E.3 Radioactive Decay Questions | IB Physics SL 2025

1E-1A-08

Isotopes·E.3 Radioactive decay

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksState

Two nuclides are isotopes of the same element.

Which quantities must be the same for both?

Show mark scheme

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

Answer: A · 1 stage of work, one mark

Every option, and why

ACorrect: isotopes have the same number of protons, which fixes the element, but different numbers of neutrons.
BIsotopes have different nucleon numbers — that is what distinguishes them.
CIf both were the same the two nuclides would be identical, not isotopes.
DNuclides with the same neutron number but different proton numbers are different elements.

Syllabus understandingE.3 — isotopes, and nuclear notation ^A_ZX where A is the nucleon number and Z is the proton number Command term: State

2E-1A-09

Changes in the nucleus·E.3 Radioactive decay

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

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

Show mark scheme

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

Answer: D · 1 stage of work, one mark

Every option, and why

AThe alpha particle takes away 2 protons and 2 neutrons: the nucleon number falls by 4 but the neutron number falls by only 2.
BIn beta-minus decay a neutron becomes a proton: the proton number rises by 1 and the nucleon number is unchanged.
CThat describes beta-minus decay; in beta-plus decay a proton changes into a neutron.
DCorrect: a gamma photon carries away energy but no charge and no nucleons.

Syllabus understandingE.3 — the changes in the state of the nucleus following alpha, beta and gamma radioactive decay Command term: Identify

3E-1A-10

Beta decay & the neutrino·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

Which equation represents this decay?

Show mark scheme

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

Answer: D · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the radioactive decay equations involving α, β⁻, β⁺, γ; the existence of neutrinos ν and antineutrinos ν̄ Command term: Deduce

4E-1A-11

Penetration & ionization·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

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

Diagram NOT accurately drawn

RadiationReason

Show mark scheme

Marking point	Mark	Notes
Step 1The radiation must be partly absorbed by the sheet: alpha particles are stopped completely by paper, and gamma rays pass through almost unaffected.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Beta particles are partly absorbed by paper, so a small change in thickness gives a measurable change in the count rate.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays (real-life context: thickness of materials) Command term: Deduce

5E-1A-12

Choosing an isotope·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

Which row describes the most suitable tracer?

Radiation emittedHalf-life

Show mark scheme

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

Answer: C · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the activity, count rate and half-life in radioactive decay (real-life context: choice of isotope in medical use) Command term: Deduce

6E-1A-13

Background radiation·E.3 Radioactive decay

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

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

Show mark scheme

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

Answer: D · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the effect of background radiation on count rate; the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Deduce

7E-1A-14

Random & spontaneous decay·E.3 Radioactive decay

Paper 1AHard1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Radioactive decay is described as a random and spontaneous process.

Which statement describes what is meant by spontaneous?

Show mark scheme

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

Answer: B · 1 stage of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the random and spontaneous nature of radioactive decay Command term: Identify

8E-1A-15

The binding energy curve·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

Which statement is correct?

Graph drawn to scale

Show mark scheme

Marking point	Mark	Notes
Step 1The curve rises steeply to a maximum near A = 56 and falls slowly beyond it, so nuclei near the peak are the most tightly bound.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Energy is released when the products have a greater total binding energy than the reactants, i.e. move towards the peak: fusion of very light nuclei and fission of very heavy ones both do this.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the variation of the binding energy per nucleon with nucleon number Command term: Deduce

9E-1A-16

The strong nuclear force·E.3 Radioactive decay

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksExplain

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

Why does the nucleus not fly apart?

Show mark scheme

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

Answer: D · 1 stage of work, one mark

Every option, and why

AGravity between two protons is about 10³⁶ times weaker than their electrostatic repulsion.
BThe electrons are far outside the nucleus and do not cancel the repulsion between protons inside it.
CThe electrostatic force between two protons is always repulsive; a different, attractive force is needed.
DCorrect: at separations of about 10⁻¹⁵ m the strong nuclear force is attractive and larger than the electrostatic repulsion, and it acts on neutrons as well as protons.

