837
lg(R / cm) against lg(E / MeV) with the line of best fit (drawn to scale).
(a)
(i)
Explain why all the tracks from one source are very nearly the same length.
(1)
(b)
(i)
Determine n.
(2)
(c)
(i)
Determine k, giving its unit.
(2)
(d)
(i)
Tracks of mean length 4.7 cm are observed from another alpha-emitting nuclide. Determine the kinetic energy of its alpha particles.
(1)
(e)
(i)
Near the end of each track the line of droplets becomes thicker. Suggest why.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The alpha particles from one nuclide are all emitted with the same kinetic energy, because the parent and daughter nuclei have discrete (definite) energy levels; each particle loses energy in the same way in the air, so all travel the same distance before stopping
✓ 1
Both ideas (same initial energy from discrete levels; same energy loss per unit length) needed.
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (0.854 − 0.401)/(0.90 − 0.60)
✓ 1
Two adjacent data points only: [1 max].
n = 1.51
✓ 1
Accept 1.45–1.55. n is a pure number: a unit is not needed.
Part (c)(i)
lg k = lg R − n lg E at any point on the line, e.g. at lg E = 0.75: lg k = 0.628 − 1.51 × 0.75 = −0.505, so k = 0.31
✓ 1
Allow ECF from (b). Accept 0.30–0.34. The intercept cannot be read directly from the graph because the axes do not start at lg E = 0.
Unit: cm MeV−1.5 (cm MeV−3/2)
✓ 1
Allow ECF (cm MeV−n with their n).
Part (d)(i)
E = (R/k)1/n = (4.7/0.31)1/1.51 = 6.0 MeV
✓ 1
Allow ECF from (b) and (c). Accept 5.8–6.2 MeV. ALT: lg 4.7 = 0.672 read from the line gives lg E ≈ 0.78.
Part (e)(i)
As the alpha particle slows down it spends longer near each air molecule, so it produces more ion pairs per millimetre (it is more strongly ionising); more ions give more droplets
✓ 1
Accept "loses more energy per unit length as it slows". Do not accept "the particle gets bigger".
Answers: (b)(i) 1.51 · (c)(i) 0.31 cm MeV−3/2 · (d)(i) 6.0 MeV (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; E.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; Tools 3 — linearising a power law with logarithms, gradient and intercept, units of a derived constant, interpolation from a model Command term: Determine
41E-1B-32
Half-life of a medical isotope·E.3 Radioactive decay
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
Gallium-68 is used in medical imaging. In a hospital radiopharmacy, 4.0 mL of a solution containing gallium-68 is placed in a vial at t = 0, and its activity A is measured every 30 min in a calibrated ionisation chamber. Each reading of A has an uncertainty of ±4 %; the uncertainty in t is negligible.
The graph shows ln(A / MBq) against t with error bars and the line of best fit.
t / min
0
30
60
90
120
150
180
A / MBq
936
702
520
376
274
204
151
ln(A / MBq)
6.842
6.554
6.254
5.930
5.613
5.318
5.017
ln(A / MBq) against time t, with error bars and the line of best fit (drawn to scale).
(a)
(i)
Show that the absolute uncertainty in ln(A / MBq) is about ±0.04.
(1)
(b)
(i)
Determine the decay constant of gallium-68, giving its unit.
(2)
(c)
(i)
Use the graph to determine the half-life of gallium-68 and its absolute uncertainty.
(2)
(d)
(i)
A patient must be given 150 MBq of gallium-68 at t = 100 min. Determine the volume of solution that must be drawn from the vial.
Either the general result or a numerical check must be seen.
Part (b)(i)
Gradient of the line of best fit from two well-separated points, e.g. (5.01 − 6.85)/(180 − 0) = −1.02 × 10−2
✓ 1
Accept −9.8 × 10−3 to −1.06 × 10−2.
λ = 1.02 × 10−2 min−1
✓ 1
Unit required (min−1, or 1.7 × 10−4 s−1).
Part (c)(i)
T½ = ln 2/1.02 × 10−2 = 68 min; steepest and shallowest lines through all the error bars have gradients of about −1.06 × 10−2 and −9.7 × 10−3 min−1
✓ 1
Allow ECF from (b). Accept a half-life of 65–71 min. Each line must pass through every error bar.
T½ from ln 2/1.06 × 10−2 = 65 min to ln 2/9.7 × 10−3 = 71 min, so T½ = (68 ± 3) min
✓ 1
Accept ±2 to ±4 min. Uncertainty to 1 s.f. (or matching the value). ALT: the same percentage uncertainty as the gradient.
Part (d)(i)
Activity of the whole vial at 100 min: ln A = 6.85 − 1.02 × 10−2 × 100 = 5.83, so A = 340 MBq
✓ 1
Allow ECF from (b). Accept 335–360 MBq (read from the line or from A0e−λt).
Volume = 150/340 × 4.0 mL = 1.8 mL
✓ 1
Allow ECF. Accept 1.7–1.8 mL. Using the activity at t = 0: [1 max].
Answers: (b)(i) 1.02 × 10−2 min−1 · (c)(i) (68 ± 3) min · (d)(i) 1.8 mL (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — the decay constant λ and the radioactive decay law as given by N = N0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; the activity as the rate of decay as given by A = λN = λN0e−λt; guidance: the determination of the half-life of a nuclide; medical isotopes; Tools 3 — linearising with logarithms, uncertainty in a logarithm, gradient with units, steepest and shallowest lines, propagating uncertainty Command term: Determine
42E-2-05
Binding energy per nucleon·E.3 Radioactive decay
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksDetermine
The graph shows the binding energy per nucleon b against the nucleon number A for stable nuclides. Three light nuclides are shown as separate points.
(1 u = 931.5 MeV c−2, mass of a proton = 1.007 276 u, mass of a neutron = 1.008 665 u)
Binding energy per nucleon against nucleon number (graph drawn to scale)
(a)
(i)
Explain, with reference to the graph, why energy is released both when a nucleus of large A splits into two nuclei of intermediate A and when two very light nuclei fuse.
(2)
(b)
(i)
Determine, using the graph, the mass defect, in u, of a lithium-6 (63Li) nucleus.
(2)
(ii)
Hence determine the mass, in u, of a lithium-6 nucleus.
(2)
(c)
In one fusion reaction a lithium-6 nucleus absorbs a hydrogen-2 nucleus and two helium-4 nuclei are formed: 63Li + 21H → 2 42He.
(i)
Using the graph, determine the energy released per nucleon in this reaction and compare it with the energy released per nucleon when a uranium-235 nucleus absorbs a neutron and undergoes fission, releasing about 200 MeV.
(3)
(ii)
Estimate the fraction of the rest energy of the reactants that is released in this fusion reaction.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
In both cases the products have a greater binding energy per nucleon than the reactants (they are closer to the peak near A ≈ 60)
✓ 1
So the total binding energy increases: the products have less mass than the reactants and the mass difference is released as energy (E = Δmc²)
✓ 1
Do not accept "energy is needed to make the bonds".
Part (b)(i)
From the graph b ≈ 5.3 MeV, so the binding energy = 6 × 5.3 = 32 MeV
✓ 1
Accept 5.2–5.4 MeV per nucleon.
Mass defect = 31.8/931.5 = 0.034 u
✓ 1
Allow ECF from the reading. Accept 0.033–0.035 u.
Part (b)(ii)
Mass of the separate nucleons = 3 × 1.007 276 + 3 × 1.008 665 = 6.047 823 u
✓ 1
Mass of the nucleus = 6.0478 − 0.0341 = 6.014 u
✓ 1
Allow ECF from (b)(i). Accept 6.013–6.015 u. Adding the mass defect (6.082 u) scores [1].
Part (c)(i)
Binding energy before = 6 × 5.3 + 2 × 1.1 = 34.0 MeV; after = 8 × 7.1 = 56.8 MeV
✓ 1
Values read from the graph, ±0.1 MeV per nucleon each.
Energy released ≈ 23 MeV, i.e. about 2.8 MeV per nucleon
✓ 1
Accept 21–24 MeV and 2.6–3.0 MeV per nucleon (22.4 MeV with the plotted values 5.33, 1.11 and 7.07 MeV).
