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Lambda -> n + pi^0 decay

  1. Jun 26, 2011 #1
    The [itex]\Lambda[/itex] baryon (quark content [itex]uds[/itex]) decays into [itex]n + \pi^0[/itex] or [itex]p + \pi^-[/itex]. In the case [itex]\Lambda \to p + \pi^-[/itex], the [itex]s[/itex] quark decays into a [itex]u[/itex] quark, releasing a [itex]W^-[/itex] in the process (which subsequently decays into a [itex]\pi^-[/itex] meson). What happens in the [itex]\Lambda \to n + \pi^0[/itex] case? (I tried Google, but couldn't find anything about this specific decay.)

    Thanks!
     
  2. jcsd
  3. Jun 26, 2011 #2

    phyzguy

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    At the quark level, these two decays are the same. The s quark decays into a u quark plus a W-, which decays into and up-bar antiquark and a down quark. In the first case, the u quark from the s decay ends up with the u and d from the original lambda, and the upbar and down pair together, so we end up with a proton and a pi-. In the second case, the d quark from the W- decay ends up with the u and d from the original lambda, and the upbar and up quark pair together, so we end up with a neutron and a pi-0.
     
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