134 free, high-quality past-paper-style questions on Radioactive decay for IB Physics HL, each with a full mark scheme. Covers: Activity and decay law; Mass defect and binding energy; Beta decay and the neutrino; and more. For when you have finished the past papers, or want to practise one topic at a time.E.3 at Standard Level →Read the E.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.638 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1E-1A-07Activity and decay lawPaper 1AEasy1 mark
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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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.
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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.
3E-1A-40Beta decay and the neutrinoPaper 1AEasy1 mark
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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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.
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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.
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.
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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.
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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Step 1Each photon energy equals the difference between two nuclear energy levels.
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A nucleus of strontium-90 undergoes β− decay. Which particles are emitted by the nucleus?
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In β− decay a neutron becomes a proton, and an electron and an electron antineutrino are emitted. The antineutrino is required by lepton number conservation: the electron has lepton number +1, so the other lepton must have −1. A positron with a neutrino is β+ decay. Leaving out the antineutrino ignores the evidence from the continuous β energy spectrum.
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Answer A
Answer: A
Common error: β⁻ goes with an antineutrino and β⁺ with a neutrino. Check lepton number to decide which.
9IBP2H-039-00-02Absorption of radiationPaper 1AEasy1 mark
A factory rolls aluminium foil about 0.05 mm thick. A radioactive source sits on one side of the foil and a detector on the other. The count rate is used to control the thickness. Why is a β− source chosen rather than an α or a γ source?
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A useful gauge needs a count rate that changes measurably when the thickness changes. α particles would be almost completely stopped by even thin foil. γ rays would pass through almost unaffected, so small thickness changes would hardly alter the count. β particles are partially absorbed, which makes the count rate sensitive to thickness. α, not β, is the most ionising radiation, and β particles are charged.
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Answer A
Answer: A
Exam tip: For thickness gauges, choose the radiation whose range is comparable to the material being measured. Partial absorption is what gives sensitivity.
10IBP2H-039-01-01Binding energy per nucleonPaper 1AEasy1 mark
A uranium-235 nucleus absorbs a slow neutron and undergoes fission. It splits into two nuclei with mass numbers of about 90 and 140, plus a few neutrons. Which statement explains why energy is released?
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Nuclei with mass numbers near 90 and 140 lie closer to the peak of the binding energy per nucleon curve than uranium-235 does. Their nucleons are more tightly bound, so the products have a greater total binding energy and a smaller total mass. The mass difference is released as energy. The nucleon number is conserved, so A is wrong. C reverses the relationship. A slow neutron carries very little kinetic energy, so D is wrong.
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Answer B
Answer: B
Common error: A larger binding energy means a more stable nucleus and less mass. It does not mean more energy is 'stored' in the nucleus. Energy is released whenever the products are more tightly bound than the reactants.
11IBP2H-039-01-04Uses of radioisotopesPaper 1AEasy1 mark
A water company wants to locate a leak in a water main buried 1 m below a road. A radioisotope is added to the water, and a detector is moved along the road surface above the pipe. Which radioisotope is most suitable?
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The radiation must pass through about 1 m of soil and road surface, so a gamma emitter is needed. Alpha and beta radiation would be absorbed before reaching the detector. The half-life must be long enough for the tracer to travel along the pipe and be detected. It must also be short enough that the drinking water does not stay contaminated for long. A gamma emitter with a half-life of 15 hours meets both conditions. A 30-year half-life would leave the water supply radioactive for decades.
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Answer C
Answer: C
Exam tip: For any tracer question, justify two choices: the type of radiation (penetration to the detector) and the half-life (long enough to measure, short enough to limit exposure).
12IBP2H-039-01-07The random nature of decayPaper 1AEasy1 mark
Consider these statements about background radiation. I. Its value varies with geographical location and with altitude. II. It is mainly due to emissions from nuclear power stations. III. It should be subtracted from measured count rates when determining the half-life of a source. Which statements are correct?
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Statement I is correct. Background radiation depends on local rock types, which affect radon levels. It also depends on altitude, because cosmic rays are stronger higher up. Statement III is correct. The background does not decay, so it must be removed before the decay of the source can be analysed. Statement II is false. Nuclear power stations contribute only a tiny fraction of the background. Most of it comes from natural sources such as radon gas, rocks, cosmic rays and food, with medical procedures as the main artificial source.
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Answer B
Answer: B
Exam tip: Learn the main natural sources of background radiation: radon, rocks and soil, cosmic rays, and food and drink. Also learn that medical uses are the largest artificial source.
13IBP2H-039-02-01Decays on the N–Z chartPaper 1AEasy1 mark
A nuclide is plotted on a chart with neutron number N on the horizontal axis and proton number Z on the vertical axis. The nuclide undergoes β+ decay. How does the point representing the nucleus move on the chart?
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In β+ decay a proton changes into a neutron, so Z decreases by 1 (one unit down) and N increases by 1 (one unit to the right). The nucleon number is unchanged. Up-and-left is β− decay, two down and two left is α decay, and no movement is γ emission.
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Answer C
Answer: C
Exam tip: Always check which quantity is on which axis before describing a move on an N–Z chart. The same decay looks different if the axes are swapped.
14IBP2H-039-02-02The random nature of decayPaper 1AEasy1 mark
A radioactive nuclide has a decay constant λ. Which statement about λ is correct?
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The decay constant is the probability per unit time that a given nucleus decays. It is a fixed property of the nuclide, with unit s−1. The half-life is a time, equal to ln 2/λ. The fraction remaining after one half-life is always 0.5. λ does not change as the sample ages. Only the activity falls, because A = λN.
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Answer A
Answer: A
Common error: Do not confuse the decay constant with the activity. Activity falls with time; λ stays constant.
15IBP3H-E3-20r6-01A cobalt-60 radiotherapy sourcePaper 1AEasy1 mark
A hospital uses a cobalt-60 source (proton number 27) for radiotherapy. Cobalt-60 undergoes β− decay. Which is the daughter nucleus?
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In β− decay a neutron becomes a proton, so the nucleon number stays 60 and the proton number rises by one to 28: nickel-60. Manganese-56 is the result of α decay (Z down 2, A down 4). Iron-60 would follow β+ decay (Z down 1). Cobalt-59 has lost a nucleon, which does not happen in β decay.
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Answer D
Answer: D
Common error: In β decay the nucleon number never changes; only the proton number does, up by one for β⁻ and down by one for β⁺.
16IBP3H-E3-20r6-02Mass defect and binding energyPaper 1AEasy1 mark
The mass of a deuteron is less than the sum of the masses of a free proton and a free neutron. Which statement explains this?
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When separate nucleons bind, energy equal to the binding energy is released, and by E = Δmc2 the system has correspondingly less mass. No neutrino is emitted in forming a deuteron, mass-energy is conserved rather than destroyed, and no electrons are involved.
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Answer A
Answer: A
Exam tip: Link every mass defect to energy: the mass difference multiplied by c² is the energy released on binding, or the energy needed to separate the nucleons.
17IBP3H-E3-20r6-06Nuclear energy levelsPaper 1AEasy1 mark
Technetium-99m (proton number 43) is used in hospital imaging. The letter m means an excited state of the nucleus. It decays by emitting a single γ photon. Which describes the nucleus that remains?
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γ emission removes energy only: the nucleus keeps Z = 43 and A = 99 but drops to a lower energy level. Z = 44 would need β− decay, Z = 41 with A = 95 would need α decay, and Z = 42 would need β+ decay.
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Answer C
Answer: C
Exam tip: If an equation shows only a γ photon on the right, the daughter has exactly the same A and Z as the parent.
18IBP3H-E3-20r15-01Modelling decay with dicePaper 1AEasy1 mark
A class models radioactive decay using 1296 identical six-sided dice. All the dice are thrown together; any die showing a six is removed and the rest are thrown again. What is the expected number of dice remaining after three throws?
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Each throw removes, on average, one sixth of the dice still present, so the number remaining is 1296 × (5/6)3 = 750. 648 wrongly removes the same number (216) in every throw; 6 is 1296 × (1/6)3, the dice that would show a six three times in a row; 1080 is the number left after only one throw.
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Answer C
Answer: C
Exam tip: A fixed probability per throw means a fixed fraction removed, not a fixed number: the model decays exponentially.
19IBP3H-E3-20r15-02The random nature of decayPaper 1AEasy1 mark
Coffee beans contain a trace of potassium-40. Which statement about one particular potassium-40 nucleus in a bean is correct?
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Decay is random with a constant probability per unit time for each nucleus, independent of its age, so B is correct. A and D treat the half-life as a fixed lifetime for one nucleus; C is the 'due to decay' misconception, as survival does not change the probability.
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Answer B
Answer: B
Common error: The half-life describes a large number of nuclei; it says nothing about when one particular nucleus will decay.
20IBP3H-E3-20r15-03Background radiationPaper 1AEasy1 mark
A gamma source used to inspect welds on a bridge contains a single radioactive nuclide. Its activity falls from 640 MBq to 40 MBq in 2.0 hours. What is the half-life of the nuclide?
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640/40 = 16 = 24, so four half-lives pass in 120 min and the half-life is 120/4 = 30 min. 24 min counts the five listed activities (640, 320, 160, 80, 40) as five half-lives; 60 min uses two half-lives; 7.5 min divides 120 min by 16.
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Answer B
Answer: B
Exam tip: Count the number of halvings, not the number of values in the sequence.
21IBP3H-E3-20r18-01Absorption of radiationPaper 1AEasy1 mark
A technician in a marine biology laboratory measures the radiation from a small sample with a Geiger–Müller tube. Which statement about background radiation is correct?
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B is correct: the tube also detects natural radiation (cosmic rays, radon, rocks), so the background count rate is measured separately and subtracted. A is wrong because most background is natural. C is wrong because shielding reduces but cannot remove all background (for example high-energy cosmic rays). D is wrong because it varies with rock type, altitude and building materials.
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Answer B
Answer: B
Exam tip: Always measure the background count over the same time interval as the source measurement before correcting.
22IBP3H-E3-20r18-02Beta-plus decayPaper 1AEasy1 mark
Sodium-22 (2211Na) is a \(\beta^+\) emitter. Which row gives the daughter nuclide and the neutral particle emitted?
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A is correct: in \(\beta^+\) decay a proton becomes a neutron, so Z falls from 11 to 10 with A unchanged, and the positron (lepton number −1) is accompanied by a neutrino (+1). B has the correct nuclide but an antineutrino, which accompanies \(\beta^-\) decay. C and D raise Z, which is what happens in \(\beta^-\) decay.
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Answer A
Answer: A
Common error: The positron carries lepton number −1, so the neutral partner must be a neutrino rather than an antineutrino.
23IBP3H-E3-20r18-03Background radiationPaper 1AEasy1 mark
A logistics company delivers a medical radioisotope that has a half-life of 3.0 hours. The activity of the consignment was 640 Bq when it was calibrated. What is its activity 12 hours later?
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12 h is four half-lives, so the activity is 640 × (1/2)4 = 40 Bq, which is C. A divides by 12, treating the decay as linear. B halves only twice. D halves only three times.
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Answer C
Answer: C
Common error: Count half-lives by dividing the elapsed time by the half-life, then halve that many times.
24IBP3H-E3-20r18-07Positron emission tomographyPaper 1AEasy1 mark
In a hospital scanner, a positron and an electron, both essentially at rest, annihilate. The rest energy of an electron is 0.511 MeV. Which statement is correct?
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A is correct: the total energy is 2 × 0.511 MeV, and zero initial momentum requires two photons of equal energy moving in opposite directions. B would violate conservation of momentum. C double counts the available energy. D is wrong because energy and momentum are carried away by photons even though net charge is zero.
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Answer A
Answer: A
Exam tip: Momentum conservation is the reason annihilation at rest cannot produce a single photon.
27E-1A-19Absorption of radiationPaper 1AMedium1 mark
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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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.
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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.
29E-1A-45Binding energy per nucleonPaper 1AMedium1 mark
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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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).
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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).
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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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.
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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.
31E-1A-65The random nature of decayPaper 1AMedium1 mark
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?
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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.
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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).
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?
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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.
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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).
33IBP2H-039-00-03Beta decay and the neutrinoPaper 1AMedium1 mark
The β− particles emitted by one particular nuclide have a continuous range of kinetic energies, from almost zero up to a maximum value. What is the best explanation for this?
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Each decay releases the same total energy Q. If only two bodies were produced, conservation of momentum and energy would give the electron a single fixed kinetic energy. The continuous spectrum shows that a third, undetected particle (the antineutrino) carries away a variable share of Q. The maximum electron kinetic energy corresponds to the antineutrino receiving almost no energy. Nuclear energy levels produce discrete energies, not a continuum. β electrons are created in the nucleus, not taken from orbital shells.
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Answer A
Answer: A
Exam tip: Link the continuous β spectrum to conservation of energy and momentum. A two-body decay always gives a single, fixed energy.
