Sub­sec­tions


14.11 Draft: Al­pha De­cay

In al­pha de­cay a nu­cleus emits an “al­pha par­ti­cle,” later iden­ti­fied to be sim­ply a he­lium-4 nu­cleus. Since the es­cap­ing al­pha par­ti­cle con­sists of two pro­tons plus two neu­trons, the atomic num­ber $Z$ of the nu­cleus de­creases by two and the mass num­ber $A$ by four. This sec­tion ex­plains why al­pha de­cay oc­curs.


14.11.1 Draft: De­cay mech­a­nism

Fig­ure 14.10 gives de­cay data for the nu­clei that de­cay ex­clu­sively through al­pha de­cay. Nu­clei are much like cher­ries: they have a vari­able size that de­pends mainly on their mass num­ber $A$, and a charge $Z$ that can be shown as dif­fer­ent shades of red. You can even de­fine a stem for them, as ex­plained later. Nu­clei with the same atomic num­ber $Z$ are joined by branches.

Fig­ure 14.10: Half-life ver­sus en­ergy re­lease for the atomic nu­clei marked in NUBASE 2003 as show­ing pure al­pha de­cay with un­qual­i­fied en­er­gies. Top: only the even val­ues of the mass and atomic num­bers cherry-picked, omit­ting ${}\fourIdx{8}{4}{}{}{\rm{Be}}$. In­set: re­ally cherry-pick­ing, only a few even mass num­bers for tho­rium and ura­nium! Bot­tom: all the nu­clei ex­cept ${}\fourIdx{8}{4}{}{}{\rm{Be}}$ (67 as, 0.092 MeV). [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,37...
...ebox(0,0)[b]{5}}
\put(-66,210){\makebox(0,0)[b]{6}}
\end{picture}
\end{figure}

Not shown in fig­ure 14.10 is the un­sta­ble beryl­lium iso­tope ${}\fourIdx{8}{4}{}{}{\rm {Be}}$, which has a half-life of only 67 as, (i.e. 67 10$\POW9,{-18}$ s), and a de­cay en­ergy of only 0.092 MeV. As you can see from the graph, these num­bers are wildly dif­fer­ent from the other, much heav­ier, al­pha-de­cay nu­clei, and in­clu­sion would make the graph very messy.

Note the tremen­dous range of half-lives in fig­ure 14.10, from mere nanosec­onds to quin­til­lions of years. And that ex­cludes beryl­lium’s at­tosec­onds. In the early his­tory of al­pha de­cay, it seemed very hard to ex­plain how nu­clei that do not seem that dif­fer­ent with re­spect to their num­bers of pro­tons and neu­trons could have such dra­mat­i­cally dif­fer­ent half-lives. The en­ergy that is re­leased in the de­cay process does not vary that much, as fig­ure 14.10 also shows.

To add to the mys­tery in those early days of quan­tum me­chan­ics, if an al­pha par­ti­cle was shot back at the nu­cleus with the same en­ergy that it came out, it would not go back in! It was re­flected by the elec­tro­sta­tic re­pul­sion of the pos­i­tively charged nu­cleus. So, it had not enough en­ergy to pass through the re­gion of high po­ten­tial en­ergy sur­round­ing the nu­cleus, yet it did pass through it when it came out.

Gamow, and in­de­pen­dently Gur­ney & Con­don, rec­og­nized that the ex­pla­na­tion was quan­tum tun­nel­ing. Tun­nel­ing al­lows a par­ti­cle to get through a po­ten­tial en­ergy bar­rier even if clas­si­cally it does not have enough en­ergy to do so, chap­ter 7.12.2.

