44 free, high-quality past-paper-style questions on Fission for IB Physics HL, each with a full mark scheme. Covers: Fission reactors; Moderation and shielding; Spontaneous fission; and more. For when you have finished the past papers, or want to practise one topic at a time.E.4 at Standard 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.208 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1E-1A-48Fission reactorsPaper 1AEasy1 mark
Three statements about a thermal nuclear fission reactor are:
I. Pushing the control rods further into the core reduces the average number of neutrons per fission that go on to cause another fission. II. The moderator slows the neutrons down, which makes them more likely to cause fission of uranium-235 nuclei. III. The thick shielding around the core is needed mainly to absorb the alpha particles emitted by the fuel.
Which statements are correct?
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Step 1I: the control rods absorb neutrons, so fewer are available to cause fission — correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: slow (thermal) neutrons are much more likely to be captured by uranium-235 and cause fission — correct.
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Step 3III: alpha particles are stopped by the fuel cladding itself; the shielding is needed for the penetrating neutrons and gamma rays — incorrect.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: control rods absorb neutrons; the moderator makes fission more likely.
BIII is wrong: alpha particles cannot even escape the fuel rods; the shielding stops neutrons and gamma radiation.
CI is correct (inserting the rods absorbs more neutrons) and III is wrong (the shielding is for neutrons and gamma rays).
DIII is wrong: alpha particles have a range of only a few centimetres in air and are stopped by the fuel cladding.
2IBP2H-041-00-01Moderation and shieldingPaper 1AEasy1 mark
A compact fission reactor supplies electricity to a remote polar research station. Its core contains enriched uranium fuel rods surrounded by graphite. What is the main role of the graphite?
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Graphite is a moderator. Fast neutrons released in fission lose kinetic energy in repeated collisions with carbon nuclei and become thermal neutrons. Thermal neutrons are much more likely to be absorbed by uranium-235 and induce fission. Absorbing neutrons is the job of the control rods, and transferring energy is the job of the coolant. Neutrons are uncharged, so they feel no electrostatic repulsion.
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Answer B
Answer: B
Common error: Do not mix up the moderator with the control rods. The moderator slows neutrons down; the control rods remove neutrons.
3IBP2H-041-00-04Spontaneous fissionPaper 1AEasy1 mark
Californium-252 sources are used to start up reactors and in some cancer therapies because the isotope undergoes spontaneous fission. Which statement about spontaneous fission is correct?
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Spontaneous fission is a random nuclear process. A very heavy nucleus splits into two fragments, usually releasing neutrons, without absorbing anything first. Absorbing a neutron is induced fission. Nuclear decay rates do not depend on temperature. Emitting an alpha particle is alpha decay, not fission.
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Answer B
Answer: B
Common error: Spontaneous fission is not alpha decay. Fission gives two fragments of comparable mass, whereas alpha decay emits a single helium-4 nucleus.
4IBP3H-E4-20r19-01Fission reactorsPaper 1AEasy1 mark
In the core of a nuclear reactor, the control rods are lowered further into the core. What is the effect?
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Control rods contain a material that absorbs neutrons. Lowering them removes more neutrons from the chain reaction, so fewer fissions occur per generation and the power falls (A). B reverses the effect. C describes a moderator, not a control rod. D confuses the rods with the coolant, which is the part that carries thermal energy away.
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Answer A
Answer: A
Common error: Keep the roles apart: the moderator slows neutrons, the control rods absorb them and the coolant removes thermal energy.
5IBP3H-E4-20r19-02Binding energy and fissionPaper 1AEasy1 mark
Uranium-235 undergoes fission and releases energy. Which statement gives the correct reason?
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The fragments are more tightly bound per nucleon than a nucleus near A = 235, so the total binding energy rises and the total rest mass falls; the difference appears as kinetic energy (D). A is wrong because a lower binding energy per nucleon would mean energy must be supplied, and nuclei do not release energy by 'breaking bonds'. B has the mass change the wrong way round. C is wrong because the total binding energy does change.
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Answer D
Answer: D
Common error: Energy is released when a more tightly bound system forms, not when bonds are broken.
6IBP3H-E4-20r19-03Fission reactorsPaper 1AEasy1 mark
Which neutrons are most likely to be captured by a uranium-235 nucleus and cause it to undergo fission in a thermal reactor?
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The capture probability of U-235 for neutrons rises sharply as the neutron speed falls, so slow (thermal) neutrons are the most effective (C). A is wrong because a neutron is uncharged and there is no repulsion to overcome. B is wrong because the probability depends strongly on energy. D invents an intermediate optimum that does not exist for U-235.
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Answer C
Answer: C
Exam tip: Fission neutrons are born fast, which is why a moderator is needed before they can be used efficiently.
7IBP3H-E4-20r19-07A natural fission reactorPaper 1AEasy1 mark
A natural fission reactor operated in a uranium deposit about 2 billion years ago, but such a reactor could not start spontaneously in the same ore today. Which statement is the best explanation?
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Uranium-235 (half-life about 0.70 billion years) decays faster than uranium-238 (about 4.5 billion years), so its fraction of natural uranium has fallen over time; a few per cent then made a chain reaction possible with water as moderator (A). B has the wrong direction: U-235 is not formed by U-238 decay. C gives a smaller fraction, which would make a chain reaction harder. D denies the change in composition, which is the key point.
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Answer A
Answer: A
Exam tip: A shorter half-life means that a nuclide's share of the mixture falls faster as time passes.
Uranium-235 has a neutron-to-proton ratio of 1.55. Its fission produces two nuclei with nucleon numbers roughly between 90 and 145. Stable nuclides in this range have neutron-to-proton ratios between about 1.2 and 1.45.
Which row gives the reason why the fission products are radioactive and the decay they usually undergo?
ReasonUsual decay
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Step 1The fragments keep roughly the neutron-to-proton ratio of the parent, about 1.55.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2This is higher than the 1.2–1.45 of stable nuclides of the same mass: the fragments are neutron-rich.
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Step 3β− decay turns a neutron into a proton, lowering N/Z; successive β− decays move the products towards stability.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: neutron-rich products undergo β− decay (often followed by gamma emission).
Bβ+ decay turns a proton into a neutron and would raise the ratio further, away from stability.
CInverts the comparison: 1.55 is above, not below, the stable range.
DAlpha decay is typical of very heavy nuclei (A > about 200); medium-mass fragments are unstable because of their neutron excess, not their size.
In a simple model of a sphere of fissile uranium, neutrons are produced by fission at a rate proportional to the volume of the sphere, and neutrons escape through its surface at a rate proportional to its surface area. The sphere is replaced by a sphere of the same material with twice the radius.