34IBP2H-039-00-04Nuclear energy levelsPaper 1AMedium1 mark
A nucleus has only two excited states, at 0.30 MeV and 0.80 MeV above its ground state. The nucleus is formed in the 0.80 MeV state and de-excites by γ emission only. Which list gives all the possible γ-ray photon energies?
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Each photon energy equals the difference between two nuclear energy levels. The possible transitions are 0.80→0 (0.80 MeV), 0.80→0.30 (0.50 MeV) and 0.30→0 (0.30 MeV). Option B gives only the level values, not the differences. 1.10 MeV wrongly adds the two levels. 'Only 0.50 MeV' ignores the direct transition and the final step to the ground state.
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Answer A
Answer: A
Common error: Photon energies are differences between levels, never sums. Count every possible pair of levels.
35IBP2H-039-00-05Activity and decay lawPaper 1AMedium1 mark
The activity of a sample of a radioactive tracer falls to \(\frac{1}{8}\) of its initial value in 36 hours. What is the decay constant of the nuclide?
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A fall to 1/8 means three half-lives, so T½ = 36/3 = 12 h. Then λ = ln2/T½ = ln2/12 h−1. Option B omits ln2. Option C treats 36 h as the half-life. Option D inverts the relation.
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Answer A
Answer: A
Exam tip: Use 1/2, 1/4, 1/8 as one, two and three half-lives. Then apply λ = ln2/T½.
36IBP2H-039-00-07Background radiationPaper 1AMedium1 mark
A detector near a radioactive source records a total count rate of 150 counts per minute, which includes background. The background count rate is 30 counts per minute. What total count rate does the detector record after three half-lives of the source?
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Only the source's contribution decays. The corrected source rate is 150 − 30 = 120 min−1. After three half-lives this falls by a factor of 23 = 8, giving 120/8 = 15 min−1. The background does not change, so the detector records 15 + 30 = 45 min−1. Option B divides the uncorrected total by 8, treating the background as if it decays. Option C gives only the source's contribution and leaves out the background. Option D divides the uncorrected total by 8 and then adds the background, so the background is counted twice.
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Answer A
Answer: A
Exam tip: Subtract the background first, then apply the half-lives to the corrected rate. Add the background back if the question asks for what the detector records.
37IBP2S-038-02-08Uses of radioisotopesPaper 1AMedium1 mark
Technetium-99m, used in medical imaging, has a half-life of 6.0 hours. What is its decay constant?
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λ = ln 2 / T½ T½ = 6.0 × 3600 = 21600 s λ = 0.693 / 21600 = 3.2×10−5 s−1. The distractors come from three mistakes: leaving the half-life in hours gives A; omitting ln 2 gives C; converting hours to minutes only gives D.
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Answer B
Answer: B
Common error: Convert the half-life to seconds before calculating λ in s⁻¹.
38IBP2H-039-01-02Activity and decay lawPaper 1AMedium1 mark
Two radioactive samples, X and Y, contain the same number of undecayed nuclei at a particular instant. The half-life of X is 3.0 h and the half-life of Y is 6.0 h. What is the ratio \(\frac{\text{activity of X}}{\text{activity of Y}}\) at that instant?
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Activity is A = λN, and the decay constant is λ = ln 2 / T1/2. With equal N, the ratio of activities equals the ratio of decay constants. Here λX/λY = TY/TX = 6.0/3.0 = 2. The shorter-lived sample decays faster, so it has the greater activity.
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Answer C
Answer: C
Exam tip: Activity depends on both N and λ. Equal numbers of nuclei do not give equal activities unless the half-lives are also equal.
39IBP2H-039-01-03Background radiationPaper 1AMedium1 mark
The background count rate in a laboratory is 0.50 s−1. A source is placed near a detector, and the measured count rate is initially 2.50 s−1. What is the measured count rate after two half-lives of the source?
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The count rate due to the source alone is 2.50 − 0.50 = 2.00 s−1. After two half-lives this falls to 2.00/4 = 0.50 s−1. The background does not decay, so the detector still records it. The measured count rate is therefore 0.50 + 0.50 = 1.0 s−1. Each wrong option comes from a specific slip: • 0.63 s−1 divides the total by 4, as if the background decayed too. • 0.50 s−1 forgets to add the background back. • 1.3 s−1 uses only one half-life.
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Answer C
Answer: C
Common error: Only the source count rate halves each half-life. Subtract the background, apply the decay, then add the background back if the question asks for the measured reading.
40IBP2H-039-01-06Activity and decay lawPaper 1AMedium1 mark
A radioactive nuclide has decay constant λ. What fraction of the original undecayed nuclei remains after a time equal to 1/λ?
Show mark scheme
Marking point
Mark
Notes
The number of undecayed nuclei is N = N0e−λt. When t = 1/λ, N/N0 = e−1 ≈ 0.37. Each wrong option comes from a specific slip: • 0.50 would be correct after one half-life, which is ln 2/λ, not 1/λ. • 0.63 is the fraction that has decayed, not the fraction remaining. • 0.69 is ln 2.
✓ 1
Answer A
Answer: A
Exam tip: The time 1/λ is longer than the half-life, because ln 2 ≈ 0.69 is less than 1. So the fraction remaining must be less than one half.
41IBP2H-039-02-03Beta-plus decayPaper 1AMedium1 mark
In a hospital PET scanner, a tracer nucleus decays by emitting a positron. Which other particle must be emitted from the nucleus in this decay?
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Notes
β+ decay: p → n + e+ + νe. The positron is an antilepton (lepton number −1), so a neutrino (lepton number +1) is needed to conserve lepton number. The antineutrino accompanies β− decay. An electron would break charge conservation. The gamma photons detected in PET come from the later annihilation of the positron, not from the nucleus.
✓ 1
Answer D
Answer: D
Common error: Pairing the positron with an antineutrino is a frequent slip. Antiparticle out (e⁺) means ordinary neutrino out, to keep lepton number zero.
42IBP2H-039-02-04Mass defect and binding energyPaper 1AMedium1 mark
A nucleus contains Z protons and N neutrons and has nuclear mass M. The mass of a free proton is mp and of a free neutron is mn. What is the binding energy per nucleon of this nucleus?
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Notes
The mass defect is the mass of the separated nucleons minus the nuclear mass: Zmp + Nmn − M. This is positive. Multiplying by c2 gives the total binding energy. Dividing by the nucleon number Z + N gives the binding energy per nucleon. The other options divide by Z only, give a negative value, or give the total rather than the per-nucleon value.
✓ 1
Answer B
Answer: B
Exam tip: Binding energy is always positive: the separated nucleons have more mass than the bound nucleus.
43IBP2H-039-02-05Decays on the N–Z chartPaper 1AMedium1 mark
A nucleus undergoes one α decay followed by two β− decays. What is the final nucleus?
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Notes
α decay: Z − 2, N − 2. Each β− decay: Z + 1, N − 1. The net change is Z + 0 and N − 4. The element is the same, with four fewer neutrons, so it is an isotope with nucleon number reduced by 4.
✓ 1
Answer A
Answer: A
Common error: β⁻ decay does not leave N unchanged. It converts a neutron into a proton, so each β⁻ removes one neutron.
44IBP2H-039-02-08Activity and decay lawPaper 1AMedium1 mark
A radioactive source used to calibrate a satellite's radiation detector has decay constant λ. What fraction of the original nuclei has decayed after a time 1/λ?
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Mark
Notes
N = N0e−λt. At t = 1/λ, N/N0 = e−1 ≈ 0.37 remain, so the fraction decayed is 1 − 0.37 = 0.63. The value 0.37 is the fraction remaining. The value 0.50 applies after one half-life, which is ln 2/λ. The value 0.69 is ln 2.
✓ 1
Answer D
Answer: D
Exam tip: Read whether the question asks for the fraction remaining or the fraction decayed. These are complementary.
45IBP2S-038-01-03Activity and decay lawPaper 1AMedium1 mark
Samples of two different radioactive nuclides X and Y contain equal numbers of undecayed nuclei. The half-life of X is twice the half-life of Y. What is \(\frac{\text{activity of X}}{\text{activity of Y}}\)?
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Notes
Activity A = λN and λ = ln2 / T1/2. Doubling the half-life halves the decay constant. With N equal, AX/AY = λX/λY = 1/2. Choosing 1 assumes activity depends only on N. Choosing 2 assumes activity is proportional to half-life. Choosing 1/4 wrongly squares the ratio.
✓ 1
Answer B
Answer: B
Exam tip: The decay constant is inversely proportional to the half-life. A longer-lived nuclide decays more slowly, so it has a lower activity for the same number of nuclei.
46IBP3H-E3-20r6-03Mass defect and binding energyPaper 1AMedium1 mark
The nucleus of an atom of a certain isotope has a mass defect of 0.0301 u and contains 4 nucleons. Use 1 u = 931.5 MeV c−2. What is its binding energy per nucleon?
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Notes
Binding energy = 0.0301 × 931.5 = 28.0 MeV; dividing by 4 nucleons gives 7.01 MeV per nucleon. 28.0 MeV forgets to divide by the number of nucleons; 112 MeV multiplies by 4 instead of dividing; 7.5 × 10−3 divides the mass defect by 4 and never converts to MeV.
✓ 1
Answer C
Answer: C
Common error: Binding energy per nucleon needs two steps, mass defect to energy and then division by the nucleon number; check you have done both before choosing an option.
47IBP3H-E3-20r6-04Decay equationsPaper 1AMedium1 mark
A nucleus of uranium-238 (proton number 92) emits one α particle. The daughter then emits two β− particles in succession. What is the proton number of the final nucleus?
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Notes
The α decay lowers Z by 2 (to 90) and each β− raises it by 1, so Z = 92, an isotope of uranium. 90 forgets the β decays. 88 treats the β particles as lowering Z. 94 forgets the proton number lost in α decay.
✓ 1
Answer A
Answer: A
Exam tip: Track A and Z in two columns after each step; α changes (A, Z) by (−4, −2), β⁻ by (0, +1).
48IBP3H-E3-20r6-05Radioisotope power sourcesPaper 1AMedium1 mark
The thermal power of a radioisotope power source on a space probe is proportional to its activity. The decay constant of the fuel is 7.9 × 10−3 year−1. After what time does the thermal power fall to 1/e of its initial value?
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Notes
Power ∝ e−λt, which equals 1/e when λt = 1, so t = 1/λ = 1/(7.9 × 10−3) = 127 years. 88 years is the half-life ln 2/λ. 63 years uses 1/(2λ). 175 years is two half-lives, which corresponds to 1/4, not 1/e.
✓ 1
Answer B
Answer: B
Common error: Time to fall to 1/e is the mean life 1/λ, not the half-life ln 2/λ; they differ by a factor of about 1.44.
49IBP3H-E3-20r6-07Beta-plus decayPaper 1AMedium1 mark
Fluorine-18 (proton number 9) is a β+ emitter used in medical scanners. Which equation correctly represents its decay?
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Notes
A proton becomes a neutron, so Z falls from 9 to 8 with A fixed at 18. The positron has lepton number −1, so it must be accompanied by a neutrino (+1). Option B has an antineutrino, which breaks lepton conservation. Option C is the pattern for β− decay. Option D does not conserve nucleon number.
✓ 1
Answer A
Answer: A
Common error: β⁺ decay emits a neutrino and β⁻ decay emits an antineutrino; the pairing is fixed by lepton number conservation.
50IBP3H-E3-20r15-04Absorption of radiationPaper 1AMedium1 mark
A small gamma source behaves as a point source. A detector 0.80 m from it records a net count rate of 450 s−1. At what distance from the source will the net count rate be 112.5 s−1? Assume no absorption.
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Notes
Count rate ∝ 1/d2. The rate falls by a factor of 4, so the distance increases by a factor of √4 = 2, giving 1.6 m. 3.2 m treats the rate as inversely proportional to distance; 6.4 m squares the factor instead of taking the root; 0.40 m inverts the relationship.
✓ 1
Answer B
Answer: B
Common error: Inverse-square: when the intensity falls by a factor n, the distance grows by √n, not by n.
51IBP3H-E3-20r15-05Activity and decay lawPaper 1AMedium1 mark
A heat source on a space probe contains two radioisotopes. Isotope X has half-life 2.0 years and isotope Y has half-life 6.0 years. Initially the two isotopes have equal activities. What is the total activity after 6.0 years as a fraction of the initial total activity?
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Notes
After 6.0 years X has gone through 3 half-lives (1/8 of its activity) and Y through 1 (1/2). Each started with half of the total, so the fraction is (1/8 + 1/2)/2 = 0.3125. 0.125 counts X only; 0.500 counts Y only; 0.625 omits the division by 2 for the initial total.
✓ 1
Answer B
Answer: B
Exam tip: Activities add, but each nuclide is halved on its own half-life: treat them separately, then sum.
