IB DP Physics: Fusion, stellar evolution, HR diagrams and stellar distances (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover E.5, Fusion and stars, in the IB Physics guide, first assessment 2025, for Standard Level and Higher Level. E.5 has no additional higher level content, so every question is suitable for both. Data: σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; L☉ = 3.83 × 10²⁶ W; R☉ = 6.96 × 10⁸ m; T☉ = 5800 K; c = 3.00 × 10⁸ m s⁻¹; 1 u = 1.661 × 10⁻²⁷ kg, equivalent to 931.5 MeV c⁻²; 1 MeV = 1.602 × 10⁻¹³ J; 1 pc = 3.09 × 10¹⁶ m = 3.26 ly; k = 8.99 × 10⁹ N m² C⁻²; e = 1.602 × 10⁻¹⁹ C; kB = 1.38 × 10⁻²³ J K⁻¹; Wien constant 2.9 × 10⁻³ m K.

The study guide teaches the topic. The revision notes have the formulas. Course page: IB DP Physics. Checklist: IB DP Physics checklist.

Questions

1. A star has a parallax angle of 0.0400 arc-second. Calculate its distance in parsecs, in light years and in metres. [3]

2. Explain why a main sequence star is stable, and why a small change in the core fusion rate is corrected. [4]

3. Two protons approach to a separation of 2.0 × 10⁻¹⁵ m. (a) Calculate their electric potential energy in MeV. [2] (b) The core of a star is at 2.0 × 10⁷ K. Calculate the mean thermal energy 3/2 kB T of a nucleus in keV. [2] (c) Use your answers to explain why fusion still occurs, and why high density is also needed. [2]

4. The table gives data for four stars.

Star	Luminosity / L☉	Surface temperature / K
A	3.0 × 10⁻³	14 000
B	1.0 × 10⁵	3400
C	1.0	5800
D	1.0 × 10⁴	25 000

(a) State the region of the Hertzsprung-Russell (HR) diagram where each star lies. [2] (b) Calculate the radius of star B in terms of R☉. [3]

5. Carbon-12 and helium-4 can fuse to form oxygen-16. The masses are carbon-12 12.000000 u, helium-4 4.002603 u and oxygen-16 15.994915 u. (a) Calculate the energy released by one reaction, in MeV and in joules. [3] (b) A star releases energy at 5.0 × 10²⁹ W by this reaction alone. Calculate the number of reactions per second. [2]

6. Compare the evolution of a star of one solar mass with that of a star of fifteen solar masses. [6]

7. The spectrum of star X has its peak at 1.45 × 10⁻⁷ m. Star Y peaks at 8.0 × 10⁻⁷ m. (a) Calculate the surface temperature of X. [2] (b) Explain how dark lines in a stellar spectrum show the composition of the star. [2] (c) Calculate the surface temperature of Y. [1]

8. A star has a parallax angle of 0.0310 arc-second and an apparent brightness of 2.6 × 10⁻⁹ W m⁻². Its surface temperature is 6200 K. (a) Calculate its distance in parsecs and in metres. [2] (b) Use b = L / 4πd² to calculate its luminosity. [2] (c) Express the luminosity in terms of L☉. [1] (d) Calculate the radius of the star in R☉. [3] (e) Suggest why the star is not on the main sequence. [1]

9. A star keeps constant luminosity while its surface temperature falls from 6000 K to 3000 K. (a) Determine the factor by which its radius changes. [2] (b) Describe how the star moves on an HR diagram. [2]

10. A star has a luminosity of 1.0 × 10²⁹ W. Take 26.7 MeV as the energy released when four hydrogen atoms fuse into one helium atom. The mass of a hydrogen atom is 1.007825 u. (a) Calculate the number of helium nuclei formed per second. [2] (b) Calculate the mass of hydrogen fused per second. [2] (c) Calculate the mass converted into energy per second. [1] (d) The star has 3.0 × 10²⁹ kg of hydrogen available in its core. Estimate how long it lasts, in years (1 year = 3.16 × 10⁷ s). [2] (e) The Sun fuses hydrogen at about 6 × 10¹¹ kg per second. Suggest what this comparison says about massive stars. [1]

11. Describe the sketch of an HR diagram that you would draw, naming the features you would include. [4]

Answers

1. d = 1/p = 1/0.0400 = 25.0 pc [1]. In light years: 25.0 × 3.26 = 81.5 ly [1]. In metres: 25.0 × 3.09 × 10¹⁶ = 7.7 × 10¹⁷ m [1]. [3] Examiner insight: Show the one-line rearrangement and then each conversion separately.

2. Gravity acts inward on every layer [1]. Outward pressure from hot gas and radiation acts against it [1]. The star is stable when these are in balance at each depth [1]. If the fusion rate rose, the core would expand and cool, reducing the rate again; if it fell, the core would contract and heat, raising it [1]. [4] Examiner insight: Name both pressure sources and say “balance”. “Fusion pushes against gravity” is too loose.

3. (a) U = k e² / r = 8.99 × 10⁹ × (1.602 × 10⁻¹⁹)² / 2.0 × 10⁻¹⁵ = 1.15 × 10⁻¹³ J [1] = 0.72 MeV [1]. (b) 3/2 × 1.38 × 10⁻²³ × 2.0 × 10⁷ = 4.14 × 10⁻¹⁶ J [1] = 2.6 keV [1]. (c) The barrier is roughly 280 times the mean thermal energy, yet fusion occurs because a small fraction of nuclei move much faster than average, and quantum effects allow fusion without the full barrier energy [1]. High density makes collisions frequent, so the fusion rate is large enough to power the star [1]. [6] Examiner insight: Convert to the same unit before comparing, and state the ratio.