By what factor does (rate at which neutrons escape)/(rate at which neutrons are produced) change?
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Step 1Doubling the radius multiplies the surface area (∝ r²) by 4 and the volume (∝ r³) by 8.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The ratio (rate of escape)/(rate of production) ∝ r²/r³ = 1/r, so it changes by a factor 4/8 = 1/2.
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Step 3A larger piece of fuel therefore loses a smaller fraction of its neutrons through the surface: a chain reaction can be sustained only if the mass of fuel is large enough.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis allows for the eightfold increase in the production rate but forgets that the escape rate also rises, by a factor of 4.
BThis squares the correct factor, taking the ratio to be proportional to 1/r² instead of 1/r.
CCorrect: the escape rate rises by 4 and the production rate by 8, so the ratio halves.
DThis inverts the ratio, giving the factor by which (rate of production)/(rate of escape) changes.
13E-1A-69Fission products and their managementPaper 1AMedium1 mark
A sample of waste from a nuclear reactor contains two radioactive nuclides: X, of half-life 30 years, and Y, of half-life 2.4 × 104 years. At a certain time the sample contains equal numbers of nuclei of X and of Y.
Which row identifies the nuclide with the greater activity at that time and the nuclide with the greater activity 600 years later?
Greater activity at that timeGreater activity 600 years later
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Step 1A = λN with λ = ln 2/T½: for equal N, AX/AY = (2.4 × 104)/30 = 800, so X is more active at first.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2600 years is 20 half-lives of X, so NX falls by 220 ≈ 106, while NY hardly changes (600 years is only 0.025 of its half-life).
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Step 3Then AX/AY ≈ 800/106 ≈ 8 × 10−4: Y is now the more active.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the short-lived nuclide dominates the activity at first, but the long-lived nuclide dominates in the long term — which is why long-term storage of waste is needed.
BThis assumes the shorter half-life always gives the greater activity, forgetting that the number of nuclei of X falls by a factor of about 106 in 600 years.
CThis confuses half-life with the rate of decay, taking the longer half-life to mean faster decay: Y is then thought to be more active at first and, having used up its nuclei, to be overtaken by X later.
DThis treats activity as proportional to half-life, so that the long-lived nuclide is always the more active.
14IBP2H-041-00-02Binding energy and fissionPaper 1AMedium1 mark
A hypothetical nucleus with mass number 240 has a binding energy per nucleon of 7.6 MeV. It splits into two identical fragments, each of mass number 120 and binding energy per nucleon 8.5 MeV. Ignore any neutrons released. What is the energy released?
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Energy released = total binding energy of products − total binding energy of parent = 240 × 8.5 − 240 × 7.6 = 240 × 0.9 = 216 MeV. The value 0.9 MeV is only the gain per nucleon. The value 2040 MeV is the total binding energy of the products. The value 108 MeV uses the nucleons of only one fragment.
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Answer B
Answer: B
Exam tip: Multiply the binding energy per nucleon by the number of nucleons for every nucleus before you subtract.
15IBP2H-041-00-03A natural fission reactorPaper 1AMedium1 mark
About two billion years ago, natural chain reactions took place in uranium deposits at Oklo in Gabon, with groundwater acting as the moderator. Why can a similar natural reactor not operate in uranium ore today?
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Uranium-235 decays faster than uranium-238 because its half-life is shorter. The proportion of U-235 in natural uranium has therefore fallen, from a few percent then to about 0.7% now. That is too low to sustain a chain reaction with a water moderator. U-238 is not fissile with thermal neutrons, and it has a very long half-life. The water's role was moderation, not cooling. Stray neutrons are always present to start a chain reaction.
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Answer A
Answer: A
Exam tip: For Oklo, compare the two half-lives. The isotope with the shorter half-life becomes rarer as time passes.
16IBP2H-041-00-06Fission fragmentsPaper 1AMedium1 mark
Which sequence best describes the main energy transfers in a nuclear power station, from fission to the electricity grid?
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Most of the energy from fission appears as kinetic energy of the fragments. The fragments collide with surrounding atoms, raising the internal energy of the fuel. This energy passes to the coolant, which produces steam. The steam turns the turbine, and the turbine drives the generator. Neutrons carry only a few percent of the energy. No chemical energy is involved, and electricity is not generated directly in the fuel.
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Answer B
Answer: B
Common error: Most fission energy is carried by the fragments, not by the neutrons.
17IBP2H-041-00-08Fission productsPaper 1AMedium1 mark
The fission products found in spent reactor fuel are usually β− emitters. Which statement best explains this?
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Fission fragments keep roughly the high N/Z ratio of uranium, about 1.5. Stable nuclei of medium mass have a lower N/Z ratio. The fragments are therefore neutron-rich. β− decay converts a neutron into a proton, which lowers N/Z. A low N/Z ratio would lead to β+ decay instead. A higher binding energy per nucleon means greater stability, not less. β decay does not change the mass number.
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Answer A
Answer: A
Common error: β decay leaves the mass number unchanged. It changes Z by one and N by one in the opposite direction.
18IBP3H-E4-20r19-04Energy from fissionPaper 1AMedium1 mark
A reactor has a thermal power output of 90 MW. Assume that each fission of uranium-235 releases 200 MeV and that all of this becomes thermal energy. The molar mass of uranium-235 is 235 g mol−1, 1 eV = 1.60×10−19 J and NA = 6.02×1023 mol−1. What mass of uranium-235 undergoes fission in one day?
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Energy per fission = 200×106×1.60×10−19 = 3.2×10−11 J. Fissions per second = 90×106 / 3.2×10−11 = 2.81×1018, so per day 2.43×1023 fissions = 0.404 mol, which is 0.404×235 = 94.9 g (D). A is a factor of 1000 too large (g treated as kg). B arises from using only half the fissions. C is the number of moles, not the mass.
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Answer D
Answer: D
Exam tip: Write the energy per event in joules first, then work through events per second, per day, moles and mass in turn.
19IBP3H-E4-20r19-05Chain reactionsPaper 1AMedium1 mark
In a reactor, the neutron multiplication factor is k = 1.02: the number of neutrons in each generation is 1.02 times the number in the previous generation. By what factor has the number of neutrons in a generation increased after 100 generations?
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The number is multiplied by 1.02 once per generation, so after 100 generations the factor is 1.02100 = 7.24 (D). A applies the factor only once. B treats the growth as linear, 1 + 0.02×100. C multiplies k by the number of generations.
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Answer D
Answer: D
Common error: Repeated multiplication by k gives a power, not a product: the growth is exponential.
20IBP3H-E4-20r19-06Fission productsPaper 1AMedium1 mark
A fission fragment 14054Xe has too many neutrons to be stable and decays by two successive β− emissions. What is the nuclide that results?
