14.2 Draft: The Sim­plest Nu­clei

This sub­sec­tion in­tro­duces the sim­plest nu­clei and their prop­er­ties.

14.2.1 Draft: The pro­ton

The sim­plest nu­cleus is the hy­dro­gen one, just a sin­gle pro­ton. It is triv­ial. Or at least it is if you ig­nore the fact that that pro­ton re­ally con­sists of a con­glom­er­ate of three quarks held to­gether by glu­ons. A pro­ton has an elec­tric charge $e$ that is the same as that of an elec­tron but op­po­site in sign (pos­i­tive). It has the same spin $s$ as an elec­tron, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Spin is the quan­tum num­ber of in­her­ent an­gu­lar mo­men­tum, chap­ter 5.4. Also like an elec­tron, a pro­ton has a mag­netic di­pole mo­ment $\mu$. In other words, it acts as a lit­tle elec­tro­mag­net.

How­ever, the pro­ton is roughly 2 000 times heav­ier than the elec­tron. On the other hand the mag­netic di­pole mo­ment of a pro­ton is roughly 700 times smaller than that of an elec­tron. The dif­fer­ences in mass and mag­netic di­pole mo­ment are re­lated, chap­ter 13.4. In terms of clas­si­cal physics, a lighter par­ti­cle cir­cles around a lot faster for given an­gu­lar mo­men­tum.

Ac­tu­ally, the pro­ton has quite a large mag­netic mo­ment for its mass. The pro­ton has the same spin and charge as the elec­tron but is roughly 2 000 times heav­ier. So log­i­cally speak­ing the pro­ton mag­netic mo­ment should be roughly 2 000 times smaller than the one of the elec­tron, not 700 times. The ex­pla­na­tion is that the elec­tron is an el­e­men­tary par­ti­cle, but the pro­ton is not. The pro­ton con­sists of two up quarks, each with charge $\frac23e$, and one down quark, with charge $-\frac13e$. All three quarks have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Since the quarks have sig­nif­i­cantly lower ef­fec­tive mass than the pro­ton, they have cor­re­spond­ingly higher mag­netic mo­ments. Even though the spins of the quarks are not all aligned in the same di­rec­tion, the re­sult­ing net mag­netic mo­ment is still un­usu­ally large for the pro­ton net charge, mass, and spin.

Key Points

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The pro­ton is the nu­cleus of a nor­mal hy­dro­gen atom.

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It re­ally con­sists of three quarks, but ig­nore that.

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It has the op­po­site charge of an elec­tron, pos­i­tive.

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It has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$.

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It is roughly 2 000 times heav­ier than an elec­tron.

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It has a mag­netic di­pole mo­ment. But this mo­ment is roughly 700 times smaller than that of an elec­tron.

14.2.2 Draft: The neu­tron

It is hard to call a lone neu­tron a nu­cleus, as it has no net charge to hold onto any elec­trons. In any case, it is some­what aca­d­e­mic, since a lone neu­tron dis­in­te­grates in on av­er­age about 10 min­utes. The neu­tron emits an elec­tron and an an­ti­neu­trino and turns into a pro­ton. That is an ex­am­ple of what is called “beta de­cay.” Neu­trons in nu­clei can be sta­ble.

A neu­tron is slightly heav­ier than a pro­ton. It too has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. And de­spite the zero net charge, it has a mag­netic di­pole mo­ment. The mag­netic di­pole mo­ment of a neu­tron is about two thirds of that of a pro­ton. It is in the di­rec­tion op­po­site to the spin rather than par­al­lel to it like for the pro­ton.

The rea­son that the neu­tron has a di­pole mo­ment is that the three quarks that make up a neu­tron do have charge. A neu­tron con­tains one up quark with a charge of $\frac23e$ and two down quarks with a charge of $-\frac13e$ each. That makes the net charge zero, but the mag­netic di­pole mo­ment can be and is nonzero.

Key Points

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The neu­tron is slightly heav­ier than the pro­ton.

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It too has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$.

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It has no charge.

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De­spite that, it does have a com­pa­ra­ble mag­netic di­pole mo­ment.

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Lone neu­trons are un­sta­ble. They suf­fer beta de­cay.

14.2.3 Draft: The deuteron

The small­est non­triv­ial nu­cleus con­sists of one pro­ton and one neu­tron. This nu­cleus is called the deuteron. (An atom with such a nu­cleus is called deu­terium). Just like the pro­ton-elec­tron hy­dro­gen atom has been crit­i­cal for de­duc­ing the struc­ture of atoms, so the pro­ton-neu­tron deuteron has been very im­por­tant in de­duc­ing knowl­edge about the in­ter­nal struc­ture of nu­clei.

