IB · E · extended response
A fusion power estimate and a star's fuel lifetime
Nuclear and quantum physics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
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(a)
Find energy per cycle, mass defect per cycle and the cycle rate supplying the stated luminosity.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Convert 26.7 MeV to joules: E = 26.7(1.60×10⁻¹³) = 4.272×10⁻¹² J.
- 2
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.
- 3
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.
(b)
Estimate the fuel-consumption rate and idealised lifetime, taking 1 yr = 3.16×10⁷ s. Explain two limits of this estimate.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
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.
- 2
Time is available fuel divided by its consumption rate: 2.0×10²⁹/(6.25×10¹¹)≈3.20×10¹⁷ s≈1.01×10¹⁰ yr.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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