Fission: 200/236 ≈ 0.85 MeV per nucleon, so this fusion releases about 3 times as much energy per nucleon (per kilogram of fuel)
✓ 1
Accept division by 235. Allow ECF from MP2. A comparison (ratio or "about three times") is required.
Part (c)(ii)
Rest energy of the reactants ≈ 8 × 931.5 = 7.45 × 103 MeV
✓ 1
ALT: (6.014 + 2.014) u × 931.5 MeV u−1 using (b)(ii); either is accepted.
Fraction ≈ 22.8/7.45 × 103 = 3 × 10−3 (0.3 %)
✓ 1
Allow ECF from (c)(i). Accept 2.8–3.2 × 10−3.
Answers: (b)(i) 0.034 u · (b)(ii) 6.014 u · (c)(i) ≈ 2.8 MeV per nucleon, about 3 times that of fission · (c)(ii) 3 × 10−3(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — nuclear binding energy and mass defect; the variation of the binding energy per nucleon with nucleon number; the mass-energy equivalence as given by E = mc² in nuclear reactions; guidance: an interpretation of binding energy curves is required; nuclear masses will be expressed in kg, in MeVc−2 and in (unified) atomic mass units u; E.4 — that energy is released in spontaneous and neutron-induced fission; E.5 — that fusion is a source of energy in stars Command term: Determine
43E-2-07
Alpha decay and nuclear energy levels·E.3 Radioactive decay
Paper 2Hard12 marks
Short answer & extended response9 steps to full marksDetermine
Thorium-228 (Z = 90) decays by alpha emission to radium-224. The graph shows the energy spectrum of the alpha particles emitted by a very thin source, measured with a detector in a vacuum. The energy of each peak is marked on the graph.
Assume that the thorium nucleus is at rest when it decays and take the mass of each nucleus to be proportional to its nucleon number. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg)
Energy spectrum of the alpha particles from thorium-228 (graph drawn to scale)
(a)
(i)
Write the equation for the decay.
(1)
(ii)
Explain how the spectrum provides evidence for nuclear energy levels.
(2)
(b)
In most decays an alpha particle of kinetic energy 5.423 MeV is emitted and the radium nucleus is left in its ground state.
(i)
Determine the speed of the recoiling radium nucleus.
(2)
(ii)
Hence determine the total energy Q released in this decay.
(2)
(iii)
In the other decays the radium nucleus is left in an excited state, from which it emits a gamma-ray photon. Determine the energy, in keV, of this photon.
(2)
(iv)
Calculate the wavelength of this photon.
(1)
(c)
(i)
Suggest why the spectrum must be measured with a very thin source and with the detector in a vacuum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
22890Th → 22488Ra + 42He
✓ 1
Nucleon and proton numbers must balance; accept α for 42He.
Part (a)(ii)
The alpha particles have only two discrete energies (two narrow peaks), not a continuous range
✓ 1
Each alpha energy corresponds to the radium nucleus being left in a state of definite energy: the ground state or an excited state
✓ 1
Do not accept answers that refer to electron energy levels.
Allow ECF from (b)(iii). Accept 1.4–1.5 × 10−11 m.
Part (c)(i)
Alpha particles are strongly ionising: they lose kinetic energy in collisions with atoms of the air or of the source material
✓ 1
Accept "they have a very short range in air".
The energy lost differs from particle to particle, so the peaks would be shifted to lower energies and broadened; the two peaks, only 0.083 MeV apart, could then not be distinguished
✓ 1
Allow ECF from (b)(iii).
Answers: (b)(i) 2.9 × 105 m s−1 · (b)(ii) 5.52 MeV · (b)(iii) 84 keV · (b)(iv) 1.5 × 10−11 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; E.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving α, γ; the penetration and ionizing ability of alpha particles; A.2 — conservation of linear momentum; A.3 — Ek = ½mv² = p²/2m; E.1 — E = hfCommand term: Determine
44E-2-10
Uranium–lead dating·E.3 Radioactive decay
Paper 2Hard10 marks
Short answer & extended response6 steps to full marksDetermine
Uranium-238 decays, through a chain of short-lived nuclides, to stable lead-206. The half-life of uranium-238 is 4.47 × 109 years. When a zircon crystal forms, it contains uranium but no lead.
(1 year = 3.16 × 107 s, NA = 6.02 × 1023 mol−1)
(a)
(i)
Show that the decay constant of uranium-238 is about 4.9 × 10−18 s−1.
(1)
(ii)
Show that, a time t after the crystal formed, the ratio of the number of lead-206 atoms to the number of uranium-238 atoms is NPb/NU = eλt − 1.
(2)
(iii)
Outline why the intermediate nuclides can be ignored in (a)(ii).
(1)
(b)
(i)
In a zircon crystal NPb/NU = 0.48. Determine the age of the crystal in years.
(2)
(ii)
Explain whether the age found in (b)(i) would be too large or too small if the crystal had contained some lead-206 when it formed.
(2)
(c)
(i)
The crystal now contains 3.0 mg of uranium-238. Determine the number of lead-206 atoms formed in the crystal each second.
Each lead-206 atom was once a uranium-238 atom, so NU + NPb = N0
✓ 1
This hidden step must be stated.
With NU = N0e−λt: NPb/NU = (N0 − NU)/NU = eλt − 1
✓ 1
Part (a)(iii)
Their half-lives are very short, so each decays almost as soon as it forms: at any time a negligible number of atoms are in intermediate states
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
eλt = 1.48, so t = ln 1.48/λ = 7.99 × 1016 s
✓ 1
Allow ECF from (a)(i) and (a)(ii).
t = 2.5 × 109 years
✓ 1
Accept 2.5 × 109 years. ALT: (ln 1.48/ln 2) × 4.47 × 109 years.
Part (b)(ii)
Some of the lead did not come from decay, so the measured ratio is larger than the ratio due to decay alone
✓ 1
A larger ratio gives a larger eλt, so the calculated age is too large
✓ 1
MP2 depends on MP1. Allow ECF from (a)(ii).
Part (c)(i)
NU = (3.0 × 10−3/238) × 6.02 × 1023 = 7.6 × 1018
✓ 1
Mass in grams divided by 238 g mol−1.
Each uranium decay leads, almost at once, to one lead atom: rate = λNU = 37 s−1
✓ 1
Allow ECF from (a)(i). Accept 36–38 s−1.
Answers: (b)(i) 2.5 × 109 years · (c)(i) 37 s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 (HL) — the radioactive decay law N = N0e−λt and T½ = ln 2/λ; the activity A = λN; application of the decay equations for arbitrary time intervals; guidance: radioactive dating based on the penetration of the decay particle and half-life Command term: Determine
45E-2-12
Beta decay and the antineutrino·E.3 Radioactive decay
Paper 2Medium12 marks
Short answer & extended response7 steps to full marksExplain
Self-illuminating exit signs contain tritium (hydrogen-3) gas sealed in glass tubes coated on the inside with a phosphor. Tritium decays by β− emission with a half-life of 12.3 years. The graph shows the distribution of the kinetic energies of the emitted electrons; their mean kinetic energy is 5.7 keV.
Atomic masses: ³H 3.016 049 u, ³He 3.016 029 u. (1 u = 931.5 MeV c−2 = 1.661 × 10−27 kg, e = 1.60 × 10−19 C, 1 year = 3.16 × 107 s)
Kinetic energies of the electrons emitted by tritium (graph drawn to scale)
(a)
(i)
Write the equation for the decay of tritium.
(1)
(ii)
Show that the energy released in each decay is about 19 keV, explaining why the atomic masses may be used.
(2)
(iii)
Explain how the graph provides evidence for the existence of the antineutrino.
(2)
(iv)
State the energy of the antineutrino emitted with an electron of kinetic energy 5.0 keV. Ignore the recoil of the nucleus.
(1)
(b)
(i)
One sign contains tritium of activity 9.0 × 1011 Bq. The phosphor converts 20 % of the kinetic energy of the electrons into light. Determine the light power emitted by the sign.
(2)
(ii)
Determine the percentage of this light power that remains after 15 years.
(2)
(iii)
Determine the mass of tritium in the sign.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
³₁H → ³₂He + ⁰₋₁e + ν̄
✓ 1
The antineutrino is required.
