Fission releases energy because the products have a higher binding energy per nucleon than the original heavy nucleus. You need to calculate the energy released from masses or binding energies and explain how neutrons sustain a chain reaction.
Reactor questions ask about the role of the moderator, control rods, heat exchanger and shielding, the energy flow from fuel to electrical output, and the handling of spent fuel.
35 questions
182 marks
Paper 1A: 20
Paper 1B: 4
Paper 2: 11
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Showing 35 of 35 questions · 182 marks
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10 practice questions on E.4 Fission
1E-1A-08
Binding energy and fission·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksEstimate
A uranium-236 nucleus (X on the graph) undergoes fission into two nuclei, each with nucleon number 117 (Y on the graph), and two free neutrons.
What is the best estimate of the energy released?
Binding energy per nucleon against nucleon number (gridlines every 0.25 MeV).Show mark scheme
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Notes
Step 1Read the graph: binding energy per nucleon ≈ 7.6 MeV at X (A = 236) and ≈ 8.5 MeV at Y (A = 117).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Total binding energy before = 236 × 7.6 ≈ 1794 MeV; after = 234 × 8.5 ≈ 1989 MeV (free neutrons have no binding energy).
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Step 3Energy released = increase in total binding energy ≈ 195 MeV ≈ 2.0 × 102 MeV.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is the increase in binding energy per nucleon (0.9 MeV) — it has not been multiplied by the number of nucleons.
BThis counts only one of the two fragments (117 × 8.5 − 118 × 7.6 ≈ 98 MeV).
CCorrect: the total binding energy rises by about 2 × 102 MeV.
DThis is the total binding energy of the two fragments (≈ 1989 MeV), not the increase in binding energy.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; calculations to determine the energy released in fission reactions; E.3 — the variation of the binding energy per nucleon with nucleon number (interpretation of binding energy curves) Command term: Estimate
2E-1A-20
Energy from fission·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksEstimate
Each fission of a uranium-235 nucleus releases about 200 MeV. (NA = 6.0 × 1023 mol−1, 1 eV = 1.6 × 10−19 J)
What is the approximate energy released by the complete fission of 1 kg of uranium-235?
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Step 1Number of nuclei in 1 kg = (1000/235) × 6.0 × 1023 = 2.6 × 1024.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy per fission = 200 × 106 × 1.6 × 10−19 = 3.2 × 10−11 J.
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Step 3Total = 2.6 × 1024 × 3.2 × 10−11 ≈ 8 × 1013 J (about a million times the energy from burning 1 kg of coal).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the energy from one mole (235 g, 6.0 × 1023 nuclei) of uranium, not from 1 kg.
BCorrect: 2.6 × 1024 × 3.2 × 10−11 J.
CThis treats 1 kg as 1000 mol (6.0 × 1026 nuclei), forgetting to divide by the molar mass 235 g mol−1.
DThis is mc² for 1 kg — the energy if the whole mass were converted; fission converts only about 0.1 %.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; guidance: calculations to determine the energy released in fission reactions are required; B.3 — the amount of substance n = N/NACommand term: Estimate
3E-1A-44
Chain reactions·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In a nuclear reactor fuelled by uranium-235, each fission releases on average k neutrons. The reactor is operating at a constant power.
What fraction of all the neutrons released must, on average, be absorbed without causing fission or escape from the core?
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Notes
Step 1At constant power the rate of fission is constant, so on average exactly one neutron from each fission causes a further fission.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Of the k neutrons released, k − 1 must be removed (by control rods, other absorbers or leakage).
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Step 3Fraction removed = (k − 1)/k.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the fraction of the neutrons that DO go on to cause fission.
BCorrect: one neutron in k sustains the chain; the other k − 1 are lost.
CThis is the number of neutrons removed per fission, not the fraction of those released.
DThis is the inverse of the fraction; it is greater than 1, which is impossible for a fraction.
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions; the role of control rods in a nuclear power plant Command term: Deduce
4E-1A-46
Fission products·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksExplain
Uranium-235 has a neutron-to-proton ratio of 1.55. Its fission produces two nuclei with nucleon numbers roughly between 90 and 145. Stable nuclides in this range have neutron-to-proton ratios between about 1.2 and 1.45.
Which row gives the reason why the fission products are radioactive and the decay they usually undergo?
ReasonUsual decay
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Step 1The fragments keep roughly the neutron-to-proton ratio of the parent, about 1.55.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2This is higher than the 1.2–1.45 of stable nuclides of the same mass: the fragments are neutron-rich.
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Step 3β− decay turns a neutron into a proton, lowering N/Z; successive β− decays move the products towards stability.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: neutron-rich products undergo β− decay (often followed by gamma emission).
Bβ+ decay turns a proton into a neutron and would raise the ratio further, away from stability.
CInverts the comparison: 1.55 is above, not below, the stable range.
DAlpha decay is typical of very heavy nuclei (A > about 200); medium-mass fragments are unstable because of their neutron excess, not their size.
Syllabus understandingE.4 — the properties of the products of nuclear fission and their management; E.3 (HL) — the role of the ratio of neutrons to protons for the stability of nuclides Command term: Explain
5E-1A-48
Fission reactors·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Three statements about a thermal nuclear fission reactor are:
I. Pushing the control rods further into the core reduces the average number of neutrons per fission that go on to cause another fission. II. The moderator slows the neutrons down, which makes them more likely to cause fission of uranium-235 nuclei. III. The thick shielding around the core is needed mainly to absorb the alpha particles emitted by the fuel.
Which statements are correct?
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Notes
Step 1I: the control rods absorb neutrons, so fewer are available to cause fission — correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: slow (thermal) neutrons are much more likely to be captured by uranium-235 and cause fission — correct.
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Step 3III: alpha particles are stopped by the fuel cladding itself; the shielding is needed for the penetrating neutrons and gamma rays — incorrect.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: control rods absorb neutrons; the moderator makes fission more likely.
BIII is wrong: alpha particles cannot even escape the fuel rods; the shielding stops neutrons and gamma radiation.
CI is correct (inserting the rods absorbs more neutrons) and III is wrong (the shielding is for neutrons and gamma rays).
DIII is wrong: alpha particles have a range of only a few centimetres in air and are stopped by the fuel cladding.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; E.3 — the penetration and ionizing ability of alpha particles, beta particles and gamma rays Command term: Identify
6E-1A-68
Chain reactions·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In a simple model of a sphere of fissile uranium, neutrons are produced by fission at a rate proportional to the volume of the sphere, and neutrons escape through its surface at a rate proportional to its surface area. The sphere is replaced by a sphere of the same material with twice the radius.
By what factor does (rate at which neutrons escape)/(rate at which neutrons are produced) change?
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Step 1Doubling the radius multiplies the surface area (∝ r²) by 4 and the volume (∝ r³) by 8.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The ratio (rate of escape)/(rate of production) ∝ r²/r³ = 1/r, so it changes by a factor 4/8 = 1/2.
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Step 3A larger piece of fuel therefore loses a smaller fraction of its neutrons through the surface: a chain reaction can be sustained only if the mass of fuel is large enough.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis allows for the eightfold increase in the production rate but forgets that the escape rate also rises, by a factor of 4.
BThis squares the correct factor, taking the ratio to be proportional to 1/r² instead of 1/r.
CCorrect: the escape rate rises by 4 and the production rate by 8, so the ratio halves.
DThis inverts the ratio, giving the factor by which (rate of production)/(rate of escape) changes.
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions Command term: Determine
7E-1A-69
Fission products and their management·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A sample of waste from a nuclear reactor contains two radioactive nuclides: X, of half-life 30 years, and Y, of half-life 2.4 × 104 years. At a certain time the sample contains equal numbers of nuclei of X and of Y.
Which row identifies the nuclide with the greater activity at that time and the nuclide with the greater activity 600 years later?
Greater activity at that timeGreater activity 600 years later
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Step 1A = λN with λ = ln 2/T½: for equal N, AX/AY = (2.4 × 104)/30 = 800, so X is more active at first.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2600 years is 20 half-lives of X, so NX falls by 220 ≈ 106, while NY hardly changes (600 years is only 0.025 of its half-life).
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Step 3Then AX/AY ≈ 800/106 ≈ 8 × 10−4: Y is now the more active.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the short-lived nuclide dominates the activity at first, but the long-lived nuclide dominates in the long term — which is why long-term storage of waste is needed.
BThis assumes the shorter half-life always gives the greater activity, forgetting that the number of nuclei of X falls by a factor of about 106 in 600 years.
CThis confuses half-life with the rate of decay, taking the longer half-life to mean faster decay: Y is then thought to be more active at first and, having used up its nuclei, to be overtaken by X later.
DThis treats activity as proportional to half-life, so that the long-lived nuclide is always the more active.