Fig­ure 14.11: Schematic po­ten­tial for an al­pha par­ti­cle that tun­nels out of a nu­cleus.
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,13...
...}
\put(-86,64){\makebox(0,0)[r]{$V_{\rm Coulomb}$}}
\end{picture}
\end{figure}

Fig­ure 14.11 gives a rough model of the bar­rier.The hor­i­zon­tal line rep­re­sents the to­tal en­ergy of the al­pha par­ti­cle. Far from the nu­cleus, the po­ten­tial en­ergy $V$ of the al­pha par­ti­cle can be de­fined to be zero. Closer to the nu­cleus, the po­ten­tial en­ergy of the al­pha par­ti­cle ramps up due to Coulomb re­pul­sion. How­ever, right at the outer edge $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R$ of the nu­cleus it­self, the strong but very short-range at­trac­tive nu­clear force pops up, and the com­bined po­ten­tial en­ergy plum­mets al­most ver­ti­cally down­wards to some low value $V_n$. In be­tween the ra­dial po­si­tion $r_1$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $R$ and some larger ra­dius $r_2$, the po­ten­tial en­ergy ex­ceeds the to­tal en­ergy that the al­pha par­ti­cle has avail­able. Clas­si­cally, the al­pha par­ti­cle can­not pen­e­trate into this re­gion. How­ever, in quan­tum me­chan­ics it re­tains a very small prob­a­bil­ity of do­ing so.

The re­gion in be­tween $r_1$ and $r_2$ is called the “Coulomb bar­rier.” It is a poorly cho­sen name, be­cause the bar­rier is only a Coulomb one for an al­pha par­ti­cle try­ing to get in the nu­cleus. For an al­pha par­ti­cle try­ing to get out, it is a nu­clear force bar­rier; here the Coulomb force as­sists the tun­nel­ing par­ti­cle to get through the bar­rier and es­cape. The term nu­clear bar­rier would avoid this am­bi­gu­ity. There­fore physi­cists do not use it.

Now, to get a rough pic­ture of al­pha de­cay, imag­ine an al­pha par­ti­cle wave packet rat­tling around in­side the nu­cleus try­ing to es­cape. Each time it hits the bar­rier at $r_1$, it has a small chance of es­cap­ing. Even­tu­ally it gets lucky.

As­sume that the al­pha par­ti­cle wave packet is small enough that the mo­tion can be as­sumed to be one-di­men­sion­al. Then the small chance of es­cap­ing each time it hits the bar­rier is ap­prox­i­mately given by the analy­sis of chap­ter 7.13 as

\begin{displaymath}
\fbox{$\displaystyle
T \approx e^{-2\gamma_{12}}
\qquad
...
...\hbar} \int_{r_1}^{r_2} \sqrt{2m_\alpha(V-E)} {\,\rm d}r
$} %
\end{displaymath} (14.12)

The fact that this prob­a­bil­ity in­volves an ex­po­nen­tial is the ba­sic rea­son for the tremen­dous range in half-lives: ex­po­nen­tials can vary greatly in mag­ni­tude for rel­a­tively mod­est changes in their ar­gu­ment.


14.11.2 Draft: Com­par­i­son with data

The pre­vi­ous sub­sec­tion ex­plained al­pha de­cay in terms of an im­pris­oned al­pha par­ti­cle tun­nel­ing out of the nu­cleus. To ver­ify whether that is rea­son­able, the next step is ob­vi­ously to put in some ball­park num­bers and see whether the ex­per­i­men­tal data can be ex­plained.

First, the en­ergy $E$ of the al­pha par­ti­cle may be found from Ein­stein’s fa­mous ex­pres­sion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, sec­tion 14.6. Just find the dif­fer­ence be­tween the rest mass of the orig­i­nal nu­cleus and the sum of that of the fi­nal nu­cleus and the al­pha par­ti­cle, and mul­ti­ply by the square speed of light. That gives the en­ergy re­lease. It comes out pri­mar­ily as ki­netic en­ergy of the al­pha par­ti­cle, ig­nor­ing any ex­ci­ta­tion en­ergy of the fi­nal nu­cleus. (A re­duced mass can be used to al­low for re­coil of the nu­cleus.) Note that al­pha de­cay can­not oc­cur if $E$ is neg­a­tive; the ki­netic en­ergy of the al­pha par­ti­cle can­not be neg­a­tive.