Part (a)(ii)
The ³He atom has one more electron than the ³H atom; this extra electron accounts for the emitted β− particle, so the electron masses balance
If only the electron and the ³He nucleus were produced, conservation of energy and momentum would give every electron the same kinetic energy (≈ 18.6 keV)
✓ 1
Allow ECF from (a)(ii).
The electrons have a continuous range of energies from 0 to 18.6 keV, so a third, undetected particle must carry away the rest of the energy
✓ 1
Part (a)(iv)
18.6 − 5.0 = 13.6 keV
✓ 1
Allow ECF from (a)(ii). Accept the maximum read from the graph, 18–19 keV, minus 5.0 keV.
Part (b)(i)
Power carried by the electrons = 9.0 × 1011 × 5.7 × 103 × 1.60 × 10−19 = 8.2 × 10−4 W
✓ 1
Using 18.6 keV (the maximum, not the mean) scores [1] max.
Light power = 0.20 × 8.2 × 10−4 = 1.6 × 10−4 W
✓ 1
Part (b)(ii)
Power ∝ activity = A0e−λt, with λt = ln 2 × 15/12.3 = 0.845
m = 5.05 × 1020 × 3.016 × 1.661 × 10−27 = 2.5 × 10−6 kg
✓ 1
Accept 2.5 mg.
Answers: (a)(iv) 13.6 keV · (b)(i) 1.6 × 10−4 W · (b)(ii) 43 % · (b)(iii) 2.5 × 10−6 kg (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the radioactive decay equations involving β−; the existence of neutrinos and antineutrinos; E.3 (HL) — the continuous spectrum of beta decay as evidence for the neutrino; A = λN = λN0e−λt; A.3 — power and efficiency Command term: Explain
46E-2-15
Binding energy of the deuteron·E.3 Radioactive decay
Paper 2Hard10 marks
Short answer & extended response6 steps to full marksDetermine
The deuteron (the hydrogen-2 nucleus) consists of one proton and one neutron. Masses: proton 1.007 276 u, neutron 1.008 665 u, deuteron 2.013 553 u.
(1 u = 931.5 MeV c−2, 1 MeV = 1.60 × 10−13 J, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1)
(a)
(i)
Determine the binding energy of the deuteron in MeV.
(2)
(ii)
Compare the binding energy per nucleon of the deuteron with that of iron-56, 8.8 MeV.
(1)
(b)
(i)
A gamma-ray photon can split a deuteron into a free proton and neutron. Determine the longest wavelength of photon that could do this if the proton and neutron were left at rest.
(2)
(ii)
Explain why a photon of exactly this wavelength cannot in fact split a deuteron that is at rest.
(2)
(c)
(i)
When a slow neutron is captured by a stationary proton, a deuteron forms and a single photon of energy Eγ is emitted. Show that the kinetic energy of the recoiling deuteron is approximately Eγ²/(2Mc²), where M is the mass of the deuteron.
(2)
(ii)
Calculate this kinetic energy in keV, taking Eγ = 2.22 MeV.
1.11 MeV per nucleon, about 8 times smaller: the deuteron is very weakly bound
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
E = 2.22 MeV = 3.56 × 10−13 J
✓ 1
Allow ECF from (a)(i).
λ = hc/E = 5.6 × 10−13 m
✓ 1
Part (b)(ii)
The photon has momentum (h/λ), which must be conserved
✓ 1
So the proton and neutron cannot both be at rest after the interaction; they must have kinetic energy, so the photon needs more than 2.22 MeV (a shorter wavelength)
✓ 1
Allow ECF from (b)(i).
Part (c)(i)
The initial momentum is (almost) zero, so the deuteron's momentum equals that of the photon: p = Eγ/c
Answers: (a)(i) 2.22 MeV · (a)(ii) 1.11 MeV · (b)(i) 5.6 × 10−13 m · (c)(ii) 1.3 keV (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — binding energy and mass defect; the variation of the binding energy per nucleon with nucleon number; the mass–energy equivalence E = mc²; E.2 (HL) — photon momentum p = h/λ; A.2 — conservation of linear momentum, Ek = p²/2mCommand term: Determine
47E-2-22
Identifying radiations and background·E.3 Radioactive decay
Paper 2Easy8 marks
Short answer & extended response6 steps to full marksDeduce
A student places a radioactive source 5.0 cm from the window of a Geiger–Müller tube and records the number of counts in 100 s with different absorbers placed between them. With the source removed, 52 counts are recorded in 100 s.
Absorber
Counts in 100 s
none
4210
paper
4195
3 mm aluminium
978
6 mm aluminium
970
(a)
(i)
Deduce which types of radiation reach the tube from the source.
(2)
(ii)
Calculate the count rate due to the gamma radiation alone.
(1)
(b)
(i)
With the 6 mm of aluminium in place, the tube is moved to 10.0 cm from the source. Predict the number of counts that will be recorded in 100 s.
(2)
(ii)
Estimate the absolute uncertainty in the count of 970 and hence comment on whether the second 3 mm of aluminium absorbs any of the gamma radiation.
(2)
(c)
(i)
Explain why the result with paper does not show that the source emits no alpha particles.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The paper causes no significant change, so no alpha particles reach the tube; 3 mm of aluminium greatly reduces the count: beta
✓ 1
Through 3 mm and 6 mm of aluminium the count is almost unchanged and far above background: gamma
✓ 1
Part (a)(ii)
(970 − 52)/100 = 9.2 s−1
✓ 1
Accept 9.2–9.3 s−1.
Part (b)(i)
Only the counts from the source follow the inverse-square law: (970 − 52)/4 = 230
✓ 1
Allow ECF from (a)(ii).
Adding the background: 230 + 52 = 282
✓ 1
Accept 280–285. 970/4 = 243 scores [1].
Part (b)(ii)
Uncertainty ≈ √970 ≈ ±31 (about 3 %)
✓ 1
The difference between 978 and 970 is far smaller than this, so there is no measurable absorption by the extra aluminium
✓ 1
MP2 depends on MP1.
Part (c)(i)
Alpha particles have a range of only a few centimetres in air, so any emitted would be absorbed by the 5.0 cm of air (and the tube window) even without the paper
✓ 1
Allow ECF from (a)(i).
Answers: (a)(ii) 9.2 s−1 · (b)(i) 282 · (b)(ii) ±31 (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the random and spontaneous nature of radioactive decay; the effect of background radiation on count rate; B.1 — apparent brightness b = L/4πd² (inverse-square law) Command term: Deduce
48E-2-23
The strong nuclear force and stability·E.3 Radioactive decay
Paper 2Hard12 marks
Short answer & extended response8 steps to full marksExplain
In a nucleus, neighbouring protons are about 2.0 × 10−15 m apart. (Mass of a proton = 1.673 × 10−27 kg, k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, G = 6.67 × 10−11 N m² kg−2)
(a)
Forces between nucleons.
(i)
Calculate the magnitude of the electric force between two protons 2.0 × 10−15 m apart.
(1)
(ii)
Calculate the magnitude of the gravitational force between the same two protons.
(1)
(iii)
Explain how these results, together with the existence of stable nuclei, provide evidence for the strong nuclear force.
(2)
(b)
The neutron-to-proton ratio.
(i)
Light stable nuclides have roughly equal numbers of neutrons and protons, but heavy stable nuclides such as lead-208 (20882Pb) have many more neutrons than protons. Explain this, with reference to the properties of the strong nuclear force and of the electric force.
(3)
(c)
Part of a natural decay chain is 21884Po → 21482Pb → 21483Bi → 21484Po.
(i)
Identify the type of decay in each of the three steps.
(1)
(ii)
Show that the first step increases the neutron-to-proton ratio.
(2)
(iii)
Using your answer to (c)(ii), explain why lead-214 decays by β− emission rather than by β+ emission.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
F = ke²/r² = 8.99 × 109 × (1.60 × 10−19)²/(2.0 × 10−15)² = 58 N
✓ 1
Accept 57–58 N.