Syllabus understandingE.4 — the properties of the products of nuclear fission and their management; guidance: the impact of the long-term storage of nuclear waste; E.3 (HL) — the activity as the rate of decay, A = λN; T½ = ln 2/λCommand term: Deduce
8E-1A-111
Balancing a fission equation·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A plutonium-239 nucleus absorbs a slow neutron and undergoes fission:
10n + 23994Pu → 13454Xe + AZX + 310n
Which row gives A and Z for nuclide X?
AZ
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Notes
Step 1Nucleon number: 1 + 239 = 134 + A + 3 × 1, so A = 240 − 137 = 103.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Proton number: 0 + 94 = 54 + Z + 3 × 0, so Z = 40 (X is zirconium-103).
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Step 3Neutrons carry nucleon number 1 but proton number 0.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis leaves out the absorbed neutron on the left-hand side: 239 − 134 − 3 = 102.
BThis gives each neutron a proton number of 1, as if it were a proton: 95 − 54 − 3 = 38.
CCorrect: nucleon number 240 − 134 − 3 = 103 and proton number 94 − 54 = 40.
DThis ignores the three neutrons released: 240 − 134 = 106.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; E.3 — decay equations; isotopes Command term: Deduce
9E-1A-112
Energy released from rest masses·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A uranium-235 nucleus of mass mU absorbs a slow neutron of mass mn. It splits into nuclei X and Y, of masses mX and mY, and three neutrons. The kinetic energy of the slow neutron is negligible.
What is the energy released in this fission?
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Step 1Mass before = mU + mn; mass after = mX + mY + 3mn.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy released = (mass before − mass after)c2 = (mU + mn − mX − mY − 3mn)c2.
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Step 3One of the three neutrons released replaces the one absorbed: (mU − mX − mY − 2mn)c2.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis leaves out the absorbed neutron on the left-hand side, so it subtracts one neutron mass too many.
BThis leaves out the three neutrons released, as if only X and Y were produced.
CThis is (mass after − mass before)c2: the sign is reversed and the expression is negative.
DCorrect: (mU + mn) − (mX + mY + 3mn) = mU − mX − mY − 2mn.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; guidance: calculations to determine the energy released in fission reactions; E.3 — mass defect and binding energy; E = mc2Command term: Determine
10E-1A-113
The heat exchanger·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In a pressurized water reactor, the thermal power produced in the core is 3.0 GW and the overall efficiency of the power station is 33 %. All of this thermal power is transferred in the heat exchanger to the water of the secondary circuit, which enters the heat exchanger at its boiling point and leaves it as steam at the same temperature. The specific latent heat of vaporization of this water is 1.6 MJ kg−1.
What mass of steam is produced each second?
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Step 1The thermal power of the core, not the electrical output, is transferred in the heat exchanger: 3.0 × 109 J each second.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The water only changes state, so Q = mL for each second.
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Step 3m = 3.0 × 109/1.6 × 106 = 1.9 × 103 kg each second.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses the electrical output, 0.33 × 3.0 GW = 0.99 GW: the efficiency has been applied to energy that has not yet reached the turbine.
BCorrect: 3.0 × 109/1.6 × 106 = 1.9 × 103 kg.
CThis divides by the efficiency, treating 3.0 GW as the electrical output: 3.0 × 109/(0.33 × 1.6 × 106).
DThis takes 1.6 MJ kg−1 as 1.6 × 103 J kg−1.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; B.1 — Q = mL; A.3 — efficiency Command term: Determine
11E-1A-114
Choosing a moderator·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Which property makes a material suitable for use as the moderator in a thermal fission reactor?
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Notes
Step 1The moderator slows fast fission neutrons so that they are likely to cause fission of uranium-235.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2A neutron transfers the largest fraction of its kinetic energy when it collides with a nucleus of similar mass, so nuclei of small mass slow neutrons in few collisions.
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Step 3The neutrons must not be absorbed by the moderator, or too few would remain to sustain the chain reaction.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis describes the material of a control rod, not a moderator.
BA neutron colliding with a heavy nucleus rebounds with almost the same speed, losing only a small fraction of its kinetic energy.
CThis describes the shielding, not the moderator.
DCorrect: light nuclei take a large share of the kinetic energy in each collision, and few neutrons are lost by absorption.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; the role of chain reactions in nuclear fission reactions; A.2 — energy considerations in elastic collisions Command term: Identify
12E-1A-115
Speed of thermal neutrons·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In a reactor core, neutrons are slowed by the water moderator until they are in thermal equilibrium with it. The temperature of the water is 290 °C. The mean kinetic energy of the neutrons is then (3/2)kBT.
Which is the best estimate of the speed of a neutron with this kinetic energy?
AThis uses the temperature in degrees Celsius (290) instead of kelvin.
BThis takes the kinetic energy as kBT, leaving out the factor 3/2.
CCorrect: Ek = 1.17 × 10−20 J, so v = √(2Ek/m) = 3.7 × 103 m s−1.
DThis is 2Ek/m = v2: the square root has not been taken.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; B.1 — Ek = (3/2)kBT; Kelvin and Celsius scales; A.3 — Ek = ½mv2Command term: Determine
13E-1A-116
Fuel consumption of two power stations·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two nuclear power stations X and Y have the same electrical power output. The data below give their overall efficiency and the percentage by mass of uranium-235 in their fuel. In both, all the uranium-235 undergoes fission and the uranium-238 releases no energy.
Power station X: efficiency 30 %, uranium-235 4.0 % Power station Y: efficiency 40 %, uranium-235 3.0 %
What is (mass of fuel used by X)/(mass of fuel used by Y) in the same time?
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Step 1Thermal power = electrical power/efficiency, so the mass of uranium-235 used ∝ 1/η.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Mass of fuel = mass of uranium-235/(fraction of uranium-235), so mass of fuel ∝ 1/(η × fraction).
AThis multiplies by the efficiency instead of dividing: (0.30 × 0.030)/(0.40 × 0.040) = 0.56.
BCorrect: X needs 4/3 as much uranium-235 as Y, but its fuel is 4/3 as rich in uranium-235, so the masses of fuel are equal.
CThis allows for the efficiencies (0.40/0.30) but forgets that the fuels contain different percentages of uranium-235.
DThis inverts the effect of the enrichment: (0.40 × 0.040)/(0.30 × 0.030) = 1.8.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; guidance: calculations to determine the energy released in fission reactions; A.3 — efficiency η = Pout/PinCommand term: Deduce
14E-1A-117
Spontaneous and neutron-induced fission·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Three statements about the fission of nuclei in the fuel of a reactor are:
I. Spontaneous fission can occur without a neutron being absorbed by the nucleus. II. Spontaneous fission of nuclei in fresh fuel can provide the first neutrons for a chain reaction. III. The rate of spontaneous fission in the fuel is reduced when the control rods are pushed further into the core.
Which statements are correct?
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Step 1I: spontaneous fission is a random decay of an unstable nucleus with no external cause — correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: each spontaneous fission releases neutrons, which can induce further fissions — correct.
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Step 3III: control rods absorb neutrons, so they reduce only neutron-induced fission; the rate of spontaneous fission depends only on the number of nuclei — incorrect.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: spontaneous fission needs no neutron and releases neutrons, but it cannot be controlled by absorbing neutrons.
BThis rejects II and accepts III: it treats spontaneous fission as if it were caused by neutrons, so it could be controlled but could not start a chain reaction.
CThis rejects I, confusing spontaneous fission with neutron-induced fission.
DIII is wrong: like other radioactive decays, spontaneous fission is not affected by the neutrons absorbed by the control rods.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of chain reactions in nuclear fission reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; E.3 — the random and spontaneous nature of radioactive decay Command term: Deduce
15E-1A-118
Managing spent fuel·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Spent fuel removed from a reactor is first stored under water for several years. It is later sealed in containers and placed in a deep geological repository.
Which row gives a main reason for each stage?
Storage under waterDeep geological repository
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Step 1Freshly removed fuel contains many short-lived fission products: their decay releases thermal energy at a high rate, and the water both cools the fuel and shields the emitted radiation.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Some products (and other nuclides in the fuel) have half-lives of thousands of years or more; the half-life cannot be changed by pressure or temperature.
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Step 3So the waste must be isolated from people and from groundwater for a very long time.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ABoth reasons are wrong: water is a moderator, which would make fission more likely, not less, and decay rates cannot be changed by pressure.
BThe first reason is wrong: water is a moderator, so it makes fission of uranium-235 more likely, not less likely.
CThe second reason is wrong: half-lives are not affected by pressure or temperature.
DCorrect: the pool cools and shields the fuel while the short-lived products decay; the repository isolates the long-lived nuclides.