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In each β− decay a neutron becomes a proton: Z rises by 1 and A is unchanged. Two decays give Z = 56 and A = 140, which is 140Ba (C). A has Z falling, as for β+ decay. B and D have lost two nucleons, which would need other decay modes.
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Answer C
Answer: C
Common error: In β⁻ decay the nucleon number stays the same and the proton number goes up by one.
21IBP2H-041-00-05Chain reactionsPaper 1AHard1 mark
In a research reactor, each fission of uranium-235 releases on average 2.4 neutrons. The reactor operates at constant power. What fraction of the released neutrons must be lost, either by escaping or by being absorbed without causing fission?
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At constant power, exactly one neutron from each fission must go on to cause another fission. The fraction causing fission is therefore 1/2.4 = 0.42. The fraction lost is 1 − 0.42 = 0.58, or about 58%. The value 42% is the fraction that causes fission, not the fraction lost. With 0% lost the reaction would grow rapidly. With 100% lost the chain reaction would stop.
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Answer C
Answer: C
Exam tip: Steady power means the multiplication factor is exactly 1. It does not mean that no fissions occur.
22IBP2H-041-00-07Fission fragmentsPaper 1AHard1 mark
A uranium-236 nucleus at rest splits into two fragments of mass numbers 140 and 94, together with two neutrons. Ignore the momentum of the neutrons. What is the ratio (kinetic energy of the 94 fragment) / (kinetic energy of the 140 fragment)?
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The total momentum is zero, so the two fragments have momenta of equal magnitude p. Kinetic energy is Ek = p2/2m, so Ek is inversely proportional to m. The ratio is 140/94 = 1.49, so the lighter fragment has more kinetic energy. The value 0.67 inverts the ratio. The value 1.00 assumes equal kinetic energies. The value 2.22 wrongly squares the mass ratio.
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Answer C
Answer: C
Exam tip: When a body at rest splits into two, use equal and opposite momenta together with Ek = p²/2m.
23IBP3H-E4-20r19-08Moderation and shieldingPaper 1AHard1 mark
In a model of neutron moderation, a neutron of mass m and speed u makes a head-on elastic collision with a stationary deuterium nucleus of mass 2m. What fraction of the neutron's initial kinetic energy does it keep after the collision?
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Momentum: mu = mv1 + 2mv2. Elastic collision: relative speed is unchanged, so v2 − v1 = u. Substituting v2 = v1 + u gives 3v1 = −u, so v1 = −u/3, so the kinetic energy fraction is (1/3)2 = 1/9 (C). A forgets to square the speed ratio. B is the fraction transferred to the deuteron. D comes from using the speed of the deuteron, 2u/3.
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Answer C
Answer: C
Common error: Kinetic energy depends on speed squared, so a speed ratio must be squared before it is quoted as an energy fraction.
In a pool-type research reactor, the thermal energy released in the core is removed by water pumped through it. With the reactor held at constant power, the operators vary the mass flow rate ṁ of the cooling water and record the rise ΔT in water temperature between the inlet and the outlet of the core. Energy losses from the core other than to the cooling water are negligible, and each fission of a uranium-235 nucleus releases 200 MeV.
The graph shows ΔT against 1/ṁ with a line of best fit through the origin. (Specific heat capacity of water = 4180 J kg−1 K−1; 1 MeV = 1.60 × 10−13 J; 1 u = 1.66 × 10−27 kg)
ṁ / kg s−1
1/ṁ / kg−1 s
ΔT / K
20.0
0.0500
28.8
25.0
0.0400
22.9
30.0
0.0333
19.2
40.0
0.0250
14.3
50.0
0.0200
11.5
60.0
0.0167
9.5
ΔT against 1/ṁ (graph drawn to scale)
(a)
(i)
Determine the gradient of the line, giving its unit.
(2)
(b)
(i)
Show that the thermal power of the reactor is about 2.4 MW.
(1)
(c)
(i)
Determine the number of fissions per second in the core.
(2)
(d)
(i)
Estimate the mass of uranium-235 that undergoes fission in one day.
(1)
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Part (a)(i)
Gradient = 28.7/0.0500 = 574
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Accept 560–590.
Unit: K kg s−1
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Part (b)(i)
P = ṁcΔT, so ΔT = (P/c)(1/ṁ) and P = c × gradient = 4180 × 574 = 2.40 × 106 W
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Allow ECF from (a). At least 3 s.f. or full substitution required.
Part (c)(i)
Energy per fission = 200 × 1.60 × 10−13 = 3.20 × 10−11 J
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Rate = 2.40 × 106/3.20 × 10−11 = 7.5 × 1016 s−1
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Allow ECF from (b). Use of 2.4 MW gives 7.5 × 1016 s−1.
Answers: (a)(i) 574 K kg s−1 · (b)(i) 2.40 × 106 W · (c)(i) 7.5 × 1016 s−1 · (d)(i) 2.5 × 10−3 kg (the remaining parts are explanations — see the table above)
25IBP2H-041-00-09Energy from fissionPaper 2Easy6 marks
A small fission reactor is proposed to power a crewed base on Mars. It must supply an electrical power of 10 kW. The overall efficiency of conversion from thermal power to electrical power is 25%. Each fission of uranium-235 releases 200 MeV. Take e = 1.60×10−19 C, 1 u = 1.66×10−27 kg and 1 year = 3.15×107 s.
(a)
State what is meant by induced fission.
(1)
(b)
Calculate the thermal power that the reactor core must produce.
(1)
(c)
Calculate the number of fissions per second in the core.
(2)
(d)
Calculate the mass of uranium-235 that undergoes fission in one year.
(2)
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Part (a)
A heavy nucleus absorbs a neutron and then splits into two smaller nuclei, releasing further neutrons and energy .
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Part (b)
P = 10 kW / 0.25 = 40 kW .
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Part (c)
Energy per fission = 200×106 × 1.60×10−19 = 3.2×10−11 J .
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Rate = 4.0×104 / 3.2×10−11 = 1.25×1015 s−1 .
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Allow error carried forward from the thermal power calculated in the previous part.
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Part (d)
Number of fissions = 1.25×1015 × 3.15×107 = 3.94×1022 .
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Mass = 3.94×1022 × 235 × 1.66×10−27 = 0.0154 kg, i.e. about 15 g (accept 0.015–0.0154 kg) .
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Allow error carried forward from the fission rate calculated in the previous part.
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Exam tip: Convert MeV to joules before dividing a power by the energy per event. Here 1 MeV = 1.60×10⁻¹³ J.
