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IB Physics Standard Level · Topic E.4

E.4 Fission

89 free, high-quality past-paper-style questions on Fission for IB Physics SL, each with a full mark scheme. Covers: Reactor components; Spontaneous fission; Fission equations; and more. For when you have finished the past papers, or want to practise one topic at a time.E.4 at Higher Level →Read the E.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.441 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-18Reactor components Paper 1AEasy1 mark

A thermal nuclear reactor contains control rods and a moderator.

Which row gives the main function of each?

A schematic thermal nuclear reactor: fuel rods and control rods in a moderator inside a shielded core, with a coolant loop to a heat exchanger.control rods (white) lower into the corereactor corefuel rods (gold) in a moderatorheat exchangerhotcoolsteam to the turbinethick concrete shielding (grey) surrounds the core
Diagram NOT accurately drawn
Control rodsModerator
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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.✓ 1Answer 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.
2E-1A-20Reactor components Paper 1AEasy1 mark

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?

A schematic thermal nuclear reactor: fuel rods and control rods in a moderator inside a shielded core, with a coolant loop to a heat exchanger.control rods (white) lower into the corereactor corefuel rods (gold) in a moderatorheat exchangerhotcoolsteam to the turbinethick concrete shielding (grey) surrounds the core
Diagram NOT accurately drawn
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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.✓ 1Answer 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.
3E-1A-89Reactor components Paper 1AEasy1 mark

The core of a nuclear reactor is surrounded by thick shielding made of steel and concrete.

Which statement about this shielding is correct?

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Step 1Alpha and beta particles are stopped within the fuel, its cladding and the reactor vessel. The radiation that leaves the core is mainly neutrons and gamma rays: they are uncharged, ionize weakly and so need a large thickness of material to absorb them.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AAlpha and beta particles have short ranges and are absorbed inside the fuel rods and the reactor vessel; they are not the reason for thick shielding.
  • BCorrect: neutrons and gamma photons carry no charge, interact weakly and travel far through matter, so a great thickness of steel and concrete is needed.
  • CThis is the job of the moderator inside the core. Neutrons that reach the shielding have left the core and must be absorbed.
  • DAfter shutdown the fission products go on decaying and emit intense gamma radiation, so the shielding is still needed.
4E-1A-95Spontaneous fission Paper 1AEasy1 mark

Some heavy nuclides, such as uranium-238 and plutonium-240, can undergo spontaneous fission.

Which statement about spontaneous fission is correct?

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Step 1Spontaneous fission is a form of radioactive decay: the nucleus splits without absorbing any particle, at a random moment.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Its rate in a sample depends only on the number of nuclei of that nuclide, not on outside conditions such as the neutrons present.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis describes neutron-induced fission. In spontaneous fission no particle is absorbed.
  • BThe products are more tightly bound than the original nucleus whatever starts the fission, so about 200 MeV is still released.
  • CFor medium-mass nuclei splitting would lower the binding energy per nucleon; spontaneous fission is observed only in the heaviest nuclei.
  • DCorrect: like other radioactive decays it is random and spontaneous, so neutrons (or control rods) do not change its rate.
5IBP2S-040-02-01Fission equations Paper 1AEasy1 mark

A university research reactor uses uranium-235 fuel. One possible fission reaction is
23592U + 10n → 14054Xe + 9438Sr + x 10n
What is x?

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Nucleon number must be conserved. The left side has 235 + 1 = 236 nucleons. The fragments have 140 + 94 = 234 nucleons, so 236 − 234 = 2 neutrons are released. Choosing 0 checks only proton numbers, which balance anyway because neutrons carry no charge. Choosing 1 forgets the incident neutron on the left. Choosing 3 recalls a 'typical' value instead of balancing this equation.✓ 1Answer C

Answer: C

Exam tip: Balance both the top numbers (nucleons) and the bottom numbers (charge). Always count the incoming neutron on the left.

6IBP2S-040-02-03Chain reactions Paper 1AEasy1 mark

A nuclear power station is producing a constant power output. On average, how many of the neutrons released by each fission go on to cause another fission?

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The power is constant, so the fission rate is constant. Each fission must therefore lead, on average, to exactly one further fission. This is the critical condition, k = 1. With zero the chain reaction stops. With more than one, including all of the neutrons, the fission rate and power grow over time.✓ 1Answer B

Answer: B

Exam tip: Constant power means k = 1. If k 1 it grows.

7IBP2S-040-02-08Chain reactions Paper 1AEasy1 mark

Demand on the electricity grid falls at night. The operator of a nuclear power station wants to reduce the rate of fission in the core. What should the operator do?

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Control rods absorb neutrons. Inserting them further removes more neutrons, so fewer neutrons are available to cause fission and the fission rate falls. Withdrawing the rods does the opposite. Increasing the coolant flow removes thermal energy faster but does not directly reduce the number of neutrons. Adding fuel increases the fission rate.✓ 1Answer B

Answer: B

Exam tip: Describe control rods by what they do: they absorb neutrons and so change the number of neutrons per fission that cause further fission.

8IBP2S-040-00-01Fission equations Paper 1AEasy1 mark

The fragments produced when a uranium-235 nucleus undergoes fission are almost always radioactive and decay by β− emission. Which statement best explains this?

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A heavy nucleus such as U-235 has a high neutron-to-proton ratio, about 1.55. The fragments keep roughly this ratio. Stable nuclei of mass number around 90–140 need a lower ratio, about 1.2–1.4. The fragments therefore have too many neutrons. β− decay turns a neutron into a proton, which moves each nucleus towards stability. Option B describes nuclei that would decay by β+ emission. Option C is false, because the fragments have a higher binding energy per nucleon than U-235. Option D confuses the kinetic energy of the fragments with the decay energy of the nucleus.✓ 1Answer A

Answer: A

Common error: Do not mix up the direction of β decay. Neutron-rich nuclei undergo β⁻ decay (n → p). Proton-rich nuclei undergo β⁺ decay.

9IBP2S-040-00-02Energy released in fission Paper 1AEasy1 mark

About 200 MeV of energy is released when a nucleus of uranium-235 undergoes fission. Most of this energy appears as

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The two fragments are both positively charged and repel each other strongly. They fly apart with a combined kinetic energy of roughly 165–170 MeV, which is over 80% of the total. This kinetic energy becomes internal energy of the fuel as the fragments collide with surrounding atoms. The two or three neutrons carry only about 5 MeV in total. Prompt gamma rays carry a similar small amount. Antineutrinos come from the later β− decays and carry only a modest fraction.✓ 1Answer C

Answer: C

Exam tip: Link the fragment kinetic energy to how a reactor heats its fuel. The fragments stop within the fuel and transfer their energy as heat.

10IBP2S-040-00-03Fission products & waste Paper 1AEasy1 mark

Spent fuel rods removed from a nuclear reactor are kept in deep pools of water for several years before they are moved to dry storage. What is the main reason for this?

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Fresh spent fuel contains many fission products with short half-lives. Its activity is therefore extremely high, and the decay energy heats the rods strongly. The water removes this heat and also shields the surroundings from the radiation. Option A is incorrect because the chain reaction has already stopped and is not the main concern. Option C is incorrect because radioactive decay rates cannot be changed by the surroundings. Option D is incorrect because the activity falls quickly at first as the short-lived nuclides decay.✓ 1Answer B

Answer: B

Common error: External conditions such as temperature, pressure or water do not change a half-life. Cooling removes heat, but it does not slow the decay.

11IBP2S-040-01-01Reactor components Paper 1AEasy1 mark

An engineer is checking a briefing sheet for a thermal fission reactor. Which component is correctly matched with its function?

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The coolant circulates through the core and carries the thermal energy released by fission away from the fuel, for example to a steam generator. Control rods absorb neutrons; they do not slow them. The moderator slows neutrons; it does not absorb them. Shielding absorbs radiation to protect people outside the reactor; returning neutrons to the core is the job of a reflector.✓ 1Answer C

Answer: C

Common error: Swapping the roles of the moderator (slows neutrons) and the control rods (absorb neutrons) is the most frequent mistake on this topic.

12IBP2S-040-01-02Spontaneous fission Paper 1AEasy1 mark

Californium-252 is used in hospitals and research laboratories as a compact neutron source because it undergoes spontaneous fission. Which statement about spontaneous fission is correct?

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In spontaneous fission an unstable heavy nucleus splits into two smaller nuclei, usually with a few neutrons, without any external trigger. Absorbing a neutron first describes induced fission. Nuclear processes do not depend on ordinary temperatures. Energy is released because the products have a greater binding energy per nucleon, not a lower one.✓ 1Answer B

Answer: B

Exam tip: To distinguish spontaneous from induced fission in one line, write: spontaneous fission needs no incoming neutron; induced fission is triggered by absorbing one.

13IBP2S-040-01-08Reactor components Paper 1AEasy1 mark

In a pressurised water reactor, what is the main function of the heat exchanger (steam generator)?

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The heat exchanger takes energy from the hot, pressurised primary coolant and boils water in a secondary circuit. The steam drives the turbines. Keeping two circuits means the radioactive primary coolant never reaches the turbines. Slowing neutrons is the moderator's role. Absorbing neutrons is the control rods' role. No reactor component converts nuclear energy directly into electrical energy.✓ 1Answer D

Answer: D

Exam tip: Mention that the two circuits are kept separate so that radioactive coolant stays inside the containment.

14E-1A-17Fission equations Paper 1AMedium1 mark

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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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.✓ 1Answer 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.
15E-1A-38Chain reactions Paper 1AMedium1 mark

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.✓ 1Answer 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.
16E-1A-39Fission products & waste Paper 1AMedium1 mark

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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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.✓ 1Answer 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.
17E-1A-54Reactor components Paper 1AMedium1 mark

The moderator in a thermal fission reactor slows down the fast neutrons released in fission.

Which row describes a good moderator?

Mass of each nucleus of the moderatorFraction of the neutrons that the moderator absorbs
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Step 1A neutron loses a large fraction of its kinetic energy in each collision only if it hits a nucleus of similar mass; it bounces off a heavy nucleus with almost the same speed.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The moderator must also absorb few neutrons, otherwise too few remain to cause further fissions and the chain reaction cannot continue.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AVery little absorption is right, but neutrons bounce off heavy nuclei with almost no loss of kinetic energy, so very many collisions would be needed.
  • BLight nuclei slow neutrons well, but a moderator that absorbed many neutrons would stop the chain reaction; absorbing neutrons is the job of the control rods.
  • CThis describes a control-rod material, which absorbs neutrons; heavy nuclei also slow neutrons very poorly.
  • DCorrect: nuclei of similar mass take a large share of the neutron's kinetic energy in each collision, and few neutrons are lost by absorption.
18E-1A-55Chain reactions Paper 1AMedium1 mark

Uranium fuel that has been enriched, so that a larger fraction of its nuclei are uranium-235, is stored in small containers that are kept well apart from one another.

What is the main reason for this?

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Step 1A chain reaction grows only if, on average, at least one neutron from each fission goes on to cause another fission.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2In a small, isolated mass a large fraction of the neutrons leave through the surface before meeting a uranium-235 nucleus, so this condition is not met; keeping the containers apart stops neutrons from one container reaching another.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe half-life is a property of the nuclide and does not depend on the amount present.
  • BThe number of neutrons released per fission is a property of the fission process (about 2 to 3) and does not depend on the size of the sample.
  • CCorrect: a small mass has a large surface area for its volume, so too many neutrons escape for a chain reaction to grow.
  • DSpontaneous fission, like all radioactive decay, is not affected by temperature; the reason is the escape of neutrons.
19E-1A-57Energy released in fission Paper 1AMedium1 mark

When a uranium-235 nucleus absorbs a neutron, the total mass of the nucleus and the neutron is about 236 u. The fission that follows releases about 200 MeV.

What fraction of the mass of the nucleus and the neutron is converted into the energy released? (1 u = 931.5 MeV c−2)

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Step 1Mass converted = 200 MeV c−2 = 200/931.5 = 0.215 u.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Fraction = 0.215/236 = 9.1 × 10−4.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: 200/(236 × 931.5) = 9.1 × 10−4, so less than 0.1 % of the mass is converted.
  • B200/931.5 = 0.21 is the mass converted in u; it has not been divided by the 236 u of the reactants.
  • C200/236 = 0.85 divides MeV by u as though they were the same unit; 1 u is equivalent to 931.5 MeV.
  • D(236 × 931.5)/200 = 1099 is the fraction inverted: it is the total mass-energy divided by the energy released.
20E-1A-90Fission products & waste Paper 1AMedium1 mark

In one neutron-induced fission of uranium-235 the fragments are 14757La and 8735Br, and two neutrons are released. The heaviest stable isotopes of these two elements are 13957La and 8135Br.

Which row compares the number of neutrons in each fragment with the number in the stable isotope, and gives the decay that the fragments are most likely to undergo?

Number of neutrons in fragmentLikely decay of fragment
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Step 1La-147 has 90 neutrons and La-139 has 82; Br-87 has 52 and Br-81 has 46. Each fragment has more neutrons than any stable isotope of its element: it is neutron-rich.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2In β− decay a neutron in the nucleus changes into a proton, so the fragments decay by β− emission towards stable nuclides.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the fragments have 8 and 6 neutrons more than the heaviest stable isotopes, and β− decay changes a neutron into a proton.
  • BThe comparison is right, but in β+ decay a proton changes into a neutron, which would make the excess of neutrons even larger.
  • CThis assumes that releasing two neutrons leaves the fragments short of neutrons; counting A − Z shows 90 and 52, more than 82 and 46.
  • DThis describes neutron-poor nuclei. The fragments come from a nucleus with many more neutrons than protons, so they are neutron-rich.
21E-1A-91Chain reactions Paper 1AMedium1 mark

A reactor is operating at a constant fission rate. At time t = 0 the control rods are withdrawn a small distance and then held in their new position, so that the average number of neutrons from each fission that cause a further fission becomes 1.002 and stays at this value.

Which graph shows how the fission rate R varies with t after t = 0?

Four sketch graphs of fission rate R against time t: A straight line rising, B rising then levelling off, C falling curve, D rising curve of increasing gradientA.Rt0B.Rt0C.Rt0D.Rt0
Sketch graphs, not to scale
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Step 1Withdrawing the control rods means fewer neutrons are absorbed, so more than one neutron per fission, on average, causes a further fission.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each generation of fissions is then 1.002 times as large as the one before: the rate is multiplied by the same factor in equal times, so it grows exponentially, with a gradient that keeps increasing, for as long as the factor stays at 1.002.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AA straight line would mean the same number of extra fissions in every generation; in fact each generation is a fixed multiple of the last, so the increase itself grows.
  • BThe rate would level off only if the factor returned to 1. With the rods held still and the factor fixed at 1.002, nothing brings the rate back to a steady value.
  • CWithdrawing the rods reduces the absorption of neutrons, so the rate rises; a falling rate would follow pushing the rods further in.
  • DCorrect: a constant factor greater than 1 per generation gives exponential growth of the fission rate.
22E-1A-93Energy released in fission Paper 1AMedium1 mark

Just after a uranium-235 nucleus has split, the two fragments 14455Cs and 9037Rb have their centres 1.25 × 10−14 m apart. Treat each fragment as a point charge at its centre.

What is the magnitude of the electric force between the fragments?

