The Deuterium–Deuterium Side Reactions
Deuterium fusing with itself produces the burner's neutrons through two nearly equal branches, one of which breeds tritium.
Two branches, both unavoidable
With deuterium in the fuel, D–D fusion happens alongside the primary D–3He reaction. It splits into two branches of nearly equal probability:
The first branch, D + D → 3He + n, emits a 2.45 MeV neutron. The second, D + D → T + p, breeds tritium in the plasma; that tritium then burns with deuterium (D + T → 4He + n) and releases a 14.1 MeV neutron. Together these set the 5.44% neutron fraction.
Controlling the tail, not eliminating it
The neutron tail scales with the deuterium density and temperature relative to the helium-3 content. A helium-3-rich mix suppresses D–D reactions but demands more of a scarce fuel; a deuterium-rich mix eases fuel supply but raises the neutron fraction. The design point balances the two — it does not remove the neutrons, because D–D cannot be switched off while deuterium is present.
Because the tritium bred in the second branch takes time to burn, part of the neutron output is delayed relative to the primary reaction, and the balance between prompt D–D neutrons and secondary D–T neutrons shifts with confinement and temperature. Accounting for both channels, and for the tritium inventory that transiently builds in the plasma, is part of computing the 5.44% figure honestly rather than quoting only the primary branch.
- D + D → 3He + n (2.45 MeV), ~50%
- D + D → T + p (~50%), then T burns → 14.1 MeV n
- Together: the 5.44% neutron fraction
- Tail is tunable by fuel mix, never zero with D present