I am watching this Khan Academy video on mass defect and binding energy, it uses helium-4 as an example to go through the concept. The person finds the predicted mass (protons + neutrons) to be 4.0318274 amu. He also finds that the experimentally measured mass for He-4 to be 4.00150608 amu.

What's baffling me is, where did he find the experimentally measured mass for He-4? I have been searching for the atomic mass of He-4 all over, and every source I found seems to agree that its mass is, in fact, 4.002602 u ± 0.000002 u, that's a 0.001 difference!

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    $\begingroup$ The difference looks mighty like the weight of two electrons. That is, you are comparing $\ce{^4He}$ to $\ce{^4He^2+}$. Also, welcome to Chem.SE. $\endgroup$ Commented Jun 8, 2016 at 4:17
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    $\begingroup$ Also note that the video shows the obsolete unit $\mathrm{amu}$ (which is deprecated since 1961), although the given numerical values rather correspond to the unit $\mathrm u$ ($1\ \mathrm{u}=1.660539040(20)\times10^{-27}\ \mathrm{kg}$). The proton mass actually is $m_\mathrm p=1.007276466879(91)\ \mathrm u$ and the neutron mass is $m_\mathrm n=1.00866491588(49)\ \mathrm u$. The difference is small but significant in this case. For example, the atomic mass of C-12 is about $12.0005\ \mathrm{amu}$ (as used in chemistry until 1961) or $12.0000\ \mathrm{u}$ (by definition, since 1961). $\endgroup$
    – user7951
    Commented Jun 8, 2016 at 17:10
  • $\begingroup$ Do protons and neutrons have exactly the same mass in an atom regardless of the combination of quarks. For example, you could conceivably have a proton consisting of Up/Red, Up/Blue, Down/Green quarks and another one U/B, U/G, D/R. In fact, given that a neutron has slightly more mass than a Proton, it seems to me that the charge/color charge combination does result in different combined masses. $\endgroup$ Commented Sep 17, 2018 at 19:03

1 Answer 1


As noted by Ivan in his comment, the difference is likely due to using the helium nucleus $\ce{^4He^2+}$ and not a helium atom. Shame on the Khan Academy video for not making this clear.

However, there is a good reason to only consider the nucleus. The mass defect and binding energy are consequences of the strong nuclear fore, and thus properties of the nucleus.

The choice of helium-4 is convenient: the helium-4 nucleus is the alpha particle - a common product of radioactive decay and thus well-characterized. Thanks to mass spectrometry, we can determine the mass of an alpha particle very precisely to be $4.00150608\ \text{u}$. Here is the source of that mysterious number. Again, shame on Khan Academy for not being explicit.

The electron has a mass of $5.48579909\times 10^{-4}\ \text{u}$, so a helium-4 atom would be expected to have a mass of one alpha particle and two electrons:

$$4.00150608\ \text{u} + 2\left(5.4858\times 10^{-4}\ \text{u}\right) = 4.00260324\ \text{u}$$

  • $\begingroup$ While it appears there might be a small mass defect from adding the electrons $\left(4.002603 > 4.002602\right)$, however the value I calculated is within the $\pm 0.000002$ error on the experimental value. $\endgroup$
    – Ben Norris
    Commented Jun 9, 2016 at 13:15

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