E.4 Fission flashcards: IB Physics SL and HL

All 75 E.4 Fission cards

1. DefinitionSL & HL

   Define nuclear fission.

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   Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of roughly comparable mass, together with the release of two or three neutrons and a large amount of energy. Exam tip: the mark scheme wants "heavy/large nucleus splits into two SMALLER nuclei of similar size" plus "neutrons released" plus "energy released". Answers saying "a nucleus breaks up" score zero because that also describes α decay; the distinguishing feature is two fragments of comparable mass. Energy is a scalar, measured in MeV per fission or J. Typical value: about 200 MeV per fission of uranium-235.

2. DefinitionSL & HL

   Define induced fission and contrast it with spontaneous fission.

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   Induced fission is fission that occurs when a nucleus absorbs a bombarding particle — in a reactor, a slow (thermal) neutron — forming a highly excited compound nucleus that immediately splits. Spontaneous fission is fission that happens without any external particle, a rare random decay mode of very heavy nuclei. Exam tip: students lose the mark by writing "the neutron hits the nucleus"; the mark is for ABSORBED/captured, forming U-236 which is unstable and splits. Reactors rely entirely on induced fission because its rate can be controlled by controlling the neutron population; spontaneous fission cannot be controlled.

3. DefinitionSL & HL

   Define a thermal (slow) neutron and state why U-235 fission requires one.

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   A thermal neutron is a neutron that has been slowed until its kinetic energy is comparable to the average thermal energy of the surrounding material, about 0.025 eV at room temperature (speed ≈ 2200 m s⁻¹). Exam tip: state a value — "slow" alone is often not enough for the mark. The reason is that the probability (cross-section) for capture by U-235 is far higher for slow neutrons: a fast neutron spends less time near the nucleus and is more likely to scatter or be captured by U-238. Neutrons produced BY fission are fast, averaging about 2 MeV, so they must be moderated.

4. DefinitionSL & HL

   Define a chain reaction in the context of nuclear fission.

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   A chain reaction is a self-sustaining sequence of fissions in which the neutrons released by one fission event go on to induce further fission events in other nuclei. Exam tip: the two marking points are (i) each fission releases two or three neutrons and (ii) at least one of these on average is absorbed by another fissile nucleus and causes a further fission. Answers that just say "the reaction keeps going" gain nothing. A controlled chain reaction (one neutron per fission continuing the chain) powers a reactor; an uncontrolled, exponentially growing one is a fission weapon.

5. DefinitionSL & HL

   Define critical mass.

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   Critical mass is the minimum mass of fissile material, in a given shape and configuration, needed to sustain a chain reaction — that is, for exactly one neutron per fission on average to go on to cause a further fission. Exam tip: the mark is for "minimum mass … to SUSTAIN a chain reaction", not "to start" one. Below it too many neutrons escape through the surface (surface area to volume ratio is too large) and the reaction dies out. Critical mass depends on enrichment, geometry (a sphere minimises leakage) and whether a neutron reflector surrounds the fuel. Unit: kg; scalar.

6. DefinitionSL & HL

   Define subcritical, critical and supercritical in terms of the neutron multiplication factor k.

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   k is the average number of neutrons from one fission that go on to induce a further fission. Subcritical: k < 1, the number of fissions per generation falls and the reaction dies out. Critical: k = 1, the fission rate is constant and the reactor delivers steady power — this is normal operating condition. Supercritical: k > 1, the fission rate grows exponentially, used briefly to raise power and deliberately achieved in a weapon. Exam tip: the mark scheme expects k to be defined per GENERATION of neutrons; saying "more neutrons are produced" without "than are used up/absorbed" is not enough.

7. DefinitionSL & HL

   Define the moderator and state the materials used.

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   A moderator is a material of low mass number placed around the fuel rods whose function is to slow the fast neutrons released by fission down to thermal energies, so that they are more likely to induce further fission in U-235. Examples: graphite, ordinary (light) water, heavy water D₂O. Exam tip: the mark scheme requires the MECHANISM — repeated elastic collisions with light nuclei, in which the neutron transfers a large fraction of its kinetic energy because the target nucleus has a similar mass. Saying the moderator "absorbs neutrons" is the classic zero-mark error; that is the job of the control rods.

8. DefinitionSL & HL

   Define control rods and state how they regulate reactor power.

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   Control rods are rods of a strongly neutron-absorbing material — boron or cadmium — that can be raised out of or lowered into the reactor core between the fuel rods. They absorb neutrons without fissioning, so lowering them further reduces the number of neutrons available to induce fission and lowers k; raising them increases k. Exam tip: the required answer is "absorb neutrons to control the rate of the chain reaction / keep k = 1". A common trap is to write that they "slow the neutrons down" — that is the moderator. Full insertion shuts the reactor down (a scram).

9. DefinitionSL & HL

   Define the coolant and the heat exchanger in a thermal fission reactor.

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   The coolant is a fluid (pressurised water, CO₂ gas or liquid sodium) pumped through the core; it removes the thermal energy produced by the kinetic energy of the fission fragments and prevents the core overheating. The heat exchanger is where the hot coolant transfers that energy to a secondary water circuit, turning it to steam that drives a turbine and generator. Exam tip: state that the two circuits are kept separate so that radioactive coolant never reaches the turbine. In a PWR the light water acts as both coolant and moderator — a self-limiting safety feature, since loss of water also stops moderation.

10. DefinitionSL & HL

    Define enrichment of nuclear fuel and state typical values.

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    Enrichment is the process of increasing the proportion of the fissile isotope U-235 in a sample of uranium above its natural abundance of 0.7%, the remainder being U-238. Reactor fuel is typically enriched to 3–5% U-235; weapons-grade uranium is above about 90%. Exam tip: the mark scheme wants the reason — natural uranium contains too little U-235 for most reactor designs to reach criticality, because U-238 captures neutrons (strongly at resonance energies, as they slow down) without fissioning. Enrichment is difficult because the isotopes are chemically identical, so physical methods (gas centrifuges on UF₆) are used; this is the origin of nuclear proliferation concerns.

11. DefinitionSL & HL

    Define fuel rods and fission fragments (daughter nuclei).

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    Fuel rods are sealed metal tubes containing pellets of enriched uranium dioxide, arranged in an array through the moderator so that neutrons pass through the moderator between rods. Fission fragments (daughter nuclei) are the two lighter nuclei produced by the split, typically with mass numbers around 95 and 140. Exam tip: state that the fragments are highly radioactive and neutron-rich, so they decay by β⁻ emission, and that about 165 MeV of the 200 MeV appears as their kinetic energy — this is the energy that ultimately heats the coolant. The cladding contains the radioactive fragments inside the rod.

12. DefinitionSL & HL

    Define mass defect.

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    The mass defect Δm of a nucleus is the difference between the total mass of its separated constituent nucleons and the mass of the assembled nucleus: Δm = Zm_p + (A − Z)m_n − m_nucleus. It is always positive. Exam tip: the mark scheme requires "mass of separate/individual nucleons MINUS mass of the nucleus"; reversing the subtraction or omitting "separated" loses the mark. Units: kg, or u (1 u = 1.66 × 10⁻²⁷ kg). Mass defect is a scalar. It arises because energy is released when nucleons bind together, and by E = Δmc² that energy corresponds to a loss of mass.

13. DefinitionSL & HL

    Define binding energy and binding energy per nucleon.

