Meson Decay in Relativistic Mechanics

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SUMMARY

The discussion focuses on the decay of the Ko meson into two charged pions, specifically the process Ko → π+ + π-. The Ko meson has a rest mass of 498 MeV/c2, while each pion has a mass of 140 MeV/c2. By applying conservation of energy and momentum principles, the energies of the two pions are determined to be equal, resulting in each pion having an energy of 249 MeV. This conclusion is derived from the fact that the total energy of the two pions equals the rest mass energy of the Ko meson.

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Homework Statement


The Ko meson is a subatomic particle of rest mass MK = 498 MeV/c2 that decays into two charged pions, Ko \rightarrow \pipositive+ \pinegative. Both pions have the same mass, m\pi = 140 Mev/c2. A Ko at rest decays into two pions. Use conservation of energy and momentum to find the energies, momenta and speeds of the two pions. (Give algebraic answers, in terms of the symbols MK and m\pi. Then plug in numbers.)


Homework Equations


Total E = Mc2

Mc2 = (K1 + K2) + (m1 + m2)*c2

\DeltaM = M - (m1 + m2)

\DeltaMc2 = K1 + K2


The Attempt at a Solution


I don't know how to find the energies of the two pions. The back of the book says 249 MeV. MK/2 = 249 MeV but I don't know why. I'm not sure how they got that answer. Any help would be appreciated. Thanks!
 
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Because the sum of the total energy of the two pions is equal to the total energy of the K (total energy includes rest mass). To conserve momentum the sum of the momenta of the two pions must be zero, since the initial momentum of the K was zero. That means the momenta are equal and opposite. So they have equal energy.
 
I got it. Thank you!
 

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