Part (a)(ii)
F = Gm²/r² = 6.67 × 10−11 × (1.673 × 10−27)²/(2.0 × 10−15)² = 4.7 × 10−35 N
✓ 1
Part (a)(iii)
The gravitational attraction is about 1036 times smaller than the electric repulsion, so it cannot hold the protons together
✓ 1
Since stable nuclei exist, there must be another attractive force between nucleons, stronger than the electric force at these separations: the strong nuclear force
✓ 1
Part (b)(i)
The strong force is attractive between all nucleons (n–n, n–p, p–p) but short-range (about 10−15 m), so each nucleon is attracted only by its nearest neighbours
✓ 1
The electric repulsion is long-range and acts between every pair of protons, so the total repulsion grows faster than Z (roughly as Z²) as Z increases
✓ 1
Extra neutrons add strong-force attraction without adding repulsion, so heavy nuclei need N > Z to be stable (lead-208: N/Z = 126/82 = 1.54)
✓ 1
Part (c)(i)
α, then β−, then β−
✓ 1
All three for the mark.
Part (c)(ii)
Polonium-218: N/Z = 134/84 = 1.595
✓ 1
Lead-214: N/Z = 132/82 = 1.610, which is larger
✓ 1
Alpha emission removes 2 protons and 2 neutrons; because N > Z the proportional loss of protons is greater.
Part (c)(iii)
Lead-214 has N/Z = 1.61, above that of stable lead-208 (1.54): it has too many neutrons for its proton number
✓ 1
β− decay turns a neutron into a proton, lowering N/Z (to 131/83 = 1.58) towards stability; β+ would raise N/Z further
✓ 1
Answers: (a)(i) 58 N · (a)(ii) 4.7 × 10−35 N · (c)(ii) 1.595 → 1.610 (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the existence of the strong nuclear force, a short-range, attractive force between nucleons; the changes in the state of the nucleus following alpha and beta decay; E.3 (HL) — the evidence for the strong nuclear force; the role of the ratio of neutrons to protons for the stability of nuclides; D.1/D.2 — F = Gm1m2/r² and F = kq1q2/r² Command term: Explain
49E-2-24
Radioisotope power sources·E.3 Radioactive decay
Paper 2Hard13 marks
Short answer & extended response9 steps to full marksDetermine
A radioisotope thermoelectric generator powers a deep-space probe. It contains 4.8 kg of plutonium-238 (Z = 94), which decays by alpha emission to uranium-234 with a half-life of 87.7 years. The generator converts 6.0 % of the thermal power released into electrical power.
Atomic masses: 238Pu 238.049 560 u, 234U 234.040 952 u, 4He 4.002 603 u. (1 u = 931.5 MeV c−2 = 1.661 × 10−27 kg; 1 year = 3.16 × 107 s)
(a)
The decay.
(i)
Write the equation for the decay of plutonium-238.
(1)
(ii)
Show that the energy released in each decay is about 9.0 × 10−13 J.
(2)
(b)
Choice of isotope.
(i)
Explain why an alpha emitter is a good choice as the heat source on a spacecraft.
(2)
(ii)
Suggest why plutonium-238 is preferred both to polonium-210 (half-life 138 days) and to uranium-238 (half-life 4.5 × 109 years) for a mission lasting decades.
(2)
(c)
Power.
(i)
Show that the decay constant of plutonium-238 is about 2.5 × 10−10 s−1.
(1)
(ii)
Determine the initial thermal power of the generator.
(3)
(iii)
The probe needs at least 140 W of electrical power. Determine how long after launch the generator can meet this requirement.
Alpha particles have a very short range (strongly ionising), so they are absorbed within the fuel and its casing and almost all their energy becomes thermal energy
✓ 1
Little penetrating radiation escapes, so only light shielding is needed to protect the probe’s electronics
✓ 1
Part (b)(ii)
Polonium-210: the half-life is far too short — the power would halve every 4.5 months and be negligible within a few years
✓ 1
Uranium-238: the half-life is so long that the activity per kilogram (∝ 1/T½) is about 5 × 107 times smaller, so an impossibly large mass would be needed
Answers: (a)(ii) 8.98 × 10−13 J · (c)(i) 2.50 × 10−10 s−1 · (c)(ii) 2.7 kW · (c)(iii) 20 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the radioactive decay equations involving α; the mass-energy equivalence E = mc² in nuclear reactions; the penetration and ionizing ability of alpha particles; E.3 (HL) — A = λN = λN0e−λt, T½ = ln 2/λ; application of the decay equations for arbitrary time intervals; A.3 — efficiency Command term: Determine
50E-2-32
Radiocarbon dating·E.3 Radioactive decay
Paper 2Easy13 marks
Short answer & extended response7 steps to full marksDetermine
Carbon-14 (146C) is formed continually in the atmosphere and decays by β− emission with a half-life of 5730 years. Living organisms take in carbon from the atmosphere, so the activity of the carbon-14 in 1.0 g of carbon from a living organism stays at 0.25 Bq. After the organism dies it takes in no more carbon.
(Molar mass of carbon = 12 g mol−1; NA = 6.02 × 1023 mol−1; 1 year = 3.16 × 107 s; nitrogen has Z = 7)
(a)
(i)
Write down the equation for the decay of carbon-14.
(1)
(b)
(i)
Show that the decay constant of carbon-14 is about 3.8 × 10−12 s−1.
(1)
(ii)
Determine the number of carbon-14 nuclei in 1.0 g of carbon from a living organism.
(2)
(iii)
Hence determine the ratio (number of carbon-14 nuclei)/(total number of carbon atoms) in living carbon.
(2)
(c)
(i)
The carbon from a wooden tool has an activity per gram that is 1/8 of the value for living carbon. Determine the age of the tool and the number of carbon-14 nuclei that remain in each gram of its carbon.
(2)
(ii)
A 4.0 g sample of carbon from an ancient fire has an activity of 0.62 Bq. Determine the age of the sample.
(2)
(iii)
Radiocarbon dating is not used for samples older than about 50 000 years. Determine the activity of the carbon-14 in a 2.0 g sample of carbon that is 6.0 × 104 years old, and explain why such a sample cannot be dated reliably.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
146C → 147N + 0−1e + 00ν̄
✓ 1
All four products correct, with nucleon and proton numbers. Accept β− for the electron. Do not accept a neutrino ν.
Part (b)(i)
λ = ln 2/(5730 × 3.16 × 107) = 3.83 × 10−12 s−1
✓ 1
Full substitution or an answer to at least 3 s.f. must be seen.
Part (b)(ii)
N = A/λ
✓ 1
N = 0.25/3.8 × 10−12 = 6.5 × 1010
✓ 1
Allow ECF from (b)(i). Accept 6.5–6.6 × 1010.
Part (b)(iii)
Number of carbon atoms in 1.0 g = 6.02 × 1023/12 = 5.0 × 1022
✓ 1
Ratio = 6.5 × 1010/5.0 × 1022 = 1.3 × 10−12
✓ 1
Allow ECF from (b)(ii).
Part (c)(i)
1/8 = (½)³, so three half-lives: age = 3 × 5730 = 1.7 × 104 years
✓ 1
Accept 17 000 or 17 200 years.
N = 6.5 × 1010/8 = 8.2 × 109
✓ 1
Allow ECF from (b)(ii). Accept 8.1–8.2 × 109.
Part (c)(ii)
Activity per gram = 0.62/4.0 = 0.155 Bq, and A = A0e−λt gives t = ln(0.25/0.155)/λ
✓ 1
Allow ECF from (b)(i). Comparing 0.62 Bq with 0.25 Bq without dividing by the mass: 0 for this mark.
t = 0.478/1.21 × 10−4 y−1 = 4.0 × 103 years
✓ 1
Accept 3.9–4.0 × 103 years, or 1.25 × 1011 s.
Part (c)(iii)
Number of half-lives = 6.0 × 104/5730 = 10.5
✓ 1
A = 0.25 × 2.0 × (½)10.5 = 3.5 × 10−4 Bq
✓ 1
Accept 3.4–3.6 × 10−4 Bq. ALT: A = 0.50 e−λt with λ from (b)(i). Allow ECF from (b)(i).
This is far smaller than the background count rate (typically of order 0.1–1 s−1), so the counts from the sample are lost in the random fluctuations of the background
✓ 1
Reference to background is needed; "too few nuclei left" alone is not enough.
