Reaction Energies When One Event Is the Whole Story
Nuclear physics is written per event. A uranium-235 nucleus splits and releases about 200 MeV; a deuterium-tritium pair fuses and releases 17.6 MeV; a radium-226 nucleus sheds an alpha and lets go of 4.87 MeV. None of those numbers can be measured with an instrument — what a calorimeter, a reactor or a power balance reports is joules. The megaelectronvolt-to-joule step is where a Q-value stops being a table entry and becomes something you can multiply by an atom count and compare against a fuel bill.
Reading a Q-value Correctly Before Converting It
One Nucleus, Not One Kilogram
Mass and Energy Share a Unit
The Q-value Is a Mass Balance
Prompt Energy Is Not All of It
Scaling a Per-Event Q-value up to a Measurable Energy
Work the nuclear physics in megaelectronvolts, convert once, and only then multiply by however many nuclei are involved. Converting first and scaling afterwards keeps the arithmetic auditable.
Settle the Q-value first
Take it from a nuclear data table, or compute it from the mass defect: sum the reactant masses in atomic mass units, subtract the products, and multiply the difference by 931.494 MeV/u.
Point the pair at the direction you need
Start on the MeV side with the swap arrows when the Q-value is the known quantity and the joule figure is what you are after; leave the pair as it stands when a measured energy has to be read back as a per-event value. Both fields respond to typing either way.
Multiply by the number of nuclei
One kilogram of uranium-235 holds 1 000 ÷ 235.04 × 6.022e23 = 2.56e24 atoms. At 3.204353e-11 J each, complete fission of that kilogram would release 8.21e13 J — about 82 TJ.
Take the digits into the yield calculation
The copy control above each field returns the value alone, without a unit or thousands spacing, so an exponent-form joule figure can be pasted into the cell that will multiply it by an atom count. Ctrl+C inside a field is equivalent.
Nuclear and Particle Processes: MeV per Event and Joules
Each row is one reaction, decay or rest mass, with the megaelectronvolt figure alongside the joule value obtained by multiplying by 1.602177e-13. The last row is a chemical reaction, included only so the scale gap is visible in the same units.
| Process | Energy (MeV) | Energy (J) |
|---|---|---|
| U-235 fission, total release | 200 | 3.204353e-11 |
| Proton rest mass | 938.272 | 1.503277e-10 |
| D-T fusion, total | 17.59 | 2.818229e-12 |
| D-T fusion, neutron share | 14.06 | 2.252660e-12 |
| D-T fusion, alpha share | 3.52 | 5.639662e-13 |
| D-D fusion, tritium branch | 4.03 | 6.456772e-13 |
| Ra-226 alpha decay | 4.87 | 7.802600e-13 |
| Electron rest mass | 0.511 | 8.187123e-14 |
| Carbon burned to CO2, per atom | 0.00000407834 | 6.534221e-19 |
The last two rows carry the point. Burning one carbon atom returns about 4.08 eV; splitting one uranium nucleus returns 200 MeV — roughly 49 million times more from a single event. Notice too that the D-T total sits far below a proton's own rest mass: fusion recovers only a small slice of the mass already present, which is why the reaction is so demanding to sustain despite the size of the number.
What This Pairing Gives a Reaction-Energy Calculation
Per-Event Results Drop Into Exponent Form
A single reaction lands near 1e-12 J, well past the threshold where the output stops printing leading zeros and switches to an exponent that can be read and checked.
eV, keV and GeV Sit Beside MeV in the List
Decay energies span chemical eV to accelerator GeV; the searchable list on each side crosses that whole range without leaving the page or rescaling by hand.
Q-value Digits Ready for a Yield Sheet
Copied output has no unit label attached, so a per-event joule value pastes cleanly into the cell where it gets multiplied by an atom count or a reaction rate.
Start From MeV When the Q-value Is Known
The arrows reverse the pair so a tabulated reaction energy can be entered directly and the SI figure read out, rather than the other way round.
Q-value Questions About Fission, Fusion and Mass Defect
How does 200 MeV per fission become joules per kilogram of fuel?
Convert once, then count nuclei. One fission is 3.204353e-11 J. A kilogram of uranium-235 contains 1 000 ÷ 235.04 moles, or 2.562e24 atoms, so fissioning all of them gives 2.562e24 × 3.204353e-11 = 8.21e13 J. That is 82 TJ per kilogram, or about 22.8 million kWh. A single gram, fully fissioned, would still yield 8.21e10 J.
Are nuclear energies really a million times chemical ones?
Two different comparisons get conflated here. Per event the ratio is far larger: 200 MeV against roughly 4.08 eV for burning one carbon atom is about 49 million to one. Per kilogram it is smaller, because a uranium nucleus is heavy and only a fraction of the material takes part — 8.21e13 J/kg for pure U-235 against roughly 24 MJ/kg for coal is about 3.4 million to one. Both numbers are correct; they answer different questions, and the per-kilogram one is the honest figure for an energy-density argument.
What does MeV/c² mean when a particle mass is quoted that way?
It is a mass expressed through the energy it is equivalent to. Since E = mc², dividing an energy by c² gives a mass, and a proton's 938.272 MeV/c² is the same statement as 1.673e-27 kg. The convention pays off in reaction bookkeeping: masses and energies can be added in one column without carrying c² around, so a Q-value falls out as a subtraction. Converting 938.272 MeV here gives 1.503277e-10 J, the energy released if a proton were entirely annihilated.
How does D-T fusion's 17.6 MeV split between the neutron and the alpha?
By momentum conservation, inversely to the masses. The neutron takes 14.06 MeV (2.252660e-12 J) and the alpha 3.52 MeV (5.639662e-13 J), which is very nearly a four-to-one split because the alpha is roughly four times heavier. The consequence is structural: the alpha is charged and stays confined to heat the plasma, while 80 per cent of the yield leaves as a neutral 14 MeV neutron that has to be caught in a blanket. Deuterium-deuterium is gentler, at 4.03 and 3.27 MeV across its two branches.
How is a Q-value worked out from a mass defect?
Add the rest masses of the reactants in unified atomic mass units, subtract the rest masses of the products, and convert the difference at 931.494 MeV per u. For deuterium plus tritium the leftover mass is about 0.01888 u, and 0.01888 × 931.494 ≈ 17.59 MeV. The missing mass has not vanished — it now sits in the kinetic energy of the products, which is precisely what a detector or a blanket eventually measures as joules.
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