Syllabus understandingE.3 — the existence of the strong nuclear force, a short-range, attractive force between nucleons Command term: Explain

10E-1B-01

Half-life & activity·E.3 Radioactive decay

Paper 1BMedium11 marks

Data-based question11 steps to full marksDetermine

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

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

t / s	N (counts in 10.0 s)	C / s⁻¹
0	591	58.7
60	433	42.9
120	355	35.1
180	257
240	206	20.2
300	142	13.8
360	122	11.8
420	87
480	75	7.1
540	50	4.6
600	43	3.9

Graph drawn to scale

(a)

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

(1)

(b)

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

(2)

(c)

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

(1)

(d)

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

(2)

(e)

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

(3)

(f)

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

(1)

(g)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the effect of background radiation on count rate. Guidance: the determination of the half-life of a nuclide is required. Command term: Determine

11E-1B-03

Half-life & activity·E.3 Radioactive decay

Paper 1BEasy9 marks

Data-based question9 steps to full marksDetermine

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

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

Isotope	Half-life
fluorine-18	1.83 h
technetium-99m	6.01 h
iodine-123	13.2 h
indium-111	67.3 h
gallium-67	78.3 h

Graph drawn to scale

(a)

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

(1)

(b)

Draw the curve of best fit for the data.

(1)

(c)

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

(2)

(d)

Identify the isotope, using the table.

(1)

(e)

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

(2)

(f)

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

(2)

Show mark scheme

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

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

Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life. Guidance: real-life contexts include the choice of isotope in medical use; the determination of the half-life of a nuclide is required. Command term: Determine

12E-1B-04

Penetration & ionization·E.3 Radioactive decay

Paper 1BMedium9 marks

Data-based question9 steps to full marksDeduce

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

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

Arrangement	Counts in 100 s
no source (background)	42
source, no absorber	3960
source, paper 0.1 mm	3885
source, aluminium 5 mm	468
source, aluminium 10 mm	451
source, lead 20 mm	176

(a)

(i)

State the absolute uncertainty in the count of 3960.

(1)

(ii)

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

(2)

(b)

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

(2)

(c)

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

(2)

(d)

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

(1)

(e)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the effect of background radiation on count rate. Command term: Deduce

13E-1B-09

Penetration & ionization·E.3 Radioactive decay

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

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

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

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

x / cm	N	C / s⁻¹	C^−½ / s^½
2.0	8077	80.47	0.111
4.0	3163	31.33	0.179
6.0	1732	17.02
8.0	1103	10.73	0.305
10.0	731	7.01	0.378
12.0	559	5.29
14.0	410	3.80	0.513

Graph drawn to scale

(a)

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

(1)

(b)

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

(2)

(c)

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

(2)

(d)

Draw the line of best fit and hence determine x₀.

(3)

(e)

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

(2)

(f)

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

(2)

Show mark scheme

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

Answers: (a) 17.02 s⁻¹ · (b) 0.242 and 0.435 s^½ · (d) x₀ ≈ 1.4 cm · (e) ≈ ±0.2 cm (the remaining parts are explanations — see the table above)

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the effect of background radiation on count rate. Command term: Determine

14E-1B-10

Half-life & activity·E.3 Radioactive decay

Paper 1BEasy8 marks

Data-based question8 steps to full marksDetermine

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

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

Throw number n	Mean number remaining	Uncertainty
0	500	0
2	344	9
4
6	170	10
8	114	8
10	82	7
12	55	6
14	40	5

Graph drawn to scale

(a)

State how the model shows that radioactive decay is random.

(1)

(b)

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

(2)

(c)

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

(1)

(d)

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

(2)

(e)

Comment on your answer to (d).