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    The binding energy of a nucleus is the energy required to completely separate the nucleus into its individual nucleons (equivalently the energy released when the nucleus is assembled from separate nucleons). Binding energy per nucleon is that binding energy divided by the nucleon number A. Exam tip: the mark scheme accepts either the "energy needed to separate" or the "energy released on formation" wording, but NOT "energy holding the nucleus together" as a stand-alone phrase. Binding energy per nucleon is the correct measure of stability: the greater it is, the more stable the nucleus. Units: MeV, or MeV per nucleon; scalar.

14. DefinitionSL & HLData booklet: Yes

    Define the unified atomic mass unit u and the mass–energy unit MeV c⁻².

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    The unified atomic mass unit u is defined as one twelfth of the mass of an unbound, ground-state atom of carbon-12: 1 u = 1.66 × 10⁻²⁷ kg. Because mass and energy are equivalent, 1 u corresponds to 931.5 MeV, so nuclear masses are often quoted in MeV c⁻²: 1 u = 931.5 MeV c⁻². Exam tip: keep the c⁻² — writing a mass as "931.5 MeV" is dimensionally wrong and is penalised in "state the unit" questions. Both constants are printed in the data booklet, so never derive them; they are the fastest route from Δm in u to energy in MeV.

15. DefinitionSL & HL

    Define high-level nuclear waste and outline the storage problem.

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    High-level waste is the intensely radioactive spent fuel and fission-fragment material removed from a reactor; it has a high activity and contains nuclides with half-lives from years to thousands of years, and it continues to generate heat after removal. Exam tip: the marking points are (i) long half-lives so it stays hazardous for thousands of years, (ii) it must be shielded and cooled, initially in water ponds, then vitrified in glass and sealed in casks, (iii) long-term plans involve deep geological repositories in stable rock. Low-level waste (gloves, clothing, coolant) is far larger in volume but much less active.

16. DefinitionSL & HL

    Define meltdown and containment in reactor safety.

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    A meltdown occurs when the coolant fails to remove the decay heat still produced by the fission fragments after shutdown, so the core temperature rises until the fuel and cladding melt. Containment is the sealed, thick steel-and-reinforced-concrete structure around the reactor vessel designed to prevent the escape of radioactive material even in an accident. Exam tip: the mark scheme looks for the key insight that shutting down the chain reaction does NOT stop the heating, because the radioactive decay of fission fragments continues — this is what drove Fukushima. Chernobyl had no full containment building, which is why activity spread over Europe.

17. DefinitionSL & HL

    Define the strong nuclear force and explain its role in fission energy release.

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    The strong nuclear force is the attractive force acting between nucleons; it is charge-independent (acting equally between p–p, n–n and p–n), very short-ranged (about 10⁻¹⁵ m, negligible beyond a few fm) and becomes repulsive below about 0.5 fm. Exam tip: because it is short-ranged, a nucleon in a large nucleus is bound only to its neighbours, while the electrostatic repulsion between protons is long-ranged and acts across the whole nucleus. This is why binding energy per nucleon falls beyond iron and why splitting a heavy nucleus into two smaller ones releases energy.

18. EquationSL & HLData booklet: Yes

    State the mass–energy equivalence equation and how it is used for fission.

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    E = Δmc². E is the energy released, in joules (J); Δm is the mass defect or mass difference between reactants and products, in kilograms (kg); c = 3.00 × 10⁸ m s⁻¹ is the speed of light in a vacuum. Data booklet: printed. Validity: applies to any process where rest mass changes. Common misuse: substituting Δm in u while leaving c in m s⁻¹ — convert with 1 u = 1.66 × 10⁻²⁷ kg first, or use 1 u = 931.5 MeV c⁻² instead. Sanity check: Δm = 0.215 u gives 0.215 × 931.5 ≈ 200 MeV, the standard per-fission value.

19. EquationSL & HLData booklet: No – memorise

    Give the equation for the mass defect of a nucleus ᴬ_Z X.

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    Δm = Zm_p + (A − Z)m_n − m_nucleus. Δm is the mass defect (kg or u); Z is the proton number; A is the nucleon number; A − Z is the neutron number; m_p = 1.007276 u, m_n = 1.008665 u are the proton and neutron rest masses; m_nucleus is the measured nuclear mass (u). Data booklet: not printed — memorise. Common misuse: using the ATOMIC mass in place of the nuclear mass without subtracting Z electron masses; if atomic masses are used throughout, the electron masses cancel in a Q-value calculation, so be consistent. Sanity check: Δm for helium-4 is about 0.0304 u ≈ 28.3 MeV.

20. EquationSL & HLData booklet: Yes

    Give the equation linking binding energy, mass defect and binding energy per nucleon.

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    E_b = Δmc², and binding energy per nucleon = E_b/A. E_b is the total binding energy (J or MeV); Δm is the mass defect (kg or u); A is the nucleon number (no unit). Data booklet: E = Δmc² is printed; the division by A is not. Validity: applies to any nuclide. Common misuse: comparing TOTAL binding energies to judge stability — U-235 has a huge total binding energy (about 1784 MeV) yet is less stable than iron-56, because per nucleon it is only ≈ 7.6 MeV against ≈ 8.8 MeV. Sanity check: 235 × 7.6 ≈ 1790 MeV.

21. EquationSL & HLData booklet: No – derive

    Give the equation for the energy released in a fission reaction using binding energies.

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    Q = Σ E_b(products) − Σ E_b(reactants), or equivalently Q ≈ A × [(E_b/A)_products − (E_b/A)_reactants]. Q is the energy released (MeV); E_b is total binding energy (MeV); A is the total nucleon number involved. Data booklet: not printed — derive from the binding-energy curve. Validity: nucleon number is conserved, so the approximation using an average A is good. Common misuse: subtracting in the wrong order — the PRODUCTS are more tightly bound, so products minus reactants is positive. Sanity check: 236 × (8.5 − 7.6) ≈ 210 MeV, close to the quoted 200 MeV.

22. EquationSL & HLData booklet: Yes

    Give the equation for the energy released in a fission reaction using particle masses.

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    Q = (Σm_reactants − Σm_products)c², with Q in MeV when the mass difference is in u and the conversion 931.5 MeV per u is used. Σm_reactants is the total rest mass before (u), Σm_products the total rest mass after (u). Data booklet: E = Δmc² and 1 u = 931.5 MeV c⁻² are both printed. Validity: rest masses only; neutrons on both sides must be counted. Common misuse: forgetting the extra 2 or 3 neutrons on the product side. Sanity check: for U-235 fission Δm ≈ 0.2 u, giving 0.2 × 931.5 ≈ 190 MeV.

23. EquationSL & HLData booklet: Yes

    State the conversion between electronvolts, MeV and joules, and apply it to a fission event.

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    1 eV = 1.60 × 10⁻¹⁹ J, so 1 MeV = 1.60 × 10⁻¹³ J. E(J) = E(eV) × e, where e = 1.60 × 10⁻¹⁹ C is the elementary charge. Data booklet: e is printed on the constants page. Validity: the electronvolt is the work done accelerating one elementary charge through 1 V. Common misuse: dividing by e instead of multiplying when going from eV to J. Sanity check: one fission releases 200 MeV = 200 × 1.60 × 10⁻¹³ = 3.2 × 10⁻¹¹ J — tiny individually, which is why huge fission rates are needed for megawatt outputs.