26IBP3H-E4-20r19-09Energy from fissionPaper 2Easy7 marks
In one fission reaction a neutron is absorbed by uranium-235: 10n + 23592U → 14156Ba + 9236Kr + 3 10n. Atomic masses: 235U 235.043930 u; 141Ba 140.914411 u; 92Kr 91.926156 u; neutron 1.008665 u. 1 u = 931.5 MeV c−2 = 1.66×10−27 kg; 1 eV = 1.60×10−19 J.
(a)
Determine the mass difference between the reactants and the products, in u.
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(b)
Calculate the energy released in this reaction, in MeV, to 3 significant figures.
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(c)
A marathon runner transfers 1.1×107 J of energy during a race. Using 173 MeV per fission, estimate the mass of uranium-235 that would have to undergo fission to release the same energy. Take the mass of a uranium-235 atom as 235 u.
(3)
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Part (a)
M1 mass before = 235.043930 + 1.008665 and mass after = 140.914411 + 91.926156 + 3(1.008665)
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A1 mass difference = 0.186 u (0.186033 u)
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Part (b)
M1 energy = 0.186033 × 931.5 (or 0.186 × 931.5)
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A1 173 MeV
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Part (c)
M1 energy per fission = 173×106×1.60×10−19 = 2.77×10−11 J
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M1 number of fissions = 1.1×107 / 2.77×10−11 = 3.97×1017 and mass = number × 235 × 1.66×10−27 kg
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A1 1.55×10−7 kg (0.155 mg)
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Exam tip: Keep a note of units at each stage: u to MeV in part (b), MeV to joules, then number of nuclei to mass.
27E-2-03Energy flow in a nuclear power stationPaper 2Medium12 marks
A pressurized-water reactor delivers 1.10 GW of electrical power with an overall efficiency of 33 %. Each fission of uranium-235 releases on average 200 MeV, and the fuel contains 4.0 % uranium-235 by mass; the rest is uranium-238.
Water pumped through the core at high pressure enters at 290 °C and leaves at 325 °C. Its specific heat capacity at these temperatures is 5.5 kJ kg−1 K−1.
The water from the core is radioactive, so it is kept in a closed primary circuit
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In the heat exchanger it transfers thermal energy to water in a separate secondary circuit, which boils to make the steam that drives the turbines; no radioactive material reaches the turbines
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Part (c)(i)
Control rods absorb neutrons (without undergoing fission)
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Do not accept "slow the neutrons down": that is the role of the moderator.
They are inserted or withdrawn so that on average exactly one neutron from each fission causes a further fission, keeping the fission rate (and so the power) constant
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Answers: (a)(ii) 1.0 × 1020 s−1 · (a)(iii) 32 t · (b)(i) 1.7 × 104 kg s−1(the remaining parts are explanations — see the table above)
28E-2-25Spent fuel and its managementPaper 2Medium10 marks
After removal from a reactor, a spent fuel assembly is stored under water. The graph shows the thermal power P produced in the assembly by radioactive decay against the time t after its removal.
Thermal power of a spent fuel assembly against time (graph drawn to scale)
(a)
(i)
Explain why the assembly continues to produce thermal energy after the chain reaction has stopped.
(2)
(ii)
Use the graph to show that the decrease in P cannot be described by a single half-life.
(2)
(iii)
At t = 100 days the cooling system of a pool containing 1.5 × 105 kg of water and one such assembly fails. Determine the initial rate of rise of the water temperature in K per day. (Specific heat capacity of water = 4.2 × 103 J kg−1 K−1)
(2)
(b)
(i)
After several decades the thermal power is due mainly to caesium-137, of half-life 30.1 years. Determine the time for the activity of the caesium-137 to fall to 0.10 % of its initial value.
(2)
(ii)
Discuss one advantage and one disadvantage of placing such waste in a deep geological repository rather than storing it at the surface.
(2)
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Part (a)(i)
The fission products are radioactive (neutron-rich nuclei that decay mainly by β− and γ emission)
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The energy of the emitted radiation is absorbed within the assembly, heating it
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Part (a)(ii)
P falls from 40 kW to 20 kW in about 46 days
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Accept 40–52 days (or any other correctly read halving).
but from 20 kW to 10 kW in about 84 days: the halving time increases, so several nuclides with different half-lives contribute
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Accept 75–95 days. A conclusion is needed.
Part (a)(iii)
P(100 days) = 12.5 kW
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Accept 12–13 kW.
ΔT/Δt = P/(mc) = 1.25 × 104/(1.5 × 105 × 4.2 × 103) = 2.0 × 10−5 K s−1 = 1.7 K per day
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Accept 1.6–1.8 K per day; the answer must be in K per day for MP2.
Advantage: isolated in stable rock far from people and groundwater, secure against misuse, and needs no active cooling or monitoring by future generations
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Disadvantage: it must stay intact for hundreds of years (300 years for caesium-137, far longer for long-lived nuclides), which cannot be verified; groundwater could carry radioactive material; it is hard to retrieve
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Allow ECF from (b)(i).
Answers: (a)(iii) 1.7 K day−1 · (b)(i) 300 years (the remaining parts are explanations — see the table above)
Californium-252 (25298Cf) has a half-life of 2.65 years. Of its nuclei that decay, 96.9 % emit an alpha particle and 3.1 % undergo spontaneous fission, which releases on average 3.8 neutrons. Small californium-252 sources are used as neutron sources to start up nuclear reactors.
Atomic masses: 252Cf 252.081 627 u, 140Xe 139.921 645 u, 108Ru 107.910 186 u; neutron 1.008 665 u. (1 year = 3.16 × 107 s)
(a)
Spontaneous fission.
(i)
Distinguish between spontaneous fission and neutron-induced fission.
(1)
(ii)
One spontaneous fission of californium-252 produces xenon-140 (Z = 54), ruthenium-108 (Z = 44) and x neutrons. Deduce x.
(1)
(iii)
Calculate the energy released in this fission.
(2)
(b)
The source.
(i)
A start-up source contains 5.0 μg of californium-252. Show that its activity is about 1.0 × 108 Bq.
(2)
(ii)
Calculate the number of neutrons emitted per second by the source.
(1)
(c)
Starting the reactor.
(i)
Explain the role of the source in starting the chain reaction in fresh uranium fuel.
(2)
(ii)
The source is replaced when its neutron emission rate has fallen to 25 % of its initial value. State the time after which it is replaced.
