Energy is released when a heavy nucleus splits, either spontaneously or after absorbing a neutron, because the products have a higher binding energy per nucleon. You need to calculate the energy released in a fission reaction from masses or binding energies.
In a nuclear power plant, chain reactions are controlled with control rods, neutrons are slowed by the moderator, energy is carried away by heat exchangers and people are protected by shielding. You also need the properties of fission products and how long-lived nuclear waste is managed.
17 questions
113 marks
Paper 1A: 8
Paper 1B: 2
Paper 2: 7
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17 practice questions on E.4 Fission
1E-1A-17
Fission equations·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A neutron-induced fission is represented by 23592U + 10n → 14055Cs + AZRb + 2 10n.
What are A and Z for the rubidium nuclide?
AZ
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Notes
Step 1Nucleon numbers balance: 235 + 1 = 140 + A + 2 × 1, so A = 94.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Proton numbers balance: 92 + 0 = 55 + Z + 0, so Z = 37.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe incident neutron has been left out of the nucleon-number balance: 235 − 140 − 2 = 93.
BCorrect: 236 = 140 + A + 2 gives A = 94, and 92 = 55 + Z gives Z = 37.
CThe incident neutron has been counted as a proton, giving Z = 93 − 55 = 38.
DThe two emitted neutrons have been left out of the nucleon-number balance: 236 − 140 = 96.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission Command term: Determine
2E-1A-18
Reactor components·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
A thermal nuclear reactor contains control rods and a moderator.
Which row gives the main function of each?
Diagram NOT accurately drawn
Control rodsModerator
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Notes
Step 1Control rods absorb neutrons, so moving them in or out changes the number of neutrons available to cause fission; the moderator slows fast neutrons in collisions so that they are more likely to cause fission of uranium-235.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: control rods (e.g. boron or cadmium) absorb neutrons to control the rate of fission, and the moderator (e.g. graphite or water) slows neutrons down.
BThe two functions have been swapped: a good moderator absorbs as few neutrons as possible.
CAbsorbing gamma radiation is the job of the shielding, not the moderator.
DControl rods absorb neutrons; returning neutrons to the fuel would increase the fission rate, not control it.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: State
3E-1A-19
Energy released in fission·E.4 Fission
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A nucleus X of nucleon number A and binding energy per nucleon BX absorbs a slow neutron and splits into two nuclei, P and Q, and three neutrons.
P has nucleon number AP and binding energy per nucleon BP; Q has nucleon number AQ and binding energy per nucleon BQ. What is the energy released?
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Notes
Step 1Total binding energy of a nucleus = nucleon number × binding energy per nucleon; free neutrons have no binding energy.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Energy released = binding energy of the products − binding energy of the reactants = APBP + AQBQ − ABX.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the products are more tightly bound; the increase in total binding energy, APBP + AQBQ − ABX, is the energy released.
BBinding energies per nucleon cannot be added and subtracted like this; each must first be multiplied by its nucleon number.
CThe sign is reversed: this treats binding energy as energy stored in the nucleus, but energy is released when the total binding energy increases.
DThe binding energies have been added instead of subtracted.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; E.3 — the variation of the binding energy per nucleon with nucleon number Command term: Deduce
4E-1A-20
Reactor components·E.4 Fission
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
In the nuclear power plant shown, a coolant carries thermal energy from the reactor core to a heat exchanger.
What is the main purpose of the heat exchanger?
Diagram NOT accurately drawnShow mark scheme
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Step 1The coolant, which has passed through the core, gives up thermal energy in the heat exchanger to water in a separate circuit; this water turns to steam that drives the turbine, and the radioactive coolant never reaches the turbine.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThat is the job of the moderator.
BCorrect: the heat exchanger passes thermal energy from the coolant to a separate water circuit, so steam can drive the turbine while the coolant stays inside the shielded circuit.
CThat is the job of the control rods.
DThat is the job of the shielding, the thick concrete and steel around the core.
Syllabus understandingE.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: State
5E-1B-02
Energy released in fission·E.4 Fission
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A student estimates the energy released in the neutron-induced fission
23592U + 10n → 14156Ba + 9236Kr + 3 10n
in two ways. First, the student plots the binding energy per nucleon of the long-lived nuclides listed in a nuclear database (graph) and reads values from a curve drawn through them. Second, the student uses the atomic masses in the table. (1 u = 931.5 MeV c−2)
Particle
Mass / u
235U
235.043930
141Ba
140.914411
92Kr
91.926156
neutron
1.008665
Graph drawn to scale
(a)
Draw the curve of best fit for the data.
(1)
(b)
Using your curve, state the binding energy per nucleon for nucleon numbers 235, 141 and 92.
(1)
(c)
Show that the energy released estimated from your readings in (b) is about 190 MeV.
