8E

Nucleosynthesis in stars

How stars make energy and new elements: applying E = mc² to the Sun's mass loss, the proton–proton chain and the CNO cycle in main-sequence stars, the triple-alpha process and later burning stages in post-main-sequence stars, why fusion stops at iron, and how the s-process, r-process, supernovae and neutron-star mergers make the heavy elements.

Each second the Sun turns about 600 million tonnes of hydrogen into helium, and in the process about 4 million tonnes of matter simply disappears, converted into energy. It has done this for 4.6 billion years and has lost less than a tenth of a percent of its mass. Until the 1920s nobody could explain what kept the Sun shining for so long. The answer came from $E = mc^2$.

What powers a star?

In the 1800s, the best idea was that the Sun was slowly shrinking, turning gravitational potential energy into heat. Lord Kelvin calculated that this could keep it shining for only about 30 million years, far less than geologists needed for the Earth's rocks and Darwin needed for evolution.

In 1920 Arthur Eddington pointed out that four hydrogen atoms have about 0.7% more mass than one helium atom. If the Sun could fuse hydrogen into helium, that mass difference would be released as energy, enough for billions of years. Cecilia Payne's 1925 result that the Sun is mostly hydrogen supplied the fuel, and in 1938–39 Hans Bethe worked out the reaction steps.

Worked example: the Sun's mass loss

The Sun's luminosity is $3.83 \times 10^{26}$ W. How much mass does it convert to energy each second?

The question

Each second the Sun radiates $E = 3.83 \times 10^{26}$ J.

Energy per second

$m = \dfrac{E}{c^2} = \dfrac{3.83 \times 10^{26}}{(3.00 \times 10^{8})^2} = 4.26 \times 10^{9}$ kg

Rearrange E = mc²

About 4.3 million tonnes per second. Since the mass converted is 0.7% of the hydrogen fused, the Sun fuses about $4.26 \times 10^{9}/0.007 \approx 6 \times 10^{11}$ kg (600 million tonnes) of hydrogen each second.

Interpret

The proton–proton chain

Fusion needs nuclei to get close enough (about $10^{-15}$ m) for the strong nuclear force to act, despite their electrostatic repulsion. That needs extremely high temperatures: about 10 to 15 million K in the Sun's core. Even then, most protons don't have enough kinetic energy to get over the repulsion. Fusion only happens because of quantum tunnelling, which gives a small chance of getting through the barrier.

In stars up to about 1.3 solar masses, including the Sun, most energy comes from the proton–proton (p–p) chain:

The main branch of the proton–proton chain. The first two steps happen twice for every helium-4 nucleus made, and two protons are returned at the end.

$$^1_1\text{H} + {}^1_1\text{H} \rightarrow {}^2_1\text{H} + {}^0_{+1}e + \nu_e$$

$$^2_1\text{H} + {}^1_1\text{H} \rightarrow {}^3_2\text{He} + \gamma$$

$$^3_2\text{He} + {}^3_2\text{He} \rightarrow {}^4_2\text{He} + 2\,{}^1_1\text{H}$$

The first step is extraordinarily slow, because one proton has to turn into a neutron by the weak interaction while the two protons are briefly close. An average proton in the Sun's core waits billions of years for it. That slowness is why the Sun lasts 10 billion years instead of exploding. The positrons annihilate with electrons immediately, adding their energy to the core.

Overall: $4\,{}^1_1\text{H} \rightarrow {}^4_2\text{He} + 2\,{}^0_{+1}e + 2\nu_e + 2\gamma$, releasing 26.7 MeV once the positrons have annihilated. A small fraction (about 2%) is carried away by the neutrinos, which pass straight out of the Sun.

