Radioactivity, decay and half-life: worked solutions

Answer: Corrected rate = (2.70 ± 0.17) s⁻¹, using the supplied standard-uncertainty model. Activity cannot be inferred without detection information; longer counting reduces relative random uncertainty.

1. rg = 420/120 = 3.50 s⁻¹;
   rb = 480/600 = 0.800 s⁻¹;
   r = rg − rb = 2.70 s⁻¹.
2. σr = √(420/120² + 480/600²)
   = 0.1746… s⁻¹ ≈ 0.17 s⁻¹.
   Report (2.70 ± 0.17) s⁻¹ with the value and uncertainty at matching decimal precision.
3. Count rate describes events registered by the detector. Activity is the mean number of nuclear decays per second in the source. Geometry, absorption, detector efficiency and the number/type of emissions per decay affect the registered fraction; they have not been specified.
4. Count for longer, or combine repeated equal-time measurements into a longer total, while keeping the geometry and source activity effectively constant. Larger total counts reduce relative fluctuations. A longer background measurement helps its contribution, although here the source-present term dominates.

Check: The corrected rate is less than the gross rate. The relative standard uncertainty is about 6.5%; this is a statistical estimate, not a guaranteed bound on the result.

Mark checklist · 6 marks

• 1 mark — Calculates both rates using their own counting intervals.
• 1 mark — Subtracts background to obtain 2.70 s⁻¹.
• 1 mark — Uses the supplied independent-count uncertainty expression with original counts.
• 1 mark — Obtains about 0.17 s⁻¹ and quotes matching precision, e.g. (2.70 ± 0.17) s⁻¹.
• 1 mark — Distinguishes source decays/activity from registered counts, with a relevant detection factor.
• 1 mark — Proposes longer total counting/repeated counting under fixed conditions to reduce relative random uncertainty.

Common mistake: Subtracting 480 from 420 would combine different durations. A larger background count can still correspond to a smaller background rate. Do not relabel a detector rate as activity without a detection model.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/3A, November 2021 — Q01.2–01.3; report pp. 3 and 6

Answer: N0 = 1.61 × 10¹⁶; λ = 7.33 × 10⁻¹⁰ s⁻¹; A0 = 1.18 × 10⁷ Bq (11.8 MBq). Undecayed P after 90 years = 0.300 μg; retained daughters remain part of the sample.

1. m = 2.40 × 10⁻⁶ g;
   n = (2.40 × 10⁻⁶)/90.0 = 2.67 × 10⁻⁸ mol;
   N0 = nNA = 1.61 × 10¹⁶ nuclei.
2. T½ = 30.0 × 3.1536 × 10⁷ = 9.4608 × 10⁸ s;
   λ = ln2/T½ = 7.33 × 10⁻¹⁰ s⁻¹.
3. A0 = λN0 = 1.18 × 10⁷ Bq = 11.8 MBq. Retain unrounded N0 and λ until the final result.
4. mP = 2.40 × (½)³ = 0.300 μg after 90.0 years.
5. This last value counts only undecayed parent P. Most of the original nuclei have become daughter nuclei rather than disappeared from the sample. With retained products and the stated negligible mass-loss approximation, the total sample mass remains approximately its initial value.

Check: After three half-lives both parent mass and parent activity are one eighth of their initial values. Dividing A0 by N0 recovers λ, independently checking the conversion.

Mark checklist · 6 marks

• 1 mark — Uses m/M and NA consistently to obtain about 1.61 × 10¹⁶ nuclei.
• 1 mark — Converts the half-life to seconds and obtains λ ≈ 7.33 × 10⁻¹⁰ s⁻¹.
• 1 mark — Uses A = λN with the number of parent nuclei, not the nucleon number.
• 1 mark — Obtains A0 ≈ 1.18 × 10⁷ with unit Bq.
• 1 mark — Uses three half-lives to obtain 0.300 μg of undecayed parent.
• 1 mark — Explains that retained daughter nuclei remain in the sample; parent mass is not total sample mass.

