HKDSE Physics Paper 2 Elective Practice Paper | 2 questions, 40 marks, 1 hour

2026-10-31

HKDSE Physics Paper 2 Electives

HKDSE Physics Paper 2 elective practice paper: 2 questions, 40 marks, 60 minutes. Section A (E3 Energy and use of energy) covers solar-array power, daily output, payback and carbon savings; Section B (E4 Medical physics) covers X-ray photon energy, attenuation and half-value layer, ultrasound wavelength and echo timing, cumulative dose and ALARA, with full worked solutions.

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Section A: E3 Energy and use of energy (20 marks)

Elective questions: this paper covers two of the Paper 2 electives (E3 and E4), 20 marks each; show your working, quote the formula used and give correct units. Total 40 marks.

#	Question	Answer and marking notes
1	A school installs a solar photovoltaic system on its roof: total panel area 12.0 m², photovoltaic efficiency 18%, average solar irradiance 850 W m⁻², effective sunshine 5.2 hours per day. The school uses 9.5 kWh per day; the system costs $42 000, electricity is priced at $1.20 per kWh and the grid emission factor is 0.70 kg CO₂ per kWh. (a) Calculate the electrical power output of the system in kW. (2 marks) (b) Calculate the energy generated per day in kWh. (2 marks) (c) Find the daily output as a percentage of the school daily consumption. (2 marks) (d) Calculate the annual (365-day) output, the annual saving in electricity cost and the payback period in years. (3 marks) (e) Calculate the mass of carbon dioxide avoided per year in kg. (2 marks) (f) If the panel efficiency falls by 0.5% of its value each year, calculate the daily output after 10 years. (2 marks) (g) A coal-fired plant of 33% efficiency loses 8% of its output in transmission. Calculate the primary energy needed to supply 9.5 kWh per day. (3 marks) (h) Explain why the real output is usually lower than the value calculated in (b). (2 marks) (i) Apart from reducing carbon emissions, state two benefits of the solar system to the school. (2 marks)	Answers: (a) Power = efficiency × irradiance × area = 0.18 × 850 W m⁻² × 12.0 m² = 1836 W = 1.84 kW. (b) Daily output = 1.84 kW × 5.2 h = 9.5472 kWh (about 9.55 kWh). (c) Percentage = 9.5472 ÷ 9.5 × 100% = 100.5%, i.e. just above one day of school use. (d) Annual output = 9.5472 × 365 = 3484.7 kWh; annual saving = 3484.7 × $1.20 = $4182; payback = $42 000 ÷ $4182 = 10.04 years (about 10 years). (e) CO₂ avoided = 3484.7 kWh × 0.70 kg kWh⁻¹ = 2439 kg (about 2.44 tonnes). (f) After 10 years = 9.5472 × (1 − 0.005)¹⁰ = 9.5472 × 0.951110 = 9.0804 kWh. (g) Primary energy = 9.5 ÷ (0.33 × 0.92) = 9.5 ÷ 0.3036 = 31.29 kWh; the solar system needs only 9.55 kWh of sunlight, avoiding about 21.74 kWh of fuel energy. (h) Dust and droppings shade the panels, cell temperature rise lowers efficiency, the inverter has losses, sunlight is not always perpendicular to the panels, and 5.2 h is only an average — cloudy days give far less. (i) It is a live teaching resource on renewables and energy efficiency; it cuts the electricity bill by reducing grid purchases; with storage it can supply emergency power.

Section B: E4 Medical physics (20 marks)

Elective questions: this paper covers two of the Paper 2 electives (E3 and E4), 20 marks each; show your working, quote the formula used and give correct units. Total 40 marks.

#	Question	Answer and marking notes
2	(a) An X-ray photon has a wavelength of 0.050 nm. Calculate its energy in J and in keV (h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹). (2 marks) (b) The linear attenuation coefficient of X-rays in soft tissue is 0.25 cm⁻¹. Calculate the percentage of intensity remaining after passing through 6.0 cm of tissue, and the thickness needed to reduce the intensity to one eighth. (3 marks) (c) Calculate the half-value layer of this tissue in cm. (2 marks) (d) An obstetric ultrasound probe operates at 3.5 MHz; the speed of sound in tissue is 1 540 m s⁻¹. Calculate the wavelength in mm. (2 marks) (e) An echo returns from an interface 4.2 cm deep. Calculate the echo return time in µs. (2 marks) (f) One scan delivers an absorbed dose of 2.5 mGy and a patient has 12 scans in a year. Calculate the annual cumulative dose in mGy and compare it with the natural background of about 2.4 mSv per year as a multiple. (3 marks) (g) Explain why ultrasound rather than X-rays is normally used in antenatal examinations. (2 marks) (h) State the ALARA principle and give one specific measure that applies it in an X-ray examination. (2 marks) (i) A CT scan reconstructs an image from projections taken at many angles. Explain why its dose is usually higher than that of a single plain radiograph. (2 marks)	Answers: (a) E = hc ÷ λ = (6.63×10⁻³⁴ × 3.00×10⁸) ÷ 0.050×10⁻⁹ = 3.978e-15 J; ÷ 1.602×10⁻¹⁶ J keV⁻¹ = 24.83 keV, in the X-ray region. (b) I ÷ I₀ = e^(−μx) = e^(−0.25 × 6.0) = e^(−1.5) = 0.2231, so 22.3% remains; one eighth corresponds to 3 half-value layers = 3 × 2.7726 = 8.32 cm. (c) Half-value layer = ln2 ÷ μ = 0.6931 ÷ 0.25 = 2.77 cm. (d) λ = v ÷ f = 1 540 ÷ 3.5×10⁶ = 0.0004 m = 0.44 mm. (e) t = 2d ÷ v = 2 × 0.042 ÷ 1 540 = 5.45e-05 s = 54.5 µs. (f) Annual total = 12 × 2.5 = 30 mGy ≈ 30 mSv (radiation weighting factor for X-rays is 1), i.e. 30 ÷ 2.4 = 12.5 times the natural background. (g) Ultrasound is a mechanical wave with no ionising radiation and no ionisation damage to DNA, so there is no known radiation risk to the fetus; X-rays are ionising and fetal cells divide rapidly, making them more radiosensitive. (h) ALARA means keeping the dose As Low As Reasonably Achievable while still obtaining the needed diagnostic information — e.g. shortest exposure time, lead shielding of gonads, collimating the beam to the region of interest, staff standing outside the beam. (i) A CT scan rotates around the patient and takes hundreds to over a thousand projections, each requiring an exposure, so the total exposure far exceeds the single exposure of a plain radiograph.

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