14.1 Fun­da­men­tal Con­cepts

This sec­tion de­scribes the most ba­sic facts about nu­clei. These facts will be taken for granted in the rest of this chap­ter.

Nu­clei con­sist of pro­tons and neu­trons. Pro­tons and neu­trons are there­fore called “nu­cle­ons.” Neu­trons are elec­tri­cally neu­tral, but pro­tons are pos­i­tively charged. In par­tic­u­lar, the elec­tric charge of a pro­ton has the same mag­ni­tude as the charge of an elec­tron, but has op­po­site sign. Since op­po­site charges at­tract, the pro­tons in a nu­cleus at­tract elec­trons. De­spite that, the elec­trons do not end up in­side the nu­cleus. They have much larger quan­tum me­chan­i­cal un­cer­tainty in po­si­tion than the much heav­ier nu­cle­ons. So the elec­trons form a cloud around the tiny nu­cleus, pro­duc­ing an atom.

Since charges of the same sign re­pel, pro­tons mu­tu­ally re­pel each other. That is due to the same elec­tric Coulomb force that al­lows them to at­tract elec­trons. By it­self, the Coulomb force be­tween the pro­tons in a nu­cleus would cause the nu­cleus to fly apart im­me­di­ately. But nu­cle­ons, both pro­tons and neu­trons, also at­tract each other through an­other force, the “nu­clear force.” It is this force that keeps a nu­cleus to­gether.

The nu­clear force is very strong, which al­lows it to dom­i­nate elec­tro­mag­netic forces like the re­pul­sive Coulomb force in sta­ble nu­clei. But the nu­clear force is also very short range, ex­tend­ing over no more than a few fem­tome­ters. (A fem­tome­ter, or fm, equals 10$\POW9,{-15}$ m. It is some­times called a fermi af­ter fa­mous nu­clear physi­cist En­rico Fermi. While not ap­proved by SI, Fermi was one of the good guys, so we should make al­lowances.) In big nu­clei, nu­cle­ons are only held to­gether to other nu­cle­ons in their im­me­di­ate neigh­bor­hood by the nu­clear force. But the pro­tons are re­pulsed by other pro­tons every­where in the nu­cleus. If the nu­cleus gets too big, this re­pul­sion be­comes so big that the nu­cleus can no longer be sta­ble. Lead, with 82 pro­tons, is the heav­i­est el­e­ment that can be sta­ble, and then only if it con­tains a suit­able num­ber of neu­trons to keep the pro­tons some­what apart.

The strength of the nu­clear force is about the same re­gard­less of the type of nu­cle­ons in­volved, pro­tons or neu­trons. That is called “charge in­de­pen­dence.”

More re­stric­tively, but even more ac­cu­rately, the nu­clear force is the same if you swap the nu­cleon types. In other words, the nu­clear force is the same if you re­place all pro­tons by neu­trons and vice-versa. That is called “charge sym­me­try.” For ex­am­ple, if you swap the nu­cleon type of a pair of pro­tons, you get a pair of neu­trons. There­fore the nu­clear force be­tween a pair of pro­tons is very ac­cu­rately the same as the one be­tween a pair of neu­trons, all else be­ing equal. (The al­ready men­tioned Coulomb re­pul­sion be­tween the pro­tons is ad­di­tional and not the same.) But if you swap the nu­cleon type of a pair of pro­tons, or of a pair of neu­trons, you do not get a pro­ton and a neu­tron. So the nu­clear force be­tween a pro­ton and a neu­tron is less ac­cu­rately the same as that be­tween two pro­tons or two neu­trons.

The nu­clear force is not a fun­da­men­tal one. It is just an ef­fect of the “color force” or “strong force” be­tween the “quarks” of which pro­tons and neu­trons con­sist. That is why the nu­clear force is also of­ten called the “resid­ual strong force.” It is much like how the Van der Waals force be­tween mol­e­cules is not a fun­da­men­tal one; that force is a resid­ual of the elec­tro­mag­netic force be­tween the elec­trons and nu­clei of which mol­e­cules ex­ist, {A.33}.