(1)

(f)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the random and spontaneous nature of radioactive decay; the activity, count rate and half-life in radioactive decay. Command term: Determine

15E-2-04

Mass–energy equivalence·E.3 Radioactive decay

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

A space probe that will orbit Saturn is powered by a radioisotope generator. The fuel is plutonium-238, which decays by alpha emission to uranium-234 with a half-life of 88 years. The kinetic energy of the alpha particles and of the recoiling uranium nuclei is absorbed inside the fuel, so the decays heat it. Thermocouples convert 6.0 % of the thermal power of the fuel into electrical power. At launch the thermal power of the fuel is 4.40 kW.

(1 u = 931.5 MeV c⁻², 1 MeV = 1.60 × 10⁻¹³ J, c = 3.00 × 10⁸ m s⁻¹, 1 year = 3.16 × 10⁷ s)

Particle	Mass / u
plutonium-238 atom	238.049558
uranium-234 atom	234.040950
helium-4 atom	4.002603

(a)

The decay of plutonium-238.

(i)

State the nuclear equation for the alpha decay of ²³⁸₉₄Pu.

(1)

(ii)

Show that the energy released in one decay is about 5.6 MeV.

(1)

(iii)

Determine the activity of the plutonium-238 at launch.

(2)

(b)

The energy released in the fuel leaves the generator as thermal radiation and as electrical energy. Assuming that the thermal power stays at 4.40 kW, calculate the decrease in the mass of the generator during the first year after launch.

(2)

(c)

The instruments on the probe need an electrical power of at least 120 W. Determine whether the generator can still supply this power 88 years after launch. Assume that the efficiency of the thermocouples does not change.

(3)

(d)

Explain why an alpha emitter is a good choice of fuel for this generator.

(2)

(e)

Saturn is 9.5 AU from the Sun. At the Earth, 1.0 AU from the Sun, the intensity of sunlight is 1360 W m⁻². Suggest, with a calculation, why solar panels are not used to power this probe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
²³⁸₉₄Pu → ²³⁴₉₂U + ⁴₂He «⁴₂α»	✓ 1	All four numbers on the right-hand side needed. Accept ⁴₂α.
Part (a)(ii)
Δm = 238.049558 − 234.040950 − 4.002603 = 0.006005 u; E = 0.006005 × 931.5 = 5.59 MeV	✓ 1	Must see full substitution OR answer to 3 s.f. Atomic masses may be used because the 94 electrons on the left balance 92 + 2 on the right.
Part (a)(iii)
Energy per decay = 5.594 × 1.60 × 10⁻¹³ = 8.95 × 10⁻¹³ J	✓ 1	Accept 8.96 × 10⁻¹³ J from 5.6 MeV.
A = 4400/8.95 × 10⁻¹³ = 4.9 × 10¹⁵ Bq	✓ 1	Award [2] for CNA. Award [1 max] for 4.9 × 10¹² Bq «4.40 W used».
Part (b)
Energy transferred in one year = 4400 × 3.16 × 10⁷ = 1.39 × 10¹¹ J	✓ 1
Δm = E/c² = 1.39 × 10¹¹/(3.00 × 10⁸)² = 1.5 × 10⁻⁶ kg	✓ 1	Accept 1.5–1.6 × 10⁻⁶ kg. Award [2] for CNA. Allow ECF for the energy.
Part (c)
Electrical power at launch = 0.060 × 4400 = 264 W	✓ 1
88 years is one half-life, so the activity and the thermal power halve «to 2200 W»	✓ 1
Electrical power = 132 W, which is greater than 120 W, so the generator can still supply the instruments	✓ 1	MP3 needs a comparison with 120 W. Allow a consistent conclusion from an ECF value.
Part (d)
Alpha particles have a very short range «a few μm in the solid fuel», so they are absorbed within the fuel and almost all of the decay energy becomes thermal energy of the fuel	✓ 1
Very little penetrating radiation leaves the fuel, so only light shielding is needed to protect the instruments «keeping the mass of the probe low»	✓ 1	OWTTE
Part (e)
Intensity ∝ 1/d²: at Saturn it is 1360/9.5² = 15 W m⁻²	✓ 1	Accept 'about 1/90 of the intensity at the Earth'.
So the panels would need an area about 90 times larger than near the Earth to give the same power — far too large and heavy for the probe	✓ 1	OWTTE. Do not accept answers based on the probe being in the shadow of Saturn alone.