It may be noted that the en­ergy re­lease $E$ in a nu­clear process is gen­er­ally called the “$Q$-​value.” The rea­son is that one of the most com­mon other quan­ti­ties used in nu­clear physics is the so-called quadru­pole mo­ment $Q$. Also, the to­tal nu­clear charge is in­di­cated by $Q$, as is the qual­ity fac­tor of ra­di­a­tion, while pro­jec­tion and ro­ta­tion op­er­a­tors, and sec­ond points are of­ten also in­dicted by $Q$. The un­der­ly­ing idea is that when you are try­ing to fig­ure out some tech­ni­cal ex­pla­na­tion, then if al­most the only math­e­mat­i­cal sym­bol used is $Q$, it pro­vides a pretty strong hint that you are prob­a­bly read­ing a book on nu­clear physics.

The nu­clear ra­dius $R$ ap­prox­i­mately de­fines the start $r_1$ of the re­gion that the al­pha par­ti­cle has to tun­nel through, fig­ure 14.11. It can be ball­parked rea­son­ably well from the num­ber of nu­cle­ons $A$; ac­cord­ing to sec­tion 14.10,

\begin{displaymath}
R \approx R_A \sqrt[3]{A} \qquad R_A = 1.23 \mbox{ fm}
\end{displaymath}

where f, femto, is 10$\POW9,{-15}$. That is a lot smaller than the typ­i­cal Bohr ra­dius over which elec­trons are spread out. Elec­trons are far away and are not re­ally rel­e­vant.

It should be pointed out that the re­sults are very sen­si­tive to the as­sumed value of $r_1$. The sim­plest as­sump­tion would be that at $r_1$ the al­pha par­ti­cle would have its cen­ter at the nu­clear ra­dius of the re­main­ing nu­cleus, com­puted from the above ex­pres­sion. But very no­tice­able im­prove­ments are ob­tained by as­sum­ing that at $r_1$ the cen­ter is al­ready half the ra­dius of the al­pha par­ti­cle out­side. (In lit­er­a­ture, it is of­ten as­sumed that the al­pha par­ti­cle is a full ra­dius out­side, which means fully out­side but still touch­ing the re­main­ing nu­cleus. How­ever, half works bet­ter and is maybe some­what less im­plau­si­ble.)

The good news about the sen­si­tiv­ity of the re­sults on $r_1$ is that con­versely it makes al­pha de­cay a rea­son­ably ac­cu­rate way to de­duce or ver­ify nu­clear radii, [30, p. 57]. You are hardly likely to get the nu­clear ra­dius no­tice­ably wrong with­out get­ting into ma­jor trou­ble ex­plain­ing al­pha de­cay.

The num­ber of es­cape at­tempts per unit time is also needed. If the al­pha par­ti­cle has a typ­i­cal ve­loc­ity $v_\alpha$ in­side the orig­i­nal nu­cleus, it will take it a time of about $2r_0$$\raisebox{.5pt}{$/$}$$v_\alpha$ to travel the $2r_0$ di­am­e­ter of the nu­cleus. So it will bounce against the bar­rier about $v_\alpha$$\raisebox{.5pt}{$/$}$$2r_0$ times per sec­ond. That is sure to be a very large num­ber of times per sec­ond, the nu­cleus be­ing so small, but each time it hits the perime­ter, it only has a minis­cule $e^{-2\gamma_{12}}$ chance of es­cap­ing. So it may well take tril­lions of years be­fore it is suc­cess­ful any­way. Even so, among a very large num­ber of nu­clei a few will get out every time. Re­mem­ber that a mol of atoms rep­re­sents in the or­der of 10$\POW9,{23}$ nu­clei; among that many nu­clei, a few al­pha par­ti­cles are likely to suc­ceed what­ever the odds against. The rel­a­tive frac­tion of suc­cess­ful es­cape at­tempts per unit time is by de­f­i­n­i­tion the rec­i­p­ro­cal of the life­time $\tau$;

\begin{displaymath}
\frac{v_\alpha}{2r_0} e^{-2\gamma_{12}} = \frac{1}{\tau}
\end{displaymath} (14.13)

Mul­ti­ply the life­time by $\ln2$ to get the half-life.

