Sub­sec­tions


14.19 Draft: Beta de­cay


14.19.1 Draft: In­tro­duc­tion

Beta de­cay is the de­cay mech­a­nism that af­fects the largest num­ber of nu­clei. It is im­por­tant in a wide va­ri­ety of ap­pli­ca­tions, such as be­ta­voltaics and PET imag­ing.

In stan­dard beta de­cay, or more specif­i­cally, beta-mi­nus de­cay, a nu­cleus con­verts a neu­tron into a pro­ton. The num­ber of neu­trons $N$ de­creases by one unit, and the num­ber of pro­tons $Z$ in­creases by one. So the neu­tron ex­cess de­creases by two. Beta de­cay moves nu­clei with too many neu­trons closer to the sta­ble range.

Un­like the neu­tron, the pro­ton has a pos­i­tive charge, so by it­self, con­vert­ing a neu­tron into a pro­ton would cre­ate charge out of noth­ing. How­ever, that is not pos­si­ble as net charge is pre­served in na­ture. In beta de­cay, the nu­cleus also emits a neg­a­tively charged elec­tron, mak­ing the net charge that is cre­ated zero as it should.

But there is an­other prob­lem with that. Now a neu­tron with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ is con­verted into a pro­ton and an elec­tron, each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. That vi­o­lates an­gu­lar mo­men­tum con­ser­va­tion. (Re­gard­less of any or­bital an­gu­lar mo­men­tum, the net an­gu­lar mo­men­tum would change from half-in­te­ger to in­te­ger.) In beta de­cay, the nu­cleus also emits a sec­ond par­ti­cle of spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, thus keep­ing the net an­gu­lar mo­men­tum half-in­te­ger. Fermi called that sec­ond par­ti­cle the neu­trino, since it was elec­tri­cally neu­tral and so small that it was ini­tially im­pos­si­ble to ob­serve. In fact, even at the time of writ­ing, al­most a cen­tury later, the mass of the neu­trino, though known to be nonzero, is too small to mea­sure.

Nowa­days the neu­trino emit­ted in beta de­cay is more ac­cu­rately iden­ti­fied as the elec­tron an­ti­neu­trino. An an­ti­neu­trino is the an­tipar­ti­cle of an or­di­nary neu­trino, just like the positron is the an­tipar­ti­cle of the elec­tron. (Par­ti­cles and an­tipar­ti­cles are ex­act op­po­sites in all prop­er­ties ex­cept mass, but in­clud­ing charge, al­low­ing a par­ti­cle and the cor­re­spond­ing an­tipar­ti­cle to an­ni­hi­late each other, leav­ing only pho­tons.)

The rea­son that an an­ti­neu­trino is emit­ted rather than a neu­trino is known as con­ser­va­tion of lep­ton num­ber. Lep­tons are el­e­men­tary par­ti­cles that do not re­spond to the “strong force,” in­clud­ing elec­trons and neu­tri­nos. The net lep­ton num­ber is de­fined as the num­ber of lep­tons, mi­nus the num­ber of an­tilep­tons. It is found that this num­ber is con­served in na­ture. So when in beta de­cay the nu­cleus emits an elec­tron, a lep­ton, and an an­ti­neu­trino, an an­tilep­ton, the lep­ton num­ber stays un­changed as it should (like net an­gu­lar mo­men­tum and net charge stay un­changed, as al­ready noted).

The an­ti­neu­trino does not af­fect the ba­sics of beta de­cay, as it has no charge and vir­tu­ally zero mass. How­ever, the an­ti­neu­trino does af­fect the de­tailed analy­sis; for one, the an­ti­neu­trino can come out with a lot of ki­netic en­ergy, thus re­duc­ing the oth­er­wise ex­pected ki­netic en­ergy of the elec­tron.

In beta de­cay, the new nu­cleus must be lighter than the orig­i­nal one. Clas­si­cal mass con­ser­va­tion would say that the re­duc­tion in nu­clear mass must equal the mass of the emit­ted elec­tron plus the (neg­li­gi­ble) mass of the an­ti­neu­trino. How­ever, Ein­stein’s mass-en­ergy re­la­tion im­plies that that is not quite right. Mass is equiv­a­lent to en­ergy, and the rest mass re­duc­tion of the nu­cleus must also pro­vide the ki­netic en­er­gies of the elec­tron and neu­trino, as well as the (much smaller) one that the nu­cleus it­self picks up dur­ing the de­cay by re­coil.

Still, the bot­tom line is that the nu­clear mass re­duc­tion must be at least the rest mass of the elec­tron (plus an­ti­neu­trino). In en­ergy units, it must be at least 0.511 MeV, the rest mass en­ergy of the elec­tron. The first sub­sec­tion be­low will graph­i­cally ex­am­ine which nu­clei have enough en­ergy to beta de­cay.

Beta-plus de­cay is the op­po­site of beta de­cay. In beta-plus de­cay, the nu­cleus con­verts a neu­tron into a pro­ton in­stead of the other way around. To con­serve charge, the nu­cleus can emit a positron, and with it, an elec­tron neu­trino to con­serve an­gu­lar mo­men­tum and lep­ton num­ber.

How­ever, while con­vert­ing a pro­ton into a neu­tron, the nu­cleus has a much eas­ier way to con­serve charge. In­stead of emit­ting a pos­i­tively charged positron, it can ab­sorb a neg­a­tively charged elec­tron from the atom it is in. The elec­tron's charge then can­cels that of the pro­ton. To pre­serve an­gu­lar mo­men­tum and lep­ton num­ber, an elec­tron neu­trino is again emit­ted. This process is called “elec­tron cap­ture” (or also K-cap­ture or L-Cap­ture de­pend­ing on the elec­tron shell name from which the elec­tron is swiped). Now the nu­clear mass re­duc­tion does not need to pro­vide the 0.512 MeV rest mass en­ergy of a positron. In­stead the nu­clear mass can in­crease up to the 0.512 MeV rest mass en­ergy of the elec­tron that dis­ap­pears.

So elec­tron cap­ture can oc­cur in cir­cum­stances where positron cre­ation is not pos­si­ble. How­ever, if the nu­clear mass re­duc­tion is plenty for both elec­tron cap­ture and positron emis­sion, the lat­ter tends to dom­i­nate. The rea­son is the large quan­tum me­chan­i­cal un­cer­tainty in po­si­tion of the low-en­ergy atomic elec­tron. This un­cer­tainty dwarfs the size of the nu­cleus. It makes it very un­likely for the elec­tron to be found in­side the nu­cleus. A high-en­ergy positron cre­ated by the nu­cleus it­self can be cre­ated in any state, in­clud­ing high en­ergy ones with short wave lengths.

Also note elec­tron cap­ture is of course not pos­si­ble if some­how the nu­cleus has been stripped of all its atomic elec­trons, like might oc­cur in space.

Elec­tron cap­ture is also called in­verse beta de­cay, be­cause an elec­tron be­ing ab­sorbed by a nu­cleus is much like a movie of an elec­tron be­ing emit­ted played back­wards in time. But there are some prob­lems with this idea. For one, the time-re­versed movie would also have an elec­tron an­ti­neu­trino go­ing into the nu­cleus, not an elec­tron neu­trino com­ing out.

Still, ab­sorp­tion of a par­ti­cle is much like the emis­sion of the cor­re­spond­ing an­tipar­ti­cle, at least as far as con­ser­va­tion laws other than en­ergy are con­cerned. For ex­am­ple, cap­ture of an elec­tron adds one unit of neg­a­tive charge, while emis­sion of a positron re­moves one unit of pos­i­tive charge. Ei­ther way, the nu­clear charge be­comes one unit more neg­a­tive. In those terms, the no­tion of “in­verse beta de­cay” may not be that far out, es­pe­cially since the neu­trino is a mi­nor ac­tor in the first place.


14.19.2 Draft: En­er­get­ics Data

Fig­ure 14.47: En­ergy re­lease in beta de­cay of even-odd nu­clei. [pdf]
\begin{figure}\centering
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\begin{picture}(405,56...
...\PB391.9,128.3,'147\,\,\,\,\ \ \ '
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Fig­ure 14.48: En­ergy re­lease in beta de­cay of odd-even nu­clei. [pdf]
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...\PB391.9,138.2,'146\,\,\,\,\ \ \ '
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Fig­ure 14.49: En­ergy re­lease in beta de­cay of odd-odd nu­clei. [pdf]
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...\PB391.9,130.3,'147\,\,\,\,\ \ \ '
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Fig­ure 14.50: En­ergy re­lease in beta de­cay of even-even nu­clei. [pdf]
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...\PB391.9,141.8,'148\,\,\,\,\ \ \ '
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As the in­tro­duc­tion ex­plained, in beta de­cay a nu­cleus con­verts a neu­tron into a pro­ton, thus chang­ing into a dif­fer­ent nu­cleus. It can only oc­cur if the nu­clear mass re­duc­tion ex­ceeds the 0.511 MeV rest mass en­ergy of the elec­tron emit­ted in the process.

Fig­ures 14.47 through 14.50 show the nu­clear mass re­duc­tion for beta de­cay as the ver­ti­cal co­or­di­nate. The re­duc­tion ex­ceeds the rest mass en­ergy of the elec­tron only above the hor­i­zon­tal cen­ter bands. The left half of each square in­di­cates the nu­cleus be­fore beta de­cay, the right half the one af­ter the de­cay. The hor­i­zon­tal co­or­di­nate in­di­cates the atomic num­bers, with the val­ues and el­e­ment sym­bols as in­di­cated. Neu­tron num­bers are listed at the square it­self. Lines con­nect pairs of nu­clei with equal neu­tron ex­cess.