Answers: (a)(ii) 5.59 MeV · (a)(iii) 4.9 × 10¹⁵ Bq · (b) 1.5 × 10⁻⁶ kg · (c) 132 W — yes, it can · (e) 15 W m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingE.3 — the mass–energy equivalence as given by E = mc² in nuclear reactions; the radioactive decay equations involving α, β⁻, β⁺, γ; the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; B.1 — the concept of apparent brightness b and luminosity L of a body as given by b = L/4πd² Command term: Determine

16E-2-05

Beta decay & the neutrino·E.3 Radioactive decay

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksState

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

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

(a)

The two beta decays.

(i)

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

(2)

(ii)

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

(2)

(iii)

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

(1)

(b)

The gamma ray.

(i)

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

(1)

(ii)

Calculate the wavelength of the gamma-ray photon.

(2)

(c)

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

(2)

(d)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the radioactive decay equations involving α, β⁻, β⁺, γ; the existence of neutrinos ν and antineutrinos ν̄; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay; the penetration and ionizing ability of alpha particles, beta particles and gamma rays Command term: State

17E-2-06

Half-life & activity·E.3 Radioactive decay

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksDraw

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

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

Graph axes drawn to scale

(a)

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

(1)

(b)

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

(3)

(c)

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

(2)

(d)

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

(2)

(e)

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

(1)

(f)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life; the random and spontaneous nature of radioactive decay; the penetration and ionizing ability of alpha particles, beta particles and gamma rays Command term: Draw

18E-2-07

Background radiation·E.3 Radioactive decay

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksEvaluate

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

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

(a)

Radon and background radiation.

(i)

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

(1)

(ii)

State one source of background radiation other than radon.

(1)

(b)

The laboratory measurement.

(i)

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

(1)

(ii)

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

(2)

(iii)

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

(1)

(c)

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

(2)

(d)

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

(2)

(e)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the effect of background radiation on count rate; the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life; the random and spontaneous nature of radioactive decay; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay (guidance: real-life contexts) Command term: Evaluate

19E-2-08

The binding energy curve·E.3 Radioactive decay

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksEstimate

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

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

Graph drawn to scale

(a)

The curve.

(i)

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

(1)

(ii)

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

(2)

(b)

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

(i)

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

(1)

(ii)

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

(2)

(c)

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

(i)

Show that about 13 MeV is released in this reaction.

(1)

(ii)

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

(3)

(iii)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the variation of the binding energy per nucleon with nucleon number; nuclear binding energy and mass defect; the mass–energy equivalence as given by E = mc² in nuclear reactions (guidance: an interpretation of binding energy curves is required); E.5 — that fusion is a source of energy in stars; the effect of stellar mass on the evolution of a star Command term: Estimate

20E-1A-31

Decay chains·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

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

Nucleon numberProton number

Show mark scheme

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

Answer: B · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the radioactive decay equations involving α, β⁻, β⁺, γ; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay Command term: Deduce

21E-1A-32

Mass defect & binding energy·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

A nucleus ^A_ZX has mass M. The mass of a proton is m_p and the mass of a neutron is m_n.

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

Show mark scheme

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

Answer: B · 2 stages of work, one mark

Every option, and why

AThis is the total binding energy of the nucleus; it has not been divided by the number of nucleons.
BCorrect: the mass defect is Zm_p + (A − Z)m_n − M; multiplying by c² gives the binding energy, and dividing by A gives the value per nucleon.
CA has been used as the number of neutrons; the neutron number is A − Z.
DThis divides by the number of protons Z instead of by the number of nucleons A.