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Step 1The charges are 55e = 8.80 × 10−18 C and 37e = 5.92 × 10−18 C.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = kq1q2/r2 = 8.99 × 109 × 55 × 37 × (1.60 × 10−19)2/(1.25 × 10−14)2 = 3.0 × 103 N.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe separation has not been squared: kq1q2/r = 3.7 × 10−11, which has the unit of energy (J), not of force.
  • BCorrect: about 3 kN acts on each fragment; this repulsion drives the fragments apart, and they gain most of the energy released as kinetic energy.
  • CThe nucleon numbers 144 and 90 have been used for the charges; neutrons are uncharged, so only the proton numbers count.
  • DThe elementary charge has been included only once: 55 × 37 × e instead of 55e × 37e.
23E-1A-94Reactor components Paper 1AMedium1 mark

Neutrons released in fission have a kinetic energy of about 2.0 MeV. The moderator of a reactor slows them until their kinetic energy is about 0.025 eV.

What is the ratio (speed of a neutron released in fission)/(speed of a slowed neutron)?

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Step 1Ek = ½mv2 with the same mass, so v ∝ √Ek.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = √(2.0 × 106 eV/0.025 eV) = √(8.0 × 107) = 8.9 × 103.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis is the slow speed divided by the fast speed: the ratio has been inverted.
  • BThe prefix of MeV has been ignored: √(2.0/0.025) treats 2.0 MeV as 2.0 eV.
  • CCorrect: the speed falls by a factor √(8.0 × 107) ≈ 8.9 × 103, from about 2 × 107 m s−1 to about 2 × 103 m s−1.
  • DThis is the ratio of the kinetic energies; the square root has been left out.
24IBP2S-040-02-02Fission equations Paper 1AMedium1 mark

Molybdenum-99 is used to make the medical tracer technetium-99m. It can be produced by the fission of uranium-235:
23592U + 10n → 9942Mo + X + 2 10n
What is nuclide X?

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Proton number: 92 = 42 + Z, so Z = 50 (tin). Nucleon number: 235 + 1 = 99 + A + 2, so A = 135. X is 13550Sn. A = 134 forgets the incident neutron. A = 137 forgets the two emitted neutrons. Z = 48 wrongly gives the neutrons a charge of +1 each.✓ 1Answer A

Answer: A

Common error: Neutrons have proton number 0. They affect only the nucleon-number balance, not the charge balance.

25IBP2S-040-02-04Reactor components Paper 1AMedium1 mark

The pressurized-water reactor in a submarine uses ordinary water as its moderator. Which property of the hydrogen nuclei in water makes water an effective moderator?

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A moderator slows neutrons through collisions. In a collision between particles of similar mass, a large fraction of the neutron's kinetic energy is transferred, so a proton (mass ≈ neutron mass) slows neutrons in few collisions. Strong absorption describes control rods and would actually make a moderator worse. High specific heat capacity is a coolant property. Neutrons are uncharged, so there is no electrostatic repulsion.✓ 1Answer B

Answer: B

Common error: Do not confuse the moderator (slows neutrons), the control rods (absorb neutrons) and the coolant (removes thermal energy).

26IBP2S-040-02-05Energy released in fission Paper 1AMedium1 mark

Why is energy released when a uranium-235 nucleus undergoes fission into two medium-mass nuclei?

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Medium-mass nuclei lie nearer the peak of the binding-energy-per-nucleon curve than uranium-235. The nucleons in the products are therefore more tightly bound, so the products have less mass and the mass difference is released as energy. A greater product mass would need energy input, not release it. Nucleon number is conserved. The final option reverses the true relationship.✓ 1Answer C

Answer: C

Exam tip: Energy is released whenever the products have a higher binding energy per nucleon, which means a lower total mass.

27IBP2S-040-02-06Energy released in fission Paper 1AMedium1 mark

In a single fission of uranium-235, about 200 MeV of energy is released. In which form is most of this energy released immediately?

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The two fragments are both positively charged and start very close together. Their strong electrostatic repulsion pushes them apart, giving them most of the energy (roughly 170 MeV) as kinetic energy. This kinetic energy becomes internal energy of the fuel. Prompt gamma rays and neutrons each carry only a few MeV. Antineutrinos come later, from the beta decay of the fragments.✓ 1Answer C

Answer: C

Common error: The released neutrons sustain the chain reaction, but they carry only a small share of the energy.

28IBP2S-040-02-07Chain reactions Paper 1AMedium1 mark

Engineers designing a compact fission reactor for a deep-space probe find that a small sphere of fuel cannot sustain a chain reaction, but a larger sphere of the same material can. What is the best explanation?

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Neutrons are produced throughout the volume but lost through the surface. The surface area to volume ratio increases as size decreases. Below the critical mass, too many neutrons escape, so fewer than one per fission causes another fission. The number density of nuclei is the same for the same material. Neutron speed and the number of neutrons per fission do not depend on the size of the sphere.✓ 1Answer A

Answer: A

Exam tip: Link critical mass to neutron leakage through the surface: losses scale with area, production scales with volume.

29IBP2S-040-02-09Energy released in fission Paper 1AMedium1 mark

A nuclear-powered icebreaker ship has a reactor with a thermal power of 320 MW. Each fission releases 200 MeV, which is 3.2 × 10−11 J. What is the rate of fission in the reactor?

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Rate = power ÷ energy per fission = 3.2 × 108 W ÷ 3.2 × 10−11 J = 1.0 × 1019 s−1. The answer 1.6 × 106 divides 320 by 200 without converting units. The answer 1.0 × 1013 uses 320 W instead of 320 MW. The answer 1.0 × 1025 treats 200 MeV as 200 eV.✓ 1Answer C

Answer: C

Common error: Convert both MW and MeV to SI units before dividing: 1 MeV = 1.6 × 10⁻¹³ J.

30IBP2S-040-00-04Energy released in fission Paper 1AMedium1 mark

The binding energy per nucleon of uranium-235 is about 7.6 MeV. The average binding energy per nucleon of its fission fragments is about 8.5 MeV. What is the best estimate of the energy released when one uranium-235 nucleus undergoes fission?

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Every nucleon ends up about 8.5 − 7.6 = 0.9 MeV more tightly bound. The energy released is therefore about 235 × 0.9 ≈ 210 MeV, which is 2.1×102 MeV. Option A is the gain for a single nucleon only. Option B (235 × 7.6) is the total binding energy of U-235. Option C (235 × 8.5) is the total binding energy of the products. Neither total is the energy released.✓ 1Answer D

Answer: D

Exam tip: The energy released is the increase in total binding energy. Multiply the change in binding energy per nucleon by the number of nucleons.

31IBP2S-040-00-05Chain reactions Paper 1AMedium1 mark

A compact fission reactor designed to power a lunar base supplies thermal energy at a steady rate of 4.0 kW for 3.2×107 s (about one year). What mass is converted into energy in this time?

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The energy supplied is E = Pt = 4.0×103 × 3.2×107 = 1.28×1011 J. The mass converted is m = E/c2 = 1.28×1011 / (9.0×1016) ≈ 1.4×10−6 kg, which is about 1.4 mg, so B is correct.
Option A uses 4.0 W instead of 4.0 kW. Option C divides by c instead of c2. Option D wrongly uses E = ½mc2, which doubles the mass.
✓ 1Answer B

Answer: B

Common error: Always square the speed of light. Dividing by c instead of c² gives an absurdly large mass. Also, mass–energy equivalence is E = mc², with no factor of ½.

32IBP2S-040-00-06Fission equations Paper 1AMedium1 mark

In the fuel of a reactor, a nucleus of \(^{238}_{92}\text{U}\) absorbs a neutron. The nucleus formed then undergoes two successive β− decays. Which nucleus is finally formed?

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Absorbing a neutron forms \(^{239}_{92}\)U. Each β− decay raises the proton number by 1 and leaves the nucleon number unchanged. The first decay gives \(^{239}_{93}\)Np and the second gives Z = 94, A = 239. This is plutonium-239. Pu-239 has a half-life of about 24 000 years, which makes it a significant long-lived nuclide in spent fuel. Option A stops after only one decay. Option B assumes β− decay lowers the proton number. Option D ignores the extra nucleon gained by neutron capture.✓ 1Answer C

Answer: C

Exam tip: Check every nuclear equation by balancing nucleon number and proton number separately at each step.

33IBP2S-040-00-07Fission products & waste Paper 1AMedium1 mark

Spent reactor fuel is stored securely for several thousand years. At the end of this time, the remaining activity of the stored waste is mainly due to

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Marking pointMarkNotes
After thousands of years, nuclides with half-lives of days or decades have decayed to negligible levels. What remains is dominated by nuclides with very long half-lives, such as Pu-239. These heavy nuclides are formed when U-238 captures neutrons. Option A is wrong because radioactive decay is exponential, so the activity never becomes exactly zero. Option B is wrong because the total activity of spent fuel falls over time. Option C is wrong because I-131 has a half-life of only 8 days and disappears within months.✓ 1Answer D

Answer: D

Exam tip: Divide waste into two groups. Short-lived fission products cause the early heat and high activity. Long-lived heavy nuclides decide how long the waste must be isolated.

34IBP2S-040-00-08Fission equations Paper 1AMedium1 mark

A nucleus of \(^{235}_{92}\text{U}\) absorbs a slow neutron and splits into two fragments X and Y. Three neutrons are released. Which pair of fragments is possible?

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Marking pointMarkNotes
There are 235 + 1 = 236 nucleons before fission. After fission the fragments must contain 236 − 3 = 233 nucleons, and 141 + 92 = 233. The proton numbers must also balance: 56 + 36 = 92. Option A has 236 nucleons in the fragments, which leaves none for the three neutrons. Option C forgets the absorbed neutron (235 − 3 = 232). Option D does not conserve charge, because 56 + 38 = 94.✓ 1Answer B

Answer: B

Common error: Include the incoming neutron on the left-hand side. The total number of nucleons before fission is 236, not 235.

35IBP2S-040-00-10Energy released in fission Paper 1AMedium1 mark

A fission event releases 200 MeV. The rest energy of a neutron plus a uranium-235 nucleus is approximately 236 × 931.5 MeV. Approximately what percentage of the rest mass of the reactants is converted into energy?

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Marking pointMarkNotes
The rest energy of the reactants is about 236 × 931.5 ≈ 2.2×105 MeV. The fraction converted is 200 / 2.2×105 ≈ 9×10−4, which is about 0.1%. Option C compares 200 MeV with the total binding energy of the nucleus (about 1800 MeV) rather than its rest energy. Option B comes from a power-of-ten error in the energy equivalent of 1 u. Option D is roughly the fraction of the released energy that goes into fragment kinetic energy, not a fraction of the mass.✓ 1Answer A

Answer: A

Exam tip: Fission releases a lot of energy per kilogram, yet only about 0.1% of the mass is converted. Even a small mass fraction multiplied by c² gives a huge energy.

36IBP2S-040-01-03Reactor components Paper 1AMedium1 mark

Some reactors run on natural (unenriched) uranium. These reactors use heavy water (D2O) as the moderator rather than ordinary water (H2O). What is the main reason for this choice?

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Marking pointMarkNotes
Natural uranium contains only about 0.7% U-235, so the reactor cannot afford to lose neutrons. Hydrogen nuclei capture neutrons quite strongly. Deuterium already contains a neutron and captures very few. With heavy water, enough thermal neutrons remain to keep the chain reaction going. Option B is false: a hydrogen nucleus (a proton) is closer in mass to a neutron. Option C is not the reason. Option D is false because deuterium does not undergo fission.✓ 1Answer A

Answer: A

Common error: A good moderator needs two properties: light nuclei, and low neutron absorption. Many answers give only the first.

37IBP2S-040-01-04Chain reactions Paper 1AMedium1 mark

A submarine reactor is operating at a steady power. The control rods are then withdrawn slightly from the core. What happens immediately afterwards?

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Marking pointMarkNotes
Control rods absorb neutrons. Withdrawing them means fewer neutrons are absorbed, so each fission leads on average to more than one further fission, and the fission rate (and so the power) rises. The energy per fission and the number of neutrons per fission are properties of the fission process and do not change. Control rods have no effect on how quickly neutrons are slowed.✓ 1Answer D

Answer: D

Exam tip: Describe control-rod effects in terms of the number of neutrons available to cause further fission. Do not say they change the energy or the neutrons produced in each fission.

38IBP2S-040-01-05Energy released in fission Paper 1AMedium1 mark

The reactor of a nuclear-powered icebreaker ship produces a thermal power of 160 MW. Each fission of uranium-235 releases about 200 MeV. Take 1 MeV = 1.6×10−13 J. What is the approximate number of fissions per second in the core?

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Marking pointMarkNotes
Energy per fission = 200 × 1.6×10−13 = 3.2×10−11 J.
Rate = 1.6×108 / 3.2×10−11 = 5.0×1018 s−1.
8.0×105 comes from dividing 160×106 by 200 without converting MeV to joules. 5.0×1024 comes from treating 200 MeV as 200 eV. 2.5×1018 comes from doubling the energy, as if 200 MeV were released by each fragment.
✓ 1Answer B

Answer: B

Exam tip: Convert MeV to joules before dividing a power by an energy. Then check that the answer is a sensible rate.

39IBP2S-040-01-06Energy released in fission Paper 1AMedium1 mark

Why does the fission of a uranium-235 nucleus release energy?

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Marking pointMarkNotes
The products of fission lie nearer the peak of the binding-energy-per-nucleon curve, so their total binding energy is greater. This difference in binding energy, which corresponds to the mass defect, is released as kinetic energy of the products. B states the reverse of what happens. The neutron is not annihilated (C). Nucleon number is conserved (D).✓ 1Answer A

Answer: A

Common error: Binding energy is the energy needed to separate a nucleus, not energy stored in it. Products with greater binding energy means that energy has been released.

40IBP2S-040-01-07Spontaneous fission Paper 1AMedium1 mark

Geologists date volcanic crystals by counting microscopic damage tracks. These tracks are left when uranium-238 nuclei in the crystal undergo spontaneous fission. The half-life of U-238 is 4.5×109 years. Two crystals, A and B, have the same uranium concentration. Crystal A has twice as many tracks per unit area as crystal B. Both crystals are a few million years old. What can be concluded?

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The ages are tiny compared with the half-life, so the number of U-238 nuclei stays almost constant. The spontaneous fission rate is therefore constant, and tracks build up in proportion to time. Twice the track density with the same uranium content means about twice the age. A ignores the build-up of tracks over time. B reverses the relationship. D wrongly treats the relationship as a square law.✓ 1Answer C

Answer: C

Exam tip: When a time is much shorter than the half-life, treat the decay rate as constant. The number of events is then proportional to the time.

41IBP2S-040-01-09Spontaneous fission Paper 1AMedium1 mark

A nucleus of californium-252, 25298Cf, undergoes spontaneous fission. It produces a barium-144 nucleus, 14456Ba, four neutrons and one other nucleus. What is the other nucleus?

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Marking pointMarkNotes
No neutron is absorbed in spontaneous fission.
Nucleon number: 252 = 144 + A + 4, so A = 104.
Proton number: 98 = 56 + Z + 0, so Z = 42 (molybdenum).
Mo-105 comes from adding an incoming neutron, as in induced fission. Mo-108 ignores the four neutrons. Pd-104 treats the neutrons as if they carried charge.
✓ 1Answer B

Answer: B

Common error: Do not add an incident neutron to the left-hand side for spontaneous fission. That step belongs only to induced fission.