The ve­loc­ity $v_\alpha$ of the al­pha par­ti­cle can be ball­parked from its ki­netic en­ergy $E-V_n$ in the nu­cleus as $\sqrt{2(E-V_n)/m_\alpha}$. Un­for­tu­nately, find­ing an ac­cu­rate value for the nu­clear po­ten­tial $V_n$ in­side the nu­cleus is not triv­ial. But have an­other look at fig­ure 14.10. For­get about en­gi­neer­ing ideas about ac­cept­able ac­cu­racy. A 50% er­ror in half-life would be in­vis­i­ble seen on the tremen­dous range of fig­ure 14.10. Be­ing wrong by a fac­tor 10, or even a fac­tor 100, two or­ders of mag­ni­tude, is ho-hum on the scale that the half-life varies. So, the po­ten­tial en­ergy $V_n$ in­side the nu­cleus can be ball­parked. The cur­rent re­sults use the typ­i­cal value of $\vphantom{0}\raisebox{1.5pt}{$-$}$35 MeV given in [30, p. 252].

That leaves the value of $\gamma_{12}$ to be found from the in­te­gral over the bar­rier in (14.12). Be­cause the nu­clear forces are so short-range, they should be neg­li­gi­ble over most of the in­te­gra­tion range. So it seems rea­son­able to sim­ply sub­sti­tute the Coulomb po­ten­tial every­where for $V$. The Coulomb po­ten­tial is in­versely pro­por­tional to the ra­dial po­si­tion $r$, and it equals $E$ at $r_2$, so $V$ can be writ­ten as $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Er_2$$\raisebox{.5pt}{$/$}$$r$. Sub­sti­tut­ing this in, and do­ing the in­te­gral by mak­ing a change of in­te­gra­tion vari­able to $u$ with $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_2\sin^2u$, pro­duces

\begin{displaymath}
\gamma_{12} = \frac{\sqrt{2m_\alpha E}}{\hbar} r_2
\left[
...
...rac{r_1}{r_2}\right)}
-\arcsin\sqrt{\frac{r_1}{r_2}}
\right]
\end{displaymath}

The last two terms within the square brack­ets are typ­i­cally rel­a­tively small com­pared to the first one, be­cause $r_1$ is usu­ally fairly small com­pared to $r_2$. Then $\gamma_{12}$ is about pro­por­tional to $\sqrt{E}r_2$. But $r_2$ it­self is in­versely pro­por­tional to $E$, be­cause the to­tal en­ergy of the al­pha par­ti­cle equals its po­ten­tial en­ergy at $r_2$;

\begin{displaymath}
E = \frac{(Z-Z_\alpha)e\,Z_\alpha e}{4\pi\epsilon_0 r_2}
\qquad Z_\alpha=2
\end{displaymath}

That makes $\gamma_{12}$ about pro­por­tional to 1$\raisebox{.5pt}{$/$}$$\sqrt{E}$ for a given atomic num­ber $Z$.

So if you plot the half-life on a log­a­rith­mic scale, and the en­ergy $E$ on an rec­i­p­ro­cal square root scale, as done in fig­ure 14.10, they should vary lin­early with each other for a given atomic num­ber. This does as­sume that the vari­a­tions in num­ber of es­cape at­tempts are also ig­nored. The pre­dicted slope of lin­ear vari­a­tion is in­di­cated by the stems on the cher­ries in fig­ure 14.10. Ide­ally, all cher­ries con­nected by branches should fall on a sin­gle line with this slope. The fig­ure shows that this is quite rea­son­able for even-even nu­clei, con­sid­er­ing the rough ap­prox­i­ma­tions made. For nu­clei that are not even-even, the de­vi­a­tions from the pre­dicted slope are more sig­nif­i­cant. The next sub­sec­tion dis­cusses the ma­jor sources of er­ror.

Fig­ure 14.12: Half-life pre­dicted by the Gamow / Gur­ney & Con­don the­ory ver­sus the true value. Top: even-even nu­clei only. Bot­tom: all the nu­clei ex­cept ${}\fourIdx{8}{4}{}{}{\rm{Be}}$ (55 as ver­sus 67 as). [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(300,21...
...,0)[b]{1Ey}}
\put(115,-9.5){\makebox(0,0)[b]{true}}
\end{picture}
\end{figure}