(1)
(d)
Each reading in (b) has an absolute uncertainty of ±0.02 MeV. Determine the absolute uncertainty in the energy released, and explain why it is a much larger fraction of the result than 0.02 MeV is of each reading.
(3)
(e)
Using the atomic masses, determine the energy released in this fission.
(3)
(f)
Discuss whether the two values of the energy released are consistent, suggesting a reason for any difference.
(2)
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Part (a)
A single smooth curve passing through, or very close to, all the points
✓ 1
Do not accept a dot-to-dot line.
Part (b)
235: 7.59 MeV; 141: 8.36 MeV; 92: 8.69 MeV
✓ 1
Accept 7.57–7.61, 8.34–8.38 and 8.67–8.71 MeV. All three needed.
Part (c)
141 × 8.36 + 92 × 8.69 − 235 × 7.59 = 194.6 MeV
✓ 1
Must see the full substitution OR an answer to at least 3 s.f. Allow ECF from (b): answers in the range 185–204 MeV. The neutrons have no binding energy.
Part (d)
Uncertainties add: (141 + 92 + 235) × 0.02
✓ 1
= ±9.4 MeV, i.e. about ±9 MeV
✓ 1
Accept ±9 MeV or ±10 MeV.
The result is the small difference between two large, nearly equal totals (≈ 1980 MeV and ≈ 1780 MeV), so their absolute uncertainties add up to a large fraction of the difference
✓ 1
OWTTE
Part (e)
Mass before = 235.043930 + 1.008665 = 236.052595 u; mass after = 140.914411 + 91.926156 + 3 × 1.008665 = 235.866562 u
✓ 1
Δm = 0.186033 u
✓ 1
Accept 0.186 u.
E = 0.186033 × 931.5 = 173 MeV
✓ 1
Award [3] for CNA. Accept 173–174 MeV.
Part (f)
Not consistent: the graph value 195 ± 9 MeV (≈ 185–204 MeV) does not include 173 MeV, so the difference is a systematic error, not a random one
✓ 1
Allow ECF from (c), (d) and (e).
The fission fragments are neutron-rich, radioactive nuclides that are less tightly bound than the stable nuclides of the same A on which the curve is based (e.g. 92Kr has 8.51 MeV per nucleon, not ≈ 8.69 MeV), so the graph overestimates the energy released
✓ 1
Accept: the curve is for stable nuclides; the fragments' binding energies per nucleon are lower. OWTTE
Answers: (b) 7.59, 8.36 and 8.69 MeV · (c) 195 MeV · (d) ±9 MeV · (e) 173 MeV (the remaining parts are explanations — see the table above)
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); E.3 — the variation of the binding energy per nucleon with nucleon number. Command term: Determine
6E-1B-08
Energy released in fission·E.4 Fission
Paper 1BHard10 marks
Data-based question10 steps to full marksDetermine
The thermal power of a small research reactor fuelled with uranium-235 is found from its cooling circuit. Water flows through the core at a mass flow rate of (42.0 ± 0.5) kg s−1. During a 6.0 hour run at constant power, the operators record the temperatures of the water entering and leaving the core every hour. Each thermometer has a resolution of 0.1 °C.
Specific heat capacity of water = 4180 J kg−1 K−1. Average energy released per fission of uranium-235 = 200 MeV. 1 u = 1.661 × 10−27 kg.
Hour
Temperature in / °C
Temperature out / °C
1
35.2
46.6
2
35.1
46.7
3
35.3
46.8
4
35.2
46.5
5
35.0
46.5
6
35.1
46.8
(a)
Determine the mean temperature rise ΔT of the water and its absolute uncertainty.
(2)
(b)
Show that the thermal power of the reactor is about 2 MW.
(1)
(c)
Determine the absolute uncertainty in the thermal power.
(2)
(d)
Determine the number of fissions per second in the core.
(2)
(e)
Determine the mass of uranium-235 that undergoes fission during the 6.0 hour run.
(2)
(f)
State and explain whether the fission rate found in (d) is likely to be larger or smaller than the true fission rate.
(1)
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Part (a)
ΔT values 11.4, 11.6, 11.5, 11.3, 11.5, 11.7 K; mean = 11.5 K
✓ 1
Uncertainty = half the range = (11.7 − 11.3)/2 = ±0.2 K, so ΔT = (11.5 ± 0.2) K
✓ 1
Accept ±0.2 K from the thermometer resolutions (0.1 + 0.1).
Part (b)
P = 42.0 × 4180 × 11.5 = 2.02 × 106 W
✓ 1
Must see the substitution OR an answer to at least 3 s.f.
Allow ECF from (a). MP2 is for matching precision.
Part (d)
Energy per fission = 200 × 106 × 1.60 × 10−19 = 3.2 × 10−11 J
✓ 1
Rate = 2.02 × 106/3.2 × 10−11 = 6.3 × 1016 s−1
✓ 1
Award [2] for CNA.