Common mistake: N in A = λN is the number of undecayed nuclei, not the isotope’s mass number. The parent mass decays exponentially; the mass of the entire retained sample does not halve.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2018 — Q06.5–06.6; report p. 6 · AQA 7408/2, June 2025 — Q03.1–03.2; report p. 6

Answer: T½ = 3.00 min = 180 s; λ = 3.85 × 10⁻³ s⁻¹. A gross rate of 22.0 s⁻¹ is reached after 12.0 min.

1. Half-life is the time for the number of undecayed nuclei, or the activity of a single parent isotope, to fall to half its initial value. With fixed detection conditions its corrected count rate has the same half-life.
2. Source-only rates: 172 − 12 = 160;
   92 − 12 = 80; 52 − 12 = 40; 32 − 12 = 20 s⁻¹.
   Each halves over 3.00 min.
3. λ = ln2/180 = 3.85 × 10⁻³ s⁻¹. Using 3 directly would give min⁻¹, not s⁻¹.
4. Target rnet = 22.0 − 12.0 = 10.0 s⁻¹;
   10/160 = 1/16 = (½)⁴;
   t = 4T½ = 12.0 min.
5. The late-time gross rate tends towards the background floor. A target below that floor cannot be reached by this expected gross curve; random measured rates can still fluctuate.

Check: At 12 min the source contribution is 10 s⁻¹ and adding 12 s⁻¹ background gives the stated 22 s⁻¹. Successive halving intervals are equal despite progressively smaller drops.

Mark checklist · 7 marks

• 1 mark — Defines half-life using the undecayed population or activity falling by half.
• 1 mark — Corrects gross rates by subtracting 12.0 s⁻¹.
• 1 mark — Finds a consistent half-life of 3.00 min from at least one halving interval.
• 1 mark — Converts the half-life to 180 s.
• 1 mark — Uses λ = ln2/T½ to obtain 3.85 × 10⁻³ s⁻¹.
• 1 mark — Corrects the target to 10.0 s⁻¹ and recognises four halvings.
• 1 mark — Obtains t = 12.0 min.

Common mistake: Half of the gross initial rate is 86 s⁻¹. That includes the unchanged background and would overestimate the half-life. Halve the rate measured above the background floor.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2018 — Q06.1; report pp. 5–6

Answer: Gradient = −2.50 × 10⁻³ s⁻¹; λ = 2.50 × 10⁻³ s⁻¹; T½ = 277 s. Intercept = 4.79, giving r0 ≈ 120 s⁻¹. The base-10 gradient is −1.09 × 10⁻³ s⁻¹.

1. The exponential law applies to the source-only rate. A gross rate r = rsource + rbackground is not a single exponential tending to zero; taking its logarithm can produce a curved graph and a biased decay constant.
2. r = r₀e^−λt ⇒ ln[r/(1 s⁻¹)] = ln[r0/(1 s⁻¹)] − λt.
   The y-axis is dimensionless and gradient = −λ when t is in seconds.
3. Gradient = (3.89 − 4.69)/(360 − 40)
   = −2.50 × 10⁻³ s⁻¹;
   λ = 2.50 × 10⁻³ s⁻¹;
   T½ = ln2/λ = 277 s.
4. Intercept = 4.69 − (−0.00250 × 40) = 4.79;
   r0 = e^4.79 × 1 s⁻¹ = 120 s⁻¹ (about 120.3 before rounding).
5. log₁₀(r/rref) = log₁₀(r0/rref) − (λ/ln10)t;
   gradient = −0.00250/ln10
   = −1.09 × 10⁻³ s⁻¹.
6. A logarithmically spaced axis labelled with actual rates is different from an axis labelled with the numerical logarithms of the rates. Read its labels before using a gradient. The given task explicitly plots natural-log numbers.

Check: A 320 s interval reduces the fitted rate by e⁻⁰·⁸ = 0.449; it must therefore be somewhat longer than one half-life. Using base 10 or e gives the same physical λ once converted.