Syllabus understandingE.3 — nuclear binding energy and mass defect; the mass–energy equivalence as given by E = mc² in nuclear reactions Command term: Deduce

22E-1A-33

Half-life & activity·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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

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

Show mark scheme

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

Answer: C · 2 stages of work, one mark

Every option, and why

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

Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Determine

23E-1A-34

Half-life & activity·E.3 Radioactive decay

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

Which row gives the ratios for Q compared with P?

Half-life of Q ÷ half-life of PActivity of Q ÷ activity of P

Show mark scheme

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

Answer: B · 2 stages of work, one mark

Every option, and why

AThe half-lives are equal, but the larger sample contains twice as many nuclei, so twice as many decay each second.
BCorrect: the half-life does not depend on the amount of the isotope, while the activity is proportional to the number of nuclei present.
CThis treats the half-life as if it depended on the amount present; each nucleus decays independently, so it does not.
DMore nuclei do not make each nucleus decay sooner, so the half-life is not shortened.

Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the random and spontaneous nature of radioactive decay Command term: Deduce

24E-1A-35

Radioactive dating·E.3 Radioactive decay

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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

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

Show mark scheme

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

Answer: C · 2 stages of work, one mark

Every option, and why

A1/8 is the fraction of the original X that remains, not the ratio of Y to X.
BThis is N_X/N_Y, the ratio the wrong way round.
CCorrect: 1/8 of X remains and 7/8 has become Y, so there are 7 atoms of Y for every atom of X.
D8 compares the original number of X atoms with the number remaining; the atoms that have decayed number 7/8, not 8/8, of the original.

Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life (real-life context: radioactive dating) Command term: Determine

25E-1A-36

Mass–energy equivalence·E.3 Radioactive decay

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksDeduce

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

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

Show mark scheme

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

Answer: B · 1 stage of work, one mark

Every option, and why

AThis is the negative of the energy released: for a decay that happens, the products have less mass than the parent, so this expression is negative.
BCorrect: the decrease in mass is M − m₁ − m₂, and E = Δmc².
CThe speed of light has not been squared; E = mc².
DMc² is the energy equivalent of the whole nucleus; only the decrease in mass is released.

Syllabus understandingE.3 — the mass–energy equivalence as given by E = mc² in nuclear reactions Command term: Deduce

26E-1B-11

Background radiation·E.3 Radioactive decay

Paper 1BEasy10 marks

Data-based question10 steps to full marksDiscuss

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

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

Room	1	2	3	4	5	6	7	8	9	10
A (granite walls)	28	35	31	40	26	33	37	29	34	38
B (brick walls)	24	19	27	21	16	25	22	28	20	23

(a)

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

(1)

(b)

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

(2)

(c)

Discuss whether the data support the hypothesis.

(2)

(d)

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

(1)

(e)

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

(2)

(f)

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

(2)

Show mark scheme

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

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

Syllabus understandingE.3 — the effect of background radiation on count rate; the random and spontaneous nature of radioactive decay. Command term: Discuss

27E-1B-12

The binding energy curve·E.3 Radioactive decay

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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

Δm = Zm_H + (A − Z)m_n − M

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

Nuclide	Z	A	M / u	Binding energy per nucleon / MeV
⁴He	2	4	4.002602	7.07
⁷Li	3	7	7.016003	5.61
¹²C	6	12	12.000000	7.68
¹⁶O	8	16	15.994915	7.98
²⁴Mg	12	24	23.985042	8.26
⁴⁰Ca	20	40	39.962591	8.55
⁵⁶Fe	26	56	55.934936	8.79
⁹⁰Zr	40	90	89.904698	8.71
¹²⁰Sn	50	120	119.902199
¹⁵⁸Gd	64	158	157.924104	8.20
²⁰⁸Pb	82	208	207.976652
²³⁸U	92	238	238.050788	7.57

Graph drawn to scale

(a)

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

(1)

(b)

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

(1)

(c)

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

(2)

(d)

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

(2)

(e)

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

(2)

(f)

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

(1)