If the left-half square is col­ored blue, beta de­cay is ob­served. Blue left-half squares are only found above the cen­ter bands, so the mass re­duc­tion is in­deed at least the mass of the elec­tron. How­ever, some blue left-half squares are right on top of the band. Their beta de­cay should be very slow.

Note that some left-half squares above the band may not be blue. The color in­di­cates the dom­i­nant de­cay process, so if the left-hand nu­cleus also ex­pe­ri­ences an­other de­cay mode in ad­di­tion to beta de­cay, like al­pha de­cay or beta-plus de­cay with an­other nu­cleus, and at a higher rate, its half square will not be blue. How­ever, there should be no left-half squares above the band that are sta­ble green.

Note here that even-even nu­clei ${}\fourIdx{48}{20}{}{}{\rm Ca}$ and ${}\fourIdx{96}{40}{}{}{\rm Zr}$ are not sta­ble, How­ever, their beta de­cay is so ex­tremely slow that dou­ble-beta de­cay dom­i­nates. Nor­mal beta-de­cay has never been ob­served for them. This is not just be­cause the en­ergy re­lease is small, but more im­por­tantly be­cause these tran­si­tions are strongly for­bid­den in the sense dis­cussed in sec­tion 14.19.6.

In beta-plus de­cay, the nu­cleus con­verts a pro­ton into a neu­tron in­stead of the other way around. To find the en­ergy re­lease in that process, the fig­ures may be read the other way around. The nu­cleus be­fore the de­cay is now the right hand one, and the de­cay is ob­served when the right-half square is red.

The en­ergy re­lease is now pos­i­tive down­ward, and it is now be­low the cen­ter bands that the nu­clear mass re­duc­tion is suf­fi­cient to pro­duce the rest mass of a positron that can carry the pro­ton’ pos­i­tive charge away. The positron, the anti-par­ti­cle of the elec­tron, has the same mass as the elec­tron but op­po­site charge.

But note that red right-half squares ex­tend to within the cen­ter bands. The rea­son is that in­stead of emit­ting a positron, the nu­cleus can cap­ture an elec­tron from the atomic elec­tron cloud sur­round­ing the nu­cleus. In that case, rather than hav­ing to come up with an elec­tron mass worth of en­ergy, the nu­cleus re­ceives an in­fu­sion of that amount of en­ergy. So the re­quired en­ergy goes down by two elec­tron masses.

It fol­lows that the left-hand nu­cleus will suf­fer beta de­cay if the square is above the top of the band, while the right-hand nu­cleus will suf­fer elec­tron cap­ture if the square is be­low the top of the band. There­fore at most one nu­cleus of each pair can be sta­ble.

Note, once more, that color in­di­cates the dom­i­nant de­cay mode. So right-half squares be­low the top of the band do not have to be red; they just should not be sta­ble green. This is es­pe­cially rel­e­vant for the odd-odd nu­clei in fig­ures 14.49 and 14.50. Odd-odd nu­clei are un­usu­ally un­sta­ble, and the even-even nu­clei they de­cay into are un­usu­ally sta­ble. So it is quite likely that an odd-odd nu­clei in the re­gion of rel­a­tively sta­ble nu­clei finds that it has enough en­ergy to both beta de­cay to the neigh­bor­ing even-even nu­cleus of higher $Z$ and beta-plus de­cay / elec­tron cap­ture to the neigh­bor­ing even-even nu­cleus of lower $Z$. Then, if one of the two processes is rel­a­tively slow, be­cause the process is just above the top of the band in fig­ure 14.49, or just be­low the top of the band in fig­ure 14.50, then the other process is likely to dom­i­nate. So the half square does not have the ex­pected color. Left-hand squares just above the top of the band in fig­ure 14.49 will be red, and right-hand squares just be­low the top of the band in fig­ure 14.49 will be blue.

One ex­am­ple is ${}\fourIdx{40}{19}{}{}{\rm {K}}$ potas­sium-40, with 21 neu­trons. It ap­pears above the band in fig­ure 14.49, in­di­cat­ing that it suf­fers beta de­cay. But it also ap­pears be­low the band in fig­ure 14.50, so that it also suf­fers elec­tron cap­ture and positron emis­sion. In this case, beta de­cay dom­i­nates beta-plus de­cay and elec­tron cap­ture 9 to 1. Re­call that elec­tron cap­ture is rel­a­tively slow and the nu­cleus is just be­low the bot­tom of the band in fig­ure 14.50, so beta-plus de­cay will be too.


14.19.3 Draft: Beta de­cay and magic num­bers

The magic neu­tron num­bers are quite vis­i­ble in fig­ures 14.47 through 14.50. For ex­am­ple, di­ag­o­nal bands at neu­tron num­bers 50, 82, and 126 are promi­nent in all four fig­ures. Con­sider for ex­am­ple fig­ure 14.48. For the 50/49 neu­tron nu­clei, beta de­cay takes the tightly bound 50th neu­tron to turn into a pro­ton. That re­quires rel­a­tively large en­ergy, so the en­ergy re­lease is re­duced. For the neigh­bor­ing 52/51 nu­clei, beta de­cay takes the much less tightly bound 52nd neu­tron, and the en­ergy re­lease is cor­re­spond­ingly higher.

The magic pro­ton num­bers tend to show up as step-downs in the curves. For ex­am­ple, con­sider the nu­clei at the ver­ti­cal $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 50 line also in fig­ure 14.48. In the In/Sn (in­dium/tin) beta de­cay, the beta de­cay neu­tron be­comes the tightly bound 50th pro­ton, and the en­ergy re­lease is cor­re­spond­ingly high. In the Sb/Te (an­ti­mony/tel­lurium) de­cay, the neu­tron be­comes the less tightly bound 52nd pro­ton, and the en­ergy re­lease is lower.

When the neu­tron and pro­ton magic num­ber lines in­ter­sect, com­bined ef­fects can be seen. One pointed out by Mayer in her No­ble prize ac­cep­tance lec­ture [[10]] is the de­cay of ar­gon-39. It has 18 pro­tons and 21 neu­trons. If you in­ter­po­late be­tween the neigh­bor­ing pairs of nu­clei on the same neu­tron ex­cess line in fig­ure 14.47, you would ex­pect ar­gon-39 to be be­low the top of the cen­ter band, hence to be sta­ble against beta de­cay. But the ac­tual en­ergy re­lease for ar­gon-39 is un­usu­ally high, and beta de­cay it does. Why is it un­usu­ally high? For the pre­vi­ous pairs of nu­clei, beta de­cay con­verts a neu­tron in the neu­tron shell that ends at magic num­ber 20 into a pro­ton in the cor­re­spond­ing pro­ton shell. For the su­vse­quent pairs, beta de­cay con­verts a neu­tron in the neu­tron shell that ends at magic num­ber 28 to a pro­ton in the cor­re­spond­ing pro­ton shell. Only for ar­gon-39, beta de­cay con­verts a neu­tron in the neu­tron shell that end at magic num­ber 28 into a pro­ton in the lower en­ergy pro­ton shell that ends at magic num­ber 20. The low­er­ing of the ma­jor shell re­leases ad­di­tional en­ergy, and the de­cay has enough en­ergy to pro­ceed.

In fig­ures 14.47 and 14.48, the low­est line for the light­est nu­clei is un­usu­ally smooth. These lines cor­re­spond to a neu­tron ex­cess of 1 or $\vphantom{0}\raisebox{1.5pt}{$-$}$1, de­pend­ing on whether it is be­fore or af­ter the de­cay. The pairs of nu­clei on these two lines are mir­ror nu­clei. Dur­ing beta de­cay the neu­tron that turns into a pro­ton trans­fers from the neu­tron shells into the ex­act same po­si­tion in the pro­ton shells. Be­cause of charge in­de­pen­dence, the nu­clear en­ergy does not change. The Coulomb en­ergy does change, but as a rel­a­tively small, long-range ef­fect, it changes fairly grad­u­ally.

These lines also show that beta-plus de­cay and elec­tron cap­ture be­come en­er­get­i­cally fa­vored when the nu­clei get heav­ier. That is to be ex­pected since this are nu­clei with al­most no neu­tron ex­cess. For the heav­ier ones, it is there­fore en­er­get­i­cally fa­vor­able to con­vert pro­tons into neu­trons, rather than the other way around.


14.19.4 Draft: Von Weizsäcker ap­prox­i­ma­tion

Since the von Weizsäcker for­mula of sec­tion 14.10.2 pre­dicts nu­clear mass, it can be used to pre­dict whether beta-mi­nus or beta-plus/elec­tron cap­ture will oc­cur.

The math­e­mat­ics is rel­a­tively sim­ple, be­cause the mass num­ber $A$ re­mains con­stant dur­ing beta de­cay. For a given mass num­ber, the von Weizsäcker for­mula is just a qua­dratic in $Z$. Like in the pre­vi­ous sub­sec­tion, con­sider again pairs of nu­clei with the same $A$ and one unit dif­fer­ence in $Z$. Set the mass dif­fer­ence equal to the elec­tron mass and solve the re­sult­ing equa­tion for $Z$ us­ing sim­ple al­ge­bra.

It is then seen that beta-mi­nus de­cay, re­spec­tively beta-plus de­cay / elec­tron cap­ture oc­curs for a pair of nu­clei de­pend­ing whether the av­er­age $Z$ value is less, re­spec­tively greater, than

\begin{displaymath}
\fbox{$\displaystyle
Z_{\rm bd} =
A \frac{4C_a +m_{\rm n}-m_{\rm p}-m_{\rm e}+C_cC_zA^{-1/3}}{8C_a + 2C_c A^{2/3}}
$} %
\end{displaymath} (14.50)

where the con­stants $C_{.}$ are as given in sec­tion 14.10.2. The nu­cleon pair­ing en­ergy must be ig­nored in the de­riva­tion, so the re­sult may be off by a pair of nu­clei for even-even and odd-odd nu­clei.

