IB DP Physics: Nuclear fission and power plants (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic E.4, Fission, from the IB Physics guide for first assessment 2025. The content is the same for SL and HL. Read the study guide and revision notes first, then use the IB DP Physics hub and printable checklist to track progress.

Data used where needed: 1 u = 1.661 × 10⁻²⁷ kg = 931.5 MeV/c²; 1 MeV = 1.602 × 10⁻¹³ J; mass of a nucleus taken as A × 1 u. Other values are given in each question.

Questions

1. A uranium-235 nucleus (proton number 92) absorbs a neutron and splits into strontium-93 (proton number 38), xenon-140 (proton number 54) and x free neutrons. (a) Find x and show that proton number is conserved. [2] (b) State one difference between spontaneous fission and neutron-induced fission. [1]

2. Explain, with reference to the curve of binding energy per nucleon, why energy is released when uranium-235 undergoes fission. [3]

3. In a fission, U-235 + n becomes two fragments plus 2 neutrons. Masses: U-235 235.043930 u; neutron 1.008665 u; the two fragments together 233.856000 u. (a) Calculate the mass difference between reactants and products. [2] (b) Calculate the energy released in MeV. [1] (c) Convert your answer to joules. [1]

4. U-235 splits into two nuclei and two free neutrons. Per nucleon, the binding energies are 7.59 MeV (U-235), 8.41 MeV (the nucleus with nucleon number 134) and 8.63 MeV (the nucleus with nucleon number 99). Estimate the energy released. [4]

5. The multiplication factor k of a reactor is the average number of neutrons from each fission that trigger a further fission. (a) State what k = 1 means for the rate of fission. [1] (b) In a reactor with k = 1.03, how many generations are needed for the number of fissions per generation to increase by a factor of 1.0 × 10⁵? [2] (c) With k = 0.92, calculate the fraction of the original number of fissions per generation left after 20 generations. [2]

6. Each fission of uranium-235 releases 190 MeV. Calculate the energy released if all the nuclei in 1.00 kg of uranium-235 undergo fission, and compare it with the 4.2 × 10⁷ J obtained from 1.00 kg of oil. [4]

7. A power plant delivers 480 MW of electrical power at an overall efficiency of 31%. Each fission supplies 185 MeV. (a) Calculate the thermal power produced. [1] (b) Calculate the power wasted as heat. [1] (c) Calculate the number of fissions per second. [2] (d) Calculate the mass of uranium-235 that undergoes fission in 365 days. [2]

8. A visitor to a power plant makes four statements. Each is wrong. Correct each one and give the true role of the part named. (a) “The moderator absorbs neutrons so the reaction cannot run away.” [2] (b) “Control rods slow neutrons down so that they cause fission.” [2] (c) “The heat exchanger sends steam from the core straight to the turbine.” [2] (d) “Shielding is what stops the chain reaction.” [2]

9. A fast neutron of mass m and speed v collides head-on and elastically with a stationary carbon-12 nucleus of mass 12m. (a) Use conservation of momentum and of kinetic energy to show that the neutron’s speed after the collision is 11v/13. [3] (b) The neutron starts with 2.0 MeV. Find its kinetic energy after five such collisions. [2] (c) Suggest why a moderator made of much lighter nuclei slows neutrons in fewer collisions. [1]

10. The primary coolant of a reactor carries 1.4 GW of thermal power out of the core and warms by 35 K. Take c = 4.2 × 10³ J kg⁻¹ K⁻¹. (a) Calculate the mass of coolant that passes through the core each second. [3] (b) Explain why the energy is passed to a second circuit in a heat exchanger. [1]

11. Spent fuel contains caesium-137 (half-life 30.1 years) and plutonium-239 (half-life 2.41 × 10⁴ years). (a) Calculate the fraction of caesium-137 left after 120 years. [2] (b) Calculate the time for the caesium-137 to fall to 0.1% of its original amount. [2] (c) Calculate the fraction of plutonium-239 left after the time in (b). [1] (d) Evaluate deep geological storage as a way of managing this waste. [3]

Answers

1. (a) Nucleon number: 235 + 1 = 236 = 93 + 140 + x, so x = 3 [1]. Proton number: 38 + 54 = 92, equal to uranium’s 92 [1]. (b) Spontaneous fission needs no trigger, whereas neutron-induced fission begins when the nucleus absorbs a neutron [1]. Examiner insight: A conservation check works best when you show both sums, nucleon number and proton number, not just the answer.

2. Uranium-235 has a lower binding energy per nucleon (about 7.6 MeV) than the mid-mass fragments (about 8.5 MeV) [1]. The fragments are therefore more tightly bound, so total binding energy increases [1]. The products have less mass than the reactants, and the mass difference is released as energy through ΔE = Δm c² [1]. Examiner insight: Say which side of the curve each nucleus sits on and what happens to total binding energy; “fragments are more stable” alone is too vague.

3. (a) Reactants: 235.043930 + 1.008665 = 236.052595 u [1]. Products: 233.856000 + 2 × 1.008665 = 235.873330 u, so Δm = 0.179265 u [1]. (b) E = 0.179265 × 931.5 = 167 MeV [1]. (c) E = 166.99 × 1.602 × 10⁻¹³ = 2.68 × 10⁻¹¹ J [1]. Examiner insight: Include every neutron on both sides of the mass sum. Keep the unrounded energy for the joule conversion.