Syllabus understandingE.4 — the properties of the products of nuclear fission and their management; guidance: the impact of the long-term storage of nuclear waste; E.3 — half-life Command term: Identify
16E-1A-119
Kinetic energies of fission fragments·E.4 Fission
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Uranium-235 nuclei undergo fission after absorbing slow neutrons. The graph shows the distribution of the kinetic energies Ek of the fragments. The two peaks correspond to the most probable light fragment and the most probable heavy fragment.
Assume that the uranium-236 nucleus is at rest when it splits, that the momentum of the released neutrons is negligible, and that the total nucleon number of the two fragments is 234.
What is the nucleon number of the most probable light fragment?
Distribution of the kinetic energies of the fission fragments (drawn to scale).Show mark scheme
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Step 1Read the peaks: the light fragment has Ek ≈ 100 MeV and the heavy fragment Ek ≈ 68 MeV.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Momentum is conserved, so the fragments have equal and opposite momenta p; with Ek = p2/2m, Ek ∝ 1/m: AL/AH = 68/100.
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Step 3AL = 234 × 68/(68 + 100) = 94.7 ≈ 95.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: equal momenta mean the lighter fragment has the larger kinetic energy in the inverse ratio of the masses: 234 × 68/168 ≈ 95.
BThis uses equal momenta but takes Ek ∝ 1/m2, as if the masses were in the inverse ratio of √Ek: 234 × √68/(√68 + √100) ≈ 106.
CThis assumes the nucleus splits into two equal fragments, ignoring the different kinetic energies at the two peaks.
DThis assumes Ek ∝ m (as if the fragments had equal speeds): 234 × 100/168 ≈ 139, the heavy fragment.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the properties of the products of nuclear fission and their management; A.2 — explosions; conservation of linear momentum; A.3 — Ek = p2/2mCommand term: Deduce
17E-1A-120
Where the energy of fission goes·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The chart shows how the average energy released per fission of uranium-235 is shared between different forms. The decay energies are those released later by the radioactive fission products.
A reactor has operated at constant power for a long time. Which is the best estimate of the percentage of the energy released per fission that is transferred to internal energy of the reactor (core, moderator, coolant and shielding)?
Average energy released per fission of uranium-235, in its different forms (bars drawn to scale).Show mark scheme
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Step 1Total energy per fission = 169 + 5 + 7 + 7 + 6 + 9 = 203 MeV.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The fragments, neutrons, β− particles and γ rays are all stopped within the reactor; the antineutrinos pass through all of it almost without interacting.
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Step 3After a long time at constant power the products decay at the same rate as they are formed, so their decay energy is deposited too: (203 − 9)/203 × 100 = 96 %.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis counts only the kinetic energy of the fragments; the neutrons, β− particles and γ rays are also absorbed in the reactor.
BThis counts only the energy released at the moment of fission, 181 MeV, leaving out the decay energy of the products, which is deposited while the reactor runs.
CCorrect: only the 9 MeV carried by the antineutrinos escapes: 194/203 = 96 %.
DThis assumes that energy conservation means all the energy stays in the reactor; the antineutrinos carry 9 MeV out of it.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the properties of the products of nuclear fission and their management; E.3 — β− decay and the antineutrino; the penetration and ionizing ability of radiation Command term: Determine
18E-1A-121
Parts of a nuclear power station·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
The diagram shows the main parts of a nuclear power station with a pressurized water reactor. Four parts are labelled P, Q, R and S.
Which part transfers thermal energy from the water that has passed through the core to the water that drives the turbine, so that no radioactive material reaches the turbine?
Main parts of a pressurized water reactor power station (not to scale).Show mark scheme
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Step 1The water pumped through the core (the primary circuit) becomes radioactive and stays in a closed loop between the pressure vessel and R.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In R this water flows through a coil and transfers thermal energy to the water of the separate secondary circuit, which boils.
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Step 3R is the heat exchanger: only the steam from the secondary circuit reaches the turbine.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AP are the control rods, which absorb neutrons to control the rate of fission; they do not transfer thermal energy.
BQ is the concrete shielding, which absorbs neutrons and gamma rays escaping from the core; it does not separate the water circuits.
CCorrect: R is the heat exchanger between the primary and secondary circuits.
DS is the condenser, which cools the steam leaving the turbine; this water has already been through the turbine.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: Identify
19E-1A-122
Changing the power of a reactor·E.4 Fission
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A reactor is operating at constant power. The control rods are withdrawn slightly, and the power starts to rise. When the power has doubled, the control rods are returned to their original positions. Ignore any effects of the changes in temperature and in the amount of fission products.
Three statements are:
I. While the control rods are withdrawn, on average more than one neutron from each fission causes a further fission. II. After the control rods are returned, the power falls back to its original value. III. After the control rods are returned, the rate of fission stays at about twice its original value.
Which statements are correct?
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Mark
Notes
Step 1With the rods withdrawn, fewer neutrons are absorbed, so on average more than one neutron per fission causes a further fission and the rate of fission grows — I is correct.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Returning the rods restores the original neutron absorption, so again exactly one neutron per fission causes a further fission.
—
Step 3Each generation then has the same number of fissions as the one before, so the rate of fission (and the power) stays at the doubled value; it does not return to the original value — II is wrong, III is correct.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AII is wrong: returning the rods stops the growth but does not reverse it; to reduce the power the rods must be pushed in further than before for a while.
BCorrect: the rods control whether the rate of fission grows, stays constant or falls, not its value.
CI is correct: the power can only rise if, on average, more than one neutron per fission causes another fission.
DII and III cannot both be correct: they predict different final powers.
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: Deduce
20E-1A-123
Growth of a chain reaction·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In a model of a chain reaction, each fission causes on average k further fissions in the next generation. The first generation contains N fissions.
How many fissions are there in generation n?
Show mark scheme
Marking point
Mark
Notes
Step 1Generation 2 has Nk fissions, generation 3 has Nk2, and so on.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each step from one generation to the next multiplies by k; there are n − 1 steps from generation 1 to generation n.
—
Step 3Generation n has Nkn − 1 fissions.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: n − 1 multiplications by k take the number from N to Nkn − 1.
BThis counts one multiplication too many: it is the number in generation n + 1.
CThis is the total number of fissions in all n generations added together, not the number in generation n.
DThis assumes the number grows linearly, adding the same number of fissions each generation.
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions Command term: Deduce
21E-1B-17
Energy from fission·E.4 Fission
Paper 1BEasy6 marks
Data-based question5 steps to full marksDetermine
In a pool-type research reactor, the thermal energy released in the core is removed by water pumped through it. With the reactor held at constant power, the operators vary the mass flow rate ṁ of the cooling water and record the rise ΔT in water temperature between the inlet and the outlet of the core. Energy losses from the core other than to the cooling water are negligible, and each fission of a uranium-235 nucleus releases 200 MeV.
The graph shows ΔT against 1/ṁ with a line of best fit through the origin. (Specific heat capacity of water = 4180 J kg−1 K−1; 1 MeV = 1.60 × 10−13 J; 1 u = 1.66 × 10−27 kg)
ṁ / kg s−1
1/ṁ / kg−1 s
ΔT / K
20.0
0.0500
28.8
25.0
0.0400
22.9
30.0
0.0333
19.2
40.0
0.0250
14.3
50.0
0.0200
11.5
60.0
0.0167
9.5
ΔT against 1/ṁ (graph drawn to scale)
(a)
(i)
Determine the gradient of the line, giving its unit.
(2)
(b)
(i)
Show that the thermal power of the reactor is about 2.4 MW.
(1)
(c)
(i)
Determine the number of fissions per second in the core.
(2)
(d)
(i)
Estimate the mass of uranium-235 that undergoes fission in one day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient = 28.7/0.0500 = 574
✓ 1
Accept 560–590.
Unit: K kg s−1
✓ 1
Part (b)(i)
P = ṁcΔT, so ΔT = (P/c)(1/ṁ) and P = c × gradient = 4180 × 574 = 2.40 × 106 W
✓ 1
Allow ECF from (a). At least 3 s.f. or full substitution required.
Part (c)(i)
Energy per fission = 200 × 1.60 × 10−13 = 3.20 × 10−11 J
✓ 1
Rate = 2.40 × 106/3.20 × 10−11 = 7.5 × 1016 s−1
✓ 1
Allow ECF from (b). Use of 2.4 MW gives 7.5 × 1016 s−1.