The bot­tom line ques­tion is whether the the­ory, rough as it may be, can pro­duce mean­ing­ful val­ues for the ex­per­i­men­tal half-lives, within rea­son. Fig­ure 14.12 shows pre­dicted half-lives ver­sus the ac­tual ones. Cher­ries on the black line in­di­cate that the cor­rect value is pre­dicted. It is clear that there is no real ac­cu­racy to the pre­dic­tions in any nor­mal sense; they are eas­ily off by sev­eral or­ders of mag­ni­tude. What can you ex­pect with­out an ac­cu­rate model of the nu­cleus it­self? How­ever, the pre­dic­tions do suc­cess­fully re­pro­duce the tremen­dous range of half-lives and they do not de­vi­ate from the cor­rect val­ues that much com­pared to that tremen­dous range. It is hard to imag­ine any other the­ory be­sides tun­nel­ing that could do the same.

The worst per­for­mance of the the­ory is for the ${}\fourIdx{209}{83}{}{}{\rm {Bi}}$ bis­muth iso­tope in­di­cated by the right­most dot in fig­ure 14.12. Its true half-life of 19 Ey, 19 10$\POW9,{18}$ years, is grossly un­der­es­ti­mated to be just 9 Py, 9 10$\POW9,{15}$ years. Then again, since the uni­verse has only ex­isted about 14 10$\POW9,{9}$ years, who is go­ing to live long enough to com­plain about it? Es­sen­tially none of the bis­muth-209 that has ever been cre­ated in the uni­verse has de­cayed. It took un­til 2003 for physi­cists to ob­serve that bis­muth-209 ac­tu­ally did de­cay; it is still listed as sta­ble in many ref­er­ences. For ${}\fourIdx{8}{4}{}{}{\rm {Be}}$, which is not shown in the fig­ure, the pre­dicted half-life is 55 as (55 10$\POW9,{-18}$ s), ver­sus a true value of 67 as.


14.11.3 Draft: For­bid­den de­cays

You may won­der why there is so much er­ror in the the­o­ret­i­cal pre­dic­tions of the half life. Or why the the­ory seems to work so much bet­ter for even-even nu­clei than for oth­ers. A de­vi­a­tion by a fac­tor 2 000 like for bis­muth-209 seems an aw­ful lot, rough as the the­ory may be.

Some of the sources of in­ac­cu­racy are self-ev­i­dent from the the­o­ret­i­cal de­scrip­tion as given. In par­tic­u­lar, there is the al­ready men­tioned ef­fect of the value of $r_1$. It is cer­tainly pos­si­ble to cor­rect for de­vi­a­tions from the Coulomb po­ten­tial near the nu­cleus by a suit­able choice of the value of $r_1$. How­ever, the pre­cise value that kills off the er­ror is un­known, and un­for­tu­nately the re­sults strongly de­pend on that value. To fix this would re­quire an ac­cu­rate eval­u­a­tion of the nu­clear force po­ten­tial, and that is very dif­fi­cult. Also, the po­ten­tial of the elec­trons would have to be in­cluded. The al­pha par­ti­cle does reach a dis­tance of the or­der of a tenth of a Bohr ra­dius from the nu­cleus at the end of tun­nel­ing. The Bohr ra­dius is here taken to be based on the ac­tual nu­clear charge, not the hy­dro­gen one.

Also, the pic­ture of a rel­a­tively com­pact wave packet of the al­pha par­ti­cle rat­tling around as­sumes that that the size of that wave packet is small com­pared to the nu­cleus. That spa­tial lo­cal­iza­tion is as­so­ci­ated with in­creased un­cer­tainty in mo­men­tum, which im­plies in­creased en­ergy. And the ki­netic en­ergy of the al­pha par­ti­cle is not re­ally known any­way, with­out an ac­cu­rate value for the nu­clear force po­ten­tial.

A very ma­jor other prob­lem is the as­sump­tion that the fi­nal al­pha par­ti­cle and nu­cleus end up in their ground states. If ei­ther ends up in an ex­cited state, the en­ergy that the al­pha par­ti­cle has avail­able for es­cape will be cor­re­spond­ingly re­duced. Now the al­pha par­ti­cle will most cer­tainly come out in its ground state; it takes over 20 MeV to ex­cite an al­pha par­ti­cle. But for most nu­clei, the re­main­ing nu­cleus can­not be in its ground state if the mech­a­nism is as de­scribed.

