|Three-body beta kinematics|
When beta decay takes place, the energy released is shared between the recoiling nucleus, the electron and its antineutrino. As the nucleus was initially at rest, the sum of the three momentum vectors must be equal to zero (for the nucleus, the momentum is equal to the mass times the velocity v). Shown above is the 120° case where all three momenta have equal values. We can see that the kinetic energy T and speed of the recoiling nucleus are negligible compared to the two other particles, which shared the decay energy (1.16 MeV in the decay of bismuth 210).