Question

What is the maximum kinetic energy of an electron emitted in the beta decay of a free neutron?

Answer 1

In these sorts of questions, it is easier to deal with mass in terms of eV/c^2.
(From E = mc^2, physicists rearrange to get m = E/c^2, where energy is given in electron-Volts, eV)

Now we have:
(n) neutron mass: 939.565560 MeV/c^2
(p) proton mass: 938.272013 MeV/c^2
(e) electron mass: 0.51099891 MeV/c^2
(v) antineutrino mass: <1eV (Negligible)

(Note: 1 MeV = 1 million eV = 1 x 10^6 eV)

The beta decay process is: n --> p + e + v
There is a mass difference between the left hand side and the right hand side.
(939.565560) - (938.272013 + 0.51099891) = 0.78254809 MeV/c^2

This missing mass is converted in to kinetic energy shared between the beta electron and the anti-neutrino. It is the maximum energy that the electron can have (i.e, when the neutrino energy is zero)
Thus, the kinetic energy released is mc^2 = 0.78254809MeV. <<<ANSWER>>>
(Multiply by the charge of an electron (1.602x10^(-19) Coulombs) to convert MeV to Joules if you wish.)