S6 Physics Practice Paper: Radioactive Decay, Nuclear Energy And Radiation Safety (12 Questions, 100 Marks)

S6 Physics Practice Paper: Radioactive Decay, Nuclear Energy and Radiation Safety (12 questions, 100 marks)

2026-11-08

S6 Physics: Radioactive Decay, Nuclear Energy and Safety

S6 Physics: radiation types, nuclear equations, half-life, decay constant, binding energy, fission energy and absorbed dose, applied to radiocarbon dating and reactor safety.

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Section A: Radioactive Decay and Half-life (4 questions, 25 marks)

Radioactive decay: alpha decay reduces the mass number by 4 and the atomic number by 2; beta-minus decay keeps the mass number and raises the atomic number by 1; gamma emission only carries away energy. Always check both mass number and atomic number conservation.

#	Question	Answer and marking notes
1	Describe the nature of alpha particles, beta particles and gamma rays, and compare their charge, penetrating power and ionising power (one sentence each).	An alpha particle is a helium nucleus (two protons and two neutrons) with charge +2; it has the weakest penetration, being stopped by a sheet of paper or a few centimetres of air, but the strongest ionising power. A beta particle is a fast electron emitted when a neutron decays, with charge -1; its penetration is moderate, being stopped by a few millimetres of aluminium, and its ionising power is moderate. A gamma ray is a very short wavelength electromagnetic wave with no charge; it is the most penetrating, needing thick lead or concrete, and the least ionising.
2	Write the complete nuclear equation for each case, showing mass numbers and atomic numbers: (a) uranium-238 emits an alpha particle; (b) carbon-14 emits a beta particle; (c) nitrogen-14 is struck by an alpha particle and emits a proton.	(a) U-238 -> Th-234 + alpha, since 238 = 234 + 4 and 92 = 90 + 2. (b) C-14 -> N-14 + beta- + antineutrino: the mass number stays 14 while the atomic number rises from 6 to 7. (c) N-14 + alpha -> O-17 + p, since 14 + 4 = 17 + 1 and 7 + 2 = 8 + 1. In every case check that the totals of mass number and atomic number balance.
3	What is meant by the half-life of a radioactive isotope? (a) Relate the half-life to the decay constant. (b) What fraction of the original nuclei remains after four half-lives? (c) Does the half-life change with temperature, pressure or chemical state? Explain.	(a) The half-life t½ is the time for the number of radioactive nuclei, or the activity, to fall to half its value; t½ = ln2 / lambda, so a larger decay constant means faster decay and a shorter half-life. (b) Half remains after each half-life, so (1/2)^4 = 1/16 remains after four. (c) It does not change. Decay comes from instability inside the nucleus and is a random nuclear process, unaffected by temperature, pressure or chemical state, so the half-life is a fixed property of the isotope.
4	A radioactive source has an initial activity of 800 Bq and a half-life of 5 days. (a) Find the activity after 15 days. (b) Find the shortest time for the activity to fall below 50 Bq. (c) Distinguish between activity and dose.	(a) 15 days is 15 / 5 = 3 half-lives, so the activity is 800 x (1/2)^3 = 100 Bq. (b) 800 / 50 = 16, that is four half-lives, so 50 Bq is reached at 20 days and slightly more than 20 days is needed to fall below it. (c) Activity is the number of decays per second (Bq) and describes the source itself, while dose (Gy or Sv) describes the energy absorbed by matter; the same source gives very different doses at different distances and with different shielding.

Section B: Nuclear Reactions, Binding Energy and Nuclear Energy (4 questions, 33 marks)

Half-life: t½ = ln2 / lambda and A = lambda N. The number of undecayed nuclei halves every half-life, so (1/2)^n remains after n half-lives. Keep the units of lambda and time consistent, and note when a question asks for falling below a value.