Part (e)
Number of fissions = 6.31 × 1016 × 21600 = 1.4 × 1021
✓ 1
Mass = 1.36 × 1021 × 235 × 1.661 × 10−27 = 5.3 × 10−4 kg (≈ 0.53 g)
✓ 1
Allow ECF from (d). Award [2] for CNA.
Part (f)
Smaller: some of the energy released escapes without heating the coolant (thermal energy lost to the surroundings, gamma rays and neutrons absorbed in the shielding, neutrinos), so the measured power and hence the fission rate are underestimates
✓ 1
Decision and reason both needed.
Answers: (a) (11.5 ± 0.2) K · (b) 2.02 × 106 W · (c) ±0.06 MW · (d) 6.3 × 1016 s−1 · (e) 5.3 × 10−4 kg (the remaining parts are explanations — see the table above)
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); the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant. Command term: Determine
7E-2-09
Chain reactions·E.4 Fission
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksExplain
About 2.1 × 109 years ago, nuclear fission chain reactions took place naturally in a rich deposit of uranium ore at Oklo, in Gabon. Groundwater filled the cracks and pores in the ore. Today natural uranium consists of 0.72 % uranium-235 and 99.28 % uranium-238 by number of atoms. The half-life of uranium-235 is 7.0 × 108 years and that of uranium-238 is 4.5 × 109 years.
(energy released per fission of uranium-235 = 200 MeV, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s)
(a)
Uranium then and now.
(i)
Show that, if the decay of uranium-238 is ignored, the ratio (number of uranium-235 atoms)/(number of uranium-238 atoms) in the ore 2.1 × 109 years ago was about 0.058.
(1)
(ii)
The true value of this ratio 2.1 × 109 years ago was smaller than 0.058. Suggest why.
(1)
(iii)
Nuclear power stations that use ordinary water as the moderator need uranium fuel enriched to about 3 % uranium-235. Explain why a natural chain reaction was possible at Oklo then, but cannot occur in uranium ores today.
(3)
(b)
Analysis of the ore shows that about 6.0 × 103 kg of uranium-235 underwent fission over a period of about 1.5 × 105 years.
(i)
Determine the total energy released by these fissions.
(2)
(ii)
Calculate the average power of the natural reactor.
(1)
(iii)
Calculate the total decrease in mass that corresponds to this energy, and comment on your answer.
(2)
(c)
Studies of the fission products suggest that each zone of the ore was active for about 30 minutes, then inactive for about 2.5 hours, and then active again, repeatedly. Explain how the groundwater could produce this cycle.
(3)
(d)
Many of the fission products have stayed within a few metres of where they were formed for about 2 × 109 years. Suggest why this is of interest to engineers who design stores for nuclear waste.
(1)
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Part (a)(i)
Ratio today = 0.72/99.28 = 7.25 × 10−3; 2.1 × 109 years = 3 half-lives of uranium-235, so the ratio was 23 × 7.25 × 10−3 = 0.058
✓ 1
Must see 23 or ×8 AND the ratio today OR answer to 3 s.f. «0.0580».
Part (a)(ii)
Uranium-238 also decays «half-life 4.5 × 109 years», so there were more uranium-238 atoms then than now «about 1.4 times as many»
✓ 1
OWTTE
Part (a)(iii)
Then the ore contained about 4–5 % uranium-235, more than the fuel of a modern water-moderated reactor
✓ 1
Accept values from 4 % to 6 %, or a comparison of 0.058 with about 0.03.
The groundwater acted as a moderator: it slowed the neutrons so that they were likely to cause further fissions of uranium-235
✓ 1
Today, with only 0.72 % uranium-235, too many neutrons are absorbed by other nuclei «mainly uranium-238», so on average fewer than one neutron from each fission causes another fission and a chain reaction cannot be sustained
✓ 1
OWTTE
Part (b)(i)
Number of nuclei = 6.0 × 103/(235 × 1.661 × 10−27) = 1.54 × 1028
P = 4.92 × 1017/(1.5 × 105 × 3.16 × 107) = 1.0 × 105 W
✓ 1
Accept 1.0–1.1 × 105 W. Allow ECF from (b)(i).
Part (b)(iii)
Δm = E/c2 = 4.92 × 1017/(3.00 × 108)2 = 5.5 kg
✓ 1
Allow ECF from (b)(i).
Only about 0.09 % of the mass of the uranium-235 that underwent fission was converted; almost all of the mass remains in the fission products and neutrons
✓ 1
Accept any valid comparison with 6.0 × 103 kg.