Fig­ure 14.51: Ex­am­ples of beta de­cay. [pdf]
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\begin{picture}(405,56...
...makebox(0,0)[l]{$\fourIdx{40}{18}{}{}{\rm Ar}$}}
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The re­sult is plot­ted as the black curve in the de­cay graph fig­ure 14.51. It gives the lo­ca­tion where the change in nu­clear mass is just enough for ei­ther beta-mi­nus de­cay or elec­tron cap­ture to oc­cur, with noth­ing to spare. The curve lo­cates the sta­ble nu­clei fairly well. For light nu­clei, the curve is about ver­ti­cal, in­di­cat­ing there are equal num­bers of pro­tons and neu­trons in sta­ble nu­clei. For heav­ier nu­clei, there are more neu­trons than pro­tons, caus­ing the curve to de­vi­ate to the right, the di­rec­tion of in­creas­ing neu­tron ex­cess.

Be­cause of the pair­ing en­ergy, sta­ble even-even nu­clei can be found well away from the curve. Con­versely, sta­ble odd-odd nu­clei are hard to find at all. In fact, there are only four: hy­dro­gen-2 (deu­terium), lithium-6, boron-10, and ni­tro­gen-14. For com­par­i­son, there are 150 sta­ble even-even ones. For nu­clei of odd mass num­ber, it does not make much dif­fer­ence whether the num­ber of pro­tons is odd or the num­ber of neu­trons: there are 49 sta­ble odd-even nu­clei and 53 sta­ble even-odd ones.

(There is also the bizarre ex­cited ${}\fourIdx{180\rm {m}}{73}{}{}{\rm {Ta}}$ nu­cleus that is sta­ble, and is odd-odd to boot. But that is an ex­cited state and an­other story, which is dis­cussed un­der gamma de­cay. The ground state ${}\fourIdx{180}{73}{}{}{\rm {Ta}}$ has a half life of only 8 hours, as a rel­a­tively heavy odd-odd nu­cleus should.)

As an ex­am­ple of the in­sta­bil­ity of odd-odd nu­clei, con­sider the cu­ri­ous case of potas­sium-40, ${}\fourIdx{40}{19}{}{}{K}$. It has both an odd num­ber of pro­tons, 19, and of neu­trons, 21. Potas­sium-40 is pretty much on top of the sta­ble line, as ev­i­dent from the fact that both its neigh­bors, odd-even iso­topes potas­sium-39 and potas­sium-41, are sta­ble. But potas­sium-40 it­self is un­sta­ble. It does have a life­time com­pa­ra­ble to the age of the uni­verse; long enough for sig­nif­i­cant quan­ti­ties to ac­cu­mu­late. About 0.01% of nat­ural potas­sium is potas­sium-40.

But de­cay it does. De­spite the two bil­lion year av­er­age life­time, there are so many potas­sium-40 nu­clei in a hu­man body that al­most 5 000 de­cay per sec­ond any­way. About 90% do so through beta de­cay and end up as the dou­bly-magic cal­cium-40. The other 10% de­cay by elec­tron cap­ture or positron emis­sion and end up as even-even ar­gon-40, with 18 pro­tons and 22 neu­trons. So potas­sium-40 suf­fers all three beta de­cay modes, the only rel­a­tively com­mon nu­cleus in na­ture that does.

Ad­mit­tedly only 0.001% de­cays through positron emis­sion. The nu­clear mass dif­fer­ence of 0.99 MeV with ar­gon-40 is enough to cre­ate a positron, but not by much. Be­fore a positron can be cre­ated, potas­sium is al­most sure to have cap­tured an elec­tron al­ready. For a nu­cleus like xenon-119 the mass dif­fer­ence with io­dine-119 is sub­stan­tially larger, 4.5 MeV, and about 4 in 5 xenon-119 nu­clei de­cay by positron emis­sion, and the fifth by elec­tron cap­ture.

It is en­er­get­i­cally pos­si­ble for the potas­sium-40 de­cay prod­uct cal­cium-40 to de­cay fur­ther into ar­gon-40, by cap­tur­ing two elec­trons from the atom. En­er­get­i­cally pos­si­ble means that this does not re­quire ad­di­tion of en­ergy, it lib­er­ates en­ergy, so it can oc­cur spon­ta­neously. Note that cal­cium-40 would have to cap­ture two elec­trons at the same time; cap­tur­ing just one elec­tron would turn it into potas­sium-40, and that re­quires ex­ter­nal en­ergy ad­di­tion. In other words, cal­cium-40 would have to skip over the in­ter­me­di­ate odd-odd potas­sium 40. While it is pos­si­ble, it is be­lieved that cal­cium-40 is sta­ble; if it de­cays at all, its half-life must be more than 5.9 zettayear (5.9 10$\POW9,{21}$ year).

But some even-even nu­clei do de­cay through “dou­ble beta-mi­nus” de­cay. For ex­am­ple, ger­ma­nium-76 with 32 pro­tons and 44 neu­trons will in a cou­ple of zettayear emit two elec­trons and so turn into even-even se­le­nium-76, skip­ping over odd-odd ar­senic-76 in the process. How­ever, since the en­tire life­time of the uni­verse is much less than the blink of an eye com­pared to a zettayear, this does not get rid of much ger­ma­nium-76. About 7.5% of nat­ural ger­ma­nium is ger­ma­nium-76.

The re­duced sta­bil­ity of odd-odd nu­clei is the main rea­son that tech­netium (Tc) and prome­thium (Pm) can end up with no sta­ble iso­topes at all while their im­me­di­ate neigh­bors have many. Both tech­netium and prome­thium have an odd-odd iso­tope sit­ting right on top of the sep­a­rat­ing line be­tween beta-mi­nus and beta-plus de­cay; tech­netium-98 re­spec­tively prome­thium-146. Be­cause of the ap­prox­i­ma­tion er­rors in the von Weizsäcker for­mula, they are not quite on the the­o­ret­i­cal curve in fig­ure 14.51. How­ever, ex­am­i­na­tion of the ex­per­i­men­tal nu­clear masses shows the ex­cess mass re­duc­tion for beta-mi­nus de­cay and elec­tron cap­ture to be vir­tu­ally iden­ti­cal for these odd-odd nu­clei. And in fact prome­thium-146 does in­deed de­cay both ways. Tech­netium-98 could too, but does not; it finds it quicker to cre­ate an elec­tron than to cap­ture one from the sur­round­ing atom.

Be­cause the the­o­ret­i­cal sta­ble line slopes to­wards the right in fig­ure 14.51, only one of the two odd-even iso­topes next to tech­netium-98 should be un­sta­ble, and the same for the ones next to prome­thium-146. How­ever, the en­ergy lib­er­ated in the de­cay of these odd-even nu­clei is only a few hun­dred keV in each case, far be­low the level for which the von Weizsäcker for­mula is any­where mean­ing­ful. For tech­netium and prome­thium, nei­ther neigh­bor­ing iso­tope is sta­ble. This is a qual­i­ta­tive fail­ure of the von Weizsäcker model. But it is rare; it hap­pens only for these two out of the low­est 82 el­e­ments. Few books even men­tion it is a fun­da­men­tal fail­ure of the for­mula.


14.19.5 Draft: Ki­netic En­er­gies

The ki­netic en­ergy of nu­clear de­cay prod­ucts is im­por­tant to un­der­stand the cor­rect na­ture of the de­cay.

His­tor­i­cally, one puz­zling ob­ser­va­tion in beta de­cay was the ki­netic en­er­gies with which the elec­trons came out. When the beta de­cay of a col­lec­tion of nu­clei of a given type is ob­served, the elec­trons come out with a range of ki­netic en­er­gies. In con­trast, in the al­pha de­cay of a col­lec­tion of nu­clei of a given type, all al­pha par­ti­cles come out with pretty much the ex­act same ki­netic en­ergy.

Con­sider the rea­son. The to­tal ki­netic en­ergy re­lease in the de­cay of a given nu­cleus is called the $Q$ value. Fol­low­ing Ein­stein’s fa­mous re­la­tion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, the $Q$ value in al­pha de­cay is given by the re­duc­tion in the net rest mass en­ergy dur­ing the de­cay:

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{\rm N1} c^2 - m_{\rm N2} c^2 - m_\alpha c^2
$} %
\end{displaymath} (14.51)

where 1 in­di­cates the nu­cleus be­fore the de­cay and 2 the nu­cleus af­ter the de­cay.