42E-1A-19Energy released in fission Paper 1AHard1 mark

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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Marking pointMarkNotes
Step 1Total binding energy of a nucleus = nucleon number × binding energy per nucleon; free neutrons have no binding energy.—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.✓ 1Answer 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.
43E-1A-37Energy released in fission Paper 1AHard1 mark

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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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.✓ 1Answer 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.
44E-1A-40Energy released in fission Paper 1AHard1 mark

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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Marking pointMarkNotes
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).✓ 1Answer 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.
45E-1A-56Energy released in fission Paper 1AHard1 mark

A neutron-induced fission is represented by

23592U + 10n → X + Y + 3 10n

The mass of the uranium-235 nucleus is mU, the mass of a neutron is mn, and the masses of the nuclei X and Y are mX and mY. What is the energy released? (c is the speed of light.)

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Marking pointMarkNotes
Step 1Mass before = mU + mn; mass after = mX + mY + 3mn.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Energy released = (mass before − mass after)c2 = (mU − mX − mY − 2mn)c2: one of the three emitted neutrons cancels the incident neutron.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe incident neutron has been left out of the mass before the fission, so one neutron mass too many has been subtracted.
  • BCorrect: (mU + mn) − (mX + mY + 3mn) = mU − mX − mY − 2mn, multiplied by c2.
  • CThe three neutrons released by the fission have been left out of the mass after it.
  • DThis is (mass after − mass before)c2, which is negative for a fission that releases energy: the sign is reversed.
46E-1A-92Energy released in fission Paper 1AHard1 mark

A neutron of kinetic energy En is absorbed by a stationary uranium-235 nucleus, which then undergoes fission. The products have a total kinetic energy Ek, and gamma photons of total energy Eγ are emitted.

What is the decrease in total rest mass? (c is the speed of light.)

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Marking pointMarkNotes
Step 1Energy is conserved: (rest energy before) + En = (rest energy after) + Ek + Eγ.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2So the decrease in rest energy is Ek + Eγ − En, and the decrease in rest mass is this divided by c2.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the energy released is the energy of the products, Ek + Eγ, less the kinetic energy En that was brought in; divide by c2 for the mass.
  • BThe kinetic energy of the incident neutron was supplied, not released, so it must be subtracted, not added.
  • CGamma photons have no rest mass, but their energy also comes from the decrease in rest mass, so Eγ must be included.
  • DMass = energy/c2; multiplying by c2 gives a quantity with the unit J m2 s−2, not a mass.
47E-1A-96Energy released in fission Paper 1AHard1 mark

The thermal power of a reactor, found from the heating of its coolant, is 2.00 GW. Each fission of uranium-235 releases a total of 202 MeV, including the energy from the decay of its fission products. Of this, 9 MeV is carried by the antineutrinos emitted in the β− decays of the fission products.

How many fissions occur each second? (1 MeV = 1.60 × 10−13 J)

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Step 1Antineutrinos interact so rarely that they pass straight out of the reactor, so the energy that heats the coolant is 202 − 9 = 193 MeV = 3.09 × 10−11 J per fission.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Fissions per second = 2.00 × 109/3.09 × 10−11 = 6.5 × 1019 s−1.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe antineutrino energy has been added to the total (211 MeV) instead of subtracted.
  • BThis assumes all 202 MeV heats the coolant; the 9 MeV carried by antineutrinos escapes from the reactor.
  • CCorrect: only 193 MeV per fission is deposited in the reactor, so the fission rate is 2.00 × 109/(193 × 1.60 × 10−13).
  • DOnly the 9 MeV carried by the antineutrinos has been used, instead of the 193 MeV that is deposited.
48IBP2S-040-02-10Energy released in fission Paper 1AHard1 mark

Fission fragments from uranium-235 are nearly always radioactive and usually decay by β− emission. Which statement explains this?

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Heavy nuclei need a high N/Z ratio (about 1.55 for U-235) to be stable. Stable intermediate-mass nuclei have a lower ratio (about 1.2–1.4). Fragments inherit roughly the parent's ratio, so they are neutron-rich. In β− decay a neutron changes into a proton, which lowers N/Z. A proton excess would lead to β+ decay, not β−. The fragments actually have a higher binding energy per nucleon than U-235. De-excitation of an excited nucleus produces gamma photons, not electrons.✓ 1Answer C

Answer: C

Exam tip: Neutron-rich nuclei undergo β⁻ decay, and proton-rich nuclei undergo β⁺ decay.

49IBP2S-040-00-09Energy released in fission Paper 1AHard1 mark

Three statements are made about the fission of uranium-235 into two fragments and some neutrons.
I. The total binding energy of the products is greater than the binding energy of the uranium-236 compound nucleus.
II. The total rest mass of the products is less than the total rest mass of the uranium-235 nucleus and the absorbed neutron.
III. The binding energy per nucleon of the fragments is less than that of uranium-235.
Which statements are correct?

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Marking pointMarkNotes
The fragments lie closer to the peak of the binding-energy-per-nucleon curve, so they are more tightly bound than U-235. Their binding energy per nucleon is greater, about 8.5 MeV compared with 7.6 MeV, so III is false. The total binding energy rises, so I is correct. This increase in binding energy is released as energy, so the products have less rest mass than the reactants, and II is correct. I and II describe the same process: an increase in binding energy matches an equal decrease in rest mass.✓ 1Answer C

Answer: C

Common error: Greater binding energy means the products are more stable and have less mass. It does not mean they 'contain more energy'.

50IBP2S-040-01-10Reactor components Paper 1AHard1 mark

In a heavy-water moderator, a fast neutron collides elastically and head-on with a stationary deuterium nucleus. Take the mass of the deuterium nucleus to be twice the mass of the neutron. What fraction of the neutron's initial kinetic energy is transferred to the deuterium nucleus?

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Marking pointMarkNotes
Momentum: m u = m v1 + 2m v2.
For an elastic collision the relative speed is unchanged: v2 − v1 = u.
Solving gives v1 = −u/3 and v2 = 2u/3.
The neutron keeps (1/3)2 = 1/9 of its kinetic energy, so 8/9 is transferred.
A total transfer (1) happens only for equal masses. 1/3 confuses the speed ratio with the energy ratio. 1/2 assumes the energy is shared equally.
✓ 1Answer C

Answer: C

Exam tip: For a head-on elastic collision, use conservation of momentum together with the fact that the relative speed of approach equals the relative speed of separation. This avoids solving a quadratic.

51E-1B-25Reactor components Paper 1BEasy8 marks

In a university laboratory, a sealed source emits fast neutrons. A detector that responds only to slow neutrons is placed a fixed distance from the source. Sheets of polyethylene, a plastic containing many hydrogen atoms, are placed between the source and the detector, and the number of counts N in 300 s is recorded for each total thickness x. The background count is negligible. The uncertainty in a number of counts N is ±√N. The graph shows the data with error bars.

x / cm02468101214
N1887091371157515141225971692
Graph of the number of counts N in 300 s against polyethylene thickness x / cm from 0 to 14 cm, with error bars; points only. N axis 0 to 1800.03691215x / cm0300600900120015001800N
Graph drawn to scale
(a)

Calculate the percentage uncertainty in N when x = 0.

(1)
(b)

Draw the curve of best fit for the data.

(1)
(c)

Determine, using your curve, the thickness for which N is greatest, and its uncertainty.

(2)
(d)

Explain why N increases as x increases from 0 to about 6 cm.

(2)
(e)

Suggest why N decreases when x is larger than about 8 cm.

(1)
(f)

Suggest one change to the procedure that would allow the thickness in (c) to be found more precisely.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
√188/188 × 100 = 7.3 %✓ 1Accept 7 % or 7.4 %.
Part (b)
Smooth curve rising to a single maximum and then falling, passing through all the error bars✓ 1Do not accept a dot-to-dot line.
Part (c)
Peak of the curve read at x ≈ 6.3 cm✓ 1Accept 5.5–7.1 cm from the candidate's curve.
Uncertainty ≈ ±1 cm, e.g. from the range of x over which the curve stays within the error bar of the highest point✓ 1Accept ±0.5 cm to ±2 cm with a sensible method.
Part (d)
Fast neutrons lose kinetic energy in collisions with hydrogen nuclei, whose mass is similar to that of a neutron✓ 1
A greater thickness means more collisions, so more of the neutrons are slow enough to be detected✓ 1OWTTE
Part (e)
In thicker polyethylene more of the slow neutrons are absorbed «captured» by nuclei or scattered away before they reach the detector✓ 1OWTTE
Part (f)
Take more readings at smaller intervals of thickness «e.g. every 0.5 cm» between about 4 cm and 9 cm / count for longer at each thickness✓ 1Do not accept 'repeat the readings' alone.

Answers: (a) 7.3 %  ·  (c) ≈ 6.3 cm ± 1 cm (the remaining parts are explanations — see the table above)

52E-1B-43Reactor components Paper 1BEasy11 marks

To model how a moderator slows neutrons, a student uses two gliders on a level air track. Glider 1, of mass 0.200 kg, represents a neutron. It moves at speed u towards glider 2, of mass M, which is at rest and represents a nucleus of the moderator. Magnetic bumpers make the collision nearly elastic, and glider 1 rebounds with speed v. A card of length (100.0 ± 0.5) mm on glider 1 passes through a light gate before and after the collision, taking times t1 and t2. Each time has an uncertainty of ±0.002 s.

For a perfectly elastic head-on collision, theory predicts v/u = (A − 1)/(A + 1), where A = M/0.200 kg.

M / kgAt1 / st2 / sv/u(A − 1)/(A + 1)
0.40020.2010.6630.3030.333
0.60030.1980.4300.4600.500
0.80040.2040.3630.5620.600
1.20060.2000.2990.6690.714
1.80090.1970.261……
2.400120.2020.2530.7980.846
v/u against (A − 1)/(A + 1) for five collisions, with error bars and the dashed line predicted for perfectly elastic collisions0.00.20.40.60.81.0(A − 1)/(A + 1)0.00.20.40.60.81.0v/uelastic theory
Graph drawn to scale
(a)

Calculate u for the collision with A = 4, and its absolute uncertainty.

(2)
(b)

Show that v/u = t1/t2, and complete the table for A = 9.

(1)
(c)

Plot the point for A = 9 and draw the line of best fit.

(2)
(d)

Determine the gradient and the intercept of your line.

(2)
(e)

Compare the results with the theory for perfectly elastic collisions, and suggest a reason for any difference.

(2)
(f)

Using the data for A = 2 and A = 12, explain why a good moderator contains nuclei of small mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
u = 0.1000/0.204 = 0.490 m s−1✓ 1
Fractional uncertainty = 0.5/100.0 + 0.002/0.204 «= 1.5 %», so u = (0.490 ± 0.007) m s−1✓ 1Accept ±0.007 or ±0.008 m s−1. MP2 is for an absolute uncertainty with matching precision.
Part (b)
v/u = (d/t2)/(d/t1) = t1/t2 AND for A = 9: v/u = 0.197/0.261 = 0.755, (A − 1)/(A + 1) = 8/10 = 0.800✓ 1All needed. Accept 2 or more decimal places.
Part (c)
Point plotted to within half a small square✓ 1Allow ECF from (b).
Single straight line passing through all the error bars✓ 1Do not credit the dashed theory line.
Part (d)
Gradient = 0.97✓ 1Accept 0.95–1.00.
Intercept = −0.02✓ 1Accept −0.05 to 0.00. Allow a value read from an extrapolated line.
Part (e)
Every point lies slightly below the theory line «gradient a little less than 1 and negative intercept», so the difference is systematic, not random✓ 1OWTTE
The collisions are not perfectly elastic: some kinetic energy is transferred to the bumpers «as internal energy / sound», so glider 1 rebounds more slowly than predicted✓ 1Accept friction or air resistance reducing the speed between the gate and the collision.
Part (f)
Fraction of kinetic energy kept by glider 1 = (v/u)2: about 0.64 for A = 12 but only 0.09 for A = 2✓ 1Accept values calculated from the candidate's line or table.
A neutron loses a much larger fraction of its kinetic energy in each collision with a light nucleus, so fewer collisions are needed to slow it down✓ 1OWTTE

Answers: (a) (0.490 ± 0.007) m s−1  ·  (b) 0.755, 0.800  ·  (d) gradient 0.97, intercept −0.02  ·  (f) 0.64 and 0.09 (the remaining parts are explanations — see the table above)

53E-1B-24Fission products & waste Paper 1BMedium10 marks

After fuel is removed from a nuclear reactor it must be stored in a cooling pond. A fuel-storage company publishes the thermal power P released per tonne of spent fuel at times t after removal, for one type of fuel. Each value of P has an uncertainty of ±4 %.

An engineer suggests the model P = kt−n, where k and n are constants. The graph shows log(P/kW t−1) against log(t/days), with error bars, for six of the times.

t / daysP / kW t−1log(t / days)log(P / kW t−1)
5049.81.6991.697
10028.32.0001.452
20017.52.3011.243
40010.22.6021.009
7006.88
10005.063.0000.704
20003.103.3010.491
Graph of log(P / kW t^−1) against log(t / days) with error bars, for six of the seven times; points only. Horizontal axis 1.6 to 3.4, vertical axis 0.2 to 1.8.1.622.42.83.2log(t / days)0.20.611.41.8log(P / kW t⁻¹)
Graph drawn to scale
(a)

Outline why spent fuel continues to release energy although no fission takes place in it.

(1)
(b)

Complete the table for t = 700 days.

(1)
(c)

Plot the point for t = 700 days and draw the line of best fit.

(2)
(d)

Determine n.

(2)
(e)

A student claims that P halves each time t doubles. Comment on this claim.

(1)
(f)

Fuel may be moved from the pond into dry storage casks once P is less than 2.0 kW t−1. Estimate, using your line and your answer to (d), the time after removal at which this occurs.

(2)
(g)

Suggest why the model may not predict P reliably at times much longer than those in the table.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The fission products are radioactive, and the energy of their decays «beta particles, gamma rays» is absorbed in the fuel✓ 1OWTTE
Part (b)
log(700) = 2.845 and log(6.88) = 0.838✓ 1Both needed. Accept 2 or more decimal places.
Part (c)
Point plotted to within half a small square✓ 1Allow ECF from (b).
Single straight line passing through all the error bars✓ 1
Part (d)
Gradient calculated from a large triangle; gradient = −n✓ 1
n = 0.75✓ 1Accept 0.70–0.80. Award [1 max] for a negative value of n.
Part (e)
Halving for each doubling would mean n = 1; the data give n ≈ 0.75 «lines through all the error bars give n between 0.74 and 0.76», so P falls by only about 20.75 ≈ 1.7 when t doubles: the claim is not supported✓ 1Allow ECF from (d). Accept a comparison of two pairs of data values, e.g. 100 → 200 days.
Part (f)
log t = (log k − log 2.0)/n, or the line extended to log P = log 2.0 = 0.301, giving log t ≈ 3.55✓ 1
t ≈ 3500 days «≈ 10 years»✓ 1Accept 3100–4200 days. Allow ECF from (d).
Part (g)
P comes from many fission products with different half-lives; as the shorter-lived ones decay away the mixture changes, so n need not stay constant «extrapolation beyond the data is unreliable»✓ 1OWTTE

Answers: (b) 2.845, 0.838  ·  (d) n ≈ 0.75  ·  (f) ≈ 3500 days (the remaining parts are explanations — see the table above)

54E-1B-40Fission products & waste Paper 1BMedium12 marks

A published database lists measurements made on samples of spent fuel from pressurized-water reactors. For each sample it gives the burn-up B, the thermal energy released per tonne of uranium in the fuel, in GW d t−1 (1 GW d is the energy released in one day at a power of 1 GW), and the mass m of caesium-137 per tonne of uranium. A student selects seven samples of one fuel type and plots m against B. The uncertainties are ±3 % in B and ±2 % in m.

