IB Physics Paper 2 · HL structured questions practice papers
HL · 90 marks · 2h 30m. Each generated paper follows the SL & HL Paper 2 · HL structured questions format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.
Sample Paper 2 · HL structured questions questions
Question 1Structured · 8 marksA* challenge
The solar constant is 1360 W m−2 and the Earth’s albedo is 0.30.
(a)
Show that the mean intensity incident per square metre of the Earth’s surface is 340 W m−2.
[1]
(b)
Calculate the mean intensity absorbed.
[1]
(c)
Calculate the equilibrium surface temperature assuming the Earth is a black body.
[2]
(d)
The actual mean surface temperature is 288 K. Determine the effective emissivity of the Earth.
[2]
(e)
Explain how greenhouse gases cause the surface temperature to be higher.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
340 W m−2
1360/4 = 340 W m−2✓
1
(b)
238 W m−2
0.70 × 340 = 238 W m−2✓
1
(c)
255 K
238 = σT4✓ T = 255 K ✓
2
(d)
0.61
e = 238/(5.67 × 10−8 × 2884) ✓ = 0.61 ✓
2
(e)
absorb and re-emit IR
greenhouse gases absorb infrared radiated by the surface ✓ and re-emit it in all directions, some back towards the surface ✓
2
Question 2Structured · 10 marksA* challenge
Ultraviolet light of wavelength 300 nm is incident on a metal surface of work function 2.3 eV.
(a)
Calculate the energy of a photon in eV.
[2]
(b)
Calculate the maximum kinetic energy of the emitted electrons and the stopping potential.
[2]
(c)
Explain why the photoelectric effect provides evidence for the particle nature of light.
[3]
(d)
Calculate the de Broglie wavelength of the fastest emitted electrons.
[2]
(e)
Suggest why these electrons would show measurable diffraction from a crystal.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
4.14 eV
E = hcλ✓ 4.14 eV ✓
2
(b)
1.84 eV; 1.84 V
Ek,max = hf − Φ = 1.84 eV ✓ Vs = 1.84 V ✓
2
(c)
threshold frequency, instant emission
there is a threshold frequency below which no electrons are emitted, whatever the intensity ✓ emission is instantaneous ✓ explained by one photon giving all its energy to one electron ✓
3
(d)
9.04 × 10−10 m
p = √2mEk✓ λ = hp = 9.04 × 10−10 m ✓
2
(e)
λ comparable to atomic spacing
their wavelength is comparable with the spacing of atoms in a crystal (~10−10 m) ✓
1
Question 3Structured · 4 marks
A star has a surface temperature of 4000 K and a luminosity 100 times that of the Sun (Sun: 5800 K).
(a)
Calculate the peak wavelength of the star.
[1]
(b)
Calculate the ratio of the star’s radius to the Sun’s radius.
[2]
(c)
Identify the type of star.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
725 nm
λ = 2.9 × 10−3/4000 = 725 nm ✓
1
(b)
21
L ∝ R2T4✓ R/R☉ = 10 × (5800/4000)2 = 21 ✓
2
(c)
red giant
red giant ✓
1
Question 4Structured · 6 marks
A satellite orbits the Earth at a height of 400 km. Earth mass = 5.97 × 1024 kg, radius = 6.37 × 106 m.
(a)
Calculate the gravitational field strength at this height.
[2]
(b)
Calculate the orbital speed.
[2]
(c)
Explain why astronauts in the satellite appear weightless.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
8.7 N kg−1
g = GM/r2✓ 8.7 N kg−1✓
2
(b)
7.7 km s−1
v = √(GM/r) ✓ 7.67 × 103 m s−1✓
2
(c)
free fall together
astronaut and satellite are both in free fall ✓ with the same acceleration, so no contact force ✓
2
Question 5Structured · 5 marks
Light of wavelength 600 nm is incident on a diffraction grating with 400 lines per mm.
(a)
Calculate the grating spacing.
[1]
(b)
Calculate the angle of the second-order maximum.
[2]
(c)
Determine the total number of maxima observed.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2.5 × 10−6 m
2.5 × 10−6 m ✓
1
(b)
28.7°
sin θ = 2λ/d ✓ 28.7° ✓
2
(c)
9
nmax = 4 ✓ 2 × 4 + 1 = 9 ✓
2
Question 6Structured · 11 marks
A battery of emf 9 V and internal resistance 0.5 Ω is connected to a resistor of 4 Ω.
(a)
Distinguish between emf and terminal potential difference.
[2]
(b)
Calculate the current and the terminal potential difference.
[2]
(c)
Calculate the efficiency of the battery in this circuit.
[2]
(d)
A second identical resistor is connected in parallel with the first. Determine the new power dissipated in the external circuit.
[3]
(e)
State the external resistance for maximum power transfer and calculate that power.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
definitions
emf is the energy transferred to electrical energy per unit charge by the source ✓ terminal pd is the energy per unit charge transferred in the external circuit (emf minus “lost volts”) ✓
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in