4. Binding energy of U-235 = 235 × 7.59 = 1783.7 MeV [1]. Fragments: 134 × 8.41 + 99 × 8.63 = 1126.9 + 854.4 = 1981.3 MeV [1]. Free neutrons add nothing [1]. Energy released = 1981.3 − 1783.7 = 198 MeV (about 2 × 10² MeV) [1]. Examiner insight: This is an estimate from rounded values, so say so; a response that gives products minus parent, in that order, avoids a negative answer.

5. (a) The fission rate stays constant (a steady chain reaction) [1]. (b) 1.03ⁿ = 1.0 × 10⁵, so n = ln(1.0 × 10⁵) ÷ ln(1.03) [1] = 389.5, so 390 generations [1]. (c) Fraction = 0.92²⁰ [1] = 0.19 [1]. Examiner insight: Round a count of generations up to a whole number, and state that k < 1 means the chain dies away.

6. Energy per fission = 190 × 1.602 × 10⁻¹³ = 3.04 × 10⁻¹¹ J [1]. Nuclei in 1.00 kg = 1.00 ÷ (235 × 1.661 × 10⁻²⁷) = 2.56 × 10²⁴ [1]. Energy = 2.56 × 10²⁴ × 3.04 × 10⁻¹¹ = 7.80 × 10¹³ J [1]. Ratio to oil = 7.80 × 10¹³ ÷ 4.2 × 10⁷ = 1.9 × 10⁶ times as much [1]. Examiner insight: Convert MeV to joules before multiplying by the number of nuclei, and make the comparison explicit as a ratio.

7. (a) Thermal power = 480 ÷ 0.31 = 1.55 × 10³ MW, 1.55 × 10⁹ W [1]. (b) Wasted = 1548 − 480 = 1.07 × 10³ MW [1]. (c) Energy per fission = 185 × 1.602 × 10⁻¹³ = 2.96 × 10⁻¹¹ J [1]. Fissions per second = 1.548 × 10⁹ ÷ 2.96 × 10⁻¹¹ = 5.2 × 10¹⁹ [1]. (d) Mass per second = 5.22 × 10¹⁹ × 235 × 1.661 × 10⁻²⁷ = 2.04 × 10⁻⁵ kg [1]. Over 365 days (3.1536 × 10⁷ s): 643 kg [1]. Examiner insight: The number of fissions depends on thermal power, not on electrical output; using 480 MW here understates the fuel used.

8. (a) A moderator slows fast neutrons by elastic collisions [1] so they are more likely to cause fission; it should absorb few neutrons [1]. (b) Control rods absorb neutrons [1]; moving them in or out changes k to hold the reaction steady or shut it down [1]. (c) Primary coolant carrying energy from the core passes it to a separate circuit [1] that makes the steam, so the steam does not carry radioactive material from the core [1]. (d) Shielding (thick concrete or steel) absorbs neutrons and gamma rays from the core to protect people [1]; the chain reaction is controlled by the control rods, not the shielding [1]. Examiner insight: Give the correct role and the mechanism for each part. “Controls the reaction” earns less than saying the rods absorb neutrons and so change k.

9. (a) Momentum: m v = m v₁ + 12m v₂, so v = v₁ + 12v₂ [1]. Kinetic energy: v² = v₁² + 12v₂² [1]. Substituting v₂ = (v − v₁)/12 gives v₁ = −11v/13, so the speed is 11v/13 with direction reversed [1]. (b) Energy fraction per collision = (11/13)² = 0.716 [1]. After five: 2.0 × 0.716⁵ = 0.38 MeV [1]. (c) A lighter nucleus takes a larger share of the neutron’s energy in each collision, so fewer collisions are needed [1]. Examiner insight: A “show that” answer needs both conservation equations written out and the substitution shown, ending on the stated result.

10. (a) P = (m/t) c ΔT [1], so m/t = 1.4 × 10⁹ ÷ (4.2 × 10³ × 35) [1] = 9.5 × 10³ kg s⁻¹ [1]. (b) The primary coolant becomes radioactive in the core, so the steam circuit is kept separate [1]. Examiner insight: Keep GW converted to W before dividing; a missing factor of 10⁹ is the usual slip here.

11. (a) 120 ÷ 30.1 = 3.99 half-lives [1]. Fraction = 0.5³·⁹⁹ = 0.063 [1]. (b) 2ⁿ = 1000, so n = 9.97 half-lives [1]. Time = 9.97 × 30.1 = 300 years [1]. (c) 300 ÷ 2.41 × 10⁴ = 0.0124 half-lives, so fraction = 0.5⁰·⁰¹²⁴ = 0.991 [1]. (d) For: sealed containers in stable rock keep radioactive material away from people for a very long time [1]. Plutonium-239 would stay almost unchanged for centuries (0.991 left), so storage must last far longer than institutions [1]. Against: leaks, ground movement and security risks remain, and future generations carry the burden; a judgement should weigh these against the alternatives [1]. Examiner insight: An evaluation reaches a conclusion; listing advantages and disadvantages without a judgment usually leaves marks unclaimed.

Where marks are usually lost

• Leaving out a free neutron when adding product masses.
• Giving the energy in MeV when joules are asked for.
• Using electrical power rather than thermal power for the fission rate.
• Stating that control rods “slow” neutrons or that the moderator “absorbs” them.
• Describing waste storage without a half-life, time scale or risk.
• Forgetting to round generations up to a whole number.
• Taking a binding-energy answer as exact when it comes from curve values.
• Mixing MW and GW.

Next steps

• Re-read the revision notes and the study guide.
• Check the IB DP Physics hub and printable checklist.
• Take the free 10-minute diagnostics to find other weak spots.
• Book a free trial class for help with calculations.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page paraphrases sub-topic E.4, Fission.