Energy released per fission = 200 MeV; 1 MeV = 1.60 × 10−13 J; 1 u = 1.661 × 10−27 kg.

SampleB / GW d t−1 (± 3 %)m / kg t−1 (± 2 %)
112.50.470
219.80.725
327.31.023
433.91.253
541.21.553
647.61.752
754.02.015
Mass of caesium-137 per tonne of uranium against burn-up for seven fuel samples, with error bars0102030405060burn-up B / GW d t⁻¹0.00.51.01.52.02.5m / kg t⁻¹
Graph drawn to scale
(a)

Outline why m is expected to be proportional to B.

(1)
(b)

Draw the line of best fit for the data.

(1)
(c)

Determine the gradient of your line. State its unit.

(2)
(d)

Show that the energy of 1 GW d corresponds to about 2.7 × 1024 fissions.

(1)
(e)

Determine the fraction of all fissions that produce a caesium-137 nucleus.

(2)
(f)

The steepest and the least steep straight lines that pass through all the error bars have gradients of 0.0398 and 0.0344 in the unit of (c). Determine the absolute uncertainty in your answer to (e).

(2)
(g)

The accepted fission yield of caesium-137 from uranium-235 is 6.2 %.

(i)

Comment on whether your result agrees with the accepted value.

(1)
(ii)

The fuel stayed in the reactor for several years. Suggest one reason why the value from the database could be smaller than the accepted yield.

(1)
(h)

Caesium-137 (half-life 30 years) is one of the main concerns in the long-term management of spent fuel. Suggest one reason why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Each fission releases «on average» the same energy, and a fixed fraction of fissions produce caesium-137, so the number of caesium-137 nuclei is proportional to the number of fissions and hence to the energy released✓ 1OWTTE
Part (b)
Single straight line passing through all the error bars «close to the origin»✓ 1Do not accept a dot-to-dot line.
Part (c)
Gradient calculated from a large triangle «over at least half the line»✓ 1
0.0372 kg GW−1 d−1 «kg per GW d»✓ 1Accept 0.0344–0.0398. Unit required: accept kg (GW d)−1.
Part (d)
1 × 109 × 86 400/(200 × 1.60 × 10−13) «= 2.70 × 1024»✓ 1Must see full substitution OR answer to 3 s.f.
Part (e)
Number of Cs-137 nuclei per GW d = gradient/(137 × 1.661 × 10−27) «= 1.64 × 1023»✓ 1Allow ECF from (c).
Fraction = 1.64 × 1023/2.70 × 1024 = 0.061 «≈ 6.1 %»✓ 1Accept 0.056–0.065. Award [2] for CNA.
Part (f)
Fractional uncertainty = half the range of gradients/gradient = (0.0398 − 0.0344)/(2 × 0.0372) «= 7 %»✓ 1Accept use of either extreme gradient minus the best gradient.
Absolute uncertainty = ± 0.004 «so the fraction is 6.1 ± 0.4 %»✓ 1Allow ECF from (e). Accept ±0.004 to ±0.005 «±0.4 % to ±0.5 %». MP2 is for an absolute value with precision matching the result.
Part (g)(i)
The accepted value 6.2 % lies within the range 5.7 % to 6.5 %, so the result agrees with it «within its uncertainty»✓ 1Allow ECF from (e) and (f). The conclusion must match the candidate's range.
Part (g)(ii)
Some caesium-137 has decayed since it was formed «during the years in the reactor and in storage» / some caesium-137 nuclei absorbed neutrons in the reactor and became a different nuclide✓ 1Accept any one. Do not accept 'measurement error' alone.
Part (h)
Its half-life is long enough for the waste to stay highly active for centuries «about 300 years for ten half-lives», and it emits penetrating gamma radiation «through its decay product» / it is soluble and chemically similar to potassium, so if released it spreads in water and the food chain✓ 1Accept any one reason linked to a property.

Answers: (c) 0.0372 kg GW−1 d−1  ·  (e) 0.061  ·  (f) ± 0.004 (the remaining parts are explanations — see the table above)

55E-1B-42Spontaneous fission Paper 1BMedium12 marks

Uranium-238 undergoes spontaneous fission, very rarely. A laboratory uses this to measure the mass of uranium in sealed drums of depleted uranium oxide (almost pure uranium-238) without opening them. Each drum is placed in a neutron counter for 1000 s and the number of counts N is recorded. Six calibration drums of the same size and type, containing known masses M of uranium, give the results below. The uncertainty in a number of counts N is ±√N; the error bars are too small to show.

M / kgN
01221
206567
4012124
6017626
8022884
10028349
Number of counts in 1000 s against mass of uranium in six calibration drums020406080100M / kg051015202530N / 10³ counts
Graph drawn to scale
(a)

Explain why the number of spontaneous fissions per second in a drum is expected to be proportional to M.

(1)
(b)

Calculate the percentage uncertainty in N for M = 20 kg.

(1)
(c)

Draw the line of best fit for the data.

(1)
(d)
(i)

Determine the gradient of the line, with its unit.

(2)
(ii)

State what the intercept on the N axis represents.

(1)
(e)

The counter detects 2.0 % of the neutrons emitted. On average 2.0 neutrons are emitted in each spontaneous fission of uranium-238. Determine the number of spontaneous fissions per second in 1 kg of uranium-238.

(2)
(f)

A sealed drum of the same type gives 19 450 counts in 1000 s. Determine the mass of uranium in the drum and its absolute uncertainty. Consider only the uncertainty in the number of counts.

(3)
(g)

Suggest why the calibration drums must be of the same size and chemical form as the drums that are measured.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The number of uranium-238 nuclei is proportional to M, and each nucleus has the same «constant» probability of spontaneous fission per second «decay is random»✓ 1OWTTE
Part (b)
√6567/6567 × 100 = 1.2 %✓ 1Accept 1 %.
Part (c)
Single straight line passing through or very close to all the points✓ 1
Part (d)(i)
Gradient calculated from a large triangle✓ 1
272 counts kg−1 «in 1000 s»✓ 1Accept 263–280. Accept 0.272 s−1 kg−1. Unit required.
Part (d)(ii)
The background count «in 1000 s», e.g. neutrons produced by cosmic rays, with no uranium present✓ 1OWTTE
Part (e)
Neutrons emitted per second per kg = 272/(1000 × 0.020) «= 13.6 s−1»✓ 1Allow ECF from (d)(i).
Fissions per second = 13.6/2.0 = 6.8 s−1✓ 1Accept 6.5–7.1 s−1. Award [2] for CNA.
Part (f)
Background subtracted: 19450 − 1217 = 18233✓ 1Allow ECF from the candidate's intercept.
M = 18233/272 = 67.0 kg✓ 1Accept 65–69 kg. Allow ECF from (d)(i).
Uncertainty = √19450/272 ≈ ±0.5 kg, so M = (67.0 ± 0.5) kg✓ 1Accept ±0.5 kg to ±0.6 kg «including the uncertainty of the background count». MP3 is for matching the precision of value and uncertainty.
Part (g)
Neutrons are absorbed or scattered by the drum and its contents, so the fraction that reaches the detector «the efficiency» depends on the size and composition of the drum✓ 1OWTTE. Accept: some neutrons may cause further fissions or be produced in other reactions in the oxide, which depend on the material.

Answers: (b) 1.2 %  ·  (d)(i) 272 counts kg−1  ·  (e) 6.8 s−1  ·  (f) (67.0 ± 0.5) kg (the remaining parts are explanations — see the table above)

56E-1B-02Energy released in fission Paper 1BHard11 marks

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)

ParticleMass / u
235U235.043930
141Ba140.914411
92Kr91.926156
neutron1.008665
Binding energy per nucleon against nucleon number A for long-lived nuclides6080100120140160180200220240nucleon number A7.47.67.88.08.28.48.68.8binding energy per nucleon / MeV
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)
Show mark scheme
Marking pointMarkNotes
Part (a)
A single smooth curve passing through, or very close to, all the points✓ 1Do not accept a dot-to-dot line.
Part (b)
235: 7.59 MeV; 141: 8.36 MeV; 92: 8.69 MeV✓ 1Accept 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✓ 1Must 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✓ 1Accept ±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✓ 1OWTTE
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✓ 1Accept 0.186 u.
E = 0.186033 × 931.5 = 173 MeV✓ 1Award [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✓ 1Allow 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✓ 1Accept: 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)

57E-1B-08Energy released in fission Paper 1BHard10 marks

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.

HourTemperature in / °CTemperature out / °C
135.246.6
235.146.7
335.346.8
435.246.5
535.046.5
635.146.8
(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)
Show mark scheme
Marking pointMarkNotes
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✓ 1Accept ±0.2 K from the thermometer resolutions (0.1 + 0.1).
Part (b)
P = 42.0 × 4180 × 11.5 = 2.02 × 106 W✓ 1Must see the substitution OR an answer to at least 3 s.f.
Part (c)
Percentage uncertainty = 0.5/42.0 + 0.2/11.5 = 1.2 % + 1.7 % = 2.9 %✓ 1Fractional uncertainties add for a product.
ΔP = ±0.06 MW, so P = (2.02 ± 0.06) MW✓ 1Allow 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✓ 1Award [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)✓ 1Allow 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✓ 1Decision 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)

58E-1B-41Chain reactions Paper 1BHard12 marks

A student uses a computer simulation to study the chain reaction in a cube of a solution of a uranium-235 salt in water. For a cube of side L the simulation follows many generations of neutrons and gives the multiplication factor k, the average number of neutrons from each fission that go on to cause another fission. Each value of k has an uncertainty of ±0.010. To save computing time, only cubes with L ≥ 25 cm were simulated.

The student tests the model k = k∞ − b/L, where k∞ and b are constants. The model assumes that the fraction of neutrons lost through the surface of the cube is proportional to surface area/volume.

L / cm(1/L) / cm−1k
250.04001.053
30…1.081
400.02501.099
500.02001.121
600.01671.126
80…1.144
1000.01001.147
Multiplication factor k against 1/L for five cube sizes, with error bars; two points still to be plotted0.000.010.020.030.040.050.060.071/L / cm⁻¹0.951.001.051.101.151.20k
Graph drawn to scale
(a)

Complete the table.

(1)
(b)

Plot the two missing points on the graph and draw the line of best fit. Extend your line to k = 1.00.

(2)
(c)

Determine b and k∞. Give the unit of b.

(2)
(d)

Determine the side length Lc of the smallest cube in which a chain reaction can be sustained.

(2)
(e)

Draw the lines of greatest and least gradient that pass through all the error bars, and hence determine the absolute uncertainty in Lc.

(2)
(f)

A technician proposes to store this solution in cubic containers of side 16 cm, kept well apart. Evaluate this proposal.

(2)
(g)

The cube is surrounded by a material that reflects escaping neutrons back into the solution. Suggest how the line on the graph would change.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.0333 AND 0.0125 «cm−1»✓ 1Both needed. Accept 0.033 and 0.0125.
Part (b)
Both points plotted to within half a small square✓ 1Allow ECF from (a).
Single straight line passing through all the error bars, extended to k = 1.00✓ 1
Part (c)
b = −gradient «from a large triangle» = 3.1 cm✓ 1Accept 2.6–3.7 cm. Unit required.
k∞ = intercept on the k axis = 1.18✓ 1Accept 1.17–1.19.
Part (d)
Uses k = 1: 1/Lc = (k∞ − 1)/b OR reads 1/L where the line crosses k = 1.00 «≈ 0.058 cm−1»✓ 1
Lc = 17 cm✓ 1Accept 16–18 cm. Allow ECF from (c). Award [2] for CNA.
Part (e)
Two lines drawn through all the error bars, one steeper and one less steep than the line of best fit, each extended to k = 1.00✓ 1
Lc from the two lines ≈ 15.5 cm and 19.0 cm, so uncertainty = ±1.7 cm «≈ ±2 cm»✓ 1Accept ±1 cm to ±2.5 cm from a correct method. Allow ECF from the candidate's lines.
Part (f)
16 cm is smaller than the best value of Lc but lies within its uncertainty range «15.5–19.0 cm», so it is not certain that k < 1 in such a container: the proposal is not safe✓ 1Allow ECF from (d) and (e). Decision must be consistent with the candidate's range.
Lc was found by extrapolating beyond the simulated range «smallest cube 25 cm», so the model may not hold there: use containers well below the lower limit «about 13 cm or less» or simulate small cubes directly✓ 1OWTTE
Part (g)
Fewer neutrons are lost, so b is smaller: the line is less steep with the same intercept k∞, so it crosses k = 1 at a larger 1/L «smaller Lc»✓ 1OWTTE. Both the smaller gradient and the unchanged intercept are needed.

Answers: (a) 0.0333, 0.0125 cm−1  ·  (c) b = 3.1 cm, k∞ = 1.18  ·  (d) Lc ≈ 17 cm  ·  (e) ± 1.7 cm (the remaining parts are explanations — see the table above)

59E-2-10Reactor components Paper 2Easy10 marks

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.

Simplified pressurised-water nuclear power station: inside a thick wall, a reactor vessel filled with water contains vertical fuel rods with control rods between them; a primary circuit with a pump carries water to a heat exchanger; a secondary circuit carries steam to a turbine and condenser. Boxes 1–4 mark the control rods, the water around the fuel, the heat exchanger and the thick wall.pumpprimary circuitturbinecondensersteamwaterfuel rods1234
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)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any two labels correct: 1 control rods, 2 moderator, 3 heat exchanger, 4 shielding✓ 1
All four labels correct✓ 1Award [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»✓ 1OWTTE
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✓ 1OWTTE
Part (e)
Neutrons and gamma rays✓ 1Both needed.
60E-2-48Chain reactions Paper 2Easy13 marks

A small fission reactor is designed to supply a base on the Moon. It must deliver an electrical power of 40 kW for 10 years. Its core contains 8.0 kg of uranium-235 and is surrounded by a reflector that scatters escaping neutrons back into the core. The overall efficiency of the conversion of thermal energy into electrical energy is 25 %. The energy that is not converted is radiated into space from radiator panels at a temperature of 400 K, with emissivity 0.90.

(energy released per fission = 200 MeV, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg, σ = 5.67 × 10−8 W m−2 K−4, 1 year = 3.16 × 107 s)

(a)

Outline what is meant by a chain reaction in the core.

(1)
(b)

Explain why the reflector allows the core to contain less uranium-235.

(2)
(c)

Show that about 5 × 1015 fissions occur in the core each second.

(1)
(d)

Calculate the total area of the radiating surfaces. Ignore any radiation absorbed by the panels.

(2)
(e)

Explain why the radiator panels are designed to work at as high a temperature as possible.

(2)
(f)

Only about 10 % of the uranium-235 in the core can undergo fission before the core must be replaced. Determine whether the core can supply the base for 10 years.