How­ever, the the­ory of the color force,“quan­tum chrom­e­dy­nam­ics,” is well be­yond the scope of this book. It is also not re­ally im­por­tant for nan­otech­nol­ogy. In fact, it is not all that im­por­tant for nu­clear en­gi­neer­ing ei­ther be­cause the de­tails of the the­ory are un­cer­tain, and nu­mer­i­cal so­lu­tion is in­tractable, [19].

De­spite the fact that the nu­clear force is poorly un­der­stood, physi­cists can say some things with con­fi­dence. First of all,

Nu­clei are nor­mally in the ground state.
The ground state is the quan­tum state of low­est en­ergy $E$. Nu­clei can also be in ex­cited states of higher en­ergy. How­ever, a bit of ther­mal en­ergy is not go­ing to ex­cite a nu­cleus. Dif­fer­ences be­tween nu­clear en­ergy lev­els are ex­tremely large on a mi­cro­scopic scale. That is why nu­clear bombs and nu­clear re­ac­tors can cre­ate so much en­ergy. Still, nu­clear re­ac­tions will typ­i­cally leave nu­clei in ex­cited states. Usu­ally such states de­cay back to the ground state very quickly. (In spe­cial cases, it may take for­ever.)

It should be noted that if a nu­clear state is not sta­ble, it im­plies that it has a very slight un­cer­tainty in en­ergy, com­pare chap­ter 7.4.1. This un­cer­tainty in en­ergy is com­monly called the “width” $\Gamma$ of the state. The dis­cus­sion here will al­most al­ways ig­nore the un­cer­tainty in en­ergy.

A sec­ond gen­eral prop­erty of nu­clei is:

Nu­clear states have def­i­nite nu­clear mass $m_{\rm N}$.
You may be sur­prised by this state­ment. It seems triv­ial. You would ex­pect that the nu­clear mass is sim­ply the sum of the masses of the pro­tons and neu­trons that make up the nu­cleus. But Ein­stein’s fa­mous re­la­tion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$ re­lates en­ergy and mass. The nu­clear mass is slightly less than the com­bined mass of the pro­tons and neu­trons from which it is made. The dif­fer­ence is the bind­ing en­ergy that keeps the nu­cleus to­gether, ex­pressed in mass units. (In other words, di­vided by the square speed of light $c^2$.) Sure, even for nu­clear en­er­gies the changes in nu­clear mass due to bind­ing en­ergy are tiny. But physi­cists can mea­sure nu­clear masses to very great ac­cu­racy. Dif­fer­ent nu­clear states have dif­fer­ent bind­ing en­er­gies. So they have slightly dif­fer­ent nu­clear masses.

(Sim­i­larly, a hy­dro­gen atom has less mass than a free pro­ton and a free elec­tron. But here the dif­fer­ence, a few eV, is far too small to note. Since nu­clear bind­ing en­er­gies are mil­lions of times big­ger, in nu­clei the ef­fect is much more im­por­tant.)

It may be noted that bind­ing en­er­gies are al­most never ex­pressed in mass units in nu­clear physics. In­stead masses are ex­pressed in en­ergy units! And not in Joule ei­ther. The en­ergy units used are al­most in­vari­ably elec­tron volts (eV). Never use an SI unit when talk­ing to nu­clear physi­cists. They will im­me­di­ately know that you are one of those de­spised non­ex­perts. Just call it a blah. In the un­likely case that they ask, tell them That is what Fermi called it.