Part (c)
The energy released heated the water, which boiled or was driven out of the ore
✓ 1
Without water there was no moderator: the neutrons stayed fast, fast neutrons rarely cause fission of uranium-235, so the chain reaction stopped
✓ 1
The ore then cooled, water seeped back, the neutrons were slowed again and the chain reaction restarted
✓ 1
OWTTE
Part (d)
It shows that fission products can stay contained in suitable rock for very long times, supporting the storage of waste deep underground in stable rock
✓ 1
OWTTE
Answers: (a)(i) 0.058 · (b)(i) 4.9 × 1017 J · (b)(ii) 1.0 × 105 W · (b)(iii) 5.5 kg (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; that energy is released in spontaneous and neutron-induced fission; the properties of the products of nuclear fission and their management; E.3 — the changes in activity and count rate during radioactive decay using integer values of half-life; the mass–energy equivalence as given by E = mc2 in nuclear reactions Command term: Explain
8E-2-10
Reactor components·E.4 Fission
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksExplain
The diagram shows a simplified nuclear power station of the pressurized-water type. Water in the reactor vessel surrounds the uranium fuel rods and is pumped round a closed primary circuit. Four parts of the power station are marked by numbered boxes.
Diagram not to scale
(a)
Annotate the diagram by writing, in each numbered box, the name of the part it marks: moderator, control rods, heat exchanger or shielding.
(2)
(b)
Explain the role of the moderator.
(3)
(c)
State the condition for the chain reaction to proceed at a constant rate, and outline how the control rods are used to maintain this condition.
(2)
(d)
Explain why the water that drives the turbine is kept separate from the water that passes through the reactor core.
(2)
(e)
State the two types of radiation from the core that the shielding must absorb.
(1)
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Notes
Part (a)
Any two labels correct: 1 control rods, 2 moderator, 3 heat exchanger, 4 shielding
✓ 1
All four labels correct
✓ 1
Award [1] for two or three correct labels.
Part (b)
Neutrons released in fission are fast «high kinetic energy»
✓ 1
Uranium-235 is far more likely to undergo fission when it absorbs a slow «thermal» neutron
✓ 1
The neutrons lose kinetic energy in repeated collisions with the light nuclei of the moderator «hydrogen nuclei in the water, of similar mass to a neutron»
✓ 1
OWTTE
Part (c)
On average, exactly one neutron from each fission goes on to cause another fission
✓ 1
The control rods absorb neutrons: they are pushed further in to reduce the number of neutrons available «and so the rate» and withdrawn to increase it
✓ 1
Part (d)
The water that passes through the core becomes radioactive «it absorbs neutrons and may carry radioactive material»
✓ 1
The heat exchanger transfers thermal energy to the separate secondary circuit without the water mixing, so the turbine and condenser remain free of radioactive contamination
✓ 1
OWTTE
Part (e)
Neutrons and gamma rays
✓ 1
Both needed.
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: Explain
9E-2-11
Fission products & waste·E.4 Fission
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksEvaluate
Spent fuel removed from a nuclear reactor contains many radioactive fission products, among them iodine-131 (half-life 8.0 days) and caesium-137 (half-life 30 years). The spent fuel is first stored under water in a cooling pond.
(a)
Spent fuel.
(i)
State one property of the fission products that makes spent fuel hazardous.
(1)
(ii)
Explain why spent fuel must be cooled for several years after it is removed from the reactor.
(2)
(b)
A cooling pond contains 1.5 × 106 kg of water at 30 °C. The fuel in the pond releases thermal energy at a rate of 2.0 MW. The cooling system of the pond fails. Estimate the time before the water begins to boil, and state one assumption that you make. (specific heat capacity of water = 4200 J kg−1 K−1)
(3)
(c)
The two isotopes.
(i)
Show that after 80 days the activity of the iodine-131 has fallen to about 0.1 % of its initial value.
(1)
(ii)
Calculate the time taken for the activity of the caesium-137 to fall to the same fraction of its initial value.
(1)
(iii)
Hence explain why the two isotopes need different methods of management.
(2)
(d)
A student claims that waste with a long half-life is less dangerous than waste with a short half-life because it has a lower activity. Evaluate this claim.
(2)
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Part (a)(i)
It is highly radioactive «has a high activity», emitting beta and gamma radiation
✓ 1
Accept: some products have very long half-lives / the decays release thermal energy.
Part (a)(ii)
The decays of the fission products release energy «kinetic energy of beta particles, gamma photons» that is absorbed in the fuel and keeps heating it
✓ 1
Without cooling the temperature would rise enough to damage the fuel and release radioactive material; the heating falls as the short-lived products decay
t = 4.41 × 1011/2.0 × 106 = 2.2 × 105 s «≈ 61 hours, about 2.6 days»
✓ 1
Award [2] for CNA.