52IBP3H-E3-20r15-06Activity and decay lawPaper 1AMedium1 mark
A thickness sensor in a coffee-roasting plant contains a sealed source with half-life 8.0 days. How long does it take for the number of undecayed nuclei to fall to 20% of the initial number?
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Notes
N/N0 = 0.20 = e−λt with λ = ln 2/8.0, so t = 8.0 × ln 5/ln 2 = 18.6 days. 16 days is two half-lives (25%); 6.4 days is 20% of the half-life; 40 days is 5 × 8.0.
✓ 1
Answer C
Answer: C
Common error: Percentages remaining are not time fractions: 20% remaining is between two and three half-lives, not 0.2 half-lives.
54E-1A-24Decay constant and half-lifePaper 1AHard1 mark
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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Mark
Notes
Step 1ln A = ln A0 − λt: the intercept gives A0 = ec Bq.
—
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.
—
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.
56IBP2H-039-00-06Activity and decay lawPaper 1AHard1 mark
Samples X and Y start with equal numbers of radioactive nuclei. The half-life of X is T and the half-life of Y is 2T. What is the ratio \(\frac{\text{activity of X}}{\text{activity of Y}}\) at time 2T?
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Notes
Activity A = λN. At time 2T, X has passed two half-lives, so NX = N0/4. Y has passed one half-life, so NY = N0/2. The decay constants are in the ratio λX = 2λY. So AX/AY = 2 × (1/4)/(1/2) = 1. Option B compares the numbers of nuclei only and ignores λ. Option C compares the decay constants only. Option D uses X's remaining fraction alone.
✓ 1
Answer A
Answer: A
Common error: Activity depends on both λ and N. A shorter half-life gives a larger λ even when fewer nuclei remain.
57IBP2H-039-00-08Beta-plus decayPaper 1AHard1 mark
Fluorine-18 decays by β+ emission. Which row correctly describes the change in quark content and the leptons emitted?
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Notes
In β+ decay a proton (uud) becomes a neutron (udd), so one up quark changes into a down quark. Charge is conserved: +1 = 0 + (+1) from the positron. The positron has lepton number −1, so an electron neutrino (lepton number +1) must also be emitted to keep the total lepton number at zero. A down-to-up change is β− decay. An antineutrino with a positron would violate lepton number conservation.
✓ 1
Answer A
Answer: A
Common error: Do not carry the β⁻ pattern across to β⁺. Check charge, baryon number and lepton number separately.
58IBP2S-038-04-07Activity and decay lawPaper 1AHard1 mark
Samples P and Q each contain the same number of undecayed nuclei. The half-life of the isotope in P is twice the half-life of the isotope in Q. What is the ratio \(\frac{\text{activity of P}}{\text{activity of Q}}\) at this instant?
Show mark scheme
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Mark
Notes
Activity is A = λN with λ = ln2 / T1/2. Doubling the half-life halves λ. With N equal, the activity of P is half the activity of Q, so the ratio is 1/2. Option D comes from thinking a longer half-life means more decays. Option C ignores λ. Option A squares the factor.
✓ 1
Answer B
Answer: B
Common error: A longer half-life means a smaller decay constant, so fewer decays per second from the same number of nuclei. Long-lived isotopes are less active per nucleus.
59IBP2H-039-01-05Binding energy per nucleonPaper 1AHard1 mark
The binding energy per nucleon of deuterium \(^{2}_{1}\text{H}\) is 1.11 MeV. The binding energy per nucleon of helium-4 \(^{4}_{2}\text{He}\) is 7.07 MeV. What is the energy released when two deuterium nuclei fuse to form one helium-4 nucleus?
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Notes
Energy released = total binding energy of the products − total binding energy of the reactants. Total binding energy of helium-4 = 4 × 7.07 = 28.28 MeV. Each deuteron has 2 nucleons, so its binding energy is 2 × 1.11 = 2.22 MeV. Two deuterons have 4.44 MeV in total. Energy released = 28.28 − 4.44 = 23.8 MeV. Each wrong option comes from a specific slip: • 5.96 MeV is only the difference per nucleon. • 26.1 MeV forgets that each deuteron contains two nucleons. • 28.3 MeV ignores the binding energy of the deuterons.
✓ 1
Answer B
Answer: B
Common error: Always multiply binding energy per nucleon by the nucleon number before finding a difference. Comparing the per-nucleon values directly gives the wrong energy.
60IBP2H-039-01-08Activity and decay lawPaper 1AHard1 mark
Equal masses of two pure radionuclides, P and Q, are prepared at the same time. The molar mass of P is 2M and its half-life is T. The molar mass of Q is M and its half-life is 4T. What is the ratio \(\frac{\text{initial activity of P}}{\text{initial activity of Q}}\)?
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Notes
For equal masses, the number of nuclei is inversely proportional to molar mass, so NP/NQ = M/2M = 1/2. The decay constant is inversely proportional to half-life, so λP/λQ = 4T/T = 4. Activity is A = λN, so AP/AQ = (1/2) × 4 = 2. Each wrong option comes from a specific slip: • 8 multiplies by the molar-mass ratio instead of dividing. • 1/2 ignores the half-lives. • 1/8 assumes a longer half-life gives a greater activity.
✓ 1
Answer C
Answer: C
Common error: Doubling the molar mass halves the number of nuclei in a given mass, because each nucleus is heavier. Check this direction before combining it with the λ ratio.
61IBP2H-039-02-06Decay constant as a probabilityPaper 1AHard1 mark
A particular nucleus of a radioactive nuclide is observed to remain undecayed for a time equal to three half-lives. What is the probability that this nucleus decays during the next half-life?
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Notes
The decay constant is a constant probability of decay per unit time. It does not depend on how long the nucleus has already existed. For any nucleus that is still undecayed, the probability of decaying within the next half-life is 1 − 2−1 = 1/2. 1/8 is the probability of surviving three half-lives from the start. 1/16 is the probability of surviving four half-lives from the start. 15/16 is the probability of decaying within four half-lives of the start. None of these applies to a nucleus already known to have survived.
✓ 1
Answer C
Answer: C
Exam tip: Radioactive nuclei have no memory. For a nucleus that has survived, decay probabilities restart from that moment.
62IBP2H-039-02-07Beta decay and the antineutrinoPaper 1AHard1 mark
A nuclide decays by β+ emission. The atomic mass of the parent atom exceeds the atomic mass of the daughter atom by 0.00210 u. The electron mass is 0.00055 u, and 1 u is equivalent to 931.5 MeV. What is the approximate energy released in the decay?
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Notes
With atomic masses, two electron masses must be subtracted. One accounts for the extra orbital electron in the parent atom. The other is the mass of the emitted positron. Δm = 0.00210 − 2(0.00055) = 0.00100 u, so the energy released ≈ 0.00100 × 931.5 ≈ 0.93 MeV. Using 0.00210 u directly gives 2.0 MeV. Subtracting one electron mass gives 1.4 MeV. Adding two gives 3.0 MeV.
✓ 1
Answer B
Answer: B
Common error: For β⁺ decay with atomic masses, subtract 2me. For β⁻ decay with atomic masses, the electron masses cancel.
63IBP2S-038-01-07Activity and decay lawPaper 1AHard1 mark
A sample contains N undecayed nuclei of a nuclide and has activity A. Which expression gives the half-life of the nuclide?
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Notes
A = λN gives λ = A/N. Then T1/2 = ln2/λ = (N ln2)/A. The first option is ln2 multiplied by λ. The second is the reciprocal of the correct expression. The third places ln2 in the denominator instead of the numerator.
✓ 1
Answer D
Answer: D
Exam tip: Check your expression with a quick sense test: more nuclei for the same activity means slower decay, so N must be in the numerator of the half-life.
64IBP3H-E3-20r6-08Binding energy of the deuteronPaper 1AHard1 mark
The binding energy of the deuteron is 2.22 MeV. A deuteron at rest absorbs a γ photon of energy 3.00 MeV and splits into a proton and a neutron. The momentum of the photon may be ignored and the proton and neutron have almost equal mass. What is the kinetic energy of the neutron immediately after the split?
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Notes
Energy left for kinetic energy = 3.00 − 2.22 = 0.78 MeV. With the photon momentum ignored the two particles have equal and opposite momenta, and since their masses are almost equal each takes half: 0.39 MeV. 0.78 MeV is the total, 2.22 MeV is the energy that went into separating the nucleons, and 3.00 MeV is the photon's energy.
✓ 1
Answer D
Answer: D
Exam tip: Subtract the binding energy first, then share what is left using momentum conservation; equal masses share the kinetic energy equally.
65IBP3H-E3-20r15-08Background radiationPaper 1AHard1 mark
A detector placed 0.50 m from a point gamma source records 2400 counts per minute, including a background of 60 counts per minute. What count rate, in counts per minute, will the same detector record at 1.50 m from the source?
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Notes
The net rate at 0.50 m is 2340. At three times the distance the net rate is 2340/9 = 260, and the background of 60 is added back to give the recorded rate of 320. 260 omits the background; 327 applies the inverse-square law to the total including background; 800 divides by 3 instead of 9.
✓ 1
Answer B
Answer: B
Exam tip: Subtract the background before applying the inverse-square law, then add it back if the recorded rate is asked for.
66IBP3H-E3-20r18-08Activity and decay lawPaper 1AHard1 mark
Samples X and Y initially contain the same number of nuclei. X has a half-life of 2.0 h and Y has a half-life of 6.0 h. What is the ratio of the activity of X to the activity of Y after 12 h?
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Notes
Remaining nuclei: X has \((1/2)^6=1/64\) and Y has \((1/2)^2=1/4\), a ratio 1/16. But \(A=\lambda N\) and \(\lambda_X/\lambda_Y=3\), so the activity ratio is \(3\times\frac1{16}=\frac{3}{16}\) (D). A forgets that activity also depends on \(\lambda\). B inverts the answer. C is the inverse ratio of the half-lives.
✓ 1
Answer D
Answer: D
Common error: Activity depends on both the number of nuclei and the decay constant, \(A=\lambda N\); equal numbers of nuclei do not mean equal activities.
67E-1B-15Absorption of beta radiationPaper 1BEasy6 marks
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)
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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)
68E-1B-21Modelling decay with dicePaper 1BEasy7 marks
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
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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)
69E-1B-01Half-life with background correctionPaper 1BMedium6 marks
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
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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)
70E-1B-08Range of alpha particles in airPaper 1BMedium7 marks
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)
71E-1B-10Inverse-square law for gamma radiationPaper 1BMedium6 marks
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].
72E-1B-16Determining a long half-lifePaper 1BHard7 marks
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
73E-1B-23Two nuclides decaying togetherPaper 1BHard7 marks
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.
74E-2-22Identifying radiations and backgroundPaper 2Easy8 marks
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
Diagram not to scale
(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)
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)
76E-2-33A cobalt-60 radiotherapy sourcePaper 2Easy13 marks
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)
Energy levels of nickel-60, not to scale
(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)
77IBP2H-039-00-09Positron emission tomographyPaper 2Easy5 marks
A hospital prepares a dose of fluorine-18 for a positron emission tomography (PET) scan. Fluorine-18 has a half-life of 110 minutes. The dose has an activity of 400 MBq when prepared at 08:00. It is injected into a patient at 11:40.
(a)
State what is meant by the half-life of a radioactive nuclide.
(1)
(b)
Calculate the activity of the dose at the time of injection.
(2)
(c)
Outline why a positron emitter is suitable for imaging inside the body, whereas an α emitter would not be.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The time taken for the number of undecayed nuclei (or the activity) of the sample to fall to half its initial value
✓ 1
Part (b)
Time elapsed = 220 min = 2 half-lives
✓ 1
A = 400/22 = 100 MBq
✓ 1
Part (c)
The positrons annihilate with electrons to produce γ photons, which are weakly absorbed and so escape the body to reach external detectors
✓ 1
α particles have a very short range in tissue, so they would be absorbed inside the body, would not be detected outside, and would cause intense local ionisation damage
✓ 1
Exam tip: Convert clock times to an elapsed time first. A whole number of half-lives lets you halve repeatedly without logarithms.
78IBP2H-039-02-09Mass defect and binding energyPaper 2Easy5 marks
Most of the lithium in rechargeable phone batteries is lithium-7, 73Li. Nuclear mass of lithium-7 = 7.01436 u Mass of proton = 1.007276 u Mass of neutron = 1.008665 u 1 u is equivalent to 931.5 MeV.
(a)
State what is meant by the mass defect of a nucleus.
(1)
(b)
Calculate the mass defect of a lithium-7 nucleus.
(2)
(c)
Determine the binding energy per nucleon of lithium-7, in MeV.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The total mass of the separated (individual) nucleons minus the mass of the nucleus
✓ 1
Part (b)
Total nucleon mass = 3(1.007276) + 4(1.008665) = 7.056488 u
79IBP3H-E3-20r6-09Beta decay and the neutrinoPaper 2Easy8 marks
Phosphorus-32 has a half-life of 14.3 days. It is used as a radioactive tracer in a study of how worker bees pass sugar syrup to other bees in a hive. Phosphorus-32 has proton number 15 and decays by β− emission to sulfur (proton number 16).
