E.4 Fission – IB Physics exam questions with mark scheme

1Core3 marks

Nuclear fission can occur spontaneously or be induced by a neutron.

(a) Distinguish between spontaneous fission and neutron-induced fission. [2]

(b) State why the neutrons released in neutron-induced fission make a chain reaction possible. [1]

(a) spontaneous fission: a (heavy) nucleus splits into two smaller nuclei without any external trigger / randomly, with no particle absorbed(1)
(a) neutron-induced fission: a heavy nucleus absorbs a (slow/thermal) neutron, becomes unstable and then splits (into two smaller nuclei plus neutrons) (do not award if the student says the neutron 'hits and breaks' the nucleus with no absorption)(1)
(b) each fission releases more than one neutron / 2–3 neutrons, which can go on to induce further fissions (allow 'the released neutrons can be absorbed by other U-235 nuclei causing further fission')(1)
Answer: (a) Spontaneous: splits with no trigger. Induced: nucleus absorbs a neutron, then splits. (b) Each fission releases 2–3 neutrons that can cause further fissions.
Examiner tip: Students lose the induced-fission mark by saying the neutron 'smashes' the nucleus; the key word is absorbed.

2Core4 marks

The diagram shows a simplified nuclear reactor in a power station. The parts labelled A, B, C and D are named below. The water in the reactor vessel acts as both moderator and coolant.

A: control rods B: moderator C: heat exchanger D: shielding

Outline the role of each of the parts A, B, C and D. [4]

A (control rods): absorb neutrons, so (by moving in or out) control the rate of fission / keep exactly one neutron per fission going on to cause another fission (allow 'slow down/stop the chain reaction'; do not accept 'slow the neutrons down')(1)
B (moderator): slows down / reduces the kinetic energy of the fast neutrons (by collisions) so that they are more likely to be absorbed by U-235 / cause fission(1)
C (heat exchanger): transfers thermal energy from the (hot) primary coolant to water in a separate circuit, turning it into steam (to drive the turbine) (allow keeps radioactive coolant separate from the turbine water)(1)
D (shielding): absorbs (ionising) radiation / neutrons and gamma rays to protect workers and the environment (accept 'thick concrete/steel stops radiation escaping')(1)
Answer: A absorbs neutrons to control the fission rate; B slows neutrons; C transfers energy from coolant to make steam; D absorbs escaping radiation.
Examiner tip: Moderators slow neutrons, control rods absorb them: swapping these two loses two marks at once.

3Standard3 marks

One possible neutron-induced fission of uranium-235 is

$$^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$$

The table gives the masses of the particles.

particle	mass / u
neutron	1.008665
uranium-235	235.043930
barium-141	140.914403
krypton-92	91.926173

Calculate, in MeV, the energy released in this fission. [3]

mass before = 235.043930 + 1.008665 = 236.052595 u and mass after = 140.914403 + 91.926173 + 3(1.008665) = 235.866571 u (allow evidence of both sums)(1)
mass defect Δm = 0.186024 u (allow 0.186 u)(1)
E = 0.186024 × 931.5 = 173 MeV (allow 172–174 MeV; allow 2.78 × 10⁻¹¹ J converted to MeV; if only one neutron is included on the right (Δm = 2.203 u, E ≈ 2.05 × 10³ MeV) award this mark only, by ecf, for the correct conversion of their Δm)(1)
Answer: 173 MeV
Examiner tip: Count all three neutrons on the right and the one on the left: forgetting the extra two neutrons changes the answer by about 1.9 GeV of 'missing' mass.

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4Standard4 marks

The graph shows how the binding energy per nucleon varies with nucleon number. Three nuclides are marked, with their binding energy per nucleon in MeV.

Uranium-235 undergoes the fission

$$^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{94}_{38}\text{Sr} + {}^{140}_{54}\text{Xe} + 2\,^{1}_{0}\text{n}$$

(a) Explain, with reference to the graph, why energy is released in this fission. [2]

(b) Determine the energy released in the fission. [2]

(a) the products have a greater binding energy per nucleon than uranium-235 / products lie nearer the peak of the curve(1)
(a) so the total binding energy of the products is greater (the products are more tightly bound / have less mass); the difference in binding energy is released (as kinetic energy of the products) (do not accept 'binding energy is released when the bonds break')(1)
(b) total binding energies: Sr = 94 × 8.59 = 807.5 MeV, Xe = 140 × 8.29 = 1160.6 MeV, U = 235 × 7.59 = 1783.7 MeV(1)
(b) energy released = 807.5 + 1160.6 − 1783.7 = 184 MeV (allow 180–190 MeV; free neutrons have zero binding energy)(1)
Answer: (a) Products have higher binding energy per nucleon, so total binding energy increases and the difference is released. (b) 184 MeV
Examiner tip: Students multiply the binding energy per nucleon by the wrong nucleon number or subtract 'per nucleon' values directly (8.59 + 8.29 − 7.59), which gives a meaningless answer.

5Hard5 marks

In a graphite-moderated reactor, fast neutrons are slowed by collisions with carbon-12 nuclei. Assume that a neutron of mass $m$ and speed $u$ makes a head-on elastic collision with a stationary carbon nucleus of mass $12m$.