Since en­ergy must be con­served, the re­duc­tion in rest mass en­ergy given by the $Q$-​value is con­verted into ki­netic en­ergy of the de­cay prod­ucts. Clas­si­cal analy­sis makes that:

\begin{displaymath}
Q = {\textstyle\frac{1}{2}} m_{\rm N2} v^2_{\rm N2} + {\textstyle\frac{1}{2}} m_\alpha v^2_\alpha
\end{displaymath}

This as­sumes that the ini­tial nu­cleus is at rest, or more gen­er­ally that the de­cay is ob­served in a co­or­di­nate sys­tem mov­ing with the ini­tial nu­cleus. Lin­ear mo­men­tum must also be con­served:

\begin{displaymath}
m_{\rm N1} \vec v_{\rm N1}
= m_{\rm N2} \vec v_{\rm N2} + m_\alpha \vec v_\alpha
\end{displaymath}

but since the ve­loc­ity of the ini­tial nu­cleus is zero,

\begin{displaymath}
m_{\rm N2} \vec v_{\rm N2} = - m_\alpha \vec v_\alpha
\end{displaymath}

Square both sides and di­vide by $2m_{\rm {N2}}$ to get:

\begin{displaymath}
{\textstyle\frac{1}{2}} m_{\rm N2}v^2_{\rm N2}
= \frac {m_\alpha}{m_{\rm N2}}{\textstyle\frac{1}{2}} m_\alpha v^2_\alpha
\end{displaymath}

Now, ex­clud­ing the spe­cial case of beryl­lium-8, the mass of the al­pha par­ti­cle is much smaller than that of the fi­nal nu­cleus. So the ex­pres­sion above shows that the ki­netic en­ergy of the fi­nal nu­cleus is much less than that of the al­pha par­ti­cle. The al­pha par­ti­cle runs away with al­most all the ki­netic en­ergy. Its ki­netic en­ergy is al­most equal to $Q$. There­fore it is al­ways the same for a given ini­tial nu­cleus, as claimed above. In the spe­cial case that the ini­tial nu­cleus is beryl­lium-8, the fi­nal nu­cleus is also an al­pha par­ti­cle, and each al­pha par­ti­cle runs away with half the ki­netic en­ergy. But still, each al­pha par­ti­cle al­ways comes out with a sin­gle value for its ki­netic en­ergy, in this case ${\textstyle\frac{1}{2}}Q$.

In beta de­cay, things would be pretty much the same if just an elec­tron was emit­ted. The elec­tron too would come out with a sin­gle ki­netic en­ergy. The fact that it did not led Pauli to pro­pose that an­other small par­ti­cle also comes out. That par­ti­cle could carry away the rest of the ki­netic en­ergy. It had to be elec­tri­cally neu­tral like a neu­tron, be­cause the nu­clear charge change is al­ready ac­counted for by the charge taken away by the elec­tron. The small neu­tral par­ti­cle was called the neu­trino by Fermi. The neu­trino was also re­quired for an­gu­lar mo­men­tum con­ser­va­tion: a pro­ton and an elec­tron each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ have net spin 0 or 1, not $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ like the orig­i­nal neu­tron.

The neu­trino that comes out in beta-mi­nus de­cay is more ac­cu­rately called an elec­tron an­ti­neu­trino and usu­ally in­di­cated by $\bar\nu$. The bar in­di­cates that it is counted as an an­tipar­ti­cle.

The analy­sis of the ki­netic en­ergy of the de­cay prod­ucts changes be­cause of the pres­ence of an ad­di­tional par­ti­cle. The $Q$-​value for beta de­cay is

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{{\rm{N}}1}c^2 - m_{{\rm{N}}2} c^2 - m_{\rm e}c^2 - m_{\bar\nu} c^2
$} %
\end{displaymath} (14.52)

How­ever, the rest mass en­ergy of the neu­trino can safely be ig­nored. At the time of writ­ing, num­bers less than a sin­gle eV are bandied around. That is im­mea­sur­ably small com­pared to the nu­clear rest mass en­er­gies which are in terms of GeV. In fact, physi­cists would love the neu­trino mass to be non­neg­li­gi­ble: then they could fig­ure out what is was!

As an aside, it should be noted that the nu­clear masses in the $Q$ val­ues are nu­clear masses. Tab­u­lated val­ues are in­vari­ably atomic masses. They are dif­fer­ent by the mass of the elec­trons and their bind­ing en­ergy. Other books there­fore typ­i­cally con­vert the $Q$-​val­ues to atomic masses, usu­ally by ig­nor­ing the elec­tronic bind­ing en­ergy. But us­ing atomic masses in a de­scrip­tion of nu­clei, not atoms, is con­fus­ing. It is also a likely cause of mis­takes. (For ex­am­ple, [30, fig. 11.5] seems to have mis­tak­enly used atomic masses to re­late iso­baric nu­clear en­er­gies.)

It should also be noted that if the ini­tial and/or fi­nal nu­cleus is in an ex­cited state, its mass can be com­puted from that of the ground state nu­cleus by adding the ex­ci­ta­tion en­ergy, con­verted to mass units us­ing $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$. Ac­tu­ally, nu­clear masses are usu­ally given in en­ergy units rather than mass units, so no con­ver­sion is needed.

Be­cause the amount of ki­netic en­ergy that the neu­trino takes away varies, so does the ki­netic en­ergy of the elec­tron. One ex­treme case is that the neu­trino comes out at rest. In that case, the given analy­sis for al­pha de­cay ap­plies pretty much the same way for beta de­cay if the al­pha is re­placed by the elec­tron. This gives the max­i­mum ki­netic en­ergy at which the elec­tron can come out to be $Q$. (Un­like for the al­pha par­ti­cle, the mass of the elec­tron is al­ways small com­pared to the nu­cleus, and the nu­cleus al­ways ends up with es­sen­tially none of the ki­netic en­ergy.) The other ex­treme is that the elec­tron comes out at rest. In that case, it is the neu­trino that pretty much takes all the ki­netic en­ergy. Nor­mally, both elec­tron and neu­trino each take their fair share of ki­netic en­ergy. So usu­ally the ki­netic en­ergy of the elec­tron is some­where in be­tween zero and $Q$.

A fur­ther mod­i­fi­ca­tion to the analy­sis for the al­pha par­ti­cle must be made. Be­cause of the rel­a­tively small masses of the elec­tron and neu­trino, they come out mov­ing at speeds close to the speed of light. There­fore the rel­a­tivis­tic ex­pres­sions for mo­men­tum and ki­netic en­ergy must be used, chap­ter 1.1.2.

Con­sider first the ex­treme case that the elec­tron comes out at rest. The rel­a­tivis­tic en­ergy ex­pres­sion gives for the ki­netic en­ergy of the neu­trino:

\begin{displaymath}
T_{\bar\nu} = \sqrt{(m_{\bar\nu} c^2)^2 + (pc)^2} - m_{\bar\nu} c^2
\end{displaymath} (14.53)

where $c$ is the speed of light and $p$ the mo­men­tum. The nu­cleus takes only a very small frac­tion of the ki­netic en­ergy, so $T_{\bar\nu}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $Q$. Also, what­ever the neu­trino rest mass en­ergy $m_{\bar\nu}c^2$ may be ex­actly, it is cer­tainly neg­li­gi­bly small. It fol­lows that ${T}_{\bar\nu}\approx{Q}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $pc$.

The small frac­tion of the ki­netic en­ergy that does end up with the nu­cleus may now be es­ti­mated, be­cause the nu­cleus has the same mag­ni­tude of mo­men­tum $p$. For the nu­cleus, the non­rel­a­tivis­tic ex­pres­sion may be used:

\begin{displaymath}
T_{\rm N2} = \frac{p^2}{2m_{\rm N2}} = pc \frac{pc}{2m_{\rm N2}c^2}
\end{displaymath} (14.54)

The fi­nal frac­tion is very small be­cause the en­ergy re­lease $pc$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $Q$ is in MeV while the nu­clear mass is in GeV. There­fore the ki­netic en­ergy of the nu­cleus is in­deed very small com­pared to that of the neu­trino. If higher ac­cu­racy is de­sired, the en­tire com­pu­ta­tion may now be re­peated, start­ing from the more ac­cu­rate value $T_{\bar\nu}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q-T_{\rm {N2}}$ for the ki­netic en­ergy of the neu­trino.

The ex­treme case that the neu­trino is at rest can be com­puted in much the same way, ex­cept that the rest mass en­ergy of the elec­tron is com­pa­ra­ble to $Q$ and must be in­cluded in the com­pu­ta­tion of $pc$. If it­er­a­tion is not de­sired, an ex­act ex­pres­sion for $pc$ can be de­rived us­ing a bit of al­ge­bra:

\begin{displaymath}
pc = \sqrt{
\frac{[E^2 - (E_{\rm N2}+E_{\rm e})^2]
[E^2 -...
...}-E_{\rm e})^2]}{4E^2}}
\qquad E = E_{\rm N2} + E_{\rm e} + Q
\end{displaymath} (14.55)

where $E_{\rm {N2}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {N2}}c^2$ and $E_{\rm {e}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${m_{\rm e}}c^2$ are the rest mass en­er­gies of the fi­nal nu­cleus and elec­tron. The same for­mula may be used in the ex­treme case that the elec­tron is at rest and the neu­trino is not, by re­plac­ing $E_{\rm {e}}$ by the neu­trino rest mass, which is to all prac­ti­cal pur­poses zero.

In the case of beta-plus de­cay, the elec­tron be­comes a positron and the elec­tron an­ti­neu­trino be­comes an elec­tron neu­trino. How­ever, an­tipar­ti­cles have the same mass as the nor­mal par­ti­cle, so there is no change in the en­er­get­ics. (There is a dif­fer­ence if it is writ­ten in terms of atomic in­stead of nu­clear masses.) In case of elec­tron cap­ture, it must be taken into ac­count that the nu­cleus re­ceives an in­fu­sion of mass equal to that of the cap­tured elec­tron. The $Q$-​value be­comes

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{{\rm{N}}1}c^2 + m_{\rm e}c^2 - m_{{\rm{N}}2} c^2 - m_{\bar\nu} c^2
- E_{\rm B,ce}
$} %
\end{displaymath} (14.56)

where $E_{\rm {B,ce}}$ is the elec­tronic bind­ing en­ergy of the cap­tured elec­tron. Be­cause this is an in­ner elec­tron, nor­mally a K or L shell one, it has quite a lot of bind­ing en­ergy, too large to be ig­nored. Af­ter the elec­tron cap­ture, an elec­tron far­ther out will drop into the cre­ated hole, pro­duc­ing an X-ray. If that elec­tron leaves a hole be­hind too, there will be more X-rays. The en­ergy in these $X$-​rays sub­tracts from that avail­able to the neu­trino.