(a)
Write the nuclear equation for the decay of phosphorus-32.
(3)
(b)
Explain why the β− particles emitted from a sample of phosphorus-32 have a continuous range of kinetic energies.
(2)
(c)
A sample of syrup has an initial P-32 activity A0. Calculate the fraction of A0 that remains after 28 days. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
sulfur nuclide written as 3216S with correct A and Z
✓ 1
β− particle (electron) emitted
✓ 1
antineutrino emitted
✓ 1
Part (b)
the energy released in each decay is shared between the β− particle and the antineutrino
✓ 1
the share varies from decay to decay, so β− energies range from nearly zero up to the maximum
✓ 1
Part (c)
number of half-lives = 28/14.3 = 1.96
✓ 1
fraction = (1/2)1.96
✓ 1
0.257
✓ 1
Exam tip: For a decay-equation mark, check that nucleon numbers and proton numbers add up to the same totals on both sides before you move on.
82IBP3H-E3-20r15-10Inverse-square law for gamma radiationPaper 2Easy6 marks
A Geiger counter next to a sealed source used for testing welds on a bridge records the number of counts in six successive 10 s intervals: 41, 52, 47, 38, 55, 43.
(a)
Calculate the mean count rate in s−1.
(2)
(b)
Explain why the counts differ from one interval to the next although the source does not change.
(2)
(c)
Suggest one change to the procedure that would reduce the percentage variation in the measured count rate, and explain why it works.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total counts 276 divided by 60 s
✓ 1
A1 4.60 s−1
✓ 1
Part (b)
R1 radioactive decay is a random process
✓ 1
R1 each nucleus has a constant probability of decaying per unit time, so the number decaying in a fixed interval varies about an average
✓ 1
Part (c)
A1 count over longer intervals (or add up the counts)
✓ 1
R1 a larger total number of counts means the random fluctuation is a smaller fraction of the total
✓ 1
Common error: Do not say the counts vary because of faults in the counter; the variation is a property of decay itself.
83IBP3H-E3-20r18-09Background radiationPaper 2Easy6 marks
A Geiger–Müller counter is used to test a rock sample in an architecture materials laboratory.
(a)
State two natural sources of background radiation.
(2)
(b)
With the sample present, the counter records 315 counts in 2.0 minutes. Without the sample it records 45 counts in 2.0 minutes. Calculate the corrected count rate due to the sample, in counts per minute.
(3)
(c)
Suggest why the background count should be measured several times.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
any valid source, e.g. cosmic rays, radon gas from rocks, potassium-40 in food or rocks
✓ 1
second valid source (different from the first)
✓ 1
Part (b)
subtract the background: 315 − 45 = 270 counts
✓ 1
divide by the time: 270/2.0
✓ 1
135 counts per minute
✓ 1
Part (c)
radioactive decay is random, so repeating and averaging reduces the effect of random fluctuations
✓ 1
Common error: Subtract the background count from the total count before dividing by time, using the same time interval for both.
84E-2-05Binding energy per nucleonPaper 2Medium11 marks
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)
85E-2-12Beta decay and the antineutrinoPaper 2Medium12 marks
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
86IBP2H-039-00-11Half-life with background correctionPaper 2Medium7 marks
In an engine wear test, piston rings are labelled with a short-lived radioactive isotope. Metal worn off the rings collects in the lubricating oil. A detector is placed next to an oil sample removed from the engine. The table shows the background-corrected count rate C of the oil sample at different times t.
t / min: 0, 10, 20, 30, 40 C / s−1: 800, 606, 459, 348, 264
Data points from the table
(a)
Show that the data are consistent with exponential decay.
(2)
(b)
Determine the decay constant of the isotope.
(2)
(c)
Calculate the half-life of the isotope.
(1)
(d)
Determine the time t at which the corrected count rate will be 100 s−1.
(2)
Show mark scheme
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Notes
Part (a)
Calculates ratios of successive values at equal time intervals, e.g. 606/800 = 0.758, 459/606 = 0.757, 348/459 = 0.758, 264/348 = 0.759
✓ 1
The ratio is constant (≈0.758) for equal time intervals, which is characteristic of exponential decay
✓ 1
Part (b)
Uses C = C0e−λt, e.g. λ = ln(800/264)/40
✓ 1
λ = 0.0277 min−1 (≈4.6×10−4 s−1)
✓ 1
Part (c)
T½ = ln2/λ = 0.693/0.0277 = 25 min ECF from (b)
✓ 1
Part (d)
100/800 = 1/8, which is three half-lives
✓ 1
t = 3 × 25 = 75 min ECF from (c)
✓ 1
Exam tip: To show a decay is exponential, use the constant-ratio test over equal time intervals. A falling trend alone is not enough.
87IBP2H-039-00-12Beta decay and the antineutrinoPaper 2Medium7 marks
Self-luminous exit signs contain sealed glass tubes filled with tritium gas (hydrogen-3). The inside of each tube is coated with a phosphor. Tritium decays by β− emission to helium-3. Atomic mass of hydrogen-3 = 3.016049 u; atomic mass of helium-3 = 3.016029 u; 1 u = 931.5 MeV c−2.
(a)
Write the nuclear equation for this decay, including nucleon and proton numbers.
(2)
(b)
Explain why the β particles emitted have a range of kinetic energies up to a maximum value.
(2)
(c)
Calculate, in keV, the maximum kinetic energy of the emitted β particles.
(2)
(d)
Outline why the sign does not present an external radiation hazard.
(1)
Show mark scheme
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Notes
Part (a)
31H → 32He + 0−1e (or 0−1β), with correct nucleon and proton numbers
✓ 1
+ electron antineutrino, ν̄e
✓ 1
Part (b)
The energy released in each decay is shared between the electron and the antineutrino; the recoil of the nucleus is negligible
✓ 1
The share varies from one decay to the next. The electron has its maximum energy when the antineutrino receives almost no energy
The β particles have very low energy (at most about 19 keV), so they are absorbed within the tube (mainly by the glass, and also by the phosphor and gas) and cannot escape
✓ 1
Common error: When using atomic masses in β⁻ decay, do not subtract an extra electron mass. The daughter atom's extra orbital electron already accounts for it.
88IBP2H-039-00-13Nuclear energy levelsPaper 2Medium7 marks
Cobalt-57 decays by electron capture to an excited state of iron-57. This state is 136.5 keV above the ground state. Iron-57 has one other excited state, at 14.4 keV above the ground state. The iron nucleus returns to its ground state by γ emission only. Such sources are used to calibrate γ-ray detectors in medical gamma cameras. h = 6.63×10−34 J s; c = 3.00×108 m s−1; 1 eV = 1.60×10−19 J.
Energy levels of iron-57, not to scale
(a)
Outline how the γ-ray spectrum from such a source provides evidence for nuclear energy levels.
(2)
(b)
Determine all the γ-ray photon energies that can be emitted by the iron-57 nucleus.
(2)
(c)
State the change, if any, in the nucleon number and the proton number of iron-57 when it emits a γ photon.
(1)
(d)
Calculate the wavelength of the lowest-energy γ photon.
(2)
Show mark scheme
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Notes
Part (a)
The γ photons are emitted only at certain discrete energies (a line spectrum)
✓ 1
Each photon energy equals the difference between two definite energy states of the nucleus
✓ 1
Part (b)
136.5 keV and 14.4 keV
✓ 1
122.1 keV (from 136.5 − 14.4)
✓ 1
Part (c)
No change: both nucleon number and proton number stay the same; the nucleus only loses energy
✓ 1
Part (d)
E = 14.4×103 × 1.60×10−19 = 2.30×10−15 J ECF from (b)
✓ 1
λ = hc/E = (6.63×10−34 × 3.00×108)/2.30×10−15 = 8.6×10−11 m
✓ 1
Exam tip: Convert keV to joules before using λ = hc/E. Forgetting the factor 10³ makes the answer 1000 times too large.
89IBP2H-039-00-15Absorption of radiationPaper 2Medium8 marks
A technician finds an unlabelled sealed source in a hospital store. She measures the count rate with different absorbers placed between the source and a detector. The source–detector distance is fixed. The background count rate is 25 min−1. The measured (uncorrected) count rates are:
No absorber: 840 min−1 One sheet of paper: 838 min−1 5 mm aluminium: 170 min−1 5 mm aluminium + 30 mm lead: 41 min−1
(a)
Deduce that the source does not emit α particles.
(2)
(b)
Deduce that the source emits both β and γ radiation.
(2)
(c)
Estimate the fraction of the source's count rate (with no absorber) that is due to β particles. Assume the paper and aluminium absorb a negligible fraction of the γ radiation.
(3)
(d)
The source is later identified as caesium-137. Its β− decay produces barium-137, which emits the γ radiation. Outline why the γ emission follows the β decay.
(1)
Show mark scheme
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Mark
Notes
Part (a)
The paper reduces the count rate only from 840 to 838 min−1, which is within random fluctuation
✓ 1
α particles would be stopped by paper, so a significant drop would be expected if they were present
✓ 1
Part (b)
5 mm of aluminium causes a large fall in count rate, consistent with β particles being absorbed by a few mm of aluminium
✓ 1
The count rate after the aluminium is still well above background and is reduced further by lead, so γ radiation is present
✓ 1
Part (c)
Background-corrected rates: 815 min−1 with no absorber and 145 min−1 after the aluminium
✓ 1
β contribution = 815 − 145 = 670 min−1
✓ 1
Fraction = 670/815 = 0.82
✓ 1
Part (d)
The β decay leaves the barium-137 nucleus in an excited energy state, and it emits a γ photon as it falls to its ground state
✓ 1
Exam tip: Correct every reading for background before comparing or subtracting count rates. The background does not depend on the absorber.
90IBP2S-038-04-16The random nature of decayPaper 2Medium6 marks
A patient with an overactive thyroid is treated with iodine-131, which has a half-life of 8.0 days. Immediately after treatment, the activity of the iodine-131 in the patient is 3.7 GBq. Ignore any removal of iodine from the body by biological processes.
(a)
State what is meant by the statement that radioactive decay is spontaneous.
(1)
(b)
Calculate the decay constant of iodine-131 in day−1.
(1)
(c)
Show that the probability that a particular iodine-131 nucleus decays within the next 24 hours is about 0.08.
(2)
(d)
The patient may go home when the activity has fallen to 0.40 GBq. Calculate the time after treatment at which this occurs.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Decay is not triggered or affected by external conditions such as temperature, pressure or chemical state .
✓ 1
Part (b)
λ = ln2 / 8.0 = 0.087 day−1 .
✓ 1
Part (c)
Probability of surviving 1 day = e−λt = e−0.0866 = 0.917 (or (1/2)1/8) . Probability of decaying = 1 − 0.917 = 0.083 ≈ 0.08 . ECF from (b).
✓ 2
Part (d)
0.40 = 3.7 e−λt, so t = ln(3.7/0.40) / λ . t = 2.22 / 0.0866 = 26 days . ECF from (b).
✓ 2
Exam tip: λ multiplied by time gives the decay probability only when that product is small. For an exact value, use 1 − e⁻λt.
91IBP2S-038-02-14Positron emission tomographyPaper 2Medium6 marks
Fluorine-18 is used as a tracer in PET scans and has a half-life of 110 minutes. A dose is prepared at 07:00 with an activity of 740 MBq and is injected into a patient at 09:20.
(a)
Calculate the decay constant of fluorine-18 in s−1.
(1)
(b)
Determine the activity of the dose at the moment of injection.
(2)
(c)
Calculate the number of undecayed fluorine-18 nuclei in the dose at the moment of injection.
(2)
(d)
Suggest one reason why a short half-life is an advantage for a medical tracer.
(1)
Show mark scheme
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Mark
Notes
Part (a)
λ = ln 2 / (110 × 60 s) = 1.05×10−4 s−1.
✓ 1
Part (b)
Time elapsed is 140 min, so A = 740 × exp(−ln 2 × 140/110), or equivalently A = 740 × (1/2)140/110.
✓ 1
A ≈ 306 MBq (about 3.1×108 Bq).
✓ 1
Part (c)
N = A/λ = 3.06×108 Bq / 1.05×10−4 s−1, with activity converted to Bq.
✓ 1
N ≈ 2.9×1012. Allow error carried forward from the decay constant and activity found in the earlier parts.
✓ 1
Part (d)
Either of the following earns the mark :
✓ 1
• The activity falls quickly after the scan, so the patient receives a smaller total radiation dose.
—
• A high enough activity for imaging can be obtained from only a small number of nuclei, so only a small quantity of material is needed.
—
Common error: In N = A/λ, the activity must be in Bq (s⁻¹) and λ must be in s⁻¹. Leaving the activity in MBq gives an answer that is wrong by a factor of 10⁶.
