Question

Show that the threshold kinetic energy in the Laboratory for the production of n pions in the collision of protons with a hydrogen target is given by T= 2nm (1 + nm,/4mp) where m, and mp are respectively the pion and proton masses. A gamma ray interacts with a stationary proton and produces a neutral pion according to the scheme y + p → P +7° Calculate the threshold energy given Mp = 940 MeV and M. = 135 MeV Calculate the threshold energy of the reaction " + p 5- + K+ + Kº The masses for "p, 3-, K+, Kº are respectively 140, 938, 1,321, 494, 498, MeV Attempts have been made to produce a hypothetical new particle, tue W+, using the reaction v + p p + +W+ where stationary protons are bombarded with neutrinos. If a neutrino energy of 5 GeV is not high enough for this reaction to proceed, estimate a lower limit for the mass of the W+ (mass of proton 938 MeV/c, mass of muon 106 MeV/c?, neutrino has zero rest mass)

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Answer 1

solo Threshold = [(me + mp + Mao) mr. tm) ]/2018 - (1321 +494 +498) (140+938); 28938 = 2,233 Mev E production, the threshold Note that for that for is much higher than as it has to be & production with two ther in association produced particles Strange