Nuclear fission vs fusion — process comparison, conditions, and applications

Question

Compare nuclear fission and fusion. What conditions are needed for each, and why does fusion release more energy per nucleon?

Solution — Step by Step

A heavy nucleus (like $^{235}\text{U}$ ) absorbs a slow neutron and splits into two medium-mass fragments, releasing energy and 2-3 neutrons.

^{235}_{92}\text{U} + ^{1}_{0}\text{n} \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\ ^{1}_{0}\text{n} + \text{energy}

Energy released: about 200 MeV per fission event. The source of this energy is the mass defect — the products have less total mass than the reactants. By Einstein’s equation:

E = \Delta m \cdot c^2

Two light nuclei (like deuterium and tritium) fuse to form a heavier nucleus, releasing energy.

^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + ^{1}_{0}\text{n} + 17.6 \text{ MeV}

Conditions: Extremely high temperature (about $10^7$ K or higher) to overcome the electrostatic repulsion between positively charged nuclei. This is why fusion is also called a thermonuclear reaction. Stars like the Sun achieve this naturally through gravitational compression.

Look at the binding energy per nucleon curve. It peaks at iron-56 (about 8.8 MeV/nucleon).

Fission: Heavy nuclei (beyond the peak) split into medium nuclei (closer to the peak), gaining binding energy. But they are already fairly well-bound, so the gain per nucleon is modest.
Fusion: Light nuclei (far left of the curve) fuse into heavier nuclei (moving toward the peak), gaining a lot of binding energy per nucleon. The jump from H to He represents the steepest part of the curve.

Energy per nucleon: Fusion produces about 6.75 MeV/nucleon vs Fission at about 0.85 MeV/nucleon.

graph TD
    A[Nuclear Reactions] --> B[Fission: Heavy to Medium]
    A --> C[Fusion: Light to Heavier]
    B --> D["U-235 + n = Ba + Kr + 3n"]
    B --> E["~200 MeV total, ~0.85 MeV/nucleon"]
    C --> F["D + T = He-4 + n"]
    C --> G["17.6 MeV total, ~6.75 MeV/nucleon"]
    B --> H[Nuclear reactors, atom bomb]
    C --> I[Stars, H-bomb, future reactors]

Why This Works

Feature	Fission	Fusion
Process	Heavy nucleus splits	Light nuclei combine
Fuel	U-235, Pu-239	H-2 (deuterium), H-3 (tritium)
Temperature needed	Room temperature (neutron-induced)	$\sim 10^7$ K
Energy per event	~200 MeV	~17.6 MeV (D-T)
Energy per nucleon	~0.85 MeV	~6.75 MeV
Chain reaction	Yes (neutrons trigger more fissions)	No natural chain reaction
Radioactive waste	Significant (fission products)	Minimal (helium is harmless)
Controlled?	Yes (nuclear reactors)	Not yet (research stage)

The binding energy curve is the master key for understanding both processes. Any nuclear reaction that moves nuclei toward the peak of the curve (iron-56) will release energy.

Alternative Method

For numerical problems, the energy released equals the mass defect times $c^2$ :

Find total mass of reactants
Find total mass of products
$\Delta m = m_{\text{reactants}} - m_{\text{products}}$ (should be positive for energy release)
$E = \Delta m \times 931.5$ MeV (since 1 amu = 931.5 MeV/ $c^2$ )

Common Mistake

Students often assume fusion is “easier” because it releases more energy per nucleon. The opposite is true from an engineering perspective. Fission is easy to initiate (just add a slow neutron) and has been controlled since the 1940s. Fusion requires temperatures of millions of degrees and remains uncontrolled for power generation. The challenge of fusion is confinement — how to hold a plasma at $10^7$ K without it touching (and melting) any container.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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