Mark checklist · 8 marks

• 1 mark — Explains that unchanged additive background spoils a single-exponential source-only log fit.
• 1 mark — Takes natural logs of the source decay law correctly.
• 1 mark — Identifies gradient = −λ with time in seconds.
• 1 mark — Uses the two fitted-line coordinates to find λ = 0.00250 s⁻¹.
• 1 mark — Finds T½ ≈ 277 s.
• 1 mark — Finds intercept 4.79 and exponentiates it to estimate r0 ≈ 120 s⁻¹.
• 1 mark — Derives the base-10 gradient as −λ/ln10.
• 1 mark — Obtains −1.09 × 10⁻³ s⁻¹, with the correct sign and unit.

Common mistake: The negative slope is not a negative decay constant. On a log₁₀ plot λ is not simply minus the gradient. The intercept is a logarithm of a rate, not the rate itself.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2024 — Q07.1–07.2; report p. 8

Answer: Use fixed alignment and known active positions, long background and repeated timed source measurements, then fit corrected rate against 1/d². A straight line through the origin within uncertainty supports the model over the tested range. Long tongs increase handling distance.

1. Clamp the source and detector in alignment. Keep their relative orientation and detector settings fixed. Change only their separation, using positions well beyond their physical size so the point-source/small-detector approximation is reasonable.
2. Use the known active positions, not arbitrary case edges. Read their ruler positions and subtract, using a set-square to project each position to the ruler and viewing square-on to limit parallax. Record any known case-to-active-position offsets.
3. Measure background with the source stored well away for a long timed interval. At each of at least five well-spaced distances, record repeated counts and their times. Combine equal-time repeats into a mean or total rate; use rg = total count / total time, then subtract rb. Longer counting is particularly useful where the net rate is small.
4. Plot rnet / s⁻¹ against 1/d² / m⁻².
   rnet = k(1/d²), so the expected fit is a straight line through the origin, with gradient k in m² s⁻¹.
5. Use a fitted line and measurement uncertainties. An intercept consistent with zero and a linear trend support the relationship over the measured range. A line that is merely straight with a substantial intercept is insufficient. Check that the detector is not saturating and that distance uncertainty is not dominant at the closest point.
6. Use long tongs for necessary source handling, following the supervised procedure. They increase distance from the source, reducing received intensity in the point-source model. Keep handling time short and return the source to its designated shielding when it is not needed. Adjust the detector rather than repeatedly approaching the source.

Check: For this geometric model doubling d quarters the source-only rate. It does not quarter the unchanged background. A half-life much longer than the measurement period keeps decay from confounding the distance change.

Mark checklist · 7 marks

• 1 mark — Keeps alignment, orientation and detector conditions fixed while varying separation in a suitable geometric range.
• 1 mark — Measures active-source to active-detector distance, accounting for offsets and reducing parallax.
• 1 mark — Measures a timed background with the source away and subtracts its rate, not arbitrary counts.
• 1 mark — Records timed source counts at multiple distances with repeats/longer counting to reduce random fluctuations.
• 1 mark — Plots corrected rate against 1/d² and tests for a fitted line through the origin.
• 1 mark — Judges agreement within uncertainty over the tested range and identifies a concrete limitation, e.g. dead time or finite geometry.
• 1 mark — Gives a suitable precaution with its exposure-reducing mechanism, e.g. long tongs increasing distance.

Common mistake: “A straight line proves inverse square” omits the origin for rnet against 1/d². “Be careful” or “avoid human error” gives neither a procedure nor a reason. Gamma radiation can reach the operator from directions other than the source holder’s opening.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/3A, November 2021 — Q01.1 and 01.4; report p. 6 · AQA 7408/3A, November 2021 — Q01.2–01.3; report pp. 3 and 6 · AQA 7408/2, June 2024 — Q07.3–07.4; report pp. 8–9

Answer: rb = 1.20 s⁻¹; corrected rates = 12.0, 5.33, 3.00 and 1.917 s⁻¹. k = 0.480, 0.480, 0.480 and 0.479 m² s⁻¹, consistent within the stated uncertainty. Log–log gradient = −2.00. Gross rate at 0.600 m = 2.53 s⁻¹.