(3)
(g)

The Moon has no atmosphere. Suggest two reasons why the reactor is placed about 1 km from the base and partly buried in lunar soil.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
neutrons released by each fission go on to cause further fissions «so the process is self-sustaining»✓ 1OWTTE
Part (b)
fewer of the neutrons produced escape from the core without causing fission✓ 1
so the chain reaction can be sustained «on average one neutron per fission causes another» with a smaller mass of fuel / a smaller core «saving launch mass»✓ 1OWTTE
Part (c)
thermal power = 40/0.25 = 160 kW; rate = 1.6 × 105/(200 × 1.60 × 10−13) «= 5.0 × 1015 s−1»✓ 1Must see the thermal power AND the substitution OR answer to 3 s.f.
Part (d)
power radiated = 160 − 40 = 120 kW = eσAT4✓ 1Award [1 max] for 160 kW used.
A = 1.2 × 105/(0.90 × 5.67 × 10−8 × 4004) = 92 m2✓ 1Award [2] for CNA. Allow ECF for the power.
Part (e)
the power radiated per unit area is proportional to T4✓ 1
so a small increase in temperature greatly reduces the area «and mass» of panel needed, e.g. doubling T reduces the area 16 times✓ 1OWTTE
Part (f)
fissions in 10 years = 5.0 × 1015 × 10 × 3.16 × 107 = 1.6 × 1024✓ 1Allow ECF from (c).
mass of uranium-235 = 1.6 × 1024 × 235 × 1.661 × 10−27 = 0.62 kg✓ 1
0.62 kg < 0.10 × 8.0 = 0.80 kg, so the core can supply the base for 10 years✓ 1The conclusion must be consistent with the calculation. Award [3] for a correct comparison of the time for 0.80 kg «13 years» with 10 years.
Part (g)
the soil absorbs neutrons and gamma rays from the core, providing shielding without bringing heavy shielding from the Earth✓ 1
the intensity of the radiation falls with distance «approximately as 1/d2» / the Moon has no atmosphere to absorb it, so distance is needed to reduce the dose✓ 1Accept any two distinct reasons.

Answers: (c) 5.0 × 1015 s−1  ·  (d) 92 m2  ·  (f) 0.62 kg needed < 0.80 kg available — yes (the remaining parts are explanations — see the table above)

61E-2-77Energy released in fission Paper 2Easy13 marks

A coal-fired power station has an electrical power output of 1.2 GW and an overall efficiency of 38 %. It is to be replaced by a nuclear power station with the same electrical power output and an overall efficiency of 34 %.

Energy density of the coal = 29 MJ kg−1. Burning 1.0 kg of this coal produces 2.6 kg of carbon dioxide. Energy released per fission of uranium-235 = 200 MeV. 1 MeV = 1.60 × 10−13 J; 1 u = 1.661 × 10−27 kg.

(a)

Outline how the carbon dioxide released by the coal-fired power station contributes to an increase in the mean temperature of the Earth's surface.

(3)
(b)
(i)

Show that the coal-fired power station burns about 9 × 106 kg of coal each day.

(1)
(ii)

Calculate the mass of carbon dioxide that it releases in one year.

(1)
(c)
(i)

Determine the mass of uranium-235 that undergoes fission each day in the nuclear power station.

(3)
(ii)

Calculate the ratio (mass of coal burned each day)/(mass of uranium-235 that undergoes fission each day).

(1)
(d)

The nuclear power station releases no carbon dioxide while it operates. Discuss whether replacing the coal-fired power station by the nuclear power station is better for the environment.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
The Earth's surface emits infrared radiation✓ 1
Carbon dioxide molecules absorb infrared photons «whose energy matches differences between molecular energy levels / resonance» and re-emit infrared radiation in all directions, some back towards the surface✓ 1
More carbon dioxide means more energy is returned to the surface, so the surface temperature rises until it again radiates as much power as it receives «enhanced greenhouse effect»✓ 1OWTTE
Part (b)(i)
1.2 × 109/0.38 × 86 400/2.9 × 107 «= 9.4 × 106 kg»✓ 1Must see full substitution OR answer to 2 s.f. with the power of ten.
Part (b)(ii)
9.4 × 106 × 2.6 × 365 = 8.9 × 109 kg✓ 1Accept 8.5–9.0 × 109 kg. Allow ECF from (b)(i).
Part (c)(i)
Thermal power = 1.2 × 109/0.34 = 3.5 × 109 W, so energy per day = 3.0 × 1014 J✓ 1
Fissions per day = 3.0 × 1014/(200 × 1.60 × 10−13) = 9.5 × 1024✓ 1
Mass = 9.5 × 1024 × 235 × 1.661 × 10−27 = 3.7 kg✓ 1Award [3] for CNA. Award [2 max] if the efficiency is ignored «1.3 kg».
Part (c)(ii)
9.4 × 106/3.7 = 2.5 × 106✓ 1Accept 2.4–2.6 × 106. Allow ECF from (b)(i) and (c)(i).
Part (d)
Advantage: about 9 × 109 kg of carbon dioxide per year is no longer released, which reduces the enhanced greenhouse effect «and global warming»✓ 1Award [1] per distinct relevant point; at least one advantage and one disadvantage needed for [4].
Second advantage: the mass of fuel is about a million times smaller, so far less mining and transport are needed / the output does not depend on the weather✓ 1
Disadvantage: the fission products are radioactive, and some have long half-lives, so the waste must be isolated «in deep geological stores» for thousands of years✓ 1
Second disadvantage: risk of an accident releasing radioactive material / carbon dioxide is still released in mining, enrichment and construction / the lower efficiency means more thermal energy is transferred to the surroundings per joule of electrical energy✓ 1Accept a reasoned overall conclusion as the fourth point.

Answers: (b)(i) 9.4 × 106 kg  ·  (b)(ii) 8.9 × 109 kg  ·  (c)(i) 3.7 kg  ·  (c)(ii) 2.5 × 106 (the remaining parts are explanations — see the table above)

62E-2-78Reactor components Paper 2Easy12 marks

In a pool-type research reactor the core is near the bottom of an open tank of water 8.0 m deep. The top of the core is 1.2 m above the floor of the tank. The water acts as the moderator, the coolant and the shielding. The reactor operates at a thermal power of 2.0 MW; a pump circulates the water through the core and through a heat exchanger.

Open tank of water 8.0 m deep with the reactor core near the bottom, its top 1.2 m above the floor; pipes lead from the tank to a heat exchanger and pump outside itcorewater8.0 m1.2 mheatexchangerand pump
Diagram drawn to scale
(a)

Outline how the water acts as (i) the moderator, (ii) the coolant and (iii) the shielding.

(3)
(b)

The most penetrating gamma radiation from the core is reduced to half its intensity by every 0.30 m of water.

(i)

Determine the least depth of water for the intensity of this radiation to be reduced to less than 1.0 × 10−6 of its value.

(2)
(ii)

Hence state whether the tank is deep enough.

(1)
(c)

Neutrons from the core convert some oxygen-16 nuclei in the water into nitrogen-16: 168O + 10n → 167N + X. Nitrogen-16 has a half-life of 7.1 s and emits penetrating gamma rays. A device above the core slows the water so that it takes 71 s to reach the surface.

(i)

Identify X.

(1)
(ii)

Show that about 0.1 % of the nitrogen-16 nuclei reach the surface.

(1)
(d)

When the reactor is shut down and the pump is switched off, the core is cooled by natural convection in the tank.

(i)

Draw arrows on the diagram to show the convection current in the water.

(1)
(ii)

Explain how this convection current arises.

(2)
(e)

State the origin of the thermal energy released by the core after it has been shut down.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Moderator: neutrons lose kinetic energy in collisions with hydrogen nuclei «of similar mass», so they are slowed and are more likely to cause fission of uranium-235✓ 1
Coolant: the water flows past the fuel, absorbs thermal energy «its temperature rises» and carries it to the heat exchanger✓ 1
Shielding: the large depth of water absorbs the neutrons and gamma radiation from the core, so the intensity at the surface is very small✓ 1
Part (b)(i)
219 = 5.2 × 105 is not enough but 220 = 1.05 × 106 is, so 20 halvings are needed✓ 1Accept log working.
Depth = 20 × 0.30 = 6.0 m✓ 1Award [1 max] for 5.7 m «19 halvings».
Part (b)(ii)
Water above the core = 8.0 − 1.2 = 6.8 m > 6.0 m, so the tank is deep enough✓ 1Allow ECF from (b)(i).
Part (c)(i)
A proton, 11p «or 11H»✓ 1Accept 'proton'.
Part (c)(ii)
71/7.1 = 10 half-lives, so fraction = 1/210 = 1/1024 «= 9.8 × 10−4 ≈ 0.1 %»✓ 1Must see 10 half-lives AND 1/1024 OR 210.
Part (d)(i)
Arrows upwards from the core through the middle of the tank, and downwards near the walls, forming closed loops back to the bottom of the core✓ 1Both directions needed.
Part (d)(ii)
Water next to the core is heated, expands and becomes less dense, so it rises «buoyancy»✓ 1
Cooler, denser water sinks near the walls and flows in at the bottom to replace it✓ 1OWTTE
Part (e)
The radioactive decay of the fission products in the fuel✓ 1Do not accept 'fission continues'.

Answers: (b)(i) 6.0 m  ·  (b)(ii) 6.8 m of water: deep enough  ·  (c)(ii) 1/1024 ≈ 0.1 % (the remaining parts are explanations — see the table above)

63IBP2S-040-02-11Chain reactions Paper 2Easy6 marks

A design study for a lunar surface power station considers plutonium-239 as the fuel. One possible fission reaction is
23994Pu + 10n → 14556Ba + 9238Sr + x 10n

(a)

Outline what is meant by nuclear fission.

(2)
(b)

Determine x.

(2)
(c)

Explain how the neutrons released can lead to a chain reaction.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A heavy nucleus splits into two smaller nuclei of comparable mass . This usually follows absorption of a neutron and releases energy and neutrons .✓ 2
Part (b)
Conserve nucleon number: 239 + 1 = 145 + 92 + x . x = 3 .✓ 2
Part (c)
The released neutrons can be absorbed by other Pu-239 nuclei, causing further fissions . Each fission releases more than one neutron, so if at least one per fission causes another fission, the process is self-sustaining .✓ 2

Exam tip: In 'explain' answers about chain reactions, state the condition: at least one neutron from each fission must go on to cause another fission.

64IBP2S-040-02-12Reactor components Paper 2Easy6 marks

A submarine is powered by a pressurized-water reactor that uses enriched uranium fuel, water as moderator and coolant, boron control rods and thick steel and concrete shielding.

(a)

State the role of the moderator.

(1)
(b)

Outline how the control rods allow the reactor to be shut down quickly in an emergency.

(2)
(c)

State the role of the coolant and give one reason why water is suitable.

(2)
(d)

Outline why the reactor is surrounded by thick shielding.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
To slow fast neutrons to thermal energies, so that they are more likely to cause fission of U-235 .✓ 1
Part (b)
The rods are made of a neutron-absorbing material such as boron . When rapidly and fully inserted into the core, they absorb so many neutrons that, on average, fewer than one neutron per fission causes another fission, so the chain reaction stops .✓ 2
Part (c)
The coolant transfers thermal energy from the core to the heat exchanger or steam generator . Any one valid reason why water is suitable: it has a high specific heat capacity, OR it is a liquid that is easily pumped, OR it also acts as the moderator, OR it is cheap and readily available .✓ 2
Part (d)
The shielding absorbs neutrons and gamma radiation that escape the core, protecting the crew .✓ 1

Common error: Saying control rods 'slow down' the reaction scores nothing. The marks come from the idea that the rods absorb neutrons.

65IBP2S-040-00-11Energy released in fission Paper 2Easy6 marks

The reactor of a nuclear-powered submarine produces thermal energy at a steady rate of 150 MW. Each fission of a uranium-235 nucleus releases 200 MeV.
Molar mass of uranium-235 = 0.235 kg mol−1.

(a)

Show that the energy released in one fission is about 3×10−11 J.

(1)
(b)

Calculate the number of fissions occurring per second in the reactor.

(2)
(c)

Determine the mass of uranium-235 that undergoes fission in one day.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 200×106 × 1.60×10−19 = 3.2×10−11 J✓ 1
Part (b)
Uses rate = P / E = 150×106 / 3.2×10−11✓ 1
Rate = 4.7×1018 s−1✓ 1
Part (c)
Allow error carried forward from the number of fissions per second.—
Fissions per day = 4.7×1018 × 86 400 ≈ 4.1×1023 (accept 4.05×1023 or 4.06×1023)✓ 1
Amount = 4.05×1023 / 6.02×1023 ≈ 0.67 mol✓ 1
Mass = 0.67 × 0.235 = 0.16 kg✓ 1
Alternative method: mass = number of fissions × 235 × 1.66×10−27 kg = 0.16 kg, for full credit.—

Exam tip: Convert MeV to joules before you divide it into a power in watts. Also convert the time to seconds: one day is 86 400 s. The Avogadro constant is in the data booklet.

66IBP2S-040-01-11Chain reactions Paper 2Easy6 marks

A small fission reactor is proposed to supply power to a crewed base on the Moon. Its compact core of uranium-235 is surrounded by a beryllium oxide reflector. A single boron carbide control rod runs through the centre of the core. The reactor delivers a thermal power of 40 kW. Each fission of U-235 releases 3.2×10−11 J.
Take 1 u = 1.661×10−27 kg and 1 year = 3.16×107 s.

core(U-235)beryllium oxidereflectorboron carbide control rod
Cross-section (schematic)
(a)

State the role of the boron carbide control rod.

(1)
(b)

Outline the role of the beryllium oxide reflector.

(2)
(c)

Calculate the number of fissions per second in the core.

(1)
(d)

Determine the mass of uranium-235 that undergoes fission in one year of continuous operation.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
It absorbs neutrons, controlling the rate of fission (the chain reaction) .✓ 1
Part (b)
It scatters neutrons that would escape from the core back into the fuel .✓ 1
This reduces neutron loss, so a chain reaction can be sustained with a smaller mass of fuel (smaller critical mass), which keeps the launch mass low .✓ 1
Part (c)
Rate = 4.0×104 / 3.2×10−11 = 1.25×1015 s−1 .✓ 1
Part (d)
Number of fissions = 1.25×1015 × 3.16×107 = 3.95×1022 .✓ 1
Mass = 3.95×1022 × 235 × 1.661×10−27 = 1.54×10−2 kg ≈ 1.5×10−2 kg (about 15 g) .✓ 1
ECF from (c).—

Exam tip: To find the mass of one nucleus, multiply the nucleon number by the unified atomic mass unit. This approximation is accurate enough for a fuel-consumption estimate.

67IBP2S-040-01-12Reactor components Paper 2Easy7 marks

A submarine is powered by a pressurised water reactor. The fuel is enriched uranium. Water in the primary circuit is pumped through the core, where it acts as both moderator and coolant. The core produces a thermal power of 150 MW. The primary water rises in temperature by 30 K as it passes through the core. At the operating temperature and pressure, the specific heat capacity of the water is 5.4×103 J kg−1 K−1.

(a)

State the role of the control rods in this reactor.

(1)
(b)

Explain how the water in the primary circuit acts as both the moderator and the coolant.

(2)
(c)

Outline why the fuel must be enriched when ordinary water is the moderator.

(2)
(d)

Calculate the mass of water that must flow through the core each second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
They absorb neutrons to control the fission rate (keep the chain reaction steady, or shut it down) .✓ 1
Part (b)
Moderator: neutrons collide with hydrogen nuclei of similar mass and lose kinetic energy, reaching thermal speeds at which U-235 is much more likely to undergo fission .✓ 1
Coolant: the water flowing through the core absorbs thermal energy from the fuel and carries it to the heat exchanger .✓ 1
Part (c)
Hydrogen nuclei in ordinary water absorb a significant fraction of the neutrons .✓ 1
A higher proportion of U-235 means enough of the remaining neutrons still cause fission, so the chain reaction can be sustained .✓ 1
Part (d)
P = (m/t) c ΔT, so m/t = 1.5×108 / (5.4×103 × 30) .✓ 1
m/t = 9.3×102 kg s−1 .✓ 1

Common error: In P = (m/t)cΔT, ΔT is the temperature rise of the coolant, not its operating temperature.

