Sub­sec­tions


14.10 Draft: Liq­uid drop model

Nu­cle­ons at­tract each other with nu­clear forces that are not com­pletely un­der­stood, but that are known to be short range. That is much like mol­e­cules in a clas­si­cal liq­uid drop at­tract each other with short-range Van der Waals forces. In­deed, it turns out that a liq­uid drop model can ex­plain many prop­er­ties of nu­clei sur­pris­ingly well. This sec­tion gives an in­tro­duc­tion.


14.10.1 Draft: Nu­clear ra­dius

The vol­ume of a liq­uid drop, hence its num­ber of mol­e­cules, is pro­por­tional to the cube of its ra­dius $R$. Con­versely, the ra­dius is pro­por­tional to the cube root of the num­ber of mol­e­cules. Sim­i­larly, the ra­dius of a nu­cleus is ap­prox­i­mately equal to the cube root of the num­ber of nu­cle­ons:

\begin{displaymath}
R \approx R_A \sqrt[3]{A} \qquad R_A = 1.23 \mbox{ fm} %
\end{displaymath} (14.9)

Here $A$ is the mass num­ber, equal to the num­ber of pro­tons $Z$ plus the num­ber of neu­trons $N$. Also fm stands for fem­tome­ter, equal to 10$\POW9,{-15}$ me­ter; it may be re­ferred to as a “fermi” in some older ref­er­ences. En­rico Fermi was a great curse for early nu­clear physi­cists, quickly do­ing all sorts of things be­fore they could.

It should be noted that the above nu­clear ra­dius is an av­er­age one. A nu­cleus does not stop at a very sharply de­fined ra­dius. (And nei­ther would a liq­uid drop if it only con­tained 100 mol­e­cules or so.) Also, the con­stant $R_A$ varies a bit with the nu­cleus and with the method used to es­ti­mate the ra­dius. Val­ues from 1.2 to 1.25 are typ­i­cal. This book will use the value 1.23 stated above.

It may be noted that these re­sults for the nu­clear radii are quite solidly es­tab­lished ex­per­i­men­tally. Physi­cists have used a wide va­ri­ety of in­ge­nious meth­ods to ver­ify them. For ex­am­ple, they have bounced elec­trons at var­i­ous en­ergy lev­els off nu­clei to probe their Coulomb fields, and al­pha par­ti­cles to also probe the nu­clear forces. They have ex­am­ined the ef­fect of the nu­clear size on the elec­tron spec­tra of the atoms; these ef­fects are very small, but if you sub­sti­tute a muon for an elec­tron, the ef­fect be­comes much larger since the muon is much heav­ier. They have dropped pi mesons on nu­clei and watched their de­cay. They have also com­pared the en­er­gies of nu­clei with $Z$ pro­tons and $N$ neu­trons against the cor­re­spond­ing mir­ror nu­clei that have with $N$ pro­tons and $Z$ neu­trons. There is good ev­i­dence that the nu­clear force is the same when you swap neu­trons with pro­tons and vice versa, so com­par­ing such nu­clei shows up the Coulomb en­ergy, which de­pends on how tightly the pro­tons are packed to­gether. All these dif­fer­ent meth­ods give es­sen­tially the same re­sults for the nu­clear radii. They also in­di­cate that the neu­trons and pro­tons are well-mixed through­out the nu­cleus, [30, pp. 44-59]


14.10.2 Draft: von Weizsäcker for­mula

The bind­ing en­ergy of nu­clei can be ap­prox­i­mated by the “von Weizsäcker for­mula,“ or “Bethe-von Weizsäcker for­mula:”

\begin{displaymath}
\fbox{$\displaystyle
E_{{\rm B,vW}} = C_v A - C_s A^{2/3} ...
... - C_d\frac{(Z-N)^2}{A} - C_p\frac{o_Z+o_N-1}{A^{C_{e}}}
$} %
\end{displaymath} (14.10)

where the $C_{.}$ are con­stants, while $o_Z$ is 1 if the num­ber of pro­tons is odd and zero if it is even, and sim­i­lar for $o_N$ for neu­trons. This book uses val­ues given by [36] for the con­stants:

\begin{displaymath}
C_v = 15.409 \;{\rm MeV}\quad
C_s = 16.873 \;{\rm MeV}\quad
C_c = 0.695 \;{\rm MeV}\quad
C_z = 1 \quad
\end{displaymath}


\begin{displaymath}
C_d = 22.435 \;{\rm MeV}\quad
C_p = 11.155 \;{\rm MeV}\quad
C_e = 0.5
\end{displaymath}

where a MeV (mega elec­tron volt) is 1.602 18 10$\POW9,{-13}$ J, equal to the en­ergy that an elec­tron picks up in a one mil­lion volt elec­tric field.