Next,

Nu­clear states have def­i­nite nu­clear spin $j_{\rm N}$.
Here the “nu­clear spin” $j_{\rm N}$ is the quan­tum num­ber of the net nu­clear an­gu­lar mo­men­tum. The mag­ni­tude of the net nu­clear an­gu­lar mo­men­tum it­self is

\begin{displaymath}
J = \sqrt{j_{\rm N}(j_{\rm N}+1)}\hbar
\end{displaymath}

Nu­clei in ex­cited en­ergy states usu­ally have dif­fer­ent an­gu­lar mo­men­tum than in the ground state.

The name nu­clear spin may seem in­ap­pro­pri­ate since net nu­clear an­gu­lar mo­men­tum in­cludes not just the spin of the nu­cle­ons but also their or­bital an­gu­lar mo­men­tum. But since nu­clear en­er­gies are so large, in many cases nu­clei act much like el­e­men­tary par­ti­cles do. Ex­ter­nally ap­plied elec­tro­mag­netic fields are not by far strong enough to break up the in­ter­nal nu­clear struc­ture. And the an­gu­lar mo­men­tum of an el­e­men­tary par­ti­cle is ap­pro­pri­ately called spin. How­ever, the fact that nu­clear spin is two words and “az­imuthal quan­tum num­ber of the net nu­clear an­gu­lar mo­men­tum” is nine might con­ceiv­ably also have some­thing to do with the ter­mi­nol­ogy.

Ac­cord­ing to quan­tum me­chan­ics, $j_{\rm N}$ must be in­te­ger or half-in­te­ger. In par­tic­u­lar, $j_{\rm N}$ must be an in­te­ger if the num­ber of nu­cle­ons is even ($j_{\rm N}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or 1 or 2 or ...). If the num­ber of nu­cle­ons is odd, $j_{\rm N}$ must be half an odd in­te­ger ($j_{\rm N}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ or ...).

The fact that nu­clei have def­i­nite an­gu­lar mo­men­tum does not de­pend on the de­tails of the nu­clear force. It is a con­se­quence of the very fun­da­men­tal ob­ser­va­tion that empty space has no build-in pre­ferred di­rec­tion. That is­sue was ex­plored in more de­tail in chap­ter 7.3.

(Many ref­er­ences use the sym­bol $J$ also for $j_{\rm N}$ for that spicy ex­tra bit of con­fu­sion. So one ref­er­ence tells you that the eigen­value [sin­gu­lar] of $J^2$ is $J(J+1)$, leav­ing the $\hbar^2$ away from con­cise­ness. No kid­ding. One pop­u­lar book uses $I$ in­stead of $j_{\rm N}$ and re­serves $J$ for elec­tronic an­gu­lar mo­men­tum. At least this ref­er­ence uses a bold face $I$ to in­di­cate the an­gu­lar mo­men­tum it­self, as a vec­tor.)

Con­sider also the com­po­nent $J_z$ of the nu­clear an­gu­lar mo­men­tum in a se­lected $z$ di­rec­tion. Ac­cord­ing to quan­tum me­chan­ics, this com­po­nent can have the mea­sur­able val­ues $-j_{\rm N}$, $-j_{\rm N}{+}1$, $-j_{\rm N}{+}2$, ..., $j_{\rm N}{-}1$, or $j_{\rm N}$. Note also that if $J_z$ has a def­i­nite value for nonzero $j_{\rm N}$, then the com­po­nents in or­thog­o­nal di­rec­tions have un­cer­tain val­ues and are there­fore not that in­ter­est­ing for analy­sis.

Fi­nally,

Nu­clear states have def­i­nite par­ity.
Here par­ity is what hap­pens to the wave func­tion when the nu­cleus is mir­rored and then ro­tated 180$\POW9,{\circ}$ around the axis nor­mal to the mir­ror, chap­ter 7.3. (Math­e­mat­i­cally, this cor­re­sponds to in­vert­ing every ${\skew0\vec r}$ po­si­tion vec­tor mea­sured from the cen­ter of grav­ity into $-{\skew0\vec r}$. Since the ro­ta­tion is al­ready cov­ered by an­gu­lar mo­men­tum, the im­por­tant step is the mir­ror­ing.) The wave func­tion can ei­ther stay the same, (called par­ity 1 or even par­ity), or it can change sign, (called par­ity $\vphantom0\raisebox{1.5pt}{$-$}$1 or odd par­ity). The fact that nu­clei have def­i­nite par­ity too does not de­pend on the de­tails of the nu­clear force. It is a con­se­quence of the fact that the forces of na­ture be­have the same way when seen in the mir­ror.