68E-2-11Fission products & waste Paper 2Medium12 marks

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)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
It is highly radioactive «has a high activity», emitting beta and gamma radiation✓ 1Accept: 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✓ 1OWTTE
Part (b)
Energy needed = mcΔT = 1.5 × 106 × 4200 × 70 = 4.4 × 1011 J✓ 1
t = 4.41 × 1011/2.0 × 106 = 2.2 × 105 s «≈ 61 hours, about 2.6 days»✓ 1Award [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 %✓ 1Must 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»✓ 1OWTTE
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✓ 1Award 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)

69E-2-25Energy released in fission Paper 2Medium12 marks

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.

NuclideBinding energy per nucleon / MeV
uranium-2357.59
barium-1448.27
krypton-898.62
A schematic thermal nuclear reactor: fuel rods and control rods in a moderator inside a shielded core, with a coolant loop to a heat exchanger.control rods (white) lower into the corereactor corefuel rods (gold) in a moderatorheat exchangerhotcoolsteam to the turbinethick concrete shielding (grey) surrounds the core
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 pointMarkNotes
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»✓ 1Free neutrons have no binding energy.
energy released = 1958 − 1784 «= 174 MeV»✓ 1Must 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»✓ 1Do 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»✓ 1Award [2] for CNA.
Part (b)(ii)
3.0 × 109 = m × 1.5 × 106✓ 1Assumes that all the thermal power is transferred to the secondary circuit.
m = 2.0 × 103 kg «each second»✓ 1Award [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✓ 1
ΔT = 2.0 × 109/(1.2 × 105 × 4.2 × 103) = 4.0 K✓ 1ALTERNATIVE: flow rate needed = 2.0 × 109/(4.2 × 103 × 3.0) = 1.6 × 105 kg s−1.
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✓ 1Conclusion 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✓ 1OWTTE

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)

70E-2-27Spontaneous fission Paper 2Medium13 marks

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)

ParticleMass / u
californium-252 atom252.081627
xenon-140 atom139.921646
ruthenium-108 atom107.910190
neutron1.008665
(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 pointMarkNotes
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✓ 1OWTTE
Part (a)(ii)
x = 252 − 140 − 108 = 4✓ 1Proton 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✓ 1Neutrons have no electrons.
Part (b)(ii)
mass of products = 139.921646 + 107.910190 + 4 × 1.008665 = 251.866496 u✓ 1Allow ECF from (a)(ii).
Δm = 252.081627 − 251.866496 = 0.215131 u✓ 1
E = 0.215131 × 931.5 = 200 MeV✓ 1Award [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✓ 1Award [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✓ 1Do 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✓ 1Accept any one.

Answers: (a)(ii) x = 4  ·  (b)(ii) 200 MeV  ·  (c) 7.8 years (the remaining parts are explanations — see the table above)

71E-2-35Energy released in fission Paper 2Medium12 marks

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 pointMarkNotes
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»✓ 1Must see Ek = p2/2m (or equivalent) AND the mass ratio.
Part (b)(iii)
EKr = 170 × 141/(141 + 92)✓ 1ALTERNATIVE: EKr + EKr/1.53 = 170.
= 103 MeV✓ 1Award [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✓ 1Allow 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»✓ 1OWTTE

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)

72E-2-46Reactor components Paper 2Medium13 marks

In a graphite-moderated reactor, neutrons released by fission with kinetic energies of about 2 MeV are slowed down by elastic collisions with carbon-12 nuclei until their kinetic energy is about 0.025 eV. Take the mass of a carbon-12 nucleus to be 12 times the mass m of a neutron. The carbon nuclei may be treated as stationary.

For a head-on elastic collision of a neutron with a stationary nucleus of nucleon number A, the fraction of its kinetic energy that the neutron keeps is f = ((A − 1)/(A + 1))2.

Beforeneutron (m)carbon-12 nucleus (12m)2.0 × 10⁷ m s⁻¹at restAfterneutroncarbon-12 nucleus1.69 × 10⁷ m s⁻¹v = ?
Head-on collision along a line; not to scale
(a)

Outline why the neutrons released by fission must be slowed down in a thermal reactor.

(1)
(b)

A neutron moving at 2.0 × 107 m s−1 collides head-on with a stationary carbon-12 nucleus and rebounds along its original line with a speed of 1.69 × 107 m s−1.

(i)

Calculate the speed of the carbon nucleus after the collision.

(2)
(ii)

Show that the collision is elastic.

(1)
(c)

Show that a neutron keeps about 72 % of its kinetic energy in each head-on collision with a carbon-12 nucleus.

(1)
(d)

Determine whether 50 head-on collisions with carbon-12 nuclei are enough to slow a neutron from 2.0 MeV to 0.025 eV.

(3)
(e)

Most collisions in the moderator are not head-on. Suggest how this affects the number of collisions needed.

(1)
(f)

For hydrogen-1, A = 1, but hydrogen-1 nuclei sometimes absorb neutrons; carbon-12 nuclei absorb very few. Reactors moderated by ordinary water need uranium fuel enriched in uranium-235, whereas some graphite-moderated reactors can use natural uranium. Discuss these observations.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
uranium-235 nuclei are much more likely to undergo fission when they absorb slow «thermal» neutrons than fast ones✓ 1OWTTE
Part (b)(i)
momentum conserved: m × 2.0 × 107 = −m × 1.69 × 107 + 12m × V✓ 1The rebound velocity must be negative.
V = 3.69 × 107/12 = 3.08 × 106 m s−1✓ 1Award [2] for CNA. Award [1 max] for 2.6 × 105 m s−1 «rebound treated as positive».
Part (b)(ii)
before: ½m(2.0 × 107)2 = 2.00 × 1014m; after: ½m(1.69 × 107)2 + ½(12m)(3.08 × 106)2 = 2.00 × 1014m, so the kinetic energy is unchanged✓ 1Must see both totals OR a comparison to 3 s.f. Allow ECF from (b)(i).
Part (c)
f = (11/13)2 = 0.716✓ 1Must see 11/13 OR (1.69/2.0)2 = 0.714. Answer to 3 s.f.
Part (d)
after 50 collisions the kinetic energy = 2.0 × 106 × 0.71650 eV✓ 1
= 0.11 eV✓ 1Accept 0.097–0.15 eV «from 0.714 or 0.72».
this is greater than 0.025 eV, so 50 collisions are not enough «55 are needed»✓ 1The conclusion must be consistent with the calculation.
Part (e)
in a glancing collision the neutron keeps a larger fraction of its kinetic energy, so more collisions «on average about twice as many» are needed✓ 1OWTTE
Part (f)
a neutron can lose all its kinetic energy in one head-on collision with hydrogen-1 «f = 0», so water slows neutrons in far fewer collisions / a much shorter distance than graphite✓ 1
but neutrons absorbed by hydrogen nuclei are lost to the chain reaction✓ 1
to keep, on average, one neutron per fission causing a further fission, a larger fraction of uranium-235 «enrichment» is needed in water-moderated reactors✓ 1
graphite absorbs so few neutrons that enough survive the many collisions to sustain the chain reaction even with the 0.7 % uranium-235 of natural uranium «at the cost of a larger core»✓ 1OWTTE. Award marks for any four distinct correct points.

Answers: (b)(i) 3.08 × 106 m s−1  ·  (c) 0.716  ·  (d) 0.11 eV — not enough (the remaining parts are explanations — see the table above)

73E-2-47Fission products & waste Paper 2Medium13 marks

High-level waste from reprocessed fuel is melted into glass, sealed in a steel canister and placed 500 m underground in a hole in stable rock. The canister is surrounded by a layer of bentonite clay 0.35 m thick, which swells when wet and keeps groundwater away from the canister. The clay must stay below 100 °C.

The thermal power of the canister comes from the decay of fission products, mainly strontium-90 and caesium-137; take the half-life of both to be 30 years. Forty years after the fuel left the reactor the thermal power of the canister is 1.2 kW. Model the clay as a flat layer of area 16 m2 and thermal conductivity 1.0 W m−1 K−1, whose outer surface is kept at 50 °C by the surrounding rock.

Axes for thermal power of one canister, 0 to 1.4 kW, against time since the fuel left the reactor, 0 to 160 years; one point is plotted at 40 years and 1.2 kW.020406080100120140160time since the fuel left the reactor / years00.20.40.60.811.21.4thermal power / kW
Graph axes drawn to scale
(a)

Strontium-90, 9038Sr, is a β− emitter. Complete the equation for its decay: 9038Sr → … Y + … + …

(2)
(b)

The canister is placed in the repository 40 years after the fuel left the reactor.

(i)

Show that the temperature difference across the clay is about 26 K.

(1)
(ii)

Hence state the temperature of the clay next to the canister.

(1)
(c)

Sketch, on the axes, the variation of the thermal power of the canister from 40 years to 160 years after the fuel left the reactor.

(2)
(d)

Determine whether the canister could have been placed in the repository only 10 years after the fuel left the reactor.

(3)
(e)

Suggest one disadvantage of waiting longer than 40 years before placing the canister in the repository.

(1)
(f)

Explain why it is important to keep groundwater away from the canister for thousands of years, although the waste is a solid.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
9039Y✓ 1Both numbers needed.
0−1e AND ν̄✓ 1Do not accept a neutrino.
Part (b)(i)
ΔQ/Δt = kAΔT/Δx, so ΔT = 1200 × 0.35/(1.0 × 16) «= 26.3 K»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
50 + 26 = 76 °C✓ 1Allow ECF from (b)(i).
Part (c)
curve through «70 years, 0.6 kW», «100 years, 0.3 kW», «130 years, 0.15 kW» and «160 years, 0.075 kW»✓ 1Within half a small square.
smooth curve starting at the plotted point with decreasing gradient, not reaching zero✓ 1Do not accept straight-line segments.
Part (d)
10 years after leaving the reactor is one half-life earlier, so the power was at least 2 × 1.2 = 2.4 kW «more, because shorter-lived products were also present»✓ 1
ΔT = 2400 × 0.35/(1.0 × 16) = 52.5 K✓ 1Allow ECF from (b)(i): 2 × 26 K.
clay temperature ≥ 50 + 52.5 = 102.5 °C > 100 °C, so the canister could not have been placed after only 10 years✓ 1The conclusion must be consistent with the working.
Part (e)
the waste must be stored and actively cooled «and guarded» at the surface for longer, at extra cost and risk✓ 1Accept any sensible disadvantage.
Part (f)
water would corrode the steel and slowly dissolve the glass, releasing the radioactive fission products✓ 1
groundwater could carry them to the surface, into drinking water or the food chain✓ 1
once inside the body there is no shielding, so even alpha and beta emitters irradiate tissue; long-lived products remain hazardous for thousands of years✓ 1OWTTE

Answers: (b)(i) 26 K  ·  (b)(ii) 76 °C  ·  (d) ≥ 102.5 °C — not allowed (the remaining parts are explanations — see the table above)

74E-2-76Fission products & waste Paper 2Medium14 marks

Molybdenum-99 is used in hospitals as the source of technetium-99m for medical imaging. It is made by neutron-induced fission: small targets containing uranium-235 are placed in a research reactor for several days, and the molybdenum is then separated chemically from the other fission products. One possible fission is

23592U + 10n → 9942Mo + AZSn + 3 10n

About 6.1 % of all fissions of uranium-235 lead to a nucleus of molybdenum-99.

(a)
(i)

Deduce A and Z.

(1)
(ii)

Each target is removed from the reactor after about one week. The half-life of molybdenum-99 is 66 hours. Suggest why leaving a target in the reactor for several months would produce very little extra molybdenum-99.

(2)
(b)

During irradiation, 3.0 × 1015 fissions occur each second in one target. Each fission releases 200 MeV.

(i)

Show that the thermal power generated in the target is about 1 × 105 W.

(1)
(ii)

The target is cooled by water that enters at 35 °C. The water must leave at no more than 60 °C. Determine the minimum mass of water that must flow past the target each second. (specific heat capacity of water = 4180 J kg−1 K−1)

(2)
(iii)

Calculate the number of molybdenum-99 nuclei produced each second in the target.

(1)
(c)

The molybdenum reaches a hospital 5.5 days after it is separated. The hospital needs an activity of 2.0 × 1011 Bq of molybdenum-99 on delivery. Determine the activity of the molybdenum-99 when it is separated.

(2)
(d)

Molybdenum-99 decays by β− emission to technetium-99m, which emits gamma photons. The molybdenum-99 is transported to the hospital in a container with thick lead walls. Explain why thick lead is needed for the gamma radiation but much less shielding is needed for the beta particles.

(2)
(e)

Suggest why hospitals cannot keep a stock of molybdenum-99 for several months.

(1)
(f)

After the molybdenum has been removed, the rest of the target is radioactive waste. Outline two properties of this waste that affect how it must be managed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Z = 92 − 42 = 50 AND A = 236 − 99 − 3 = 134✓ 1Both needed.
Part (a)(ii)
Molybdenum-99 decays while it is being produced, and the number of nuclei decaying each second is proportional to the number present «activity ∝ N»✓ 1OWTTE
After a few half-lives «one week ≈ 2.5 half-lives» the rate of decay is almost equal to the rate of production, so the number of molybdenum-99 nuclei hardly increases further «it levels off»✓ 1OWTTE. Accept a sketch of N against time levelling off, with the reason.
Part (b)(i)
3.0 × 1015 × 200 × 1.60 × 10−13 «= 9.6 × 104 W»✓ 1Must see full substitution OR answer to 2 s.f. with power of ten.
Part (b)(ii)
P = (m/t)cΔT with ΔT = 25 K✓ 1
m/t = 9.6 × 104/(4180 × 25) = 0.92 kg «each second»✓ 1Accept 0.9–1.0 kg. Award [2] for CNA. Allow ECF from (b)(i).
Part (b)(iii)
0.061 × 3.0 × 1015 = 1.8 × 1014 «s−1»✓ 1
Part (c)
5.5 days = 132 h = 2 half-lives✓ 1
Activity = 22 × 2.0 × 1011 = 8.0 × 1011 Bq✓ 1Award [2] for CNA. Award [1 max] for 1.0 × 1011 Bq «halved».
Part (d)
Beta particles are charged and strongly ionizing, so they lose their energy quickly and are stopped by a few mm of metal «the inner container»✓ 1
Gamma photons are uncharged and weakly ionizing, so they penetrate far; a large thickness of a dense material such as lead is needed to reduce their intensity to a safe level✓ 1OWTTE. Accept 'gamma intensity is only reduced, never stopped completely'.
Part (e)
Its half-life is short «66 h»: after a few weeks almost all of it has decayed «after 10 half-lives, about 28 days, the activity is about 0.1 % of its starting value»✓ 1OWTTE
Part (f)
High activity / emits penetrating gamma radiation, so it must be shielded and handled remotely✓ 1Any two distinct properties, each linked to management, [1] each.
Releases thermal energy from its decays, so it must be cooled / contains long-lived products «such as caesium-137 and strontium-90» and uranium, so it must be isolated for a very long time «vitrified and stored deep underground»✓ 1

Answers: (a)(i) A = 134, Z = 50  ·  (b)(i) 9.6 × 104 W  ·  (b)(ii) 0.92 kg s−1  ·  (b)(iii) 1.8 × 1014 s−1  ·  (c) 8.0 × 1011 Bq (the remaining parts are explanations — see the table above)

75E-2-81Energy released in fission Paper 2Medium12 marks

A nuclear-powered icebreaker has two pressurized-water reactors, each with a thermal power of 171 MW. Steam turbines drive generators that supply electric motors, which turn the propellers. The total power delivered to the propellers is 54 MW.

