Mass-Energy Equivalence

When a uranium-235 atom splits in a reactor, the pieces weigh slightly less than the original atom. The difference is about 0.1% of the starting mass. That "missing" mass did not disappear. It became 200 million electron-volts of energy. Einstein's equation tells you exactly how much.

E = mc² states that mass and energy are two forms of the same thing. The conversion factor is c², the speed of light squared: approximately 9 x 10¹⁶ m²/s². This number is so large that even tiny mass conversions produce enormous energy. One kilogram of matter, if fully converted, would yield 9 x 10¹⁶ joules, roughly the energy output of a large power plant running for three years.

No technology achieves full mass-to-energy conversion. Fission converts about 0.1% of fuel mass. Fusion converts about 0.7%. The rest remains as mass in the reaction products. Even so, the 0.1% fission conversion makes uranium roughly a million times more energy-dense than coal (which converts zero measurable mass).

The equation also works in reverse: creating mass requires energy input. Particle accelerators convert kinetic energy into new particles, confirming that mass and energy are interchangeable.

Worked Example

A nuclear reactor fissions 1 kg of U-235 over 18 months. About 0.1% of the mass converts to energy.

• Find converted mass. 1 kg x 0.001 = 0.001 kg.
• Apply E = mc². E = 0.001 x (3 x 10⁸)² = 0.001 x 9 x 10¹⁶ = 9 x 10¹³ J = 90 TJ.

How does that compare to burning coal at 24 MJ/kg?

9 x 10¹³ / 24 x 10⁶ = 3.75 x 10⁶ kg, or about 3,750 tonnes of coal. (The precise reactor-physics figure, based on 200 MeV released per U-235 fission, comes to about 82 TJ per kilogram rather than this rounded 90 TJ; either way, one kilogram of uranium replaces well over 3,000 tonnes of coal.) The entire difference comes from the c² multiplier acting on that fraction of a percent of mass.

Every debate about nuclear waste volume, refueling schedules, or reactor siting traces back to this one multiplier: a few kilograms of mass, converted at less than a tenth of a percent, outrun a freight train of coal.

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