HKDSE Physics Paper 2 Practice Paper (Elective Part) with Answers

HKDSE Physics Paper 2 Practice Paper (Elective Part) with Answers | 30 questions, 60 minutes, 100 marks

2026-10-29

Practice paper · HKDSE Physics Paper 2

HKDSE Physics Paper 2 elective-part practice paper: 25 multiple-choice and 5 structured questions on astronomy and atomic world, 60 minutes, 100 marks, full answers and PDFs.

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Paper details　Level: S6 (HKDSE) | Subject: Physics (Paper 2, Elective Part) | Time: 60 minutes | Total: 100 marks | Questions: 30

Section A: Astronomy and Space Science (30 marks)

1. Two marks each. Choose the best answer.

#	Question	Answer and marking notes
1	The gravitational field strength at the Earth’s surface (radius R) is 9.8 N kg⁻¹. What is the field strength at a distance 2R from the Earth’s centre?	Field strength follows an inverse-square law, g ∝ 1/r². Doubling the distance from R to 2R quarters the value: 9.8 ÷ 4 = 2.45 N kg⁻¹ (about 2.5). (2 marks)
2	A satellite moves in a circular orbit 400 km above the Earth’s surface (Earth radius 6.37×10⁶ m, GM = 3.98×10¹⁴ m³ s⁻²). Which value is closest to its orbital speed?	r = 6.37×10⁶ + 4.0×10⁵ = 6.77×10⁶ m; v = √(GM/r) = √(3.98×10¹⁴ ÷ 6.77×10⁶) ≈ 7.67×10³ m s⁻¹ (about 7.7 km s⁻¹). (2 marks)
3	For the same satellite, which value is closest to its orbital period?	T = 2πr/v = 2π × 6.77×10⁶ ÷ 7669 ≈ 5.55×10³ s ≈ 92.4 minutes. (2 marks; 1 mark for “about 1.5 hours” only.)
4	Two satellites orbit the same planet in circular orbits with radii in the ratio 1 : 4. What is the ratio of their periods?	Kepler’s third law: T² ∝ r³, so T ∝ r^{3/2}. Increasing r four times multiplies T by 4^{3/2} = 8, i.e. 1 : 8. (2 marks; 1 mark for quoting T² ∝ r³ only.)
5	The orbital speed just above the Earth’s surface is about 7.9 km s⁻¹. What is the escape speed from the Earth’s surface?	v_esc = √(2GM/R) = √2 × v_orb = 1.414 × 7.9 ≈ 11.2 km s⁻¹. (2 marks)
6	An astronaut of mass 60 kg stands on the Moon, where g = 1.62 N kg⁻¹. What is his weight?	W = mg = 60 × 1.62 = 97.2 N (about 97 N). Mass is unchanged; the weight is smaller because the Moon’s g is smaller. (2 marks)
7	A star has a parallax of 0.04 arcsecond. How far is it from the Earth?	d(pc) = 1 / p(arcsec) = 1 ÷ 0.04 = 25 pc, i.e. 25 × 3.26 ≈ 81.5 light years. (2 marks; 1 mark for the formula only.)
8	A star has apparent magnitude 8 and absolute magnitude 3. What is its distance?	m − M = 5 log(d/10) → 8 − 3 = 5 = 5 log(d/10) → d/10 = 10^1.0 = 10 → d = 100 pc. (2 marks)
9	Take the Hubble constant as 70 km s⁻¹ Mpc⁻¹. What is the recession speed of a galaxy at 200 Mpc?	v = H₀d = 70 × 200 = 14000 km s⁻¹ (1.4×10⁴ km s⁻¹). (2 marks)
10	A spectral line of rest wavelength 486 nm is observed at 495.7 nm in a galaxy. What is its recession speed (take c = 3.0×10⁵ km s⁻¹)?	z = Δλ/λ₀ = 9.72 / 486 = 0.020; v = zc = 0.020 × 3.0×10⁵ ≈ 6000 km s⁻¹. (2 marks)
11	A telescope of aperture 0.20 m observes light of wavelength 550 nm. What is its angular resolution in rad?	θ = 1.22λ/D = 1.22 × 550×10⁻⁹ ÷ 0.20 ≈ 3.35×10⁻⁶ rad (about 0.69 arcseconds). (2 marks)
12	Two telescopes have apertures of 0.30 m and 0.06 m. How many times greater is the light gathering power of the larger one?	Light gathering power ∝ D²: (0.30 ÷ 0.06)² = 5² = 25 times. (2 marks)
13	A star has surface temperature about 5800 K. Using Wien’s law λ_max = 2.9×10⁻³ / T, what is its peak wavelength?	λ_max = 2.9×10⁻³ ÷ 5800 = 5.0×10⁻⁷ m = 500 nm, in the visible band. (2 marks)
14	Two stars have the same radius but one has twice the surface temperature of the other. How many times greater is the luminosity of the hotter star?	L = 4πR²σT⁴ with the same radius means L ∝ T⁴, so 2⁴ = 16 times. (2 marks; 1 mark for L ∝ T⁴ only.)
15	Which of the following is a necessary condition for a geostationary satellite?	Its period must equal the Earth’s rotation period (about 24 hours), the orbit must lie in the equatorial plane, and it must move west to east; all three are needed for it to stay fixed above one point. (2 marks; 1 mark for “24-hour period” only.)

