14.6 Draft: 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}= \mbox{1.672\,621~10$\POW9,{-27}$\ kg}
\qquad m_{\rm n}= \mbox{1.674\,927~10$\POW9,{-27}$\ 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}=\mbox{0.510\,998\,91 MeV} \quad
A_e=\mbox{1.443\...
...9,{-5}$\ MeV} \quad
B_e=\mbox{1.554\,68~10$\POW9,{-12}$\ 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 (but not in the data pre­sented here).