Nuclear fusion in stars — pp chain and CNO cycle basics

Question

Explain the proton-proton (pp) chain and the CNO cycle as sources of stellar energy. What is the net nuclear reaction in each case? Calculate the energy released per cycle given that the mass defect for $4p \rightarrow \text{He-4}$ is 0.0287 u.

(JEE Main 2022, similar pattern)

Solution — Step by Step

The pp chain dominates in stars like our Sun (core temperature ~ $1.5 \times 10^7$ K). The net reaction fuses four protons into one helium-4 nucleus:

4\,^1_1\text{H} \rightarrow \,^4_2\text{He} + 2e^+ + 2\nu_e + \text{energy}

The chain proceeds in three steps: (1) two protons fuse to form deuterium, (2) deuterium fuses with a proton to form He-3, (3) two He-3 nuclei fuse to form He-4 and release two protons.

In stars hotter than the Sun (core temperature > $1.8 \times 10^7$ K), the CNO cycle dominates. Carbon, nitrogen, and oxygen act as catalysts — they are recycled. The net reaction is the same:

4\,^1_1\text{H} \rightarrow \,^4_2\text{He} + 2e^+ + 2\nu_e + \text{energy}

Carbon-12 starts the cycle and is regenerated at the end. It facilitates the fusion but is not consumed.

Mass defect: $\Delta m = 0.0287$ u

Using $E = \Delta m \times c^2$ and $1\text{ u} \times c^2 = 931.5$ MeV:

E = 0.0287 \times 931.5 = 26.73 \text{ MeV}

\boxed{E \approx 26.7 \text{ MeV per fusion cycle}}

Of this, about 2 MeV is carried away by neutrinos (undetectable), so the effective energy deposited in the star is about 24.7 MeV.

Why This Works

Nuclear fusion releases energy because the binding energy per nucleon increases when light nuclei combine. Helium-4 is exceptionally stable — its binding energy per nucleon (7.07 MeV) is much higher than hydrogen’s. The “missing mass” (mass defect) is converted to energy via $E = mc^2$ .

The Coulomb barrier between protons is enormous, but quantum tunnelling allows fusion to occur even at temperatures where classical physics would say it’s impossible. This is why fusion requires temperatures of millions of kelvin but not billions.

Alternative Method

You can also calculate the energy from individual masses. Mass of 4 protons = $4 \times 1.00728$ u = $4.02912$ u. Mass of He-4 = $4.00260$ u. Mass of 2 positrons = $2 \times 0.00055$ u = $0.00110$ u.

$\Delta m = 4.02912 - 4.00260 - 0.00110 = 0.02542$ u, giving $E = 23.7$ MeV (after accounting for neutrino energy). The slight difference from the simpler calculation arises from how the positron annihilation energy is counted.

For JEE, the key facts: (1) pp chain dominates in the Sun, CNO in hotter stars, (2) net reaction is the same for both, (3) energy per cycle is ~26.7 MeV, (4) $1\text{ u} \times c^2 = 931.5$ MeV. These four points cover all typical MCQ patterns on stellar fusion.

Common Mistake

Students sometimes confuse fusion with fission and write that heavy nuclei are formed in stars. Stars form light elements (up to iron) through fusion. Elements heavier than iron are formed in supernovae via rapid neutron capture (r-process), not in normal stellar fusion. Also, do not confuse the pp chain with beta decay — the pp chain involves proton-proton collisions, not spontaneous nuclear decay.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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