S6 Physics Paper: Radioactivity, Half-life And Nuclear Energy (100 Marks)

S6 Physics Paper: Radioactivity, Half-life and Nuclear Energy (100 marks)

2026-11-11

S6 Physics – radioactivity and nuclear energy

S6 Physics paper on radioactivity: half-life, decay equations, mass defect and binding energy, counting statistics, fission reactor power and radiation protection.

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Paper details　S6 Physics practice paper: six questions on radioactivity, half-life, decay equations, binding energy, counting statistics, fission reactor power and radiation protection.

Question 1 Half-life and decay constant (14 marks)

A radionuclide has a half-life of 5.0 days. A sample has an initial activity of 8.0 x 10^5 Bq. Answer the following.

#	Question	Answer and marking notes
1	Find the decay constant of the nuclide in per day and in per second.	Answer: the decay constant is ln 2 divided by the half-life, giving 0.139 per day; dividing by 86400 gives 1.60 x 10^-6 per second. (Marking: 1 mark for the formula, 1 for the per-day value, 1 for the conversion, 1 for the units.)
2	Find the fraction of atoms that have not decayed after 15.0 days.	Answer: 15.0 days is three half-lives, so the remaining fraction is one half cubed, which is 0.125, that is 12.5 per cent. (Marking: 1 mark for the number of half-lives, 1 for the power expression, 1 for the value.)
3	Find the activity of the sample after 15.0 days.	Answer: activity is proportional to the number of undecayed atoms, so it falls to one eighth, giving 1.0 x 10^5 Bq. (Marking: 2 marks for the proportional reasoning, 1 for substitution, 1 for the value with unit.)
4	How many days are needed for the activity to fall to one sixteenth of its initial value?	Answer: one sixteenth is one half to the fourth power, so four half-lives are needed, that is 20.0 days. (Marking: 1 mark for the power expression, 1 for the number of half-lives, 1 for the value with unit.)

Question 2 Decay equations and changes in nuclides (16 marks)

The following parts involve alpha, beta and gamma emission. In the notation used, the mass number is the superscript and the atomic number the subscript.

#	Question	Answer and marking notes
5	Radium 226 (mass number 226, atomic number 88) emits one alpha particle. State the mass number and atomic number of the daughter nucleus and write the nuclear equation.	Answer: an alpha particle is helium 4, so the daughter has mass number 226 minus 4, which is 222, and atomic number 88 minus 2, which is 86, that is radon 222. The equation balances radium 226 giving radon 222 plus helium 4. (Marking: 1 mark for the alpha particle figures, 1 for the mass number, 1 for the atomic number, 1 for a balanced equation.)
6	Carbon 14 (mass number 14, atomic number 6) decays by beta minus emission. State the mass number and atomic number of the product and explain why the mass number is unchanged.	Answer: the product is nitrogen 14 with atomic number 7 and mass number 14. In beta minus decay a neutron changes into a proton with emission of an electron and an antineutrino, so the total number of nucleons stays the same while the atomic number rises by one. (Marking: 2 marks for the product, 1 for the neutron to proton change, 1 for the reason.)
7	A nuclide emits one alpha particle and then two beta minus particles. Find the net change in atomic number and in mass number.	Answer: the atomic number changes by minus two then plus one twice, a net change of zero; the mass number falls by 4. The final nuclide therefore has the same atomic number as the parent and a mass number four less. (Marking: 1 mark for the alpha change, 1 for the beta changes, 1 for each net value.)
8	How does gamma emission affect the atomic number and the mass number, and why does it usually accompany alpha or beta decay?	Answer: gamma emission changes neither the atomic number nor the mass number. After alpha or beta decay the daughter nucleus is often left in an excited state, and the surplus energy leaves as a gamma photon as the nucleus drops to a lower energy state. (Marking: 2 marks for the unchanged quantities, 1 for the excited state, 1 for the energy release.)

Question 3 Mass defect, binding energy and stability (18 marks)

A nucleus has mass number 56 and a mass defect of 0.5284 u. Take 1 u as 931.5 MeV and 1 MeV as 1.602 x 10^-13 J. Answer the following.

#	Question	Answer and marking notes
9	Find the binding energy of the nucleus in MeV.	Answer: the binding energy is the mass defect multiplied by 931.5, giving 492.2 MeV. (Marking: 2 marks for the expression, 1 for substitution, 1 for the value, 1 for the unit.)
10	Find the binding energy per nucleon.	Answer: dividing the binding energy by the mass number gives 8.79 MeV per nucleon. (Marking: 2 marks for dividing by the mass number, 1 for the value, 1 for the unit.)
11	Convert the binding energy of the nucleus into joules.	Answer: multiplying 492.2 MeV by 1.602 x 10^-13 gives 7.89 x 10^-11 J. (Marking: 1 mark for the conversion factor, 2 for the product, 1 for the significant figures and unit.)
12	The binding energy per nucleon of uranium 238 is about 7.57 MeV. Which nucleus, this one or uranium 238, is more stable, and why?	Answer: the nucleus with mass number 56 is more stable. Its binding energy per nucleon of 8.79 MeV is higher than the 7.57 MeV of uranium 238, so its nucleons are held more tightly and more energy is needed to separate them. (Marking: 1 mark for the conclusion, 2 for the comparison, 2 for the explanation.)