Answers: (b)(ii) 6.5 × 1010 · (b)(iii) 1.3 × 10−12 · (c)(i) 1.7 × 104 years; 8.2 × 109 · (c)(ii) 4.0 × 103 years · (c)(iii) 3.5 × 10−4 Bq (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the radioactive decay equations involving β−; the existence of antineutrinos; the changes in activity and count rate during radioactive decay using integer values of half-life; the effect of background radiation on count rate; guidance: radioactive dating based on the penetration of the decay particle and half-life; E.3 (HL) — A = λN = λN0e−λt, T½ = ln 2/λ, decay equations for arbitrary time intervals; B.3 — amount of substance n = N/NACommand term: Determine
51E-2-33
A cobalt-60 radiotherapy source·E.3 Radioactive decay
Paper 2Easy13 marks
Short answer & extended response7 steps to full marksDetermine
Cobalt-60 (6027Co) decays by β− emission, with a half-life of 5.27 years, to an excited state of nickel-60 (Z = 28) that is 2.51 MeV above the nickel ground state. The nickel nucleus reaches its ground state by emitting two gamma-ray photons in succession, passing through a single intermediate level 1.33 MeV above the ground state.
In a radiotherapy unit, a sealed cobalt-60 source of activity 2.2 × 1014 Bq is held in a thick tungsten container. A narrow opening lets a beam of gamma rays reach a tumour deep inside a patient.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg, 1 year = 3.16 × 107 s)
(a)
(i)
Write down the equation for the β− decay of cobalt-60.
(1)
(ii)
Determine the energies of the two gamma-ray photons, and the wavelength of the more energetic photon.
(2)
(b)
(i)
Show that the power carried away from the source by gamma radiation is about 90 W.
(2)
(ii)
Determine the mass of cobalt-60 in the source.
(3)
(iii)
The source is replaced 10.5 years after it is installed. Deduce the power carried away by gamma radiation at that time.
(1)
(c)
(i)
Explain why the gamma radiation, and not the beta radiation, from cobalt-60 is used to treat a tumour deep inside the body.
(2)
(ii)
Assume that the source emits gamma radiation equally in all directions and that none is absorbed. Determine the intensity of the gamma radiation at a distance of 0.80 m from the source if the container were opened completely.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
6027Co → 6028Ni + 0−1e + 00ν̄
✓ 1
Accept Ni* for the excited nickel nucleus. The antineutrino is required.
Using 1.18 MeV (1.05 × 10−12 m) scores 0 for this mark; using 2.51 MeV likewise.
Part (b)(i)
Gamma energy per decay = 1.18 + 1.33 = 2.51 MeV = 2.51 × 1.60 × 10−13 J
✓ 1
Allow ECF from (a)(ii).
P = 2.2 × 1014 × 2.51 × 1.60 × 10−13 = 88.4 W
✓ 1
An answer to at least 3 s.f. or the full substitution must be seen. Using one photon per decay (47 W or 42 W) scores [1 max].
Part (b)(ii)
λ = ln 2/(5.27 × 3.16 × 107) = 4.16 × 10−9 s−1
✓ 1
N = A/λ = 2.2 × 1014/4.16 × 10−9 = 5.3 × 1022
✓ 1
m = 5.29 × 1022 × 60 × 1.661 × 10−27 = 5.3 × 10−3 kg
✓ 1
Accept 5.2–5.3 × 10−3 kg. Allow ECF from MP2.
Part (b)(iii)
10.5 years is two half-lives, so the activity and the power fall to one quarter: 88.4/4 = 22 W
✓ 1
Allow ECF from (b)(i). Using 90 W gives 23 W; accept 22–23 W.
Part (c)(i)
Beta particles are much more strongly ionising than gamma rays and are absorbed within a few millimetres of tissue (or in the source's own capsule), so they cannot reach a deep tumour
✓ 1
Gamma rays are weakly ionising and very penetrating, so a significant fraction passes through the overlying tissue and deposits energy in the tumour
✓ 1
Comparison of penetration is needed for both marks.
Part (c)(ii)
I = P/4πr² = 88.4/(4π × 0.80²)
✓ 1
Allow ECF from (b)(i). Using 90 W gives 11 W m−2.
I = 11 W m−2
✓ 1
Accept 11 W m−2. Using 2πr² or πr² scores [1 max].
Answers: (a)(ii) 1.18 MeV and 1.33 MeV; 9.3 × 10−13 m · (b)(ii) 5.3 × 10−3 kg · (b)(iii) 22 W · (c)(ii) 11 W m−2(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving β− and γ; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the changes in activity using integer values of half-life; guidance: the choice of isotope in medical use; E.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; A = λN; E.1 — E = hf; A.3 — power; B.1 — intensity from a point source, I = P/4πr² Command term: Determine
52E-2-47
Nitrogen-16 in reactor cooling water·E.3 Radioactive decay
Paper 2Easy13 marks
Short answer & extended response10 steps to full marksExplain
In a pressurised-water nuclear reactor, the water that cools the core is bombarded by fast neutrons. A few oxygen-16 nuclei absorb a neutron and emit a proton, forming nitrogen-16 (Z = 7). Nitrogen-16 decays by β− emission, with a half-life of 7.1 s, to an excited state of oxygen-16, which reaches its ground state by emitting a gamma-ray photon of energy 6.13 MeV. The only stable isotopes of nitrogen are nitrogen-14 and nitrogen-15.
State one similarity and one difference between the nuclei of nitrogen-14 and nitrogen-16.
(1)
(ii)
Explain, with reference to the neutron-to-proton ratio, why nitrogen-16 decays by β− emission.
(2)
(iii)
Write down the equation for the β− decay of nitrogen-16.
(1)
(b)
(i)
Calculate the wavelength of the gamma-ray photon.
(1)
(ii)
The pipes that carry the cooling water out of the reactor are surrounded by thick concrete, although the β− particles from nitrogen-16 cannot escape from the steel pipes. Explain why the concrete is needed.
(2)
(c)
(i)
Show that the decay constant of nitrogen-16 is about 0.098 s−1.
(1)
(ii)
The water takes 12 s to travel from the core to the steam generator outside the reactor. Calculate the fraction of the nitrogen-16 formed in the core that remains when the water reaches the steam generator.
(1)
(iii)
When the reactor is shut down, the cooling water continues to circulate. Determine the time after shutdown at which the activity of the nitrogen-16 has fallen to 1.0 × 10−4 of its value at shutdown.
(2)
(d)
(i)
Explain why workers may enter the room containing the steam generator a few minutes after the reactor has been shut down, but not while it is operating.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Both have 7 protons; nitrogen-14 has 7 neutrons and nitrogen-16 has 9 neutrons
✓ 1
Both needed. Accept "different nucleon numbers (14 and 16)" for the difference.
Part (a)(ii)
N/Z for nitrogen-16 is 9/7 = 1.29, greater than for the stable isotopes (1.00 and 1.14): it has too many neutrons for stability
✓ 1
A comparison with a stable isotope (or with N/Z ≈ 1 for stable light nuclides) is needed.
β− decay changes a neutron into a proton, lowering N/Z to 8/8 = 1.00 (oxygen-16, a stable nuclide)
✓ 1
Do not accept "β− decay removes a neutron".
Part (a)(iii)
167N → 168O + 0−1e + ν̄
✓ 1
Accept O* for the excited oxygen nucleus; accept β− for the electron. The antineutrino is required.
β− particles are moderately ionising (far more than gamma photons) and lose their energy within a short range, so they are absorbed by a few millimetres of metal (the pipe wall)
✓ 1
The 6.13 MeV gamma-ray photons are weakly ionising and very penetrating: they pass through the steel, so a large thickness of dense material is needed to absorb them
✓ 1
Comparison of the penetration of the two radiations is needed for both marks.
Part (c)(i)
λ = ln 2/7.1 = 0.0976 s−1
✓ 1
An answer to at least 3 s.f. (0.0976) or the full substitution must be seen.
Allow ECF from (c)(i). ALT: number of half-lives = lg(104)/lg 2 = 13.3.
t = 9.21/0.0976 = 94 s
✓ 1
Accept 93–95 s.
Part (d)(i)
After shutdown the control rods absorb the neutrons and fission stops, so no more fast neutrons are produced and no new nitrogen-16 is formed
✓ 1
Reference to the loss of neutrons (no new production) is needed.
