A free neutron at rest decays into a proton and other fundamental particles. Given the rest masses of the particles as follows: m_ n = 1.00866 u m_ p = 1.00727 u m_ e = 0.00055 u Take 1 u = 931.5 MeV /c^2 . The maximum possible kinetic energy of the emitted beta particle is approximately:

Step-by-step solution

The fundamental decay equation of a free neutron is: n p + e ⁻ + v The mass defect m for this process is given by the difference between the mass of the neutron and the sum of the masses of the proton and the electron (the mass of the antineutrino is negligible). m = m_ n - (m_ p + m_ e ) m = 1.00866 - (1.00727 + 0.00055) m = 1.00866 - 1.00782 = 0.00084 u The total energy released ( Q -value) is: Q = m 931.5 MeV Q = 0.00084 931.5 0.782 MeV The maximum kinetic energy of the beta particle (electron) corresponds to th