ExaminerPrep › Question bank › IAL Physics › 5.4 Radioactive decay
Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
40 free, high-quality past-paper-style questions on Radioactive decay for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which row correctly describes alpha radiation?
| Answer | Mark | Notes |
|---|---|---|
| An alpha particle is a helium nucleus (charge +2e). It ionises strongly, so it loses its energy quickly and is absorbed by a sheet of paper or a few cm of air. | ✓ 1 | Answer D |
Answer: D
Exam tip: Strong ionisation and short range go together: the more ionisations per mm, the sooner the radiation runs out of energy.
A nucleus of americium-241, 24195Am, emits an alpha particle.
What are the nucleon number and proton number of the nucleus formed?
| Answer | Mark | Notes |
|---|---|---|
| An alpha particle is 42α: the nucleon number falls by 4 and the proton number by 2, giving 237 and 93 (neptunium-237). | ✓ 1 | Answer C |
Answer: C
Exam tip: Check any nuclear equation by adding the top numbers and the bottom numbers on each side.
A GM tube records 1350 counts in 5.0 minutes when a source is in front of it. The background count rate is 36 counts per minute.
What is the count rate due to the source?
| Answer | Mark | Notes |
|---|---|---|
| Measured rate = 1350 ÷ 300 s = 4.5 s−1. Background = 36 ÷ 60 s = 0.60 s−1. Corrected = 4.5 − 0.60 = 3.9 s−1. | ✓ 1 | Answer A |
Answer: A
Exam tip: Put both rates in the same unit before subtracting the background.
A sample contains 80 mg of a radioactive isotope with a half-life of 15 hours.
What mass of the isotope remains after 60 hours?
| Answer | Mark | Notes |
|---|---|---|
| 60 h = 4 half-lives. 80 × (½)4 = 5.0 mg. | ✓ 1 | Answer B |
Answer: B
Exam tip: Count the half-lives first, then halve that many times: 80 → 40 → 20 → 10 → 5.0.
A student measures the radiation from a small sample of a radioactive mineral, using a GM tube placed 1 cm from the sample.
State two sources of background radiation.
With no sample present the counter records 252 counts in 10.0 minutes. With the sample in place it records 690 counts in 5.0 minutes. Calculate the count rate due to the sample, in s−1.
A sheet of paper placed between the sample and the GM tube reduces the count rate to the background value. State what this shows about the radiation emitted.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any two from: radon gas from rocks and soil / rocks and building materials / cosmic rays / food, drink or the human body / medical sources / nuclear industry or fallout | ✓ 2 | |
| Part (b) | ||
| background = 25.2 min−1 and with sample = 138 min−1 | ✓ 1 | |
| (138 − 25.2) ÷ 60 = 1.9 s−1 (1.88 s−1) | ✓ 1 | |
| Part (c) | ||
| the sample emits only alpha radiation, since only alpha is absorbed by paper | ✓ 1 | |
Exam tip: Background is random, so it is measured over a long time; convert both counts to rates in the same unit before subtracting.
Radon-222, 22286Rn, is a radioactive gas that seeps out of the ground. It decays by alpha emission to an isotope of polonium (Po), with a half-life of 3.8 days.
Complete the nuclear equation: 22286Rn → Po + 42α
A sealed container of air holds some radon-222. What fraction of the radon remains after 15.2 days?
Alpha particles cannot pass through the outer layer of skin. Explain why radon is nevertheless a health hazard when breathed in.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 21884Po: nucleon number 218, proton number 84 | ✓ 1 | |
| Part (b) | ||
| 15.2 ÷ 3.8 = 4 half-lives, so (½)4 = 1/16 = 0.0625 | ✓ 1 | |
| Part (c) | ||
| inside the lungs the alpha particles are emitted right next to living cells | ✓ 1 | |
| alpha is strongly ionising and deposits all its energy in a small volume of tissue, damaging cells / DNA | ✓ 1 | |
Exam tip: Alpha is the least hazardous radiation outside the body but the most hazardous inside it — always say where the source is.
A source emits alpha, beta-minus and gamma radiation into a vacuum. The radiation passes between two parallel plates that have a potential difference between them. The diagram shows the three tracks P, Q and R.
Which row identifies the radiation in each track?
| Answer | Mark | Notes |
|---|---|---|
| Gamma radiation is uncharged, so it passes straight through: Q. A β− particle is negative, so it is attracted towards the positive plate: P. An α particle is positive, so it bends towards the negative plate: R. Its much greater mass makes its deflection smaller. The answer is P: β−; Q: γ; R: α. | ✓ 1 | Answer D |
Answer: D
Exam tip: Decide the charge of each radiation first (α: +2e, β−: −e, γ: 0), then use the direction of the bend. The size of the bend is a second clue, not the first.
With no source present, a GM tube and counter record the background count in ten successive 10 s intervals, shown in the bar chart. A radioactive source is then placed in front of the tube and 1240 counts are recorded in 200 s.