Assumption: no energy is lost to the surroundings «or: the power stays constant / the water is at a uniform temperature»
✓ 1
Part (c)(i)
80 days = 10 half-lives, so the fraction remaining = 1/210 = 1/1024 = 9.8 × 10−4 ≈ 0.1 %
✓ 1
Must see 210 OR 1/1024 OR 0.098 %.
Part (c)(ii)
10 × 30 = 300 years
✓ 1
Part (c)(iii)
Iodine-131 becomes negligible within months, so it can be stored on site until it has decayed
✓ 1
Caesium-137 remains hazardous for centuries, so it must be isolated for a very long time «e.g. vitrified and stored deep underground in stable rock, away from groundwater»
✓ 1
OWTTE
Part (d)
For the same number of nuclei, a longer half-life does give a lower activity «so the claim is partly correct»
✓ 1
But the waste remains radioactive for much longer — thousands of years for some isotopes — so it must be kept isolated for longer than any building or institution can be relied on; the claim is not valid overall
✓ 1
Award MP2 for any valid counter-argument linked to the duration of the hazard.
Answers: (b) 2.2 × 105 s «≈ 61 h» · (c)(i) 1/1024 ≈ 0.098 % · (c)(ii) 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 should be considered); E.3 — the changes in activity during radioactive decay using integer values of half-life; B.1 — quantitative analysis of thermal energy transfers Q = mcΔTCommand term: Evaluate
10E-1A-37
Energy released in fission·E.4 Fission
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The fission of one uranium-235 nucleus releases about 200 MeV. The fusion reaction 21H + 31H → 42He + 10n releases 17.6 MeV.
Taking the mass of each nucleus to be its nucleon number × 1 u, what is (energy released per kilogram of fusion fuel)/(energy released per kilogram of uranium-235)?
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Notes
Step 1Fusion: 17.6 MeV from 2 + 3 = 5 u of fuel, i.e. 3.52 MeV per u.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Fission: 200 MeV from 235 u of fuel, i.e. 0.851 MeV per u.
—
Step 3Ratio = 3.52/0.851 = 4.1.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis compares the energy per reaction, 17.6/200; the fusion fuel has a far smaller mass per reaction.
BThis is the fission value per kilogram divided by the fusion value — the ratio has been inverted.
CCorrect: 3.52 MeV per u for fusion against 0.851 MeV per u for fission gives a ratio of 4.1.
DThis compares the energy per reaction and inverts it, 200/17.6; the masses of the fuels have been ignored.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; E.5 — that fusion is a source of energy in stars Command term: Determine
11E-1A-38
Chain reactions·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
In a sample of uranium-235, k is the average number of neutrons from each fission that go on to cause another fission.
Which row describes the reaction when k = 1.5 and gives the value of k for a reactor working at steady power?
When k = 1.5k for steady power
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Step 1Each generation of fissions is k times as large as the one before, so with k = 1.5 the rate grows rapidly (1.510 ≈ 58 after ten generations).
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2For steady power each fission must lead on average to exactly one further fission: k = 1.0, which the control rods maintain.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: with k > 1 the number of fissions multiplies each generation, and a steady rate needs exactly one neutron per fission to continue the chain.
BA value of k above 1 makes the reaction grow; it dies away only when k < 1.
CA constant rate needs each fission to cause exactly one more; with k = 1.5 every generation is 1.5 times larger than the last.
DAbout 2 to 3 neutrons are released per fission, but for steady power only one of them, on average, may go on to cause another fission.
Syllabus understandingE.4 — the role of chain reactions in nuclear fission reactions Command term: Deduce
12E-1A-39
Fission products & waste·E.4 Fission
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
Spent fuel removed from a reactor is first kept for several years in a pool of water and is later sent for long-term storage deep underground.
Which row gives a correct reason for each stage?
Kept in water at first becauseStored deep underground because
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Notes
Step 1Newly removed fuel contains many short-lived fission products whose decays release a great deal of thermal energy, so it must be cooled (the water also absorbs much of the radiation).
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Some fission products have half-lives of thousands of years, so the waste must be isolated from the environment for a very long time.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: decay heat must be removed at first, and long-lived products mean the waste stays hazardous for thousands of years.
BRadioactive decay is spontaneous: cooling does not change its rate. The water removes the thermal energy the decays release.
CThe activity of the waste falls with time as the products decay; it is the long half-lives that make long-term storage necessary.
DOnce the fuel is removed from the reactor there is no chain reaction; the thermal energy comes from the radioactive decay of the fission products.
Syllabus understandingE.4 — the properties of the products of nuclear fission and their management (long-term storage of nuclear waste) Command term: Identify
13E-1A-40
Energy released in fission·E.4 Fission
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A nuclear power station has an electrical output power P and an overall efficiency η. Each fission in the reactor releases energy E.
How many fissions occur in the reactor each second?