92IBP2H-039-01-10Positron emission tomographyPaper 2Medium8 marks
Technetium-99m is used in hospital imaging. It emits gamma rays and has a half-life of 6.0 h. A dose is prepared in the hospital pharmacy at 09:00. The dose must have an activity of 400 MBq when it is injected into a patient at 13:30.
(a)
Outline two reasons why technetium-99m is suitable for medical imaging.
(2)
(b)
Calculate the activity that the dose must have at 09:00.
(2)
(c)
Determine the number of technetium-99m nuclei in the dose at the moment of injection.
(2)
(d)
Ignoring any removal of the substance from the body, calculate the time after injection at which the activity falls to 5.0% of its value at injection.
(2)
Show mark scheme
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Mark
Notes
Part (a)
It emits gamma rays, which penetrate the body and can be detected by a camera outside the patient
✓ 1
Its half-life is long enough to allow preparation and scanning, but short enough to keep the patient's total radiation dose low
✓ 1
Part (b)
The elapsed time is 4.5 h, so A0 = 400 × 24.5/6.0
✓ 1
A0 = 670 MBq
✓ 1
Part (c)
λ = ln 2 / (6.0 × 3600) = 3.21×10−5 s−1
✓ 1
N = A/λ = 4.00×108 / 3.21×10−5 = 1.2×1013
✓ 1
Part (d)
Set 0.050 = 2−t/6.0, giving t = 6.0 × ln 20 / ln 2
✓ 1
t = 26 h
✓ 1
Common error: When working backwards in time, the earlier activity must be larger. If your activity at 09:00 is less than 400 MBq, the sign of the exponent is wrong.
93IBP2H-039-01-12Mass defect and binding energyPaper 2Medium7 marks
The following data refer to the oxygen-16 nucleus, \(^{16}_{8}\text{O}\). mass of proton = 1.007276 u mass of neutron = 1.008665 u mass of oxygen-16 nucleus = 15.990523 u 1 u = 931.5 MeV c−2
(a)
Define the binding energy of a nucleus.
(1)
(b)
Calculate the mass defect of the oxygen-16 nucleus.
(2)
(c)
Determine the binding energy per nucleon of oxygen-16, in MeV.
(2)
(d)
The binding energy per nucleon of iron-56 is 8.79 MeV. Explain why fusion reactions involving oxygen nuclei in massive stars release energy, but fusion reactions that produce nuclei heavier than iron do not.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The energy required to separate a nucleus completely into its individual protons and neutrons (or the energy released when the nucleus forms from them)
✓ 1
Part (b)
Total mass of separate nucleons = 8(1.007276) + 8(1.008665) = 16.127528 u
✓ 1
Mass defect Δm = 16.127528 − 15.990523 = 0.137005 u
✓ 1
Part (c)
Binding energy = 0.137005 × 931.5 = 127.6 MeV
✓ 1
Binding energy per nucleon = 127.6 / 16 = 7.98 MeV
✓ 1
Allow error carried forward from the mass defect calculated in the previous part.
—
Part (d)
Fusing oxygen produces nuclei closer to iron, which have a greater binding energy per nucleon. Total binding energy increases, so mass decreases and energy is released
✓ 1
Beyond iron, binding energy per nucleon decreases as nucleon number increases. Fusion would therefore lower the binding energy per nucleon, so it would require an energy input
✓ 1
Common error: Keep at least six significant figures until the subtraction is done. The mass defect is a small difference between two nearly equal numbers, so early rounding destroys it.
94IBP2H-039-01-14Decay constant and half-lifePaper 2Medium7 marks
Radon-222 seeps into buildings from the ground. In a basement room measuring 4.0 m × 5.0 m × 2.5 m, the activity of radon in the air is 300 Bq per cubic metre. The half-life of radon-222 is 3.8 days.
(a)
State what is meant by background radiation, and identify one artificial source of background radiation.
(2)
(b)
Calculate the total activity of radon in the air of the room.
(1)
(c)
Determine the number of radon-222 nuclei in the room.
(2)
(d)
A sealed floor membrane is installed, which stops any further radon entering the room. Assuming radon is lost from the room only by radioactive decay, calculate the time taken for the activity to fall to 100 Bq per cubic metre.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Background radiation is the ionising radiation that is always present in the environment, from natural and artificial sources other than the one being studied
✓ 1
Any one artificial source, for example medical X-rays or scans, fallout from past weapons tests, or nuclear industry discharges
✓ 1
Part (b)
Volume = 4.0 × 5.0 × 2.5 = 50 m3, so A = 300 × 50 = 1.5×104 Bq
✓ 1
Part (c)
λ = ln 2 / (3.8 × 86400 s) = 2.1×10−6 s−1
✓ 1
N = A/λ = 1.5×104 / 2.1×10−6 = 7.1×109
✓ 1
Allow error carried forward from the total activity found in the previous part.
—
Part (d)
Set 100/300 = 2−t/3.8, or equivalently 1/3 = e−λt
✓ 1
t = 3.8 × ln 3 / ln 2 = 6.0 days
✓ 1
Exam tip: When the fraction remaining is not a power of one half, use logarithms: t = T₁/₂ × ln(A₀/A) / ln 2.
95IBP2H-039-01-15Absorption of radiationPaper 2Medium7 marks
In a factory that rolls aluminium foil, the foil passes between a strontium-90 source and a detector. Strontium-90 is a beta emitter with a half-life of 28.8 years. The detector count rate is used to adjust the pressure on the rollers automatically.
(a)
Explain why a beta source is used rather than an alpha or a gamma source.
(2)
(b)
Calculate the percentage decrease in the activity of the source over one year.
(2)
(c)
The count rate rises above its set value. Outline what this indicates, and how the rollers should respond.
(2)
(d)
Suggest why a beta source with a half-life of a few days would be unsuitable.
(1)
Show mark scheme
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Mark
Notes
Part (a)
Alpha particles would be completely absorbed by the foil, so the count rate would not change with thickness
✓ 1
Gamma rays would pass through almost unaffected, so small changes in thickness would produce no measurable change in count rate. Beta particles are partially absorbed, so the count rate depends on thickness
✓ 1
Part (b)
A/A0 = 2−1/28.8 = 0.976
✓ 1
Percentage decrease = 2.4%
✓ 1
Part (c)
The foil is too thin, because fewer beta particles are being absorbed
✓ 1
The roller pressure should be reduced, or the gap between the rollers increased
✓ 1
Part (d)
The activity would fall significantly during production. The falling count rate would be wrongly interpreted as the foil becoming thicker, unless the system were recalibrated frequently
✓ 1
Common error: A higher count rate means less absorption, so the material is thinner, not thicker. Think about what the detector actually receives.
96IBP2H-039-02-10Positron emission tomographyPaper 2Medium8 marks
Fluorine-18, 189F, is used in PET scans. It decays by β+ emission to oxygen-18, with a half-life of 110 minutes. A patient is injected with a dose of fluorine-18 with an activity of 370 MBq at 08:00. The scan begins at 10:45.
(a)
Write the nuclear equation for the decay of fluorine-18.
(2)
(b)
On a chart of proton number Z (vertical) against neutron number N (horizontal), outline the positions of the parent and daughter nuclei and the move between them.
(2)
(c)
Calculate the decay constant of fluorine-18, in s−1.
(1)
(d)
Determine the number of undecayed fluorine-18 nuclei in the patient when the scan begins. Assume none is excreted.
(3)
Show mark scheme
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Mark
Notes
Part (a)
189F → 188O + 0+1e, with all nucleon and proton numbers correct
✓ 1
+ νe (electron neutrino) included
✓ 1
Part (b)
Parent at N = 9, Z = 9; daughter at N = 10, Z = 8
✓ 1
The move is one unit right and one unit down, a diagonal step with A constant
✓ 1
Part (c)
λ = ln 2 / (110 × 60) = 1.05 × 10−4 s−1
✓ 1
Part (d)
Elapsed time = 165 min = 1.5 half-lives, so A = 370 × 2−1.5 = 131 MBq
✓ 1
Use of N = A/λ
✓ 1
N = 1.31 × 108 / 1.05 × 10−4 = 1.25 × 1012
✓ 1
ECF from (c)
—
Exam tip: When using A = λN, put the activity in Bq and λ in s⁻¹. Mixing minutes and seconds is a quick way to lose the final mark.
97IBP2H-039-02-14Radioisotope power sourcesPaper 2Medium7 marks
A deep-space probe uses a radioisotope heater containing plutonium-238. Plutonium-238 is an α emitter with a half-life of 87.7 years. Each decay releases 5.59 MeV, all of which is absorbed in the heater. At launch the heater must produce a thermal power of 2.0 kW. 1 year = 3.16 × 107 s; 1 u = 1.66 × 10−27 kg; mass of a plutonium-238 atom = 238 u; 1 MeV = 1.60 × 10−13 J.
(a)
Show that the decay constant of plutonium-238 is about 2.5 × 10−10 s−1.
(2)
(b)
Determine the mass of plutonium-238 needed at launch.
(3)
(c)
Calculate the thermal power of the heater 20 years after launch.
(2)
Show mark scheme
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Mark
Notes
Part (a)
T½ = 87.7 × 3.16 × 107 = 2.77 × 109 s
✓ 1
λ = ln 2 / 2.77 × 109 = 2.50 × 10−10 s−1
✓ 1
Part (b)
Energy per decay = 5.59 × 1.60 × 10−13 = 8.94 × 10−13 J, so A = 2000 / 8.94 × 10−13 = 2.24 × 1015 Bq
✓ 1
N = A/λ = 8.9 × 1024
✓ 1
Mass = 8.9 × 1024 × 238 × 1.66 × 10−27 = 3.5 kg
✓ 1
ECF from (a)
—
Part (c)
P = 2.0 × 2−20/87.7 kW (or 2.0 e−λt)
✓ 1
= 1.7 kW (1.71 kW)
✓ 1
Exam tip: Power ∝ activity ∝ number of undecayed nuclei, so the power falls by the same exponential factor as N.
98IBP2H-039-02-16Decay constant as a probabilityPaper 2Medium6 marks
An iridium-192 source is used for industrial radiography of pipeline welds. Its activity is 1.50 TBq on delivery and 1.13 TBq exactly 30 days later. The company replaces the source when its activity falls to 0.40 TBq.
(a)
State what is meant by the decay constant.
(1)
(b)
Determine the decay constant of iridium-192, in day−1.
(2)
(c)
Calculate the half-life of iridium-192.
(1)
(d)
Determine how many days after delivery the source must be replaced.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The probability per unit time that a nucleus decays
✓ 1
Part (b)
ln(1.50/1.13) = λ × 30
✓ 1
λ = 0.283/30 = 9.44 × 10−3 day−1
✓ 1
Part (c)
T½ = ln 2 / λ = 73.4 days
✓ 1
ECF from (b)
—
Part (d)
t = ln(1.50/0.40) / λ
✓ 1
= 1.322 / 9.44 × 10−3 = 140 days
✓ 1
ECF from (b)
—
Exam tip: You can keep λ in day⁻¹ throughout, as long as every time is also in days. There is no need to convert to seconds when no activity–number calculation is required.
99IBP2S-038-05-13Radioisotope power sourcesPaper 2Medium9 marks
A radioisotope generator on a space probe is powered by the alpha decay of plutonium-238: 23894Pu → 23492U + 42He Atomic masses: plutonium-238 = 238.049 560 u; uranium-234 = 234.040 952 u; helium-4 = 4.002 603 u. 1 u = 931.5 MeV c−2. The half-life of plutonium-238 is 87.7 years.
(a)
Outline why an alpha emitter is a good choice as a heat source on a probe that carries sensitive electronics.
(2)
(b)
Show that the energy released in one decay is about 5.6 MeV.
(2)
(c)
At launch, the generator produces 2.0 kW of thermal power. Calculate the activity of the fuel at launch.
(2)
(d)
Determine the thermal power of the generator 20.0 years after launch.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Alpha particles are strongly ionising with a very short range, so they are absorbed within the fuel and its casing and almost all of their kinetic energy becomes thermal energy.
✓ 1
Little radiation escapes, so only light shielding is needed to protect the electronics.
E = 0.006 005 × 931.5 = 5.59 MeV, which is about 5.6 MeV
✓ 1
Part (c)
Energy per decay = 5.59 × 1.60 × 10−13 = 8.94 × 10−13 J
✓ 1
A = 2.0 × 103 / 8.94 × 10−13 = 2.2 × 1015 Bq
✓ 1
Allow follow-through from the energy per decay found earlier; using 5.6 MeV also gives 2.2 × 1015 Bq.
—
Part (d)
λ = ln 2 / 87.7 = 7.90 × 10−3 yr−1
✓ 1
P = 2.0 e−λt = 2.0 × e−7.90 × 10−3 × 20.0
✓ 1
P = 1.7 kW
✓ 1
Alternative method: P = 2.0 × (1/2)20.0/87.7 earns the first two marks, and 1.7 kW earns the third.