What is the count rate due to the source?
| Answer | Mark | Notes |
|---|---|---|
| Background: 4 + 6 + 3 + 5 + 7 + 4 + 5 + 6 + 3 + 7 = 50 counts in 100 s = 0.50 s−1. With the source: 1240 ÷ 200 s = 6.2 s−1. Count rate due to the source = 6.2 − 0.50 = 5.7 s−1. | ✓ 1 | Answer A |
Answer: A
Exam tip: Average the background over all the intervals: a single 10 s reading fluctuates too much to be a reliable value.
A source containing an isotope with a half-life of 30 years is placed in front of a GM tube. The graph shows the number of counts recorded in 20 successive 10 s intervals. The background count is negligible.
Which statement explains the variation in the counts?
| Answer | Mark | Notes |
|---|---|---|
| Over a few minutes, a 30-year half-life source has a constant mean activity, and the graph shows no trend. The counts scatter about the mean because decay is random: the number of nuclei that happen to decay in each interval varies. | ✓ 1 | Answer A |
Answer: A
Exam tip: Random: you cannot predict when a particular nucleus will decay. Spontaneous: decay is not affected by external conditions such as temperature.
Figure 1 shows a system that controls the thickness of aluminium foil as it is made. Beta radiation from a source below the foil is detected by a GM tube above it. The graph in Figure 1 shows the count rate recorded during one minute of production.
Explain why a beta source is used, rather than an alpha source or a gamma source.
Use the graph to describe what happened to the foil at t = 30 s, and state how the control unit should respond.
Suggest why the source should have a long half-life.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| alpha would be completely absorbed by the foil, so no radiation would reach the tube / the count would not depend on thickness | ✓ 1 | |
| gamma would pass through almost unaffected, so small changes in thickness would cause no measurable change in count rate | ✓ 1 | |
| Part (b) | ||
| the count rate fell from 120 s−1 to 90 s−1, so more beta radiation was absorbed: the foil became thicker | ✓ 1 | |
| the control unit should move the rollers closer together / increase the roller pressure | ✓ 1 | |
| Part (c) | ||
| so that its activity does not fall noticeably during use; otherwise a fall in count rate due to decay would be mistaken for the foil becoming thicker | ✓ 1 | |
Exam tip: In a thickness gauge the radiation must be partly absorbed. Choose the radiation whose range is comparable with the thickness being measured.
A student measures the count rate from an alpha source as a GM tube is moved away from it through air. Figure 1 shows the results, with no correction for background.
Use Figure 1 to estimate the range of the alpha particles in air.
Use Figure 1 to state the background count rate.
Explain why the count rate decreases between 2.0 cm and 3.75 cm, and why it then falls to the background value within a further 0.5 cm.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| about 4.2 cm (allow 4.0–4.4 cm) | ✓ 1 | |
| Part (b) | ||
| about 0.6 s−1 (allow 0.5–0.7 s−1) | ✓ 1 | |
| Part (c) | ||
| decrease up to 3.75 cm: the radiation spreads out as it travels, so a smaller fraction of the alpha particles enters the GM tube at greater distances | ✓ 1 | |
| fall to background: the alpha particles all start with about the same energy and lose it by ionising the air at a similar rate, so they are all stopped at about the same distance (the range) | ✓ 1 | |
Exam tip: Alpha particles have a definite range in air because they ionise strongly and all lose their energy over about the same distance.
Engineers suspect that a water pipe buried 1 m underground is leaking. They add a small amount of a radioactive isotope to the water and, a few hours later, move a detector along the ground above the pipe. Figure 1 shows the count rate recorded.
State the position of the leak.
State the background count rate.
Explain why the isotope must emit gamma radiation.
Suggest a suitable half-life for the isotope, and explain your choice.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at 34 m along the pipe (allow 33–35 m) | ✓ 1 | |
| Part (b) | ||
| about 0.8 s−1 (allow 0.7–0.9 s−1) | ✓ 1 | |
| Part (c) | ||
| the radiation has to pass through about 1 m of soil to reach the detector | ✓ 1 | |
| alpha and beta radiation would be absorbed by the soil (and the pipe), only gamma is penetrating enough | ✓ 1 | |
| Part (d) | ||
| a few hours to a few days | ✓ 1 | |
| long enough for the activity to remain detectable while the water spreads and the survey is done, but short enough that the activity left in the ground and water soon becomes negligible | ✓ 1 | |
Exam tip: For a tracer, justify the half-life both ways: long enough to do the job, short enough not to leave radioactive material behind.
A nucleus decays by β+ emission.
What are the changes in its proton number and its neutron number?
| Answer | Mark | Notes |
|---|---|---|
| In β+ decay a proton changes into a neutron, emitting a positron and an electron neutrino: proton number −1, neutron number +1, nucleon number unchanged. | ✓ 1 | Answer B |
Answer: B
Common error: Mixing up β− and β+. The sign tells you the charge carried away: a positive positron leaves, so the nucleus loses one positive charge.