A typical fission in the fuel is

23592U + 10n → 14455Cs + 9037Rb + x 10n

The table gives some atomic masses. 1 u = 931.5 MeV c−2; 1 MeV = 1.60 × 10−13 J.

ParticleAtomic mass / u
uranium-235235.043928
caesium-144143.932075
rubidium-9089.914798
neutron1.008665
Blank square grid, 26 by 20 small squares, for a Sankey diagram; the side of one small square represents 20 MWside of 1 small square = 20 MW
Grid for (b)(ii)
(a)
(i)

Deduce x.

(1)
(ii)

Determine the energy released in this fission, in MeV.

(3)
(b)
(i)

Calculate the overall efficiency with which the thermal power of the two reactors is transferred to the propellers.

(1)
(ii)

Draw, on the grid, a Sankey diagram for the energy transfers in the two reactors and the propulsion system together.

(2)
(c)

The icebreaker moves through thin ice at a constant speed of 6.0 m s−1. Assume that all the power delivered to the propellers is used to overcome the resistive forces. Calculate the total resistive force.

(1)
(d)

Using your answer to (a)(ii), calculate the number of fissions each second in one reactor.

(2)
(e)

The fuel is uranium with a much higher fraction of uranium-235 than the fuel of a land-based power station, so that the core can be small. Explain why a higher fraction of uranium-235 allows a chain reaction to be sustained in a smaller core.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
x = 236 − 144 − 90 = 2✓ 1
Part (a)(ii)
Mass before = 236.052593 u; mass after = 235.864203 u✓ 1The electron masses cancel «92 electrons on each side».
Δm = 0.188390 u✓ 1Allow ECF from (a)(i).
E = 0.188390 × 931.5 = 175 MeV✓ 1Accept 174–176 MeV. Award [3] for CNA.
Part (b)(i)
54/342 = 0.16 «16 %»✓ 1
Part (b)(ii)
Single input arrow «342 MW thermal» dividing into two labelled outputs: useful energy to the propellers and energy transferred to the surroundings «sea water, air»✓ 1
Widths in proportion: input ≈ 17.1 squares, useful ≈ 2.7 squares, wasted ≈ 14.4 squares «288 MW»✓ 1Accept ± half a square for each width. Allow ECF from (b)(i).
Part (c)
F = P/v = 54 × 106/6.0 = 9.0 × 106 N✓ 1
Part (d)
Energy per fission = 175 × 1.60 × 10−13 = 2.80 × 10−11 J✓ 1Allow ECF from (a)(ii).
Rate = 1.71 × 108/2.80 × 10−11 = 6.1 × 1018 s−1✓ 1Award [2] for CNA. Award [1 max] for 1.2 × 1019 s−1 «both reactors».
Part (e)
Fewer neutrons are absorbed by uranium-238 without causing fission, so each neutron is more likely to cause fission of uranium-235 before it leaves the core✓ 1
A small core loses a larger fraction of its neutrons through the surface, but with more uranium-235 an average of one neutron per fission can still cause a further fission «k = 1»✓ 1OWTTE

Answers: (a)(i) x = 2  ·  (a)(ii) 175 MeV  ·  (b)(i) 0.16  ·  (c) 9.0 × 106 N  ·  (d) 6.1 × 1018 s−1 (the remaining parts are explanations — see the table above)

76IBP2S-040-02-13Fission equations Paper 2Medium6 marks

A reactor that produces medical isotopes uses the fission reaction
23592U + 10n → 13755Cs + 9637Rb + 3 10n
The table gives binding energy per nucleon data.

Nuclide | Binding energy per nucleon / MeV
uranium-235 | 7.59
caesium-137 | 8.39
rubidium-96 | 8.61

(a)

State why no binding energy is listed for the neutron.

(1)
(b)

Calculate the energy released in one fission.

(3)
(c)

The reactor operates at a thermal power of 10 MW. Calculate the number of fissions per second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A free neutron is a single nucleon, so it is not bound to any other nucleon and its binding energy is zero .✓ 1
Part (b)
Binding energy of products = 137 × 8.39 + 96 × 8.61 = 1976.0 MeV . Binding energy of U-235 = 235 × 7.59 = 1783.7 MeV . Energy released = 1976.0 − 1783.7 = 192.3 MeV ≈ 192 MeV .✓ 3
Part (c)
Allow error carried forward from the previous part. Energy per fission = 192.3 × 106 × 1.60 × 10−19 ≈ 3.08 × 10−11 J (accept 3.07 × 10−11 J from 192 MeV) . Rate = 1.0 × 107 ÷ 3.08 × 10−11 ≈ 3.2 × 1017 s−1 (accept any answer from 3.2 × 1017 to 3.3 × 1017 s−1) .✓ 2

Exam tip: Multiply each binding energy per nucleon by its nucleon number before subtracting. Subtracting the per-nucleon values alone gives a meaningless result. Keep unrounded values until the final step.

77IBP2S-040-02-14Chain reactions Paper 2Medium8 marks

A compact uranium-235 fission reactor is planned to power a deep-space probe. The reactor must supply 1.0 kW of electrical power for 10 years. The conversion of thermal energy to electrical energy has an efficiency of 25%. Each fission releases 200 MeV.

(a)

Calculate the thermal power of the reactor.

(1)
(b)

Using your thermal power, calculate the rate of fission in the reactor.

(2)
(c)

Using your fission rate, determine the mass of uranium-235 that undergoes fission during the 10-year mission. The molar mass of uranium-235 is 235 g mol−1.

(3)
(d)

The reactor actually contains many kilograms of uranium. Explain why the fuel mass is much greater than the mass you calculated for the 10-year mission.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
P = 1.0 kW ÷ 0.25 = 4.0 kW .✓ 1
Part (b)
Error carried forward from the thermal power is allowed. 200 MeV = 200 × 106 × 1.60 × 10−19 J = 3.2 × 10−11 J . Rate = 4.0 × 103 ÷ 3.2 × 10−11 = 1.25 × 1014 s−1 .✓ 2
Part (c)
Error carried forward from the fission rate is allowed. Time = 10 × 365 × 24 × 3600 s ≈ 3.15 × 108 s, so the number of fissions = 1.25 × 1014 × 3.15 × 108 ≈ 3.9 × 1022 . Amount = 3.9 × 1022 ÷ 6.02 × 1023 ≈ 0.065 mol . Mass = 0.065 mol × 235 g mol−1 ≈ 15 g, which is 15.4 g before rounding .✓ 3
Part (d)
The fuel must exceed the critical mass, so that enough neutrons cause further fissions rather than escaping, and the chain reaction is sustained. Accept any equivalent reference to limiting neutron leakage . Only a small fraction of the fuel can fission before the chain reaction can no longer be sustained. Accept as the reason: fission products building up and absorbing neutrons; neutron absorption by uranium-238 or control rods; or the fuel not being pure uranium-235 .✓ 2

Common error: Use the thermal power, not the electrical output, to find the fission rate. The energy released by fission is thermal.

78IBP2S-040-00-12Energy released in fission Paper 2Medium7 marks

An engineering team modelling a reactor core uses the following fission reaction:
\(^{1}_{0}\text{n} + ^{235}_{92}\text{U} \rightarrow ^{144}_{56}\text{Ba} + ^{89}_{36}\text{Kr} + x\,{}^{1}_{0}\text{n}\)
The table gives the masses used.

ParticleMass / u
neutron1.008665
uranium-235235.043930
barium-144143.922953
krypton-8988.917630
1 u = 931.5 MeV c−2

(a)

Determine the value of x.

(1)
(b)

Calculate the mass defect for this reaction in u.

(2)
(c)

Calculate the energy released in one fission, in MeV.

(2)
(d)

In a working reactor, the total thermal energy obtained per fission is closer to 200 MeV. Explain this difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
1 + 235 = 144 + 89 + x, so x = 3✓ 1
Part (b)
Allow ECF from (a).—
Mass before = 235.043930 + 1.008665 = 236.052595 u and mass after = 143.922953 + 88.917630 + 3 × 1.008665 = 235.866578 u✓ 1
Δm = 0.1860 u✓ 1
Part (c)
Allow ECF from (b).—
E = 0.1860 × 931.5✓ 1
E = 173 MeV✓ 1
Part (d)
Identifies an extra energy source, for example: the fission fragments are neutron-rich and undergo β− decays with gamma emission, OR the emitted neutrons are captured by nuclei in the core, which then emit gamma rays✓ 1
This extra energy is absorbed in the reactor as thermal energy, adding to the 173 MeV released in the fission itself✓ 1

Common error: Count every emitted neutron in the product mass, and count the absorbed neutron in the reactant mass. Leaving either out changes the mass defect by about 1 u, which is far larger than the true answer.

79IBP2S-040-00-13Fission products & waste Paper 2Medium7 marks

A university research reactor operates at a thermal power of 20 MW. It is shut down by fully inserting its control rods. Immediately after shutdown, the fuel still produces thermal energy at 6.5% of the operating power. Cooling water enters the core at 30 °C and leaves at 45 °C.
Specific heat capacity of water = 4200 J kg−1 K−1.

(a)

Outline why the fuel continues to produce thermal energy after the chain reaction has stopped.

(2)
(b)

Calculate the minimum mass of cooling water that must flow through the core per second immediately after shutdown.

(3)
(c)

One day after shutdown, the thermal power has fallen to about 0.5% of the operating power. It then decreases much more slowly. Explain this pattern.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The fuel contains a large quantity of radioactive fission products✓ 1
Their decay releases energy (β and γ radiation) that is absorbed in the fuel and heats it✓ 1
Part (b)
Decay power = 0.065 × 20 MW = 1.3 MW✓ 1
Uses mass flow rate = P / (cΔT) = 1.3×106 / (4200 × 15)✓ 1
Mass flow rate = 21 kg s−1✓ 1
Part (c)
Many fission products have short half-lives (seconds to hours) and decay away quickly, so the power falls rapidly at first✓ 1
The remaining nuclides have long half-lives, so their activity is lower and decreases only slowly✓ 1

Exam tip: Decay heat is the reason reactors need cooling even after shutdown. Explain it in terms of the half-lives of the fission products.

80IBP2S-040-01-14Reactor components Paper 2Medium6 marks

A design team compares four possible moderator materials for a new research reactor. The table below gives data for each material.
Material | Mass number of moderating nucleus | Average number of collisions to slow a neutron from 2 MeV to 0.025 eV | Neutron absorption per nucleus (relative to carbon = 1)
Ordinary water (H) | 1 | 18 | 95
Heavy water (D) | 2 | 25 | 0.15
Beryllium | 9 | 86 | 2.2
Graphite (C) | 12 | 114 | 1

(a)

Outline why fewer collisions are needed when the moderating nucleus has a smaller mass number.

(2)
(b)

Suggest, using the data, which material is the most suitable moderator for a reactor fuelled by natural (unenriched) uranium.

(2)
(c)

Assume that every collision in graphite reduces the neutron's kinetic energy by the same fraction. Determine the fraction of kinetic energy that a neutron keeps in each collision.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
In an elastic collision, a neutron transfers the largest share of its kinetic energy to a target of similar mass .✓ 1
So each collision with a light nucleus removes a larger fraction of the neutron's energy, and fewer collisions are needed .✓ 1
Part (b)
Heavy water .✓ 1
It has by far the lowest neutron absorption, so enough neutrons survive to cause fission of the scarce U-235, and it needs only a few collisions .✓ 1
Part (c)
f114 = 0.025 / (2×106) = 1.25×10−8 .✓ 1
f = (1.25×10−8)1/114 = 0.85 .✓ 1

Exam tip: When a quantity is multiplied by the same factor many times, write it as fⁿ = final/initial and use logarithms or a fractional power to find f.

81E-2-09Chain reactions Paper 2Hard14 marks

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)
Show mark scheme
Marking pointMarkNotes
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✓ 1Must 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»✓ 1OWTTE
Part (a)(iii)
Then the ore contained about 4–5 % uranium-235, more than the fuel of a modern water-moderated reactor✓ 1Accept 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✓ 1OWTTE
Part (b)(i)
Number of nuclei = 6.0 × 103/(235 × 1.661 × 10−27) = 1.54 × 1028✓ 1
E = 1.54 × 1028 × 200 × 1.60 × 10−13 = 4.9 × 1017 J✓ 1Accept 4.9–5.0 × 1017 J. Award [2] for CNA.
Part (b)(ii)
P = 4.92 × 1017/(1.5 × 105 × 3.16 × 107) = 1.0 × 105 W✓ 1Accept 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✓ 1Allow 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✓ 1Accept 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✓ 1OWTTE
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✓ 1OWTTE

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)

82E-2-26Chain reactions Paper 2Hard11 marks

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.

A neutron causes the fission of a uranium-235 nucleus, producing two fission fragments and three neutrons. Further right are two uranium-235 nuclei, a uranium-238 nucleus, part of a control rod and, at the far right, a dashed line marking the edge of the core.nU-235fission fragmentfission fragmentnnnU-235U-235U-238control rodedge ofcore
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 pointMarkNotes
Part (a)
exactly one of the three neutrons drawn going on to cause the fission of a uranium-235 nucleus✓ 1Do 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»✓ 1Must 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✓ 1Accept 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»✓ 1OWTTE
Part (c)(i)
Energy released in one generation = 60 × 106 × 0.10 = 6.0 × 106 J✓ 1
N = 6.0 × 106/(200 × 1.60 × 10−13) = 1.88 × 1017 «≈ 1.9 × 1017»✓ 1Award [2] for CNA.
Part (c)(ii)
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✓ 1Allow 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)

83E-2-79Chain reactions Paper 2Hard12 marks

In a pressurized-water reactor, ordinary water is both the moderator and the coolant. The water is kept at a high pressure so that it does not boil. When the temperature of the water rises, its density decreases.

The multiplication factor k is the average number of neutrons from each fission that go on to cause a further fission.

Axes of thermal power P against time t, with the steady power P0 drawn before t = 00time t0P₀2P₀thermal power P
Axes for (c)(i)
(a)

Explain why a decrease in the density of the water reduces k.

(2)
(b)

At a steady thermal power of 3.0 GW the water enters the core at 291 °C and leaves it at 325 °C. The power then rises to 3.3 GW, with the same flow rate and inlet temperature. Assume that the specific heat capacity of the water does not change. Show that the water now leaves the core at about 328 °C.

(1)
(c)

The reactor is operating at a steady power P0. At t = 0 the control rods are withdrawn a small distance and are then held in their new position. The flow rate of the water and its temperature when it enters the core do not change.

(i)

Sketch, on the axes, the variation of the thermal power P with time t.

(2)
(ii)

Explain the shape of your sketch.

(2)
(d)

Later, the demand for electrical energy rises. More steam is drawn from the heat exchanger, so the water returns to the core at a lower temperature. The control rods are not moved. Explain how the thermal power of the reactor responds.

