2) A pi0(meson) of kinetic energy 350 MeV decays in flight into 2 gamma rays of

Question

2) A pi0(meson) 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.

Not sure where I am going wrong but my answer is not correct.

Energy of pi0 meson
E = 350 MeV

Rest mass Energy of pi0 meson
E0 = 135 MeV

Initial momentum of pi00 meson
P = sqrt E^2 - Eo^2 / c
P = sqrt(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 degrees

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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