24. EquationSL & HLData booklet: No – derive

    Give the equation for the number of fissions per second needed for a given reactor power.

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    R = P/E, where R is the fission rate (fissions per second, s⁻¹), P is the thermal power output (W = J s⁻¹) and E is the energy released per fission (J). Data booklet: not printed — derive from power = energy per event × rate. Validity: requires E in joules, so convert from MeV first. Common misuse: using the ELECTRICAL output when the question quotes an efficiency — divide by η first to get thermal power. Sanity check: a 1000 MW thermal reactor needs R = 1.0 × 10⁹/3.2 × 10⁻¹¹ = 3.1 × 10¹⁹ fissions per second.

25. EquationSL & HLData booklet: Yes

    Give the equation linking number of nuclei, amount of substance and mass of fuel.

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    n = N/N_A and m = nM, combined as N = mN_A/M. N is the number of nuclei; n is the amount of substance (mol); N_A = 6.02 × 10²³ mol⁻¹; m is the mass of fuel (g when M is in g mol⁻¹); M is the molar mass, 235 g mol⁻¹ for U-235. Data booklet: n = N/N_A is printed. Common misuse: mixing kg with g mol⁻¹ — a factor of 1000 error. Sanity check: 1.0 kg of U-235 contains (1000/235) × 6.02 × 10²³ = 2.6 × 10²⁴ nuclei.

26. EquationSL & HLData booklet: No – derive

    Give the equation for the mass of fuel consumed per unit time in a reactor.

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    Rate of fuel use = (P/E) × (M/N_A), giving mass per second in grams when M is in g mol⁻¹. P is thermal power (W), E is energy per fission (J), M is molar mass (g mol⁻¹), N_A = 6.02 × 10²³ mol⁻¹. Data booklet: not printed — build it from R = P/E and N = mN_A/M. Common misuse: quoting the answer per second when the question asked per year — multiply by 3.15 × 10⁷ s. Sanity check: 3.1 × 10¹⁹ fissions s⁻¹ × 235/6.02 × 10²³ ≈ 1.2 × 10⁻² g s⁻¹ ≈ 380 kg per year.

27. EquationSL & HLData booklet: No – derive

    Give the equation for the energy density of nuclear fuel and compare it with a fossil fuel.

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    Energy per kilogram = (N_A/M) × E, where M is the molar mass in kg mol⁻¹ (0.235 kg mol⁻¹ for U-235) so that N_A/M is the number of nuclei per kilogram, and E is the energy per fission (J). Data booklet: not printed — derive. Validity: assumes complete fission of the fissile isotope, so real fuel yields far less because it is only 3–5% U-235 and is removed before full burn-up. Common misuse: quoting this figure for natural uranium. Sanity check: 2.6 × 10²⁴ × 3.2 × 10⁻¹¹ ≈ 8 × 10¹³ J kg⁻¹, about two million times the ≈ 3 × 10⁷ J kg⁻¹ of coal.

28. EquationSL & HLData booklet: Yes

    State the efficiency equation as used in reactor energy questions.

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    η = useful output energy/total input energy = P_electrical/P_thermal, a ratio with no unit (often × 100 for a percentage). P_electrical is the electrical power delivered by the generator (W); P_thermal is the rate of energy release in the core (W). Data booklet: printed in the energy section. Validity: the thermal-to-electrical step is limited by the second law, so typical values are 30–40%. Common misuse: taking the quoted "1000 MW power station" as thermal — it is almost always the electrical output. Sanity check: at η = 0.33 a 1000 MW electrical station releases 3000 MW thermal in the core.

29. EquationSL & HLData booklet: No – memorise

    State the neutron multiplication factor and the equation for neutron numbers in successive generations.

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    k = (number of neutrons causing fission in one generation)/(number in the previous generation), and N_g = N₀kᵍ, where N_g is the number of fission neutrons in generation g and N₀ is the initial number. Data booklet: not printed — memorise. Validity: describes an idealised, uniform core. Common misuse: confusing k with the 2–3 neutrons emitted per fission; most of those are lost to leakage, absorption in U-238 and capture in the control rods, so a steady reactor has k = 1.000 exactly. Sanity check: k = 1.01 for 100 generations gives 1.01¹⁰⁰ ≈ 2.7, a doubling of power.

30. EquationSL & HLData booklet: No – memorise

    State the conservation rules that a fission equation must satisfy, using U-235 as the example.

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    ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n. Nucleon number A is conserved: 235 + 1 = 236 = 141 + 92 + 3. Proton number Z (charge) is conserved: 92 + 0 = 92 = 56 + 36. Mass–energy and momentum are also conserved, the mass difference appearing as kinetic energy of the fragments. Data booklet: not printed — memorise a typical equation. Common misuse: forgetting the incident neutron on the left, which makes the numbers fail to balance by one. Sanity check: always add up A and Z on both sides before writing anything else.

31. EquationSL & HLData booklet: No – derive

    Give the equation for the kinetic energy shared by the two fission fragments and how momentum fixes the split.

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    By conservation of momentum for a nucleus initially almost at rest, m₁v₁ = m₂v₂, and since E_k = p²/2m the energies satisfy E₁/E₂ = m₂/m₁. E₁ and E₂ are the fragment kinetic energies (J or MeV); m₁ and m₂ are the fragment masses (kg or u); p is the common magnitude of momentum (kg m s⁻¹). Data booklet: not printed — derive. Common misuse: assuming the energy splits equally. Sanity check: for Ba-141 and Kr-92 sharing about 165 MeV, the lighter Kr carries the larger share, roughly 165 × 141/233 ≈ 100 MeV.

32. EquationSL & HLData booklet: Yes

    Give the equation for the thermal energy of a neutron in equilibrium with the moderator.

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    E_k = (3/2)k_BT, with k_B = 1.38 × 10⁻²³ J K⁻¹ and T the absolute temperature in kelvin (K); E_k is the average translational kinetic energy in joules. Data booklet: printed. Validity: applies once the neutron has been thermalised by many elastic collisions and shares the moderator's temperature. Common misuse: using Celsius for T. Sanity check: at T = 293 K, E_k = 1.5 × 1.38 × 10⁻²³ × 293 = 6.1 × 10⁻²¹ J = 0.038 eV, of the order of the quoted 0.025 eV for a thermal neutron.

33. Graph/diagramSL & HL

    Describe the graph of binding energy per nucleon against nucleon number.

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    Axes: x is nucleon number A (0 to about 240, no unit), y is binding energy per nucleon (MeV per nucleon, 0 to about 9). Shape: a steep rise from ¹H at 0 through ²H (1.1) and ⁴He (7.1, a local peak), a broad maximum of about 8.8 MeV near iron-56, then a slow decline to about 7.6 MeV at uranium-235. Extracting a quantity: total binding energy = (value read off) × A; the energy released in any reaction is the total binding energy of the products minus that of the reactants. Fission of nuclei to the right of the peak and fusion of nuclei to the left both move the products UPWARDS on this curve, which is why both release energy.

34. Graph/diagramSL & HL

    Describe how the binding energy per nucleon curve is used to obtain the 200 MeV released per fission.

