Subsections


14.10 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 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 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 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 Accuracy of the formula

Figure 14.7 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 14.7: Error in the von Weizsäcker formula. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
... 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.