(1)
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Part (a)(i)
Spontaneous fission happens randomly without any external cause, like other radioactive decays; neutron-induced fission occurs when a nucleus splits after absorbing a (slow) neutron
A chain reaction must be started by free neutrons inducing fission of uranium-235, but fresh fuel contains very few (spontaneous fission of uranium is extremely rare)
✓ 1
The source supplies a steady, known flux of neutrons; each can induce a fission whose neutrons cause further fissions, so the chain reaction builds up reliably and can be monitored as the control rods are withdrawn
✓ 1
Part (c)(ii)
25 % = two half-lives = 2 × 2.65 = 5.3 years
✓ 1
Answers: (a)(ii) 4 · (a)(iii) 200 MeV · (b)(i) 9.9 × 107 Bq · (b)(ii) 1.2 × 107 s−1 · (c)(ii) 5.3 years (the remaining parts are explanations — see the table above)
When a uranium-235 nucleus absorbs a slow neutron, the resulting uranium-236 nucleus splits into two fragments and a few neutrons. The graph shows the percentage of fissions, the fission yield, that produce a fragment of each nucleon number A.
(1 u = 1.661 × 10−27 kg, 1 MeV = 1.60 × 10−13 J)
Fission yield against nucleon number of the fragment for uranium-235 (graph drawn to scale)
(a)
(i)
State what the graph shows about the relative sizes of the two fragments in most fissions.
(1)
(ii)
The yields shown on the graph add up to 200 %. Suggest why.
(1)
(b)
In one fission, the fragments are strontium-94 (Z = 38) and xenon-140 (Z = 54): 10n + 23592U → 9438Sr + 14054Xe + x10n Binding energies per nucleon: uranium-235 7.591 MeV, strontium-94 8.594 MeV, xenon-140 8.291 MeV.
(i)
Deduce the value of x.
(1)
(ii)
Show that the energy released in this fission is about 180 MeV.
(2)
(iii)
Just after the nucleus splits, the two fragments are momentarily at rest with their centres 17 fm apart. Treating each fragment as a point charge, show that their electric potential energy is about 170 MeV.
(2)
(iv)
Hence state the origin of the kinetic energy of the fragments.
(1)
(v)
The fragments eventually share 168 MeV of kinetic energy. Suggest why this is less than the energy released in (b)(ii).
(2)
(c)
(i)
In a fuel rod, the fission fragments stop within about 10 μm of the point where fission occurs, whereas the neutrons and gamma rays travel much further. Explain why, and state the importance of this for the transfer of energy in a reactor.
(2)
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Part (a)(i)
The fragments are usually of unequal size, with nucleon numbers near 95 and 139; splitting into two equal fragments (A ≈ 117) is very rare
✓ 1
Both the unequal split and an approximate pair of values (accept 90–100 and 134–144) needed.
Part (a)(ii)
Each fission produces two fragments, so every fission is counted twice
✓ 1
Part (b)(i)
Nucleon number: 235 + 1 = 94 + 140 + x, so x = 2
✓ 1
Accept a check of proton numbers 38 + 54 = 92 as well.
Part (b)(ii)
Energy released = 94 × 8.594 + 140 × 8.291 − 235 × 7.591
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The free neutrons have no binding energy.
= 807.8 + 1160.7 − 1783.9 = 185 MeV
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An answer to at least 3 s.f. must be seen. Using 236 nucleons for uranium scores [1 max].
An answer to at least 3 s.f. (or 2.78 × 10−11 J with the conversion) must be seen.
Part (b)(iv)
The electric repulsion between the positively charged fragments: their electric potential energy is converted into kinetic energy as they fly apart
✓ 1
Allow ECF from (b)(iii). Do not accept "the strong force" or "binding energy" alone.
Part (b)(v)
The neutrons released carry away kinetic energy (about 2 MeV each)
✓ 1
The fragments are left in excited states and emit gamma-ray photons (and later decay by β− emission, with energy carried by antineutrinos)
✓ 1
Any two distinct ways for [2]. Allow ECF from (b)(ii).
Part (c)(i)
The fragments are heavy and highly charged, so they ionise very strongly and lose their kinetic energy in a very short distance (neutrons are uncharged; gamma rays ionise weakly)
✓ 1
Accept "large charge" with "strong ionisation".
So most of the energy released (the fragments' share, 168 of about 185 MeV) becomes internal energy of the fuel itself, which is then transferred to the coolant
✓ 1
Allow ECF from (b)(ii).
Answers: (b)(i) 2 · (b)(iii) 174 MeV (2.78 × 10−11 J) (the remaining parts are explanations — see the table above)
31IBP2H-041-00-10Binding energy and fissionPaper 2Medium8 marks
One possible fission reaction of uranium-235 is 23592U + 10n → 14156Ba + AZKr + 310n The table gives binding energies per nucleon. Nuclide | Binding energy per nucleon / MeV uranium-235 | 7.591 barium-141 | 8.326 krypton-A | 8.513
(a)
Determine A and Z for the krypton nucleus.
(2)
(b)
Explain why energy is released in this reaction even though the total number of nucleons is unchanged.
(2)
(c)
Calculate the energy released in one fission.
(2)
(d)
Each emitted neutron has a kinetic energy of about 2.0 MeV. Estimate the fraction of the released energy carried by the neutrons.
(2)
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Part (a)
Nucleon number: 235 + 1 = 141 + A + 3, so A = 92 .
✓ 1
Proton number: 92 = 56 + Z, so Z = 36 .
✓ 1
Part (b)
The fragments have a greater binding energy per nucleon than uranium-235, so the total binding energy of the products is greater .
✓ 1
The products therefore have less mass than the reactants, and the mass difference is released as energy (E = mc2) .
✓ 1
Part (c)
Binding energy of products = 141 × 8.326 + 92 × 8.513 = 1957 MeV, and of uranium-235 = 235 × 7.591 = 1784 MeV (free neutrons have zero binding energy) .
✓ 1
Energy released = 1957 − 1784 ≈ 173 MeV .
✓ 1
ECF for A from (a).
—
Part (d)
Energy carried by the neutrons = 3 × 2.0 = 6.0 MeV .
✓ 1
Fraction = 6.0 / 173 ≈ 0.035, i.e. about 3.5% .
✓ 1
ECF from (c).
—
Common error: Free neutrons have zero binding energy. Do not include them in the products' binding energy, and do not use the nucleon number 236 for the uranium.
32IBP2H-041-00-12Energy flow in a nuclear power stationPaper 2Medium9 marks
A coastal nuclear power station delivers an electrical power of 1.2 GW to the grid. Its overall efficiency is 34%. Each fission releases 200 MeV. All of the waste thermal energy is removed by sea water pumped through the condensers. Environmental rules limit the temperature rise of this water to 6.0 K. The specific heat capacity of sea water is 4180 J kg−1 K−1.
(a)
Outline the energy transfers that take place between a fission event in the core and the production of electrical energy.
