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Q & A: neutron decay

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Most recent answer: 10/22/2007
Q:
In neutron decay to proton, electron, and neutrino, the rest mass difference(0.7823 MeV) between neutron and (proton + electron) is converted to the kinetic energy of proton, electron and neutrino. However, the "newly born" proton and electron, which should be very close to each other just after beta decay (for example, 2 fermi ?), should have very big negative electrostatic potential energy(-0.7190 MeV), and therefore, I think that most of the rest mass difference(0.7823 MeV) should be used to overcome the attractive electrostatic force between proton and electron. But, I cannot find this point in the explanation of beta decay in the textbook, and I would like to know why. Thank you so much.
- Josh
Boston College, Chestnut Hill, MA
A:
Great question. Actually, the quantum mechanical nature of the proton and especially the electron (which has less mass) prevent them from scrunching down into small spaces, so they really can't get all that close. Trying to scrunch an electron into a 2 fermi space would require that it have an enormous range of momentum states (by the uncertainty relation) and hence an enormous expected kinetic energy.

Think of the lowest energy state of a proton-electron combination, i.e. the ground state of a hydrogen atom. Its energy is only 13.6 eV below that of fully separated particles with near-zero speeds. Even the electrostatic potential energy taken by itself is only 27.2 eV below that of the fully separated particles. Those energies don't amount to much on the MeV scale.

However, you're right that before the electron and proton fly apart, their kinetic energy must be a little higher than afterwards. Since the standard rest masses of particles refer to the state where they are far from other particles, the kinetic energies calculated are those they end up with, not the slightly higher ones found before separation.

Mike W.

(published on 10/22/2007)

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