# Mass conservation during $\beta^+$ and $\beta^-$ decay?

During the above mentioned decays the following happens $$n\to p+e^{-} \ and \ p\to n+e^{+}$$ then i assumed that this implied $$m_n=m_p+m_{e^{-}} \ and \ m_p=m_n+m_{e^{+}}$$ but by adding both sides of the equations and simplifying we obtain $$m_{e^{-}}=-m_{e^{+}}$$ So obviously my assumptions are wrong, " mass isn't conserved ", the only way that equation would make sense is if the mass of a positron and of an electron were identical and eq. to 0. So where I'm wrong?

• A proton on itself does not decay to a neutron. (Also, there is a second particle emitted during $\beta$-decay, an antineutrino ($\beta^-$) or neutrino ($\beta^+$)). Also, if the mass were conserved, the emitted electron would have no momentum, what is conserved is total energy (and for a particle $E = \sqrt{(cp)^2 + (mc^2)^2}$), not mass. – Sebastian Riese Oct 21 '15 at 17:24

Mass-energy is conserved, not "mass".

In beta decay, the particles have kinetic energy. The electrons and neutrinos are ejected in the MeV range. So your equation must be roughly:

$m_n + KE_n=m_p+KE_p+KE_{e-}+m_{e-}+ KE_{neutrino} + m_{neutrino}$

where KE is kinetic energy. If you take that into account you will see that equation holds true. The neutrino mass is very close to zero, but usually both the neutrino and the electron have kinetic energies in the MeV range.

This is a simplification because the exact calculation would be with the relativistic energy dispersion equation. It holds approximately true because in the center of mass frame, the neutrons and protons are subrelativistic and most of neutrinos and electron energy is kinetic energy.

• Assuming units $c = 1$. – Sebastian Riese Oct 21 '15 at 17:27
• thxx $.$ $;$ – user153330 Oct 21 '15 at 17:30
• for the proton to neutron positron you should add that the energy is taken from the rest of the nucleus, as energy conservation ensures that a lower mass (proton) cannot decay to a higher mass( neutron) particle. – anna v Oct 21 '15 at 18:30