Final answer:
In the decay of an omega minus particle to a lambda zero and a kaon minus, strangeness changes by +3, baryon number and charge are conserved, and lepton numbers are unaffected since no leptons are involved in the process.
Step-by-step explanation:
The decay mode of the omega minus particle (Ω−) into a lambda zero (Λ0) and a kaon minus (K−) is an example of a weak decay process in particle physics. A student is tasked to find the change in strangeness, verify the conservation of baryon number and charge, and confirm that lepton numbers remain unaffected in this decay. Additionally, the student is asked to illustrate the decay at the quark level with a Feynman diagram.
In terms of strangeness change, the parent particle Ω− has a strangeness of -3, while the Λ0 has a strangeness of -1 and the K− has a strangeness of +1. Adding the strangeness of the decay products (-1 + 1 = 0), we see that the change in strangeness is ΔS = 0 - (-3) = +3.
Conservation of baryon number is seen as both Ω− and Λ0 have a baryon number of +1 and the K− being a meson, has a baryon number of 0. Hence, the total baryon number before and after the decay is +1. Charge conservation is also observed as the omega minus has a charge of -1, which matches the sum of charges of the lambda zero (0) and the kaon minus (-1). Since there are no leptons involved in this process, lepton numbers are unaffected.