1. rb = 1800/1500 = 1.20 s⁻¹.
   rnet = C/120 − 1.20 gives 12.0, 5.333…, 3.00 and 1.9167… s⁻¹.
2. k = rnet d² gives 0.480, 0.480, 0.480 and 0.47917… m² s⁻¹.
   The last differs from 0.480 by about 0.17%, far below the supplied conservative 10% rate uncertainty. All four rows support the inverse-square model over this distance range.
3. ln(rnet/rref) = ln[k/(rref dref²)] − 2ln(d/dref),
   where rref = 1 s⁻¹ and dref = 1 m.
   The expected log–log gradient is −2.
4. Two-point gradient = ln(3.00/12.0)/ln(0.400/0.200)
   = ln(¼)/ln2 = −2.00.
5. Gross rate equals k/d² + rb. Its product with d² is k + rb d², which increases with distance even when the source obeys the model. Background must be removed before the product or logarithmic test.
6. At 0.600 m, rnet = 0.480/0.600² = 1.333… s⁻¹;
   rgross = 1.333… + 1.20 = 2.53 s⁻¹.

Check: The 0.400 m reading has one quarter of the 0.200 m corrected rate. The corrected-rate versus 1/d² line and the log–log line test the same model; their gradients have different meanings and units.

Mark checklist · 8 marks

• 1 mark — Finds rb = 1.20 s⁻¹ and applies C/120 − rb.
• 1 mark — Obtains all four corrected rates, retaining sufficient precision.
• 1 mark — Calculates rnet d² for all four rows with appropriate units.
• 1 mark — Compares the approximately 0.17% spread with the supplied 10% uncertainty and concludes support over the tested range.
• 1 mark — Derives a log–log line with expected gradient −2.
• 1 mark — Calculates −2.00 from the two stated rate/distance ratios.
• 1 mark — Explains the unchanged additive background changes gross-rate products/logs.
• 1 mark — Predicts gross rate ≈ 2.53 s⁻¹ at 0.600 m.

Common mistake: A straight line on log–log axes shows a power law, but inverse square specifically needs gradient −2. Use all the table entries and quantify differences rather than merely saying the products look similar.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2018 — Q05.3–05.4; report p. 5 · AQA 7408/3A, November 2021 — Q01.2–01.3; report pp. 3 and 6 · AQA 7408/3A, November 2021 — Q01.6; report p. 6; checked marking p. 10

Answer: Using x gives products 0.500, 0.640 and 0.698 m² s⁻¹. Using d = x + 0.030 m gives 0.845, 0.846 and 0.845 m² s⁻¹, consistent within 1%. Random closely spaced photons can be missed during dead time.

1. rnet x² = 0.500, 0.640 and 0.6984 m² s⁻¹.
   These increase by about 40% across the table, far beyond 1% uncertainty; x is not an adequate active separation.
2. d = x + 0.030 m ⇒ d = 0.130, 0.230, 0.330 m.
   rnet d² = 0.8450, 0.8464 and 0.8451 m² s⁻¹.
   The spread is only about 0.17%, so the corrected geometry supports inverse square within the stated uncertainty.
3. The radiation spreads from the active source before reaching the detecting region. Using the front face omits an internal path length. This is a systematic distance offset; collecting more photons does not remove it.
4. If another photon arrives during the stated blind interval following a registered event, it is not recorded. At high arrival rates such close arrivals are more common, so measured rate can understate the rate entering the detector.
5. Radioactive decay is random. A constant mean arrival rate describes an average over many intervals; photons may cluster, and other intervals may contain none. They are not emitted at perfectly regular spacings. The photons themselves are not undergoing radioactive decay.

Check: The offset fraction is 30% of the smallest x but 10% of the largest. That changing fractional error explains the apparent drift in the uncorrected products.

Mark checklist · 7 marks

• 1 mark — Obtains all three rnet x² products.
• 1 mark — Quantifies their variation as far larger than 1% and rejects x as the appropriate separation.
• 1 mark — Uses d = x + 0.030 m and calculates all three corrected products.
• 1 mark — Obtains products near 0.845 m² s⁻¹ and judges their approximately 0.17% spread consistent with 1%.
• 1 mark — Explains the active region lies behind the face, creating a systematic distance offset.
• 1 mark — Explains missed close arrivals during the supplied dead time reduce a high measured rate.
• 1 mark — Links random decay to uneven photon arrivals despite a constant mean rate.

