Two deuterons with almost no kinetic energy fuse in the reaction ²H + ²H → ³He + n…

Step 1. The Q value. Before: $2 \times 2.014102\ \text{u} = 4.028204$ u. After: $3.016029\ \text{u} + 1.008665\ \text{u} = 4.024694$ u. The loss is $0.003510$ u, so $Q = 0.003510\ \text{u} \times 931.494\ \text{MeV/u} = 3.270$ MeV.

Step 2. The sharing. The momentum before is zero, so the neutron and the helium-3 nucleus have equal momenta. The neutron is lighter, so it takes the larger share: $K_n = Q\,\dfrac{m_{He}}{m_{He} + m_n} = 3.270\ \text{MeV} \times \dfrac{3.016029}{4.024694} = 2.450$ MeV.

Step 3. Helium-3. $K_{He} = 3.270\ \text{MeV} - 2.450\ \text{MeV} = 0.819$ MeV.