Final answer:
The sigma-zero (Σ0) decaying into the lambda-zero (Λ0) plus a photon implies that the Σ0 is an excited state of the Λ0 because they have identical quark compositions (uds), and the decay conserves charge, baryon number, and strangeness, with energy being released as a photon.
Step-by-step explanation:
The sigma-zero particle (Σ0) and the lambda-zero particle (Λ0) are both baryons that are part of the baryon octet in particle physics. The quark composition of the two particles is key to understanding their relationship. The Σ0 particle is made up of one up quark (u), one down quark (d), and one strange quark (s), exactly like the Λ0. However, the Σ0 has more mass than the Λ0, which indicates that the Σ0 is in a higher energy state and hence can be considered an excited state of the Λ0.
When the Σ0 decays into the Λ0 plus a photon (γ), the energy difference between the two particles is released in the form of the photon's energy. Since the only difference between the initial and final states of the decay is the presence of a photon and that Σ0 and Λ0 share the same quark composition, we can conclude that no quarks are gained or lost, and the decay is a transition from a high energy state to a lower energy state of the same quark configuration. This transition conserves charge, baryon number, and strangeness, supporting our interpretation of the Σ0 as an excited state of the Λ0.