Common mistake: The rates are already background corrected, so subtracting background again is not a remedy. “Human error” names no mechanism. Longer counting reduces fluctuations but does not move the active detector region.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2018 — Q05.3–05.4; report p. 5 · AQA 7408/3A, November 2021 — Q01.1 and 01.4; report p. 6

Answer: The source emits alpha, beta and gamma under the stated absorption model. Beta is appropriate for the foil monitor because its transmission is partial and sensitive to thickness. A source-only reference separates declining activity from changing foil attenuation.

1. Net counts = 2400, 1500, 120 and 12;
   net rates = 40.0, 25.0, 2.00 and 0.200 s⁻¹.
2. Paper removes a component while beta/gamma are unaffected in the supplied model: this identifies alpha. Aluminium removes a further component that passed through paper: this identifies beta. A remaining above-background component passes through aluminium and is reduced by lead: this identifies gamma. Lead attenuates rather than automatically eliminating all gamma radiation.
3. Choose beta for these supplied foil data. Its partial transmission allows a measurable rate change when thickness changes. Alpha is already fully absorbed; the stated gamma transmission changes too little around this thickness to give a sensitive monitor.
4. A lower registered rate can arise from either more attenuation by thicker foil or lower source activity. At fixed geometry measure a background-corrected reference with no foil, or use an equivalent fixed reference channel. Compare foil/reference corrected rates taken at approximately the same time.
5. In a simple stable-efficiency model rfoil = εA × transmission and rreference = εA. Their ratio isolates transmission; a lower source activity changes both rates. Calibration is needed to convert transmission into thickness.

Check: All counting times match, so subtracting 30 directly is legitimate. The 12-count excess after lead is small: the independent-count uncertainty estimate is √(42 + 30) ≈ 8.5 counts. Repeat or count longer before claiming a statistically secure residual. Gamma identification here comes from the much larger aluminium signal and its reduction by lead.

Mark checklist · 7 marks

• 1 mark — Identifies alpha from the component removed by paper under the supplied assumptions.
• 1 mark — Identifies beta from the further component removed by aluminium.
• 1 mark — Identifies gamma from the residual above-background signal reduced by lead.
• 1 mark — Chooses beta for the supplied foil-monitor data.
• 1 mark — Explains partial, thickness-sensitive transmission and why zero/nearly complete transmission is unsuitable here.
• 1 mark — Explains thicker foil and declining source activity can both reduce the reading.
• 1 mark — Uses a background-corrected no-foil/fixed-channel reference or equivalent activity-normalised comparison to separate the effects.

Common mistake: A large drop is evidence to interpret with absorber properties; it is not a universal identification rule without assumptions about energies and materials. “Beta is medium penetrating” needs to be linked to useful partial absorption by this foil.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2018 — Q05.1; report p. 5

Answer: N0 = NP + ND; parent fraction = 0.400. Age = 1.59 × 10⁹ years. Closed-system retention and no initial daughter are essential; uncorrected inherited daughter overestimates age.

1. N0 = NP + ND.
   Take NP = 2 units and ND = 3 units: N0 = 5 units.
   NP/N0 = 2/5 = 0.400.
2. 0.400 = 2^(−t/T½);
   t = T½ × ln(1/0.400)/ln2
   = 1.59 × 10⁹ years.
3. The system must be closed to parent and daughter since formation. Otherwise present numbers need not represent the original parent ledger. There must also be no inherited daughter, or its initial number must be known and subtracted. The stated one-daughter-per-original-parent relation and effectively rapid intermediate steps are also needed.
4. Inherited daughter would be wrongly counted as parent that decayed after formation. That makes the inferred N0 too large, the remaining parent fraction too small and the calculated age too large.

Check: 40% lies below one half but above one quarter, so the age must lie between one and two half-lives. The answer is 1.322 half-lives, within that interval.

