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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.
Boston College, Chestnut Hill, MA
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
(republished on 07/20/06)
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