Answers: (a)(i) 574 K kg s−1 · (b)(i) 2.40 × 106 W · (c)(i) 7.5 × 1016 s−1 · (d)(i) 2.5 × 10−3 kg (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in neutron-induced fission; the role of heat exchangers in a nuclear power plant; guidance: calculations to determine the energy released in fission reactions; B.1 — quantitative analysis of thermal energy transfers with Q = mcΔT; Tools 3 — gradient with its unit Command term: Determine
22E-1B-33
A model heat exchanger·E.4 Fission
Paper 1BEasy8 marks
Data-based question6 steps to full marksDetermine
A student models the heat exchanger of a nuclear power station. Hot water, representing the primary circuit, flows at a constant mass flow rate of (2.00 ± 0.04) g s−1 through a copper coil immersed in 2.00 kg of cold water, representing the secondary circuit, in an insulated beaker. The secondary water is stirred and its temperature θ is recorded every 60 s.
During the 600 s of the experiment, the mean temperatures of the hot water entering and leaving the coil are 75.0 °C and 51.0 °C. Each thermometer reading has an uncertainty of ±0.5 °C. The specific heat capacity of water is 4180 J kg−1 K−1.
t / s
0
60
120
180
240
300
360
420
480
540
600
θ / °C
17.9
19.3
20.6
21.7
23.1
24.2
25.6
26.9
28.2
29.7
30.9
Temperature θ of the secondary water against time t, with the line of best fit (drawn to scale).
(a)
(i)
Determine the gradient of the graph. State its unit.
(2)
(ii)
Show that the rate at which the secondary water gains thermal energy is about 180 W.
(1)
(b)
(i)
Calculate the rate at which the hot water loses thermal energy in the coil.
(1)
(ii)
Determine the absolute uncertainty in your answer to (b)(i).
(2)
(c)
(i)
Deduce, with reference to your answers, whether all the energy lost by the hot water is gained by the secondary water, and suggest one change to the apparatus that would make the two powers agree more closely.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from a large triangle on the line, e.g. (30.84 − 17.90)/600 = 0.0216
✓ 1
Accept 0.0205–0.0225. Two adjacent points only: [0] for this mark.
Unit: K s−1 (or °C s−1)
✓ 1
Part (a)(ii)
Q/t = 2.00 × 4180 × 0.0216 = 181 W
✓ 1
Allow ECF from (a)(i). Full substitution or an answer to at least 3 s.f. required.
Part (b)(i)
Q/t = 2.00 × 10−3 × 4180 × (75.0 − 51.0) = 201 W
✓ 1
Mass flow rate in g s−1 without conversion: [0].
Part (b)(ii)
ΔT = 24.0 ± 1.0 K (two readings, ±0.5 K each), i.e. 4.2 %; flow rate 2.0 %
✓ 1
Using ±0.5 K for the difference: [0] for this mark.
Total = 6.2 %, so uncertainty = ±12.4 W ≈ ±12 W
✓ 1
Allow ECF from (b)(i). Accept ±12 W or ±13 W.
Part (c)(i)
181 W lies outside the range 189–213 W (201 ± 12 W), so not all the energy lost by the hot water is gained by the secondary water: some is transferred to the surroundings (e.g. from the pipes or the beaker)
✓ 1
Allow ECF from (a)(ii) and (b)(ii). Comparison with the uncertainty range and the conclusion both needed.
Improvement: insulate the pipes leading to and from the beaker and add an insulating lid to the beaker (or lag the coil inlet/outlet tubes)
✓ 1
"Repeat the experiment" or "use more precise thermometers": [0].
Answers: (a)(i) 0.0216 K s−1 · (a)(ii) 181 W · (b)(i) 201 W · (b)(ii) ±12 W · (c)(i) no: 181 W is outside 201 ± 12 W (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; B.1 — quantitative analysis of thermal energy transfers with Q = mcΔT; Tools — gradient with units, propagation of uncertainties, comparing a result with its uncertainty, systematic error Command term: Determine
23E-1B-34
Approach to criticality·E.4 Fission
Paper 1BMedium7 marks
Data-based question6 steps to full marksDeduce
A new research reactor is loaded with fuel for the first time. A neutron source is placed at the centre of the core and fuel elements are added in batches. After each batch a neutron detector counts for 200 s. The count rate due to background is negligible.
Let k be the average number of neutrons from each fission that cause a further fission. A model predicts that the count rate C is proportional to 1/(1 − k), as long as k < 1. The table and the graph show the results, where n is the number of fuel elements in the core.
n
counts in 200 s
C / s−1
(1/C) / 10−3 s
12
23994
120.0
8.33
24
28862
144.3
6.93
36
36640
183.2
5.46
48
50138
250.7
3.99
60
80095
400.5
66
111319
556.6
1.80
Reciprocal of the count rate, 1/C, against the number of fuel elements n, with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of 1/C.
(1)
(ii)
Explain why the count rate increases as fuel elements are added.
(2)
(b)
(i)
Use the graph to predict the number of fuel elements for which a self-sustaining chain reaction would occur. Explain your method.
(2)
(c)
(i)
The uncertainty in a count N is √N. Calculate the percentage uncertainty in C for n = 66.
(1)
(ii)
The operators add fewer elements in each batch as the number of elements approaches your answer to (b)(i). Suggest why.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
1/C = 1/400.5 = 2.50 × 10−3 s
✓ 1
Accept 2.50 × 10−3 s or 0.00250 s.
Part (a)(ii)
A larger core loses a smaller fraction of its neutrons through its surface, so k increases (towards 1)
✓ 1
OWTTE.
Each neutron from the source then starts a longer chain of fissions, so more neutrons reach the detector
✓ 1
Reference to fission chains (or to 1/(1 − k) increasing) is required.
Part (b)(i)
A self-sustaining chain reaction needs k = 1, when C → ∞, i.e. 1/C = 0
✓ 1
Extrapolating the line to 1/C = 0 gives n ≈ 81
✓ 1
Accept 77–84. Allow ECF from (a)(i).
Part (c)(i)
√111 319/111 319 × 100 = 0.3 %
✓ 1
Accept 0.3 %.
Part (c)(ii)
Near criticality C rises very steeply, and the prediction (an extrapolation) is reliable only close to the data; small batches avoid accidentally exceeding k = 1, when the power would grow without limit
✓ 1
Allow ECF from (b)(i). A reference to the uncertainty of the extrapolation or to safety (avoiding k > 1) is needed.
Answers: (a)(i) 2.50 × 10−3 s · (b)(i) n ≈ 81 · (c)(i) 0.3 % (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; Tools — linearization, extrapolation of a graph to an intercept, random uncertainty in counting; E.3 — count rate Command term: Deduce
24E-1B-35
Neutron shielding by water·E.4 Fission
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
Water is used as a shield around research reactors. To test its effect, a student places tanks of water of total thickness x between a source of fast neutrons and a detector of slow neutrons. For x ≥ 10 cm, the model predicts that the count rate due to the source is C − B = C0e−μx, where C is the measured count rate and B is the background count rate.
With the source removed, the student measures B = 0.60 s−1. The table shows the results; the graph shows ln((C − B)/s−1) against x for all six thicknesses, with the line of best fit.
x / cm
C / s−1
ln((C − B)/s−1)
10
422
6.04
20
144
4.97
30
51.2
3.92
40
18.9
2.91
50
6.81
1.83
60
2.77
ln((C − B)/s−1) against water thickness x, with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value in the table.
(1)
(ii)
Explain why it is important to subtract the background before taking logarithms, especially for the largest thicknesses.
(1)
(b)
(i)
Determine μ from the graph. State its unit.
(2)
(ii)
Determine the thickness of water needed to reduce the count rate due to the source by a factor of 1.0 × 106.
(2)
(c)
(i)
The student counted for the same time at every thickness. Suggest why the uncertainty in ln((C − B)/s−1) is greatest for x = 60 cm.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ln(2.77 − 0.60) = ln 2.17 = 0.77
✓ 1
ln 2.77 = 1.02 (background not subtracted): [0].
Part (a)(ii)
At large x the count rate from the source is comparable with B; ln C would level off towards ln B, so the graph would curve and its gradient (and μ) would be too small
✓ 1
OWTTE. Reference to the relative size of the background at large x is required.
Part (b)(i)
Gradient from a large triangle = (0.78 − 6.03)/(60 − 10) = −0.105, so μ = 0.105
✓ 1
Accept 0.100–0.110. μ = −gradient; a negative value of μ: [0] for this mark.
Unit: cm−1
✓ 1
Accept m−1 with the value converted (10.5 m−1).
Part (b)(ii)
e−μΔx = 1.0 × 10−6, so Δx = ln(1.0 × 106)/μ
✓ 1
Allow ECF from (b)(i).
Δx = 13.8/0.105 = 131 cm ≈ 1.3 m
✓ 1
Accept 125–138 cm. Using log10: [1 max].
Part (c)(i)
Fewest counts are recorded at 60 cm, so the random (fractional) uncertainty in the counts is greatest, and it is increased further because B is a large fraction of C there
✓ 1
Allow ECF from (a)(ii). Either idea, linked to counting statistics.