(a) Show that the speed of the neutron after the collision is $\dfrac{11}{13}u$. [3]

(b) A neutron is released by fission with kinetic energy 2.0 MeV. It must be slowed to 0.025 eV before it is likely to cause further fission. Estimate the minimum number of such collisions needed. [2]

(a) momentum conserved: $mu = mv_1 + 12mv_2$(1)
(a) kinetic energy conserved (elastic): $\tfrac12 mu^2 = \tfrac12 mv_1^2 + \tfrac12(12m)v_2^2$ (allow use of 'relative speed of approach = relative speed of separation': $u = v_2 - v_1$)(1)
(a) eliminating $v_2$ to give $v_1 = -\tfrac{11}{13}u$, so speed $=\tfrac{11}{13}u$ (neutron rebounds) (must see algebraic steps; do not award for simply substituting the given answer)(1)
(b) fraction of kinetic energy kept per collision $=(11/13)^2 = 0.716$, so $2.0\times10^{6}\,(0.716)^n = 0.025$ (allow working with speeds: $(11/13)^n = \sqrt{0.025/2.0\times10^6}$)(1)
(b) $n = \dfrac{\ln(1.25\times10^{-8})}{\ln 0.716} = 54.5$, so about 55 collisions (allow 54 or 55)(1)
Answer: (a) shown; (b) about 55 collisions
Examiner tip: Students apply the factor 11/13 to the kinetic energy; it applies to the speed, so the energy factor is (11/13)² ≈ 0.72.

6Standard4 marks

A nuclear power station has an electrical power output of 1.2 GW and an overall efficiency of 33%. Each fission of uranium-235 releases 200 MeV of energy.

(a) Calculate the number of fissions per second in the reactor. [2]

(b) Calculate the mass of uranium-235 that undergoes fission in one year. (1 year = 3.16 × 10⁷ s) [2]

(a) thermal power = 1.2 × 10⁹ ÷ 0.33 = 3.6 × 10⁹ W(1)
(a) rate = 3.64 × 10⁹ ÷ (200 × 10⁶ × 1.60 × 10⁻¹⁹) = 1.1 × 10²⁰ s⁻¹ (allow 1.13–1.14 × 10²⁰; award [1 max] for 3.75 × 10¹⁹ from using 1.2 GW directly)(1)
(b) number of fissions in a year = 1.13 × 10²⁰ × 3.16 × 10⁷ = 3.6 × 10²⁷(1)
(b) mass = 3.6 × 10²⁷ × 235 × 1.66 × 10⁻²⁷ kg = 1.4 × 10³ kg (allow use of molar mass: 3.6 × 10²⁷/6.02 × 10²³ × 0.235 kg; allow ecf from (a))(1)
Answer: (a) 1.1 × 10²⁰ fissions per second; (b) 1.4 × 10³ kg (about 1.4 tonnes)
Examiner tip: The 1.2 GW is electrical output: divide by the efficiency first to get the rate of energy release in the core.

7Hard6 marks

In a reactor, the multiplication factor $k$ is the average number of neutrons from one fission that go on to cause another fission. The table shows the number of neutrons in successive generations shortly after some control rods were partly withdrawn.

generation	0	1	2	3
number of neutrons	5000	5100	5202	5306

(a) Outline what is meant by a chain reaction. [1]

(b) Determine $k$ from the data. [1]

(c) Calculate the number of generations after which the number of neutrons has doubled, assuming $k$ stays constant. [2]

(d) Explain how the operators should use the control rods to return the reactor to a steady power output. [2]

(a) neutrons released by one fission go on to cause further fissions, so the process becomes self-sustaining(1)
(b) $k = 5100/5000 = 1.02$ (ratio of any consecutive pair)(1)
(c) $2 = 1.02^n$ so $n = \ln 2/\ln 1.02$(1)
(c) $n = 35$ generations (allow 35.0)(1)
(d) insert the control rods further (lower them) so that they absorb more neutrons(1)
(d) until $k = 1$ / each fission leads on average to exactly one further fission, so the number of neutrons (and fission rate/power) stays constant (do not accept 'until the reaction stops')(1)
Answer: (a) neutrons from one fission cause further fissions; (b) k = 1.02; (c) 35 generations; (d) push rods in to absorb more neutrons until k = 1
Examiner tip: For (d), the aim is k = 1, not k = 0: lowering the rods too far shuts the reactor down rather than giving steady power.

8Hard6 marks

The used (spent) fuel removed from a fission reactor contains fission products such as caesium-137 and strontium-90, together with unused uranium.