The bind­ing en­ergy may be ball­parked from the hy­dro­gen ground state en­ergy, chap­ter 4.3, by sim­ply re­plac­ing $e^2$ in it by $e^2Z$. That gives:

\begin{displaymath}
\fbox{$\displaystyle
E_{\rm B,ce} \sim 13.6\, Z^2\mbox{ eV}
$} %
\end{displaymath} (14.57)

The ball­parks for elec­tron cap­ture in fig­ure 14.54 use
\begin{displaymath}
E_{\rm B,ce} \sim {\textstyle\frac{1}{2}} (\alpha Z)^2 m_{\rm e}c^2
\left(1+ {\textstyle\frac{1}{4}}(\alpha Z)^2\right)
\end{displaymath} (14.58)

in an at­tempt to par­tially cor­rect for rel­a­tivis­tic ef­fects, which are sig­nif­i­cant for heav­ier nu­clei. Here $\alpha$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1/137 is the so-called fine struc­ture con­stant. The sec­ond term in the paren­the­ses is the rel­a­tivis­tic cor­rec­tion. With­out that term, the re­sult is the same as (14.57). See ad­den­dum {A.39} for a jus­ti­fi­ca­tion.


14.19.6 Draft: For­bid­den de­cays

En­er­get­ics is not all there is to beta de­cay. Some de­cays are en­er­get­i­cally fine but oc­cur ex­tremely slowly or not at all. Con­sider cal­cium-48 in fig­ure fig:bet­dec2e. The square is well above the cen­ter band, so en­ergy-wise there is no prob­lem at all for the de­cay to scan­dium-48. But it just does not hap­pen. The half life of cal­cium-48 is 64 10$\POW9,{18}$ years, more than three bil­lion times the en­tire life­time of the uni­verse. And when de­cay does hap­pen, it is due to dou­ble beta de­cay; as of 2016, nor­mal beta de­cay of cal­cium-48 has never been ob­served.

The big prob­lem is an­gu­lar mo­men­tum con­ser­va­tion. As an even-even nu­cleus, cal­cium-48 has zero spin, while scan­dium-48 has spin 6 in its ground state. To con­serve an­gu­lar mo­men­tum dur­ing the de­cay, the elec­tron and the an­ti­neu­trino must there­fore take six units of spin along. But to the ex­tend that the nu­clear size can be ig­nored, the elec­tron and an­ti­neu­trino come out of a math­e­mat­i­cal point. That means that they come out with zero or­bital an­gu­lar mo­men­tum. They have half a unit of spin each, and there is no way to pro­duce six units of net spin from that. The de­cay is for­bid­den by an­gu­lar mo­men­tum con­ser­va­tion.

Of course, cal­cium-48 could de­cay to an ex­cited state of scan­dium-48. Un­for­tu­nately, only the low­est two ex­cited states are en­er­get­i­cally pos­si­ble, and these have spins 5 and 4. They too are for­bid­den.


14.19.6.1 Draft: Al­lowed de­cays

To un­der­stand what beta de­cays are for­bid­den, the first step is to ex­am­ine what de­cays are al­lowed.

Con­sider the spins of the elec­tron and an­ti­neu­trino. They could com­bine into a net spin of zero. If they do, it is called a “Fermi de­cay.” Since the elec­tron and an­ti­neu­trino take no spin away, in Fermi de­cays the nu­clear spin can­not change.

The only other pos­si­bil­ity al­lowed by quan­tum me­chan­ics is that the spins of elec­tron and an­ti­neu­trino com­bine into a net spin of one; that is called a “Gamow-Teller de­cay.” The rules of quan­tum me­chan­ics for the ad­di­tion of an­gu­lar mo­men­tum vec­tors im­ply:

\begin{displaymath}
\fbox{$\displaystyle
\vert j_{\rm N1} - j_{{\rm e}\bar\nu}...
...ebox{-.7pt}{$\leqslant$}}j_{\rm N1} + j_{{\rm e}\bar\nu}
$} %
\end{displaymath} (14.59)

where $j_{\rm {N1}}$ in­di­cates the spin of the nu­cleus be­fore the de­cay, $j_{\rm {N2}}$ the one af­ter it, and $j_{{\rm {e}}\bar\nu}$ is the com­bined an­gu­lar mo­men­tum of elec­tron and an­ti­neu­trino. Since $j_{{\rm {e}}\bar\nu}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for al­lowed Gamow-Teller de­cays, spin can change one unit or stay the same. There is one ex­cep­tion; if the ini­tial nu­clear spin is zero, the fi­nal spin can­not be zero but must be one. Tran­si­tions from spin zero to zero are only al­lowed if they are Fermi ones. But they are al­lowed.

Putting it to­gether, the an­gu­lar mo­men­tum can change by up to one unit in an al­lowed beta de­cay. Also, if there is no or­bital an­gu­lar mo­men­tum, the par­i­ties of the elec­tron and an­ti­neu­trino are even, so the nu­clear par­ity can­not change. In short

\begin{displaymath}
\fbox{$\displaystyle
\mbox{allowed:}\qquad
\vert\Delta j_...
...7pt}{$\leqslant$}}1 \qquad \Delta \pi_{\rm N}= \mbox{no}
$} %
\end{displaymath} (14.60)

where $\Delta$ in­di­cates the nu­clear change dur­ing the de­cay, $j_{\rm N}$ the spin of the nu­cleus, and $\pi_{\rm N}$ its par­ity.

One sim­ple ex­am­ple of an al­lowed de­cay is that of a sin­gle neu­tron into a pro­ton. Since this is a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ de­cay, both Fermi and Gamow-Teller de­cays oc­cur. The neu­tron has a half-life of about ten min­utes. It can be es­ti­mated that the de­cay is 18% Fermi and 82% Gamow-Teller, [30, p. 290].

Some dis­claimers are in or­der. Both the dis­cus­sion above and the fol­low­ing one for for­bid­den de­cays are non­rel­a­tivis­tic. But neu­tri­nos are very light par­ti­cles that travel at very close to the speed of light. For such rel­a­tivis­tic par­ti­cles, or­bital an­gu­lar mo­men­tum and spin get mixed-up. That is much like they get mixed-up for the pho­ton. That was such a headache in de­scrib­ing elec­tro­mag­netic tran­si­tions in chap­ter 7.4.3. For­tu­nately, neu­tri­nos turn out to have some mass. So the given ar­gu­ments ap­ply at least un­der some con­di­tions, even if such con­di­tions are never ob­served.

A much big­ger prob­lem is that neu­tri­nos and an­ti­neu­tri­nos do not con­serve par­ity. That is dis­cussed in more de­tail a later sub­sec­tion, Above, this book sim­ply told you a bla­tant lie when it said that the elec­tron-an­ti­neu­trino sys­tem, (or the positron-neu­trino sys­tem in beta-plus de­cay), comes off with zero par­ity. A sys­tem in­volv­ing a sin­gle neu­trino or an­ti­neu­trino does not have def­i­nite par­ity. And par­ity is not con­served in the de­cay process any­way. But the ini­tial and fi­nal nu­clear states do have def­i­nite par­ity (to within very high ac­cu­racy). For­tu­nately, it turns out that you get the right an­swers for the change in nu­clear par­ity if you sim­ply as­sume that the elec­tron and an­ti­neu­trino come off with the par­ity given by their or­bital an­gu­lar mo­men­tum.

No you can­not have your money back. You did not pay any.

A rel­a­tivis­tic de­scrip­tion of neu­tri­nos can be found in {A.44}.


14.19.6.2 Draft: For­bid­den de­cays al­lowed

As noted at the start of this sub­sec­tion, beta de­cay of cal­cium-48 re­quires a spin change of at least 4 and that is solidly for­bid­den. But for­bid­den is not quite the same as im­pos­si­ble. There is a small loop­hole. A nu­cleus is not re­ally a math­e­mat­i­cal point, it has a nonzero size.

Clas­si­cally that would not make a dif­fer­ence, be­cause the or­bital an­gu­lar mo­men­tum would be much too small to make up the deficit in spin. A rough ball­park of the an­gu­lar mo­men­tum of, say, the elec­tron would be $pR$, with $p$ its lin­ear mo­men­tum and $R$ the nu­clear ra­dius. Com­pare this with the quan­tum unit of an­gu­lar mo­men­tum, which is $\hbar$. The ra­tio is

\begin{displaymath}
\frac{pR}{\hbar} = \frac{pc\, R}{\hbar c} = \frac{pc\, R}{197\mbox{ MeV fm}}
\end{displaymath}

with $c$ the speed of light. The prod­uct $pc$ is com­pa­ra­ble to the en­ergy re­lease in the beta de­cay and can be ball­parked as on the or­der of 1 MeV. The nu­clear ra­dius ball­parks to 5 fm. As a re­sult, the clas­si­cal or­bital mo­men­tum is just a few per­cent of $\hbar$.

But quan­tum me­chan­ics says that the or­bital mo­men­tum can­not be a small frac­tion of $\hbar$. An­gu­lar mo­men­tum is quan­tized to val­ues $\sqrt{l(l+1)}\hbar$ where $l$ must be an in­te­ger. For $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 the an­gu­lar mo­men­tum is zero, for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 the an­gu­lar mo­men­tum is $\sqrt{2}\hbar$. There is noth­ing in be­tween. An an­gu­lar mo­men­tum that is a small frac­tion of $\hbar$ is not pos­si­ble. In­stead, what is small in quan­tum me­chan­ics is the prob­a­bil­ity that the elec­tron has an­gu­lar mo­men­tum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. If you try long enough, it may hap­pen.

