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A stationary pion decays into a muon and a neutrino: pi^(+)tomu^(+)""v_(mu) Find the ratio of the energy of the neutrino to the kinetic energy of the muon. |
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Answer» `Deltaepsi=(m_(pi)-m_(mu))xx931.5=140-106=34MeV` The rest ENERGY of a muon is `epsi_(0mu)=106MeV`, the rest mass of a neutrino is zero. Assuming the dacaying PION to be at rest, we obtain that the muon and the neutrino momenta are equal in magnitude and opposite in direction. We have `p_(V)=epsi_(v)/c,p_(mu)=1/csqrt(K_(mu)(2epsi_(0mu)+K_(mu)))` which yields `epsi_(v)=sqrt(K_(mu)(2epsi_(0mu)+K_(mu)))` But, from the law of conservation of energy `epsi_(v)=Deltaepsi-K_(mu)`, so `Deltaepsi-K_(mu)=sqrt(K_(mu)(2epsi_(0mu)+K_(mu)))`. Hence it FOLLOWS that the kinetic energy of the muon is `K_(mu)=((Deltaepsi)^(2))/(2(epsi_(0mu)+Deltaepsi))` |
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