Discuss the properties of fission products that make spent fuel hazardous and how this waste is managed. [6]

fission products are neutron-rich (have too many neutrons for stability) because heavy nuclei have a higher neutron-to-proton ratio than stable medium-mass nuclei(1)
so they are unstable and decay mainly by β⁻ emission, often followed by γ emission; many undergo a series of decays(1)
some have short half-lives so spent fuel has a very high activity when first removed(1)
the high activity means the spent fuel releases a lot of energy and is very hot (decay heat)(1)
some products (e.g. Cs-137, Sr-90, about 30 years) or other products in the waste have long half-lives, so the waste stays hazardous for hundreds to thousands of years(1)
ionising radiation can damage cells/DNA; γ is penetrating, and products such as Sr-90 can be absorbed into the body (e.g. bones) if released(1)
management: spent fuel first stored under water in cooling ponds (water cools it and shields radiation)(1)
then stored in dry casks/steel containers or vitrified (sealed in glass) and placed in shielded storage(1)
long-term: deep geological disposal in stable rock, isolated from groundwater; site monitored; reprocessing can recover unused uranium/plutonium(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, logical discussion of at least three properties of fission products (e.g. neutron-rich, β/γ emitters, high initial activity/decay heat, long half-lives) each linked to the hazard and to a suitable management method (cooling, shielding, long-term isolation). Correct physics terminology throughout.
Level 2	3–4	Describes some properties and some management methods, but links between them are incomplete or one side is thin. Minor errors in terminology.
Level 1	1–2	Isolated relevant points, e.g. 'waste is radioactive' or 'it is buried underground', with little or no explanation.
No marks	0	Nothing relevant.

Answer: Neutron-rich products are β⁻/γ emitters with high activity (hot) and some long half-lives; managed by cooling ponds, then shielded dry storage/vitrification and deep geological disposal.
Examiner tip: Top-band answers link each property to a management step (hot → cooling ponds; long half-life → deep disposal); a list of storage methods with no physics stays in the middle band.

9Hard4 marks

Californium-252 can undergo spontaneous fission. In one such event the nucleus, initially at rest, splits into two fragments of nucleon numbers 106 and 142 and four neutrons. The total kinetic energy of the two fragments is 185 MeV.

(a) Suggest why spontaneous fission is observed only in very heavy nuclides. [1]

(b) Ignoring the momentum of the neutrons, show that the kinetic energy of the lighter fragment is about 106 MeV. [3]

(a) in very heavy nuclei the (long-range) electrostatic repulsion between the many protons nearly overcomes the (short-range) strong nuclear force, so the nucleus can split without input of energy (allow 'large nuclei are only weakly bound / have lower binding energy per nucleon than the products')(1)
(b) momentum conserved: fragments have equal and opposite momenta, $p_1 = p_2$(1)
(b) $E_k = \dfrac{p^2}{2m}$ so $E_k \propto \dfrac1m$: $\dfrac{E_{106}}{E_{142}} = \dfrac{142}{106}$(1)
(b) $E_{106} = 185 \times \dfrac{142}{248} = 105.9$ MeV ≈ 106 MeV (must see more sig figs than given or a clear ratio method; heavier fragment gets 79 MeV)(1)
Answer: (a) Coulomb repulsion between protons nearly overcomes the strong force; (b) 185 × 142/248 = 105.9 MeV ≈ 106 MeV
Examiner tip: The lighter fragment gets the larger share of the kinetic energy; students who share it in proportion to mass get the two values the wrong way round.

10Stretch6 marks

A pressurised-water reactor has a thermal power of 3.0 GW. Its core holds 80 tonnes of uranium fuel enriched to 4.0% uranium-235 by mass. Each fission releases 200 MeV.

(a) Fuel is replaced when 75% of its uranium-235 has undergone fission. Estimate, in years, how long the fuel lasts at full power. [4]

(b) The primary coolant enters the core at 290 °C and leaves at 320 °C. Its specific heat capacity is 5.4 kJ kg⁻¹ K⁻¹. Calculate the mass of coolant that must flow through the core each second. [2]

(a) mass of U-235 that fissions = 80 × 10³ × 0.040 × 0.75 = 2400 kg(1)
(a) number of nuclei = 2400 ÷ (235 × 1.66 × 10⁻²⁷) = 6.15 × 10²⁷ (allow 6.1–6.2 × 10²⁷; allow molar-mass route: 2400/0.235 × 6.02 × 10²³)(1)
(a) fission rate = 3.0 × 10⁹ ÷ (200 × 1.60 × 10⁻¹³) = 9.4 × 10¹⁹ s⁻¹ (or total energy available = 6.1 × 10²⁷ × 3.2 × 10⁻¹¹ = 2.0 × 10¹⁷ J)(1)
(a) time = 6.15 × 10²⁷ ÷ 9.38 × 10¹⁹ = 6.6 × 10⁷ s ≈ 2.1 years (allow 2.0–2.1 years; ecf within part)(1)
(b) $P = \dfrac{m}{t}c\Delta T$ with $\Delta T = 30$ K(1)
(b) $\dfrac{m}{t} = \dfrac{3.0\times10^{9}}{5.4\times10^{3}\times30} = 1.9\times10^{4}$ kg s⁻¹ (award [1 max] for 1.9 × 10⁷ from using c in kJ without conversion)(1)
Answer: (a) about 2.1 years; (b) 1.9 × 10⁴ kg s⁻¹
Examiner tip: Only 4.0% of the fuel is U-235 and only 75% of that is used: applying both factors before converting to nuclei is where most marks go.

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