(3)
(b)
Calculate the number of fissions per second in the reactor.
(2)
(c)
Describe the features of a Sankey diagram for the power station, stating the power represented by each arrow.
(2)
(d)
Calculate the minimum mass of sea water that must flow through the condensers each second.
(2)
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Part (a)
Energy released in fission appears mainly as kinetic energy of the fission fragments .
✓ 1
The fragments collide with atoms in the fuel, raising its internal energy. This energy is transferred to the coolant and used to produce steam .
✓ 1
The steam does work on the turbine, giving it kinetic energy, and the turbine drives the generator to produce electrical energy .
A single input arrow represents 3.5 GW from fission. It splits into a narrower useful output arrow of 1.2 GW electrical, with width proportional to 34% .
✓ 1
The other branch is a wider waste output arrow of 2.3 GW (66%), representing thermal energy transferred to the cooling water .
✓ 1
Allow error carried forward from the thermal power found in the fission-rate calculation.
—
Part (d)
Waste power = 3.53 − 1.2 = 2.33 GW .
✓ 1
Mass flow rate = P / (cΔT) = 2.33×109 / (4180 × 6.0) ≈ 9.3×104 kg s−1 .
✓ 1
Allow error carried forward from the thermal power found in the fission-rate calculation.
33IBP2H-041-00-14Energy flow in a nuclear power stationPaper 2Medium8 marks
A submarine reactor has a maximum thermal power of 150 MW. It is designed to run for 25 years without refuelling, at an average of 20% of its maximum power. The fuel is highly enriched uranium-235. Each fission releases 200 MeV. Take 1 year = 3.15×107 s and 1 u = 1.66×10−27 kg.
(a)
Outline how control rods allow the crew to reduce the reactor's power output.
(2)
(b)
Suggest why highly enriched fuel allows a small core that runs for many years.
(2)
(c)
Calculate the total thermal energy produced over the 25 years.
(2)
(d)
Estimate the mass of uranium-235 that undergoes fission in this time.
(2)
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Part (a)
Control rods are made of a material that absorbs neutrons .
✓ 1
Inserting them further removes more neutrons, so fewer are available to induce fission and the fission rate (power) decreases .
✓ 1
Part (b)
A greater proportion of the nuclei are fissile U-235, so more energy is available from a given mass or volume of fuel .
✓ 1
A chain reaction can be sustained in a smaller core, even though a greater fraction of neutrons escapes .
✓ 1
Part (c)
Average power = 0.20 × 150 = 30 MW .
✓ 1
E = 3.0×107 × 25 × 3.15×107 ≈ 2.4×1016 J .
✓ 1
Part (d)
Number of fissions = 2.36×1016 / 3.2×10−11 ≈ 7.4×1026 .
✓ 1
Mass = 7.4×1026 × 235 × 1.66×10−27 ≈ 290 kg .
✓ 1
ECF from (c).
—
Exam tip: Use the average power, not the maximum power, when finding the total energy over a long operating period.
34IBP2S-040-01-13Spontaneous fissionPaper 2Medium7 marks
In neutron brachytherapy, small capsules containing californium-252 are placed inside a tumour. Cf-252 has a half-life of 2.65 years. 3.1% of its decays are spontaneous fissions, and the rest are alpha decays. Each spontaneous fission releases on average 3.8 neutrons. A capsule contains 1.0 µg of Cf-252. Take 1 u = 1.661×10−27 kg and 1 year = 3.16×107 s.
(a)
Distinguish between spontaneous fission and induced fission.
(1)
(b)
Show that the capsule contains approximately 2.4×1015 Cf-252 nuclei.
(1)
(c)
Calculate the number of spontaneous fissions per second in the capsule.
(3)
(d)
Determine the rate at which the capsule emits neutrons.
(1)
(e)
Outline why the neutron output of the capsule is only about one quarter of its initial value after 5.3 years.
(1)
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Part (a)
Spontaneous fission happens with no external particle, whereas induced fission is triggered when the nucleus absorbs a neutron .
✓ 1
Part (b)
N = 1.0×10−9 / (252 × 1.661×10−27) = 2.39×1015 .
✓ 1
Part (c)
λ = ln2 / (2.65 × 3.16×107) = 8.28×10−9 s−1 .
✓ 1
A = λN = 8.28×10−9 × 2.39×1015 = 1.98×107 Bq .
✓ 1
Fission rate = 0.031 × 1.98×107 = 6.1×105 s−1 .
✓ 1
Part (d)
3.8 × 6.1×105 = 2.3×106 s−1 .
✓ 1
ECF from (c).
—
Part (e)
5.3 years is two half-lives, so the number of Cf-252 nuclei, and therefore the fission and neutron emission rates, has halved twice .
✓ 1
Common error: The activity counts every decay, whether alpha or fission. Multiply by the 3.1% fission branch before multiplying by the number of neutrons per fission.
35IBP3H-E4-20r19-11Fission productsPaper 2Medium9 marks
Fission of uranium-235 produces fragments such as 14055Cs and 9038Sr. Most fission fragments are radioactive.
(a)
140Cs decays by β− emission to barium. Write a nuclear equation for this decay.
(2)
(b)
Explain why fission fragments tend to be β− emitters.
(2)
(c)
The half-life of strontium-90 is 28.8 years. Calculate the time taken for the activity of a sample of strontium-90 to fall to 5.0% of its initial value. Give your answer to 3 significant figures and ignore any activity from decay products.
(3)
(d)
Outline why spent fuel rods remain hot for a long time after they are removed from the reactor.
(2)
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Part (a)
14055Cs → 14056Ba with correct Z and A for the barium
✓ 1
+ 0−1e (electron) + antineutrino
✓ 1
Part (b)
uranium-235 has a high neutron-to-proton ratio and the fragments inherit a similar ratio
✓ 1
stable nuclei of mass about 90 to 140 have a lower neutron-to-proton ratio, so the fragments have too many neutrons and a neutron changes to a proton in β− decay
36IBP3H-E4-20r19-13Energy from fissionPaper 2Medium8 marks
A coal-fired steam locomotive makes a long journey that requires 1.45×1012 J of energy from its fuel. Coal has a specific energy of 2.9×107 J kg−1. Assume that each fission of uranium-235 releases 200 MeV and that a uranium-235 atom has a mass of 235 u, where 1 u = 1.66×10−27 kg and 1 eV = 1.60×10−19 J.
(a)
Calculate the energy released per kilogram of uranium-235 that undergoes fission.
(3)
(b)
Determine the mass of coal and the mass of uranium-235 that would supply the journey's energy.
(3)
(c)
Suggest two reasons why the mass of fuel loaded into a real reactor must be much greater than the mass found in (b).
