The Sun turns hydrogen into helium, as you saw in How the Sun Shines. The energy comes from the difference in mass, worked out as in energy in nuclear reactions. But four protons never fuse in a single collision; the change happens through a chain of two-particle reactions.
Why it works
Using atomic masses (which include electrons, so the energy of the positrons annihilating is counted for you): $4 \times 1.007825\text{ u} = 4.031300$ u before, $4.002603$ u after. The missing $0.028697$ u is $0.71\%$ of the mass, and at 931.5 MeV per u it is $26.7$ MeV. A small part of that, on average about $0.6$ MeV in the Sun's main branch, leaves with neutrinos.
The proton–proton chain (main branch)
- $p + p \to {}^{2}\text{H} + e^+ + \nu_e$ (needs the weak force to turn a proton into a neutron, so it is very slow)
- $^{2}\text{H} + p \to {}^{3}\text{He} + \gamma$ (fast)
- $^{3}\text{He} + {}^{3}\text{He} \to {}^{4}\text{He} + 2p$ (each of the first two steps happens twice)
Key idea
Net: $4p \to {}^{4}\text{He} + 2e^+ + 2\nu_e$, releasing $26.7$ MeV. In the CNO cycle, carbon-12 captures four protons in turn (with two beta-plus decays) and ends as carbon-12 plus a helium-4 nucleus: carbon, nitrogen and oxygen are catalysts. The CNO cycle is very sensitive to temperature and dominates in stars above about 1.3 solar masses; in the Sun it supplies about 1% of the energy.
Worked examples
Worked example: Energy per helium nucleus
Using $m(^{1}\text{H}) = 1.007825$ u, $m(^{4}\text{He}) = 4.002603$ u and 931.5 MeV per u, find the energy released per helium-4 nucleus.
- $\Delta m = 4(1.007825) - 4.002603 = 0.028697$ u.
- $Q = 0.028697 \times 931.5$ MeV.
Answer: $Q = 26.7$ MeV.
Worked example: The Sun's fuel budget
How many helium nuclei does the Sun make each second, and how much hydrogen does it use? ($L = 3.83 \times 10^{26}$ W, $1\text{ MeV} = 1.602 \times 10^{-13}$ J.)
- Energy per reaction: $26.7 \times 1.602 \times 10^{-13} = 4.28 \times 10^{-12}$ J.
- Reactions per second: $3.83 \times 10^{26}/4.28 \times 10^{-12} = 8.95 \times 10^{37}$.
- Hydrogen: $4 \times 1.67 \times 10^{-27}\text{ kg} \times 8.95 \times 10^{37} = 6.0 \times 10^{11}$ kg/s.
Answer: About $9 \times 10^{37}$ helium nuclei and 600 million tonnes of hydrogen per second; $L/c^2 = 4.3 \times 10^{9}$ kg of mass becomes energy each second.
Worked example: One step of the CNO cycle
Find $Q$ for $^{12}\text{C} + {}^{1}\text{H} \to {}^{13}\text{N} + \gamma$, with $m(^{13}\text{N}) = 13.005739$ u.
- $\Delta m = 12.000000 + 1.007825 - 13.005739 = 0.002086$ u.
- $Q = 0.002086 \times 931.5$ MeV.
Answer: About $1.94$ MeV, carried off by the gamma ray.
Common mistake
Thinking stars fuse four protons at once in a single collision. Four particles meeting at one point at one instant essentially never happens. Fusion goes through two-particle steps, and the first step of the pp chain is so slow (a proton in the Sun's core waits billions of years on average to make one) that the Sun burns its fuel gradually instead of exploding. Check any proposed reaction: count only two nuclei on the left.