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Mark
Notes
Step 1Only a fraction η of the energy released becomes electrical energy, so fission must release energy at the rate P/η.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Fissions per second = (P/η)/E = P/(ηE).
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis takes the electrical output as the rate at which fission releases energy; the reactor must release more than P, because the efficiency is less than 1.
BThe efficiency has been applied the wrong way: the fission power is P/η, larger than P, not ηP.
CCorrect: the fission power is P/η, and dividing by the energy per fission gives P/(ηE).
DThis is the reciprocal of the correct answer, (ηE)/P: the average time between fissions, which has the unit of time, not of a rate.
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission; A.3 — efficiency Command term: Deduce
14E-2-25
Energy released in fission·E.4 Fission
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine whether
In a pressurized-water reactor, a typical neutron-induced fission is
23592U + 10n → 14456Ba + 8936Kr + x 10n
The table gives the binding energy per nucleon of the three nuclides. The diagram shows the reactor core and the heat exchanger of the power station.
Nuclide
Binding energy per nucleon / MeV
uranium-235
7.59
barium-144
8.27
krypton-89
8.62
Diagram NOT accurately drawn
(a)
(i)
Deduce the value of x.
(1)
(ii)
Show that the energy released in this fission is about 170 MeV.
(2)
(iii)
The total energy eventually released per fission is about 200 MeV. Suggest why this is greater than the value in (a)(ii).
(1)
(b)
The reactor has a thermal power output of 3.0 GW. Water in the primary circuit enters the reactor core at 292 °C and leaves it at 326 °C. The specific heat capacity of this water is 5.4 × 103 J kg−1 K−1.
(i)
Determine the mass of water that must flow through the core each second.
(2)
(ii)
In the heat exchanger, the primary water turns water in a separate secondary circuit into steam. The secondary water enters the heat exchanger at its boiling point, and its specific latent heat of vaporization at the pressure in the heat exchanger is 1.5 × 106 J kg−1. Calculate the mass of steam produced each second.
(2)
(c)
The power station generates 1.0 GW of electrical power. The rest of the thermal power is transferred to a river whose flow rate is 1.2 × 105 kg s−1. Regulations do not allow the temperature of the river to rise by more than 3.0 K. The specific heat capacity of river water is 4.2 × 103 J kg−1 K−1.
(i)
Determine whether the power station can meet the regulation using the river alone.
(3)
(ii)
Suggest one reason why the temperature rise of the river is limited.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
x = 236 − 144 − 89 = 3
✓ 1
Part (a)(ii)
binding energy of products = 144 × 8.27 + 89 × 8.62 «= 1958 MeV» AND binding energy of uranium-235 = 235 × 7.59 «= 1784 MeV»
✓ 1
Free neutrons have no binding energy.
energy released = 1958 − 1784 «= 174 MeV»
✓ 1
Must see full substitution OR answer to 3 s.f. (174 MeV).
Part (a)(iii)
the fission products are radioactive and release further energy as they decay «by β− and γ emission»
✓ 1
Do not accept 'the neutrons gain energy'.
Part (b)(i)
3.0 × 109 = m × 5.4 × 103 × 34
✓ 1
m = 1.6 × 104 kg «each second»
✓ 1
Award [2] for CNA.
Part (b)(ii)
3.0 × 109 = m × 1.5 × 106
✓ 1
Assumes that all the thermal power is transferred to the secondary circuit.
m = 2.0 × 103 kg «each second»
✓ 1
Award [2] for CNA. Award [0] for an answer using mcΔT.
Part (c)(i)
power transferred to the river = 3.0 − 1.0 = 2.0 GW
4.0 K > 3.0 K «OR 1.6 × 105 kg s−1 > 1.2 × 105 kg s−1», so the regulation cannot be met
✓ 1
Conclusion must be consistent with the calculation. If 3.0 GW is used (ΔT = 6.0 K), award MP3 by ECF but not MP1.
Part (c)(ii)
warmer water holds less dissolved oxygen, which harms fish and other river life
✓ 1
OWTTE
Answers: (a)(i) x = 3 · (a)(ii) 174 MeV · (b)(i) 1.6 × 104 kg s−1 · (b)(ii) 2.0 × 103 kg s−1 · (c)(i) ΔT = 4.0 K, so the regulation is not met (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; E.3 — the variation of the binding energy per nucleon with nucleon number; B.1 — Q = mcΔT and Q = mL Command term: Determine whether
15E-2-26
Chain reactions·E.4 Fission
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDraw
In a thermal fission reactor each fission of a uranium-235 nucleus releases two or three fast neutrons. The multiplication factor k is the average number of neutrons from one fission that go on to cause another fission.
The diagram shows one fission and the three neutrons that it releases. Nearby are two uranium-235 nuclei, a uranium-238 nucleus and part of a control rod. The dashed line is the edge of the reactor core.