(2)
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Part (a)
M1 energy per fission = 200×106×1.60×10−19 = 3.2×10−11 J
✓ 1
M1 mass of atom = 235 × 1.66×10−27 = 3.90×10−25 kg
✓ 1
A1 8.20×1013 J kg−1
✓ 1
Part (b)
coal: 1.45×1012 / 2.9×107 = 5.0×104 kg
✓ 1
M1 uranium mass = 1.45×1012 / 8.20×1013
✓ 1
A1 0.0177 kg (17.7 g)
✓ 1
Part (c)
reactor fuel contains only a few per cent uranium-235, the rest being mainly uranium-238
✓ 1
not all uranium-235 is fissioned before the fuel is removed / the energy conversion is less than 100% efficient / a reserve is needed for control (any one further valid reason)
✓ 1
Exam tip: Compare the two fuels in a single sentence at the end, for example as a ratio of masses.
37IBP3H-E4-20r19-14Moderation and shieldingPaper 2Medium9 marks
A neutron of kinetic energy 2.0 MeV is slowed in a graphite moderator. Model each collision as a head-on elastic collision between the neutron (mass m) and a stationary carbon-12 nucleus (mass 12m).
(a)
Show that after one collision the neutron retains a fraction of 0.716 of its kinetic energy.
(4)
(b)
Using the fraction (11/13)2 for every collision, calculate the kinetic energy of the neutron, in eV, after 40 collisions. Give your answer to 2 significant figures.
(2)
(c)
A thermal neutron has kinetic energy 0.025 eV. Deduce whether 40 collisions are enough to slow the neutron to this energy.
(1)
(d)
Explain why a moderator containing hydrogen nuclei would need fewer collisions than graphite.
(2)
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Part (a)
M1 momentum conservation: mu = mv1 + 12mv2
✓ 1
M1 elastic: relative speed of separation equals relative speed of approach, v2 − v1 = u (or kinetic energy conserved)
✓ 1
A1 v1 = −(11/13)u
✓ 1
A1 fraction = (11/13)2 = 0.716
✓ 1
Part (b)
M1 energy = 2.0×106 × (11/13)80
✓ 1
A1 3.1 eV
✓ 1
Part (c)
3.1 eV is greater than 0.025 eV, so more than 40 collisions are needed
✓ 1
Part (d)
a hydrogen nucleus has almost the same mass as a neutron
✓ 1
so a larger fraction of the neutron's kinetic energy is transferred in each collision
✓ 1
Common error: Write the fraction as a squared ratio of speeds and keep the unrounded fraction until the final step.
39E-2-35A natural fission reactorPaper 2Hard13 marks
About 2.0 × 109 years ago, in a uranium ore deposit in Gabon, groundwater allowed natural chain reactions of neutron-induced fission of uranium-235 to run for a long period. Today, uranium-235 makes up 0.720 % of the uranium atoms in natural uranium; almost all the rest is uranium-238. Modern reactors that use ordinary water need fuel in which at least about 3 % of the uranium atoms are uranium-235.
Show that the decay constant of uranium-235 is about 9.8 × 10−10 year−1.
(1)
(ii)
Show that the ratio R = (number of uranium-235 atoms)/(number of uranium-238 atoms) a time t ago was Rnowe(λ5 − λ8)t, where Rnow is the present ratio and λ5 and λ8 are the decay constants.
(2)
(iii)
Hence determine the percentage of uranium atoms that were uranium-235 at the time the natural reactor operated.
(2)
(b)
(i)
State the role of the groundwater in the chain reaction.
(1)
(ii)
Evidence suggests that the natural reactor switched itself on for about 30 minutes and then off for about 2.5 hours, repeatedly. Suggest an explanation, and outline why the chain reaction could not run out of control.
(3)
(c)
(i)
It is estimated that about 6.0 × 103 kg of uranium-235 underwent fission over a period of about 1.5 × 105 years, with about 200 MeV released per fission. Estimate the average power of the natural reactor.
(3)
(ii)
Explain, using your answer to (a)(iii), why a natural reactor of this kind cannot form today.
(1)
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Part (a)(i)
λ = ln 2/7.04 × 108 = 9.85 × 10−10 year−1
✓ 1
An answer to at least 3 s.f. must be seen.
Part (a)(ii)
Each isotope decays exponentially from its earlier number: N5,now = N5e−λ5t and N8,now = N8e−λ8t
✓ 1
Dividing: Rnow = Re−(λ5 − λ8)t, so R = Rnowe(λ5 − λ8)t
Allow ECF from (a)(i) and (a)(ii). Using 0.00720 for Rnow changes the answer only in the third s.f.
R = 0.0381, so the percentage is 0.0381/1.0381 × 100 = 3.7 %
✓ 1
Accept 3.6–3.8 %. Decaying uranium-235 alone (0.72 × 7.2 = 5.2 %) or using λ5 + λ8 scores [1 max].
Part (b)(i)
Moderator: it slowed the fast fission neutrons (by collisions with hydrogen nuclei) so that they were likely to cause further fissions of uranium-235
✓ 1
Do not accept "water absorbs neutrons" or "water cools the ore".
Part (b)(ii)
The energy released heated the ore until the water boiled away (or was driven out), so the neutrons were no longer slowed and the chain reaction stopped
✓ 1
Allow ECF from (b)(i).
When the rock cooled, water seeped back, moderation resumed and the chain reaction restarted
✓ 1
Negative feedback: any increase in power heats the ore and drives out more water, which reduces the moderation and hence the fission rate, so the power cannot keep rising
✓ 1
OWTTE. Self-regulation must be linked to the loss of moderation.
Part (c)(i)
Number of fissions = 6.0 × 103/(235 × 1.661 × 10−27) = 1.5 × 1028
P = 4.92 × 1017/(1.5 × 105 × 3.16 × 107) = 1.0 × 105 W
✓ 1
Allow ECF from MP2. Accept about 100 kW (0.95–1.1 × 105 W). Forgetting to convert years to seconds (3.3 × 1012 W) scores [2 max].
Part (c)(ii)
The uranium-235 fraction is now only 0.72 %, well below the ≈ 3 % needed for a chain reaction with water as moderator, whereas 2.0 × 109 years ago it was 3.7 %, above it
✓ 1
Allow ECF from (a)(iii). The comparison with ≈ 3 % at both times is required.
Answers: (a)(iii) 3.7 % · (c)(i) 1.0 × 105 W (the remaining parts are explanations — see the table above)
40IBP2H-041-00-13A natural fission reactorPaper 2Hard9 marks
At Oklo in Gabon, natural fission chain reactions took place about 2.0×109 years ago, with groundwater as the moderator. Today, the ratio of uranium-235 nuclei to uranium-238 nuclei in natural uranium is 0.0073. Half-life of U-235 = 7.04×108 years. Half-life of U-238 = 4.47×109 years.
