Calculate the small rest-mass difference

Find the small difference before multiplying

Put four hydrogen atoms on one side of a paper balance and one helium atom on the other. The helium contains almost all the initial rest mass, but the totals are slightly different. The difference supplies released energy through E = Δmc². Here E is energy in joules, Δm is the reduction of rest mass in kilograms, and c is light speed in meters per second.

This is conservation of total mass-energy, not matter vanishing without an energy destination. It also does not mean that every kilogram of processed hydrogen becomes a kilogram’s worth of radiation. Most becomes helium rest mass. Your task is to distinguish the fuel processed from the much smaller rest mass converted.

Use isotope-specific neutral masses

The fixed reference uses NIST hydrogen-1 mass 1.00782503223 u and NIST helium-4 mass 4.00260325413 u. The symbol u is an atomic mass unit. These are selected isotope masses, not chemical atomic weights averaged over an isotope mixture.

Calculate four times the hydrogen mass: 4.03130012892 u. Subtract the helium mass to obtain 0.02869687479 u. Keep several digits during subtraction because the small difference is sensitive to premature rounding. Report the final energy with reasonable precision, rather than treating every calculator digit as experimental accuracy.

Both masses describe neutral atoms. Four initial electrons are accounted for by two final helium electrons and two electrons involved in annihilating the produced positrons. Do not add annihilation rest energy again. A calculation using neutral hydrogen and a bare helium nucleus would use incompatible electron inventories.

Convert units and interpret the result

The NIST CODATA constants table supplies u approximately 1.66053906892 × 10⁻²⁷ kg and c = 299792458 m/s. Multiplying the mass difference by u gives about 4.76523 × 10⁻²⁹ kg. Multiplying that by c² gives about 4.28277 × 10⁻¹² J per completed chain.

Nuclear energies are often written in megaelectronvolts: one MeV is one million electronvolts, and one electronvolt equals 1.602176634 × 10⁻¹⁹ J. The released total is therefore approximately 26.73 MeV. This total includes energy subsequently carried away by neutrinos; it is not all available for photon luminosity.

Now divide the mass difference by the initial four-atom mass. The fraction is approximately 0.0071185. Multiply by 100 to obtain about 0.712 percent. The denominator is initial fuel rest mass, not released energy. Write that denominator beside your percentage so you cannot confuse it with the fraction of released energy deposited in the star.

Compare a reaction with a reservoir

At 10 percent available core region, the hypothetical hydrogen inventory is 1.40 × 10²⁹ kg. Increase the region to 20 percent while keeping photon demand fixed. Predict unchanged per-chain mass defect and doubled total available energy. Use paired notebook captures and compare both rows of the ledger.

The tabulated masses and atomic mass constant have source uncertainties; this calculation uses their central values without uncertainty propagation. It omits plasma ionization corrections and tiny neutrino rest-mass corrections. Printed decimal places do not establish a confidence interval. The neutral-atom convention already includes atomic binding-energy contributions in those tabulated masses.

A useful cross-check is initial fuel mass equals final helium mass plus total released energy divided by c². Use total released energy here, including the neutrino channel. Substituting only deposited energy would leave part of the mass-energy accounting unexplained.

Transfer: two inconsistent corrections

A student uses the correct neutral atomic masses, calculates Δmc², and then adds two electron-rest-mass energies as an annihilation bonus. Another student instead replaces helium’s atomic mass with a bare alpha-particle mass while leaving hydrogen unchanged. Explain why neither is the same consistent calculation.

Check the fusion-mass transfer

The first calculation double-counts electron bookkeeping already included in the neutral-atom difference. The second changes the final electron inventory without changing the initial convention. Use the supplied neutral hydrogen-1 and helium-4 masses together; their difference gives approximately 26.73 MeV total. A fully nuclear convention could be constructed, but all electrons and positrons would have to be handled consistently.