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Benzoate
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Homework Statement
A pion spontaneously decays into a muon and an antineutrino according to pion^1- => muon^1- +antineutrino.. Current experimental evidence indicates that the mass m of the antineutrino is no greater than about 190 keV and may , in fact, be zero. Assuming that the pion decays at rest in the laboratory , compute the energies and momenta of the muon and muon anti-neutrino a) if the mass of the anti-neutrino is zero and b) if its mass is 190 keV . The mass of the pion is 139.56755 MeV/c^2 and the mass of the muon is 105.65839 MeV/c^2
Homework Equations
E=pc when (m=0)
E^2=(pc)^2 +(mc)^2
E(pion)=E(muon)+E(antineutrino)
p(pion)=p(muon)+E(antineutrino)
The Attempt at a Solution
For part a) since the mass of the anti-neutrino is zero , I apply use the equation E=pc to find the momentum and the Energy for both the muon and anti neutrino particles. To Find E(muon), E(muon)=E(pion)-E(antineutrino) = m(pion)*c^2 - m(antineutrino)*c^2= (139.56755 MeV/c^2 )*c^2 - (105.65839 MeV/c^2)(c^2)= 39.9092 MeV => (E(muon)^2)/c^2= c^2((p(muon))^2) +(m(muon))^2, c^2 cancel out and so I'm left with (p(muon))^2 = (E(muon))^2-(m(muon))^2) => p(muon)=sqrt((E(muon))^2-(m(muon))^2) ). I don't understand how to obtain the momentum and energy of the neutrino .
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