# Doubt in Kaon decay problem

Reading the CP symmetry section from Griffiths elementary particles book, 2nd revised edition, page 146.

As mentioned in the above picture the neutral kaon and its antiparticle form linear combinations that are eigenstates of CP operation. To conserve CP, $$K_1$$ must decay to 2 pions and $$K_2$$ must decay to 3 pions. I don’t understand why the 2 pion decay is much faster and why more energy is released in this process.

• Two πs soak up less of the energy available, in mass, so leave more of it as kinetic energy (momentum). What do you know about phase space contribution to decay rates? Write down the energy balance of the two cases, in your question! Commented Jan 3 at 17:00
• Can you please elaborate and write it as answer. With simple language. I always find your answers hard to understand as a beginner. Commented Jan 3 at 17:04

Now, the $$2\pi$$ decay is much faster, because the energy released is greater.
Given the mass of the $$K$$ meson ($$498$$ MeV) and the mass of a $$\pi$$ meson ($$135$$ MeV), we can calculate the released kinetic energy for both decays:
$$K_1 \to 2\pi + 228\text{ MeV}$$ $$K_2 \to 3\pi + 93\text{ MeV}$$
You see, the $$K_1$$ decay releases more kinetic energy than the $$K_2$$ decay.