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    Read the y-value at A = 235 (about 7.6 MeV per nucleon) and at the fragment mass numbers A ≈ 141 and A ≈ 92 (both about 8.5 MeV per nucleon). The vertical rise is Δ(E_b/A) ≈ 0.9 MeV per nucleon. Multiply by the total number of nucleons, 236: energy released ≈ 236 × 0.9 ≈ 210 MeV, quoted as "about 200 MeV". Extracting a quantity: the energy released is the AREA-free product (height difference) × (nucleon number), not a gradient or an area. If the fragments were chosen closer in A to uranium the vertical rise would be smaller and less energy would be released.

35. Graph/diagramSL & HL

    Describe the labelled diagram of a thermal fission reactor expected in an IB answer.

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    The core is drawn as a block of moderator (graphite or water) pierced by an array of fuel rods (enriched UO₂) with control rods (boron or cadmium) inserted between them from above. The coolant flows through the core and out to a heat exchanger, where a separate water circuit boils to steam, drives a turbine and generator, and is condensed and returned. Thick concrete shielding and a containment vessel surround the core. Extracting a quantity: the thermal power is set by the fission rate in the core; the electrical power is that multiplied by the efficiency. Withdrawing control rods raises k and hence the steady power level.

36. Graph/diagramSL & HL

    Describe the branching tree diagram of a chain reaction and what it shows.

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    The diagram shows one neutron entering a U-235 nucleus, which splits into two fragments plus (say) three neutrons; each of those neutrons enters another U-235 nucleus, giving nine neutrons in the next generation, and so on. Axes are not used — the horizontal direction is generation number. Extracting a quantity: the number of fissions in generation g is kᵍ, so a plot of the number of fissions against g is exponential and a plot of ln(number) against g is a straight line of gradient ln k through ln N₀. If k is reduced to 1 by inserting control rods the tree becomes a single unbranched chain and the power is constant; if k < 1 the tree dies out.

37. Graph/diagramSL & HL

    Describe the graph of reactor power (or neutron number) against time for k < 1, k = 1 and k > 1.

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    Axes: x is time (s), y is neutron number or thermal power (W), often on a logarithmic y-scale. Shape: for k > 1 an exponential rise, for k = 1 a horizontal line, for k < 1 an exponential decay towards zero. Extracting a quantity: on a logarithmic y-axis all three are straight lines, and the gradient is positive, zero or negative in proportion to ln k, so k can be found from the gradient and the generation time. Inserting control rods further reduces k and steepens the downward line; withdrawing them steepens the upward one. Note that after shutdown the power falls to a small non-zero decay-heat level, not to zero.

38. Graph/diagramSL & HL

    Describe the graph of neutron number N against proton number Z and explain why fission fragments are β⁻ emitters.

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    Axes: x is proton number Z (0–100), y is neutron number N (0–150). Shape: the band of stability follows N = Z for light nuclei then curves above it, reaching N/Z ≈ 1.5 at uranium, because extra neutrons are needed to dilute the long-range proton repulsion. Extracting a quantity: read N/Z at any point as the gradient of the line from the origin. Uranium splits into fragments that inherit its high N/Z, so they land ABOVE the band for their new Z — neutron-rich — and decay by β⁻ (n → p + e⁻ + ν̄_e), moving diagonally down-right towards the band, usually through several successive decays.

39. Graph/diagramSL & HL

    Describe the distribution of the ≈ 200 MeV released per fission as a labelled bar chart.

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    The bars, in MeV, are: kinetic energy of the two fission fragments ≈ 165 (by far the largest), prompt γ-rays ≈ 7, kinetic energy of the prompt neutrons ≈ 5, β particles from fragment decay ≈ 7, delayed γ-rays ≈ 6, and antineutrinos ≈ 10. Extracting a quantity: the usable thermal power is the total minus the antineutrino bar, since antineutrinos escape the reactor entirely — about 190 MeV per fission. The fragment bar is the one that matters for reactor engineering: the fragments stop within microns of their origin, so essentially all of that energy appears immediately as thermal energy in the fuel rod and must be removed by the coolant.

40. Graph/diagramSL & HL

    Describe the graph of fission probability (cross-section) against neutron kinetic energy for U-235.

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    Axes: x is neutron kinetic energy, plotted logarithmically from about 0.01 eV to 10 MeV; y is the fission cross-section in barns, also logarithmic, from about 1 to 1000. Shape: a high value of several hundred barns at thermal energies, falling steadily as roughly 1/√E through a region of sharp resonance spikes near 1–100 eV, down to a few barns at MeV energies. Extracting a quantity: read the cross-section at 0.025 eV (≈ 580 barns) and at 2 MeV (≈ 1 barn) — a factor of several hundred. This is the quantitative justification for a moderator: neutrons born fast at the right of the graph must be moved to the far left before they are likely to cause fission.

41. Concept/explainSL & HL

    Explain why induced fission of uranium-235 is brought about by thermal (slow) neutrons rather than by fast neutrons.

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    • A neutron is uncharged, so it feels no Coulomb repulsion and can reach the nucleus at any speed
    • The capture probability (cross-section) rises sharply as neutron speed falls, because a slow neutron spends far longer within range of the nuclear force
    • Capture forms an excited U-236 nucleus; the binding energy released on capture, about 6.5 MeV, already exceeds the roughly 6.2 MeV needed to deform U-236 past its fission barrier, so no extra kinetic energy is needed
    • The excited nucleus oscillates like a charged liquid drop until electrostatic repulsion overcomes the surface nuclear force and it splits
    • Fission neutrons are born fast (~2 MeV), so a moderator is needed for the next generation. Exam tip: incomplete answers say slow neutrons "have more time to be attracted in", omitting cross-section and the excitation energy of U-236.
42. Concept/explainSL & HL

    Using the binding energy per nucleon curve, explain why about 200 MeV is released when a U-235 nucleus undergoes fission.

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    • Binding energy per nucleon rises from about 7.6 MeV for uranium to a peak near 8.8 MeV at iron, then falls again
    • The fission fragments (A ≈ 90–145) lie closer to the peak, at about 8.5 MeV per nucleon
    • The products are therefore more tightly bound, so energy is released: ΔE ≈ 236 × (8.5 − 7.6) ≈ 2 × 10² MeV
    • Equivalently the total rest mass of the products is less than that of U-235 plus the neutron, and this mass defect appears as energy through ΔE = Δmc²
    • Most of it (~165 MeV) is kinetic energy of the two fragments, the rest neutron kinetic energy, prompt γ-rays and later β-decay. Exam tip: students subtract the wrong way round, or multiply by the difference in nucleon number instead of by the total 236 nucleons.
43. Concept/explainSL & HL

    Explain what is meant by a chain reaction and by critical mass, and state the condition for a reactor to run at steady power.

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    • Each fission of U-235 releases on average 2 to 3 fast neutrons, so one fission can induce further fissions in a self-sustaining chain reaction
    • The multiplication factor k is the mean number of neutrons from one generation that go on to cause fission in the next
    • k < 1 sub-critical, the reaction dies out; k = 1 critical, steady power; k > 1 super-critical, power rises exponentially
    • Critical mass is the minimum mass of fissile material, for a given shape and enrichment, for which k = 1
    • Below it too many neutrons escape through the surface, since surface-to-volume ratio is larger for a smaller sample; a neutron reflector lowers the critical mass
    • Control rods hold k at exactly 1. Exam tip: defining critical mass as "the mass needed to start a reaction" omits neutron leakage, which is the reason a minimum mass exists.
44. Concept/explainSL & HL

    Explain the function of the moderator in a thermal fission reactor and justify the choice of material.

