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E.5

Fusion and stars

Theme E · Nuclear and quantum physics · SL and HL

This is all that survives of the old Option D, Astrophysics, and it is much smaller than that option was. Stellar structure, the mass–luminosity relationship and the Hertzsprung–Russell diagram remain. Cosmology — Hubble’s law, the expanding universe, dark matter and the fate of the universe — is no longer in the syllabus. It remains excellent Extended Essay territory, but it will not be examined.

🎯What you need to be able to do

  • Explain nuclear fusion and why it requires extreme temperatures.
  • Describe the equilibrium that keeps a main-sequence star stable.
  • Use luminosity and apparent brightness.
  • Interpret the Hertzsprung–Russell diagram and the mass–luminosity relationship.
  • Outline stellar evolution and the possible end states.

📚The physics

Fusion needs heat because nuclei repel. Two positively charged nuclei must be brought within about \( 10^{-15} \) m for the strong nuclear force to take over, and to get that close they must overcome the electric repulsion — the Coulomb barrier. Only at temperatures of order \( 10^{7} \) K do nuclei move fast enough. This is why fusion happens in stellar cores and is so difficult to sustain on Earth.

In the Sun the proton–proton chain converts four hydrogen nuclei into one helium-4 nucleus, releasing about 26 MeV. Referring back to E.4: helium-4 sits higher on the binding energy per nucleon curve than hydrogen, so the reaction runs downhill in energy terms.

Stellar equilibrium is the balance a main-sequence star holds for most of its life: gravity pulling inward against radiation and gas pressure from fusion pushing outward. The balance is self-correcting. If the core cools, gravity wins slightly, the core compresses, the temperature rises and fusion speeds up, pushing back out. A star is a thermostat, and this is why a star’s life is stable rather than explosive.

Luminosity and apparent brightness are different quantities and the distinction is examined. Luminosity \(L\) is the total power the star radiates, from Stefan’s law \( L = \sigma A T^{4} \). Apparent brightness \(b\) is the power we receive per square metre:

\[ b = \frac{L}{4\pi d^{2}} \]

which falls off as the inverse square of distance. A dim-looking star may be an intrinsically bright one that is very far away — the reason apparent brightness alone tells you almost nothing about a star.

The Hertzsprung–Russell diagram plots luminosity against surface temperature, with temperature increasing to the left, which catches people out. Most stars lie on a diagonal band, the main sequence, running from hot bright stars at the top left to cool dim ones at the bottom right. Off that band sit the red giants (cool but luminous, so they must be enormous) and the white dwarfs (hot but faint, so they must be tiny). Reading a star’s size off its position is exactly the reasoning \( L = \sigma A T^{4} \) supports.

The mass–luminosity relationship for main-sequence stars is roughly \( L \propto M^{3.5} \). That steep power has a consequence worth stating: a star ten times the Sun’s mass is around three thousand times more luminous, so it burns through its fuel far faster and lives a far shorter life. Massive stars die young.

Stellar evolution. When core hydrogen runs out, the core contracts and heats while the outer layers expand and cool — a red giant. What follows depends on mass. A star like the Sun sheds its outer layers as a planetary nebula and leaves a white dwarf, supported by electron degeneracy pressure, but only if its remnant mass is below the Chandrasekhar limit of about 1.4 solar masses. Above that the collapse continues to a neutron star, and beyond roughly 3 solar masses of remnant, to a black hole. Fusion in massive stars builds elements only up to iron, because iron sits at the peak of the binding energy curve and fusing beyond it absorbs energy rather than releasing it — which is why the core collapse happens at all.

✏️Worked example

A star has surface temperature 9600 K and radius \( 1.8 \times 10^{9} \) m. Take \( \sigma = 5.67 \times 10^{-8} \) W m\(^{-2}\) K\(^{-4}\), and the Sun’s luminosity as \( 3.85 \times 10^{26} \) W.

(a) Luminosity. Surface area \( A = 4\pi r^{2} = 4\pi \times (1.8 \times 10^{9})^{2} = 4.07 \times 10^{19} \) m\(^{2}\). Then \( L = \sigma A T^{4} = 5.67 \times 10^{-8} \times 4.07 \times 10^{19} \times 9600^{4} \). Since \( 9600^{4} = 8.50 \times 10^{15} \), \( L = 1.96 \times 10^{28} \) W.

(b) Compared with the Sun. \( 1.96 \times 10^{28}/3.85 \times 10^{26} = \) about 51 times the Sun’s luminosity.

(c) Apparent brightness at 25 light-years. \( d = 25 \times 9.46 \times 10^{15} = 2.37 \times 10^{17} \) m. Then \( b = L/(4\pi d^{2}) = 1.96 \times 10^{28}/(4\pi \times 5.60 \times 10^{34}) = 2.8 \times 10^{-8} \) W m\(^{-2}\).

(d) Where does it sit on the HR diagram? Hot and about fifty times more luminous than the Sun — upper left, on the main sequence. These are roughly the figures for Sirius A.

The trap in (a). The fourth power. Getting \( 9600^{4} \) wrong by a factor of ten wrecks the answer, and it is easy to slip when entering it. Compute \(T^{4}\) separately and check its order of magnitude before multiplying.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A star has radius \( 7.0 \times 10^{8} \) m and surface temperature 5800 K. Find its luminosity. Take \( \sigma = 5.67 \times 10^{-8} \) W m\(^{-2}\) K\(^{-4}\).
\( A = 4\pi r^{2} = 4\pi (7.0 \times 10^{8})^{2} = 6.16 \times 10^{18} \) m\(^{2}\), and \( 5800^{4} = 1.13 \times 10^{15} \). So \( L = \sigma A T^{4} = 4.0 \times 10^{26} \) W — these are the Sun’s figures.
2. Find the apparent brightness of that star at a distance of \( 1.5 \times 10^{11} \) m (1 AU).
\( b = \dfrac{L}{4\pi d^{2}} = \dfrac{3.95 \times 10^{26}}{4\pi (1.5 \times 10^{11})^{2}} = \dfrac{3.95 \times 10^{26}}{2.83 \times 10^{23}} = 1.4 \times 10^{3} \) W m\(^{-2}\) — the solar constant, as it should be.
3. A main-sequence star has five times the mass of the Sun. Estimate its luminosity relative to the Sun, and comment on its lifetime.
\( L \propto M^{3.5} \), so \( L/L_\odot = 5^{3.5} \approx 280 \). It is roughly 280 times more luminous while having only five times the fuel, so it burns through that fuel far faster and lives a much shorter life. Massive stars die young.
4. Find the wavelength at which a 5800 K star radiates most strongly, using \( \lambda_{\max}T = 2.9 \times 10^{-3} \) m K.
\( \lambda_{\max} = \dfrac{2.9 \times 10^{-3}}{5800} = 5.0 \times 10^{-7} \) m, or 500 nm — green-yellow, in the middle of the visible band.
5. Explain why fusion in a massive star builds elements only as far as iron.
Iron sits at the peak of the binding energy per nucleon curve, so it is the most tightly bound nucleus. Fusing anything beyond iron moves nucleons down the curve to a less tightly bound state, which absorbs energy rather than releasing it. The core can therefore no longer generate the outward pressure that balances gravity, and it collapses.
6. State what the Chandrasekhar limit is and what it determines.
It is a remnant mass of about 1.4 solar masses. Below it, electron degeneracy pressure is enough to support the remnant and it becomes a white dwarf. Above it that support fails and the collapse continues to a neutron star, or beyond roughly 3 solar masses of remnant, to a black hole.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — stellar structure, fusion and the HR diagram
  • NASA — stellar evolution and the life cycles of stars