Or ac­tu­ally, there is one force of na­ture, the still un­men­tioned so-called weak force that does not be­have the same way when seen in the mir­ror. But the weak force is, like it says, weak. On nu­clear scales, it is many or­ders of mag­ni­tude smaller than the nu­clear and elec­tro­mag­netic forces. So, while the weak force in­tro­duces some quan­tum-me­chan­i­cal un­cer­tainty in the par­ity of nu­clei, this un­cer­tainty is usu­ally neg­li­gi­bly small. The chances of find­ing a nu­cleus in a given en­ergy state with the wrong par­ity can be ball­parked at 10$\POW9,{-14}$, [30, pp. 313ff]. That is al­most al­ways neg­li­gi­ble. Only if, say, a nu­clear process is strictly im­pos­si­ble solely be­cause of par­ity, then the un­cer­tainty in par­ity might give it a very slight pos­si­bil­ity of oc­cur­ring any­way.

Par­ity is com­monly in­di­cated by $\pi$ be­cause $\pi$ is the Greek let­ter p and is not used for any­thing else in sci­ence. And physi­cists usu­ally list the spin and par­ity of a nu­cleus to­gether in the form $J^\pi$. If you have two quan­ti­ties like spin and par­ity that have noth­ing to do with one an­other, what is bet­ter than show one as a su­per­script of the other? But do not start rais­ing $J$ to the power $\pi$! You should be trans­lat­ing this into com­mon sense as fol­lows:

\begin{displaymath}
J^\pi
\quad\Rightarrow\quad
j_{\rm N}^{\pm}
\quad\Rightarrow\quad
j_{\rm N}\mbox{ and }\pm
\end{displaymath}

As a nu­mer­i­cal ex­am­ple, 3$\POW9,{-}$ means a nu­cleus with spin 3 and odd par­ity. It does not mean a nu­cleus with spin 1/3, (which is not even pos­si­ble; spins can only be in­te­ger or half-in­te­ger.)


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­clei form the cen­ters of atoms.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­clei con­sist of pro­tons and neu­trons. There­fore pro­tons and neu­trons are called nu­cle­ons.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Pro­tons and neu­trons them­selves con­sist of quarks. But for prac­ti­cal pur­poses, you may as well for­get about that.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Neu­trons are elec­tri­cally neu­tral. Pro­tons are pos­i­tively charged.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­cle­ons are held to­gether by the so-called nu­clear force.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The nu­clear force is ap­prox­i­mately in­de­pen­dent of whether the nu­cle­ons are pro­tons or neu­trons. That is called charge in­de­pen­dence. Charge sym­me­try is a more ac­cu­rate, but also more lim­ited ver­sion of charge in­de­pen­dence.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­clear states, in­clud­ing the ground state, have def­i­nite nu­clear en­ergy $E$. The dif­fer­ences in en­ergy be­tween nu­clear states are so large that they pro­duce small but mea­sur­able dif­fer­ences in the nu­clear mass $m_{\rm N}$.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­clear states also have def­i­nite nu­clear spin $j_{\rm N}$. Nu­clear spin is the az­imuthal quan­tum num­ber of the net an­gu­lar mo­men­tum of the nu­cleus. Many ref­er­ences in­di­cate it by $J$ or $I$.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Nu­clear states have def­i­nite par­ity $\pi$. At least they do if the so-called weak force is ig­nored.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Never use an SI unit when talk­ing to a nu­clear physi­cist.