(g)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — nuclear binding energy and mass defect; the variation of the binding energy per nucleon with nucleon number. Guidance: an interpretation of binding energy curves is required. Command term: Determine

28E-1B-17

Penetration & ionization·E.3 Radioactive decay

Paper 1BMedium9 marks

Data-based question9 steps to full marksDetermine

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

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

Thickness / μm	N	R / s⁻¹
0	8537	853.7
50	5943	594.3
100	4266	426.6
150	2958	295.8
200	2119	211.9
250	1446	144.6
300	1051	105.1

Graph drawn to scale

(a)

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

(2)

(b)

Draw the curve of best fit for the data.

(1)

(c)

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

(2)

(d)

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

(2)

(e)

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

(1)

(f)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays. Guidance: real-life contexts include the thickness of materials. Command term: Determine

29E-1B-20

Mixtures of isotopes·E.3 Radioactive decay

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

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

The graph shows the data, with error bars.

t / s	C / s⁻¹
0	301.1
40	171.0
80	106.8
120	69.1
160	50.1
200	37.9
300	21.9
400	15.0
500	10.3
600	7.4
700	5.6
800	3.8

Graph drawn to scale

(a)

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

(1)

(b)

Draw the curve of best fit for the data.

(1)

(c)

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

(1)

(d)

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

(2)

(e)

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

(3)

(f)

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

(3)

(g)

Predict the count rate at t = 1000 s.

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the changes in activity and count rate during radioactive decay using integer values of half-life; the activity, count rate and half-life in radioactive decay. Command term: Determine

30E-2-19

Decay chains·E.3 Radioactive decay

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksDraw

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

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

Chart drawn to scale

(a)

(i)

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

(2)

(ii)

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

(1)

(b)

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

(i)

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

(2)

(ii)

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

(1)

(iii)

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

(2)

(c)

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

(2)

Show mark scheme

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

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

Syllabus understandingE.3 — the radioactive decay equations involving α, β⁻, β⁺, γ; isotopes; the changes in the state of the nucleus following alpha, beta and gamma radioactive decay Command term: Draw

31E-2-20

Background radiation·E.3 Radioactive decay

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine whether

A food-safety laboratory tests dried mushrooms for contamination by caesium-137. Caesium-137 is a β⁻ emitter with a half-life of 30 years; most of its decays are followed by the emission of a gamma-ray photon.

A 0.40 kg sample of mushrooms is sealed in a plastic pot and placed on a gamma-ray detector inside a thick lead shield. On average the detector records one count for every 25 decays of caesium-137 in the sample. With an empty pot in place, five background counts were taken, each over 600 s. With the sample in place, 6450 counts were recorded in 600 s.

The legal limit for caesium-137 in food is 600 Bq per kilogram.

Background count over 600 s	1	2	3	4	5
Number of counts	198	221	204	230	197

(a)

Outline why the detector is surrounded by a thick lead shield.

(1)

(b)

Explain why the detector records gamma rays from the sample but no β⁻ particles.

(2)

(c)

(i)

Show that the count rate due to the caesium-137 in the sample is about 10 s⁻¹.

(1)

(ii)

Determine whether these mushrooms may be sold.

(3)

(d)

The mushrooms were picked in a forest that was contaminated by caesium-137.

(i)

Predict the activity per kilogram of mushrooms picked in the same forest 60 years later, assuming that they take up caesium-137 from the soil in the same way.

(1)

(ii)

Suggest one reason why the actual activity could be lower than your prediction.

(1)

(e)

A second 0.40 kg sample gives 226 counts in 600 s. A technician concludes that this sample contains no caesium-137. Discuss whether the technician's conclusion is justified.