#	Question	Answer and marking notes
5	Cobalt-60 has a half-life of 5.27 years and a sample contains 1.0 x 10^18 cobalt-60 nuclei. (a) Calculate the decay constant in y^-1. (b) Use A = lambda N to find the activity in Bq. (c) Explain why the activity is not zero after 10 years.	(a) lambda = ln2 / t½ = 0.693 / 5.27 years = 0.132 per year. (b) In SI units lambda = 0.132 / (365.25 x 24 x 3600) = 4.2 x 10^-9 s^-1, so A = lambda N = 4.2 x 10^-9 x 1.0 x 10^18 = 4.17 x 10^9 Bq. (c) 10 years is 10 / 5.27 = 1.9 half-lives, so about (1/2)^1.9 = 27% of the nuclei have not decayed and the activity is roughly a quarter of the original; decay is random and never stops at one fixed time.
6	The mass defect of a helium-4 nucleus is 0.0304 u and 1 u is equivalent to 931.5 MeV. (a) Calculate the binding energy of helium-4 in MeV. (b) Calculate the binding energy per nucleon. (c) Explain why a larger binding energy per nucleon means a more stable nucleus.	(a) Binding energy = mass defect x 931.5 = 0.0304 x 931.5 = 28.3 MeV. (b) Helium-4 has four nucleons, so the binding energy per nucleon is 28.3 / 4 = 7.08 MeV. (c) The binding energy per nucleon is the average energy needed to remove one nucleon, so a larger value means the nucleons are bound more tightly and the nucleus is more stable; the value peaks near iron-56 at about 8.8 MeV, which is why both fission of heavy nuclei and fusion of light nuclei release energy.
7	Compare nuclear fission and nuclear fusion: (a) which nuclei are involved in each? (b) give one example equation for each; (c) both release energy through mass defect, so why do both release energy? (d) why is fusion hard to sustain on Earth?	(a) Fission splits one heavy nucleus such as U-235 or Pu-239 into two lighter nuclei; fusion joins two very light nuclei such as deuterium and tritium into a heavier one. (b) Fission: U-235 + n -> Ba-141 + Kr-92 + 3n. Fusion: H-2 + H-3 -> He-4 + n. (c) The products have a higher binding energy per nucleon than the reactants, so the total mass falls and the lost mass is released as energy by E = mc^2. (d) Both nuclei are positively charged and need extremely high temperatures (about 10^8 K) to overcome electrostatic repulsion, and the hot plasma must be confined long enough by magnetic or inertial confinement; at present the energy input exceeds the output, so continuous generation is not yet possible.
8	Each fission of uranium-235 releases on average 200 MeV (1 eV = 1.6 x 10^-19 J). (a) Calculate the energy released by the complete fission of 1 g of U-235 in joules. (b) Given 1 kWh = 3.6 x 10^6 J, express this in kWh. (c) State the functions of the moderator and the control rods in a reactor.	(a) Moles = 1 / 235, so the number of nuclei is (1/235) x 6.02 x 10^23 = 2.56 x 10^21; each fission gives 200 x 1.6 x 10^-19 = 3.2 x 10^-11 J, so the total is 2.56 x 10^21 x 3.2 x 10^-11 = 8.20e+10 J. (b) 8.20e+10 / 3.6 x 10^6 = 2.28e+04 kWh, about 2.3 x 10^4 kWh, equivalent to burning several tonnes of coal. (c) The moderator (water or graphite) slows fast neutrons into thermal neutrons so that fission of U-235 is more likely; the control rods (boron or cadmium) absorb neutrons, and the deeper they are inserted the more they absorb, which regulates the rate and prevents a runaway reaction.

Section C: Radiation Safety, Dose and Evaluation (4 questions, 42 marks)

Nuclear energy and dose: binding energy = mass defect x 931.5 MeV, and a larger binding energy per nucleon means a more stable nucleus; both fission and fusion release energy because the products are more tightly bound. Absorbed dose D = E/m in Gy and equivalent dose H = D x weighting factor in Sv.