    Show answer

    • Neutrons released by fission are fast, around 2 MeV, at which the fission cross-section of U-235 is small
    • The moderator slows them to thermal energies (~0.025 eV) so that they are readily captured and sustain the chain reaction
    • Slowing occurs by repeated elastic collisions between neutrons and moderator nuclei
    • The fractional kinetic energy transferred in an elastic collision is greatest when the two masses are similar, so light nuclei are used: water, heavy water or graphite
    • The material must also have a low neutron-absorption cross-section, or it would remove the neutrons it is meant to slow; ordinary water absorbs enough to require enriched fuel, while heavy water and graphite permit natural uranium. Exam tip: the commonest error is to say the moderator "absorbs neutrons" or "controls the reaction" — that is the control rod; the moderator only removes kinetic energy.
45. Concept/explainSL & HL

    Explain how control rods regulate the power output of a nuclear reactor.

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    • Control rods are made of boron or cadmium, nuclei with very large absorption cross-sections for thermal neutrons
    • Inserting them further into the core removes neutrons before they can cause fission, reducing the multiplication factor k
    • Withdrawing them raises k, so the fission rate and hence the thermal power rise
    • Steady operation requires the rods to hold k at exactly 1; in an emergency they are dropped fully in to make the core strongly sub-critical (a scram)
    • Control is feasible because about 0.65% of fission neutrons are delayed by seconds from decaying fission products, stretching the effective generation time and slowing the core's response to a change in k. Exam tip: answers confuse absorbing with slowing — state explicitly that control rods capture neutrons and change k, whereas the moderator changes neutron speed.
46. Concept/explainSL & HL

    Outline the sequence of energy transfers in a fission power station, from the fuel rod to the electrical output.

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    • A thermal neutron induces fission in U-235; the mass defect appears mainly as kinetic energy of the two charged fragments
    • The fragments travel less than a millimetre before being stopped in the fuel, so their kinetic energy becomes internal (thermal) energy of the fuel rod
    • A coolant (pressurised water, CO₂ or liquid sodium) is pumped through the core and carries this energy away by forced convection
    • In the heat exchanger it transfers energy to a separate water circuit, raising steam while keeping radioactive coolant out of the turbine hall
    • The steam drives a turbine and hence a generator, producing electrical energy
    • Waste energy is rejected in the condenser, so overall efficiency is only about 30–40%. Exam tip: students omit the safety role of the heat exchanger and forget that the second law limits thermal-to-electrical conversion.
47. Concept/explainSL & HL

    Explain why natural uranium usually has to be enriched before use in a reactor, and what enrichment involves.

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    • Natural uranium is 99.3% U-238 and only 0.7% U-235, and only U-235 fissions readily with thermal neutrons
    • U-238 preferentially captures neutrons of intermediate energy (resonance capture) without fissioning, competing for the neutrons that sustain the chain
    • With ordinary water as moderator, which itself absorbs neutrons, k cannot reach 1 at 0.7%, so fuel is enriched to about 3–5% U-235
    • Enrichment exploits the tiny mass difference between UF₆ molecules containing the two isotopes, using gas centrifuges or gaseous diffusion; chemical separation is impossible because isotopes are chemically identical
    • The same technology taken to ~90% gives weapons-grade material, which is why enrichment is the central proliferation concern. Exam tip: incomplete answers say enrichment "makes the uranium more radioactive"; it changes the isotopic ratio, not the decay properties of either isotope.
48. Concept/explainSL & HL

    Explain why the fragments produced by fission are radioactive, and what type of decay they undergo.

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    • Heavy nuclei are stable only with a large neutron excess: N/Z is about 1.55 for uranium but only about 1.3 for stable nuclei of mass 90–140
    • When U-235 splits, the fragments inherit that high neutron-to-proton ratio, so they lie above the band of stability
    • They therefore decay by β⁻ emission, a neutron becoming a proton, an electron and an antineutrino, usually through a chain of several decays with γ emission as each daughter de-excites
    • Some fragments also emit delayed neutrons, which are essential for reactor control
    • Half-lives range from fractions of a second to millions of years, so spent fuel stays hazardous for a very long time and the core still produces decay heat after shutdown. Exam tip: many students state α decay; heavy fission fragments are neutron-rich, so β⁻ decay is the rule.
49. Concept/explainSL & HL

    Discuss the problem of high-level nuclear waste from fission reactors.

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    • Spent fuel contains neutron-rich fission fragments that are intensely radioactive and continue to generate decay heat for years after removal
    • It also contains transuranic nuclides such as plutonium-239 (half-life 2.4 × 10⁴ years) formed by neutron capture in U-238, giving α-active waste hazardous for 10⁵ years or more
    • Short-term management is storage in water-filled cooling ponds, which shield the radiation and remove decay heat
    • Long-term proposals are vitrification in borosilicate glass, encapsulation in corrosion-resistant containers and burial in a deep, geologically stable repository
    • The volume is small compared with fossil-fuel waste, but the containment time exceeds the lifetime of any human institution, and Pu-239 is weapons-usable. Exam tip: "it is dangerous and must be buried" scores little; the mark scheme wants half-life, decay heat, volume and proliferation as separate points.
50. Concept/explainSL & HL

    Explain what a reactor meltdown is and how reactor design reduces the risk, referring to a named incident.

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    • Even after control rods are fully inserted and fission stops, decaying fission products release decay heat at several per cent of full power
    • If coolant flow or inventory is lost, this heat cannot be removed and the fuel and cladding melt: a meltdown
    • Molten fuel can breach the pressure vessel, and hot zirconium cladding reacts with steam to release hydrogen, which may explode and disperse volatile fission products such as I-131 and Cs-137
    • Design responses are redundant, passively driven emergency core cooling, a thick steel and concrete containment building, and a negative temperature coefficient
    • At Fukushima (2011) a tsunami disabled the backup generators, so cooling was lost; at Chernobyl (1986) a positive void coefficient and no full containment allowed a power excursion. Exam tip: this is a risk-assessment question — weigh probability against consequence rather than asserting that nuclear power is unsafe.
51. Concept/explainSL & HL

    Compare the generation of electrical energy by nuclear fission with the burning of fossil fuels.

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    • Energy density: fission of U-235 gives about 8 × 10¹³ J kg⁻¹ against 3 × 10⁷ J kg⁻¹ for coal, a factor of order 10⁶, so mining and transport volumes are tiny
    • Emissions: fission releases no CO₂, SO₂ or particulates in operation, though mining and construction do
    • Waste: fossil waste is huge in volume but short-lived; nuclear waste is small in volume but radioactive for 10⁵ years
    • Both are non-renewable and both give reliable base-load power, unlike wind and solar
    • Risks unique to fission are meltdown and diversion of enriched uranium or plutonium to weapons
    • Capital and decommissioning costs are far higher for nuclear. Exam tip: "compare" requires similarities as well as differences, and each point must be genuinely two-sided rather than a list of nuclear features.
52. Concept/explainSL & HL

    Compare nuclear fission with nuclear fusion as sources of energy.