The nitrogen-16 already formed decays quickly: its activity falls below 10−4 of its initial value in about 94 s (about 1.5 min), so the gamma radiation from the pipes becomes negligible within a few minutes
✓ 1
Allow ECF from (c)(iii).
Answers: (b)(i) 2.0 × 10−13 m · (c)(ii) 0.31 · (c)(iii) 94 s (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — isotopes; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the radioactive decay equations involving α, β−, β+, γ; the existence of neutrinos ν and antineutrinos ν̄; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; E.3 (HL) — the role of the ratio of neutrons to protons for the stability of nuclides; the decay constant λ and the radioactive decay law as given by N = N0e−λt; guidance: application of the decay equations for arbitrary time intervals; E.1 — photon energy E = hf; E.4 — the role of control rods in a nuclear power plant Command term: Explain
53E-2-48
Why heavy nuclei emit alpha particles·E.3 Radioactive decay
Paper 2Hard14 marks
Short answer & extended response7 steps to full marksDetermine
Uranium-238 (Z = 92) decays by alpha emission to thorium-234 with a half-life of 4.47 × 109 years. The table gives the binding energies of four nuclei.
(NA = 6.02 × 1023 mol−1, 1 MeV = 1.60 × 10−13 J, 1 year = 3.16 × 107 s)
Nucleus
uranium-238
thorium-234
uranium-237
helium-4
Binding energy / MeV
1801.7
1777.7
1795.6
28.3
(a)
(i)
Calculate the energy released when a uranium-238 nucleus emits an alpha particle.
(2)
(ii)
A uranium-238 nucleus could, in principle, emit a single neutron instead. Determine the energy change in this process and hence explain why uranium-238 does not decay in this way.
(2)
(b)
(i)
Explain, with reference to the strong nuclear force and to your answers to (a), why a heavy nucleus can emit an alpha particle but not a single nucleon.
(3)
(c)
(i)
Show that the thermal power P released per unit mass of pure uranium-238 by alpha emission is P/m = (ln 2 × NA × Q)/(T½ × M), where Q is the energy released per decay, T½ is the half-life and M is the molar mass in kg mol−1. Assume that all the kinetic energy of the alpha particles and recoiling nuclei is absorbed in the uranium.
(2)
(ii)
Calculate P/m for uranium-238 in W kg−1.
(2)
(iii)
In a granite, uranium-238 makes up 4.0 × 10−6 of the mass. In old rock, each uranium-238 decay is followed by a chain of decays of short-lived nuclides that deposit a further 43 MeV in the rock. The specific heat capacity of granite is 790 J kg−1 K−1. Estimate the temperature rise of the granite in 1.0 × 106 years if no energy were lost from it.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Energy released = (total binding energy of the products) − (binding energy of the parent)
✓ 1
The idea that the products are more tightly bound (have less mass) must be seen, in words or in the substitution.
1777.7 + 28.3 − 1801.7 = 4.3 MeV
✓ 1
Reversing the subtraction (−4.3 MeV, "energy absorbed"): [1 max].
Part (a)(ii)
Binding energy of the products − binding energy of the parent = 1795.6 − 1801.7 = −6.1 MeV (a free neutron has no binding energy)
✓ 1
Accept "6.1 MeV must be supplied".
The products would have more mass (less binding energy) than the parent: 6.1 MeV would have to be supplied, so the process cannot happen spontaneously, whereas alpha emission releases energy
✓ 1
Allow ECF from (a)(i). A comparison with (a)(i) is needed.
Part (b)(i)
The strong force is short-range, so each nucleon is bound only to its nearest neighbours; for A > 60 every nucleon contributes roughly the same binding energy (about 7.6 MeV for uranium-238), and this is roughly the energy needed to remove any single nucleon
✓ 1
Allow ECF from (a).
Removing four nucleons costs about 4 × 7.6 ≈ 30 MeV, but the alpha particle is itself very tightly bound (28.3 MeV, about 7.1 MeV per nucleon), so almost all of this is regained when the four nucleons leave as an alpha particle
✓ 1
The electric repulsion between the many protons lowers the binding energy per nucleon of heavy nuclei, so the daughter has a slightly larger binding energy per nucleon (thorium-234: 7.597 MeV; uranium-238: 7.570 MeV); this small gain makes the total energy released positive
✓ 1
[3 max]. Accept: a single nucleon carries away no binding energy, so its whole binding energy (≈ 6 MeV) has to be supplied.
Part (c)(i)
Number of nuclei per unit mass N/m = NA/M
✓ 1
Activity per unit mass = λN/m = (ln 2/T½)(NA/M); each decay releases Q, so P/m = (ln 2 × NA × Q)/(T½ × M)
Allow ECF from (c)(i). Accept 8.4–8.6 × 10−6 W kg−1. Using M = 238 kg mol−1 (8.5 × 10−9): [1 max].
Part (c)(iii)
Power per kilogram of granite = 8.5 × 10−6 × (4.3 + 43)/4.3 × 4.0 × 10−6 = 3.8 × 10−10 W kg−1
✓ 1
Allow ECF from (c)(ii). Omitting the chain (factor 11): [2 max].
Energy per kilogram in 1.0 × 106 years = 3.8 × 10−10 × 3.16 × 1013 = 1.2 × 104 J kg−1
✓ 1
ΔT = 1.2 × 104/790 = 15 K
✓ 1
Accept 14–16 K. The amount of uranium-238 is essentially constant, since 106 years ≪ 4.47 × 109 years.
Answers: (a)(i) 4.3 MeV · (a)(ii) −6.1 MeV (6.1 MeV must be supplied) · (c)(ii) 8.5 × 10−6 W kg−1 · (c)(iii) 15 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — nuclear binding energy and mass defect; the variation of the binding energy per nucleon with nucleon number; the mass–energy equivalence as given by E = mc2 in nuclear reactions; the existence of the strong nuclear force, a short-range, attractive force between nucleons; E.3 (HL) — the approximate constancy of binding energy curve above a nucleon number of 60; the evidence for the strong nuclear force; the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; B.1 — specific heat capacity Q = mc ΔT; B.3 — molar mass and the Avogadro constant Command term: Determine
54E-2-49
The ionisation smoke detector·E.3 Radioactive decay
Paper 2Medium14 marks
Short answer & extended response8 steps to full marksDetermine
A household smoke detector contains a source of americium-241 (Z = 95) of activity 33 kBq. Americium-241 decays by alpha emission to neptunium (Np), with a half-life of 432 years; each alpha particle has a kinetic energy of 5.49 MeV. Half of the alpha particles enter the air between two metal plates connected to a battery, where each stops after producing ion pairs; on average, 34 eV is needed to produce one ion pair. The ions move to the plates and form a small current.
(e = 1.60 × 10−19 C, 1 year = 3.16 × 107 s)
(a)
(i)
Write down the equation for the decay of americium-241.
(1)
(ii)
Explain why an alpha emitter, rather than a beta or gamma emitter, is used in the detector.
(2)
(b)
(i)
Show that about 2.7 × 109 ion pairs are produced per second.
(2)
(ii)
Each ion pair transfers a charge e round the circuit. The current passes through a resistor of resistance 2.0 × 109 Ω. Calculate the maximum current and the potential difference across the resistor.
(2)
(c)
(i)
The detector has a working life of 10 years. Determine the probability that a particular americium-241 nucleus in the source decays during this time, and explain why the product λt may be used as this probability here.
(3)
(ii)
Hence determine the maximum current at the end of the working life of the detector.
(2)
(d)
(i)
The alarm sounds when the current falls by 30 %. Explain why smoke entering the space between the plates makes the alarm sound, and why the decay of the americium does not.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
24195Am → 23793Np + 42He
✓ 1
Accept α for the helium nucleus. Nucleon and proton numbers must balance.
Part (a)(ii)
Alpha particles are the most strongly ionising, so they produce many ion pairs and a measurable current from a small, low-activity source
✓ 1
Alpha particles have a range of only a few centimetres in air and are stopped by the casing, so people near the detector receive almost no radiation
✓ 1
Do not accept "alpha particles are not dangerous" without reference to their short range.
Part (b)(i)
Ion pairs per alpha particle = 5.49 × 106/34 = 1.61 × 105
Must see the factor 0.5 and an answer to at least 3 s.f., or the full substitution.