An isotope has a half-life of 2.5 minutes.
What is its decay constant?
| Answer | Mark | Notes |
|---|---|---|
| λ = ln 2/t½ = 0.693 ÷ (2.5 × 60 s) = 4.6 × 10−3 s−1. | ✓ 1 | Answer A |
Answer: A
Common error: Leaving the half-life in minutes when the answer is wanted in s−1.
A sealed source contains 3.0 × 1012 nuclei of cobalt-60, which has a half-life of 5.3 years. 1 year = 3.15 × 107 s.
What is the activity of the source?
| Answer | Mark | Notes |
|---|---|---|
| λ = 0.693 ÷ (5.3 × 3.15 × 107) = 4.15 × 10−9 s−1. A = λN = 4.15 × 10−9 × 3.0 × 1012 = 1.2 × 104 Bq. | ✓ 1 | Answer A |
Answer: A
Exam tip: For an activity in becquerels, the decay constant must be in s−1: convert the half-life to seconds.
Which statement about radioactive decay is correct?
| Answer | Mark | Notes |
|---|---|---|
| Decay is random: every undecayed nucleus has the same constant probability per unit time of decaying (the decay constant λ). It is spontaneous, so heating has no effect, and nuclei do not “age”. | ✓ 1 | Answer D |
Answer: D
Exam tip: Random: you cannot predict which nucleus decays or when. Spontaneous: external conditions do not affect it.
Bismuth-212, 21283Bi, decays in two ways. About two-thirds of its nuclei emit a β− particle and form an isotope of polonium (Po); the rest emit an α particle and form an isotope of thallium (Tl).
Write a nuclear equation for the β− decay.
Write a nuclear equation for the α decay.
The polonium isotope then decays by α emission and the thallium isotope decays by β− emission. Show that both routes end with the same nuclide, and state its nucleon number and proton number.
The β− decay of the thallium is followed by the emission of a gamma photon. Explain why this does not change the nuclide.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 21283Bi → 21284Po + … | ✓ 1 | |
| … + 0−1β + νe (electron and antineutrino) | ✓ 1 | |
| Part (b) | ||
| 21283Bi → 20881Tl + 42α | ✓ 1 | |
| Part (c) | ||
| Po: 212 − 4 = 208 and 84 − 2 = 82; Tl: 208 and 81 + 1 = 82 | ✓ 1 | |
| both give nucleon number 208, proton number 82 (lead-208) | ✓ 1 | |
| Part (d) | ||
| a gamma photon has no charge and no nucleons, so A and Z are unchanged; the nucleus only loses energy as it drops from an excited state | ✓ 1 | |
Model sketch
Exam tip: Balance the top numbers (nucleons) and the bottom numbers (charge) separately, and include the antineutrino in β− decay.
To find which radiations a source emits, a student places different absorbers between the source and a GM tube 2 cm away and records the count rate. The background count rate is 30 min−1.
no absorber: 512 min−1 · paper: 509 min−1 · 3 mm aluminium: 140 min−1 · 10 mm aluminium: 138 min−1 · 10 mm aluminium and 25 mm lead: 70 min−1
Explain whether the source emits alpha radiation.
Explain what the readings show about beta and gamma radiation from the source.
Calculate the percentage of the gamma radiation reaching the lead that passes through the 25 mm of lead.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| no: the paper causes no significant fall in count rate (512 → 509 is within random variation), and at 2 cm alpha would reach the tube | ✓ 1 | |
| Part (b) | ||
| 3 mm of aluminium causes a large fall in count rate, so beta is emitted | ✓ 1 | |
| the count rate is the same (within random variation) for 3 mm and 10 mm of aluminium, so all the beta is absorbed by 3 mm | ✓ 1 | |
| the count rate behind aluminium is still well above background and is reduced further by lead, so gamma is emitted | ✓ 1 | |
| Part (c) | ||
| corrected rates: 138 − 30 = 108 min−1 and 70 − 30 = 40 min−1 | ✓ 1 | |
| 40 ÷ 108 × 100 = 37% | ✓ 1 | |
Common error: Comparing uncorrected count rates. Subtract the background from both readings before finding a fraction or percentage.
Iodine-123 is used as a tracer to image the thyroid gland. It emits gamma radiation and has a half-life of 13 hours. A patient is given iodine-123 with an activity of 20 MBq.
Explain two reasons why a gamma emitter with a half-life of several hours is suitable for this use.
Calculate the decay constant of iodine-123 in s−1.
Calculate the number of iodine-123 nuclei given to the patient.
