I want to calculate the energy released from the beta decay of Carbon-14:

$${}_6^{14}\mathrm{C} \longrightarrow{}_7^{14}\mathrm{N}+{}_{-1}^{0}\beta+\overline\nu$$

The following masses are given:

${}_6^{14}\mathrm{C} = 14.00324 u$

${}_7^{14}\mathrm{N} = 14.00307 u$

From these I can find the mass difference which is $1.7 \times 10^{-4}u$, and form there I can calculate the correct energy released.

What I don't understand is why I don't get the same result by looking at the mass difference from the number of protons, neutrons and electrons in the reaction (ignoring the anti-neutrino since its mass is so small). In this case I get:

6 protons + 8 neutrons + 7 electrons $\longrightarrow$ 7 protons + 7 neutrons + 8 electrons

Which gives a net gain of one proton and one electron and a net loss of one neutron, so the change in mass is:

$m_n-m_p-m_e = 1.008665 - 1.007276-0.000549=8.25\times10^{-4}u$.

This is almost 5 times larger than the change in mass calculated from the data given. I have considered that in the data given the ${}_7^{14}\mathrm{N}$ would only have 7 electrons since the emitted electron (beta particle) doesn't form part of the atom, but that makes the two results diverge even more.

How should the mass difference (and therefore the energy released) from a beta decay be calculated?


1 Answer 1


It's incorrect to think that the mass of an atom is given by the sum of its constituents since when particles bind they do so to go into a lower energy state of the entire system. This difference in mass (energy) is called binding energy.

The energy released by a reaction is at the level of the reactants and not the constituents.

  • $\begingroup$ That makes a lot of sense, thank you! Does that mean that the only way to find the mass difference is essentially by looking up empirical values of either the atmoic mass or the binding energy (or binding energy per nucleon)? $\endgroup$
    – Kael
    Apr 23, 2021 at 11:47
  • $\begingroup$ @Kael From a purely theoretical point, if you know the binding energy you could add it to the calculation to find the "real" mass difference. But in practice one always uses the experimental value of the mass of the atom. $\endgroup$
    – Quiver
    Apr 23, 2021 at 15:09

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