The Liquid-Drop Model of Fission

In the liquid drop model of the nucleus, formulated by Niels Bohr, the nucleons are imagined to interact strongly with each other, just like molecules in a drop of liquid. This constant jiggling around permits us to correlate many facts about nuclear masses and energies, and provides a useful model for understanding a large class of nuclear reactions, including fission.

In this model the idea is to regard the nucleus as a classical drop, with volume energy, and surface tension. On this basis the normal modes of vibration of a liquid drop may represent the states of excitation of the nucleus. A good analogy is the emission of nuclear particles from the excited nucleus. This can be thought of as an evaporating process, the binding energies of the emitted particles being analogous to the heat of vaporisation.

Suppose now we have a nucleus, being a small spherical drop with radius $R=r_oA^{1/3}$ cm. Ignoring the quantum mechanical effects, there seems to be a static equilibrium. The attractive force that keeps the nucleus coherent, just like in a liquid drop, is short range. Each nucleon interacts with its nearest neighbour. Then the binding energy, which is denoted as $-E_B$, being negative because it is stored, is a function of the density. But density is constant, so the volume energy alone gives $-E_B=-\alpha_VA$. But like any liquid drop, the nucleus has surface. For any nucleus, there will be nucleons that lie on the surface. Their bonds with the rest of the nucleus are not as strong, and therefore there is a deficiency in binding energy. This is a positive surface energy. With that in mind then, we expect the binding energy to be $-E_B\sim\alpha(volume)+b(surface)$.

There is also the mutual Coulomb repulsion of all the protons in the nucleus. On the contrary , it is a long-range force to which all the protons contribute. On the assumption that the charge density is constant, the Coulomb energy of a spherical drop of radius $R=r_oA^{\frac{1}{3}}$ is the integral given in [3] as

$$E_C=\left[\frac{Ze}{Vol}\right]\left[\frac{Ze} {Vol}\right]{\int\... ...t}}= \frac{3}{5}\frac{Z^2e^2}{R}=\frac{3}{5} \frac{Z^2e^2}{5r_oA^{\frac{1}{3}}}$$ (1)

These very simple ideas can explain fission. Suppose that our classical liquid drop is charged. If we slowly deform it by elongating it, the surface area is increased. This costs surface energy, and the two halves of the not-so-spherical drop are now found at a larger distance apart, compared to the very spherical drop before.

Having disturbed the drop, and changed it's shape, two things might happen. It might return to its original shape, meaning the two parts get back together, or it may undergo fission, in which case the Coulomb repulsion takes in charge. The first case happens in the limit where the charge density is small compared to the surface energy, and thus outweighs the Coulomb repulsion. In this scenario, the drop will try to minimize its surface at all deformations, and the spherical shape will be stable for every distortion up to a limit where the drop is broken into two spheres, and the two spheres are moved apart, just out of the range of the surface forces.

Now consider the limiting case where the original droplet is highly charged, that the slightest displacement from the spherical shape will produce fission. The Coulomb repulsion will dominate over the surface energy, which will try to bring the shape back to spherical, and the two parts of the drop will be violently repelled from each other. Figure 3, at the end of this paper, shows the energy diagram of a fissionable nucleus. It is the Coulomb repulsion that actually makes the nuclei move apart violently, and release energy.

By doing the calculation of the classical area and Coulomb energy by considering small constant volume distortions, axially symmetric, and without motion of the center of mass, and by neglecting all terms of higher order in the small coefficient $\alpha_n$ we obtain the following results for the surface and Coulomb energies

$$E_C-{E_C}^{sphere}=-\frac{3}{5}\frac{Z^2e^2}{A^ {\frac{1}{3}}r_o}\left[\frac{{\alpha_2}^2}{5}+...\right]$$ (2)

$$E_{surf}-{E_{surf}}^{sphere}=4\pi A^{\frac{2}{3}}{r_o}^2 {\sigma}\left[\frac{2{\alpha_2}^2}{5}+...\right]$$ (3)

where $\sigma=$ nuclear surface energy/ $cm^2$

The radius of the drop surface is represented as a function of the co-latitude angle $\gamma(\theta)$, by a series of orthonormal set of Legendre polynomials as given in [3].

Then the fission threshold energy, i.e. the energy that fission can occur, measured in units of the undistorded surface energy becomes

$$\frac{E_{th}}{4\pi{A^{\frac{2}{3}}}{\gamma_o}^2\sigma}= \frac{{\alpha_2}^2}{5}\left[2-\frac{{E_C}^{sphere}}{{E_{surf}} ^{sphere}}+...\right]$$ (4)

This treatment of fission is wholly classical, and it neglects two important quantum mechanical effects. First, fission may take place for excitation energies below the fission threshold by means of the tunneling effect. In addition, the vibration of the drop in the distorted mode will have zero point energy.

There is a very interesting aspect in fission that cannot be explained by theory. The fission fragments do not appear as two equal nuclei, each 1/2 the mass of the fissioning nucleus. The distribution in figure 4 at the end of this paper,is what is known experimentally. No clear explanation of this fact has yet been given!!