In par­tic­u­lar, $pR$$\raisebox{.5pt}{$/$}$$\hbar$ gives a rough ball­park for the quan­tum am­pli­tude of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state. (The so-called Fermi the­ory of beta de­cay, {A.45}, can be used to jus­tify this and other as­sump­tions in this sec­tion.) The prob­a­bil­ity is the square mag­ni­tude of the quan­tum am­pli­tude, so the prob­a­bil­ity of get­ting $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is roughly $(pR/\hbar)^2$ smaller than get­ting $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is about 3 or 4 or­ders of mag­ni­tude less. It makes de­cays that have $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 that many or­ders of mag­ni­tude slower than al­lowed de­cays, all else be­ing the same. But if the de­cay is en­er­get­i­cally pos­si­ble, and al­lowed de­cays are not, it will even­tu­ally hap­pen. (As­sum­ing of course that some com­pletely dif­fer­ent de­cay like al­pha de­cay does not hap­pen first.)

De­cays with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are called “first-for­bid­den de­cays.” The elec­tron and neu­trino can then take up to 2 units of an­gu­lar mo­men­tum away through their com­bined or­bital an­gu­lar mo­men­tum and spin. So the nu­clear spin can change up to two units. Or­bital an­gu­lar mo­men­tum has neg­a­tive par­ity if $l$ is odd, so the par­ity of the nu­cleus must change dur­ing the de­cay. There­fore the pos­si­ble changes in nu­clear spin and par­ity are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{first-forbidden:}\qquad
\vert\...
...pt}{$\leqslant$}}2 \qquad \Delta \pi_{\rm N}= \mbox{yes}
$} %
\end{displaymath} (14.61)

That will not do for cal­cium-48, be­cause at least 4 units of spin change is needed. In sec­ond-for­bid­den de­cays, the elec­tron and neu­trino come out with a net or­bital an­gu­lar mo­men­tum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. Sec­ond for­bid­den de­cays are an­other 3 or 4 or­der of mag­ni­tude slower still than first for­bid­den ones. The nu­clear par­ity re­mains un­changed like in al­lowed de­cays. Where both al­lowed and sec­ond for­bid­den de­cays are pos­si­ble, the al­lowed de­cay should be ex­pected to have oc­curred long be­fore the sec­ond for­bid­den one has a chance. There­fore, the in­ter­est­ing sec­ond-for­bid­den cases cases are those that are not al­lowed ones:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{second-forbidden:}\qquad
\vert...
...rt = 2\mbox{ or } 3 \qquad \Delta \pi_{\rm N}= \mbox{no}
$} %
\end{displaymath} (14.62)

In third for­bid­den de­cays, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. The tran­si­tions that be­come pos­si­ble that were not in first for­bid­den ones are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{third-forbidden:}\qquad
\vert\...
... = 3\mbox{ or } 4
\qquad \Delta \pi_{\rm N}= \mbox{yes}
$} %
\end{displaymath} (14.63)

These tran­si­tions are still an­other 3 or 4 or­ders of mag­ni­tude slower than sec­ond for­bid­den ones. And they do not work for cal­cium-48, as both the cal­cium-48 ground state and the three reach­able scan­dium-48 states all have equal, pos­i­tive, par­ity.

Beta de­cay of cal­cium-48 is pos­si­ble through fourth-for­bid­den tran­si­tions:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{fourth-forbidden:}\qquad
\vert...
...rt = 4\mbox{ or } 5 \qquad \Delta \pi_{\rm N}= \mbox{no}
$} %
\end{displaymath} (14.64)

This al­lows de­cay to ei­ther the 5$\POW9,{+}$ and 4$\POW9,{+}$ ex­cited states of scan­dium-48. How­ever, fourth for­bid­den de­cays are gen­er­ally im­pos­si­bly slow.


14.19.6.3 Draft: The en­ergy ef­fect

There is an ad­di­tional ef­fect slow­ing down the beta de­cay of the 0$\POW9,{+}$ cal­cium-48 ground state to the 5$\POW9,{+}$ ex­cited scan­dium-48 state. The en­ergy re­lease, or $Q$-​value, of the de­cay is only about 0.15 MeV.

One rea­son that is bad news, (or good news, if you like cal­cium-48), is be­cause it makes the mo­men­tum of the elec­tron and neu­trino cor­re­spond­ingly small. The ra­tio $pR$$\raisebox{.5pt}{$/$}$$\hbar$ is there­fore quite small at about 0.01. And be­cause this is a fourth for­bid­den de­cay, the tran­si­tion is slowed down by a ball­park $((pR/\hbar)^{-2})^4$; that means a hu­mon­gous fac­tor 10$\POW9,{16}$ for $pR$$\raisebox{.5pt}{$/$}$$\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.01. If a 1 MeV al­lowed beta de­cay may take on the or­der of a day, you can see why cal­cium-48 is ef­fec­tively sta­ble against beta de­cay.

Fig­ure 14.52: The Fermi in­te­gral. It shows the ef­fects of en­ergy re­lease and nu­clear charge on the beta de­cay rate of al­lowed tran­si­tions. Other ef­fects ex­ists. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,28...
...beta^-$}}
\put(-50,92){\makebox(0,0)[l]{$\beta^+$}}
\end{picture}
\end{figure}

There is an­other, smaller, ef­fect. Even if the fi­nal nu­cleus is the 5$\POW9,{+}$ ex­cited scan­dium-48 state, with a sin­gle value for the mag­netic quan­tum num­ber, there is still more than one fi­nal state to de­cay to. The rea­son is that the rel­a­tive amounts of en­ergy taken by the elec­tron and neu­trino can vary. Ad­di­tion­ally, their di­rec­tions of mo­tion can also vary. The ac­tual net de­cay rate is an in­te­gral of the in­di­vid­ual de­cay rates to all these dif­fer­ent states. If the $Q$-​value is low, there are rel­a­tively few states avail­able, and this re­duces the to­tal de­cay rate too. The amount of re­duc­tion is given by the so-called “Fermi in­te­gral” shown in fig­ure 14.52. A de­cay with a $Q$ value of about 0.15 MeV is slowed down by roughly a fac­tor thou­sand com­pared to one with a $Q$ value of 1 MeV.

The Fermi in­te­gral shows beta plus de­cay is ad­di­tion­ally slowed down, be­cause it is more dif­fi­cult to cre­ate a positron at a strongly re­pelling pos­i­tively charged nu­cleus. The rel­a­tivis­tic Fermi in­te­gral also de­pends on the nu­clear ra­dius, hence a bit on the mass num­ber. Fig­ure 14.52 used a ball­park value of the mass num­ber for each $Z$ value, {A.45}.

The Fermi in­te­gral ap­plies to al­lowed de­cays, but the gen­eral idea is the same for for­bid­den de­cays. In fact, half-lives $\tau_{1/2}$ are com­monly mul­ti­plied by the Fermi in­te­gral $f$ to pro­duce a “com­par­a­tive half-life,” or “$ft$-​value” that is rel­a­tively in­sen­si­tive to the de­tails of the de­cay be­sides the de­gree to which it is for­bid­den. The $ft$-​value of a given de­cay can there­fore be used to ball­park to what ex­tent the de­cay is for­bid­den.

You see how cal­cium-48 can re­sist beta-de­cay for 64 10$\POW9,{18}$ years. (Zir­co­nium-96 with a half-life of 24 10$\POW9,{18}$ years has sim­i­lar re­sis­tance to plain beta de­cay.)


14.19.7 Draft: Data and Fermi the­ory

Fig­ure 14.53 shows nu­clei that de­cay pri­mar­ily through beta-mi­nus de­cay in blue. Nu­clei that de­cay pri­mar­ily through elec­tron cap­ture and beta-plus de­cay are shown in red. The sizes of the squares in­di­cate the de­cay rates. For ref­er­ence, the sta­ble and dou­ble-beta de­cay nu­clei are shown as full-size green squares.

Note the tremen­dous range of de­cay rates. It cor­re­sponds to half-lives rang­ing from mil­lisec­onds to 10$\POW9,{17}$ years. This is much like the tremen­dous range of half-lives in al­pha de­cay. De­cays last­ing more than about twenty years are shown as a min­i­mum-size dot in fig­ure 14.53; many would be in­vis­i­ble shown on the true scale.

Fig­ure 14.53: Beta de­cay rates. [pdf][con]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,56...
...ote: Largest decay rate is about 136 or so, does not exceed 1000.
\end{figure}

The de­cay rates in fig­ure 14.53 are color coded ac­cord­ing to a guessti­mated value for how for­bid­den the de­cay is. Darker red or blue in­di­cate more for­bid­den de­cays. Note that more for­bid­den de­cays tend to have much lower de­cay rates. (Lightly col­ored squares in­di­cate nu­clei for which the de­gree to which the de­cay is for­bid­den could not be guessti­mated by the au­to­mated pro­ce­dures used.)

Fig­ure 14.54: Beta de­cay rates as frac­tion of a ball­parked value. [pdf][con]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,56...
...akebox(0,0)[r]{$\fourIdx{236}{93}{}{}{\rm Np}$}}
}
\end{picture}
\end{figure}

Fig­ure 14.54 shows the de­cay rates nor­mal­ized with a the­o­ret­i­cal guessti­mate for them. Note the greatly re­duced range of vari­a­tion that the guessti­mate achieves, crude as it may be. One ma­jor suc­cess story is for for­bid­den de­cays. These are of­ten so slow that they must be shown as min­i­mum-size dots in fig­ure 14.53 to be vis­i­ble. How­ever, in fig­ure 14.54 they join the al­lowed de­cays as full-size squares. Con­sider in par­tic­u­lar the three slow­est de­cays among the data set. The slow­est of all is vana­dium-50, with a half-life of 150 10$\POW9,{15}$ year, fol­lowed by cad­mium-113 with 8 10$\POW9,{15}$ year, fol­lowed by in­dium-115 with 441 10$\POW9,{12}$ year. (Tel­lurium-123 has only a lower bound on its half life listed and is not in­cluded.) These de­cay times are long enough that all three iso­topes oc­cur nat­u­rally. In fact, al­most all nat­u­rally oc­cur­ring in­dium is the un­sta­ble iso­tope in­dium-115. Their dots in fig­ure 14.53 be­come full squares in fig­ure 14.54.