Calculate the activity of the iodine-123 in the patient 2.0 days later.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gamma passes out of the body to the detector / is weakly ionising so causes little damage to tissue | ✓ 1 | |
| half-life long enough to complete the scan but short enough that the activity in the body soon becomes very small | ✓ 1 | |
| Part (b) | ||
| λ = 0.693 ÷ (13 × 3600) = 1.5 × 10−5 s−1 (1.48 × 10−5 s−1) | ✓ 1 | |
| Part (c) | ||
| N = A/λ = 2.0 × 107 ÷ 1.48 × 10−5 | ✓ 1 | |
| = 1.4 × 1012 (1.35 × 1012) | ✓ 1 | |
| Part (d) | ||
| A = 20 e−λt with t = 48 h = 1.73 × 105 s, or 20 × (½)48/13 | ✓ 1 | |
| = 1.5 MBq (1.55 MBq) | ✓ 1 | |
Exam tip: Use λ in s−1 with A in Bq. For A = A0e−λt, hours work too, as long as λ and t match.
Carbon-14 is a β− emitter with a half-life of 5730 years. Archaeologists use it to date organic remains.
Explain why the activity per gram of carbon in a living tree stays constant, but falls after the tree dies.
Charcoal from an ancient hearth has one quarter of the activity per gram of carbon of living wood. State its age.
A wooden bowl has 62% of the activity per gram of carbon of living wood. Calculate its age.
Suggest why carbon-14 cannot be used to date a sample thought to be 200 000 years old.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| while alive the tree keeps taking in carbon (as carbon dioxide) with a constant proportion of carbon-14, replacing what decays | ✓ 1 | |
| after death no new carbon-14 is taken in, so the carbon-14 decays and is not replaced | ✓ 1 | |
| Part (b) | ||
| two half-lives: 2 × 5730 = 1.15 × 104 years (11 460 years) | ✓ 1 | |
| Part (c) | ||
| λ = 0.693 ÷ 5730 = 1.21 × 10−4 year−1; 0.62 = e−λt | ✓ 1 | |
| t = ln(1/0.62) ÷ 1.21 × 10−4 = 4.0 × 103 years (3950 years) | ✓ 1 | |
| Part (d) | ||
| after about 35 half-lives the remaining activity is far too small to measure against the background count / random fluctuations | ✓ 1 | |
Exam tip: In A = A0e−λt you can keep the half-life in years, as long as λ is in year−1 and the answer is in years.
A sealed beta source contains 1.5 µg of strontium-90. The mass of one strontium-90 atom is 90 u and its half-life is 28.8 years. 1 u = 1.66 × 10−27 kg; 1 year = 3.15 × 107 s.
Show that the number of strontium-90 nuclei in the source is about 1 × 1016.
Calculate the activity of the source.
Calculate the activity of the strontium-90 10.0 years later.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mass of one atom = 90 × 1.66 × 10−27 = 1.49 × 10−25 kg | ✓ 1 | |
| N = 1.5 × 10−9 ÷ 1.49 × 10−25 = 1.0 × 1016 | ✓ 1 | |
| Part (b) | ||
| λ = 0.693 ÷ (28.8 × 3.15 × 107) = 7.64 × 10−10 s−1 | ✓ 1 | |
| A = λN = 7.7 × 106 Bq (7.67 × 106 Bq) | ✓ 1 | |
| Part (c) | ||
| A = 7.67 × 106 × e−(0.693 × 10.0/28.8) = 6.0 × 106 Bq (allow 6.1 × 106 Bq from the rounded 7.7 × 106 Bq) | ✓ 1 | |
Common error: Forgetting that 1 µg = 10−9 kg, not 10−6 kg.
The graph shows how the activity A of a sample of a radioactive isotope varies with time t.
What is the decay constant of the isotope?
| Answer | Mark | Notes |
|---|---|---|
| The activity falls from 96 kBq to 48 kBq in 6.0 min, and from 48 kBq to 24 kBq in the next 6.0 min, so t½ = 6.0 min = 360 s. λ = ln 2 / t½ = 0.693 ÷ 360 s = 1.9 × 10−3 s−1. | ✓ 1 | Answer B |
Answer: B
Common error: Using the half-life in minutes in λ = ln 2 / t½ and then quoting the answer in s−1. Convert to seconds first.
The chart shows a nuclide X and four other points P, Q, R and S on a graph of proton number Z against neutron number N. X decays by alpha emission, and the nucleus formed then undergoes two β− decays, one after the other.
Which point represents the final nuclide?
| Answer | Mark | Notes |
|---|---|---|
| X is at N = 130, Z = 86. Alpha decay: N − 2, Z − 2, giving (128, 84). Each β− decay turns a neutron into a proton: N − 1, Z + 1. After two of them: (126, 86), which is Q. Q has the same Z as X, so it is an isotope of the same element. | ✓ 1 | Answer B |
Answer: B
Exam tip: On an N–Z chart: α moves 2 left and 2 down, β− moves 1 left and 1 up, β+ moves 1 right and 1 down, γ does not move.
In an investigation of the absorption of gamma radiation by lead, a student plots the corrected count rate against the thickness x of lead between the source and the GM tube.
