Question

A neutron (mass 1.00866 u) disintegrates into a proton (mass $1.00728 \mathrm{u}\( ), an electron (mass \)0.00055 \mathrm{u}$ ), and an antineutrino (mass 0 ). What is the sum of the kinetic energies of the particles produced, if the neutron is at rest? $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2} .\right)$

Short answer

The sum of the kinetic energies of the particles produced in the decay is 0.8 MeV.

Step-by-step solution

Calculate the initial energy of the neutron

The neutron is initially at rest, so its energy is entirely mass-energy and no kinetic energy is involved. Convert the mass of the neutron from atomic mass units to MeV/c² using the given conversion factor: $$E_{neutron} = m_{neutron}(\mathrm{u}) * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 1.00866 \mathrm{u} * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 939.6 \mathrm{MeV}/\mathrm{c}^{2}$$

Calculate the final energy of the proton and electron

The final energy of the proton and electron consists of both their mass-energy and their kinetic energy. Convert the masses of the proton and electron from atomic mass units to MeV/c² using the conversion factor: $$E_{proton} = m_{proton}(\mathrm{u}) * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 1.00728 \mathrm{u} * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 938.3 \mathrm{MeV}/\mathrm{c}^{2}$$ $$E_{electron} = m_{electron}(\mathrm{u}) * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 0.00055 \mathrm{u} * 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2} \mathrm{u}} = 0.5 \mathrm{MeV}/\mathrm{c}^{2}$$ Notice that since the antineutrino has no mass, its energy is purely kinetic and does not need to be calculated in this step.

Use Conservation of Energy

Using the conservation of energy, the initial energy must be equal to the final energy. We already found the initial energy while the final energy includes the kinetic energies we need. Let the kinetic energy of the proton, electron, and antineutrino be KE_proton, KE_electron, and KE_antineutrino, respectively. The total energy after the decay is the sum of the kinetic energies and mass-energy of the proton and electron: $$E_{neutron} = E_{proton} + E_{electron} + KE_{proton} + KE_{electron} + KE_{antineutrino}$$ Substitute the values from Step 1 and Step 2 in the equation: $$939.6 \mathrm{MeV}/\mathrm{c}^{2} = 938.3 \mathrm{MeV}/\mathrm{c}^{2} + 0.5 \mathrm{MeV}/\mathrm{c}^{2} + KE_{proton} + KE_{electron} + KE_{antineutrino}$$

Calculate the sum of the kinetic energies

From Step 3, to find the sum of the kinetic energies of the decay products, we need to solve for the right part of the equation: $$KE_{proton} + KE_{electron} + KE_{antineutrino} = 939.6 \mathrm{MeV}/\mathrm{c}^{2} - 938.3 \mathrm{MeV}/\mathrm{c}^{2} - 0.5 \mathrm{MeV}/\mathrm{c}^{2}$$ $$KE_{proton} + KE_{electron} + KE_{antineutrino} = 0.8 \mathrm{MeV}$$ Thus, the sum of the kinetic energies of the particles produced in the decay is 0.8 MeV.