Please help due in the next 12 hours A pi 0 of kinetic energy 350 MeV decays in

Question

Please help due in the next 12 hours

A pi0 of kinetic energy 350 MeV decays in flight into 2 gamma rays of equal energies. Determine the angles of the gamma rays from the incident pi0 direction.

This is the work and answer I got but is incorect not sure where I went wrong (may have to do witht eh rest mass energy)

Energy of pi0 meson

E = 350 MeV

Rest mass Energy of pi0 meson

E0 = 135 MeV

Initial momentum of pi0 meson

P = squroot E2 - E20 / c

P = squroot( 350 MeV)2 -(135 MeV)2 / c

P = 322.9163978 MeV/c

Law of conservation of Energy

E = Egamma + Egamma

Egamma = E/2

Egamma = 350MeV/2

Egamma = 175 MeV

Angle of gamma ray

angle = cos-1(Pc / 2E)

angle = cos-1(322.9163978 MeV/c) c / (2)(175 MeV)

angle = cos-1(0.922618279)

angle = 22.688%

Explanation / Answer

First conserve momentum. ppi = 2p?cos@ if each makes ? with incident direction. Conserving energy, 350 MeV = 2*E, E = p*c => p= 175MeV / c ppi = sqrt((2Mpi*Epi) = sqrt(2*139.572MeV/c^2 * 350 MeV) = ?(279.144*350) MeV/c => 2*350cos@ =312.5706320177889 => @ = 63.5126637578

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