What thickness of lead reduces the corrected count rate to one-eighth of its value with no lead?
| Answer | Mark | Notes |
|---|---|---|
| With no lead the corrected count rate is 120 s−1; one-eighth of this is 15 s−1. The graph reaches 15 s−1 at 24 mm. Check: the count rate halves every 8.0 mm (120 → 60 → 30 → 15), and three halvings give one-eighth. | ✓ 1 | Answer C |
Answer: C
Exam tip: A constant halving thickness is the absorption version of half-life. One-eighth = (½)3, so three halving thicknesses.
A student places different absorbers, in turn, between a radioactive source and a GM tube 1 cm away. The bar chart shows the count rates after correcting for background.
Which radiations does the source emit?
| Answer | Mark | Notes |
|---|---|---|
| Paper reduces the count rate from 50 to 18 s−1, so alpha is present. 5 mm of aluminium causes no further fall, so there is no beta. The 18 s−1 that passes through aluminium is reduced, but not stopped, by 25 mm of lead, so it is gamma. The source emits alpha and gamma only. | ✓ 1 | Answer C |
Answer: C
Exam tip: Look at the change between neighbouring bars: the fall caused by each absorber tells you what that absorber has stopped.
Carbon-14 has a half-life of 5730 years. The graph shows how the activity per gram of carbon in dead organic material, as a fraction of the value in living material, changes with the age of the material.
A wooden tool has 0.30 of the activity per gram of carbon of living wood. What is the age of the tool?
| Answer | Mark | Notes |
|---|---|---|
| Reading across from 0.30 to the curve gives an age of about 10 × 103 years. Check: e−λt = 0.30 with λ = ln 2 / 5730 y, so t = ln (1/0.30) ÷ (ln 2 / 5730) = 9950 years ≈ 1.0 × 104 years. | ✓ 1 | Answer C |
Answer: C
Exam tip: A decay graph can be read in either direction: go across from the fraction to the curve, then down to the time axis.
Figure 1 shows part of a decay chain on a chart of proton number Z against neutron number N. Nuclide 1 is thorium (Th), nuclide 2 is radium (Ra) and nuclide 3 is actinium (Ac).
Use Figure 1 to identify the type of decay in the step 1 → 2 and in the step 2 → 3. Explain how you decided.
Write the nuclear equation for the step 2 → 3.
State the nucleon number and proton number of nuclide 5.
Nuclides 1 and 4 are both isotopes of thorium. Explain how Figure 1 shows this.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1 → 2: N and Z both fall by 2 (nucleon number falls by 4), so alpha decay | ✓ 1 | |
| 2 → 3: N falls by 1 and Z rises by 1 (nucleon number unchanged), so β− decay | ✓ 1 | |
| Part (b) | ||
| 22888Ra → 22889Ac (both nuclides with correct nucleon and proton numbers read from the chart) | ✓ 1 | |
| + 0−1β + νe (electron and electron antineutrino) | ✓ 1 | |
| Part (c) | ||
| nucleon number 224 and proton number 88 | ✓ 1 | |
| Part (d) | ||
| both have Z = 90 (the same number of protons) but different neutron numbers, 142 and 138 | ✓ 1 | |
Common error: Reading nucleon number straight off the N axis. The chart gives neutron number; nucleon number A = N + Z.
Fluorine-18 is used as a medical tracer in hospital scans. A dose of fluorine-18 is prepared with an activity of 400 MBq. Figure 1 shows how its activity A changes with time t after preparation.
Use Figure 1 to show that the half-life of fluorine-18 is about 110 minutes.
Calculate the decay constant of fluorine-18, in s−1.
The scan needs an activity of at least 150 MBq. Calculate the latest time after preparation, in minutes, at which the scan can start.
Fluorine-18, 189F, decays by β+ emission to an isotope of oxygen (O). Write the nuclear equation for this decay.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the activity falls from 400 MBq to 200 MBq at t = 110 min | ✓ 1 | |
| and from 200 MBq to 100 MBq between 110 min and 220 min — the same time interval again | ✓ 1 | |
| Part (b) | ||
| λ = ln 2 ÷ (110 × 60 s) = 1.05 × 10−4 s−1 | ✓ 1 | |
| Part (c) | ||
| 150 = 400 e−λt, so t = ln (400/150) ÷ λ | ✓ 1 | |
| t = 0.981 ÷ 1.05 × 10−4 = 9340 s = 155.7 min, so the latest start is 155 min (rounded down so that the activity is still at least 150 MBq; do not accept 156 min, when the activity has fallen below 150 MBq) | ✓ 1 | |
| Part (d) | ||
| 189F → 188O + 0+1β + νe | ✓ 1 | |
Exam tip: Showing a half-life from a graph is more convincing if you check two successive halvings take the same time.
To model radioactive decay, a class throws 600 dice. After each throw, every die showing a six is removed and the rest are thrown again. Figure 1 shows the class's results (crosses) and the number expected (dashed curve).