How­ever, the deuteron is not by far as sim­ple a two-par­ti­cle sys­tem as the hy­dro­gen atom. It is also much harder to an­a­lyze. For the hy­dro­gen atom, spec­tro­scopic analy­sis of its ex­cited quan­tum states pro­vided a gold mine of in­for­ma­tion. Un­for­tu­nately, it turns out that the deuteron is so weakly bound that it has no ex­cited quan­tum states. If you try to ex­cite it by adding en­ergy, it falls apart.

The ex­per­i­men­tal bind­ing en­ergy of the deuteron is only about 2.22 MeV. Here a MeV is the en­ergy that an elec­tron would pick up in a one-mil­lion volt­age dif­fer­ence. For an elec­tron, that would be a gi­gan­tic en­ergy. But for a nu­cleus it is ho-hum in­deed. A typ­i­cal sta­ble nu­cleus has a bind­ing en­ergy on the or­der of 8 MeV per nu­cleon.

In any case, it is lucky that that 2.22 MeV of bind­ing en­ergy is there at all. If the deuteron would not bind, life as we know it would not ex­ist. The for­ma­tion of nu­clei heav­ier than hy­dro­gen, in­clud­ing the car­bon of life, be­gins with the deuteron.

The lack of ex­cited states makes it hard to un­der­stand the deuteron. In ad­di­tion, spin has a ma­jor ef­fect on the force be­tween the pro­ton and neu­tron. In the hy­dro­gen atom, that ef­fect ex­ists but it is ex­tremely small. In par­tic­u­lar, in the true hy­dro­gen atom ground state the elec­tron and pro­ton align their spins in op­po­site di­rec­tions. That pro­duces the so-called sin­glet state of zero net spin, chap­ter 5.5.6. How­ever, the elec­tron and pro­ton can also align their spins in the same di­rec­tion, at least as far as an­gu­lar mo­men­tum un­cer­tainty al­lows. That pro­duces the so-called triplet state of unit net spin. For the hy­dro­gen atom, it turns out that the triplet state has very slightly higher en­ergy than the sin­glet state, {A.39}.

In case of the deuteron, how­ever, the triplet state has the low­est en­ergy. And the sin­glet state has so much more en­ergy that it is not even bound. Al­most bound maybe, but def­i­nitely not bound. For the pro­ton and neu­tron to bind to­gether at all, they must align their spins into the triplet state.

As a re­sult, a nu­cleus con­sist­ing of two pro­tons (the dipro­ton) or of two neu­trons (the dineu­tron) does not ex­ist. That is de­spite the fact that two pro­tons or two neu­trons at­tract each other al­most the same as the pro­ton and neu­tron in the deuteron. The prob­lem is the an­ti­sym­metriza­tion re­quire­ment that two iden­ti­cal nu­cle­ons must sat­isfy, chap­ter 5.6. A spa­tial ground state should be sym­met­ric. (See ad­den­dum {A.40} for more on that.) To sat­isfy the an­ti­sym­metriza­tion re­quire­ment, the spin state of a dipro­ton or dineu­tron must then be the an­ti­sym­met­ric sin­glet state. But only the triplet state is bound.

(You might guess that the dipro­ton would also not ex­ist be­cause of the Coulomb re­pul­sion be­tween the two pro­tons. But if you ball­park the Coulomb re­pul­sion us­ing the mod­els of {A.41}, it is less than a third of the al­ready small 2.22 MeV bind­ing en­ergy. In gen­eral, the Coulomb force is quite small for light nu­clei.)

There is an­other qual­i­ta­tive dif­fer­ence be­tween the hy­dro­gen atom and the deuteron. The hy­dro­gen atom has zero or­bital an­gu­lar mo­men­tum in its ground state. In par­tic­u­lar, the quan­tum num­ber of or­bital an­gu­lar mo­men­tum $l$ equals zero. That makes the spa­tial struc­ture of the atom spher­i­cally sym­met­ric.

But or­bital an­gu­lar mo­men­tum is not con­served in the deuteron. In terms of clas­si­cal physics, the forces be­tween the pro­ton and neu­tron are not ex­actly along the line con­nect­ing them. They de­vi­ate from the line based on the di­rec­tions of the nu­cleon spins.