—
Common error: Power is proportional to activity, so it decays exponentially with the same half-life as the number of nuclei. Do not subtract the same amount of power each year, as if the decrease were linear.
100IBP2S-038-01-15Beta decay and the antineutrinoPaper 2Medium7 marks
A self-powered emergency exit sign contains tritium gas (\(^3_1\)H) sealed inside phosphor-coated glass tubes. Tritium decays by beta-minus emission with a half-life of 12.3 years. The sign has an initial activity of 9.0 × 1011 Bq. Take 1 year = 3.156 × 107 s. Assume that the brightness of the sign is proportional to the activity of the tritium.
(a)
Calculate the decay constant of tritium in year−1.
(1)
(b)
Calculate the initial number of tritium nuclei in the sign.
(2)
(c)
The sign must be replaced when its brightness falls to 30% of the initial value. Determine the age of the sign at replacement.
(2)
(d)
Outline why the tritium inside the intact sign presents almost no external radiation hazard.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = ln2 / 12.3 = 0.0564 year−1
✓ 1
Part (b)
λ = 0.0564 / 3.156 × 107 = 1.79 × 10−9 s−1
✓ 1
N = A/λ = 9.0 × 1011 / 1.79 × 10−9 = 5.0 × 1020
✓ 1
ECF from (a)
—
Part (c)
0.30 = e−λt, so t = ln(1/0.30)/λ
✓ 1
t = 1.204/0.0564 = 21 years
✓ 1
ECF from (a)
—
Part (d)
Tritium beta particles have very low energy, so their range is very short
✓ 1
They are absorbed by the gas and the glass tube, so none escape from the sign
✓ 1
Exam tip: For 'how long until X% remains', rearrange A = A₀e⁻λt to give t = ln(A₀/A)/λ. Keep λ and t in matching time units.
101IBP3H-E3-20r6-11A cobalt-60 radiotherapy sourcePaper 2Medium9 marks
In a Formula 1 engine laboratory, wear of a piston ring is measured by activating a thin surface layer with radioactive cobalt-56 (proton number 27), which has a half-life of 77.2 days. Cobalt-56 decays by β+ emission to iron. The activity of the activated layer is initially 2.0 MBq. As the ring wears, activated metal is carried away and the remaining activity is proportional to the remaining activated mass.
(a)
Write the nuclear equation for the decay of cobalt-56.
(3)
(b)
Calculate the decay constant of cobalt-56 in s−1.
(2)
(c)
Calculate the activity, in MBq, that an unworn activated layer would have after 120 days. Give your answer to 2 significant figures.
(2)
(d)
After 120 days the measured activity of the ring is 0.40 MBq. Determine the fraction of the activated layer that has been worn away. Give your answer to 2 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
iron nuclide written as 5626Fe
✓ 1
positron emitted
✓ 1
neutrino emitted
✓ 1
Part (b)
λ = ln 2/T½ with T½ = 77.2 × 86 400 s
✓ 1
1.04 × 10−7 s−1
✓ 1
Part (c)
A = A0e−λt (or A02−t/T½) with t = 120 days
✓ 1
0.68 MBq
✓ 1
Part (d)
fraction = 1 − 0.40/0.68 (ECF from (c))
✓ 1
0.41 (41%)
✓ 1
Exam tip: Time and half-life must be in the same units. When you use λ in s⁻¹, convert days to seconds; with half-lives, keep both in days.
102IBP3H-E3-20r6-12Mass defect and binding energyPaper 2Medium7 marks
Use the following data. Mass of proton = 1.00728 u, mass of neutron = 1.00866 u, mass of helium-4 nucleus = 4.00151 u, 1 u = 931.5 MeV c−2. The binding energy of a deuteron is 2.224 MeV.
(a)
Calculate the mass defect of the helium-4 nucleus in u.
(2)
(b)
Calculate the binding energy of the helium-4 nucleus in MeV. Give your answer to 4 significant figures.
(2)
(c)
Determine the binding energy per nucleon of helium-4. Give your answer to 3 significant figures.
(1)
(d)
Two deuterons fuse to form one helium-4 nucleus. Calculate the energy released, using binding energies only.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
total mass of 2 protons and 2 neutrons = 2(1.00728 + 1.00866) = 4.03188 u
✓ 1
mass defect = 0.03037 u
✓ 1
Part (b)
binding energy = 0.03037 × 931.5 (ECF from (a))
✓ 1
28.29 MeV
✓ 1
Part (c)
7.07 MeV (ECF from (b))
✓ 1
Part (d)
energy released = BE(He-4) − 2 × BE(deuteron) = 28.29 − 2 × 2.224
✓ 1
23.84 MeV
✓ 1
Exam tip: For a reaction using binding energies, energy released equals binding energy of the products minus binding energy of the reactants, not the other way round.
103IBP3H-E3-20r6-15Positron emission tomographyPaper 2Medium8 marks
Technetium-99m is produced in a hospital generator by β− decay of molybdenum-99 (proton number 42, nucleon number 99). Technetium-99m (proton number 43) has a half-life of 6.0 hours and decays by γ emission only. A patient is injected with technetium-99m of activity 600 MBq.
(a)
Write the nuclear equation for the decay of molybdenum-99 to technetium-99m.
(3)
(b)
Determine the time after injection at which the activity of technetium-99m in the patient has fallen to 50 MBq. Ignore the removal of the isotope from the body by biological processes. Give your answer to 3 significant figures.
(3)
(c)
Suggest two reasons why technetium-99m is suitable as a medical tracer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9943Tc with correct A and Z
✓ 1
β− particle (electron) emitted
✓ 1
antineutrino emitted
✓ 1
Part (b)
600/50 = 12 = 2n, where n is the number of half-lives
✓ 1
n = ln 12/ln 2 = 3.58
✓ 1
t = 3.58 × 6.0 = 21.5 hours
✓ 1
Part (c)
it emits γ photons, which are weakly ionising and pass out of the body to be detected, whereas α or β particles would be absorbed
✓ 1
its half-life of 6.0 hours is short enough that the activity, and so the dose to the patient, falls quickly
✓ 1
Exam tip: For a 'time to fall to' question, find how many half-lives are needed first and multiply by the half-life; it avoids exponential errors.
104IBP3H-E3-20r6-16Radiocarbon datingPaper 2Medium9 marks
An archaeologist finds a candle made of beeswax in a sealed ancient hive. Carbon-14 (proton number 6) decays by β− emission with a half-life of 5730 years. Living material has a fixed ratio of carbon-14 to carbon-12; this ratio in the wax is 0.62 of the ratio in living material. Use 1 year = 3.156 × 107 s.
(a)
Write the nuclear equation for the decay of carbon-14.
(3)
(b)
Determine the age of the wax. Give your answer to 3 significant figures.
(3)
(c)
A 1.0 g sample of carbon from living material has a carbon-14 activity of 0.230 Bq. Determine the number of carbon-14 nuclei in the sample. Give your answer to 2 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
147N with correct A and Z
✓ 1
β− particle (electron) emitted
✓ 1
antineutrino emitted
✓ 1
Part (b)
0.62 = (1/2)t/5730
✓ 1
t = 5730 × ln(1/0.62)/ln 2
✓ 1
3950 years
✓ 1
Part (c)
λ = ln 2/(5730 × 3.156 × 107) = 3.83 × 10−12 s−1
✓ 1
N = A/λ = 0.230/3.83 × 10−12
✓ 1
6.0 × 1010
✓ 1
Common error: The decay constant must be in s⁻¹ when the activity is in becquerel; leaving the half-life in years gives a number that is wrong by a factor of about 3 × 10⁷.
106IBP3H-E3-20r15-12Radioisotope power sourcesPaper 2Medium8 marks
A radioisotope thermoelectric generator on a space probe contains two radioactive isotopes X and Y. X has a half-life of 3.0 years and Y has a half-life of 12 years. At launch the activity of X is 6.0 × 109 Bq and the activity of Y is 2.0 × 109 Bq.
(a)
Calculate the decay constant of each isotope in yr−1.
(2)
(b)
Calculate the total activity 6.0 years after launch.
(3)
(c)
Determine the time after launch at which the two isotopes have equal activity.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 λ = ln 2 / T½
✓ 1
A1 λX = 0.231 yr−1 and λY = 0.0578 yr−1
✓ 1
Part (b)
M1 uses A = A0 2−t/T or A0e−λt for each isotope
✓ 1
A1 X: 1.5 × 109 Bq and Y: 1.41 × 109 Bq
✓ 1
A1 total 2.91 × 109 Bq
✓ 1
Part (c)
M1 6.0 × 109 × 2−t/3 = 2.0 × 109 × 2−t/12
✓ 1
M1 2t/4 = 3 so t = 4 log2 3
✓ 1
A1 6.34 years
✓ 1
Common error: Activities of different nuclides add, but half-lives do not: never average the two half-lives.
107IBP3H-E3-20r15-13A cobalt-60 radiotherapy sourcePaper 2Medium8 marks
A sealed cobalt-60 source used to inspect a bridge cable has a half-life of 5.27 years and a present activity of 3.2 × 1011 Bq. Take 1 year = 3.16 × 107 s. Take the mass of a cobalt-60 nucleus as 60 u, where u = 1.66 × 10−27 kg.
(a)
Calculate the decay constant of cobalt-60 in s−1.
(2)
(b)
Determine the number of undecayed cobalt-60 nuclei in the source now. Use the unrounded value of λ.
(2)
(c)
Calculate the activity of the source after a further 8.0 years.
(2)
(d)
Estimate the mass of cobalt-60 in the source now. Use the unrounded value of N.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 λ = ln 2 / (5.27 × 3.16 × 107 s)
✓ 1
A1 4.16 × 10−9 s−1
✓ 1
Part (b)
M1 N = A/λ
✓ 1
A1 7.69 × 1019
✓ 1
Part (c)
M1 A = 3.2 × 1011 × 2−8.0/5.27 or the exponential form with λ
✓ 1
A1 1.12 × 1011 Bq
✓ 1
Part (d)
M1 mass = N × 60 × 1.66 × 10−27 kg
✓ 1
A1 7.66 × 10−6 kg
✓ 1
Exam tip: Convert the half-life to seconds before using A = λN with an activity in becquerels.
108IBP3H-E3-20r15-14Identifying radiations and backgroundPaper 2Medium7 marks
A point source of gamma rays has a half-life of 2.6 years. A detector fixed 2.0 m from the source records a net count rate of 840 s−1 today. Assume the count rate is proportional to the activity of the source, that the radiation obeys an inverse-square law and that absorption is negligible.
(a)
Calculate the net count rate at the same position after 3.9 years.
(2)
(b)
After 3.9 years the detector is moved so that it again records 840 s−1. Calculate its new distance from the source.
(2)
(c)
Calculate the time from today at which the net count rate at 2.0 m will have fallen to 100 s−1.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 840 × 2−3.9/2.6
✓ 1
A1 297 s−1
✓ 1
Part (b)
M1 rate ∝ 1/r2 so r = 2.0 × √(297/840)
✓ 1
A1 1.19 m
✓ 1
Part (c)
M1 100 = 840 × 2−t/2.6 or 840e−λt
✓ 1
M1 t = 2.6 × ln 8.4 / ln 2
✓ 1
A1 7.98 years
✓ 1
Common error: Apply the decay law and the inverse-square law separately: one changes the activity, the other changes how much of it reaches the detector.
111IBP3H-E3-20r18-13Radiocarbon datingPaper 2Medium9 marks
A marine biologist labels plankton with phosphorus-32, which has a half-life of 14.3 days. A detector registers only 25% of the decays from the sample. The corrected count rate measured by the detector is 4.2 × 103 s−1.
(a)
Determine the activity of the sample.
(2)
(b)
Calculate the decay constant of phosphorus-32 in s−1.
(2)
(c)
Calculate the number of phosphorus-32 nuclei in the sample. Use the unrounded value of the decay constant.
(2)
(d)
The detector is used again when its corrected count rate has fallen to 5.0 × 102 s−1. Calculate the time that has passed, in days.
112E-2-07Alpha decay and nuclear energy levelsPaper 2Hard12 marks
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)
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.
114E-2-15Binding energy of the deuteronPaper 2Hard10 marks
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
115E-2-23The strong nuclear force and stabilityPaper 2Hard12 marks
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)
116E-2-24Radioisotope power sourcesPaper 2Hard13 marks
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)
117E-2-38Positron emission tomographyPaper 2Hard19 marks
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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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)
118IBP2H-039-00-10Radioisotope power sourcesPaper 2Hard10 marks
A space probe is powered by a radioisotope thermoelectric generator containing plutonium-238 (half-life 87.7 years). Each Pu-238 decay releases 5.59 MeV. Assume all of this energy is absorbed in the fuel as thermal energy. At launch, the fuel must deliver a thermal power of 2.0 kW. 1 year = 3.156×107 s; molar mass of Pu-238 = 0.238 kg mol−1; NA = 6.02×1023 mol−1; 1 eV = 1.60×10−19 J.
