3
$\begingroup$

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!

$\endgroup$
  • 6
    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Jun 8 '16 at 4:17
  • 1
    $\begingroup$ 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). $\endgroup$ – Loong Jun 8 '16 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. $\endgroup$ – Joseph Hirsch Sep 17 '18 at 19:03
4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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 Jun 9 '16 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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