(2)
(e)

In a different type of reactor, graphite is the moderator and the water that cools the fuel is allowed to boil. In this design the water absorbs a significant number of neutrons, but is not needed to slow them down. Suggest why this design can become unstable if its power rises.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Fewer hydrogen nuclei per unit volume, so the neutrons make fewer collisions and are slowed less effectively «travel further while still fast»✓ 1
More neutrons are absorbed by uranium-238 or escape from the core before they are slow enough to cause fission of uranium-235, so fewer go on to cause fission✓ 1OWTTE
Part (b)
ΔT ∝ P for the same flow rate «P = (m/t)cΔT»: ΔT = 34 × 3.3/3.0 = 37.4 K, so the outlet temperature = 291 + 37.4 «= 328.4 °C»✓ 1Must see the ratio 3.3/3.0 «or 37.4 K» AND the addition to 291 °C, OR answer to 4 s.f.
Part (c)(i)
Curve starting at P0 at t = 0 and rising✓ 1Do not credit a vertical jump at t = 0.
Gradient decreasing so that P levels off at a new constant value greater than P0✓ 1Do not credit a curve that keeps rising. A small overshoot before levelling off is acceptable.
Part (c)(ii)
Withdrawing the rods means fewer neutrons are absorbed, so k > 1 and the power rises✓ 1
A greater power gives a greater temperature rise of the water in the core «ΔT ∝ P, as in (b)», so the water is hotter and less dense and k falls; the power stops rising when k has returned to 1 «at a higher power»✓ 1OWTTE
Part (d)
Colder water is denser, so the neutrons are slowed more effectively and k becomes greater than 1✓ 1
The power increases until the water is hot enough for k to return to 1: the reactor power follows the demand without the control rods being moved✓ 1OWTTE
Part (e)
A rise in power makes more water boil, so there is less liquid water «more steam bubbles» in the core✓ 1
Fewer neutrons are absorbed by the water, while the graphite still slows them down, so k increases above 1✓ 1
The power then rises further, which boils even more water: positive feedback, so the power can grow rapidly unless the control rods are inserted quickly✓ 1OWTTE

Answers: (b) 328.4 °C (the remaining parts are explanations — see the table above)

84E-2-80Chain reactions Paper 2Hard13 marks

In a fast breeder reactor the core contains plutonium-239 and is surrounded by a blanket of uranium-238. The chain reaction is maintained by fast neutrons, which must not be slowed down, so there is no moderator. The coolant is liquid sodium (nucleon number 23).

Some neutrons are captured by uranium-238 nuclei in the blanket. Each uranium-239 nucleus formed decays by two successive β− decays to plutonium-239.

Grid of neutron number N from 144 to 148 against proton number Z from 91 to 95, with uranium-238 marked at N = 146, Z = 92144145146147148neutron number N9192939495proton number ZU-238
Grid for (a)(i)
(a)
(i)

On the grid, draw arrows to show the capture of a neutron by uranium-238 and the two β− decays that follow. Label the position of plutonium-239.

(2)
(ii)

Write the equation for the first of the two β− decays. Neptunium has proton number 93.

(1)
(b)

Explain why water cannot be used as the coolant in this reactor, but liquid sodium can.

(2)
(c)

The thermal power of the reactor is 1.20 GW. The sodium enters the core at 395 °C and leaves it at 545 °C. The specific heat capacity of liquid sodium is 1.27 × 103 J kg−1 K−1. Calculate the mass of sodium that flows through the core each second.

(2)
(d)

Some sodium-23 nuclei in the core absorb neutrons and form sodium-24, which emits beta particles and gamma rays. In this reactor, the sodium from the core gives its energy in a heat exchanger to a second, separate sodium circuit, which then heats water to make steam in another heat exchanger. Suggest two reasons for this arrangement.

(2)
(e)

Each fission of a plutonium-239 nucleus releases on average 2.9 neutrons. For each fission, on average 0.5 neutrons escape from the reactor or are absorbed without producing plutonium; all the other neutrons that do not cause fission are captured by uranium-238 nuclei. Assume that every fission is of a plutonium-239 nucleus. Determine whether the reactor produces more plutonium-239 than it uses when it operates at constant power.

(3)
(f)

Suggest one advantage of this type of reactor for the long-term use of the world's uranium.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Arrow from U-238 one square in the direction of increasing N «to N = 147, Z = 92»✓ 1
Two arrows each one square of increasing Z and decreasing N, ending at N = 145, Z = 94 labelled Pu-239✓ 1Allow ECF for the β− arrows from a wrong starting square.
Part (a)(ii)
23992U → 23993Np + 0−1e + ν̄✓ 1Antineutrino required; accept 00ν̄. Do not accept a neutrino.
Part (b)
Hydrogen nuclei in water have about the same mass as a neutron, so a neutron loses a large fraction of its kinetic energy in each collision: water would moderate the neutrons✓ 1
Sodium nuclei are about 23 times as massive as a neutron, so a neutron keeps most of its kinetic energy in a collision and stays fast✓ 1OWTTE
Part (c)
m/t = P/(cΔT) = 1.20 × 109/(1.27 × 103 × 150)✓ 1
= 6.3 × 103 kg «each second»✓ 1Award [2] for CNA.
Part (d)
The core sodium becomes radioactive «sodium-24, gamma emitter», so it is kept inside the shielded reactor building, away from the turbine and steam plant✓ 1Any two, [1] each.
Sodium reacts violently with water: if a leak occurs in the steam generator only non-radioactive sodium is involved / the steam and turbine are not contaminated✓ 1
Part (e)
At constant power exactly one neutron per fission must cause a further fission «k = 1»✓ 1
Neutrons captured by U-238 per fission = 2.9 − 1 − 0.5 = 1.4, each producing one Pu-239 nucleus✓ 1
1.4 Pu-239 nuclei are produced for each Pu-239 nucleus used, 1.4 > 1, so the reactor produces more plutonium-239 than it uses✓ 1Conclusion must be consistent with the working.
Part (f)
It turns uranium-238, about 99 % of natural uranium «including depleted uranium already stored», into fuel, so far more energy is obtained from each tonne of uranium mined✓ 1Accept: the plutonium is used as fuel instead of being stored as long-lived waste.

Answers: (c) 6.3 × 103 kg s−1  ·  (e) 1.4 Pu-239 produced per Pu-239 used: yes (the remaining parts are explanations — see the table above)

85IBP2S-040-02-15Chain reactions Paper 2Hard9 marks

A small modular reactor supplies a remote mining site. In this reactor each fission releases on average 2.5 neutrons. Of these neutrons:
• 8% escape from the core.
• Of those that do not escape, a fraction f is absorbed by the control rods.
• Of the neutrons that remain, 70% cause a further fission. The rest are absorbed without causing fission.

(a)

Calculate the neutron multiplication factor k (the average number of fissions caused by the neutrons from one fission) when the control rods are fully withdrawn, so that f = 0.

(2)
(b)

Determine the value of f needed for the reactor to operate at constant power.

(2)
(c)

A small fraction of fission neutrons are delayed. They are emitted by fission products some seconds after the fission. Outline why these delayed neutrons make the reactor controllable.

(2)
(d)

During start-up, k is set to 1.005. The effective time between generations is 0.10 s. Calculate the time taken for the fission rate to double.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Neutrons not escaping = 2.5 × 0.92 = 2.30 . k = 2.30 × 0.70 = 1.61 .✓ 2
Part (b)
ECF from (a). Constant power requires k = 1, so 1.61(1 − f) = 1 . f = 1 − 1/1.61 ≈ 0.38 .✓ 2
Part (c)
The delayed neutrons greatly increase the average time between one generation of fissions and the next . As a result, the power changes slowly enough for the control rods to be moved in response .✓ 2
Part (d)
After n generations, rate ∝ kn, so 1.005n = 2 . n = ln 2 ÷ ln 1.005 ≈ 139 generations . t = 139 × 0.10 ≈ 14 s .✓ 3

Exam tip: Growth in a chain reaction is geometric: after n generations the rate is multiplied by kⁿ, not by n × k.

86IBP2S-040-02-16Energy released in fission Paper 2Hard8 marks

A nuclear-powered aircraft carrier uses uranium-235 fuel. One fission reaction in its reactors is
23592U + 10n → 13250Sn + 101ZMo + x 10n
Atomic masses:
uranium-235: 235.043930 u
tin-132: 131.917822 u
molybdenum-101: 100.910347 u
neutron: 1.008665 u
1 u = 931.5 MeV c−2

(a)

Determine Z and x.

(2)
(b)

Calculate, in MeV, the energy released in one fission.

(3)
(c)

Determine the energy released when 1.0 kg of uranium-235 undergoes fission completely.

(2)
(d)

Jet fuel releases about 43 MJ kg−1 when burned. Calculate the ratio of the energy per kilogram from uranium-235 fission to that from jet fuel.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Z = 92 − 50 = 42 . x = 236 − 132 − 101 = 3 .✓ 2
Part (b)
ECF from (a). Mass before = 235.043930 + 1.008665 = 236.052595 u . Mass after = 131.917822 + 100.910347 + 3 × 1.008665 = 235.854164 u, so Δm = 0.198431 u . E = 0.198431 × 931.5 ≈ 185 MeV .✓ 3
Part (c)
ECF from (b). Energy per fission = 185 × 1.60 × 10−13 = 2.96 × 10−11 J . Number of nuclei = 6.02 × 1023 ÷ 0.235 = 2.56 × 1024, so E = 7.6 × 1013 J .✓ 2
Part (d)
ECF from (c). Ratio = 7.6 × 1013 ÷ 4.3 × 107 ≈ 1.8 × 106 .✓ 1

Exam tip: Atomic masses can be used directly here. Both sides contain 92 electrons (50 + 42 = 92), so the electron masses cancel. Keep all six decimal places until the subtraction.

87IBP2S-040-00-16Energy released in fission Paper 2Hard8 marks

Detectors placed beside a thin uranium foil record fragments from the reaction
\(^{1}_{0}\text{n} + ^{235}_{92}\text{U} \rightarrow ^{94}_{38}\text{Sr} + ^{140}_{54}\text{Xe} + 2\,{}^{1}_{0}\text{n}\)
Binding energies per nucleon:

NucleusBinding energy per nucleon / MeV
uranium-236 (compound nucleus)7.59
strontium-948.59
xenon-1408.29
1 u = 1.66×10−27 kg

(a)

Determine the energy released in this fission, using the uranium-236 compound nucleus as the starting point.

(3)
(b)

The two fragments share a total kinetic energy of 165 MeV. Assume the compound nucleus is at rest and the momenta of the neutrons are negligible. Show that the ratio of the kinetic energy of the Sr-94 fragment to that of the Xe-140 fragment is 140 : 94.

(2)
(c)

Calculate the kinetic energy of the Sr-94 fragment.

(1)
(d)

Estimate the speed of the Sr-94 fragment.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Binding energy of U-236 = 236 × 7.59 = 1791 MeV✓ 1
Binding energy of products = 94 × 8.59 + 140 × 8.29 = 1968 MeV (free neutrons have zero binding energy)✓ 1
Energy released = 1968 − 1791 = 177 MeV✓ 1
Part (b)
By conservation of momentum, the fragments have momenta of equal magnitude p in opposite directions✓ 1
Ek = p2/2m, so Ek ∝ 1/m, giving ESr/EXe = mXe/mSr = 140/94✓ 1
Part (c)
ESr = 165 × 140/234 = 99 MeV✓ 1
Part (d)
ECF from (c).—
Ek = 98.7 × 1.60×10−13 = 1.58×10−11 J and m = 94 × 1.66×10−27 = 1.56×10−25 kg✓ 1
v = √(2Ek/m) = 1.4×107 m s−1✓ 1

Exam tip: The lighter fragment gains the larger share of the kinetic energy. When momenta are equal in size, kinetic energy is inversely proportional to mass.

88IBP2S-040-01-15Energy released in fission Paper 2Hard9 marks

A city is considering running its electric bus fleet on electricity from a fission power station. One possible fission of uranium-235 is:
23592U + 10n → 14054Xe + 9438Sr + 2 10n
Atomic masses: U-235 = 235.043930 u; Xe-140 = 139.921646 u; Sr-94 = 93.915361 u; neutron = 1.008665 u.
Take 1 u = 931.5 MeV c−2 = 1.661×10−27 kg and 1 MeV = 1.602×10−13 J.

(a)

Outline, with reference to binding energy per nucleon, why this reaction releases energy.

(2)
(b)

Show that the energy released in this fission is about 185 MeV.

(3)
(c)

Determine the energy released when 1.0 kg of U-235 undergoes fission in this way.

(2)
(d)

A bus uses 1.2 kW h of electrical energy per km and travels 60 000 km per year. The power station converts thermal energy to electrical energy with an efficiency of 33%. Determine the mass of U-235 that must undergo fission to power one bus for one year.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Xe-140 and Sr-94 have a greater binding energy per nucleon than U-235, because they are closer to the peak of the curve .✓ 1
So the total binding energy increases, and the difference is released as kinetic energy of the products .✓ 1
Part (b)
Mass before = 235.043930 + 1.008665 = 236.052595 u; mass after = 139.921646 + 93.915361 + 2(1.008665) = 235.854337 u .✓ 1
Δm = 0.198258 u .✓ 1
E = 0.198258 × 931.5 = 184.7 MeV ≈ 185 MeV .✓ 1
Part (c)
Number of nuclei = 1.0 / (235.04 × 1.661×10−27) = 2.56×1024 .✓ 1
E = 2.56×1024 × 184.7 × 1.602×10−13 = 7.6×1013 J .✓ 1
ECF from (b).—
Part (d)
Electrical energy = 1.2 × 60 000 × 3.6×106 = 2.59×1011 J, so thermal energy = 2.59×1011 / 0.33 = 7.9×1011 J .✓ 1
Mass = 7.9×1011 / 7.6×1013 = 1.0×10−2 kg (about 10 g) .✓ 1
ECF from (c).—

Common error: Always divide by the efficiency to find the thermal energy needed. Multiplying by 0.33 gives a mass of fuel that is far too small.

89IBP2S-040-01-16Spontaneous fission Paper 2Hard7 marks

A power reactor is restarted after refuelling with fresh uranium fuel. A californium-252 neutron source is placed near the core during start-up. Later, to raise the output, the control rods are withdrawn slightly. This makes the multiplication factor k = 1.005, where k is the average number of neutrons from each fission that go on to cause another fission. The effective time between successive generations of fission, including the effect of delayed neutrons, is 0.10 s.

(a)

State the value of k when the reactor operates at a steady power, and the name given to this condition.

(1)
(b)

The U-238 in the fresh fuel undergoes spontaneous fission. Explain why a separate neutron source is still placed near the core during start-up.

(2)
(c)

Show that about 140 generations are needed for the fission rate to double.

(2)
(d)

Determine the time taken for the reactor power to double.

(1)
(e)

Outline how the operators then hold the reactor at its new, higher power.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
k = 1 and the reactor is described as critical; both are needed for the mark .✓ 1
Part (b)
Spontaneous fission of U-238 is extremely rare, so the neutron population from the fuel alone is very small and unreliable .✓ 1
The Cf-252 source supplies a steady, measurable neutron flux, so detectors can monitor the approach to criticality and the chain reaction builds up in a controlled way. Accept equivalent wording .✓ 1
Part (c)
Uses 1.005n = 2 .✓ 1
n = ln 2 / ln 1.005 = 139, which is about 140 .✓ 1
Part (d)
t = 139 × 0.10 ≈ 14 s. Accept 13.9 s, or 14 s from using 140 generations .✓ 1
Part (e)
Insert the control rods slightly so that more neutrons are absorbed, until k returns to 1 .✓ 1

Exam tip: If each generation multiplies the neutron number by k, then after n generations the factor is kⁿ. Take logarithms to find n.