14.7 Mass and energy

Nuclear masses are not what you would naively expect. For example, since the deuterium nucleus consists of one proton and one neutron, you might assume its mass is the sum of that of a proton and a neutron. It is not. It is less.

This weird effect is a consequence of Einstein’s famous relation $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, in which $E$ is energy, $m$ mass, and $c$ the speed of light, chapter 1.1.2. When the proton and neutron combine in the deuterium nucleus, they lower their total energy by the binding energy that keeps the two together. According to Einstein’s relation, that means that the mass goes down by the binding energy divided by $c^2$. In general, for a nucleus with $Z$ protons and $N$ neutrons,

\begin{displaymath}
\fbox{$\displaystyle
m_{\rm nucleus} = Z m_{\rm p}+ N m_{\rm n}- \frac{E_{\rm{B}}}{c^2}
$} %
\end{displaymath} (14.7)

where

\begin{displaymath}
m_{\rm p}= 1.672{,}621\;10^{-27}\mbox{ kg}
\qquad m_{\rm n}= 1.674{,}927\;10^{-27}\mbox{ kg}
\end{displaymath}

are the mass of a lone proton respectively a lone neutron at rest, and $E_{\rm {B}}$ is the binding energy. This result is very important for nuclear physics, because mass is something that can readily be measured. Measure the mass accurately and you know the binding energy.

In fact, even a normal hydrogen atom has a mass lower than that of a proton and electron by the 12.6 eV (electron volt) binding energy between proton and electron. But scaled down by $c^2$, the associated change in mass is negligible.

In contrast, nuclear binding energies are on the scale of MeV instead of eV, a million times higher. It is the devastating difference between a nuclear bomb and a stick of dynamite. Or between the almost limitless power than can be obtained from peaceful nuclear reactors and the limited supply of fossil fuels.

At nuclear energy levels the changes in mass become noticeable. For example, deuterium has a binding energy of 2.224,5 MeV. The proton has a rest mass that is equivalent to 938.272,013 MeV in energy, and the neutron 939.565,561 MeV. (You see how accurately physicists can measure masses.) Therefore the mass of the deuteron nucleus is lower than the combined mass of a proton and a neutron by about 0.1%. It is not big, but observable. Physicists are able to measure masses of reasonably stable nuclei extremely accurately by ionizing the atoms and then sending them through a magnetic field in a mass spectrograph or mass spectrometer. And the masses of unstable isotopes can be inferred from the end products of nuclear reactions involving them.

As the above discussion illustrates, in nuclear physics masses are often expressed in terms of their equivalent energy in MeV instead of in kg. To add further confusion and need for conversion factors, still another unit is commonly used in nuclear physics and chemistry. That is the “unified atomic mass unit“ (u), also called “Dalton,” (Da) or “universal mass unit” to maximize confusion. The “atomic mass unit” (amu) is an older virtually identical unit, or rather two virtually identical units, since physicists and chemists used different versions of it in order to achieve that supreme perfection in confusion.

These units are chosen so that atomic or nuclear masses expressed in terms them are approximately equal to the number of nucleons, (within a percent or so.) The current official definition is that a carbon-12, $\fourIdx{12}{6}{}{}{\rm {C}}$, atom has a mass of exactly 12 u. That makes 1 u equivalent 931.494,028 MeV. That is somewhat less than the mass of a free proton or a neutron.

One final warning about nuclear masses is in order. Almost always, it is atomic mass that is reported instead of nuclear mass. To get the nuclear mass, the rest mass of the electrons must be subtracted, and a couple of additional correction terms applied to compensate for their binding energy, [36]:

\begin{displaymath}
\fbox{$\displaystyle
m_{\rm nucleus} = m_{\rm atom} - Z m_{\rm e}+ A_e Z^{2.39} + B_e Z^{5.35}
$} %
\end{displaymath} (14.8)


\begin{displaymath}
m_{\rm e}=0.510{,}998{,}91\mbox{ MeV} \quad
A_e=1.443{,}...
...^{-5}\mbox{ MeV} \quad
B_e=1.554{,}68\;10^{-12}\mbox{ MeV}
\end{displaymath}

The nuclear mass is taken to be in MeV. So it is really the rest mass energy, not the mass, but who is complaining? Just divide by $c^2$ to get the actual mass. The final two correction terms are really small, especially for light nuclei, and are often left away.