S6 Physics Practice Paper: Atomic Physics, Nuclear Energy And Radioactivity (12 Questions, 100 Marks, 75 Minutes, With Answers)

S6 Physics Practice Paper: Atomic Physics, Nuclear Energy and Radioactivity (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-06

HKDSE Physics: Atomic Physics, Nuclear Energy and Radioactivity

S6 physics practice paper, 12 questions, 100 marks, 75 minutes: photoelectric effect and stopping voltage, photon energy, half-life and exponential decay, carbon-14 dating, Bohr levels and

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Section A: Photons, Photoelectric Effect and Atomic Models (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	A metal has a work function of 2.3 eV and is illuminated by monochromatic light of wavelength 400 nm. Take hc = 1240 eV nm. (a) Find the energy of each photon. (b) Find the maximum kinetic energy of the photoelectrons. (c) Find the stopping voltage.	(a) E = hc/lambda = 1240/400 = 3.10 eV. (b) KEmax = E – W = 3.10 – 2.3 = 0.80 eV. (c) Stopping voltage = 0.80 V. Marking: 1 mark per value; 2 marks for stating that KEmax depends on frequency but not on intensity; 1 mark for correct units.
2	For the metal in Question 1, find the threshold wavelength and the threshold frequency, and decide whether 400 nm light can eject photoelectrons.	Threshold wavelength = hc/W = 1240/2.3 = 539 nm; threshold frequency = c/lambda0 = 5.56e+14 Hz. Since 400 nm is shorter than the threshold wavelength, each photon carries more than the work function, so photoelectrons are emitted. Marking: 2 marks each value; 2 marks for the comparison and conclusion.
3	A radioactive source has an initial activity of 6400 Bq and a half-life of 8 days. Find the activity after 24 days and state how activity relates to the number of undecayed nuclei.	(a) 24 days is 3 half-lives, so A = A0/2^n = 6400/8 = 800 Bq. (b) Activity is proportional to the number of undecayed nuclei (A = lambda N), so it follows the same decay law. Marking: 2 marks for the number of half-lives; 2 marks for the division; 2 marks for the A = lambda N relation.
4	A wooden artefact has a carbon-14 content equal to one eighth of that in living wood. Given a half-life of 5730 years, estimate the age of the artefact.	One eighth equals 1/2^3, so 3 half-lives have elapsed. Age = 3 x 5730 = 17,190 years. Marking: 3 marks for identifying three half-lives; 2 marks for the multiplication; 1 mark for significant figures and unit.

Section B: Radioactive Decay, Detection and Radiation Protection (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	For hydrogen the energy levels are En = -13.6/n^2 eV. Find the photon energy and wavelength emitted when an electron falls from n = 3 to n = 2, and name the spectral series.	E3 = -13.6/9 = -1.51 eV and E2 = -13.6/4 = -3.40 eV, so the photon energy is 1.89 eV. Wavelength = hc/E = 1240/1.89 = 656 nm, which belongs to the Balmer series and appears red. Marking: 1.5 marks per level; 2 marks for the energy difference; 2 marks for the wavelength; 2 marks for series and colour.
6	A nucleus has a mass defect of 0.0304 u (1 u = 931.5 MeV). Find its binding energy, and the binding energy per nucleon if the nucleus contains 2 nucleons.	Binding energy = 0.0304 x 931.5 = 28.3 MeV. Binding energy per nucleon = 28.3/2 = 14.2 MeV. Marking: 2 marks for the conversion constant; 3 marks for the product; 2 marks for the per-nucleon value; 2 marks for explaining that binding energy is the energy needed to separate the nucleus completely.
7	Each fission of uranium-235 releases about 200 MeV. Find the total energy released when 1.0 kg of uranium-235 undergoes fission completely, in joules.	Number of nuclei = (1000/235) x 6.02e+23 = 2.56e+24. Energy = 2.56e+24 x 200 MeV = 5.12e+26 MeV = 8.2e+13 J. Marking: 3 marks for the mole conversion; 2 marks for the number of nuclei; 2 marks for converting MeV to joules; 2 marks for order of magnitude and unit.
8	An electron is accelerated through a potential difference of 2000 V. Find its de Broglie wavelength (electron mass 9.11e-31 kg, charge 1.6e-19 C).	Momentum p = sqrt(2 m e V) = sqrt(2 x 9.11e-31 x 1.6e-19 x 2000) = 2.41e-23 kg m/s. Wavelength = h/p = 6.63e-34 / 2.41e-23 = 2.75e-11 m. Marking: 3 marks for the kinetic-energy and momentum relation; 2 marks for substitution; 2 marks for the wavelength; 2 marks for units and significant figures.

Section C: Nuclear Energy, Mass Defect and Fusion (40 marks)

Each question carries 10 marks; give full explanations or evaluations.

#	Question	Answer and marking notes
9	A sample initially contains 8.0e+20 radioactive nuclei with a half-life of 5 days. (a) Find the number of nuclei left after 20 days. (b) Explain why a graph of ln N against time is a straight line.	(a) 20 days is 4 half-lives, so N = 8.0e+20/16 = 5.0e+19 nuclei. (b) Decay is random with dN/dt = -lambda N, hence N = N0 exp(-lambda t) and ln N = ln N0 – lambda t, a straight line of gradient -lambda. Marking: 2 marks for the number of half-lives; 3 marks for the division; 3 marks for the exponential law; 2 marks for the straight line and gradient.
10	A counter records 320 counts per minute for a sample together with background, and 20 counts per minute for the background alone. The half-life of the sample is 8 days. Find the counter reading after 16 days.	Net count rate = 320 – 20 = 300 per minute. After 16 days, two half-lives, the net rate is 300/4 = 75 per minute. Counter reading = 75 + 20 = 95 per minute. Marking: 3 marks for subtracting background; 3 marks for two half-lives; 2 marks for adding background back; 2 marks for naming sources of background radiation.
11	Body tissue of mass 0.50 kg absorbs 3.0 mJ of energy from alpha particles. Find the absorbed dose and the equivalent dose (radiation weighting factor wR = 20 for alpha particles).	Absorbed dose D = E/m = 3.0e-03/0.50 = 0.006 Gy. Equivalent dose H = wR x D = 20 x 0.006 = 0.12 Sv. Marking: 3 marks for D; 3 marks for H; 2 marks for noting the high LET of alpha particles; 2 marks for unit conversions.
12	A deuterium-tritium fusion reaction releases 17.6 MeV. (a) Write the reaction equation and explain why an extremely high temperature is needed. (b) Find the mass defect. (c) Find the energy released when 1.0 g of an equimolar deuterium-tritium mixture fuses completely.	(a) 2H + 3H -> 4He + n + 17.6 MeV; the nuclei are positively charged, so they must overcome Coulomb repulsion, which requires a very high temperature (about 1e8 K). (b) Mass defect = 17.6/931.5 = 0.0189 u. (c) 5.0 g of the equimolar mixture is 1 mol, so 1.0 g contains 1.20e+23 pairs; energy = 1.20e+23 x 17.6 MeV = 3.4e+11 J. Marking: 2 marks for the equation; 2 marks for the high-temperature reason; 2 marks for the mass defect; 4 marks for the energy per reaction and the total.

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