(2)

(f)

Suggest one change to the method that would allow smaller activities to be detected.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
to absorb gamma radiation from the surroundings «rocks, building materials», so that the background count is reduced	✓ 1	Accept 'to reduce the background count'. Do not accept 'to protect the operator'.
Part (b)
β⁻ particles have a short range in solids and are absorbed within the mushrooms or the wall of the pot «a few mm»	✓ 1
gamma rays are only weakly absorbed «most penetrating», so many pass through the sample and the pot to the detector	✓ 1
Part (c)(i)
background = 1050/5 = 210 counts, so rate = (6450 − 210)/600 «= 10.40 s⁻¹»	✓ 1	Must see full substitution OR answer to 3 s.f. (10.4 s⁻¹).
Part (c)(ii)
activity of the sample = 10.4 × 25 = 260 Bq	✓ 1	Allow ECF from (c)(i): 10 × 25 = 250 Bq.
activity per kilogram = 260/0.40 = 650 Bq kg⁻¹	✓ 1	Accept 625 Bq kg⁻¹ from 10 s⁻¹.
650 Bq kg⁻¹ > 600 Bq kg⁻¹, so the mushrooms may not be sold	✓ 1	The conclusion must be consistent with the candidate's value. Do not award MP3 for comparing 260 Bq «not divided by the mass» with the limit.
Part (d)(i)
60 years = 2 half-lives, so 650/4 ≈ 160 Bq kg⁻¹	✓ 1	Allow ECF from (c)(ii). Accept 160–163 Bq kg⁻¹.
Part (d)(ii)
caesium is washed deeper into the soil «out of reach of the mushrooms» / is carried away by water / is taken up less over time	✓ 1	Accept any sensible physical reason. OWTTE
Part (e)
226 counts lies within the range of the background counts «197 to 230», so the sample cannot be distinguished from background «random variation»	✓ 1
so the conclusion is not justified: the sample may contain caesium-137 with an activity too small to detect «although any activity is far below the legal limit»	✓ 1	MP2 only scores if MP1 scores. OWTTE
Part (f)
count for a longer time / use a larger sample / take more background counts / use a more efficient detector	✓ 1	Accept any one.

Answers: (c)(i) 10.4 s⁻¹ · (c)(ii) 260 Bq; 650 Bq kg⁻¹ — above the limit, may not be sold · (d)(i) ≈ 160 Bq kg⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingE.3 — the effect of background radiation on count rate; the activity, count rate and half-life in radioactive decay; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Determine whether

32E-2-21

The strong nuclear force·E.3 Radioactive decay

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksExplain

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

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

Particle	Mass / u
proton	1.007276
neutron	1.008665
deuterium nucleus	2.013553

(a)

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

(1)

(b)

(i)

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

(1)

(ii)

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

(2)

(c)

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

(2)

(d)

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

(i)

Calculate the gravitational force between the proton and the neutron.

(1)

(ii)

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

(1)

(iii)

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

(2)

(e)

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

(2)

Show mark scheme

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

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

Syllabus understandingE.3 — nuclear binding energy and mass defect; the existence of the strong nuclear force, a short-range, attractive force between nucleons; the variation of the binding energy per nucleon with nucleon number; the mass–energy equivalence as given by E = mc² in nuclear reactions (guidance: masses in u and MeV c⁻²); E.1 — that the frequency of the photon depends on its energy as given by E = hf; D.1 — Newton's universal law of gravitation as given by F = Gm₁m₂/r²; A.3 — that work done by a force is equivalent to a transfer of energy Command term: Explain

33E-2-22

Choosing an isotope·E.3 Radioactive decay

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDiscuss

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

Isotope	Radiation emitted	Half-life
P	alpha	6.0 hours
Q	gamma	6.0 hours
R	gamma	2.0 minutes
S	gamma	30 years

(a)

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

(4)

(b)

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

(i)

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

(1)

(ii)

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

(2)

(c)

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

(2)

(d)

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

(3)

(e)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the penetration and ionizing ability of alpha particles, beta particles and gamma rays; the changes in activity and count rate during radioactive decay using integer values of half-life (guidance: the choice of isotope in medical use) Command term: Discuss

34E-2-23

Radioactive dating·E.3 Radioactive decay

Paper 2Easy9 marks

Short answer & extended response9 steps to full marksDetermine

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

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

Graph drawn to scale

(a)

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

(2)

(b)

(i)

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

(1)

(ii)

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

(2)

(c)

Carbon-14 dating cannot be used for every object.

