The sigma-zero particle decays mostly via the reaction \[{{\rm{\Sigma }}^{\rm{0}}} \to {{\rm{\Lambda }}^{\rm{0}}}{\rm{ + \gamma }}\]. Explain how this decay and the respective quark compositions imply that the \[{{\rm{\Sigma }}^{\rm{0}}}\]is an excited state of the\[{{\rm{\Lambda }}^{\rm{0}}}\].

Short Answer

Expert verified

Baryons that are composed of the same quarks represents different states of the same particle thus having the same quark structure, \[{{\rm{\Sigma }}^{\rm{0}}}\] represents excited state of \[{{\rm{\Delta }}^{\rm{0}}}\]. The \[{{\rm{\Sigma }}^{\rm{0}}}\] decays to \[{{\rm{\Delta }}^{\rm{0}}}\] without violating the conservation of strangeness with \[{{\rm{\Delta }}^{\rm{0}}}\] the only less massive strange quark.

Step by step solution

01

Definition of Concept

The sigma zero decay relation in terms of quarks is given by,

\(uds \to uds + \gamma \)

02

Explain how this decay and the respective quark compositions imply

Sigma is a type of baryon with an unusual quark structure, uds, similar to how \[{{\rm{\Delta }}^{\rm{0}}}\] has the same quark structure, uds. Baryons made up of the same quarks represent different states of the same particle, resulting in the same quark structure; \[{{\rm{\Sigma }}^{\rm{0}}}\] represents the excited state of \[{{\rm{\Delta }}^{\rm{0}}}\]. The \[{{\rm{\Sigma }}^{\rm{0}}}\] decays to \[{{\rm{\Delta }}^{\rm{0}}}\], without breaking the law of conservation of strangeness, with \[{{\rm{\Delta }}^{\rm{0}}}\] being the only less massive strange quark.

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