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    • Both release energy because the products have a greater binding energy per nucleon than the reactants, so both involve a mass defect and ΔE = Δmc²
    • Fission splits a nucleus heavier than iron; fusion joins nuclei lighter than iron, so they sit on opposite sides of the binding-energy peak
    • Energy per reaction is larger for fission (~200 MeV against ~17.6 MeV for D–T), but energy per unit mass is several times greater for fusion
    • Fusion fuel (deuterium from sea water, lithium for tritium) is effectively unlimited, whereas fissile U-235 is scarce
    • Fusion products are far less hazardous and no runaway chain reaction is possible
    • Fusion needs temperatures of order 10⁸ K and confinement not yet giving sustained net gain. Exam tip: students claim fusion "produces no radiation at all"; the 14 MeV neutrons activate the reactor structure.
53. Concept/explainSL & HL

    Explain how the conservation rules are applied to a fission equation, and what they tell you about the energy released.

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    • Nucleon number is conserved: totals of A on each side are equal, e.g. 235 + 1 = 141 + 92 + 3(1) = 236
    • Proton number (charge) is conserved: 92 + 0 = 56 + 36 + 0
    • Momentum is conserved: the fragments recoil in opposite directions in the frame of the parent nucleus, so the lighter fragment carries the greater kinetic energy
    • Mass–energy is conserved overall, but rest mass is not: the total rest mass of the products is less than that of the reactants and the deficit Δm appears as kinetic energy and photon energy, ΔE = Δmc²
    • Conserving nucleon number therefore does not imply conserving mass, because the mass of a bound nucleon depends on the binding energy of its nucleus. Exam tip: the secure wording is that rest mass is transferred to other forms of energy, with total mass–energy conserved.
54. Concept/explainSL & HL

    Explain why a nuclear reactor cannot explode like a fission bomb, even though both rely on a chain reaction.

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    • A bomb requires uranium enriched to about 90% U-235 (or plutonium) so that fast neutrons alone sustain a very rapid super-critical chain; reactor fuel is only 3–5% enriched and can never reach that condition
    • In a reactor the chain relies on thermalised neutrons, so if the fuel overheats and the moderator boils away the neutrons stay fast and the chain shuts itself down — a negative void coefficient
    • Fuel is dispersed in rods separated by moderator and coolant, never assembled into one compact super-critical mass
    • Control depends on delayed neutrons, whose seconds-long timescale makes the power response manageable
    • Reactor accidents are therefore steam or hydrogen explosions and meltdowns driven by decay heat, not nuclear explosions. Exam tip: citing "control rods" alone is not enough; the decisive points are low enrichment and dependence on thermalised and delayed neutrons.
55. Worked problemSL & HL

    Complete the fission reaction ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x ¹₀n and determine x.

    Show answer

    Principle: nucleon number A and proton number Z are separately conserved. Proton number: 92 + 0 = 56 + 36 + 0, so 92 = 92, consistent for any x since neutrons are uncharged. Nucleon number: 235 + 1 = 141 + 92 + x(1), so 236 = 233 + x and x = 3. The reaction is ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n, and since 3 > 1 a chain reaction is possible. Check/Trap: the incident neutron must be included on the left — omitting it gives x = 2, the standard Paper 1 distractor. Check Z first, as it immediately eliminates options with the wrong daughter charge.

56. Worked problemSL & HLData booklet: Yes

    For ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n, masses are U-235 = 235.0439 u, Ba-141 = 140.9144 u, Kr-92 = 91.9262 u, n = 1.0087 u. Determine the energy released, in MeV and in J.

    Show answer

    Principle: ΔE = Δmc², with 1 u ≡ 931.5 MeV c⁻². Initial mass = 235.0439 + 1.0087 = 236.0526 u. Final mass = 140.9144 + 91.9262 + 3(1.0087) = 235.8667 u. Mass defect Δm = 0.1859 u. Energy released E = 0.1859 × 931.5 = 173 MeV. In joules E = 173.2 × 1.60 × 10⁻¹³ = 2.77 × 10⁻¹¹ J. Cross-check in SI: Δm = 0.1859 × 1.66 × 10⁻²⁷ = 3.09 × 10⁻²⁸ kg, and Δmc² = 3.09 × 10⁻²⁸ × (3.00 × 10⁸)² = 2.78 × 10⁻¹¹ J. Check/Trap: subtract three neutrons, not one; and because this is a difference of near-equal numbers, keep every digit until the final step or the significant figures collapse.

57. Worked problemSL & HL

    The binding energy per nucleon of U-235 is 7.6 MeV and that of the fission fragments averages 8.5 MeV. Estimate the energy released per fission.

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    Principle: the energy released equals the increase in total binding energy of the nucleons involved. Increase per nucleon = 8.5 − 7.6 = 0.9 MeV. Number of nucleons = 235 + 1 = 236. Energy released E = 236 × 0.9 = 212 ≈ 2 × 10² MeV. Since the two binding energies are given to 2 s.f., their difference carries only 1 s.f., so 200 MeV is the honest answer. Converting, E = 200 × 1.60 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Check/Trap: quoting 212.4 MeV is over-precise and is penalised. Multiply by the total nucleon number, not by any difference in nucleon number, and take products minus reactants so the sign shows a release.

58. Worked problemSL & HL

    A reactor has a thermal power output of 2.4 GW. Each fission releases 3.2 × 10⁻¹¹ J. Determine the number of fissions occurring each second.

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    Principle: power = energy per fission × fission rate, so R = P/E. P = 2.4 GW = 2.4 × 10⁹ J s⁻¹. R = (2.4 × 10⁹)/(3.2 × 10⁻¹¹) = 7.5 × 10¹⁹ fissions per second. Sanity check: this consumes 7.5 × 10¹⁹ × 235 × 1.66 × 10⁻²⁷ = 2.9 × 10⁻⁵ kg of U-235 per second, about 2.5 kg per day — plausible for a large reactor. Check/Trap: use the thermal (core) power, not the electrical output; if given electrical power and an efficiency, divide by the efficiency first. Watch the prefix: GW = 10⁹ W, not 10⁶ W.

59. Worked problemSL & HL

    A nuclear power station delivers 1.0 GW of electrical power at an overall efficiency of 35%. Each U-235 fission releases 200 MeV. Estimate the mass of U-235 consumed in one year.

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    Principle: work back from electrical output to thermal energy, then to number of fissions, then to mass. Thermal power P_th = P_elec/η = (1.0 × 10⁹)/0.35 = 2.86 × 10⁹ W. Energy per year = 2.86 × 10⁹ × 3.15 × 10⁷ = 9.0 × 10¹⁶ J. Energy per fission = 200 × 1.60 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Number of fissions N = (9.0 × 10¹⁶)/(3.2 × 10⁻¹¹) = 2.8 × 10²⁷. Mass of one U-235 atom = 235 × 1.66 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg. Mass consumed m = 2.8 × 10²⁷ × 3.90 × 10⁻²⁵ = 1.1 × 10³ kg, about a tonne a year. Check/Trap: divide by the efficiency, never multiply — the core must release more energy than is delivered. The mass of fuel loaded is far larger, since only a few per cent is U-235.

60. Worked problemSL & HL

    Show that fission of U-235 releases of the order of 10⁶ times more energy per kilogram than burning coal, for which the specific energy is 3.0 × 10⁷ J kg⁻¹.