Part (b)(ii)
I = 2.7 × 109 × 1.60 × 10−19 = 4.3 × 10−10 A
✓ 1
Allow ECF from (b)(i). Using 2.7 × 109 s−1 gives 4.3 × 10−10 A.
V = IR = 4.3 × 10−10 × 2.0 × 109 = 0.85 V
✓ 1
Allow ECF. Accept 0.85–0.86 V.
Part (c)(i)
λ = ln 2/(432 × 3.16 × 107) = 5.08 × 10−11 s−1
✓ 1
λt = 5.08 × 10−11 × 10 × 3.16 × 107 = 0.016
✓ 1
Accept 0.016 (1.6 %). ALT: 1 − (½)10/432 = 0.016.
λt ≪ 1 (the time is much shorter than the half-life), so the probability of decay is approximately λ × time; the exact value 1 − e−λt = 0.0159 agrees
✓ 1
The condition λt ≪ 1 (or t ≪ T½) is needed for this mark.
Part (c)(ii)
The current is proportional to the activity, i.e. to the number of undecayed nuclei, which falls by the fraction 0.016
✓ 1
Allow ECF from (b)(ii) and (c)(i).
I = 4.3 × 10−10 × (1 − 0.016) = 4.2 × 10−10 A
✓ 1
Allow ECF. Accept 4.2 × 10−10 A (a fall of only 1.6 %).
Part (d)(i)
Smoke particles absorb alpha particles and ions attach to the large, slow-moving smoke particles (and recombine), so fewer ions reach the plates and the current falls
✓ 1
OWTTE. Do not accept "smoke blocks the current".
The decay reduces the current by only about 1.6 % in 10 years (from 4.3 to 4.2 × 10−10 A), far less than 30 %
✓ 1
Allow ECF from (c).
Answers: (b)(ii) 4.3 × 10−10 A; 0.85 V · (c)(i) 0.016 · (c)(ii) 4.2 × 10−10 A (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the radioactive decay equations involving α, β−, β+, γ; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; E.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; that the decay constant approximates the probability of decay in unit time only in the limit of sufficiently small λt; guidance: application of the decay equations for arbitrary time intervals; B.5 — electric current I = Δq/Δt and V = IRCommand term: Determine
55E-2-50
Radon in buildings·E.3 Radioactive decay
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
Radon-222 (Z = 86) is a gas that seeps into buildings from the ground and is the largest single source of background radiation for many people. It decays by alpha emission to polonium-218. A sample of air containing radon is sealed in a flask, and the activity A of the radon alone is determined over 16 days. The graph shows the results.
(kB = 1.38 × 10−23 J K−1, 1 day = 8.64 × 104 s)
Activity A of the radon-222 in the sealed flask against time t (drawn to scale).
(a)
(i)
Write down the equation for the decay of radon-222.
(1)
(ii)
Use the graph to determine the half-life of radon-222.
(2)
(b)
(i)
The air in the basement of a house contains radon with an activity of 250 Bq per cubic metre. Determine the number of radon-222 atoms in each cubic metre of this air.
(2)
(ii)
Show that the fraction of the gas particles in the air that are radon atoms is f = AVkBT/(λp), where AV is the activity per unit volume, p is the pressure and T the temperature of the air.
(2)
(iii)
Calculate f for the basement air at a temperature of 20 °C and a pressure of 1.01 × 105 Pa.
(1)
(c)
(i)
Alpha particles from radon outside the body cannot penetrate the outer layer of dead skin. Explain why breathing in radon is nevertheless a health risk.
(3)
(d)
(i)
A sample of the basement air is sealed in a flask so that no more radon can enter. Determine the time for the activity of the radon in the sample to fall from 250 Bq m−3 to 20 Bq m−3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
22286Rn → 21884Po + 42He
✓ 1
Accept α for the helium nucleus.
Part (a)(ii)
Time for the activity to halve read from the curve at least twice, e.g. 480 → 240 Bq and 240 → 120 Bq, or the time for a fall to one quarter halved
✓ 1
A single halving: [1 max].
T½ = 3.8 days
✓ 1
Accept 3.6–4.0 days.
Part (b)(i)
λ = ln 2/(3.8 × 8.64 × 104) = 2.11 × 10−6 s−1
✓ 1
Allow ECF from (a)(ii).
N = A/λ = 250/2.11 × 10−6 = 1.2 × 108 m−3
✓ 1
Accept 1.1–1.3 × 108 m−3.
Part (b)(ii)
Number of radon atoms per unit volume = AV/λ (from A = λN)
✓ 1
Number of gas particles per unit volume = p/(kBT) (from pV = NkBT); dividing gives f = AVkBT/(λp)
Allow ECF from (b)(i) or (b)(ii). Accept 4.4–5.0 × 10−18. Using 20 instead of 293 K: [0].
Part (c)(i)
Inhaled radon decays inside the lungs, next to living cells, so the alpha particles reach living tissue
✓ 1
Alpha particles are strongly ionising: in their short range they deposit all their energy in a small volume of tissue, damaging cells (e.g. the DNA)
✓ 1
The decay product, polonium-218 (from (a)(i)), is a solid that is itself radioactive; it can stick to the lining of the lungs and emit further alpha particles there
✓ 1
Allow ECF from (a)(i). [3 max]. Accept: the daughter nuclides are not breathed out.
Part (d)(i)
250/20 = 12.5 = 2n, so the number of half-lives is ln 12.5/ln 2 = 3.64
✓ 1
Allow ECF from (a)(ii). ALT: t = ln 12.5/λ.
t = 3.64 × 3.8 = 14 days
✓ 1
Accept 13–15 days.
Answers: (a)(ii) 3.8 days · (b)(i) 1.2 × 108 m−3 · (b)(iii) 4.7 × 10−18 · (d)(i) 14 days (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the effect of background radiation on count rate; the radioactive decay equations involving α, β−, β+, γ; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life; E.3 (HL) — the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; guidance: application of the decay equations for arbitrary time intervals; B.3 — the equation of state for an ideal gas pV = NkBTCommand term: Determine
56E-2-51
Caesium-137: beta spectra and a gamma photon·E.3 Radioactive decay
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDeduce
Caesium-137 (Z = 55) decays by β− emission to barium-137 with a half-life of 30.1 years. In 94.4 % of decays the barium-137 nucleus is formed in an excited state, which then emits a gamma-ray photon; in the other 5.6 % it is formed directly in its ground state, as shown in the diagram.
The kinetic energies of the β− particles in the two branches form two continuous spectra, with maximum kinetic energies of 0.514 MeV (branch β1) and 1.176 MeV (branch β2). The recoil energy of the barium nucleus is negligible.
(c = 3.00 × 108 m s−1, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg, 1 year = 3.16 × 107 s)
Energy levels involved in the decay of caesium-137 (not to scale).
(a)
(i)
Write down the equation for the β− decay of caesium-137.
(1)
(b)
(i)
Deduce the energy of the gamma-ray photon.
(2)
(ii)
Explain how the beta spectra and the gamma-ray photon together provide evidence both for the antineutrino and for discrete nuclear energy levels.
(2)
(c)
(i)
A school laboratory source had an activity of 3.70 × 105 Bq when it was supplied, 25.0 years ago. Determine the mass of caesium-137 that it contains now.
(3)
(ii)
Calculate the number of gamma-ray photons emitted per second by the source now. Assume that every excited barium-137 nucleus emits one photon.
(1)
(iii)
Determine the speed of recoil of a barium-137 nucleus that emits one gamma-ray photon while at rest.
(2)
(d)
(i)
Caesium-137 and iodine-131 (half-life 8.0 days) were both released in large amounts by a reactor accident in 1986. Explain, with calculations, why caesium-137, and not iodine-131, still contaminates the land 40 years later.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
13755Cs → 13756Ba + 0−1e + ν̄
✓ 1
The antineutrino is required. Accept Ba*.
Part (b)(i)
Both branches start from the same caesium-137 level; the maximum β− energy is the energy released in that branch (when the antineutrino takes almost none), so the difference between the two maxima equals the energy of the excited state above the ground state
✓ 1
Eγ = 1.176 − 0.514 = 0.662 MeV
✓ 1
Adding the maxima (1.690 MeV): [0] for this mark.