State the probability that a particular die is removed on any one throw, and state the quantity in radioactive decay that this probability represents.
Use the dashed curve to find the number of throws needed for the number of dice to halve.
Calculate the number of dice expected to remain after 10 throws.
Explain why the crosses do not lie exactly on the dashed curve.
Explain why, for a real radioactive sample, the scatter about the expected curve is much less noticeable than in this model.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1/6, which represents the decay constant / probability that a nucleus decays per unit time | ✓ 1 | |
| Part (b) | ||
| the curve reaches 300 at about 3.8 throws (allow 3.6–4.0) | ✓ 1 | |
| Part (c) | ||
| 600 × (5/6)10 = 97 | ✓ 1 | |
| Part (d) | ||
| whether a die shows a six is random, so the number removed in each throw varies about the expected value | ✓ 1 | |
| Part (e) | ||
| a real sample contains a very large number of nuclei, so the random fluctuations are a tiny fraction of the number decaying | ✓ 1 | |
Exam tip: The model matches decay because every die has the same fixed chance of 'decaying' on each throw, whatever has happened before.
A short-lived radioactive sample is placed near a GM tube. Figure 1 shows the measured count rate R, which includes the background, over the next 400 s.
State the background count rate, and explain how Figure 1 shows this value.
Determine the half-life of the sample. Show your working.
Another student halves the measured count rate at t = 0 without subtracting the background. State the half-life this student obtains and whether it is too large or too small.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2.0 s−1 (allow 2.0–2.2 s−1) | ✓ 1 | |
| at late times the sample has almost all decayed, so the count rate levels off at a constant value that is due to background alone | ✓ 1 | |
| Part (b) | ||
| corrected count rate at t = 0: 34 − 2.0 = 32 s−1 | ✓ 1 | |
| half of this is 16 s−1, which is a measured count rate of 16 + 2.0 = 18 s−1 | ✓ 1 | |
| R = 18 s−1 at t = 50 s (allow 48–52 s) | ✓ 1 | |
| Part (c) | ||
| R = 17 s−1 at about 55 s (allow 53–57 s), which is too large | ✓ 1 | |
Common error: Halving the measured count rate. Background does not decay, so subtract it first, halve, then add it back to read the graph.
A graph of ln (A/Bq) against time t for a radioactive sample is a straight line. The value of ln (A/Bq) falls from 9.20 at t = 0 to 6.43 at t = 12.0 hours.
What is the half-life of the sample?
| Answer | Mark | Notes |
|---|---|---|
| Gradient = (6.43 − 9.20) ÷ 12.0 = −0.231 h−1, so λ = 0.231 h−1. t½ = 0.693 ÷ 0.231 = 3.0 hours. | ✓ 1 | Answer B |
Answer: B
Exam tip: On a graph of ln A against t, the gradient is −λ. Then use t½ = ln 2/λ.
In an investigation of gamma absorption, every 12 mm of lead halves the corrected count rate. With no lead the corrected count rate is 160 s−1.
What is the corrected count rate with 30 mm of lead?
| Answer | Mark | Notes |
|---|---|---|
| 30 mm = 30 ÷ 12 = 2.5 half-thicknesses. C = 160 × (½)2.5 = 28 s−1. | ✓ 1 | Answer C |
Answer: C
Exam tip: Absorption by lead follows the same exponential pattern as decay with time, so a fractional number of “halvings” is allowed: use (½)n with n = 2.5.
A student measures the count rate R from a short-lived radioactive sample at 30 s intervals. The background count rate is 0.6 s−1.
t / s: 0, 30, 60, 90, 120, 150
R / s−1: 25.6, 18.8, 14.0, 10.3, 7.7, 5.7
Explain why the background count rate must be subtracted before the data are analysed.
Calculate ln (C / s−1) for t = 90 s, where C is the corrected count rate.
Explain why a graph of ln C against t should be a straight line, and state what its gradient represents.
Plot a graph of ln C against t and use it to determine the half-life of the sample.
Explain why the points do not all lie exactly on the line.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the background does not decay with the sample, so including it would make the count rate appear to fall more slowly / the ln graph would not be a straight line | ✓ 1 | |
| Part (b) | ||
| C = 10.3 − 0.6 = 9.7 s−1 | ✓ 1 | |
| ln 9.7 = 2.27 | ✓ 1 | |
| Part (c) | ||
| C = C0e−λt, so ln C = ln C0 − λt, which has the form y = mx + c | ✓ 1 | |
| the gradient is −λ (minus the decay constant) | ✓ 1 | |
| Part (d) | ||
| straight best-fit line through the points; gradient ≈ −0.0106 s−1 (accept −0.0100 to −0.0112 s−1) | ✓ 1 | |
| t½ = 0.693 ÷ 0.0106 = 65 s (accept 62–69 s) | ✓ 1 | |
| Part (e) | ||
| radioactive decay is random, so the number of decays in each counting period fluctuates about the expected value | ✓ 1 | |
Model sketch
Common error: Taking the log of the measured count rate instead of the corrected one. Background is constant, so it must come off every reading first.
