IB Diploma Physics · topic E.5
Fusion and stars
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A fusion power estimate and a star's fuel lifetime
In an idealised net stellar fusion cycle, four hydrogen nuclei ultimately produce one helium-4 nucleus plus other particles and radiation, releasing 26.7 MeV per completed cycle. Use 1 MeV = 1.60×10⁻¹³ J and c = 3.0×10⁸ m s⁻¹. The supplied graph gives cumulative energy from completed cycles. A star emits 4.0×10²⁶ W and has 2.0×10²⁹ kg of available hydrogen fuel; take four hydrogen nuclei as 6.68×10⁻²⁷ kg. Supplied data (Completed cycles / 10³⁸; Energy / 10²⁶ J): (0, 0), (1, 4.272), (2, 8.544), (3, 12.816). (a) Find energy per cycle, mass defect per cycle and the cycle rate supplying the stated luminosity. [3 marks] (b) Estimate the fuel-consumption rate and idealised lifetime, taking 1 yr = 3.16×10⁷ s. Explain two limits of this estimate. [3 marks]
Fusion energy release
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Energy / 10²⁶ J
Horizontal axis (x): Completed cycles / 10³⁸
Read the exact data as a table
| Completed cycles / 10³⁸ | Energy / 10²⁶ J |
|---|---|
| 0 | 0 |
| 1 | 4.272 |
| 2 | 8.544 |
| 3 | 12.816 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find energy per cycle, mass defect per cycle and the cycle rate supplying the stated luminosity.
Reveal the worked solution
- Convert 26.7 MeV to joules: E = 26.7(1.60×10⁻¹³) = 4.272×10⁻¹² J.
- The decrease in total rest mass is Δm = E/c² = 4.272×10⁻¹²/(9.0×10¹⁶)≈4.75×10⁻²⁹ kg per net cycle.
- Cycle rate is L/E = 4.0×10²⁶/(4.272×10⁻¹²)≈9.36×10³⁷ s⁻¹. Nuclear binding-energy differences, rather than chemical combustion, power this process.
- Each cycle uses about 6.68×10⁻²⁷ kg of hydrogen, so consumption is (9.36×10³⁷)(6.68×10⁻²⁷)≈6.25×10¹¹ kg s⁻¹.
- Time is available fuel divided by its consumption rate: 2.0×10²⁹/(6.25×10¹¹)≈3.20×10¹⁷ s≈1.01×10¹⁰ yr.
- Luminosity changes as the star evolves, only part of its hydrogen is accessible to core fusion, and some fusion energy can escape in neutrinos. A full stellar model must account for structure, composition and evolving reactions.
Answer (a) Energy = 4.272×10⁻¹² J; mass defect≈4.75×10⁻²⁹ kg; rate≈9.36×10³⁷ cycles s⁻¹. (b) Fuel consumption≈6.25×10¹¹ kg s⁻¹; idealised lifetime≈1.01×10¹⁰ years.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Two stars at the same temperature with very different luminosities
A main-sequence reference star and a red giant both have surface temperature 4000 K. The giant has luminosity 400 times the reference. A third star is a white dwarf with temperature 12000 K and luminosity 0.01 times the reference. Use L = 4πR²σT⁴. The supplied data place the three stars on temperature–luminosity axes; luminosities are relative to the reference star. Supplied data (Surface temperature / K; Relative luminosity): (4000, 1), (4000, 400), (12000, 0.01). (a) Determine the radius of the giant and dwarf relative to the reference. Explain why colour alone cannot determine luminosity. [3 marks] (b) Relate these observations to an H–R diagram and distinguish low-mass from high-mass stellar endpoints. [3 marks]
Three stellar observations
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Relative luminosity
Horizontal axis (x): Surface temperature / K
Read the exact data as a table
| Surface temperature / K | Relative luminosity |
|---|---|
| 4000 | 1 |
| 4000 | 400 |
| 12000 | 0.01 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Determine the radius of the giant and dwarf relative to the reference. Explain why colour alone cannot determine luminosity.
Reveal the worked solution
- At equal temperature, L∝R², so the giant radius ratio is √400 = 20.
- For the dwarf, Rw/Rref = √(Lw/Lref)(Tref/Tw)² = √0.01×(1/3)² = 1/90.
- Colour indicates a characteristic surface temperature, while luminosity also depends on emitting area. A hot small star may be less luminous than a cool large one.
- A standard H–R diagram places hotter temperatures toward the left and often uses logarithmic luminosity. The supplied linear data plot must be relabelled or transformed before being read as a standard H–R diagram.
- A Sun-like star can evolve into a red giant and leave a white dwarf after shedding outer layers. A white dwarf is not a newly formed small main-sequence star.
- Sufficiently massive stars can undergo core-collapse supernovae and leave neutron stars or black holes. Initial mass, mass loss and interactions matter; a single universal life-cycle chain is inadequate.
Answer (a) Giant radius = 20 times reference; dwarf radius = 1/90 of reference. (b) An H–R diagram compares intrinsic stellar properties, not spatial locations; stellar mass strongly influences evolution and endpoint.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.