(a)
Outline why natural uranium ore today cannot sustain a chain reaction using ordinary water as the moderator.
(2)
(b)
Show that the ratio of U-235 nuclei to U-238 nuclei 2.0×109 years ago was about 0.038.
(3)
(c)
Hence determine the percentage of uranium atoms that were U-235 at that time.
(1)
(d)
The Oklo reactors regulated themselves, operating in cycles. Explain how groundwater produced this behaviour.
(3)
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Part (a)
The proportion of fissile U-235 (about 0.7%) is too low .
✓ 1
Too many neutrons are absorbed by U-238 or by hydrogen in the water without causing fission, so on average fewer than one neutron per fission causes another fission .
41IBP2H-041-00-15Spontaneous fissionPaper 2Hard9 marks
Californium-252 is used as a portable neutron source, for example to start up reactors and in neutron brachytherapy for some cancers. Data for Cf-252: half-life = 2.65 years fraction of decays that are spontaneous fission = 3.1% (the rest are alpha decays) mean number of neutrons per spontaneous fission = 3.8 Take 1 year = 3.16×107 s and 1 u = 1.66×10−27 kg.
(a)
Distinguish between spontaneous fission and induced fission.
(2)
(b)
Calculate the total activity of a 1.0 μg sample of Cf-252.
(3)
(c)
Determine the rate at which the sample emits neutrons.
(2)
(d)
A clinic replaces its source when the neutron emission rate falls to 60% of its initial value. Calculate how long the source can be used.
(2)
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Part (a)
Spontaneous fission happens randomly, without the nucleus absorbing any particle .
✓ 1
Induced fission happens only after the nucleus absorbs a particle, usually a neutron .
✓ 1
Part (b)
N = 1.0×10−9 / (252 × 1.66×10−27) = 2.39×1015 .
✓ 1
λ = ln 2 / (2.65 × 3.16×107) = 8.28×10−9 s−1 .
✓ 1
A = λN = 1.98×107 Bq ≈ 2.0×107 Bq .
✓ 1
Part (c)
Spontaneous fission rate = 0.031 × 1.98×107 ≈ 6.1×105 s−1 (6.2×105 s−1 if the rounded activity is used) .
42IBP2S-040-00-14Fission productsPaper 2Hard8 marks
Molybdenum-99 is used to make the medical tracer technetium-99m. It is extracted from uranium-235 targets that undergo fission in a reactor. In one target, fission occurs at a rate of 5.0×1014 s−1. A molybdenum-99 nucleus is produced in 6.1% of fissions. The half-life of molybdenum-99 is 66 hours. The target is irradiated for 2.0 hours. Assume that decay of molybdenum-99 during the irradiation is negligible.
(a)
Explain why molybdenum-99, like most fission products, is a β− emitter.
(2)
(b)
Calculate the number of molybdenum-99 nuclei present at the end of the irradiation.
(2)
(c)
Determine the activity of the molybdenum-99 at the end of the irradiation.
(2)
(d)
After the molybdenum is removed, the remaining target material is treated as high-level waste. Outline why.
(2)
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Part (a)
Fission fragments keep roughly the neutron-to-proton ratio of uranium, which is higher than that of stable nuclei of their (lower) mass, so they are neutron-rich
✓ 1
β− decay converts a neutron into a proton, reducing the neutron excess and moving the nucleus towards stability
✓ 1
Part (b)
Production rate = 0.061 × 5.0×1014 = 3.05×1013 s−1
✓ 1
N = 3.05×1013 × 7200 = 2.2×1017
✓ 1
Part (c)
Allow an error carried forward from the number of nuclei calculated in the previous part.
—
λ = ln 2 / (66 × 3600) = 2.92×10−6 s−1
✓ 1
A = λN = 2.92×10−6 × 2.2×1017 = 6.4×1011 Bq
✓ 1
Part (d)
It still contains many other fission products, so its activity is very high. Credit any of: it emits intense ionising radiation, it generates significant heat, or it needs cooling and heavy shielding
✓ 1
Some of the nuclides present have long half-lives (e.g. Cs-137, Sr-90, or transuranic nuclides such as Pu-239), so the material must be isolated from the environment for a very long time
✓ 1
Common error: Convert the half-life to seconds before calculating λ. Otherwise the activity will not come out in becquerels.
43IBP3H-E4-20r19-12A natural fission reactorPaper 2Hard10 marks
About 2.00×109 years ago, a deposit of natural uranium at a site in Africa supported a chain reaction for a long period. Today natural uranium contains 0.720% uranium-235 and 99.28% uranium-238. Half-lives: uranium-235 7.04×108 years; uranium-238 4.47×109 years. Assume that these were the only two isotopes present and that the ore has been undisturbed.
(a)
State why the proportion of uranium-235 was greater in the past.
(1)
(b)
Determine the factor by which the number of uranium-235 nuclei in the ore was greater 2.00×109 years ago than it is today. Give your answer to 3 significant figures.
(2)
(c)
Determine the percentage of uranium-235 in the natural uranium 2.00×109 years ago. Give your answer to 3 significant figures.
(4)
(d)
Water seeped into the ore and acted as a moderator. Explain how the presence of water allowed the reaction to regulate itself.
(3)
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Part (a)
uranium-235 has a shorter half-life than uranium-238 so it has decayed by a larger fraction
✓ 1
Part (b)
M1 factor = 2(2.00×109 / 7.04×108)
✓ 1
A1 7.16
✓ 1
Part (c)
M1 factor for uranium-238 = 2(2.00×109 / 4.47×109) = 1.36
44IBP3H-E4-20r19-16Energy from fissionPaper 2Hard7 marks
A research group in astroparticle physics places a detector 25 m from the core of a reactor that has a thermal power of 3.2 GW. Assume that each fission releases 200 MeV and that on average 6 antineutrinos are emitted per fission, spreading out equally in all directions. 1 eV = 1.60×10−19 J.
(a)
Outline the origin of the antineutrinos emitted by a reactor.
(2)
(b)
Calculate the number of fissions per second in the reactor.
(2)
(c)
Determine the number of antineutrinos passing per second through each square metre of the detector's surface, treating the reactor as a point source. Give your answer to 3 significant figures.
(3)
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Part (a)
fission fragments are neutron-rich and undergo β− decay
✓ 1
in which a neutron changes into a proton, an electron and an antineutrino
✓ 1
Part (b)
M1 energy per fission = 200×106×1.60×10−19 = 3.2×10−11 J
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