Mark checklist · 7 marks

• 1 mark — Uses N0 = NP + ND under the stated assumptions.
• 1 mark — Obtains the surviving parent fraction 2/5 = 0.400.
• 1 mark — Uses an exponential/half-life equation with a consistent year time unit.
• 1 mark — Obtains age ≈ 1.59 × 10⁹ years.
• 1 mark — Explains retention/closed-system requirement for both parent and daughter.
• 1 mark — Explains no initial daughter, or correction for an independently known initial daughter amount.
• 1 mark — Explains inherited daughter counted as newly produced makes the inferred age too large.

Common mistake: D/P is not the fraction of parent remaining. Use P/(P + D), not P/D or D/(P + D). A measured isotope ratio only becomes an age when the model’s starting and retention assumptions hold.

Exam technique: Set up the parent/daughter ledger before using logs. State the starting and retention assumptions that turn an isotope ratio into an age.

Related examiner evidence: AQA 7408/2, June 2024 — Q32; report p. 9; checked marking p. 26

Answer: At 16 years: AX = 5.00 MBq, AY = 11.5 MBq, total = 16.5 MBq, above the criterion. At 20 years: 2.50 + 10.0 = 12.5 MBq, below the criterion. The mixture has no single constant half-life.

1. At 16.0 years:
   AX = 80.0 × 2⁻⁴ = 5.00 MBq;
   AY = 20.0 × 2^−0.8 = 11.5 MBq;
   Atotal = 16.5 MBq.
2. At 20.0 years:
   AX = 80.0 × 2⁻⁵ = 2.50 MBq;
   AY = 20.0 × 2⁻¹ = 10.0 MBq;
   Atotal = 12.5 MBq.
3. 16.5 MBq is above 15.0 MBq, so the given activity criterion is not met at 16 years. 12.5 MBq is below 15.0 MBq, so it is met at 20 years. Both activities decrease continuously, so the crossing occurs between those times; these calculations do not locate its exact time.
4. The total is the sum of two exponentials with different decay constants. X decreases faster, so Y becomes an increasing fraction of the remaining activity. A constant fractional reduction of the whole mixture over every 4-year interval does not follow.
5. Using 4 years for all 100 MBq would predict 100 × 2⁻⁵ = 3.125 MBq at 20 years, far below the correct 12.5 MBq. The longer-lived component must be calculated separately.

Check: Initially X contributes 80% of activity; at 20 years it contributes only 20%. Changing component shares independently explain why the mixture’s apparent half-life changes.

Mark checklist · 7 marks

• 1 mark — Obtains AX = 5.00 MBq at 16 years.
• 1 mark — Obtains AY ≈ 11.5 MBq and total ≈ 16.5 MBq at 16 years.
• 1 mark — Obtains AX = 2.50 MBq at 20 years.
• 1 mark — Obtains AY = 10.0 MBq and total = 12.5 MBq at 20 years.
• 1 mark — Explicitly compares both totals with 15.0 MBq and judges fail/pass at the stated times.
• 1 mark — Explains the total is a sum of exponentials with different decay constants.
• 1 mark — Explains Y becomes a larger fraction, so the total does not have a constant 4-year half-life.

Common mistake: Activities add; half-lives do not add or average to a universal mixture half-life. Passing the stipulated activity criterion does not by itself calculate dose or decide every aspect of a storage design.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Answer: N0 = 1.20 × 10²⁴; λ = 5.49 × 10⁻¹⁰ s⁻¹; A0 = 6.62 × 10¹⁴ Bq; initial electrical power = 23.8 W. The 14.0 W threshold is reached after 30.7 years.

1. n = 0.400/0.200 = 2.00 mol;
   N0 = nNA = 1.20 × 10²⁴ nuclei.
2. T½ = 40.0 × 3.1536 × 10⁷ = 1.26144 × 10⁹ s;
   λ = ln2/T½ = 5.49 × 10⁻¹⁰ s⁻¹;
   A0 = λN0 = 6.62 × 10¹⁴ Bq.
3. Thermal power = A0E = 397 W;
   P0 = 0.0600 × A0E = 23.8 W.
4. P = P₀e^−λt, since efficiency and energy per decay are constant.
   t = ln(P0/14.0)/λ = 9.67 × 10⁸ s
   = 30.7 years, using unrounded P0 and λ.
5. The beacon can meet 14.0 W from the start up to this threshold time in the stated ideal model. Real changes in conversion efficiency or load would require a new budget; they are excluded here.

