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Physics · Ch 14 — Nuclei

Nuclear Fusion – Energy Generation in Stars

14.7.2

Nuclear Fusion – Energy Generation in Stars

The Energy Source of Stars: Nuclear Fusion

When two light nuclei merge to form a heavier nucleus, the process is called nuclear fusion. Because the heavier nucleus is more tightly bound (as seen from the binding energy per nucleon curve), the mass of the product is less than the sum of the masses of the reactants. This missing mass is converted into energy, which is released.

This is the fundamental process that powers the Sun and all other stars. The fuel is primarily hydrogen, and the product is helium.

Examples of Fusion Reactions

The textbook gives three specific examples of energy-releasing fusion reactions:

  1. 11H+11H→12H+e++ν+0.42 MeV^{1}_{1}\text{H} + ^{1}_{1}\text{H} \rightarrow ^{2}_{1}\text{H} + e^{+} + \nu + 0.42 \text{ MeV}

    Two protons fuse to form a deuteron (a proton and a neutron), a positron, and a neutrino.

  2. 12H+12H→23He+n+3.27 MeV^{2}_{1}\text{H} + ^{2}_{1}\text{H} \rightarrow ^{3}_{2}\text{He} + n + 3.27 \text{ MeV}

    Two deuterons fuse to form the light isotope of helium, 23He^{3}_{2}\text{He}, and a neutron.

  3. 12H+12H→13H+11H+4.03 MeV^{2}_{1}\text{H} + ^{2}_{1}\text{H} \rightarrow ^{3}_{1}\text{H} + ^{1}_{1}\text{H} + 4.03 \text{ MeV}

    Two deuterons fuse to form a triton (a hydrogen isotope with two neutrons) and a proton.

Note

The energy released in each reaction (0.42 MeV, 3.27 MeV, 4.03 MeV) is the kinetic energy of the products. This energy is much smaller than the energy released in a typical fission reaction (≈200 MeV), but the number of fusion reactions possible in a star is astronomically large.

The Coulomb Barrier and the Need for High Temperature

For fusion to occur, two positively charged nuclei must come close enough for the attractive, short-range strong nuclear force to act. However, they are both positively charged, so they experience a strong electrostatic repulsion (the Coulomb force). This repulsion creates an energy barrier that the nuclei must overcome.

The height of this Coulomb barrier depends on the charges (Z1eZ_1 e and Z2eZ_2 e) and the radii of the two nuclei. For two protons, the barrier height is approximately 400 keV. For nuclei with higher charges, the barrier is even higher.

Watch out

A common mistake is to think that the kinetic energy of the nuclei must be greater than the barrier height. In reality, quantum mechanical tunneling allows a small fraction of nuclei with energies below the barrier to fuse. However, the probability of tunneling increases dramatically as the kinetic energy approaches the barrier height.

Estimating the Required Temperature

We can estimate the temperature required for fusion in a gas of protons. The average kinetic energy of a particle in a gas at temperature TT is given by 32kT\frac{3}{2}kT, where kk is the Boltzmann constant. If we set this equal to the barrier height of 400 keV, we get:

32kT≈400 keV\frac{3}{2}kT \approx 400 \text{ keV}

Solving for TT:

T≈23×400×103 eV8.617×10−5 eV/K≈3×109 KT \approx \frac{2}{3} \times \frac{400 \times 10^3 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K}} \approx 3 \times 10^9 \text{ K}

This is a temperature of about 3 billion Kelvin.

Important

When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the Coulomb repulsion, it is called thermonuclear fusion. This is the source of energy in the interior of stars.

The Sun's Temperature: A Puzzle and Its Solution

The interior of the Sun has a temperature of only about 1.5×1071.5 \times 10^7 K (15 million Kelvin). This is considerably less than the 3 billion Kelvin estimated above. How, then, can fusion occur in the Sun?

The answer lies in the fact that the 3 billion Kelvin estimate is for particles of average energy. In any gas at a given temperature, the particles have a wide distribution of kinetic energies (a Maxwell-Boltzmann distribution). A tiny fraction of protons have energies far above the average. It is these high-energy protons, in the tail of the distribution, that are able to overcome the Coulomb barrier and initiate fusion. Furthermore, quantum tunneling allows some protons with energies even lower than the barrier to fuse.