An­other eye-catch­ing suc­cess story is ${}\fourIdx{3}{1}{}{}{\rm {H}}$, the tri­ton, which suf­fers beta de­cay into ${}\fourIdx{3}{2}{}{}{\rm {He}}$, the sta­ble he­lion. The de­cay is al­lowed, but be­cause of its minis­cule en­ergy re­lease, or $Q$-​value, it takes 12 years any­way. Scaled with the ball­park, this slow de­cay too be­comes a full size square.

The ball­parks were ob­tained from the Fermi the­ory of beta de­cay, as dis­cussed in de­tail in ad­den­dum {A.45}. Un­like the rel­a­tively sim­ple the­ory of al­pha de­cay, the Fermi the­ory is elab­o­rate even in a crude form. Tak­ing beta-mi­nus de­cay as an ex­am­ple, the Fermi the­ory as­sumes a point­wise in­ter­ac­tion be­tween the wave func­tions of the neu­tron that turns into a pro­ton and those of the elec­tron/an­ti­neu­trino pair pro­duced by the de­cay. (Quan­tum me­chan­ics al­lows the neu­tron be­fore the de­cay to in­ter­act with the elec­tron and neu­trino that would ex­ist if it had al­ready de­cayed. That is a twi­light ef­fect, as dis­cussed in chap­ter 5.3 and more specif­i­cally in ad­den­dum {A.24} for gamma de­cay.) The strength of the in­ter­ac­tion is given by em­pir­i­cal con­stants.

Note that for many nu­clei no ball­parks were found. One ma­jor rea­son is that the pri­mary de­cay mech­a­nism is not nec­es­sar­ily to the ground state of the fi­nal nu­cleus. If de­cay to the ground state is for­bid­den, de­cay to a less-for­bid­den ex­cited state may dom­i­nate. There­fore, to cor­rectly es­ti­mate the de­cay rate for a given nu­cleus re­quires de­tailed knowl­edge about the ex­cited en­ergy states of the fi­nal nu­cleus. The en­er­gies of these ex­cited states must be suf­fi­ciently ac­cu­rately known, and they may not be. In par­tic­u­lar, for a few nu­clei, the en­ergy re­lease of the de­cay, or $Q$-​value, was com­puted to be neg­a­tive even for the ground state. This oc­curred for the elec­tron cap­ture of ${}\fourIdx{163}{67}{}{}{\rm {Ho}}$, ${}\fourIdx{193}{78}{}{}{\rm {Pt}}$, ${}\fourIdx{194}{80}{}{}{\rm {Hg}}$, ${}\fourIdx{202}{82}{}{}{\rm {Pb}}$, and ${}\fourIdx{205}{82}{}{}{\rm {Pb}}$, and for the beta de­cay of ${}\fourIdx{187}{75}{}{}{\rm {Re}}$ and ${}\fourIdx{241}{94}{}{}{\rm {Pu}}$. Ac­cord­ing to the Fermi the­ory, the de­cay can­not oc­cur if the $Q$-​value is neg­a­tive. How­ever, the $Q$-​val­ues in ques­tion are much smaller than the es­ti­mated elec­tronic bind­ing en­ergy (14.8). In fact they are com­pa­ra­ble to the dif­fer­ence in elec­tronic bind­ing en­ergy be­tween ini­tial and fi­nal nu­cleus or less. Since the bind­ing en­ergy is just an es­ti­mate, the com­puted $Q$-​val­ues, and there­fore the guessti­mated de­cay rate, should not be trusted.

In ad­di­tion to the en­ergy of the ex­cited states, their spins and par­i­ties must also be ac­cu­rately known. The rea­son is that they de­ter­mine to what level the de­cay is for­bid­den, hence slowed down. The com­puter pro­gram that pro­duced fig­ures 14.53 and 14.54 as­sumed con­ser­v­a­tively that if no unique value for spin and/or par­ity was given, it might be any­thing. Also, while there was ob­vi­ously no way for the pro­gram to ac­count for any ex­cited states whose ex­is­tence is not known, the pro­gram did al­low for the pos­si­bil­ity that there might be ad­di­tional ex­cited states above the high­est en­ergy level known. This is es­pe­cially im­por­tant well away from the sta­ble line where the ex­cited data are of­ten sparse or miss­ing al­to­gether. All to­gether, for about one third of the nu­clei processed, the un­cer­tainty in the ball­parked de­cay rate was judged too large to be ac­cepted. For the re­main­ing nu­clei, the level to which the de­cay was for­bid­den was taken from the ex­cited state that gave the largest con­tri­bu­tion to the de­cay rate.

The Fermi ball­parks were con­structed such that the true de­cay rate should not be sig­nif­i­cantly more than the ball­parked one. In gen­eral they met that re­quire­ment, al­though for about 1% of the nu­clei, the true de­cay rate was more ten times the ball­parked ones, reach­ing up to 370 times for ${}\fourIdx{253}{100}{}{}{\rm {Fm}}$. All these cases were for first-for­bid­den de­cays with rel­a­tively low $Q$-​val­ues. Since they in­cluded both beta mi­nus and elec­tron cap­ture de­cays, a plau­si­ble ex­pla­na­tion may be poor $Q$-​val­ues. How­ever, for for­bid­den de­cays, the cor­rec­tion of the elec­tron/positron wave func­tion for the ef­fect of the nu­clear charge is also sus­pect.

Note that while the true de­cay rate should not be much more than the ball­parked one, it is very pos­si­ble for it to be much less. The ball­park does not con­sider the de­tails of the nu­clear wave func­tion, be­cause that is in gen­eral pro­hib­i­tively dif­fi­cult. The ball­park sim­ply hopes that if a de­cay is not strictly for­bid­den by spin or par­ity at level $l$, the nu­clear wave func­tion change will not for some other rea­son make it al­most for­bid­den. But in fact, even if the de­cay is the­o­ret­i­cally pos­si­ble, the part of the Hamil­ton­ian that gives rise to the de­cay may pro­duce a nu­clear wave func­tion that has lit­tle prob­a­bil­ity of be­ing the right one. In that case the de­cay is slowed down pro­por­tional to that prob­a­bil­ity.

As an ex­am­ple, com­pare the de­cay processes of scan­dium-41 and cal­cium-47. Scan­dium-41, with 21 pro­tons and 20 neu­trons, de­cays into its mir­ror twin cal­cium-41, with 20 pro­tons and 21 neu­trons. The de­cay is al­most all due to beta-plus de­cay to the ground state of cal­cium-41. Ac­cord­ing to the shell model, the lone pro­ton in the 4f$_{7/2}$ pro­ton shell turns into a lone neu­tron in the 4f$_{7/2}$ neu­tron shell. That means that the nu­cleon that changes type is al­ready in the right state. The only thing that beta de­cay has to do is turn it from a pro­ton into a neu­tron. And that is in fact all that the de­cay Hamil­ton­ian does in the case of Fermi de­cay. Gamow-Teller de­cays also change the spin. The nu­cleon does not have to be moved around spa­tially. De­cays of this type are called “su­per­al­lowed.” (More gen­er­ally, su­per­al­lowed de­cays are de­fined as de­cays be­tween iso­baric ana­log states, or isospin mul­ti­plets. Such states dif­fer only in nu­cleon type. In other words, they dif­fer only in the net isospin com­po­nent $T_3$.) Su­per­al­lowed de­cays pro­ceed at the max­i­mum rate pos­si­ble. In­deed the de­cay of scan­dium-41 is at 1.6 times the ball­parked value.

All the elec­tron cap­ture / beta-plus de­cays of the nu­clei im­me­di­ately to the left of the ver­ti­cal $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $N$ line in fig­ures 14.53 and 14.54 are be­tween mir­ror nu­clei, and all are su­per­al­lowed. They are full-size squares in fig­ure 14.54. Su­per­al­lowed beta-mi­nus de­cays oc­cur for the tri­ton men­tioned ear­lier, as well as for a lone neu­tron.

But now con­sider the beta-mi­nus de­cay process of cal­cium-47 to scan­dium-47. Cal­cium-47 has no pro­tons in the 4f$_{7/2}$ pro­ton shell, but it has 7 neu­trons in the 4f$_{7/2}$ neu­tron shell. That means that it has a 1-neu­tron hole in the 4f$_{7/2}$ neu­tron shell. Beta de­cay to scan­dium-47 will turn one of the 7 neu­trons into a lone pro­ton in the 4f$_{7/2}$ pro­ton shell.

At least one source claims that in the odd-par­ti­cle shell model “all odd par­ti­cles are treated equiv­a­lently,” so that we might ex­pect that the cal­cium-47 de­cay is su­per­al­lowed just like the scan­dium-41 one. That is of course not true. The odd-par­ti­cle shell model does em­phat­i­cally not treat all odd par­ti­cles equiv­a­lently. It only says that, ef­fec­tively, an even num­ber of nu­cle­ons in the shell pair up into a state of zero net spin, leav­ing the odd par­ti­cle to pro­vide the net spin and elec­tro­mag­netic mo­ments. That does not mean that the sev­enth 4f$_{7/2}$ neu­tron can be in the same state as the lone pro­ton af­ter the de­cay. In fact, if the sev­enth neu­tron was in the same state as the lone pro­ton, it would bla­tantly vi­o­late the an­ti­sym­metriza­tion re­quire­ments, chap­ter 5.7. What­ever the state of the lone pro­ton might be, 7 neu­trons re­quire 6 more in­de­pen­dent states. And each of the 7 neu­trons must oc­cupy all these 7 states equally. It shows. The nu­clear wave func­tion of cal­cium-47 pro­duced by the de­cay Hamil­ton­ian matches up very poorly with the cor­rect fi­nal wave func­tion of scan­dium-47. The true de­cay rate of cal­cium-47 is there­fore about 10 000 times smaller than the ball­park.