The main rea­son is an­gu­lar mo­men­tum con­ser­va­tion. The al­pha par­ti­cle has no net in­ter­nal an­gu­lar an­gu­lar mo­men­tum. Also, it was as­sumed that the al­pha par­ti­cle comes out ra­di­ally, which means that there is no or­bital an­gu­lar mo­men­tum ei­ther. So the an­gu­lar mo­men­tum of the nu­cleus af­ter emis­sion must be the same as that of the nu­cleus be­fore the emis­sion. That is no prob­lem for even-even nu­clei, be­cause it is the same; even-even nu­clei all have zero in­ter­nal an­gu­lar mo­men­tum in their ground state. So even-even nu­clei do not suf­fer from this prob­lem.

How­ever, al­most all other nu­clei do. All even-odd and odd-even nu­clei and al­most all odd-odd ones have nonzero an­gu­lar mo­men­tum in their ground state. Usu­ally the ini­tial and fi­nal nu­clei have dif­fer­ent val­ues. That means that al­pha de­cay that leaves the fi­nal nu­cleus in its ground state vi­o­lates con­ser­va­tion of an­gu­lar mo­men­tum. The de­cay process is called “for­bid­den.” The fi­nal nu­cleus must be ex­cited if the process is as de­scribed. That en­ergy sub­tracts from that of the al­pha par­ti­cle. There­fore the al­pha par­ti­cle has less en­ergy to tun­nel through, and the true half-life is much longer than com­puted.

Note in the bot­tom half of fig­ure 14.12 how many nu­clei that are not even-even do in­deed have half-lifes that are or­ders of mag­ni­tude larger than pre­dicted by the­ory. Con­sider the ex­am­ple of bis­muth-209, with a half-life 2 000 times longer than pre­dicted. Bis­muth-209 has a spin, i.e.  an az­imuthal quan­tum num­ber, of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. How­ever, the de­cay prod­uct thal­lium-205 has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ in its ground state. If you check out the ex­cited states of thal­lium-205, there is an ex­cited state with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, but its ex­ci­ta­tion en­ergy would re­duce the en­ergy of the al­pha par­ti­cle from 3.2 MeV to 1.7 MeV, mak­ing the tun­nel­ing process very much slower.

And there is an­other prob­lem with that. The de­cay to the men­tioned ex­ited state is not pos­si­ble ei­ther, be­cause it vi­o­lates con­ser­va­tion of par­ity, chap­ter 7.3 and 7.4. Say­ing the al­pha par­ti­cle comes out ra­di­ally, as done above is not re­ally cor­rect. The proper quan­tum way to say that the al­pha par­ti­cle comes out with no or­bital an­gu­lar mo­men­tum is to say that its wave func­tion varies with an­gu­lar lo­ca­tion as the spher­i­cal har­monic $Y_0^0$, chap­ter 4.2.3. In spec­tro­scopic terms, it comes out in an s-wave. Now the ini­tial bis­muth atom has odd par­ity; its com­plete wave func­tion changes sign if you every­where re­place ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. But the al­pha par­ti­cle, the ex­cited thal­lium state, and the $Y_0^0$ or­bital mo­tion all have even par­ity; there is no change of sign. That means that the to­tal fi­nal par­ity is even too, so the fi­nal par­ity is not the same as the ini­tial par­ity. That vi­o­lates con­ser­va­tion of par­ity so the process can­not oc­cur.

Thal­lium-205 does not have ex­cited states be­low 3.2 MeV that have been solidly es­tab­lished to have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ and odd par­ity, so you may start to won­der whether al­pha de­cay for bis­muth-209 is pos­si­ble at all. How­ever, the al­pha par­ti­cle could of course come out with or­bital an­gu­lar mo­men­tum. In other words it could come out with a wave func­tion that has an an­gu­lar de­pen­dence ac­cord­ing to $Y_l^m$ with the az­imuthal quan­tum num­ber $l$ equal to one or more. These states have even par­ity if $l$ is even and odd par­ity when $l$ is odd. Quan­tum me­chan­ics then al­lows the thal­lium-205 ex­cited state to have any spin $j$ in the range $\vert\frac92-l\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $j$ $\raisebox{-.3pt}{$\leqslant$}$ $\frac92+l$ as long as its par­ity is odd or even when­ever $l$ is even or odd.