#	Question	Answer and marking notes
9	Radiation safety: (a) state three precautions when using a gamma source and give a reason for each; (b) why is lead used for gamma sources rather than paper? (c) give two sources of background radiation; (d) distinguish a Geiger counter from a cloud chamber.	(a) Shorten the exposure time, because dose is proportional to time; increase the distance, because intensity falls as the inverse square of distance; use thick shielding of lead or concrete to absorb photons; also wear a personal dosimeter to monitor the accumulated dose. (b) Gamma rays carry no charge and are extremely penetrating, so paper or aluminium absorbs almost nothing; lead has a high atomic number and density, hence a large cross-section for absorbing and scattering photons, so a few centimetres attenuates the beam strongly. (c) Radon gas from soil and granite, cosmic rays especially at altitude, medical X-rays and fallout from tests or accidents. (d) A Geiger counter amplifies each ionisation into a pulse and measures count or dose rate but cannot identify the particle type or energy; a cloud chamber shows visible tracks, so the range and curvature of alpha and beta tracks can be observed, which is a qualitative method.
10	A worker absorbs 0.020 J of beta radiation in 0.50 kg of tissue. (a) Calculate the absorbed dose in Gy and in mGy. (b) The weighting factor for beta is 1; find the equivalent dose in Sv. (c) If the same energy came from alpha radiation (weighting factor 20), what would the equivalent dose be, and why the difference?	(a) Absorbed dose = energy / mass = 0.020 / 0.50 = 0.040 Gy = 40 mGy. (b) Equivalent dose = absorbed dose x weighting factor = 0.040 x 1 = 0.040 Sv, that is 40 mSv. (c) With alpha radiation: 0.040 x 20 = 0.80 Sv. Alpha particles are heavy and doubly charged, ionising densely along a short track and damaging DNA far more efficiently than beta, hence the weighting factor of 20; however alpha penetrates very poorly, so the risk arises mainly on inhalation or ingestion.
11	Carbon-14 has a half-life of 5730 years; a living tree stays in equilibrium with the atmosphere and takes up no more carbon after death. (a) An ancient wooden artefact has only 25% of the carbon-14 activity of living wood; find its age. (b) Write the formula used. (c) State two limitations of the method.	(a) 25% is (1/2)^2, so two half-lives have passed and the age is 2 x 5730 = 11460 years. (b) A = A0 (1/2)^(t/t½), so t = (t½ / ln2) x ln(A0/A) = (5730 / 0.693) x ln4 = 11460 years. (c) Limits: it applies only to carbon-containing organic remains, not to rock; the useful range is about 50000 years because older samples have too little activity left; contamination by modern carbon or groundwater makes samples look younger, so calibration is needed.
12	Evaluating nuclear power: (a) give two advantages and two disadvantages, one sentence each; (b) explain why high-level waste needs deep geological disposal; (c) state three safety features of a modern reactor and their functions; (d) for Hong Kong, should the share of nuclear power be greatly increased? Support your view with two reasons.	(a) Advantages: generation emits no carbon dioxide or sulphur dioxide, so the air pollution is far less than from coal; the energy density is enormous, since 1 g of U-235 rivals several tonnes of coal, so fuel transport and storage are cheap. Disadvantages: a severe accident can release large amounts of radioactive material over a wide area that is hard to clean up; high-level waste contains long-lived nuclides that need long-term monitoring and a disposal site that is hard to find. (b) The long-lived nuclides stay radioactive for tens of thousands of years, so the waste is vitrified and sealed in metal and concrete containers and buried several hundred metres down in stable rock, isolating it from groundwater and the biosphere through multiple barriers with long-term monitoring. (c) Control rods absorb neutrons and regulate the rate; multiple independent cooling systems with emergency power remove decay heat after shutdown; a thick concrete containment building keeps radioactive material inside even under accident pressure; a negative temperature coefficient makes the reaction weaken as the core heats up. (d) Sample view: increase slightly, because nuclear power is low-carbon and gives a stable supply that could replace some coal; or oppose a large increase, because Hong Kong has no suitable site or waste-disposal route and is densely populated, so the cost of an accident is extreme and renewables with regional cooperation should come first.

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