The mass of carbon-12 is $\pu{12 u}$ by definition. However, one carbon-12 atom comprises 6 neutrons (each weighing $\pu{1.0087 u}$), 6 protons (each weighing $\pu{1.0072 u}$), and 6 electrons (each weighing $\pu{0.0005 u}$), which all add up to $\pu{12.0894 u}$.

Where does the $0.7\%$ difference in mass come from?

  • 6
    $\begingroup$ In very simple and short terms: In bound states, the mass of the involved particles changes, i.e. mass defect. The loss of bonding energy translates to a loss of mass, basically demonstrated by $E=m\cdot\mathcal{c}^2$. $\endgroup$ Mar 17, 2015 at 8:37
  • $\begingroup$ Exactly that mass difference is used in nuclear reactions to "generate" energy: for light nuclei (smaller than iron), energy is released by fusing them to heavier ones; for large nuclei (larger than iron), it is released by splitting them. $\endgroup$
    – DK2AX
    Mar 17, 2015 at 10:07

1 Answer 1


This is due to the mass-energy equivalence and a phenomenon called binding energy.

Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.

The binding energy of carbon-12 is quoted on Wikipedia as $\pu{92.162 MeV}$. Therefore we can estimate the mass defect of a carbon-12 atom, $\Delta m$, using $ E = (\Delta m)c^2$:

$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$

The difference in the mass of carbon-12 to the mass of its constituent particles is $\pu{0.08940 u}$, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.

  • $\begingroup$ The small difference should be binding of the electrons to the nucleus, though I cannot see the value above (yellow boxes) $\endgroup$
    – Alchimista
    Jan 11, 2020 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.