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This may just be beyond the grasp of the everyman, but I'm trying, and failing, to grasp how conservation of mass-energy works in cases of beta decay and electron capture.

A neutron has a mass of one dalton. The combination of a proton and an electron also has a mass of one dalton. This allows us to assign an overall "mass number" to elements and their specific isotopes. We can say, for instance, that because Carbon-14 and Nitrogen-14 have the same mass number, they have equivalent masses, even though nitrogen has one more proton; it also, in its charge-balanced state, has one more electron than the carbon, and so is equivalent in mass to the carbon atom, just distributed differently.

However, Carbon-14 directly decays into Nitrogen-14 by $\beta^-$ decay of one of its neutrons. The "reactant" is one neutron from the carbon nucleus, and the products are one proton, one electron, and one electron antineutrino. This extra particle, a lepton similar to the electron, is not massless (and mass is an absolute quantity so you can't have "negative mass"), and therefore it would seem that the combined mass of a proton and electron are not in fact equal to that of a neutron.

It gets even weirder when you see it work in reverse. Atoms that have too few neutrons can lower their atomic number by electron capture; the electron is "absorbed" by a proton in the nucleus, forming a neutron, and this interaction results in the emission of an electron neutrino (also not massless).

So, if you were to see a neutron decay to a proton, then capture an electron to become a neutron again, you would have seen it emit a lepton and anti-lepton, both of them having a small but nonzero mass, and yet because a neutron is a neutron, the mass of the particle before these two transformations should be equivalent to the one after.

Obviously I'm missing something. Is there an energy input required (possibly from the neutrino-antineutrino annihilation but it could come from anywhere) that is omitted from the basic nuclear chemistry equations, that because of mass-energy equivalence is re-adding the mass lost as leptons? Is there indeed such a thing as "negative mass", and the mass of the anti-lepton is somehow modified by its spin sign? Or is this a subatomic case of the Second Law of Thermodynamics, and in fact not all neutrons are created equal?

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1) The combined mass of the proton and electron are not equal to the mass of the neutron, though it happens to be close.

2) C14 and N14 don't have masses of exactly 14 amu. These are rounded numbers.

3) It seems like you might be aware of this and are just using different terminology, but just to clarify - mass isn't conserved in nuclear processes. Conservation of mass isn't a universal law; it just seemed like that for a while before we could access / study high energy processes. What is a universal law is conservation of energy. Since a particle's mass is part of its energy ($E_{mass} = mc^2$), the sum of the masses of particles before / after a nuclear or particle process may be different as long as the total energy is the same.

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  • $\begingroup$ (2) follows naturally from (1). If for no other reason than the proton and electron don't weigh the same as the neutron, then C-14 is lighter than N-14 by the mass of the electron antineutrino. $\endgroup$ – KeithS Oct 11 '13 at 21:05
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    $\begingroup$ However, I think one question remains from my OP: If $n \to p^+ + e^- + \bar{v}_{e}$, and $p^+ + e^- \to n + v_{e}$, then if you hypothetically started with a neutron, decayed to a proton, electron and antineutrino, then recaptured the electron releasing a neutrino, you would produce a neutron of the same mass as the starting particle, when in the process two leptons of nonzero mass have been lost. The math only adds up if there is an influx of energy to these processes that is not shown in the basic equations, or if the mass of a neutrino is somehow opposite that of an antineutrino. $\endgroup$ – KeithS Oct 11 '13 at 21:07
  • $\begingroup$ Your reasoning is correct. The conclusion you should reach is that electron capture of the same electron that is produced by the neutron decay can't happen. The electron needs to have a minimum amount of energy for p+ + e- => n ve to be possible; the electron produced by n -> p+ e- ve doesn't have that minimum energy. $\endgroup$ – jwimberley Oct 12 '13 at 0:19

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