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Question
I am having difficulty with this problem. Any help would greatly be appreciated.

A pion at rest (mπ = 273 me) decays to a muon (mμ = 207 me) and an antineutrino (m~0). The reaction is written:
π^- --> μ^- + v bar (i think this is an antineutrino)

What is the total kinetic energy of the muon and the antineutrino?

Answer
The kinetic energy of the resulting particles can be found by determining the mass defect, the difference between the masses of the pion and the resulting muon plus neutrino.
The mass of the pion is:
pion=134.98MeV/c^2=2.40x10^-28kg
The mass of the muon is:
muon=105.66MeV/c^2=1.88x10^-28kg
The mass of the neutrino is, for all practical purposes, zero.
Therefore, the mass defect is:
defect=pion-muon=2.40x10^-28kg-1.88x10^-28kg=0.52x10^-28kg
Therefore, the kinetic energy released will be:
KE=defect*c^2=0.52x10^-28kg*(3.0x10^8)^2=4.68x10^-12J

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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