Mass of an anti electron neutrino


by adimantium
Tags: anti, down quark, electron, mass, neutrino, up quark
adimantium
adimantium is offline
#1
Oct4-13, 09:01 PM
P: 13
I can't seem to find the mass of an anti electron neutrino in MeV. I found that in beta radiation one down quark breaks into an up quark, an electron, and an anti electron neutrino. The mass in MeV of a down quark is 4.8, the mass of an up quark is 2.4 MeV, the mass in MeV of an electron is 0.511 MeV. So the mass of an anti electron neutrino should be 1.889 MeV. However I can't find a place to make sure I am right. Thank you and forgive me if i'm wrong.
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fzero
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#2
Oct4-13, 10:25 PM
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PF Gold
P: 2,606
Your calculation ignores the fact that the beta decay products typically carry kinetic energy. This makes it hard to use beta decay to measure the neutrino mass, since the neutrino masses are very small, probably around a few hundredths of an eV. The kinetic energy of the electron, for example, is typically at least a few % of its mass, so around a few thousand eV, which is around 5 orders of magnitude larger than the expected neutrino mass. An experiment would need a high sensitivity to measure the small neutrino mass.

Furthermore, because of neutrino oscillations, current experiments have not been able to measure the value of the electron neutrino flavor eigenstate.

There is a proposal for a new experiment sensitive to a mass of around ##0.2~\mathrm{eV}##. A review of this and other experiments is here. It may be possible in the next few years to have a direct measurement of a neutrino mass.
mfb
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#3
Oct5-13, 07:22 AM
Mentor
P: 10,813
In addition to the good post of fzero: you cannot consider the quarks as isolated particles. The decay is a process of the whole nucleon or even the whole nucleus (if you don't have a free neutron). The released energy corresponds to the mass difference between initial and final nucleus (or neutron->proton for a free neutron).

Those light quark masses are problematic anyway - it is hard to measure them as you cannot see them as isolated quarks, and the values have a large uncertainty.


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