4. (a) A is a white dwarf (faint but hot); B is a red super giant (very luminous but cool); one mark for these two [1]. C and D are main sequence stars, D near the hot, luminous top left [1]. (b) R/R☉ = √(L/L☉) × (T☉/T)² [1] = √(1.0 × 10⁵) × (5800/3400)² = 316 × 2.91 [1] = 920 R☉ [1]. [5] Examiner insight: Support a region with a reason from the data, such as “low L, high T so small”.

5. (a) Δm = 12.000000 + 4.002603 − 15.994915 = 0.007688 u [1]. E = 0.007688 × 931.5 = 7.16 MeV [1]. In joules: 7.16 × 1.602 × 10⁻¹³ = 1.15 × 10⁻¹² J [1]. (b) Rate = P / E = 5.0 × 10²⁹ / 1.147 × 10⁻¹² (unrounded energy) [1] = 4.4 × 10⁴¹ s⁻¹ [1]. [5] Examiner insight: Keep unrounded masses until after the subtraction.

6. One solar mass: lower core temperature and a slow fusion rate, so a lifetime of billions of years [1]. It leaves the main sequence and swells to a red giant [1]. It sheds its outer layers and the core remains as a white dwarf [1]. Fifteen solar masses: hotter core and far higher luminosity, so the fuel is used much faster and the lifetime is far shorter [1]. It becomes a super giant and fuses heavier elements up to iron [1]. The core collapses in a supernova, leaving a neutron star or a black hole [1]. [6] Examiner insight: A comparison needs both stars addressed on the same feature, such as lifetime or end state.

7. (a) T = 2.9 × 10⁻³ / λmax [1] = 2.9 × 10⁻³ / 1.45 × 10⁻⁷ = 2.0 × 10⁴ K [1]. (b) Atoms in the cooler outer layers absorb photons whose energies match transitions between their energy levels [1]. The missing wavelengths appear as dark lines, and each element has its own pattern, so matching the pattern identifies the element [1]. (c) T = 2.9 × 10⁻³ / 8.0 × 10⁻⁷ = 3.6 × 10³ K [1]. [5] Examiner insight: Mention the transition between energy levels in part (b). “Different elements give different lines” alone does not explain why.

8. (a) d = 1/0.0310 = 32.3 pc [1] = 32.3 × 3.09 × 10¹⁶ = 9.97 × 10¹⁷ m [1]. (b) L = 4πd²b = 4π × (9.97 × 10¹⁷)² × 2.6 × 10⁻⁹ [1] = 3.2 × 10²⁸ W [1]. (c) 3.25 × 10²⁸ / 3.83 × 10²⁶ = 85 L☉ [1]. (d) R = √(L / 4πσT⁴) [1]; 4π × 5.67 × 10⁻⁸ × 6200⁴ = 1.053 × 10⁹ so R² = 3.09 × 10¹⁹ [1]; R = 5.6 × 10⁹ m = 8.0 R☉ [1]. (e) It is only slightly hotter than the Sun yet about 85 times as luminous, so it is much larger than a main sequence star of that temperature, and lies above the main sequence [1]. [9] Examiner insight: Carry unrounded values between parts. Part (e) needs a comparison with a reference star, not just “it is big”.

9. (a) At constant L, R² ∝ 1/T⁴ so R ∝ 1/T² [1]. Halving T makes R larger by a factor of 4 [1]. (b) The star moves horizontally to the right (to lower temperature at the same luminosity) [1], into the region of red giants and super giants [1]. [4] Examiner insight: Give the proportionality before substituting. Stating “R is 4 times larger” with no link to L and T is a bare answer.

10. (a) E = 26.7 × 1.602 × 10⁻¹³ = 4.28 × 10⁻¹² J [1]. Rate = 1.0 × 10²⁹ / 4.28 × 10⁻¹² = 2.3 × 10⁴⁰ s⁻¹ [1]. (b) Each fusion uses 4 × 1.007825 × 1.661 × 10⁻²⁷ = 6.70 × 10⁻²⁷ kg [1]; mass = 2.34 × 10⁴⁰ × 6.70 × 10⁻²⁷ = 1.6 × 10¹⁴ kg s⁻¹ [1]. (c) m = L / c² = 1.0 × 10²⁹ / (3.00 × 10⁸)² = 1.1 × 10¹² kg s⁻¹ [1]. (d) t = 3.0 × 10²⁹ / 1.56 × 10¹⁴ = 1.9 × 10¹⁵ s [1] = 6.1 × 10⁷ years [1]. (e) Massive stars fuse hydrogen at a rate hundreds of times higher than the Sun, so they exhaust their fuel far sooner [1]. [8] Examiner insight: Part (c) is smaller than part (b) because only a small fraction of the mass becomes energy; saying so shows the link.

11. Axes: luminosity on the vertical axis and surface temperature on the horizontal axis, with temperature increasing towards the left [1]. A diagonal main sequence band from top left to bottom right [1]. Red giants and super giants in the upper right, white dwarfs in the lower left [1]. The near-vertical instability strip through the giant region, plus straight lines of constant radius, with larger radii towards the upper right [1]. [4] Examiner insight: Label axes and regions on the sketch itself, so the temperature direction can be credited.

Where marks are usually lost

• Putting hot stars on the right of the HR diagram.
• Using the parallax angle in degrees, or leaving it as p when d is asked.
• Using L in solar units alongside σ in SI units.
• Writing fusion “balances” gravity rather than outward pressure balancing gravity.

Next steps

• Review the revision notes and the study guide.
• Visit the IB DP Physics course page and the checklist.
• Take one of the free 10-minute diagnostics to find weak areas.
• Book a free trial class for help with any topic.

Official syllabus

Source: IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. These questions are original and are not an IB publication.