Answers: (a)(i) 0.77 · (b)(i) 0.105 cm−1 · (b)(ii) 131 cm (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; E.3 — background radiation; count rate; Tools — linearization of an exponential relationship with logarithms, gradient with units, correction for background Command term: Determine
25E-2-03
Energy flow in a nuclear power station·E.4 Fission
Paper 2Medium12 marks
Short answer & extended response7 steps to full marksEstimate
A pressurized-water reactor delivers 1.10 GW of electrical power with an overall efficiency of 33 %. Each fission of uranium-235 releases on average 200 MeV, and the fuel contains 4.0 % uranium-235 by mass; the rest is uranium-238.
Water pumped through the core at high pressure enters at 290 °C and leaves at 325 °C. Its specific heat capacity at these temperatures is 5.5 kJ kg−1 K−1.
The water from the core is radioactive, so it is kept in a closed primary circuit
✓ 1
In the heat exchanger it transfers thermal energy to water in a separate secondary circuit, which boils to make the steam that drives the turbines; no radioactive material reaches the turbines
✓ 1
Part (c)(i)
Control rods absorb neutrons (without undergoing fission)
✓ 1
Do not accept "slow the neutrons down": that is the role of the moderator.
They are inserted or withdrawn so that on average exactly one neutron from each fission causes a further fission, keeping the fission rate (and so the power) constant
✓ 1
Answers: (a)(ii) 1.0 × 1020 s−1 · (a)(iii) 32 t · (b)(i) 1.7 × 104 kg s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in neutron-induced fission; the role of chain reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; A.3 — efficiency; B.1 — Q = mcΔTCommand term: Estimate
26E-2-13
Moderation and shielding·E.4 Fission
Paper 2Hard11 marks
Short answer & extended response6 steps to full marksDeduce
In a thermal reactor, fission neutrons of kinetic energy about 2.0 MeV are slowed down by collisions with the carbon-12 nuclei of a graphite moderator.
In a head-on elastic collision between a neutron of mass m, moving with speed v, and a stationary nucleus of mass M, the neutron rebounds with speed v(M − m)/(M + m). Take M = 12m for carbon.
(a)
(i)
Show that the neutron keeps about 72 % of its kinetic energy in a head-on collision with a carbon nucleus.
(1)
(ii)
Determine the minimum number of collisions needed to reduce the kinetic energy of a neutron from 2.0 MeV to 0.040 eV.
(2)
(iii)
Suggest why more collisions than this are needed in practice.
(1)
(b)
(i)
Show, using conservation of momentum, that in a head-on collision the carbon nucleus moves off with speed 2v/13.
(2)
(ii)
Deduce, without numerical substitution, the speed of the neutron after a head-on collision with a stationary hydrogen nucleus (M = m), and hence suggest why water slows neutrons in far fewer collisions than graphite.
(2)
(c)
(i)
Explain why slowing the neutrons is necessary to sustain the chain reaction in fuel that is mostly uranium-238.
(2)
(ii)
State why the reactor core is surrounded by thick concrete.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Fraction = [(M − m)/(M + m)]² = (11/13)² = 0.716
✓ 1
Must see the square of the speed ratio.
Part (a)(ii)
2.0 × 106 × 0.716N = 0.040, so N = ln(2.0 × 10−8)/ln 0.716
✓ 1
Allow ECF from (a)(i).
N = 53.1, so 54 collisions
✓ 1
Accept 53 or 54.
Part (a)(iii)
Most collisions are not head-on (glancing), and in these the neutron loses a smaller fraction of its kinetic energy
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
mv = m(−11v/13) + 12mV, taking the neutron's initial direction as positive
✓ 1
The rebound direction must be shown by the sign.
12V = v + 11v/13 = 24v/13, so V = 2v/13
✓ 1
Part (b)(ii)
(M − m)/(M + m) = 0: the neutron stops, giving all its kinetic energy to the hydrogen nucleus
✓ 1
A hydrogen nucleus can take most or all of the neutron's kinetic energy in one collision, whereas a carbon nucleus takes at most about 28 %
✓ 1
Allow ECF from (a)(i).
Part (c)(i)
Slow (thermal) neutrons are much more likely than fast neutrons to be captured by uranium-235 nuclei and cause fission
✓ 1
Without moderation too few of the neutrons from each fission would cause a further fission (many escape or are absorbed by uranium-238), so the chain reaction would die out
✓ 1
Part (c)(ii)
Shielding: it absorbs the neutrons and gamma radiation escaping from the core, protecting workers and the environment
✓ 1
Answers: (a)(ii) 54 (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — the role of chain reactions in nuclear fission; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; A.2 — conservation of linear momentum and elastic collisions Command term: Deduce
27E-2-25
Spent fuel and its management·E.4 Fission
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine
After removal from a reactor, a spent fuel assembly is stored under water. The graph shows the thermal power P produced in the assembly by radioactive decay against the time t after its removal.
Thermal power of a spent fuel assembly against time (graph drawn to scale)
(a)
(i)
Explain why the assembly continues to produce thermal energy after the chain reaction has stopped.
(2)
(ii)
Use the graph to show that the decrease in P cannot be described by a single half-life.
(2)
(iii)
At t = 100 days the cooling system of a pool containing 1.5 × 105 kg of water and one such assembly fails. Determine the initial rate of rise of the water temperature in K per day. (Specific heat capacity of water = 4.2 × 103 J kg−1 K−1)
(2)
(b)
(i)
After several decades the thermal power is due mainly to caesium-137, of half-life 30.1 years. Determine the time for the activity of the caesium-137 to fall to 0.10 % of its initial value.
(2)
(ii)
Discuss one advantage and one disadvantage of placing such waste in a deep geological repository rather than storing it at the surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The fission products are radioactive (neutron-rich nuclei that decay mainly by β− and γ emission)
✓ 1
The energy of the emitted radiation is absorbed within the assembly, heating it
✓ 1
Part (a)(ii)
P falls from 40 kW to 20 kW in about 46 days
✓ 1
Accept 40–52 days (or any other correctly read halving).
but from 20 kW to 10 kW in about 84 days: the halving time increases, so several nuclides with different half-lives contribute
✓ 1
Accept 75–95 days. A conclusion is needed.
Part (a)(iii)
P(100 days) = 12.5 kW
✓ 1
Accept 12–13 kW.
ΔT/Δt = P/(mc) = 1.25 × 104/(1.5 × 105 × 4.2 × 103) = 2.0 × 10−5 K s−1 = 1.7 K per day
✓ 1
Accept 1.6–1.8 K per day; the answer must be in K per day for MP2.
Advantage: isolated in stable rock far from people and groundwater, secure against misuse, and needs no active cooling or monitoring by future generations
✓ 1
Disadvantage: it must stay intact for hundreds of years (300 years for caesium-137, far longer for long-lived nuclides), which cannot be verified; groundwater could carry radioactive material; it is hard to retrieve
✓ 1
Allow ECF from (b)(i).
Answers: (a)(iii) 1.7 K day−1 · (b)(i) 300 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — the properties of the products of nuclear fission and their management; guidance: the impact of the long-term storage of nuclear waste; E.3 (HL) — the decay law for arbitrary time intervals; B.1 — Q = mcΔTCommand term: Determine
28E-2-26
Spontaneous fission·E.4 Fission
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksDetermine
Californium-252 (25298Cf) has a half-life of 2.65 years. Of its nuclei that decay, 96.9 % emit an alpha particle and 3.1 % undergo spontaneous fission, which releases on average 3.8 neutrons. Small californium-252 sources are used as neutron sources to start up nuclear reactors.
Atomic masses: 252Cf 252.081 627 u, 140Xe 139.921 645 u, 108Ru 107.910 186 u; neutron 1.008 665 u. (1 year = 3.16 × 107 s)
(a)
Spontaneous fission.
(i)
Distinguish between spontaneous fission and neutron-induced fission.
(1)
(ii)
One spontaneous fission of californium-252 produces xenon-140 (Z = 54), ruthenium-108 (Z = 44) and x neutrons. Deduce x.
(1)
(iii)
Calculate the energy released in this fission.
(2)
(b)
The source.
(i)
A start-up source contains 5.0 μg of californium-252. Show that its activity is about 1.0 × 108 Bq.
(2)
(ii)
Calculate the number of neutrons emitted per second by the source.
(1)
(c)
Starting the reactor.
(i)
Explain the role of the source in starting the chain reaction in fresh uranium fuel.