A student investigates how the count rate of gamma radiation from a sealed source depends on the thickness of lead placed between the source and a GM tube.
Describe a method the student should use, including how to make the results reliable.
State two safety precautions the student should take.
The corrected count rate is 36.0 s−1 with no lead and 11.5 s−1 with 20.0 mm of lead. Assuming the count rate falls exponentially with thickness, calculate the thickness of lead that halves the count rate.
Explain why the GM tube must not be moved closer to the source as sheets are added.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| measure the background count over several minutes with the source away, and subtract it from every reading | ✓ 1 | |
| keep the source–GM tube distance fixed and record the count over a fixed, long enough time for each thickness | ✓ 1 | |
| measure the lead thickness with a micrometer at several points and take the mean; add sheets to give a range of thicknesses | ✓ 1 | |
| repeat each count and average, then plot ln (corrected count rate) against thickness | ✓ 1 | |
| Part (b) | ||
| any two from: handle the source with tongs or a holder / keep it pointing away from people and at a distance / return it to its lead-lined box when not in use / minimise the time of exposure / wear gloves or wash hands after handling lead | ✓ 2 | |
| Part (c) | ||
| μ = ln(36.0 ÷ 11.5) ÷ 20.0 = 0.0571 mm−1 | ✓ 1 | |
| half-thickness = ln 2 ÷ 0.0571 = 12.1 mm | ✓ 1 | |
| Part (d) | ||
| the count rate also depends on the distance from the source (gamma intensity falls with distance), so the distance must be a control variable for thickness to be the only factor changing | ✓ 1 | |
Exam tip: In a method description, name the measuring instrument for each quantity, the control variables, and how background and randomness are dealt with.
Radioactive decay is described as both random and spontaneous, yet a large sample of a radioactive isotope always has the same half-life.
Explain what is meant by the random and spontaneous nature of radioactive decay, and explain how a large sample can nevertheless decay with a constant half-life.
| Answer | Mark | Notes |
|---|---|---|
| Indicative content: | — | |
| random: it is not possible to predict which nucleus will decay next or when a particular nucleus will decay | ✓ 1 | |
| spontaneous: decay is not affected by external factors such as temperature, pressure or chemical state | ✓ 1 | |
| each nucleus has the same constant probability of decaying per unit time, the decay constant λ | ✓ 1 | |
| so the expected number decaying per second is proportional to the number left: dN/dt = −λN / A = λN | ✓ 1 | |
| this gives exponential decay, N = N0e−λt, in which the time to halve, ln 2/λ, is the same whatever the starting number | ✓ 1 | |
| with very large numbers of nuclei the random fluctuations are tiny compared with the total, so the sample follows the law closely; small samples show visible scatter | ✓ 1 |
Exam tip: In an extended answer, define both words separately, then link probability per unit time to A = λN and to the constant half-life.
For a radioactive sample, the graph shows ln (A / Bq) against time t, where A is the activity.
What is the activity of the sample at t = 4.0 h?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, at t = 4.0 h, ln (A / Bq) = 10.1 (the line falls 0.35 per hour from 11.5). A = e10.1 Bq = 2.4 × 104 Bq. | ✓ 1 | Answer A |
Answer: A
Common error: Undoing a natural log with 10x. The inverse of ln is ex; the inverse of lg is 10x.
A nucleus 14158X decays by β− emission to an excited state of a nucleus Y. Nucleus Y then reaches its ground state by emitting gamma photons. The diagram shows the energy levels of Y.
Which row gives the nucleon number and proton number of Y and the highest frequency of the gamma photons emitted?
| Answer | Mark | Notes |
|---|---|---|
| In β− decay a neutron becomes a proton: the nucleon number stays 141 and the proton number rises from 58 to 59. Emitting gamma photons changes neither number. The highest-energy photon comes from the 1.17 MeV level straight to the ground state: E = 1.17 × 106 × 1.60 × 10−19 = 1.87 × 10−13 J; f = E/h = 1.87 × 10−13 ÷ 6.63 × 10−34 = 2.8 × 1020 Hz. The answer is 141, 59, 2.8 × 1020 Hz. | ✓ 1 | Answer D |
Answer: D
Common error: Lowering the proton number in β− decay. A β− particle carries charge −e away, so the nucleus gains a proton: Z rises by 1 and A is unchanged. Gamma emission changes neither.
A student investigates the absorption of gamma radiation by lead. For each thickness x of lead, the student records the count, calculates the count rate and subtracts the background count rate of 0.45 s−1 to give the corrected count rate C. Figure 1 shows the student's graph of ln (C / s−1) against x, with a best-fit line.
