# Is mass defect of atoms considered when the mass of electron is measured within atoms?

First of all, here considering the mass defect as the binding energy between an electron and the rest of the atom; i.e. taking the mass of nucleus as it is. That since nuclear mass defect is present even before an electron comes to the respective ion. And asking specifically on electron, since contrary to nucleons, it is not a composite particle.

When imagining mass defect of an atom naively, it is some negative mass that is either distributed among the positive masses of constituents of the considered atom (nucleus and electrons), or it is living separately in there.

When the mass of an electron is measured at such a system (i.e. residing in an atom), it is possible to imagine that the measured mass is affected by the mass defect; like being somewhat decreased due to the mass defect of the atom.

I expect that this is something that is routinely considered when measurements of the electron mass are done. And that either some corrections are applied for it, or it is supposed (and hopefully tested) that it has no effect on the measured electron mass.

UPDATE: Since the question is apparently unclear, notice please, that I know that the electron-mass measurements are based on spectroscopy.

The thing is that the spectroscopy measurements are based on atoms/ions that have some binding energies. And if a measured value is related to mass of electron, then it could be affected by the binding energy if the binding energy affects the effective electron mass within the measured system.

Stating that the formulas used for $$m_e$$ derivation do not know about binding energies, is alike stating that the binding energy does not affect the effective value of $$m_e$$, etc. within the bound systems. If it is that way, and it is checked by measurements, then OK.

• Electron binding energies are small relative to the electron mass. Eg, the hydrogen ionisation energy is 13.6 eV, but the electron mass is 511 keV. Jul 20, 2022 at 14:43
• Electron mass is measured with $10^{-10}$ accuracy. pdglive.lbl.gov/Particle.action?node=S003&init=0
– M.S.
Jul 20, 2022 at 14:51
• High-precision measurements of the electron mass are made using free electrons. High-precision measurements of atomic masses, where the part-per-billion correction due to valence electron binding energies would be accessible, use different technology. You might be interested in mass spectroscopy of highly-charged heavy ions, where the electron binding energies may be 100 keV.
– rob
Jul 20, 2022 at 15:13
• Oh, you’re right. I have been out of touch with this literature, and the Penning trap measurements have been superseded.
– rob
Jul 20, 2022 at 16:37
• Well, older Penning trap measurements used free electrons too, as you've suggested. journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.3598 My wrong that I only knew about the newer experiments that are based on ions.
– M.S.
Jul 20, 2022 at 17:35

It's correct that the mass of a hydrogen atom is different from the mass of a proton plus the mass of an electron, but this doesn't matter since this is not how the current value for the electron mass was measured.

It is determined from the hydrogen atom by measuring the Rydberg constant, $$R$$, and the fine structure constant, $$\alpha$$, then using the equation:

$$m_e = \frac{2Rh}{c\alpha^2}$$

• But these two constants, R and α, are measured by spectroscopy on atoms/ions where there are some binding energies. And as far as I know, Rydberg constant is the derived value, based among others on the mass of electron.
– M.S.
Jul 20, 2022 at 16:04
• @M.S. $R_\infty$ is obtained from $R_\infty = R_H(m_e + m_p)/m_e$, where $R_H$ is the ionisation energy of a hydrogen atom. So only the rest masses $m_e$ and $m_p$ are needed. The mass of the hydrogen atom is not involved. Jul 20, 2022 at 16:21
• I've updated my question, so that it's hopefully clearer why I have issues with the content of your answer. The ions used in the measurements know nothing about unbound electrons, since they only contain the bound ones. Our use of formulas then implicitly state whether some effective property is affected or not. I wonder whether the implicit independence was checked at all.
– M.S.
Jul 20, 2022 at 16:49
• @M.S. the masses used in the equation above for $R_\infty$ are the rest masses or invariant masses and these are always the same and are not affected by the mass deficit in an atom. Jul 21, 2022 at 5:38