What is the maximum energy of the Λ particle after the Σ0 baryon decay?

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SUMMARY

The discussion centers on the decay of a Σ0 baryon with an energy of 2 GeV into a Λ particle and a photon. The maximum energy of the Λ particle occurs when it moves in the opposite direction to the emitted photon, ensuring conservation of momentum and energy. The calculations reveal that the energy of the Λ particle in the rest frame of the Σ0 baryon is approximately 1.155 GeV. In the lab frame, where the Σ0 baryon is moving, the Λ particle achieves its maximum energy by maintaining its forward motion while the photon moves backward.

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  • Understanding of relativistic energy-momentum relations (E^2 = m^2 + p^2)
  • Knowledge of baryon decay processes and conservation laws
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  • #31
i had 0.034c as the velocity of the lambda in the sigma rest frame. then using 2gev as the energy of the sigma in the lab frame, I calculated the momentum and used the relativistic momentum equation to find the sigma velocity. Then I used u' = (u-v)/(1-uv/c^2) to where u = 0.034c and v = 0.69c (sigma velocity in lab frame) to get the lambda velocity and then the energy. Or should I have used u = (u' + v)/(1 + u'v/c^2) since I am transforming from a rest frame to the actual lab frame
 
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  • #32
One will lead to a value that doesn't make sense, the other one is the right answer.
 

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