For ex­am­ple, bis­muth-209 could de­cay to the ground state of thal­lium-205 if the or­bital an­gu­lar mo­men­tum of the al­pha par­ti­cle is $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5. Or it could de­cay to an ex­cited $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state with an ex­ci­ta­tion en­ergy of 0.9 MeV if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The prob­lem is that the ki­netic en­ergy in the an­gu­lar mo­tion sub­tracts from that avail­able for the ra­dial mo­tion, mak­ing the tun­nel­ing, once again, much slower. In terms of the ra­dial mo­tion, the an­gu­lar mo­men­tum in­tro­duces an ad­di­tional ef­fec­tive po­ten­tial $l(l+1)\hbar^2$$\raisebox{.5pt}{$/$}$$2m_{\alpha}r^2$, com­pare the analy­sis of the hy­dro­gen atom in chap­ter 4.3.2. Note that this ef­fect in­creases rapidly with $l$. How­ever, the de­cay of bis­muth-209 ap­pears to be to the ground state any­way; the mea­sured en­ergy of the al­pha par­ti­cle turns out to be 3.14 MeV. The pre­dicted half-life in­clud­ing the ef­fec­tive po­ten­tial is found to be 4.6 Ey, much bet­ter than the one com­puted in the pre­vi­ous sec­tion.

One fi­nal source of er­ror should be men­tioned. Of­ten al­pha de­cay can pro­ceed in a num­ber of ways and to dif­fer­ent fi­nal ex­ci­ta­tion en­er­gies. In that case, the spe­cific de­cay rates must be added to­gether. This ef­fect can make the true half-life shorter than the one com­puted in the pre­vi­ous sub­sec­tion. But clearly, this ef­fect should be mi­nor on the scale of half-lifes of fig­ure 14.12. In­deed, while the pre­dicted half-lifes of many nu­clei are way be­low the true value in the fig­ure, few are sig­nif­i­cantly above it.


14.11.4 Draft: Why al­pha de­cay?

The fi­nal ques­tion that begs an an­swer is why do so many nu­clei so specif­i­cally want to eject an he­lium-4 nu­cleus? Why none of the other nu­clei? Why not the less tightly bound, but lighter deuteron, or the more tightly bound, but heav­ier car­bon-12 nu­cleus? The an­swer is sub­tle.

To un­der­stand the rea­son, re­con­sider the analy­sis of the pre­vi­ous sub­sec­tion for a more gen­eral ejected nu­cleus. As­sume that the ejected par­ti­cle has an atomic num­ber $Z_1$ and mass $m_1$. As men­tioned, the pre­cise num­ber of es­cape at­tempts is not re­ally that im­por­tant for the half life; al­most all the vari­a­tion in half-life is through the quan­tity $\gamma_{12}$. Also, to a first ap­prox­i­ma­tion the ra­tio of start to end of the tun­nel­ing do­main, $r_1$$\raisebox{.5pt}{$/$}$$r_2$, can be ig­nored. Un­der those con­di­tions, $\gamma_{12}$ is pro­por­tional to

\begin{displaymath}
\gamma_{12} \propto \sqrt{\frac{m_1}{E}} Z_1(Z-Z_1)
\end{displaymath}

It is pretty much all in there.

As long as the ejected par­ti­cle has about the usual 8 MeV bind­ing en­ergy per nu­cleon, the square root in the ex­pres­sion above does not vary that much. In such cases the en­ergy re­lease $E$ is about pro­por­tional to the amount of nu­cle­ons ejected. Ta­ble 14.3 gives some ex­am­ple num­bers. That makes $\gamma_{12}$ about pro­por­tional to $Z_1$, and the great­est chance of tun­nel­ing out then oc­curs by far for the light­est nu­clei. It ex­plains why the al­pha par­ti­cle tun­nels out in­stead of heav­ier nu­clei. It is not that a heav­ier nu­cleus like car­bon-14 can­not be emit­ted, it is just that an al­pha par­ti­cle has al­ready done so long be­fore car­bon-14 gets the chance. In fact, for ra­dium-223 it has been found that one car­bon-14 nu­cleus is ejected for every bil­lion al­pha par­ti­cles. That is about con­sis­tent with the com­puted half-lifes of the events as shown in ta­ble 14.3.