(a)
Show that the decay constant of Pu-238 is about 2.5×10−10 s−1.
(2)
(b)
Calculate the activity of the fuel needed at launch.
(2)
(c)
Estimate the mass of Pu-238 required at launch.
(2)
(d)
Determine the thermal power of the fuel 20.0 years after launch.
(2)
(e)
Explain why an α emitter is a good choice for this generator.
(2)
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Notes
Part (a)
T½ = 87.7 × 3.156×107 = 2.77×109 s
✓ 1
λ = ln2/T½ = 0.693/2.77×109 = 2.50×10−10 s−1
✓ 1
Part (b)
Energy per decay = 5.59×106 × 1.60×10−19 = 8.94×10−13 J
✓ 1
A = P/E = 2000/8.94×10−13 = 2.24×1015 Bq
✓ 1
Part (c)
N = A/λ = 2.24×1015/2.50×10−10 = 8.9×1024 nuclei ECF from (a) and (b)
✓ 1
m = (8.9×1024/6.02×1023) × 0.238 = 3.5 kg
✓ 1
Part (d)
Power is proportional to activity, so P = 2.0 × 2−20.0/87.7 (or 2.0 e−λt)
✓ 1
P = 1.7 kW (1.71 kW)
✓ 1
Part (e)
α particles have a very short range, so almost all of their kinetic energy is absorbed within the fuel and becomes thermal energy
✓ 1
Very little radiation escapes the fuel, so only light shielding is needed to protect the probe's electronics, which saves mass
✓ 1
Exam tip: Power ∝ activity ∝ N, so the decay law applies to power directly. You do not need to recalculate N for part (d).
119IBP2H-039-00-14Uses of radioisotopesPaper 2Hard8 marks
A water company suspects a leak in a buried pipe. A solution containing sodium-24 is added to the water upstream. Sodium-24 is a γ emitter with a half-life of 15.0 hours. Workers walk along the route of the pipe with a detector. The solution has an initial activity of 5.0 MBq.
(a)
Explain why a γ emitter with a half-life of several hours is suitable for this purpose.
(2)
(b)
Show that the solution initially contains about 3.9×1011 sodium-24 nuclei.
(2)
(c)
The detector cannot distinguish the leak from background once the total activity of the sodium-24 falls below 0.20 MBq. Determine the time available for the survey.
(2)
(d)
Calculate the number of sodium-24 nuclei that decay during the first 24.0 hours.
(2)
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Notes
Part (a)
γ radiation is weakly absorbed, so it can pass through the soil above the pipe and reach the detector at the surface (α and β would be absorbed)
✓ 1
A half-life of hours is long enough to complete the survey but short enough that the activity in the water supply soon becomes negligible
✓ 1
Part (b)
λ = ln2/(15.0 × 3600) = 1.28×10−5 s−1
✓ 1
N = A/λ = 5.0×106/1.28×10−5 = 3.9×1011
✓ 1
Part (c)
0.20/5.0 = e−λt, so t = T½ ln25/ln2
✓ 1
t = 15.0 × 3.22/0.693 = 70 h (69.7 h)
✓ 1
Part (d)
Number remaining = 3.9×1011 × 2−24.0/15.0 = 1.28×1011 to 1.29×1011, OR fraction decayed = 1 − e−λt = 0.670 ECF from (b)
✓ 1
Number decayed = 3.9×1011 − 1.29×1011 = 2.6×1011
✓ 1
Common error: The number that have decayed is N₀ − N, not N. Read carefully whether the question asks for the nuclei remaining or the nuclei decayed.
120IBP2H-039-00-16Beta decay and the neutrinoPaper 2Hard9 marks
Outside a nucleus, a free neutron is unstable and decays by β− emission with a half-life of 610 s. At a research reactor, a beam of slow neutrons travels at 2.20×103 m s−1 along a 15.0 m guide tube to an experiment. Mass of neutron = 1.008665 u; mass of proton = 1.007276 u; mass of electron = 0.000549 u; 1 u = 931.5 MeV c−2.
(a)
Write the decay equation for a free neutron.
(2)
(b)
Calculate the energy released in the decay of one free neutron.
(2)
(c)
The emitted electrons have kinetic energies anywhere from zero to almost the value in (b). Explain how this observation led to the proposal that a third particle is emitted.
(2)
(d)
Determine the fraction of the neutrons in the beam that decay while travelling along the guide tube.
(3)
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Notes
Part (a)
n → p + e− (proton and electron as products)
✓ 1
+ electron antineutrino, νe
✓ 1
Part (b)
Δm = 1.008665 − 1.007276 − 0.000549 = 0.000840 u
✓ 1
Q = 0.000840 × 931.5 = 0.782 MeV
✓ 1
Part (c)
If only two particles were produced, conservation of energy and momentum would fix the electron's share of the energy, giving a single kinetic energy
✓ 1
The continuous spectrum means a third, undetected, neutral particle of very small mass (the antineutrino) carries away the rest of the energy in varying amounts
✓ 1
Part (d)
Time in tube t = 15.0/2.20×103 = 6.82×10−3 s
✓ 1
λ = ln2/610 = 1.14×10−3 s−1
✓ 1
Fraction = 1 − e−λt ≈ λt = 7.7×10−6
✓ 1
Exam tip: When λt is very small, 1 − e⁻λt ≈ λt. Stating this approximation avoids rounding problems on your calculator.
121IBP2S-038-04-14Radioisotope power sourcesPaper 2Hard9 marks
A radioisotope thermoelectric generator (RTG) on a deep-space probe uses plutonium-238. Plutonium-238 is an alpha emitter with a half-life of 87.7 years. Each decay releases 5.6 MeV, which is converted to thermal energy. At launch the thermal power is 2.0 kW. Use 1 year = 3.16×107 s, molar mass of plutonium-238 = 0.238 kg mol−1 and NA = 6.02×1023 mol−1.
(a)
Explain why the mission controllers cannot switch off or adjust the thermal output of the RTG.
(1)
(b)
Calculate the decay constant of plutonium-238 in s−1.
(2)
(c)
Show that the activity of the plutonium at launch is about 2×1015 Bq.
(2)
(d)
Determine the mass of plutonium-238 in the RTG at launch.
(2)
(e)
Calculate the thermal power of the RTG 20 years after launch.
(2)
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Notes
Part (a)
Decay is spontaneous and is not affected by external factors such as temperature or pressure, so the decay rate cannot be controlled .
122IBP2S-038-02-16Uses of radioisotopesPaper 2Hard8 marks
A water engineer detects a leak in a buried pipe by adding a solution containing sodium-24, a gamma emitter. A sample of the solution is monitored in the laboratory, and a background count rate of 30 counts min−1 is measured.
State why the background count rate must be subtracted before analysing the data.
(1)
(b)
Show that the half-life of sodium-24 is about 15 h.
(2)
(c)
Determine the decay constant of sodium-24 in h−1.
(1)
(d)
Determine the time after t = 0 at which the measured count rate will be 60 counts min−1.
(2)
(e)
Explain why sodium-24 is suitable for this application.
(2)
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Notes
Part (a)
Background radiation does not come from the source and does not decay, so leaving it in would make the source appear to decay more slowly and give a half-life that is too large.
✓ 1
Part (b)
Corrected count rates are 1200 counts min−1 at 0 h and 300 counts min−1 at 30 h (or other correctly corrected pairs).
✓ 1
1200 → 300 is a fall by a factor of 4, which is two half-lives in 30 h, so T½ ≈ 15 h.
✓ 1
Part (c)
λ = ln 2 / 15 = 0.046 h−1 (allow error carried forward from the previous part).
✓ 1
Part (d)
The corrected rate must fall to 60 − 30 = 30 counts min−1, so 30 = 1200 e−λt.
✓ 1
t = ln(40)/0.0462 ≈ 80 h (allow error carried forward from the previous part).
✓ 1
Part (e)
It emits gamma radiation, which is penetrating enough to pass through the soil and be detected at the surface above the leak.
✓ 1
Its half-life of about 15 h is long enough for the test to be completed, but short enough that the activity in the water supply becomes negligible within a few days.
✓ 1
Exam tip: When a target measured count rate is given, subtract the background first and only then solve the exponential equation.
123IBP2H-039-01-13Radioisotope power sourcesPaper 2Hard10 marks
A deep-space probe is powered by a radioisotope thermoelectric generator (RTG). The RTG contains plutonium-238, an alpha emitter. The following data are given. half-life of plutonium-238 = 87.7 years energy released per decay = 5.59 MeV molar mass of plutonium-238 = 238 g mol−1 initial thermal power = 4.40 kW efficiency of conversion to electrical power = 6.0% minimum electrical power needed by the probe = 200 W Use 1 year = 3.16×107 s, e = 1.60×10−19 C and NA = 6.02×1023 mol−1. Assume that all the decay energy is converted to thermal energy in the RTG.
(a)
Show that the initial activity of the plutonium is about 4.9×1015 Bq.
(2)
(b)
Determine the initial mass of plutonium-238 in the RTG.
(3)
(c)
Calculate the time, in years, after which the electrical output falls to the minimum needed by the probe. Assume the efficiency stays constant.
(3)
(d)
Outline two reasons why an alpha emitter is preferred for an RTG.
(2)
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Notes
Part (a)
Energy per decay = 5.59×106 × 1.60×10−19 = 8.94×10−13 J
✓ 1
A = 4400 / 8.94×10−13 = 4.92×1015 Bq
✓ 1
Part (b)
λ = ln 2 / (87.7 × 3.16×107) = 2.50×10−10 s−1
✓ 1
N = A/λ = 1.97×1025
✓ 1
m = (1.97×1025 / 6.02×1023) × 238 g = 7.8 kg
✓ 1
ECF from (a).
—
Part (c)
Initial electrical power = 0.060 × 4400 = 264 W
✓ 1
Set 200 = 264 × 2−t/87.7
✓ 1
t = 87.7 × ln(1.32) / ln 2 = 35 years
✓ 1
Part (d)
Alpha particles are absorbed within the fuel and its casing, so almost all of the decay energy is deposited as thermal energy
✓ 1
Alpha particles are not penetrating, so little heavy shielding is needed to protect the probe's electronics, which reduces the launch mass
✓ 1
Exam tip: Power output is proportional to activity, so it decays with the same half-life. You can apply the decay law directly to the power without first finding N.
124IBP2H-039-01-16Determining a long half-lifePaper 2Hard8 marks
Volcanic rock can be dated using potassium-40, which has a half-life of 1.25×109 years. Of the potassium-40 nuclei that decay, 11% become argon-40. The rest become calcium-40. When the rock solidified, it contained no argon-40, and all argon-40 produced since then has been trapped in the rock. A sample now contains 0.050 argon-40 atoms for every potassium-40 atom.
(a)
State what is meant by the decay constant.
(1)
(b)
Show that the decay constant of potassium-40 is about 5.5×10−10 y−1.
(1)
(c)
Show that the ratio of the original number of potassium-40 nuclei to the present number, N0/N, is about 1.45.
(2)
(d)
Determine the age of the rock.
(2)
(e)
Some argon-40 has actually escaped from the rock. Explain the effect on the calculated age.
(2)
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Notes
Part (a)
The decay constant is the probability per unit time that a given nucleus will decay
✓ 1
Part (b)
λ = ln 2 / 1.25×109 = 5.55×10−10 y−1
✓ 1
Part (c)
The argon formed is NAr = 0.11(N0 − N)
✓ 1
So 0.050 = 0.11(N0/N − 1), which gives N0/N = 1 + 0.050/0.11 = 1.45
✓ 1
Part (d)
t = ln(N0/N) / λ = ln(1.455) / 5.55×10−10
✓ 1
t = 6.8×108 years (accept 6.7×108 years if the rounded value 1.45 is used)
✓ 1
ECF from the values of λ and N0/N found in the earlier parts.
—
Part (e)
The measured argon to potassium ratio is smaller than it should be, so N0/N is underestimated
✓ 1
The calculated age is therefore less than the true age
✓ 1
Common error: Not every decayed potassium nucleus becomes argon, so dividing by the branching fraction of 0.11 is essential. Ignoring it gives N₀/N = 1.05 and an age of about 8.8×10⁷ years. That is nearly eight times too small.
125IBP2H-039-02-11The random nature of decayPaper 2Hard9 marks
A short-lived isotope is produced in a cyclotron for wear-testing of engine parts. A detector system records the number N of undecayed nuclei in a sample at 10 s intervals.
Show that the data are consistent with exponential decay.
(2)
(b)
Determine the decay constant λ of the isotope.
(2)
(c)
Estimate the probability that a particular nucleus in the sample decays during a 1.0 s interval.
(1)
(d)
Calculate the probability that a nucleus present at t = 0 is still undecayed at t = 100 s.
(2)
(e)
Explain why your answer to (c) also applies to a nucleus that has already survived until t = 40 s.
