Answer
769 MeV
Work Step by Step
$P^{0} \rightarrow \pi^+ + \pi^-$
Rest Energy of Pions : 139.6 MeV
According to relativistic energy-momentum equation.
$E^2 = P_\pi^2C^4 + m_\pi^2 C^4$
where $P_\pi^2C^4$ is the momentum and $m_\pi^2 C^4$ is the kinetic energy
Rearrage the equation
$E = \sqrt {(P_\pi^2C^4) + (m_\pi^2 C^4)}$
$E = \sqrt {(P_\pi C^2)^2 + (m_\pi C^2)^2}$
Referring to the question, energy of pions is 139.6 MeV and momentum is 358.3 MeV/C
Substitute value into the equation
$E = \sqrt {(358.3 MeV/C)^2 + (139.6 MeV)}$
$E = 384.5 MeV$
According to law of conservation energy,
$m_p C^2 = 2E_\pi = 2 (384.5 MeV) = 769 MeV $