(2)
(ii)
The source is replaced when its neutron emission rate has fallen to 25 % of its initial value. State the time after which it is replaced.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Spontaneous fission happens randomly without any external cause, like other radioactive decays; neutron-induced fission occurs when a nucleus splits after absorbing a (slow) neutron
A chain reaction must be started by free neutrons inducing fission of uranium-235, but fresh fuel contains very few (spontaneous fission of uranium is extremely rare)
✓ 1
The source supplies a steady, known flux of neutrons; each can induce a fission whose neutrons cause further fissions, so the chain reaction builds up reliably and can be monitored as the control rods are withdrawn
✓ 1
Part (c)(ii)
25 % = two half-lives = 2 × 2.65 = 5.3 years
✓ 1
Answers: (a)(ii) 4 · (a)(iii) 200 MeV · (b)(i) 9.9 × 107 Bq · (b)(ii) 1.2 × 107 s−1 · (c)(ii) 5.3 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of chain reactions in nuclear fission reactions; guidance: calculations to determine the energy released in fission reactions; E.3 — activity and half-life; E.3 (HL) — A = λN and T½ = ln 2/λCommand term: Determine
29E-2-34
Fission fragments·E.4 Fission
Paper 2Medium12 marks
Short answer & extended response7 steps to full marksDetermine
When a uranium-235 nucleus absorbs a slow neutron, the resulting uranium-236 nucleus splits into two fragments and a few neutrons. The graph shows the percentage of fissions, the fission yield, that produce a fragment of each nucleon number A.
(1 u = 1.661 × 10−27 kg, 1 MeV = 1.60 × 10−13 J)
Fission yield against nucleon number of the fragment for uranium-235 (graph drawn to scale)
(a)
(i)
State what the graph shows about the relative sizes of the two fragments in most fissions.
(1)
(ii)
The yields shown on the graph add up to 200 %. Suggest why.
(1)
(b)
In one fission, the fragments are strontium-94 (Z = 38) and xenon-140 (Z = 54): 10n + 23592U → 9438Sr + 14054Xe + x10n Binding energies per nucleon: uranium-235 7.591 MeV, strontium-94 8.594 MeV, xenon-140 8.291 MeV.
(i)
Deduce the value of x.
(1)
(ii)
Show that the energy released in this fission is about 180 MeV.
(2)
(iii)
Just after the nucleus splits, the two fragments are momentarily at rest with their centres 17 fm apart. Treating each fragment as a point charge, show that their electric potential energy is about 170 MeV.
(2)
(iv)
Hence state the origin of the kinetic energy of the fragments.
(1)
(v)
The fragments eventually share 168 MeV of kinetic energy. Suggest why this is less than the energy released in (b)(ii).
(2)
(c)
(i)
In a fuel rod, the fission fragments stop within about 10 μm of the point where fission occurs, whereas the neutrons and gamma rays travel much further. Explain why, and state the importance of this for the transfer of energy in a reactor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The fragments are usually of unequal size, with nucleon numbers near 95 and 139; splitting into two equal fragments (A ≈ 117) is very rare
✓ 1
Both the unequal split and an approximate pair of values (accept 90–100 and 134–144) needed.
Part (a)(ii)
Each fission produces two fragments, so every fission is counted twice
✓ 1
Part (b)(i)
Nucleon number: 235 + 1 = 94 + 140 + x, so x = 2
✓ 1
Accept a check of proton numbers 38 + 54 = 92 as well.
Part (b)(ii)
Energy released = 94 × 8.594 + 140 × 8.291 − 235 × 7.591
✓ 1
The free neutrons have no binding energy.
= 807.8 + 1160.7 − 1783.9 = 185 MeV
✓ 1
An answer to at least 3 s.f. must be seen. Using 236 nucleons for uranium scores [1 max].
An answer to at least 3 s.f. (or 2.78 × 10−11 J with the conversion) must be seen.
Part (b)(iv)
The electric repulsion between the positively charged fragments: their electric potential energy is converted into kinetic energy as they fly apart
✓ 1
Allow ECF from (b)(iii). Do not accept "the strong force" or "binding energy" alone.
Part (b)(v)
The neutrons released carry away kinetic energy (about 2 MeV each)
✓ 1
The fragments are left in excited states and emit gamma-ray photons (and later decay by β− emission, with energy carried by antineutrinos)
✓ 1
Any two distinct ways for [2]. Allow ECF from (b)(ii).
Part (c)(i)
The fragments are heavy and highly charged, so they ionise very strongly and lose their kinetic energy in a very short distance (neutrons are uncharged; gamma rays ionise weakly)
✓ 1
Accept "large charge" with "strong ionisation".
So most of the energy released (the fragments' share, 168 of about 185 MeV) becomes internal energy of the fuel itself, which is then transferred to the coolant
✓ 1
Allow ECF from (b)(ii).
Answers: (b)(i) 2 · (b)(iii) 174 MeV (2.78 × 10−11 J) (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the properties of the products of nuclear fission; guidance: calculations to determine the energy released in fission reactions; E.3 — the variation of the binding energy per nucleon with nucleon number; the penetration and ionizing ability of radiation; D.2 — electric potential energy Ep = kq1q2/rCommand term: Determine
30E-2-35
A natural fission reactor·E.4 Fission
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksDeduce
About 2.0 × 109 years ago, in a uranium ore deposit in Gabon, groundwater allowed natural chain reactions of neutron-induced fission of uranium-235 to run for a long period. Today, uranium-235 makes up 0.720 % of the uranium atoms in natural uranium; almost all the rest is uranium-238. Modern reactors that use ordinary water need fuel in which at least about 3 % of the uranium atoms are uranium-235.
Show that the decay constant of uranium-235 is about 9.8 × 10−10 year−1.
(1)
(ii)
Show that the ratio R = (number of uranium-235 atoms)/(number of uranium-238 atoms) a time t ago was Rnowe(λ5 − λ8)t, where Rnow is the present ratio and λ5 and λ8 are the decay constants.
(2)
(iii)
Hence determine the percentage of uranium atoms that were uranium-235 at the time the natural reactor operated.
(2)
(b)
(i)
State the role of the groundwater in the chain reaction.
(1)
(ii)
Evidence suggests that the natural reactor switched itself on for about 30 minutes and then off for about 2.5 hours, repeatedly. Suggest an explanation, and outline why the chain reaction could not run out of control.
(3)
(c)
(i)
It is estimated that about 6.0 × 103 kg of uranium-235 underwent fission over a period of about 1.5 × 105 years, with about 200 MeV released per fission. Estimate the average power of the natural reactor.
(3)
(ii)
Explain, using your answer to (a)(iii), why a natural reactor of this kind cannot form today.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λ = ln 2/7.04 × 108 = 9.85 × 10−10 year−1
✓ 1
An answer to at least 3 s.f. must be seen.
Part (a)(ii)
Each isotope decays exponentially from its earlier number: N5,now = N5e−λ5t and N8,now = N8e−λ8t
✓ 1
Dividing: Rnow = Re−(λ5 − λ8)t, so R = Rnowe(λ5 − λ8)t
Allow ECF from (a)(i) and (a)(ii). Using 0.00720 for Rnow changes the answer only in the third s.f.
R = 0.0381, so the percentage is 0.0381/1.0381 × 100 = 3.7 %
✓ 1
Accept 3.6–3.8 %. Decaying uranium-235 alone (0.72 × 7.2 = 5.2 %) or using λ5 + λ8 scores [1 max].
Part (b)(i)
Moderator: it slowed the fast fission neutrons (by collisions with hydrogen nuclei) so that they were likely to cause further fissions of uranium-235
✓ 1
Do not accept "water absorbs neutrons" or "water cools the ore".
Part (b)(ii)
The energy released heated the ore until the water boiled away (or was driven out), so the neutrons were no longer slowed and the chain reaction stopped
✓ 1
Allow ECF from (b)(i).
When the rock cooled, water seeped back, moderation resumed and the chain reaction restarted
✓ 1
Negative feedback: any increase in power heats the ore and drives out more water, which reduces the moderation and hence the fission rate, so the power cannot keep rising
✓ 1
OWTTE. Self-regulation must be linked to the loss of moderation.
Part (c)(i)
Number of fissions = 6.0 × 103/(235 × 1.661 × 10−27) = 1.5 × 1028
P = 4.92 × 1017/(1.5 × 105 × 3.16 × 107) = 1.0 × 105 W
✓ 1
Allow ECF from MP2. Accept about 100 kW (0.95–1.1 × 105 W). Forgetting to convert years to seconds (3.3 × 1012 W) scores [2 max].
Part (c)(ii)
The uranium-235 fraction is now only 0.72 %, well below the ≈ 3 % needed for a chain reaction with water as moderator, whereas 2.0 × 109 years ago it was 3.7 %, above it
✓ 1
Allow ECF from (a)(iii). The comparison with ≈ 3 % at both times is required.