Diagram NOT accurately drawn
(a)
The reactor is operating at a constant power. Draw on the diagram to show what happens to each of the three neutrons.
(2)
(b)
The average time between one generation of fissions and the next is 0.10 s. The control rods are raised slightly so that k = 1.005.
(i)
Show that the power of the reactor increases by a factor of about 1.6 in 10 s.
(1)
(ii)
The time between generations is as long as 0.10 s only because some neutrons are released by fission products a short time after the fission. If every neutron were released at the instant of fission, the time between generations would be about 1 × 10−4 s. Determine the factor by which the power would then increase in 0.10 s with k = 1.005, and comment on your answer.
(3)
(c)
The reactor is later run at a constant thermal power of 60 MW, with the same time between generations. Each fission releases 200 MeV (1 MeV = 1.60 × 10−13 J), and on average 2.4 neutrons are released per fission.
(i)
Calculate the number of fissions in one generation.
(2)
(ii)
Deduce the number of neutrons released in one generation that do not cause a further fission.
(1)
(d)
A small sphere of uranium-235 cannot sustain a chain reaction, but a larger sphere of the same material can. Explain why.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
exactly one of the three neutrons drawn going on to cause the fission of a uranium-235 nucleus
✓ 1
Do not award MP1 if two neutrons cause fission.
each of the other two neutrons drawn being absorbed «by the control rod or by the uranium-238 nucleus» OR leaving the core
✓ 1
Part (b)(i)
10 s = 100 generations, so factor = 1.005100 «= 1.65»
✓ 1
Must see 100 generations AND 1.005100 OR answer to 3 s.f. (1.65).
Part (b)(ii)
0.10 s = 1000 generations
✓ 1
factor = 1.0051000 ≈ 150
✓ 1
Accept 146–147. Award [2] for CNA.
the power would rise about 150 times in a tenth of a second, far too fast for the control rods to respond «the delayed neutrons make the reactor controllable»
✓ 1
OWTTE
Part (c)(i)
Energy released in one generation = 60 × 106 × 0.10 = 6.0 × 106 J
At constant power k = 1, so on average 2.4 − 1 = 1.4 neutrons per fission do not cause fission: 1.4 × 1.88 × 1017 = 2.6 × 1017
✓ 1
Allow ECF from (c)(i). Accept 2.6–2.7 × 1017. Award [0] for 4.5 × 1017 «all 2.4 neutrons».
Part (d)
neutrons are produced throughout the volume but escape through the surface
✓ 1
a larger sphere has a smaller surface-area-to-volume ratio, so a smaller fraction of the neutrons escape and k can reach 1
✓ 1
Answers: (b)(i) 1.65 · (b)(ii) ≈ 150 · (c)(i) 1.9 × 1017 · (c)(ii) 2.6 × 1017(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 Command term: Draw
16E-2-27
Spontaneous fission·E.4 Fission
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
Californium-252, 25298Cf, is used as a neutron source to start up nuclear reactors. Most californium-252 nuclei decay by alpha emission, but about 3 % undergo spontaneous fission, for example
25298Cf → 14054Xe + 10844Ru + x 10n
The table gives some masses. (1 u = 931.5 MeV c−2)
Particle
Mass / u
californium-252 atom
252.081627
xenon-140 atom
139.921646
ruthenium-108 atom
107.910190
neutron
1.008665
(a)
(i)
Outline the difference between spontaneous fission and neutron-induced fission.
(1)
(ii)
Deduce the value of x.
(1)
(b)
(i)
Explain why atomic masses, rather than nuclear masses, can be used to calculate the energy released in this fission.
(1)
(ii)
Determine the energy released in this fission, in MeV.
(3)
(c)
A new source emits 2.0 × 109 neutrons per second. The half-life of californium-252 is 2.6 years. The source must be replaced when it emits fewer than 2.5 × 108 neutrons per second. Determine the time for which the source can be used.
(2)
(d)
The source is stored in a container made of a thick layer of polyethylene, which contains many hydrogen nuclei, surrounded by a layer of lead. Explain the purpose of each material, and suggest why no special shielding is needed for the alpha particles.
(3)
(e)
Suggest why spontaneous fission could not replace neutron-induced fission as the energy source in a nuclear power station.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
spontaneous fission happens without the nucleus absorbing any particle «randomly, like other decays», whereas neutron-induced fission follows the absorption of a neutron
✓ 1
OWTTE
Part (a)(ii)
x = 252 − 140 − 108 = 4
✓ 1
Proton numbers already balance: 98 = 54 + 44.
Part (b)(i)
the californium atom has 98 electrons and the xenon and ruthenium atoms together have 54 + 44 = 98 electrons, so the electron masses cancel
✓ 1
Neutrons have no electrons.
