The reaction given is: [ \pi^+ + n \rightarrow \pi^0 + p^+ ] This is a charge-exchange reaction involving a pion ((\pi^+)) and a neutron ((n)) producing a neutral pion ((\pi^0)) and a proton ((p^+)). To solve this numerically, we will calculate the kinematics, assuming the neutron is initially at rest. --- Step 1: Rest masses * (m_{\pi^+} = 139.57 , \text{MeV/c}^2) * (m_{\pi^0} = 134.98 , \text{MeV/c}^2) * (m_n = 939.57 , \text{MeV/c}^2) * (m_p = 938.27 , \text{MeV/c}^2) --- Step 2: Energy-momentum conservation Let the incoming pion have kinetic energy (T_{\pi^+}) and momentum (p_{\pi^+}). The neutron is initially at rest. Total energy and momentum conservation give: [ E_{\pi^+} + m_n c^2 = E_{\pi^0} + E_p ] [ \vec{p}*{\pi^+} = \vec{p}*{\pi^0} + \vec{p}_p ] Where [ E_{\pi^+} = \sqrt{p_{\pi^+}^2 c^2 + m_{\pi^+}^2 c^4}, \quad E_{\pi^0} = \sqrt{p_{\pi^0}^2 c^2 + m_{\pi^0}^2 c^4}, \quad E_p = \sqrt{p_p^2 c^2 + m_p^2 c^4} ] --- Step 3: Threshold energy The minimum kinetic energy of the incoming (\pi^+) occurs