Check: 14.0/23.8 ≈ 0.588, so the threshold is reached before one 40-year half-life. Bq × J per decay gives J/s, the correct power unit. The output is much smaller than the thermal input.

Mark checklist · 8 marks

• 1 mark — Finds n = 2.00 mol using consistent mass units.
• 1 mark — Finds N0 ≈ 1.20 × 10²⁴ nuclei.
• 1 mark — Converts half-life to seconds and finds λ ≈ 5.49 × 10⁻¹⁰ s⁻¹.
• 1 mark — Uses A = λN to obtain A0 ≈ 6.62 × 10¹⁴ Bq.
• 1 mark — Uses activity × energy per decay to find thermal power ≈ 397 W.
• 1 mark — Applies efficiency 0.0600 to obtain electrical power ≈ 23.8 W.
• 1 mark — Uses an exponential power/activity ratio to find threshold time with a positive logarithm.
• 1 mark — Obtains about 30.7 years with the correct time conversion and interprets the threshold.

Common mistake: The full parent mass is not converted to energy on each decay. Use the supplied energy per decay. N means the number of undecayed nuclei, and 6.00% means multiply by 0.0600, not 6 or 0.600.

Exam technique: Build the chain N → A → thermal power → electrical power before solving for time. An explicit comparison with the required load completes the argument.

Related examiner evidence: AQA 7408/2, June 2025 — Q03.1–03.2; report p. 6

Answer: Expected survivors after eight throws ≈ 275; an actual integer count can differ. λeff = 0.134 throw⁻¹. Expected losses fall as the number of surviving dice falls; real spontaneous decay requires no external trigger.

1. Real decay is random: the exact decay time of a particular nucleus cannot be predicted. It is spontaneous: it occurs without an external trigger. The dice represent random removal with a fixed per-survivor probability; the need to throw them is a limitation of the analogy for spontaneity.
2. ⟨N8⟩ = 800(7/8)⁸ = 274.887… ≈ 275.
3. This is an ensemble expectation, not the guaranteed result of one run. Different throws and runs give different integer survivor counts. A larger starting population usually produces a smaller relative fluctuation about the expected curve.
4. e^−λ_eff = 7/8;
   λeff = −ln(7/8) = 0.1335… ≈ 0.134 throw⁻¹.
5. The exact discrete coefficient is not p = 0.125. The approximation λeff ≈ p only works well when the probability per step is small. The supplied definitions make the exact relation clear.
6. Expected removals on the first throw: 800/8 = 100.
   Expected survivors then: 700; expected second-throw removals: 700/8 = 87.5.
   The same probability acts on a declining population.

Check: The corresponding continuous interpolation has a half-life of ln2/λeff = 5.19 throws, although observations occur at whole throws. An expectation can be non-integer; one actual run cannot contain a fraction of a die.

Mark checklist · 6 marks

• 1 mark — Explains unpredictable individual decay time/randomness.
• 1 mark — Explains no external trigger/spontaneity and recognises dice mainly model random probability.
• 1 mark — Uses 800(7/8)⁸ to obtain expected survivors ≈ 275.
• 1 mark — Explains actual runs fluctuate and do not have to equal the ensemble expectation.
• 1 mark — Uses the supplied discrete/exponential equivalence to obtain λeff = −ln(7/8) ≈ 0.134 throw⁻¹.
• 1 mark — Explains constant probability gives fewer expected removals as the survivor population falls.

Common mistake: λeff = 1/8 is only a small-step approximation, not the exact coefficient in the equation given here. Do not put removed dice back into the next throw or expect exactly one eighth to be removed each time.

Exam technique: State what each quantity measures, correct background before processing the data, and keep time units consistent. These are original questions with PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/2, June 2024 — Q07.1–07.2; report p. 8