(i)

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

(1)

(ii)

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

(2)

(d)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the activity, count rate and half-life in radioactive decay; the changes in activity and count rate during radioactive decay using integer values of half-life; the radioactive decay equations involving α, β⁻, β⁺, γ (guidance: radioactive dating) Command term: Determine

35E-2-24

Penetration & ionization·E.3 Radioactive decay

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksExplain

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

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

Diagram NOT accurately drawn

(a)

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

(1)

(b)

Show that one alpha particle can produce about 1.6 × 10⁵ ion pairs.

(1)

(c)

Half of the alpha particles emitted by the source enter the chamber, and each gives all of its kinetic energy to the air in the chamber. Determine the current in the ammeter.

(3)

(d)

Explain why the current falls when smoke enters the chamber.

(2)

(e)

Explain why an alpha source is used rather than a beta source of the same activity.

(3)

(f)

In a real smoke detector the current is much smaller than your answer to (c). Suggest one reason for this.

(1)

(g)

The half-life of americium-241 is about 430 years. Suggest why a source with a long half-life is chosen.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
nucleon number 237 AND proton number 93	✓ 1
Part (b)
5.5 × 10⁶/34 «= 1.62 × 10⁵»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (c)
alpha particles entering per second = 0.5 × 3.7 × 10⁴ = 1.85 × 10⁴ s⁻¹	✓ 1
ion pairs produced per second = 1.85 × 10⁴ × 1.62 × 10⁵ = 3.0 × 10⁹ s⁻¹	✓ 1	Allow ECF from (b).
I = 3.0 × 10⁹ × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁰ A	✓ 1	Each ion pair transfers a charge e round the circuit. Award [3] for CNA. Accept 4.7–4.8 × 10⁻¹⁰ A.
Part (d)
smoke particles absorb alpha particles / ions attach to the «much more massive» smoke particles	✓ 1
so fewer ions are produced or fewer ions reach the plates per second, and the current falls	✓ 1
Part (e)
alpha particles are much more strongly ionizing, so each one produces a large number of ion pairs within the small chamber	✓ 1
beta particles are weakly ionizing and travel much further in air, so they would produce few ion pairs in the chamber and the current would be far too small «and hardly affected by smoke»	✓ 1
alpha particles are absorbed by a few centimetres of air or by the casing, so no radiation escapes from the detector	✓ 1
Part (f)
the alpha particles have a range of a few cm in air, so they reach the plates/walls before giving all their energy to the air in the small chamber / some ions recombine before reaching the plates	✓ 1	Accept any one.
Part (g)
the activity, and so the current, stays almost constant over the working life of the detector	✓ 1

Answers: (a) A = 237, Z = 93 · (b) 1.62 × 10⁵ · (c) 4.8 × 10⁻¹⁰ A (the remaining parts are explanations — see the table above)

Syllabus understandingE.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays; B.5 — electric current I = Δq/Δt Command term: Explain

36E-2-34

Random & spontaneous decay·E.3 Radioactive decay

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksExplain

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

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

Interval	1	2	3	4	5	6	7	8	9	10	11	12
Counts in 10 s	318	296	341	305	287	329	312	300	334	291	322	309

Graph drawn to scale

(a)

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

(2)

(b)

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

(2)

(c)

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

(1)

(d)

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

(2)

(e)

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

(1)

(f)

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

(2)

(g)

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

(1)

Show mark scheme

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

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

Syllabus understandingE.3 — the random and spontaneous nature of radioactive decay; the activity, count rate and half-life in radioactive decay; the effect of background radiation on count rate; the changes in activity and count rate during radioactive decay using integer values of half-life Command term: Explain

E.3 Radioactive decay: IB Physics SL exam-style questions

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