In terms of quan­tum me­chan­ics, this gets phrased a bit dif­fer­ently. The po­ten­tial does not com­mute with the or­bital an­gu­lar mo­men­tum op­er­a­tors. There­fore the ground state is not a state of def­i­nite or­bital an­gu­lar mo­men­tum. The an­gu­lar mo­men­tum is still lim­ited by the ex­per­i­men­tal ob­ser­va­tions that the deuteron has spin 1 and even par­ity. That re­stricts the or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$ to the pos­si­ble val­ues 0 or 2, {A.40}. Var­i­ous ev­i­dence shows that there is a quan­tum prob­a­bil­ity of about 95% that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and 5% that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2.

One con­se­quence of the nonzero or­bital an­gu­lar mo­men­tum is that the mag­netic di­pole strength of the deuteron is not ex­actly what would be ex­pected based on the di­pole strengths of pro­ton and neu­tron. Since the charged pro­ton has or­bital an­gu­lar mo­men­tum, its acts like a lit­tle elec­tro­mag­net not just be­cause of its spin, but also be­cause of its or­bital mo­tion.

An­other con­se­quence of the nonzero or­bital an­gu­lar mo­men­tum is that the charge dis­tri­b­u­tion of the deuteron is not ex­actly spher­i­cally sym­met­ric. This asym­met­ric charge dis­tri­b­u­tion al­lows the deuteron to in­ter­act with gra­di­ents in an ex­ter­nal elec­tric field. It is said that the deuteron has a nonzero “elec­tric quadru­pole mo­ment.”

Roughly speak­ing, you may think of the charge dis­tri­b­u­tion of the deuteron as elon­gated in the di­rec­tion of its spin. That is not quite right, quan­tum-me­chan­i­cally speak­ing, since an­gu­lar mo­men­tum has un­cer­tainty in di­rec­tion. There­fore, in­stead con­sider the quan­tum state in which the deuteron spin has its max­i­mum com­po­nent, $\hbar$, in the cho­sen $z$-​di­rec­tion. In that state, the charge dis­tri­b­u­tion is elon­gated in the $z$-​di­rec­tion.

The nonzero or­bital mo­men­tum also shows up in ex­per­i­ments where var­i­ous par­ti­cles are scat­tered off deuterons.

To be sure, the pre­cise prob­a­bil­ity of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state has never been es­tab­lished. How­ever, as­sume that the deuteron is mod­eled as com­posed of a pro­ton and a neu­tron. (Al­though in re­al­ity it is a sys­tem of 6 quarks.) And as­sume that the pro­ton and neu­tron have the same prop­er­ties as they have in free space. (That is al­most cer­tainly not a good as­sump­tion; com­pare the next sec­tion.) For such a model the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state needs to have about 4% prob­a­bil­ity to get the mag­netic mo­ment right. Sim­i­lar val­ues can be de­duced from the quadru­pole mo­ment and scat­ter­ing ex­per­i­ments, [31].

Key Points

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The deuteron con­sists of a pro­ton and a neu­tron.

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The deuteron is the sim­plest non­triv­ial nu­cleus. The dipro­ton and the dineu­tron do not ex­ist.

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The deuteron has spin 1 and even par­ity. The bind­ing en­ergy is 2.225 MeV.

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There are no ex­cited states. The ground state of low­est en­ergy is all there is.

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The deuteron has a nonzero mag­netic di­pole mo­ment.

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It also has a nonzero elec­tric quadru­pole mo­ment.

14.2.4 Draft: Prop­erty sum­mary

Ta­ble 14.1 gives a sum­mary of the prop­er­ties of the three sim­plest nu­clei. The elec­tron is also in­cluded for com­par­i­son.

Ta­ble 14.1: Prop­er­ties of the elec­tron and of the sim­plest nu­clei.

The first data col­umn gives the mass. Note that nu­clei are thou­sands of times heav­ier than elec­trons. As far as the units are con­cerned, what is re­ally listed is the en­ergy equiv­a­lent of the masses. That means that the mass is mul­ti­plied by the square speed of light fol­low­ing the Ein­stein mass-en­ergy re­la­tion. The re­sult­ing en­er­gies in Joules are then con­verted to MeV. An MeV is the en­ergy that an elec­tron picks up in a 1 mil­lion volt­age dif­fer­ence. Yes it is crazy, but that is how you will al­most al­ways find masses listed in nu­clear ref­er­ences. So you may as well get used to it.