Answers: (a)(iii) 3.7 % · (c)(i) 1.0 × 105 W (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in neutron-induced fission; the role of chain reactions in nuclear fission reactions; the role of moderators in a nuclear power plant; guidance: calculations to determine the energy released in fission reactions; E.3 (HL) — the radioactive decay law N = N0e−λt; T½ = ln 2/λ; application of the decay equations for arbitrary time intervals; A.3 — power Command term: Deduce
31E-2-52
A fission power system for a spacecraft·E.4 Fission
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksDetermine
A spacecraft travelling to the outer planets is powered by a small uranium-235 fission reactor. The thermal energy from the core is used by heat engines, which deliver an electrical power of 10.0 kW with an overall efficiency of 25.0 %. The hot end of the engines is at 1070 K and the cold end at 420 K. All the energy not converted to electrical energy is transferred to a radiator at 420 K, which emits it into space.
Each fission releases 200 MeV. (1 u = 1.661 × 10−27 kg, 1 MeV = 1.60 × 10−13 J, 1 year = 3.16 × 107 s)
(a)
(i)
Calculate the thermal power produced in the core.
(1)
(ii)
Determine the maximum possible efficiency of heat engines working between 1070 K and 420 K, and suggest why the overall efficiency is lower.
(2)
(b)
(i)
The reactor runs at this power for 12 years. Determine the mass of uranium-235 that undergoes fission in this time.
(3)
(c)
(i)
The emissivity of the radiator is 0.90. Determine the total radiating area that the radiator needs. Ignore any radiation absorbed by the radiator.
(2)
(ii)
State why, in space, the energy must be removed from the spacecraft by thermal radiation.
(1)
(iii)
A designer suggests raising the temperature of the radiator and of the cold end of the engines to 470 K to make the radiator smaller. Discuss this suggestion, with reference to your answers to (a)(ii) and (c)(i).
(2)
(d)
(i)
The reactor has a thick shield only on the side facing the rest of the spacecraft. Identify the radiation that this shield must absorb and suggest why the reactor is not shielded on all sides.
(2)
Show mark scheme
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Notes
Part (a)(i)
Pth = 10.0/0.250 = 40.0 kW
✓ 1
Part (a)(ii)
ηCarnot = 1 − 420/1070 = 0.607 (61 %)
✓ 1
Temperatures in °C: [0] for this mark.
Real engines are irreversible (friction, energy transfers across finite temperature differences) and there are further losses in the generators
There is no surrounding material (vacuum), so energy cannot be transferred to the surroundings by conduction or convection
✓ 1
Part (c)(iii)
Power radiated per unit area ∝ T4, so the radiator area for the same power would fall by a factor (420/470)4 ≈ 0.64
✓ 1
Allow ECF from (c)(i).
But the maximum (Carnot) efficiency falls to 1 − 470/1070 = 56 %, so the engines are likely to be less efficient: more thermal power (more fuel) is needed for the same electrical output and more waste power must be radiated, partly offsetting the smaller area
✓ 1
Allow ECF from (a)(ii). Both effects needed for [2].
Part (d)(i)
Neutrons and gamma rays from the core (they are the penetrating radiations escaping from the reactor)
✓ 1
Alpha and beta are stopped within the core: [0] if listed as the main radiations.
In a vacuum the radiation travels in straight lines with nothing to scatter it back, so a shield is needed only in the direction of the crew/instruments; shielding all sides would add a large mass
✓ 1
Any valid reason linked to mass or to the geometry.
Answers: (a)(i) 40.0 kW · (a)(ii) 61 % · (b)(i) 0.19 kg · (c)(i) 19 m2(the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; guidance: calculations to determine the energy released in fission reactions; the role of chain reactions in nuclear fission reactions; A.3 — efficiency; B.4 — Carnot efficiency ηCarnot = 1 − Tc/Th; B.1 — conduction, convection and thermal radiation; B.2 — emissivity; B.1 — Stefan–Boltzmann law Command term: Determine
32E-2-53
A fission chamber·E.4 Fission
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A fission chamber monitors the neutron flux in a reactor. It contains argon gas between two parallel metal plates 5.0 mm apart, with a potential difference of 250 V between them. One plate is coated with a thin layer of uranium-235. When a slow neutron causes fission of a uranium-235 nucleus in the coating, one of the two fragments enters the gas and loses all its kinetic energy there by ionizing argon atoms. The energy needed to produce one ion pair in argon is 26 eV.
(1 u = 1.661 × 10−27 kg, 1 year = 3.16 × 107 s)
(a)
(i)
Explain, using conservation of momentum, why only one of the two fragments enters the gas.
(2)
(b)
(i)
A fragment has an initial kinetic energy of 98 MeV. Show that the charge of either sign that it releases in the gas is about 6 × 10−13 C.
(2)
(ii)
The average current in the chamber is 1.5 μA. Assuming every fragment releases the charge in (b)(i), determine the rate of fission in the coating.
(1)
(iii)
Gamma-ray photons from the core also pass through the chamber. A photon of energy 2.0 MeV gives all its energy to the gas. Compare the charge it releases with that released by a fragment, and suggest how the chamber can tell the two apart.
(1)
(c)
(i)
Calculate the magnitude of the electric force on a singly charged argon ion between the plates.
(2)
(d)
(i)
The reactor is meant to run at constant power. The current in the chamber rises from 1.5 μA to 1.8 μA. Determine the new rate of fission in the coating and state what the operators should do with the control rods.
(2)
(e)
(i)
The coating contains 1.0 mg of uranium-235. Determine the time, in years, for 1.0 % of its nuclei to undergo fission at the rate found in (b)(ii), and comment on your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The slow neutron has negligible momentum, so the total momentum before the fission is (almost) zero
✓ 1
So the fragments move off in opposite directions: if one moves into the gas, the other moves into the plate
Charge = 2.0 × 106/26 × 1.60 × 10−19 = 1.2 × 10−14 C, about 2 % of that from a fragment; each pulse due to a photon is much smaller, so pulses below a threshold size are ignored
✓ 1
Allow ECF from (b)(i). Both the comparison and the method needed.
Part (c)(i)
E = 250/5.0 × 10−3 = 5.0 × 104 V m−1
✓ 1
F = qE = 1.60 × 10−19 × 5.0 × 104 = 8.0 × 10−15 N
✓ 1
Part (d)(i)
Rate ∝ current: 1.8/1.5 × 2.5 × 106 = 3.0 × 106 s−1 (the neutron flux has risen by 20 %)
✓ 1
Allow ECF from (b)(ii).
Push the control rods further in, so that they absorb more neutrons and fewer than one neutron per fission causes a further fission until the power is back to its set value
✓ 1
Then return the rods to keep exactly one neutron per fission causing further fission.
Part (e)(i)
Number of nuclei = 1.0 × 10−6/(235 × 1.661 × 10−27) = 2.56 × 1018
✓ 1
Time = 0.010 × 2.56 × 1018/2.5 × 106 = 1.02 × 1010 s = 324 years
✓ 1
Allow ECF from (b)(ii). Accept 320–330 years (unrounded rate 2.49 × 106 s−1 gives 326 years).
So the coating is used up extremely slowly: the sensitivity of the chamber stays (almost) constant over its working life
✓ 1
Allow ECF: comment must match the candidate's time.
Answers: (b)(i) 6.0 × 10−13 C · (b)(ii) 2.5 × 106 s−1 · (b)(iii) 1.2 × 10−14 C · (c)(i) 8.0 × 10−15 N · (d)(i) 3.0 × 106 s−1; insert the rods · (e)(i) 324 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of chain reactions in nuclear fission reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; A.2 — conservation of linear momentum, explosions; D.2 — uniform electric field strength E = V/d, electric force; B.5 — electric current I = ΔQ/Δt; E.3 — the ionizing ability of radiation Command term: Determine
33E-2-54
Antineutrinos from a reactor·E.4 Fission
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A reactor has a thermal power of 3.2 GW. Each fission of uranium-235 releases 200 MeV. One possible fission is
10n + 23592U → 14156Ba + 9236Kr + 310n
The fragments are radioactive and decay by a series of β− decays: krypton-92 decays to the stable nuclide zirconium-92 (Z = 40) and barium-141 to the stable nuclide praseodymium-141 (Z = 59). On average, the products of one fission emit 6.0 antineutrinos as they decay. (1 MeV = 1.60 × 10−13 J)
(a)
(i)
Deduce the number of antineutrinos emitted when both fragments of this fission decay to stable nuclides.
(2)
(ii)
Write the equation for the first β− decay of krypton-92.
(1)
(b)
(i)
Show that the rate of fission in the reactor is about 1 × 1020 s−1.
(1)
(ii)
Treating the core as a point source, determine the number of antineutrinos passing per second through 1 m2 perpendicular to their direction, 30 m from the core.