Question 4 Background counting and statistical uncertainty (16 marks)

With no sample present a counter records 40 counts in 100 s. With the sample in place it records 1240 counts in 200 s, background included. Answer the following.

#	Question	Answer and marking notes
13	Find the background count rate.	Answer: the background count rate is 40 divided by 100, that is 0.40 per second. (Marking: 2 marks for the division, 1 for the value, 1 for the unit.)
14	Find the net count rate of the sample.	Answer: the rate with background is 1240 divided by 200, that is 6.20 per second, so the net rate is 6.20 minus 0.40, that is 5.80 per second. (Marking: 2 marks for the raw rate, 1 for the subtraction, 1 for the value with unit.)
15	Estimate the statistical uncertainty from the square root of the sample count, expressed per second.	Answer: the standard deviation of the count is about the square root of 1240, that is 35.2, which is 0.18 per second after dividing by 200. The net rate is therefore 5.80 plus or minus 0.18 per second. (Marking: 1 mark for the square root, 2 for dividing by the time, 1 for the result.)
16	If the sample is counted again after three days and the half-life is one day, what net rate is expected, and why do individual measurements fluctuate?	Answer: three days is three half-lives, so the expected net rate is 5.80 divided by 8, that is 0.73 per second. Radioactive decay is a random process, so the number decaying in a given interval is not fixed and a single count departs from the expected value; repeated or longer measurements reduce the uncertainty. (Marking: 2 marks for the division, 1 for the value, 1 for the randomness explanation.)

Question 5 Power and fuel in a fission reactor (18 marks)

A reactor generates electricity from uranium 235 fission, releasing about 200 MeV per fission. Its electrical output is 600 MW at an overall efficiency of 33 per cent. Take the Avogadro constant as 6.02 x 10^23 per mole and the molar mass of uranium 235 as 235 g/mol.

#	Question	Answer and marking notes
17	Find the energy released per fission in joules.	Answer: 200 MeV multiplied by 1.602 x 10^-13 J per MeV gives 3.20 x 10^-11 J. (Marking: 2 marks for the conversion factor, 2 for the product, 1 for the significant figures and unit.)
18	Find the thermal power of the reactor.	Answer: the thermal power is the electrical output divided by the efficiency, that is 600 MW divided by 0.33, or about 1.82 x 10^9 W. (Marking: 2 marks for the expression, 1 for the value, 1 for the unit.)
19	Find the number of fissions per second in the reactor.	Answer: dividing the thermal power by the energy per fission gives about 5.7 x 10^19 fissions per second. (Marking: 2 marks for the division, 2 for the value and order of magnitude.)
20	If the reactor runs continuously for one day, find the mass of uranium 235 consumed in kilograms.	Answer: in one day there are about 4.9 x 10^24 fissions, which is 8.1 moles, and multiplying by 235 g per mole gives about 1900 g, that is roughly 1.9 kg of uranium 235. (Marking: 1 mark for the daily count, 1 for the moles, 2 for the mass, 1 for the unit conversion.)

Question 6 Equivalent dose and radiation shielding (18 marks)

A worker absorbs 0.40 mGy of alpha radiation, for which the radiation weighting factor is 20. Separately, the half-value thickness of lead for a gamma source is 4.0 mm. Answer the following.

#	Question	Answer and marking notes
21	Find the equivalent dose received by the worker in mSv.	Answer: the equivalent dose is the absorbed dose multiplied by the weighting factor, that is 0.40 mGy times 20, giving 8.0 mSv. (Marking: 2 marks for the expression, 1 for substitution, 1 for the value, 1 for the unit.)
22	Find the fraction of the incident gamma intensity transmitted through 12.0 mm of lead.	Answer: 12.0 mm is three half-value thicknesses, so the transmitted fraction is one half cubed, that is 0.125, or 12.5 per cent. (Marking: 2 marks for the number of layers, 1 for the power expression, 1 for the fraction.)
23	If the incident count rate is 640 per second, find the count rate after 12.0 mm of lead.	Answer: multiplying 640 per second by 0.125 gives 80 per second. (Marking: 2 marks for using the fraction, 1 for the value, 1 for the unit.)
24	Why is an alpha source especially dangerous once inhaled or swallowed, yet of relatively low risk outside the body?	Answer: alpha particles have a very short range but a high ionising power, so outside the body the outer skin or a sheet of paper stops them and little energy reaches living tissue. Once inhaled or swallowed they irradiate the cells of the airways or gut at close range with dense ionisation, causing greater damage. (Marking: 2 marks for external shielding, 2 for close-range irradiation, 1 for ionising density.)

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