Ta­ble 14.3: Can­di­dates for nu­clei ejected by ura­nium-238, ra­dium-223, and fer­mium-256.
\begin{table}\begin{displaymath}
\renewedcommand{arraystretch}{1.2}
\begin{arr...
...0^{42}&\;2~10^{31} \\
\hline\hline
\end{array} \end{displaymath}
\end{table}


But the ar­gu­ment that $Z_1$ should be as small as pos­si­ble should make pro­tons or neu­trons, not the al­pha par­ti­cle, the ones that can es­cape most eas­ily. How­ever, these do not have any bind­ing en­ergy. While pro­tons or neu­trons are in­deed ejected from nu­clei that have a very large pro­ton, re­spec­tively neu­tron ex­cess, nor­mally the en­ergy re­lease for such emis­sions is neg­a­tive. There­fore the emis­sion can­not oc­cur. Beta de­cay oc­curs in­stead to ad­just the ra­tio be­tween pro­tons and neu­trons to the op­ti­mum value. Near the op­ti­mum value, you would still think it might be bet­ter to eject a deuteron than an al­pha. How­ever, be­cause the bind­ing en­ergy of the deuteron is only a sin­gle MeV per nu­cleon, the en­ergy re­lease is again neg­a­tive. Among the light nu­clei, the al­pha is unique in hav­ing al­most the full 8 MeV of bind­ing en­ergy per nu­cleon. It is there­fore the only one that pro­duces a pos­i­tive en­ergy re­lease.

The fi­nal prob­lem is that the ar­gu­ments above seem to show that spon­ta­neous fis­sion can­not oc­cur. For, is the fis­sion of say fer­mium-256 into two tin-128 nu­clei not just ejec­tion of a tin-128 nu­cleus, leav­ing a tin-128 nu­cleus? The ar­gu­ments above say that al­pha de­cay should oc­cur much be­fore this can hap­pen.

The prob­lem is that the analy­sis of al­pha de­cay is in­ap­plic­a­ble to fis­sion. The num­bers for fis­sion-scale half-lifes in ta­ble 14.3 are all wrong. Fis­sion is in­deed a tun­nel­ing event. How­ever, it is one in which the en­ergy bar­rier is dis­in­te­grat­ing due to a global in­sta­bil­ity of the nu­clear shape. That in­sta­bil­ity mech­a­nism strongly fa­vors large scale di­vi­sion over short scale ones. The only hint of this in ta­ble 14.3 are the large val­ues of $r_1$$\raisebox{.5pt}{$/$}$$r_2$ for fis­sion-scale events. When $r_1$$\raisebox{.5pt}{$/$}$$r_2$ be­comes one, the tun­nel­ing re­gion is gone. But long be­fore that hap­pens, the re­gion is so small com­pared to the size of the ejected nu­cleus that the ba­sic ideas un­der­ly­ing the analy­sis have be­come mean­ing­less. Even ig­nor­ing the fact that the nu­clear shapes have been as­sumed spher­i­cal and they are not in fis­sion.

Thus, un­like ta­ble 14.3 sug­gests, fer­mium-256 does fis­sion. The two frag­ments are usu­ally of dif­fer­ent size, but not vastly so. About 92% of fer­mium-256 nu­clei spon­ta­neously fis­sion, while the other 8% ex­pe­ri­ence al­pha de­cay. Ura­nium-238 de­cays for 99.999 95% through $\alpha$ de­cay, and for only 0.000 05% through spon­ta­neous fis­sion. Al­though the amount of fis­sion is very small, it is not by far as small as the num­bers in ta­ble 14.3 im­ply. Fis­sion is not known to oc­cur for ra­dium-223; this nu­cleus does in­deed show pure al­pha de­cay ex­cept for the men­tioned rare car­bon-14 emis­sion.