A 620 Mev pion decays into two equal gammas. Determinethe angle of the gamma fro

Question

A 620 Mev pion decays into two equal gammas. Determinethe angle of the gamma from incidence. I calcd pion momentum = (6202)-(1352)= 605.123MeV/c each photon momentum = E/C= 620/2= 310 Mev/C pion momentum / 2 is x component of each photon =302.56 photon momentum is hypotenus= 320 cos= 302.56/320 = 19 degrees Why is this wrong? A 620 Mev pion decays into two equal gammas. Determinethe angle of the gamma from incidence. I calcd pion momentum = (6202)-(1352)= 605.123MeV/c each photon momentum = E/C= 620/2= 310 Mev/C pion momentum / 2 is x component of each photon =302.56 photon momentum is hypotenus= 320 cos= 302.56/320 = 19 degrees Why is this wrong?

Explanation / Answer

relativistic energy of each gamma pion                    E^2 = p^2c^2 + mo^2c^4 = = 620MeV .     from the abv relation we can sovle formometum of photon along the direction of propagation, let us sayalong x- axis.     from the abv relation , mometum of gamma photon ' p ' along x-axis                mometum of each gamma along the x-axis                                   P' = (310 MeV/C )sin. total momentum of the photons                      P =2p' frm the abv relations we can solve for angle made bythe photons with the direction of     propagation .                                                                                       

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