The Proton-Proton (p-p) Cycle

The fusion of hydrogen into helium in the Sun is not a single reaction but a multi-step process called the proton-proton (p-p) cycle. The textbook presents this cycle as a set of four reactions:

  1. 11H+11H→12H+e++ν+0.42 MeV^{1}_{1}\text{H} + ^{1}_{1}\text{H} \rightarrow ^{2}_{1}\text{H} + e^{+} + \nu + 0.42 \text{ MeV}

    (Two protons fuse to form deuterium, a positron, and a neutrino.)

  2. e++e−→γ+γ+1.02 MeVe^{+} + e^{-} \rightarrow \gamma + \gamma + 1.02 \text{ MeV}

    (The positron from the first reaction annihilates with an ambient electron, producing two gamma-ray photons.)

  3. 12H+11H→23He+γ+5.49 MeV^{2}_{1}\text{H} + ^{1}_{1}\text{H} \rightarrow ^{3}_{2}\text{He} + \gamma + 5.49 \text{ MeV}

    (A deuteron fuses with another proton to form the light helium isotope 23He^{3}_{2}\text{He} and a gamma ray.)

  4. 23He+23He→24He+11H+11H+12.86 MeV^{3}_{2}\text{He} + ^{3}_{2}\text{He} \rightarrow ^{4}_{2}\text{He} + ^{1}_{1}\text{H} + ^{1}_{1}\text{H} + 12.86 \text{ MeV}

    (Two 23He^{3}_{2}\text{He} nuclei fuse to form ordinary helium-4 and two protons.)

Note

For the fourth reaction to occur, the first three reactions must occur twice. This is because the fourth reaction requires two 23He^{3}_{2}\text{He} nuclei, and each 23He^{3}_{2}\text{He} nucleus is produced by one set of reactions (i), (ii), and (iii).

The Net Effect of the p-p Cycle

If we combine two of each of the first three reactions with the fourth reaction, we get the net effect:

2×(i)+2×(ii)+2×(iii)+(iv)2 \times (\text{i}) + 2 \times (\text{ii}) + 2 \times (\text{iii}) + (\text{iv})

Doing this bookkeeping carefully — the two positrons produced in step (i) each immediately annihilate with an ambient electron (step (ii)), so the net process never actually has free positrons left over — this yields:

411H+2e−→24He+2ν+6γ+26.7 MeV4^{1}_{1}\text{H} + 2e^{-} \rightarrow ^{4}_{2}\text{He} + 2\nu + 6\gamma + 26.7 \text{ MeV}

or, written to show explicitly how many electrons are consumed by annihilation versus how many remain in the final helium atom (2 electrons are needed to neutralise the He nucleus into a neutral atom):

(411H+4e−)→(24He+2e−)+2ν+6γ+26.7 MeV\left(4^{1}_{1}\text{H} + 4e^{-}\right) \rightarrow \left(^{4}_{2}\text{He} + 2e^{-}\right) + 2\nu + 6\gamma + 26.7 \text{ MeV}

411H+2e−→24He+2ν+6γ+26.7 MeV4^{1}_{1}\text{H} + 2e^{-} \rightarrow ^{4}_{2}\text{He} + 2\nu + 6\gamma + 26.7 \text{ MeV}

Note

Common pitfall: it's tempting to write the net reaction with the two positrons from step (i) still present as products ("41H→4He+2e++2ν+2γ4^1\text{H} \to {}^4\text{He} + 2e^+ + 2\nu + 2\gamma") — but that form is wrong on two counts: it isn't what actually happens (both positrons annihilate before the process is 'done'), and if you tried to energy-balance THAT specific equation you'd have to leave out the annihilation energy from step (ii), giving 2(0.42)+2(5.49)+12.86=24.68 MeV2(0.42)+2(5.49)+12.86 = 24.68\ \text{MeV} — not 26.7 MeV26.7\ \text{MeV}. The 26.7 MeV26.7\ \text{MeV} figure is only correct for the fully annihilated, electron-consuming form above.

Thus, four hydrogen atoms (protons) combine to form one helium-4 atom, with a release of 26.7 MeV of energy. This is the fundamental energy-generating process in the Sun.

Beyond Helium: Synthesis of Heavier Elements …

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