Section B: Atomic World (20 marks)

2. Two marks each. Choose the best answer.

#	Question	Answer and marking notes
16	What is the energy of a photon of wavelength 500 nm?	E = hc/λ = 6.63×10⁻³⁴ × 3.0×10⁸ ÷ 500×10⁻⁹ = 3.98×10⁻¹⁹ J = 2.49 eV. (2 marks)
17	A metal has a threshold wavelength of 620 nm. What is its work function?	φ = hc/λ₀ = 6.63×10⁻³⁴ × 3.0×10⁸ ÷ 620×10⁻⁹ = 3.21×10⁻¹⁹ J = 2.01 eV. (2 marks)
18	Photons of energy 3.10 eV strike a metal of work function 2.00 eV. What is the maximum kinetic energy of the emitted photoelectrons?	KE_max = hf − φ = 3.10 − 2.00 = 1.10 eV. (2 marks)
19	For the same experiment, what is the stopping potential?	The stopping potential in volts equals the maximum kinetic energy in eV: 1.10 V. (2 marks)
20	In a photoelectric experiment the frequency is unchanged but the intensity is doubled. Which statement is correct?	Doubling the intensity doubles the number of photons per second, so the saturation current doubles; each photon still carries the same energy, so the maximum kinetic energy and the stopping potential are unchanged. (2 marks; 1 mark for “current doubles” only.)
21	An electron is accelerated through a potential difference of 200 V. What is its de Broglie wavelength?	KE = 200 eV = 3.2×10⁻¹⁷ J; p = √(2mKE) = 7.64×10⁻²⁴ kg m s⁻¹; λ = h/p = 8.68×10⁻¹¹ m. (2 marks; 1 mark if only h/p is quoted without p.)
22	A hydrogen atom makes a transition from n = 3 (-1.51 eV) to n = 2 (-3.40 eV). What is the wavelength of the emitted photon?	ΔE = -1.51 − (-3.40) = 1.89 eV = 3.02×10⁻¹⁹ J; λ = hc/ΔE = 658 nm, red light of the Balmer series. (2 marks)
23	How much energy is needed to ionise a hydrogen atom that is in the n = 2 state?	Ionisation means reaching E = 0: 0 − (-3.40) = 3.40 eV. (2 marks)
24	A discharge tube gives a line spectrum rather than a continuous spectrum. What does this show?	Atoms can only occupy certain discrete energy levels; each transition releases a photon of one fixed energy, so the wavelengths appear as separate lines. (2 marks; 1 mark for “quantised energy” only.)
25	An electron is confined to a region of Δx = 1.0×10⁻¹⁰ m. Using Δx·Δp ≥ h/4π, what is the minimum uncertainty in its momentum?	Δp ≥ h/(4πΔx) = 6.63×10⁻³⁴ ÷ (4π × 1.0×10⁻¹⁰) = 5.28×10⁻²⁵ kg m s⁻¹, equivalent to Δv ≈ 5.8×10⁵ m s⁻¹. (2 marks)

Section C: Structured questions (50 marks)