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    Principle: find the number of U-235 nuclei in 1 kg, then multiply by the energy per fission. Mass of one U-235 atom = 235 × 1.66 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg. Number per kilogram n = 1.00/(3.90 × 10⁻²⁵) = 2.56 × 10²⁴. Energy per fission = 200 × 1.60 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Specific energy = 2.56 × 10²⁴ × 3.2 × 10⁻¹¹ = 8.2 × 10¹³ J kg⁻¹. Ratio = (8.2 × 10¹³)/(3.0 × 10⁷) = 2.7 × 10⁶, of order 10⁶ as required. Check/Trap: this is for pure U-235; for 3% enriched fuel the specific energy as loaded is about 2 × 10¹² J kg⁻¹, still 10⁵ times coal. Using N_A/0.235 kg mol⁻¹ gives the same nuclei per kilogram.

61. Worked problemSL & HLData booklet: No – derive

    In the fission ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n the two fragments share 165 MeV of kinetic energy. Assuming they recoil back-to-back from a stationary nucleus, determine the kinetic energy of each.

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    Principle: momentum conservation gives equal and opposite momenta p, and E_k = p²/2m, so E_k ∝ 1/m — the lighter fragment takes the larger share. Hence E_Ba/E_Kr = m_Kr/m_Ba = 92/141. Let E_Ba = 92x and E_Kr = 141x, so 233x = 165 MeV and x = 0.708 MeV. E_Ba = 92 × 0.708 = 65 MeV; E_Kr = 141 × 0.708 = 1.0 × 10² MeV. Check: 65 + 100 = 165 MeV ✓, and the lighter krypton fragment does carry more. Check/Trap: the fragments neither share the energy equally nor have equal speeds — only their momenta are equal in magnitude. Neglecting the neutrons' momenta is a stated approximation; they actually carry about 5 MeV.

62. Worked problemSL & HLData booklet: Yes

    A ⁹²Kr fission fragment is produced with kinetic energy 1.0 × 10² MeV. Estimate its speed.

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    Principle: E_k = ½mv², so v = √(2E_k/m), valid while v is well below c. Mass m = 92 × 1.66 × 10⁻²⁷ = 1.53 × 10⁻²⁵ kg. Energy E_k = 1.0 × 10² × 1.60 × 10⁻¹³ = 1.60 × 10⁻¹¹ J. v = √(2 × 1.60 × 10⁻¹¹ / 1.53 × 10⁻²⁵) = √(2.09 × 10¹⁴) = 1.4 × 10⁷ m s⁻¹. Sanity check: v/c = 0.048, under 5% of the speed of light, so the non-relativistic formula is justified (γ ≈ 1.001). Check/Trap: convert MeV to joules before using SI masses, and use the mass of the whole fragment in kilograms — students often substitute 92 u as if u were a kilogram, or use a single nucleon mass.

63. Worked problemSL & HLData booklet: No – memorise

    A reactor is operating with a multiplication factor k = 1.02. If the fission rate is initially 5.0 × 10¹⁹ s⁻¹, determine the fission rate after 100 generations of neutrons.

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    Principle: each generation multiplies the number of fissions by k, so R_n = R_0 k^n — geometric growth. R_100 = 5.0 × 10¹⁹ × (1.02)^100. Evaluate the factor: (1.02)^100 = e^(100 ln 1.02) = e^(100 × 0.01980) = e^1.980 = 7.24. R_100 = 5.0 × 10¹⁹ × 7.24 = 3.6 × 10²⁰ fissions per second. The power rises about sevenfold in 100 generations; with a prompt generation time of ~10⁻³ s that would take 0.1 s, which is why delayed neutrons, stretching the effective generation time to ~0.1 s, are essential. Check/Trap: k multiplies per generation, it is not a percentage added 100 times. k = 1 gives constant power, k < 1 exponential decay.

64. Worked problemSL & HLData booklet: No – derive

    A 2.0 MeV neutron is moderated by head-on elastic collisions with carbon-12 nuclei. Determine the fraction of kinetic energy lost per collision and estimate the number of collisions needed to reach 0.025 eV.

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    Principle: for a head-on elastic collision of mass m with a stationary mass M, the rebound speed is v' = |(m − M)/(m + M)|v. With M = 12m: v'/v = 11/13 = 0.846. Fraction of kinetic energy remaining = 0.846² = 0.716, so 28% is lost per collision. After n collisions E_n = E_0(0.716)^n. Require (0.716)^n = 0.025/(2.0 × 10⁶) = 1.25 × 10⁻⁸, so n = ln(1.25 × 10⁻⁸)/ln(0.716) = (−18.2)/(−0.334) = 55 collisions. Check/Trap: this is an idealised minimum — real collisions occur at all angles, so carbon needs roughly 110, while hydrogen (M = m) can in principle stop a neutron in one head-on hit, which is why water moderates so effectively. Convert both energies to the same unit before taking the ratio.

65. Worked problemSL & HLData booklet: Yes

    Estimate the speed of a thermal neutron in a moderator at 293 K, given that the mass of a neutron is 1.67 × 10⁻²⁷ kg.

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    Principle: a neutron in thermal equilibrium has mean translational kinetic energy E_k = (3/2)k_B T, and E_k = ½mv², so v = √(3k_B T/m). E_k = 1.5 × 1.38 × 10⁻²³ × 293 = 6.07 × 10⁻²¹ J, i.e. (6.07 × 10⁻²¹)/(1.60 × 10⁻¹⁹) = 0.038 eV, consistent with the usual quoted 0.025 eV (which is k_B T rather than (3/2)k_B T). Speed v = √(2 × 6.07 × 10⁻²¹ / 1.67 × 10⁻²⁷) = √(7.27 × 10⁶) = 2.7 × 10³ m s⁻¹. Check/Trap: a 2 MeV fission neutron travels at about 2 × 10⁷ m s⁻¹, so moderation cuts the speed by ~10⁴ and the energy by ~10⁸. Never use the Celsius temperature in a kinetic-theory expression.

66. Worked problemSL & HLData booklet: Yes

    A reactor operates continuously at a thermal power of 3.0 GW. Determine the total mass equivalent of the energy released in one day.

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    Principle: the energy released comes from a decrease in rest mass, so Δm = ΔE/c². Energy in one day ΔE = Pt = 3.0 × 10⁹ × (24 × 3600) = 3.0 × 10⁹ × 8.64 × 10⁴ = 2.59 × 10¹⁴ J. Mass equivalent Δm = (2.59 × 10¹⁴)/(3.00 × 10⁸)² = (2.59 × 10¹⁴)/(9.00 × 10¹⁶) = 2.9 × 10⁻³ kg, about 2.9 g per day. Check/Trap: 2.9 g is the mass that disappears, not the fuel consumed — roughly 3 kg of U-235 must fission each day, because the mass defect is only about 0.08% of the reacting mass. Confusing the two is the standard error here.

67. Worked problemSL & HLData booklet: No – memorise

    Show that only about 0.08% of the rest mass involved is transferred to other forms in a U-235 fission, and compare with a chemical reaction releasing 4 eV per molecule of relative molecular mass 30.

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    Principle: fractional mass loss = Δm/m_total. For fission Δm ≈ 0.186 u out of 236.05 u, so Δm/m = 0.186/236.05 = 7.9 × 10⁻⁴, i.e. 0.079% ≈ 0.08%. For the chemical reaction, the rest energy of the molecule = 30 × 931.5 = 2.79 × 10⁴ MeV = 2.79 × 10¹⁰ eV, so the fractional loss = 4/(2.79 × 10¹⁰) = 1.4 × 10⁻¹⁰, about 1.4 × 10⁻⁸ %. Ratio ≈ (7.9 × 10⁻⁴)/(1.4 × 10⁻¹⁰) ≈ 6 × 10⁶. Check/Trap: this is why nuclear fuels have specific energies ~10⁶ times chemical fuels and why the mass change in a chemical reaction is unmeasurable. Use 931.5 MeV per u consistently and never mix u with kg in one ratio.