As an­other ex­am­ple, con­sider the beta-plus de­cay of oxy­gen-14 to ni­tro­gen-14. Their iso­baric ana­log states were iden­ti­fied in fig­ure 14.46. De­cay to the ground state is al­lowed by spin and par­ity, at a ball­parked de­cay rate of 0.23/s. How­ever, the true de­cay pro­ceeds at a rate 0.01/s, which just hap­pens to be 1.6 times the ball­parked de­cay rate to the 0$\POW9,{+}$ ex­cited iso­baric ana­log state. One source notes ad­di­tion­ally that over 99% of the de­cay is to the ana­log state. So de­cay to the ground state must be con­tribut­ing less than a per­cent to the to­tal de­cay. And that is de­spite the fact that de­cay to the ground state is al­lowed too and has the greater $Q$-​value. The ef­fect gets even clearer if you look at the car­bon-14 to ni­tro­gen-14 beta-mi­nus de­cay. Here the de­cay to the iso­baric ana­log state vi­o­lates en­ergy con­ser­va­tion. The de­cay to the ground state is al­lowed, but it is more than 10 000 times slower than ball­park.

Su­per­al­lowed de­cays like the one of oxy­gen-14 to the cor­re­spond­ing iso­baric ana­log state of ni­tro­gen-14 are par­tic­u­larly in­ter­est­ing be­cause they are 0$\POW9,{+}$ to 0$\POW9,{+}$ de­cays. Such de­cays can­not oc­cur through the Gamow-Teller mech­a­nism, be­cause in Gamow-Teller de­cays the elec­tron and neu­trino take away one unit of an­gu­lar mo­men­tum. That means that de­cays of this type can be used to study the Fermi mech­a­nism in iso­la­tion.

The hor­ror story of a poor match up be­tween the nu­clear wave func­tion pro­duced by the de­cay Hamil­ton­ian and the fi­nal nu­clear wave func­tion is lutetium-176. Lutetium-176 has a 7$\POW9,{-}$ ground state, and that solidly for­bids de­cay to the 0$\POW9,{+}$ hafnium-176 ground state. How­ever, hafnium has en­er­get­i­cally al­lowed 6$\POW9,{+}$ and 8$\POW9,{+}$ ex­cited states that are only first-for­bid­den. There­fore you would not re­ally ex­pect the de­cay of lutetium-176 to be par­tic­u­larly slow. But the spin of the ex­cited states of hafnium is due to col­lec­tive nu­clear ro­ta­tion, and these states match up ex­tremely poorly with the ground state of lutetium-176 in which the spin is in­trin­sic. The de­cay rate is a stun­ning 12 or­ders of mag­ni­tude slower than ball­park. While tech­ni­cally the de­cay is only first-for­bid­den, lutetium is among the slow­est de­cay­ing un­sta­ble nu­clei, with a half-life of al­most 40 10$\POW9,{12}$ year. As a re­sult, it oc­curs in sig­nif­i­cant quan­ti­ties nat­u­rally. It is com­monly used to de­ter­mine the age of me­te­orites. No other ground state nu­cleus gets any­where close to that much be­low ball­park. The run­ner up is nep­tu­nium-236, which is 8 or­ders of mag­ni­tude be­low ball­park. Its cir­cum­stances are sim­i­lar to those of lutetium-176.

The dis­cussed ex­am­ples show that the Fermi the­ory does an ex­cel­lent job of pre­dict­ing de­cay rates if the dif­fer­ences in nu­clear wave func­tions are taken into ac­count. In fact, if the nu­clear wave func­tion can be ac­cu­rately ac­counted for, like in 0$\POW9,{+}$ to 0$\POW9,{+}$ su­per­al­lowed de­cays, the the­ory will pro­duce de­cay rates to 3 dig­its ac­cu­rate, [30, ta­ble 9.2]. The the­ory is also able to give ac­cu­rate pre­dic­tions for the dis­tri­b­u­tion of ve­loc­i­ties with which the elec­trons or positrons come out. Data on the ve­loc­ity dis­tri­b­u­tions can in fact be used to solidly de­ter­mine the level to which the de­cay is for­bid­den by plot­ting them in so-called “Fermi-Kurie plots.” These and many other de­tails are out­side the scope of this book.


14.19.8 Draft: Par­ity vi­o­la­tion

For a long time, physi­cists be­lieved that the fun­da­men­tal laws of na­ture be­haved the same when seen in the mir­ror. The strong nu­clear force, elec­tro­mag­net­ism, and grav­ity all do be­have the same when seen in the mir­ror. How­ever, in 1956 Lee and Yang pointed out that the claim had not been tested for the weak force. If it was un­true there, it could ex­plain why what seemed to be a sin­gle type of K-me­son could de­cay into end prod­ucts of dif­fer­ent par­ity. The sym­me­try of na­ture un­der mir­ror­ing leads to the law of con­ser­va­tion of par­ity, chap­ter 7.3. How­ever, if the weak force is not the same un­der mir­ror­ing, par­ity can change in weak processes, and there­fore, the de­cay prod­ucts could have any net par­ity, not just that of the orig­i­nal K-me­son.

Fig­ure 14.55: Par­ity vi­o­la­tion. In the beta de­cay of cobalt-60, left, the elec­tron pref­er­en­tially comes out in the di­rec­tion that a left-handed screw ro­tat­ing with the nu­clear spin would move. Seen in the mir­ror, right, that be­comes the di­rec­tion of a right-handed screw.
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Wu and her cowork­ers there­fore tested par­ity con­ser­va­tion for the beta de­cay of cobalt-60 nu­clei. These nu­clei were cooled down to ex­tremely low tem­per­a­tures to cut down on their ther­mal mo­tion. That al­lowed their spins to be aligned with a mag­netic field, as in the left of fig­ure 14.55. It was then ob­served that the elec­trons pref­er­en­tially came out in the di­rec­tion of mo­tion of a left-handed screw ro­tat­ing with the nu­clear spin. Since a left-handed screw turns into a right-handed one seen in the mir­ror, it fol­lowed that in­deed the weak force is not the same seen in the mir­ror. The physics in the mir­ror is not the cor­rect physics that is ob­served.

Since the weak force is weak, this does not af­fect par­ity con­ser­va­tion in other cir­cum­stances too much. For­mally it means that eigen­func­tions of the Hamil­ton­ian are not eigen­func­tions of the par­ity op­er­a­tor. How­ever, nu­clear wave func­tions still have a sin­gle par­ity to very good ap­prox­i­ma­tion; the am­pli­tude of the state of op­po­site par­ity mixed in is of the or­der of 10$\POW9,{-7}$, [30, p. 313]. The prob­a­bil­ity of mea­sur­ing the op­po­site par­ity is the square of that, much smaller still. Still, if a de­cay is ab­solutely for­bid­den when par­ity is strictly pre­served, then it might barely be pos­si­ble to ob­serve the rare de­cays al­lowed by the com­po­nent of the wave func­tion of op­po­site par­ity.

An ad­di­tional op­er­a­tion can be ap­plied to the mir­ror im­age in 14.55 to turn it back into a phys­i­cally cor­rect de­cay. All par­ti­cles can be re­placed by their an­tipar­ti­cles. This op­er­a­tion is called “charge con­ju­ga­tion,” be­cause among other things it changes the sign of the charge for each charged par­ti­cle. In physics, you are al­ways lucky if a name gets some of it right. Some of the par­ti­cles in­volved may ac­tu­ally be charged, and con­ju­ga­tion is a so­phis­ti­cated-​sound­ing term to some peo­ple. It is also a vague term that quite con­ceiv­ably could be taken to mean re­ver­sal of sign by peo­ple naive enough to con­sider con­ju­ga­tion so­phis­ti­cated. Charge con­ju­ga­tion turns the elec­trons go­ing around in the loops of the elec­tro­mag­net in fig­ure 14.55 into positrons, so the cur­rent re­verses di­rec­tion. That must re­verse the sign of the mag­netic field if the physics is right. But so will the mag­netic mo­ment of an­ti­cobalt-60 nu­cleus change sign, so it stays aligned with the mag­netic field. And physi­cist be­lieve the positrons will pref­er­en­tially come out of an­ti­cobalt-60 nu­clei along the mo­tion of a right-handed screw.

Be­sides this com­bined charge con­ju­ga­tion plus par­ity (CP) sym­me­try of na­ture, time sym­me­try is also of in­ter­est here. Phys­i­cal processes should re­main phys­i­cally cor­rect when run back­wards in time, the same way you can run a movie back­wards. It turns out that time sym­me­try too is not com­pletely ab­solute, and nei­ther is CP sym­me­try for that mat­ter. How­ever, if all three op­er­a­tions, charge con­ju­ga­tion (C), mir­ror­ing (P), and time in­ver­sion (T), to­gether are ap­plied to a phys­i­cal process, the re­sult­ing process is be­lieved to al­ways be phys­i­cally cor­rect. There is a the­o­rem, the CPT the­o­rem, that says so un­der rel­a­tively mild as­sump­tions.