Worked example: energy from the p–p chain

Use atomic masses to calculate the energy released when four hydrogen atoms become one helium-4 atom. ($^1$H = 1.007825 u, $^4$He = 4.002603 u)

The question

$\Delta m = 4 \times 1.007825 - 4.002603 = 4.031300 - 4.002603 = 0.028697$ u

Mass defect

$E = 0.028697 \times 931.5 = 26.73$ MeV

Convert

Atomic masses include the electrons. Four H atoms bring 4 electrons; the He atom keeps 2. The other 2 are exactly the electrons annihilated by the 2 positrons, so this answer already includes the annihilation energy. Fraction converted: $0.028697/4.0313 = 0.71\%$.

Why atomic masses work

The CNO cycle

In stars more massive than about 1.3 solar masses, with core temperatures above about 17 million K, hydrogen is fused mainly through the carbon–nitrogen–oxygen (CNO) cycle. Carbon-12 acts as a catalyst: it captures protons one at a time, goes through nitrogen and oxygen isotopes with two β⁺ decays, and is returned at the end along with a new helium-4 nucleus.

The CNO cycle. Four protons go in, one helium-4 nucleus comes out, and the carbon-12 is returned. The net energy release is the same 26.7 MeV as the p–p chain.

$$^{12}\text{C} \xrightarrow{+p} {}^{13}\text{N} \xrightarrow{\beta^+} {}^{13}\text{C} \xrightarrow{+p} {}^{14}\text{N} \xrightarrow{+p} {}^{15}\text{O} \xrightarrow{\beta^+} {}^{15}\text{N} \xrightarrow{+p} {}^{12}\text{C} + {}^{4}\text{He}$$

The net reaction is the same as for the p–p chain, $4\,{}^1\text{H} \rightarrow {}^4\text{He} + 2e^+ + 2\nu_e$, so the same mass is converted and the same total energy is released. The difference is the rate. Each step has to push a proton into a nucleus with charge +6 or +7, so the electrostatic barrier is much higher. The CNO rate rises extremely steeply with temperature (roughly as $T^{17}$ or more, compared with about $T^4$ for the p–p chain). Below about 17 million K, the p–p chain wins; above it, the CNO cycle takes over.

In the Sun the CNO cycle supplies only about 1% of the energy. In 2020 the Borexino detector in Italy detected neutrinos from the Sun's CNO cycle for the first time, confirming Bethe's prediction directly.

Proton–proton chainCNO cycle
Dominant inMain-sequence stars below about 1.3 M☉ (including the Sun)Main-sequence stars above about 1.3 M☉
Core temperaturebelow about 17 million Kabove about 17 million K
Net reaction$4\,^1\text{H} \rightarrow {}^4\text{He} + 2e^+ + 2\nu_e$Same
Energy per He-426.7 MeV26.7 MeV
Catalyst neededNoYes: C, N and O nuclei, so the star must contain some heavier elements from earlier stars
Temperature sensitivityModerateExtreme
InteractiveStep through the p–p chain and the CNO cycle
Try this
  • With the p–p chain selected, press Next step until all seven reactions are done. Check the total: 26.7 MeV.
  • Which single step releases the most energy? Which is the slowest, and why?
  • Switch to the CNO cycle and step through it. Does the total come out the same? Which nucleus is used and then returned?
  • Set the core temperature to 15.7 MK (the Sun's core). What share does the CNO cycle supply? (About 1%.)
  • Find the temperature where the two processes contribute equally (about 18 MK in this simplified model). Then try 25 MK. What does this tell you about massive stars?

Post-main-sequence nucleosynthesis

When a star's core hydrogen runs out, the core contracts and heats up. At about 100 million K, helium nuclei can fuse by the triple-alpha process:

The triple-alpha process. Beryllium-8 falls apart in about 10⁻¹⁶ s, so a third helium nucleus must arrive almost at once. Fred Hoyle predicted an energy level of carbon-12 that makes this possible, and it was then found.

$$^4_2\text{He} + {}^4_2\text{He} \rightleftharpoons {}^8_4\text{Be}, \qquad {}^8_4\text{Be} + {}^4_2\text{He} \rightarrow {}^{12}_{\ 6}\text{C} + \gamma$$

Some carbon captures another helium nucleus to make oxygen: $^{12}\text{C} + {}^4\text{He} \rightarrow {}^{16}\text{O} + \gamma$. This is where almost all the carbon and much of the oxygen in the universe was made, including the atoms in your body.