(2)
(c)
(i)
Outline how the energy spectrum of the β− particles emitted by the fission products provides evidence for the antineutrino.
(2)
(ii)
When the reactor is shut down, the rate of emission of antineutrinos falls very rapidly at first and then more and more slowly, and remains detectable for months. Explain this.
(3)
(d)
(i)
A detector 30 m from this reactor records 1500 antineutrino events per day. An identical detector placed 60 m from the core of a second reactor records 450 events per day. Determine the thermal power of the second reactor.
(2)
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Notes
Part (a)(i)
Each β− decay increases Z by 1 and emits one antineutrino
Omitting the distance factor (0.96 GW) or squaring it the wrong way (0.24 GW): [1 max].
Answers: (a)(i) 7 · (b)(i) 1.0 × 1020 s−1 · (b)(ii) 5.3 × 1016 m−2 s−1 · (d)(i) 3.8 GW (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the properties of the products of nuclear fission and their management; guidance: calculations to determine the energy released in fission reactions; E.3 — β− decay equations; neutrinos and antineutrinos; E.3 (HL) — the continuous beta spectrum as evidence for the neutrino; B.1 — the inverse square law for intensity (b = L/4πd2) Command term: Determine
34E-2-55
The thorium fuel cycle·E.4 Fission
Paper 2Medium14 marks
Short answer & extended response8 steps to full marksDetermine
Thorium-232 does not undergo fission when it absorbs a slow neutron. Instead it forms thorium-233, which decays by β− emission (half-life 22 minutes) to protactinium-233. Protactinium-233 decays by β− emission (half-life 27.0 days) to uranium-233, which undergoes fission with slow neutrons. In a thorium reactor some protactinium-233 is removed from the core and stored while it decays.
Atomic masses: uranium-233 233.039634 u; xenon-139 138.918792 u; strontium-93 92.914024 u; neutron 1.008665 u. (1 u = 1.661 × 10−27 kg = 931.5 MeV c−2, 1 MeV = 1.60 × 10−13 J, 1 year = 3.16 × 107 s)
(a)
(i)
Write the equations for the two β− decays that convert thorium-233 (Z = 90) into uranium-233.
(2)
(ii)
Determine the fraction of a sample of protactinium-233 that has decayed to uranium-233 after 60 days of storage.
(2)
(b)
One fission of uranium-233 is: 10n + 23392U → 13954Xe + 9338Sr + 210n
(i)
Show that the energy released in this fission is about 180 MeV.
(2)
(ii)
Each thorium-232 nucleus that absorbs a neutron eventually provides one uranium-233 nucleus. Assuming every fission releases the energy in (b)(i), estimate the energy that can be obtained from 1.0 kg of thorium-232, and compare it with the energy density of coal, 3.0 × 107 J kg−1. (Mass of a thorium-232 atom = 232.04 u)
(3)
(iii)
A thorium reactor produces a thermal power of 1.0 GW. Estimate the mass of thorium-232 that it uses in one year.
(1)
(c)
(i)
Explain why a thorium reactor must contain some uranium-235 (or another nuclide that undergoes fission with slow neutrons) when it is first started.
(2)
(ii)
Suggest why protactinium-233 is removed from the core while it decays. Deduce whether 60 days of storage is enough for at least 95 % of it to decay.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
23390Th → 23391Pa + 0−1e + ν̄e
✓ 1
23391Pa → 23392U + 0−1e + ν̄e
✓ 1
Antineutrino needed in at least one equation for [2]; penalize its omission once.
Allow ECF from (b)(i). Using 180 MeV gives 7.5 × 1013 J.
About 2.6 × 106 times the energy density of coal
✓ 1
Allow ECF. Accept 2.5–2.6 × 106.
Part (b)(iii)
Mass = 1.0 × 109 × 3.16 × 107/7.7 × 1013 = 412 kg
✓ 1
Allow ECF from (b)(ii). Accept 400–420 kg.
Part (c)(i)
Thorium-232 does not undergo fission, so it cannot itself sustain a chain reaction; neutrons are needed to convert it to uranium-233
✓ 1
At first there is no uranium-233, so the fission of uranium-235 must supply the neutrons both to sustain the chain reaction and to breed uranium-233
✓ 1
OWTTE.
Part (c)(ii)
In the core it could absorb a neutron before decaying, so it would not become fissile uranium-233 (and the neutron would be lost to the chain reaction)
✓ 1
Either idea.
No: only a fraction 0.79 has decayed after 60 days; 95 % needs t = 27.0 × ln 20/ln 2 = 117 days
✓ 1
Allow ECF from (a)(ii). The conclusion must be consistent with their fraction.
Answers: (a)(ii) 0.79 · (b)(i) 185 MeV · (b)(ii) 7.7 × 1013 J kg−1; ≈ 2.6 × 106 times · (b)(iii) 412 kg · (c)(ii) no (≈ 117 days needed) (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of chain reactions in nuclear fission reactions; the properties of the products of nuclear fission and their management; guidance: calculations to determine the energy released in fission reactions; E.3 — β− decay equations; E.3 (HL) — the decay law N = N0e−λt for arbitrary time intervals; A.3 — energy density of fuel sources Command term: Determine
35E-2-56
A pulsed research reactor·E.4 Fission
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksDetermine
In a pulsed research reactor the uranium-235 fuel is mixed with the moderator. A control rod is suddenly fired out of the core, and the chain reaction grows very rapidly. The fuel heats up, and when its temperature rises the neutrons are slowed down less effectively. The graph shows the power P of the reactor against time t during one pulse.
Power P of a pulsed reactor against time t (drawn to scale).
(a)
(i)
Estimate the energy released during the pulse.
(2)
(ii)
Determine the number of fissions during the pulse.
(1)
(iii)
The pulse is too short for energy to be transferred from the fuel during it. The fuel has a mass of 150 kg and a specific heat capacity of 340 J kg−1 K−1. Calculate the rise in temperature of the fuel.
(2)
(b)
Early in the pulse the power rises exponentially: it is 7.6 MW at t = 5.0 ms and 40 MW at t = 10.0 ms.
(i)
Show that the power doubles about every 2 ms during this stage.
(2)
(ii)
In a simple model, each generation of fissions follows the previous one after a time ℓ, and each fission causes on average k further fissions. Show that the doubling time T2 is given by T2 = ℓ ln 2/ln k.
(2)
(iii)
For this reactor ℓ = 4.0 × 10−5 s. Determine k.
(1)
(c)
(i)
Explain why the power falls again after 25 ms, although the control rod is still out of the core, and why the chain reaction does not start to grow again soon afterwards.
(3)
Show mark scheme
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Mark
Notes
Part (a)(i)
Area under the graph, e.g. by counting squares or by a triangle of height 1500 MW and base ≈ 2 × 10.6 ms
✓ 1
Method mark.
E ≈ 18 MJ
✓ 1
Accept 15–20 MJ.
Part (a)(ii)
1.8 × 107/(200 × 1.60 × 10−13) = 5.6 × 1017
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
ΔT = E/(mc) = 1.8 × 107/(150 × 340)
✓ 1
Allow ECF from (a)(i).
= 353 K
✓ 1
Accept 290–390 K.
Part (b)(i)
Ratio = 40/7.6 = 5.26 in 5.0 ms
✓ 1
Doubling time = 5.0 × ln 2/ln 5.26 = 2.09 ms
✓ 1
Answer to at least 2 s.f. (2.1) required.
Part (b)(ii)
In time t there are t/ℓ generations, so P = P0kt/ℓ
✓ 1
Doubling: kT2/ℓ = 2, so (T2/ℓ) ln k = ln 2 and T2 = ℓ ln 2/ln k
✓ 1
Answer given: both steps must be seen.
Part (b)(iii)
ln k = ℓ ln 2/T2 = 4.0 × 10−5 × 0.693/2.09 × 10−3 = 0.0133, so k = 1.013
✓ 1
Allow ECF from (b)(i) and (b)(ii). Accept 1.012–1.015.
Part (c)(i)
The energy released raises the temperature of the fuel (by several hundred kelvin, as in (a)(iii)), so the neutrons are slowed less effectively and fewer of them cause fission
✓ 1
Allow ECF from (a)(iii).
k falls below 1, so each generation contains fewer fissions than the one before and the power falls
✓ 1
The fuel stays hot (its energy cannot be removed during the pulse and is lost only slowly afterwards), so k stays below 1 until the fuel cools
✓ 1
Answers: (a)(i) 18 MJ · (a)(ii) 5.6 × 1017 · (a)(iii) 353 K · (b)(i) 2.09 ms · (b)(iii) 1.013 (the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; the role of chain reactions in nuclear fission reactions; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; guidance: calculations to determine the energy released in fission reactions; B.1 — Q = mcΔT; Tools — area under a graph Command term: Determine
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