Plugged into the mass-en­ergy re­la­tion, the von Weizsäcker for­mula pro­duces the so-called “semi-em­pir­i­cal mass for­mula:”

\begin{displaymath}
m_{\rm nucleus,SE} = Z m_{\rm p}+ N m_{\rm n}- \frac{E_{{\rm B,vW}}}{c^2}
\end{displaymath} (14.11)


14.10.3 Draft: Ex­pla­na­tion of the for­mula

The var­i­ous terms in the von Weizsäcker for­mula of the pre­vi­ous sub­sec­tion have quite straight­for­ward ex­pla­na­tions. The $C_v$ term is typ­i­cal for short-range at­trac­tive forces; it ex­presses that the en­ergy of every nu­cleon is low­ered the same amount by the pres­ence of the at­tract­ing nu­cle­ons in its im­me­di­ate vicin­ity. The clas­si­cal ana­logue is that the en­ergy needed to boil away a drop of liq­uid is pro­por­tional to its mass, hence to its num­ber of mol­e­cules.

The $C_s$ term ex­presses that nu­cle­ons near the sur­face are not sur­rounded by a com­plete set of at­tract­ing nu­cle­ons. It raises their en­ergy. This af­fects only a num­ber of nu­cle­ons pro­por­tional to the sur­face area, hence pro­por­tional to $A^{2/3}$. The ef­fect is neg­li­gi­ble for a clas­si­cal drop of liq­uid, which may have a mil­lion mol­e­cules along a di­am­e­ter, but not for a nu­cleus with maybe ten nu­cle­ons along it. (Ac­tu­ally, the ef­fect is im­por­tant for a clas­si­cal drop too, even if it does not af­fect its over­all en­ergy, as it gives rise to sur­face ten­sion.)

The $C_c$ term ex­presses the Coulomb re­pul­sion be­tween pro­tons. Like the Coulomb en­ergy of a sphere with con­stant charge den­sity, it is pro­por­tional to the square net charge, so to $Z^2$ and in­versely pro­por­tional to the ra­dius, so to $A^{1/3}$. How­ever, the em­pir­i­cal con­stant $C_c$ is some­what dif­fer­ent from that of a con­stant charge den­sity. Also, a cor­rec­tion $C_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 has been thrown in to en­sure that there is no Coulomb re­pul­sion if there is just one pro­ton.

The last two terms cheat; they try to de­vi­ously in­clude quan­tum ef­fects in a sup­pos­edly clas­si­cal model. In par­tic­u­lar, the $C_d$ term adds an en­ergy in­creas­ing with the square of the dif­fer­ence in num­ber of pro­tons and neu­trons. It sim­u­lates the ef­fect of the Pauli ex­clu­sion prin­ci­ple. As­sume first that the num­ber of pro­tons and neu­trons is equal, each $A$$\raisebox{.5pt}{$/$}$​2. In that case the pro­tons will be able to oc­cupy the low­est $A$$\raisebox{.5pt}{$/$}$​2 pro­ton en­ergy lev­els, and the neu­trons the low­est $A$$\raisebox{.5pt}{$/$}$​2 neu­tron lev­els. How­ever, if then, say, some of the pro­tons are turned into neu­trons, they will have to move to en­ergy lev­els above $A$$\raisebox{.5pt}{$/$}$​2, be­cause the low­est $A$$\raisebox{.5pt}{$/$}$​2 neu­tron lev­els are al­ready filled with neu­trons. There­fore the en­ergy goes up if the num­ber of pro­tons and neu­trons be­comes un­equal.

The last $C_p$ term ex­presses that nu­cle­ons of the same type like to pair up. When both the num­ber of pro­tons and the num­ber of neu­trons is even, all pro­tons can pair up, and all neu­trons can, and the en­ergy is lower than av­er­age. When both the num­ber of pro­tons is odd and the num­ber of neu­trons is odd, there will be an un­paired pro­ton as well as an un­paired neu­tron, and the en­ergy is higher than av­er­age.


14.10.4 Draft: Ac­cu­racy of the for­mula

Fig­ure 14.9 shows the er­ror in the von Weizsäcker for­mula as col­ors. Blue means that the ac­tual bind­ing en­ergy is higher than pre­dicted, red that it is less than pre­dicted. For very light nu­clei, the for­mula is ob­vi­ously use­less, but for the re­main­ing nu­clei it is quite good. Note that the er­ror is in the or­der of MeV, to be com­pared to a to­tal bind­ing en­ergy of about $8A$ MeV. So for heavy nu­clei the rel­a­tive er­ror is small.

Fig­ure 14.9: Er­ror in the von Weizsäcker for­mula. [pdf][con]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,56...
...]{$>$14 MeV more than estimated}}
}
\end{picture}}
\end{picture}
\end{figure}

Near the magic num­bers the bind­ing en­ergy tends to be greater than the pre­dicted val­ues. This can be qual­i­ta­tively un­der­stood from the quan­tum en­ergy lev­els that the nu­cle­ons oc­cupy. When nu­cle­ons are suc­ces­sively added to a nu­cleus, those that go into en­ergy lev­els just be­low the magic num­bers have un­usu­ally large bind­ing en­ergy, and the to­tal nu­clear bind­ing en­ergy in­creases above that pre­dicted by the von Weizsäcker for­mula. The de­vi­a­tion from the for­mula there­fore tends to reach a max­i­mum at the magic num­ber. Just above the magic num­ber, fur­ther nu­cle­ons have a much lower en­ergy level, and the de­vi­a­tion from the von Weizsäcker value de­creases again.