In stars above about 8 solar masses the sequence continues, each stage needing a higher temperature and lasting a shorter time:

Burning stageMain productsCore temperatureDuration (25 M☉ star)
HydrogenHe40 million K7 million years
HeliumC, O200 million K700 000 years
CarbonNe, Na, Mg800 million Kabout 600 years
NeonO, Mg1.5 billion Kabout 1 year
OxygenSi, S2 billion Kabout 6 months
SiliconFe, Ni3.5 billion Kabout 1 day

Fusion stops at iron and nickel because nuclei around mass number 56 have the highest binding energy per nucleon (page 8K). Fusing them further would absorb energy, not release it.

Making elements heavier than iron

Elements beyond iron are made by adding neutrons, which have no charge and so feel no repulsion. A nucleus captures neutrons until it becomes unstable, then β⁻ decays, turning a neutron into a proton and moving up one element.

  • The s-process (slow neutron capture) happens in AGB stars over thousands of years. Each capture is usually followed by a β⁻ decay before the next capture. It makes about half the heavy elements, including much of the strontium, barium and lead.
  • The r-process (rapid neutron capture) needs an enormous flood of neutrons, so that many are captured within seconds. It happens in neutron-star mergers and possibly some rare supernovae. In 2017 the merger GW170817 was detected by gravitational waves and then observed by telescopes worldwide, including Australian ones; its glow showed signs of freshly made heavy elements. Much of the gold, platinum and uranium in the universe was made this way.

Supernovae and stellar winds return all these elements to space, where they mix into new gas clouds. The Sun, formed 4.6 billion years ago, inherited elements made by generations of earlier stars. As astronomers like to say, we are made of star stuff.

The Sun in false-colour orange, its surface mottled and bright active regions near the edge
The Sun imaged in extreme ultraviolet by NASA's Solar Dynamics Observatory. Its light comes from the p–p chain in a core about a quarter of its radius across.Photo: NASA/SDO (AIA), Public domain

Test yourself

  1. Which of these is the net result of both the p–p chain and the CNO cycle?

  2. Why does the CNO cycle dominate only in stars with hotter cores?

  3. A star of 0.8 solar masses is on the main sequence. Its energy comes mainly from

  4. In the triple-alpha process, carbon-12 is formed from

  5. A star converts $2.0 \times 10^{9}$ kg of mass to energy each second. Its luminosity is

Label the nuclei in the CNO cycle.

+¹H, out γ¹²Cout e⁺ + ν¹³N+¹H, out γ¹³C+¹H, out γ¹⁴Nout e⁺ + ν¹⁵O+¹H, out ⁴He¹⁵Ncarbon is acatalystCNO cycleSame net result as thep–p chain: 4 ¹H → ⁴He,about 26.7 MeV.Needs core temperaturesabove about 17 million K,because protons mustovercome the repulsion ofcarbon (6+) and nitrogen(7+) nuclei.Main energy source instars above ≈ 1.3 M☉.Only about 1% of theSun’s energy.

In which stage does each process happen?

In the CNO cycle, carbon-12 acts as a . Each helium-4 nucleus made releases about MeV. Fusion in stars stops at because it has the highest binding energy per nucleon. Elements heavier than this are made by .

5 marks

Compare the proton–proton chain and the CNO cycle as sources of energy in main-sequence stars.

3 marks

The CNO step $^{14}\text{N} + {}^1\text{H} \rightarrow {}^{15}\text{O} + \gamma$ involves the masses $^{14}$N = 14.003074 u, $^1$H = 1.007825 u and $^{15}$O = 15.003066 u. Calculate the energy released, in MeV and in joules.