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When K-mesons decay

  1. Jan 12, 2005 #1
    ok say a K- meson decays to... Oh i dont know say... A pion 0 and a beta+ and a... what is the third particle in the series. If it was to decay like this would u not need the third particle to have a charge of 2 (dont worry i know this is impossible... and by impossible i mean without doubly charged particles which is definately not the answer) to balance the equation.

    Could this decay series be a typo or could it be valid...

    If you have a site containing info on baryons decaying by releasing leptons it would also be a great help because the next two questons also pertain to these types of decays

    After a few hours online I got no closer to an answer
    In need of help!
  2. jcsd
  3. Jan 12, 2005 #2


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    The decay should be something like that:

    [tex] K^{-}\rightarrow \pi^{0} + e^{-} +\bar{\nu}_{e} [/tex]

    There are several reasons for considering this reaction.Actually they are called conservation laws.4-momentum,electric charge,spin,lepton number,isospin,color,...all of them must be conserved in elementary processes.


    PS.Hopefully someone else will give a link where u could read more into it,though i still think you ought to read a good book.Aitchinson &Hey is a good one.
  4. Jan 12, 2005 #3
    Thanks for that.

    Now all i need to know is why, but I'll try to work that out for myself.

    Thanks again!
  5. Jan 13, 2005 #4
    ok I got it wrong turns out it was a K+ meson thus the whole "charge of 2" thing... little help here!
  6. Jan 13, 2005 #5


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    Well using Daniel's equation

    [tex] K^{+}\rightarrow \pi^{0} + e^{+} + {\nu}_{e} [/tex]

    But this is only one of several possibilities of decay modes.

    The most probable decay mode (perhaps Daniel may confirm) is [itex] K^{+}\rightarrow \mu^+ + \nu_\mu [/itex].
  7. Jan 15, 2005 #6


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    Yes, the decay [itex] K^{+}\rightarrow \mu^+ + \nu_\mu [/itex] is observed about 63.5% of the time.

    The next most observed decay is to [itex] \pi^+ \pi^0 [/itex] (about 21% of the time)

    Then ther is to [itex] \pi^+ \pi^- \pi^+ [/itex] (5.6%)

    and then [itex] \pi^0 e^+ \nu_e [/itex] (4.8%)

    and then [itex] \pi^0 \mu^+ \nu_\mu [/itex] (3.2%)

    and so on. (source : Particle Properties data Booklet, but that's a very old edition (1992) so these numbers may have changed a bit)

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