3. Answer all questions. Show your working and units.

#	Question	Answer and marking notes
26	A communications satellite orbits the Earth in a circle of radius r = 4.2×10⁷ m (GM = 3.98×10¹⁴ m³ s⁻²). (a) Find its orbital speed. (4 marks) (b) Find its period in hours. (4 marks) (c) Explain why a broadcast satellite must be above the equator at about this radius. (2 marks)	(a) v = √(GM/r) = √(3.98×10¹⁴ ÷ 4.2×10⁷) = 3.08×10³ m s⁻¹. (4 marks; 2 marks for the formula alone.) (b) T = 2πr/v = 2π × 4.2×10⁷ ÷ 3079 = 8.57×10⁴ s ≈ 23.8 hours. (4 marks) (c) Only a 24-hour period in the equatorial plane keeps the satellite fixed above one point so the dish need not track it; the computed 23.8 hours is close to 24, so the radius must be near this value. (2 marks)
27	A star has a parallax of 0.025 arcsecond and apparent magnitude 5.0. (a) Find its distance in pc and in light years. (3 marks) (b) Find its absolute magnitude. (4 marks) (c) The Sun has absolute magnitude 4.8. Compare their luminosities. (Hint: M₁ − M₂ = 2.5 log(L₂/L₁)) (3 marks)	(a) d = 1/p = 1 ÷ 0.025 = 40 pc = 40 × 3.26 ≈ 130 light years. (3 marks) (b) M = m − 5 log(d/10) = 5.0 − 5 log(40/10) = 5.0 − 5 × 0.60 = 2.0. (4 marks; 1 mark if the log step is missing.) (c) A smaller M means a brighter star: 2.0 < 4.8, so this star is brighter than the Sun, with L/L☉ = 10^((4.8 − 2.0)/2.5) ≈ 13 times. (3 marks)
28	Light of wavelength 400 nm falls on a metal of work function 2.20 eV. (a) Find the photon energy in eV. (3 marks) (b) Find the maximum kinetic energy of the photoelectrons in eV and in J. (3 marks) (c) Find the stopping potential. (2 marks) (d) Find the threshold wavelength of the metal. (2 marks)	(a) E = hc/λ = 6.63×10⁻³⁴ × 3.0×10⁸ ÷ 400×10⁻⁹ = 4.97×10⁻¹⁹ J = 3.11 eV. (3 marks) (b) KE_max = 3.11 − 2.20 = 0.91 eV = 1.45×10⁻¹⁹ J. (3 marks; 2 marks for eV only.) (c) Stopping potential = 0.91 V. (2 marks) (d) λ₀ = hc/φ = 6.63×10⁻³⁴ × 3.0×10⁸ ÷ (2.20 × 1.6×10⁻¹⁹) = 565 nm. The incident 400 nm is shorter than this, so emission occurs. (2 marks)
29	Electrons accelerated through 1500 V are fired at a crystal and form a diffraction pattern. (a) Find their kinetic energy in J. (2 marks) (b) Find their momentum. (3 marks) (c) Find their de Broglie wavelength. (3 marks) (d) Explain why an electron microscope resolves far better than an optical one. (2 marks)	(a) KE = eV = 1.6×10⁻¹⁹ × 1500 = 2.4×10⁻¹⁶ J. (2 marks) (b) p = √(2mKE) = √(2 × 9.11×10⁻³¹ × 2.4×10⁻¹⁶) = 2.09×10⁻²³ kg m s⁻¹. (3 marks) (c) λ = h/p = 6.63×10⁻³⁴ ÷ 2.09×10⁻²³ = 3.17×10⁻¹¹ m, comparable with atomic spacing, which is why a crystal diffracts the beam. (3 marks) (d) Resolution is limited by the wavelength; 3×10⁻¹¹ m is far shorter than the 5×10⁻⁷ m of visible light, so much finer detail is resolved. (2 marks)
30	Take H₀ = 70 km s⁻¹ Mpc⁻¹ (1 Mpc = 3.09×10²² m). (a) State Hubble’s law and find the recession speed of a galaxy at 100 Mpc. (2 marks) (b) Estimate the age of the universe as 1/H₀ in years. (4 marks) (c) Explain how galactic redshift supports an expanding universe. (2 marks) (d) If the true H₀ were larger than 70, how would the 1/H₀ age change? (2 marks)	(a) v = H₀d, so v = 70 × 100 = 7000 km s⁻¹. (2 marks) (b) H₀ = 70×10³ ÷ 3.09×10²² = 2.27×10⁻¹⁸ s⁻¹; t = 1/H₀ = 4.41×10¹⁷ s = 1.4×10¹⁰ years. (4 marks; 1 mark if the unit conversion is wrong.) (c) Spectral lines from galaxies in every direction are shifted to longer wavelengths, and the shift grows with distance, which matches space itself expanding and galaxies receding from one another. (2 marks) (d) A larger H₀ makes 1/H₀ smaller, so the estimated age is smaller. (2 marks)

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