(2)
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Notes
Part (a)
Successive ratios are calculated, e.g. 6550/8000, 5363/6550, 4390/5363 and 3595/4390, each ≈ 0.819
✓ 1
A constant ratio over equal time intervals shows exponential decay
✓ 1
Part (b)
ln(8000/3595) = 0.800, or an equivalent method such as ln(1/0.819) per 10 s
✓ 1
λ = 0.800/40 = 0.0200 s−1
✓ 1
Part (c)
≈ λΔt = 0.020 (accept 0.0198 from 1 − e−0.02)
✓ 1
ECF from (b)
—
Part (d)
Probability of survival = e−λt
✓ 1
= e−0.0200 × 100 = e−2.0 = 0.135
✓ 1
ECF from (b)
—
Part (e)
λ is constant for the nuclide, and decay is random and spontaneous
✓ 1
The decay probability per unit time does not depend on the age or history of the nucleus. Every surviving nucleus has the same chance of decaying in the next second
✓ 1
Exam tip: For 'show that' data questions, test the ratio over several equal intervals, not just one pair of readings.
126IBP2H-039-02-13Beta decay and the antineutrinoPaper 2Hard9 marks
Carbon-11, 116C, is a positron emitter used to label drug molecules for medical imaging. It decays to boron-11. Atomic mass of carbon-11 = 11.011434 u Atomic mass of boron-11 = 11.009305 u Electron mass = 0.000549 u 1 u is equivalent to 931.5 MeV.
(a)
Write the nuclear equation for this decay.
(2)
(b)
Explain why two electron masses must be subtracted from the difference in atomic masses to find the mass converted to energy.
(2)
(c)
Explain why the positrons emitted have a continuous range of kinetic energies.
(2)
(d)
Calculate the maximum kinetic energy of an emitted positron, in MeV.
(3)
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Notes
Part (a)
116C → 115B + 0+1e, with all numbers correct
✓ 1
+ νe (electron neutrino)
✓ 1
Part (b)
Atomic masses include orbital electrons. The carbon atom has 6 electrons and the boron atom only 5, so the difference in atomic masses contains one extra electron mass
✓ 1
The emitted positron also has a mass equal to one electron mass. This must also be subtracted, giving 2me in total
✓ 1
Part (c)
The energy released is shared between the positron and the neutrino (with negligible recoil of the nucleus)
✓ 1
The share varies randomly from decay to decay, so the positron can have any kinetic energy from zero up to a maximum
✓ 1
Part (d)
Δm = 11.011434 − 11.009305 − 2(0.000549)
✓ 1
= 0.001031 u, and the energy = 0.001031 × 931.5
✓ 1
= 0.960 MeV
✓ 1
Common error: Using the atomic mass difference without subtracting 2me roughly doubles the answer here. Decide first whether the masses given are atomic or nuclear.
Common error: Multiply the binding energy per nucleon by the nucleon number before comparing products and reactants. Subtracting the per-nucleon values directly gives a meaningless answer.
128IBP2S-038-01-13Radioisotope power sourcesPaper 2Hard9 marks
A radioisotope power system for a space probe contains 4.8 kg of plutonium-238 at launch. Plutonium-238 is an alpha emitter with a half-life of 87.7 years. Its molar mass is 0.238 kg mol−1. Each decay releases 5.6 MeV of energy, all of which is converted into thermal energy in the source. 1 year = 3.156 × 107 s; NA = 6.02 × 1023 mol−1; 1 MeV = 1.60 × 10−13 J.
(a)
Calculate the decay constant of plutonium-238 in s−1.
(2)
(b)
Calculate the number of plutonium-238 nuclei present at launch.
(2)
(c)
Determine the activity of the source at launch.
(1)
(d)
Show that the thermal power of the source at launch is about 2.7 kW.
(2)
(e)
Calculate the thermal power of the source 20 years after launch.
129IBP3H-E3-20r6-13Alpha decay and nuclear energy levelsPaper 2Hard7 marks
Americium-241 (proton number 95) is used as a calibration source in a hospital physics department. It decays by α emission to neptunium (proton number 93). The total mass of the neptunium nucleus and the α particle is 0.00614 u less than the mass of the americium nucleus. Use 1 u = 931.5 MeV c−2. An americium nucleus decays while at rest.
(a)
Write the nuclear equation for the decay.
(2)
(b)
Calculate the total kinetic energy released in the decay. Give your answer to 4 significant figures.
(2)
(c)
Explain why the α particle receives most of the kinetic energy, and determine its kinetic energy. Treat the ratio of the masses as the ratio of the nucleon numbers, and give your answer to 3 significant figures.
(3)
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Notes
Part (a)
23793Np with correct A and Z
✓ 1
42He (or α) written on the right
✓ 1
Part (b)
energy = 0.00614 × 931.5
✓ 1
5.719 MeV
✓ 1
Part (c)
momentum is conserved, so the α particle and the neptunium nucleus have equal and opposite momenta, and since Ek = p2/2m the lighter particle has the larger kinetic energy
✓ 1
Eα/ENp = 237/4, so Eα = 5.719 × 237/241 (ECF from (b))
✓ 1
5.62 MeV
✓ 1
Common error: Equal and opposite momenta do not mean equal kinetic energies; energy is shared in inverse proportion to mass.
130IBP3H-E3-20r6-14Radioisotope power sourcesPaper 2Hard10 marks
A radioisotope power source on a space probe contains plutonium-238, which has a half-life of 87.7 years and decays by α emission. Each decay releases 5.59 MeV, all of which is converted to thermal energy in the source. The initial thermal power of the source is 2000 W. Use: 1 eV = 1.602 × 10−19 J, 1 year = 3.156 × 107 s, Avogadro constant = 6.02 × 1023 mol−1, molar mass of plutonium-238 = 0.238 kg mol−1. Give answers to 3 significant figures.
(a)
Calculate the initial activity of the source.
(2)
(b)
Calculate the decay constant of plutonium-238 in s−1.
(1)
(c)
Determine the mass of plutonium-238 in the source initially.
(3)
(d)
Calculate the thermal power of the source after 20 years, assuming it is proportional to the activity.
(2)
(e)
Suggest why an α emitter with a half-life of the order of decades is suitable for this use.
(2)
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Notes
Part (a)
energy per decay = 5.59 × 106 × 1.602 × 10−19 = 8.955 × 10−13 J
✓ 1
activity = 2000/8.955 × 10−13 = 2.23 × 1015 Bq
✓ 1
Part (b)
λ = ln 2/(87.7 × 3.156 × 107) = 2.50 × 10−10 s−1
✓ 1
Part (c)
N = A/λ (ECF from (a) and (b))
✓ 1
N = 8.92 × 1024
✓ 1
mass = (8.92 × 1024/6.02 × 1023) × 0.238 = 3.53 kg
✓ 1
Part (d)
P = 2000 × (1/2)20/87.7 (or 2000 e−λt)
✓ 1
1.71 × 103 W
✓ 1
Part (e)
α particles are absorbed within the source, so their energy is converted to thermal energy (and little shielding is needed)
✓ 1
a half-life of decades means the power falls slowly over the years of the mission
✓ 1
Exam tip: Thermal power equals activity multiplied by the energy released per decay; convert MeV to joules first or the numbers will not match.
131IBP3H-E3-20r15-15Decay constant and half-lifePaper 2Hard10 marks
A sample contains two radioactive nuclides X and Y, with half-lives 1.0 h and 6.0 h. The products of both decays are stable. The total activity of the sample is 1200 Bq at t = 0 and 212.5 Bq at t = 6.0 h. Let the initial activities of X and Y be AX and AY.
(a)
Write down two equations relating AX and AY.
(2)
(b)
Determine AX and AY.
(3)
(c)
Explain why a graph of ln(total activity) against time becomes a straight line at late times, and state what its gradient equals.
(2)
(d)
Determine the time at which the activity of X is 5% of the activity of Y.
(3)
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Notes
Part (a)
M1 AX + AY = 1200
✓ 1
M1 AX/64 + AY/2 = 212.5 (six half-lives of X, one of Y)
✓ 1
Part (b)
M1 eliminates one variable
✓ 1
A1 AX = 800 Bq
✓ 1
A1 AY = 400 Bq
✓ 1
Part (c)
R1 the activity of X has decayed away much faster, leaving essentially only Y
✓ 1
A1 so ln A = ln AY − λYt, a straight line with gradient −λY
✓ 1
Part (d)
M1 800 × 2−t / (400 × 2−t/6) = 0.05
✓ 1
M1 2−5t/6 = 0.025 so t = (6/5) log2 40
✓ 1
A1 6.39 h
✓ 1
Exam tip: With two unknowns you need two independent pieces of data; write one equation for each measurement before solving.
132IBP3H-E3-20r15-16Modelling decay with dicePaper 2Hard10 marks
A student models radioactive decay with 3000 ten-sided dice numbered 1 to 10. In each throw all remaining dice are thrown and any die showing 1 or 2 is removed.
(a)
Calculate the half-life of the model in throws, to 3 significant figures.
(2)
(b)
Each throw represents 20 minutes of real time. Use the unrounded half-life from (a) to calculate the decay constant of the model in s−1.
(2)
(c)
Calculate the expected number of dice remaining after 10 throws, to 3 significant figures.
(2)
(d)
Use the unrounded value from (c) to calculate the expected number of dice removed in the 11th throw, to 3 significant figures.
(2)
(e)
The probability of removal in one throw is 0.20, but the decay constant of the model per throw is 0.223. Explain why these differ.
(2)
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Notes
Part (a)
M1 0.8n = 0.5 or n = ln 0.5 / ln 0.8
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A1 3.11 throws
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Part (b)
M1 λ = ln 2 / (3.106 × 1200 s)
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A1 1.86 × 10−4 s−1
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Part (c)
M1 3000 × 0.810
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A1 322
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Part (d)
M1 0.2 × 322.12 (ECF from (c))
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A1 64.4
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Part (e)
R1 the probability applies to the dice present at the start of a throw; the decay constant is defined for an infinitesimally short interval, during which the number present is continuously falling
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A1 λ = −ln(1 − 0.20) = 0.223 per throw, which is larger than 0.20
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Exam tip: A decay constant is a rate per unit time for very short intervals; a probability per throw is not the same quantity unless it is very small.
133IBP3H-E3-20r18-15Two nuclides decaying togetherPaper 2Hard9 marks
A radioactive mixture contains two isotopes. Isotope X has a half-life of 2.0 h and isotope Y has a half-life of 8.0 h. At time zero each isotope contributes an activity of 400 Bq. Assume that the decay products are not radioactive.
(a)
State the total activity at time zero.
(1)
(b)
Determine the total activity after 8.0 h.
(3)
(c)
Determine the time at which the activity of X is 1.0% of the activity of Y.
(4)
(d)
Explain why, for large times, the total activity appears to decay with a half-life of 8.0 h.
(1)
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Notes
Part (a)
800 Bq
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Part (b)
X: 4 half-lives, 400 × (1/2)4 = 25 Bq
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Y: 1 half-life, 400/2 = 200 Bq
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total = 225 Bq
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Part (c)
\(A_X/A_Y=2^{-t/2}/2^{-t/8}=2^{-3t/8}\)
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set \(2^{-3t/8}=0.010\)
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\(\frac{3t}{8}=\log_2100=6.64\)
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17.7 h
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Part (d)
the activity of X becomes negligible, so the total is essentially that of Y
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Common error: Add the activities of the two isotopes, not their half-lives, and apply each half-life to its own isotope.
134IBP3H-E3-20r18-16Range of alpha particles in airPaper 2Hard12 marks
A sealed basement of volume 85 m3 is under assessment. Radon-222 (half-life 3.82 days) enters the air at a constant rate of 4.0 × 103 atoms per second. Assume that radon is lost only by radioactive decay unless stated otherwise, and that it is well mixed.
(a)
Calculate the decay constant of radon-222 in s−1.
(2)
(b)
Explain why the activity of radon in the basement approaches a constant value, and state this value.
(2)
(c)
Determine the number of radon atoms in the basement at this equilibrium. ECF from (a).
(2)
(d)
Calculate the activity concentration of radon, in Bq m−3.
(2)
(e)
A ventilation system removes radon atoms from the air with a rate constant of 1.5 × 10−5 s−1, in addition to radioactive decay. Determine the new activity concentration of radon at equilibrium. Use the unrounded value of the decay constant.
(4)
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Notes
Part (a)
3.82 days = 3.30 × 105 s
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\(\lambda=\ln2/T_{1/2}=2.10\times10^{-6}\) s−1
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Part (b)
the number of radon atoms stops changing when the decay rate equals the input rate
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activity = 4.0 × 103 Bq
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Part (c)
\(N=A/\lambda\)
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1.90 × 109
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Part (d)
activity per unit volume = 4.0 × 103/85
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47.1 Bq m−3
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Part (e)
at equilibrium input = total loss, \(R=(\lambda+k)N\)
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