Part (b)(ii)
mass of products = 139.921646 + 107.910190 + 4 × 1.008665 = 251.866496 u
✓ 1
Allow ECF from (a)(ii).
Δm = 252.081627 − 251.866496 = 0.215131 u
✓ 1
E = 0.215131 × 931.5 = 200 MeV
✓ 1
Award [3] for CNA. Accept 200–201 MeV.
Part (c)
2.0 × 109/2.5 × 108 = 8 = 23, so 3 half-lives
✓ 1
3 × 2.6 = 7.8 years
✓ 1
Award [2] for CNA.
Part (d)
polyethylene: the neutrons lose energy in collisions with the light hydrogen nuclei, so they are slowed down and absorbed
✓ 1
lead: absorbs the gamma radiation emitted by the source and by its radioactive fission products
✓ 1
Do not accept 'lead absorbs the alpha particles'.
alpha particles have a very short range and are stopped by the source's own casing
✓ 1
Part (e)
the rate of spontaneous fission is fixed by the half-life «and is random»; it cannot be increased or reduced to match demand, unlike neutron-induced fission, which is controlled with control rods
✓ 1
the nuclides found in nature in large amounts «such as uranium-238» undergo spontaneous fission extremely rarely, so the power would be negligible / nuclides with a high rate of spontaneous fission, such as californium-252, do not occur naturally and are very scarce and expensive to produce
✓ 1
Accept any one.
Answers: (a)(ii) x = 4 · (b)(ii) 200 MeV · (c) 7.8 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 control rods, moderators, heat exchangers and shielding in a nuclear power plant; E.3 — nuclear binding energy and mass defect (guidance: masses in u) Command term: Determine
17E-2-35
Energy released in fission·E.4 Fission
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A slow neutron is absorbed by a stationary uranium-235 nucleus, which then splits:
23592U + 10n → 14156Ba + 9236Kr + x 10n
Immediately after the fission the two fragments have a total kinetic energy of 170 MeV. Take the masses of the barium and krypton nuclei to be 141 u and 92 u, and ignore the momentum of the neutrons. (1 u = 1.661 × 10−27 kg, e = 1.60 × 10−19 C)
(a)
Deduce the value of x.
(1)
(b)
(i)
Explain why the two fragments move off in opposite directions with momenta of equal magnitude.
(2)
(ii)
Show that the ratio kinetic energy of the krypton fragment/kinetic energy of the barium fragment is about 1.5.
(1)
(iii)
Determine the kinetic energy of the krypton fragment, in MeV.
(2)
(iv)
Calculate the speed of the krypton fragment.
(2)
(c)
Explain how the kinetic energy of the fragments becomes thermal energy in the fuel, and how this energy reaches the coolant.
(3)
(d)
Suggest why it is reasonable to ignore the momentum of the neutrons.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
x = 236 − 141 − 92 = 3
✓ 1
Part (b)(i)
the total momentum before the fission is «approximately» zero «slow neutron, stationary nucleus»
✓ 1
momentum is conserved, so the momenta of the two fragments must add to zero: equal in magnitude and opposite in direction
✓ 1
Part (b)(ii)
Ek = p2/2m with the same p, so the ratio = mBa/mKr = 141/92 «= 1.53»
✓ 1
Must see Ek = p2/2m (or equivalent) AND the mass ratio.
Part (b)(iii)
EKr = 170 × 141/(141 + 92)
✓ 1
ALTERNATIVE: EKr + EKr/1.53 = 170.
= 103 MeV
✓ 1
Award [2] for CNA. Accept 102–103 MeV.
Part (b)(iv)
Ek = 103 × 1.60 × 10−13 = 1.65 × 10−11 J AND m = 92 × 1.661 × 10−27 = 1.53 × 10−25 kg
✓ 1
Allow ECF from (b)(iii).
v = √(2Ek/m) = 1.5 × 107 m s−1
✓ 1
Part (c)
the fragments collide with «and ionize» atoms of the fuel and are stopped within a very short distance
✓ 1
their kinetic energy is shared among the particles of the fuel as random kinetic energy, so the internal energy and the temperature of the fuel increase
✓ 1
thermal energy is conducted through the fuel and the fuel-rod casing to the cooler coolant flowing past the rods
✓ 1
Part (d)
each neutron has a much smaller mass «and only a few MeV of kinetic energy», so its momentum is small compared with that of a fragment «and the neutrons move in different directions»
✓ 1
OWTTE
Answers: (a) x = 3 · (b)(ii) 1.53 · (b)(iii) 103 MeV · (b)(iv) 1.5 × 107 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.4 — that energy is released in spontaneous and neutron-induced fission (guidance: calculations to determine the energy released in fission reactions); A.2 — conservation of momentum in explosions; A.3 — Ek = p2/2m; B.1 — internal energy Command term: Determine
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