Part (b)(ii)
Each β− spectrum is continuous below a sharp maximum equal to the energy released, so the rest of the energy in each decay must be carried away by another, undetected particle: the antineutrino
✓ 1
The gamma photons all have the single energy 0.662 MeV, exactly equal to the difference between the two maximum β energies: the barium-137 nucleus can exist only in states of definite energy
m = 2.85 × 1014 × 137 × 1.661 × 10−27 = 6.5 × 10−11 kg
✓ 1
Accept 6.4–6.6 × 10−11 kg. Using the original activity (1.2 × 10−10 kg): [2 max].
Part (c)(ii)
0.944 × 2.08 × 105 = 2.0 × 105 s−1
✓ 1
Allow ECF from (c)(i).
Part (c)(iii)
Photon momentum p = E/c = 0.662 × 1.60 × 10−13/3.00 × 108 = 3.5 × 10−22 kg m s−1
✓ 1
Allow ECF from (b)(i).
Momentum is conserved from zero: v = p/m = 3.5 × 10−22/(137 × 1.661 × 10−27) = 1.6 × 103 m s−1
✓ 1
Accept 1.5–1.6 × 103 m s−1.
Part (d)(i)
Caesium-137: (½)40/30.1 = 0.40, so about 40 % of the caesium-137 remains
✓ 1
Accept 0.40 (40 %).
Iodine-131: 40 years is about 1826 half-lives, so the fraction remaining is (½)1826 ≈ 10−550: none remains
✓ 1
Accept any argument that the number of half-lives is very large (about 1800) and the iodine-131 has completely decayed.
Answers: (b)(i) 0.662 MeV · (c)(i) 6.5 × 10−11 kg · (c)(ii) 2.0 × 105 s−1 · (c)(iii) 1.6 × 103 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the radioactive decay equations involving α, β−, β+, γ; the existence of neutrinos ν and antineutrinos ν̄; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; E.3 (HL) — that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels; the continuous spectrum of beta decay as evidence for the neutrino; the activity as the rate of decay as given by A = λN = λN0e−λt; the relationship between half-life and the decay constant as given by T½ = ln 2/λ; E.2 (HL) — photon momentum p = h/λ; A.2 — conservation of linear momentum; E.4 — the properties of the products of nuclear fission Command term: Deduce
Short answer & extended response11 steps to full marksDetermine
In positron emission tomography (PET), a patient is injected with a compound containing fluorine-18 (Z = 9), which decays by β+ emission to oxygen-18. Each positron travels about 1 mm in tissue, slows down and annihilates with an electron. The two gamma-ray photons produced are recorded by a ring of detectors around the patient. The graph shows the activity of a calibration sample of fluorine-18 against time.
Atomic masses: ¹⁸F 18.000 938 u, ¹⁸O 17.999 160 u, ¹H 1.007 825 u; neutron 1.008 665 u; electron 0.000 549 u = 9.11 × 10−31 kg. (1 u = 931.5 MeV c−2, c = 3.00 × 108 m s−1, 1 MeV = 1.60 × 10−13 J)
Activity of a fluorine-18 sample against time (graph drawn to scale)
(a)
(i)
Write the equation for the decay of fluorine-18.
(1)
(ii)
Show that the energy released in the decay is about 0.63 MeV.
(2)
(iii)
State why the positrons are emitted with a range of kinetic energies up to a maximum value.
(1)
(b)
(i)
Determine, using the graph, the half-life of fluorine-18 in minutes.
(2)
(ii)
A patient is injected with fluorine-18 of activity 350 MBq. Determine the number of fluorine-18 nuclei injected.
(2)
(iii)
The scan starts 60 min after the injection and lasts 30 min. Estimate the number of positrons emitted in the patient during the scan.
(2)
(iv)
State one assumption made in your estimate in (b)(iii).
(1)
(c)
(i)
A positron and an electron, both at rest, annihilate. Show that two photons are produced, each of energy mec², and that they travel in opposite directions.
(2)
(ii)
Calculate the momentum of each photon, giving an appropriate SI unit.
(1)
(iii)
The two photons from one annihilation are recorded by two detectors on opposite sides of the ring, and one photon arrives 0.50 ns before the other. Determine the distance of the annihilation point from the midpoint between the two detectors.
(2)
(d)
(i)
Fluorine-18 is made by bombarding oxygen-18 with protons: ¹⁸O + ¹H → ¹⁸F + ¹n. Determine the energy that must be supplied for this reaction to occur, and explain why the protons must in fact have more kinetic energy than this.
(3)
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Marking point
Mark
Notes
Part (a)(i)
¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ν
✓ 1
The neutrino (not antineutrino) is required.
Part (a)(ii)
The atomic masses include 9 and 8 electrons; for β+ decay two electron masses must be subtracted (one surplus atomic electron plus the created positron): Δm = 18.000 938 − 17.999 160 − 2 × 0.000 549 = 0.000 680 u
✓ 1
Without the 2me correction (1.66 MeV) scores 0.
E = 0.000 680 × 931.5 = 0.633 MeV
✓ 1
Part (a)(iii)
The energy released is shared randomly between the positron and the neutrino; the maximum is ≈ 0.63 MeV
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
The activity halves, e.g. from 800 MBq to 400 MBq, in the time read from the curve
✓ 1
Any halving read from the best-fit curve.
T½ = 110 min
✓ 1
Accept 105–115 min.
Part (b)(ii)
λ = ln 2/(110 × 60 s) = 1.05 × 10−4 s−1
✓ 1
Allow ECF from (b)(i).
N = A/λ = 3.50 × 108/1.05 × 10−4 = 3.3 × 1012
✓ 1
Accept 3.2–3.5 × 1012.
Part (b)(iii)
N(60 min) = N0e−λ × 3600 s and N(90 min) = N0e−λ × 5400 s
✓ 1
Allow ECF from (b)(ii). ALT: average activity during the scan × 1800 s.
Number emitted = N(60) − N(90) = 3.9 × 1011
✓ 1
Accept 3.7–4.1 × 1011.
Part (b)(iv)
None of the fluorine-18 leaves the patient's body during this time (or: every decay emits one positron)
✓ 1
Any one.
Part (c)(i)
The initial momentum is zero, so the total momentum of the photons must be zero: a single photon is impossible, and two photons must have equal and opposite momenta
✓ 1
Equal momenta mean equal energies (E = pc); the total energy is the rest energy 2mec², so each photon has energy mec²
✓ 1
Part (c)(ii)
p = E/c = mec = 9.11 × 10−31 × 3.00 × 108 = 2.7 × 10−22 kg m s−1
✓ 1
Allow ECF from (c)(i). The unit (kg m s−1 or N s) is required for the mark.
Part (c)(iii)
Difference in path lengths = cΔt = 3.00 × 108 × 0.50 × 10−9 = 0.15 m
✓ 1
The point is displaced by half of this, 0.075 m (7.5 cm), towards the detector that records the first photon
✓ 1
0.15 m scores [1]. Allow ECF from (c)(i): the photons travel in opposite directions.
Part (d)(i)
Δm = (17.999 160 + 1.007 825) − (18.000 938 + 1.008 665) = −0.002 618 u (the electron masses balance)
✓ 1
Compare (a)(ii), where they do not.
Energy that must be supplied = 0.002 618 × 931.5 = 2.44 MeV
✓ 1
The incoming proton has momentum, which must be conserved, so the products cannot be at rest: some of the proton's kinetic energy remains as kinetic energy of the products
✓ 1
Answers: (b)(i) 110 min · (b)(ii) 3.3 × 1012 · (b)(iii) 3.9 × 1011 · (c)(ii) 2.7 × 10−22 kg m s−1 · (c)(iii) 0.075 m · (d)(i) 2.44 MeV (the remaining parts are explanations — see the table above)
Syllabus understandingE.3 — the radioactive decay equations involving β+ and neutrinos; the mass–energy equivalence in nuclear reactions; E.3 (HL) — the decay law for arbitrary time intervals, A = λN, T½ = ln 2/λ; E.2 (HL) — photon momentum; A.2 — conservation of momentum; A.1 — motion at constant speed Command term: Determine
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