It can be ver­i­fied from the given num­bers that the deuteron mass is in­deed smaller than the sum of pro­ton and neu­tron masses by the 2.225 MeV of bind­ing en­ergy. It is a tenth of a per­cent, but it is very ac­cu­rately mea­sur­able.

The sec­ond col­umn gives the charge. Note that all these charges are whole mul­ti­ples of the pro­ton charge $e$. How­ever, that is not a fun­da­men­tal re­quire­ment of physics. In par­tic­u­lar, up quarks have charge ${\textstyle\frac{2}{3}}e$ while down quarks have charge $-{\textstyle\frac{1}{3}}e$. The pro­ton con­tains two up quarks and a down one, pro­duc­ing net charge $e$. The neu­tron con­tains one up quark and two down ones, pro­duc­ing zero net charge.

The third col­umn gives the charge ra­dius. That is a mea­sure of the spa­tial ex­tent of the charge dis­tri­b­u­tion. The elec­tron is, as far as is known, a point par­ti­cle with no in­ter­nal struc­ture. For the neu­tron, with no net charge, it is not re­ally clear what to de­fine as charge ra­dius.

The fourth col­umn shows the quan­tum num­ber of net an­gu­lar mo­men­tum. For the first three par­ti­cles, that is sim­ply their spin. For the deuteron, it is the nu­clear spin. That in­cludes both the spins and the or­bital an­gu­lar mo­menta of the pro­ton and neu­tron that make up the deuteron.

The fifth col­umn is par­ity. It is even in all cases. More com­pli­cated nu­clei can have neg­a­tive par­ity.

The sixth col­umn is the mag­netic di­pole mo­ment. It is ex­pressed in terms of the so-called nu­clear mag­ne­ton $\mu_{\rm N}$. A pro­ton cir­cling around with one quan­tum unit of or­bital an­gu­lar mo­men­tum has one nu­clear mag­ne­ton of mag­netic mo­ment due to its mo­tion. (Which would be in ad­di­tion to the in­trin­sic mag­netic mo­ment listed in the ta­ble. Note that mag­netic mo­ments are vec­tors like an­gu­lar mo­menta; they may can­cel each other when summed.)

As the ta­ble shows, nu­clei have much smaller mag­netic mo­ments than elec­trons. That is due to their much larger masses. How­ever, us­ing a mag­netic field of just the right fre­quency, nu­clear mag­netic mo­ments can be ob­served. That is how nu­clear mag­netic res­o­nance works, chap­ter 13.6. There­fore nu­clear mag­netic mo­ments are im­por­tant for many ap­pli­ca­tions, in­clud­ing med­ical ones like MRI.

The last col­umn lists the elec­tric quadru­pole strength. That is a mea­sure for the de­vi­a­tion of the nu­clear charge dis­tri­b­u­tion from a spher­i­cally sym­met­ric shape. It is a com­pli­cat­ing fac­tor in nu­clear mag­netic res­o­nance. Or an ad­di­tional source of in­for­ma­tion, de­pend­ing on your view point. Nu­clei with spin less than 1 do not have elec­tric quadru­pole mo­ments. (That is an im­plied con­se­quence of the re­la­tion be­tween an­gu­lar mo­men­tum and sym­me­try in quan­tum me­chan­ics.)

Note that the SI length unit of fem­tome­ter works very nicely for nu­clei. So, since physi­cists hate per­fec­tion, they de­fine a new non-SI unit called the barn b. A barn is 100 fm$\POW9,{2}$. So you will likely find the quadru­pole mo­ment of the deuteron listed as 0.00286 $e$b. Note the ad­di­tional lead­ing ze­ros. Some physi­cists do not like them, for good rea­son, and then use mil­libarn, giv­ing 2.86 $e$mb. How­ever, the quadru­pole mo­ments for many heavy nu­clei are quite large in terms of mil­libarn. For ex­am­ple, ein­steinium-253 has around 6 700 $e$mb. Any­time now, physi­cists are bound to fig­ure out that cen­tibarn works even bet­ter than mil­libarn. When that hap­pens, let's all agree that we will not tell them that a cen­tibarn is the same as that hated fm$\POW9,{2}$. The charge $e$ is com­monly left away from the de­f­i­n­i­tion of the quadru­pole mo­ment, giv­ing it units of area.

Key Points

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The prop­er­ties of the sim­plest nu­clei are sum­ma­rized in ta­ble 14.1.