630 Particle Physics and CosmologyP46.25(a)Kp ?p++→ +The strong interaction conserves everything.Baryon number,0 11BsoB=0Charge,++ → +111QsoQ=+1Lepton numbers,000+→+LsoLL Le===µτ0Strangeness,100SsoS=1The conclusion is that the particle must be positively charged, a non-baryon, withstrangeness of +1. Of particles in Table 46.2, it can only be the K+. Thus, this is an elasticscattering process.The weak interaction conserves all but strangeness, and ∆S=±1.(b)Ω−−→+?πBaryon number,BsoB=1Charge,−→ −QsoQ=0Lepton numbers,00Lsoe0Strangeness,−→ +30Sso∆S=1: S=−2The particle must be a neutral baryon with strangeness of –2. Thus, it is the
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