With 8.0 mm of lead the GM tube recorded 2790 counts in 120 s. Show that the plotted value of ln (C / s−1) at 8.0 mm is correct.
Use the best-fit line to determine the absorption coefficient μ of lead, where C = C0e−μx.
Calculate the thickness of lead that halves the corrected count rate.
Explain why the student plotted ln C against x rather than C against x.
Calculate the thickness of lead needed to reduce the corrected count rate to 1.0% of its value with no lead.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| count rate = 2790 ÷ 120 = 23.25 s−1; C = 23.25 − 0.45 = 22.8 s−1 | ✓ 1 | |
| ln 22.8 = 3.13, which matches the point at 8.0 mm | ✓ 1 | |
| Part (b) | ||
| gradient from a large triangle using points read from the line, e.g. (0, 3.96) and (22.0, 1.60) | ✓ 1 | |
| gradient = −0.107 mm−1, so μ = 0.107 mm−1 (allow 0.103–0.112 mm−1) | ✓ 1 | |
| Part (c) | ||
| x½ = ln 2 / μ = 0.693 ÷ 0.107 = 6.5 mm (allow 6.2–6.7 mm; allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| ln C = ln C0 − μx is a straight line, so a linear best-fit line can be drawn and μ found from its gradient / the exponential relationship is confirmed if the points lie on a straight line | ✓ 1 | |
| Part (e) | ||
| e−μx = 0.010, so x = ln 100 / μ | ✓ 1 | |
| x = 4.61 ÷ 0.107 = 43 mm (allow 41–45 mm; allow error carried forward) | ✓ 1 | |
Exam tip: Take the gradient from the best-fit line, not from two data points, and use a triangle that covers most of the line.
Figure 1 shows how the number N of undecayed nuclei in a sample of a radioactive isotope changes with time t. A tangent to the curve has been drawn at t = 20 h.
Use Figure 1 to show that the half-life of the isotope is about 14 hours.
Use the tangent to determine the activity of the sample at t = 20 h.
Use your answer to (b) and a value of N read from Figure 1 to determine the decay constant, and hence the half-life. Comment on your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| initial value of N = 8.0 × 1018, so half of it is 4.0 × 1018 | ✓ 1 | |
| N reaches 4.0 × 1018 at t ≈ 14 h (allow 13–15 h) | ✓ 1 | |
| Part (b) | ||
| gradient = −5.9 × 1018 ÷ 40 h = −1.48 × 1017 h−1 | ✓ 1 | |
| A = 1.48 × 1017 ÷ 3600 s = 4.1 × 1013 Bq (allow 4.0–4.2 × 1013 Bq) | ✓ 1 | |
| Part (c) | ||
| N at 20 h = 2.9 × 1018 (allow 2.9–3.0 × 1018); λ = A/N = 4.1 × 1013 ÷ 2.9 × 1018 = 1.4 × 10−5 s−1 | ✓ 1 | |
| t½ = ln 2 / λ = 4.9 × 104 s ≈ 14 h (allow 13–15 h) | ✓ 1 | |
| this agrees with the value from (a), within the uncertainty of drawing and reading the tangent | ✓ 1 | |
Exam tip: The gradient of an N–t graph is dN/dt = −λN. Its size is the activity, but convert hours to seconds to get becquerels.
Some rocks are dated using rubidium-87, 8737Rb, which decays by β− emission to a stable isotope of strontium (Sr). Figure 1 shows how the numbers of rubidium-87 and strontium-87 nuclei in a closed sample change with time, as fractions of the initial number of rubidium-87 nuclei.
Write the nuclear equation for the decay of rubidium-87.
Explain why the two curves cross at the half-life, and use Figure 1 to state the half-life of rubidium-87.
A rock contains 0.050 strontium-87 nuclei for every rubidium-87 nucleus. Assuming the rock contained no strontium-87 when it formed, calculate its age.
State one other assumption made in this calculation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8737Rb → 8738Sr | ✓ 1 | |
| + 0−1β + νe | ✓ 1 | |
| Part (b) | ||
| each rubidium nucleus that decays becomes one strontium nucleus, so when half have decayed the two numbers are equal (both 0.50) | ✓ 1 | |
| t½ = 4.9 × 1010 years (allow 4.8–5.0 × 1010 years) | ✓ 1 | |
| Part (c) | ||
| NRb/N0 = 1/(1 + 0.050) = 0.952 | ✓ 1 | |
| λ = ln 2 ÷ 4.9 × 1010 = 1.41 × 10−11 year−1 | ✓ 1 | |
| t = ln (1.050) ÷ λ = 3.4 × 109 years (allow 3.5 × 109 years from the rounded value of λ; allow error carried forward from (b)) | ✓ 1 | |
| Part (d) | ||
| no rubidium-87 or strontium-87 has entered or left the rock since it formed | ✓ 1 | |
Common error: Taking the measured ratio 0.050 as the fraction decayed. The fraction remaining is NRb/(NRb + NSr).
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