68. Exam technique/trapSL & HL

    Explain how the command terms state, outline, explain, discuss and estimate change what is required in a fission question.

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    State: give the fact with no reasoning — "a moderator slows neutrons" earns the mark. Outline: a brief account of the main points, usually 2 marks, e.g. the function of control rods plus the material. Explain: give the mechanism — for the moderator, why slow neutrons are needed (larger fission cross-section) and how slowing happens (elastic collisions with light nuclei). Discuss: a balanced argument with both sides, as in nuclear versus fossil-fuel generation; a one-sided list caps the mark. Estimate: a calculation to 1 significant figure with the assumptions stated. Deduce: reach a conclusion showing the reasoning, e.g. x = 3 from nucleon-number conservation. Trap: students write a full explanation for a 1-mark "state" and lose time; match the length to the command term and the marks available.

69. Exam technique/trapSL & HLData booklet: Yes

    Identify the unit-conversion traps in fission energy calculations and give the safe procedure.

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    The trap: four currencies — u, MeV c⁻², MeV and J — appear in one question and get mixed. Why students fall for it: 1 u = 1.66 × 10⁻²⁷ kg and 1 u ≡ 931.5 MeV c⁻² look interchangeable, and 1 eV = 1.60 × 10⁻¹⁹ J is applied in the wrong direction. Correct approach: decide at the start to work either in nuclear units (u and MeV, via Δm in u × 931.5) or in SI (kg and J, via Δm in kg × c²), and never switch mid-question. MeV to J: multiply by 1.60 × 10⁻¹³; J to MeV: divide. Check magnitude — one fission is about 200 MeV ≈ 3.2 × 10⁻¹¹ J, so 10⁻¹⁹ J or 10⁸ MeV is wrong by orders of magnitude. Remember 931.5 MeV c⁻² is a mass, needing c² only to become an energy.

70. Exam technique/trapSL & HL

    Explain the distinction the mark scheme looks for between the moderator, the control rods and the coolant.

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    The trap: all three sit in the core, so students blur them and write that the moderator "controls" the reaction or that control rods "slow" neutrons. Why: all three do reduce the reaction rate in some loose sense, and diagrams show them side by side. Correct approach: tie each to its distinct physics. Moderator (water, heavy water, graphite): light nuclei, elastic collisions, reduces neutron kinetic energy so the fission cross-section rises — it makes the chain reaction possible. Control rods (boron, cadmium): large capture cross-section, absorb neutrons and change k — they set the power. Coolant (water, CO₂, liquid sodium): removes thermal energy to the heat exchanger and does nothing to the neutrons. In a PWR water is both moderator and coolant, so say which role you mean. "Outline the function" needs the action and its purpose, not just the material.

71. Exam technique/trapSL & HLData booklet: Yes

    Give the wording the mark scheme accepts for the origin of the energy released in fission, and the phrases that lose marks.

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    The trap: writing "mass is converted into energy" as the whole explanation. Why students fall for it: E = mc² is usually taught with that slogan. Correct approach: state that the total rest mass of the products is less than that of the reactants, and that this mass defect Δm is transferred to kinetic energy of the fragments and neutrons and to photon energy, via ΔE = Δmc², with mass–energy conserved overall. Add the binding-energy version: the fragments have greater binding energy per nucleon than U-235, so energy is released. Phrases that lose marks: "energy is created", "the nucleus loses nucleons", and "binding energy is released from the nucleus" — binding energy is not stored energy but the energy needed to separate the nucleons. A 3-mark "explain" wants binding energy, mass defect and the equation.

72. Exam technique/trapSL & HL

    Explain the precise use of the terms chain reaction, critical mass, critical, sub-critical and super-critical.

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    The trap: treating "critical mass" as the mass needed to start a chain reaction, and calling a working reactor "super-critical". Why: everyday usage of "critical" implies danger. Correct approach: a chain reaction is self-sustaining fission in which neutrons from one fission induce further fissions. The multiplication factor k is the mean number of neutrons per fission that cause another fission. k < 1 sub-critical (power falls), k = 1 critical (steady power, the normal operating state), k > 1 super-critical (power rises). Critical mass is the minimum mass, for a given shape, enrichment and reflector, at which k = 1; it depends on geometry because neutron leakage is a surface effect. Multiple-choice trap: a sphere has the smallest critical mass, having the smallest surface-to-volume ratio. Never say a reactor runs super-critical in steady operation.

73. Exam technique/trapSL & HL

    Outline how to avoid arithmetic and uncertainty errors in a fuel-consumption calculation, such as the mass of U-235 used per year by a 1.0 GW station of efficiency 33%.

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    The trap: multiplying by the efficiency instead of dividing, dropping an SI prefix, and quoting an over-precise answer. Correct approach: write the chain P_elec → P_th = P_elec/η → E = P_th t → N = E/E_fission → m = N m_U, checking units line by line and reasoning that thermal power must exceed electrical power. Uncertainties: convert each to a fractional (percentage) uncertainty, add fractional uncertainties for a product or quotient and absolute uncertainties for a sum or difference. With η = 33 ± 2 % (6%) and P = 1.00 ± 0.02 GW (2%), m carries about 8%, so m = 1.2 × 10³ ± 1 × 10² kg — writing 1163.47 kg is wrong. Quote the final answer to the significant figures of the least precise datum, usually 2, always with a unit.

74. Exam technique/trapSL & HL

    Describe a laboratory model of a chain reaction and critical mass using mousetraps and corks, and evaluate it.

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    Method: set many loaded mousetraps in a transparent box, each holding two corks, so a released trap flings corks that spring further traps. Drop one cork in to start it, film at high frame rate and count releases in successive 0.1 s intervals. Independent variable: number of traps per unit area (analogous to fuel density, hence to k). Dependent variable: growth rate of releases. Controlled: box dimensions, trap type, cork mass, release height. Expect exponential growth above a threshold density and rapid die-out below it, showing why leakage at the walls sets a minimum size. Limitations: traps cannot reset, so fuel is used once; there is no moderator or control analogue; corks are stopped by the walls rather than escaping. Improvements: vary box size at fixed density to reveal the surface-to-volume dependence, and repeat each density five times, plotting means with error bars.

75. Exam technique/trapSL & HL

    Describe an experiment using colliding trolleys to model neutron moderation, and explain how the data are analysed.

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    Method: run a low-friction trolley of fixed mass m (the neutron) into a stationary target trolley of mass M on a level track, using a spring bumper or magnets to make the collision approximately elastic. Measure incident and rebound speeds with two light gates and interrupt cards, repeating for M/m from about 1 to 12. Independent variable: mass ratio M/m; dependent variable: fraction of kinetic energy retained, (v'/v)². Controlled: track level, ramp release height, card width, bumper type. Analysis: compare with ((M − m)/(M + m))² and plot retained fraction against M/m, showing the fraction is least when M ≈ m — so light nuclei make the best moderators. Uncertainties: light-gate timing is random and reduced by repeats; friction and imperfect elasticity are systematic and always lower v', so add error bars and take maximum and minimum gradients. Improvement: use an air track and a motion sensor.

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