Nuclear Engineering

Liquid Drop Model: Nuclear Stability

To understand nuclear properties various theoretical models have been proposed by various nuclear physicists. The simplest model to understand some of the basic properties is ‘Liquid drop model’. In these two lectures we are going to discuss this model, which in spite of its simplicity can explain number of salient features of a nucleus.



The Liquid drop model of nucleus

Weizsacher in 1935 recognized some experimental evidences and found resemblance of nucleus with a liquid drop. Later on, in 1939 Bohr and Wheeler employed a liquid drop model of the nucleus in their theory of nuclear fission. Considerable information about nuclear behaviour across the periodic table is found in a plot between neutron numbers against proton numbers for various stable nuclei, called Segre chart as depicted in Fig. m1.8.


FIGURE m1.8 Segre chart of stable nuclides



This figure illustrates that the points representing stable nuclei goes away from the line N/Z = 1 for low Z towards the direction of higher number of neutrons as one takes heavier elements. For A = 238, N/Z = 1.6. We will see later that this trend is ascribed to an increase in Coulomb repulsion energy among protons with increase in Z. To counteract this, heavy nuclei tend to be made up of more neutrons than protons. The curve depicted in Fig. m1.8 is known as stability curve and the isotopes lying above and below this curve are radioactive. They decay by emitting alpha and beta particles (discussed in module 4) and reach towards stability curve. In order to understand this in detail, let us look at how binding energy of a nucleus varies with nuclear mass and what all other factors influence it.

If the mass number of a nucleus is A, binding energy per nucleon is defined by,



When this quantity is plotted against mass number, the plot obtained is shown in Fig. m1.9.


The decrease of for large A is due to the coulomb repulsion between the protons, which clearly makes the nuclei increasingly less stable.


	FIGURE m1.9 Binding energy per nucleon versus mass number of nuclei for different elements of the periodic table

From the graph it is evident that the elements in the middle of the periodic table are most stable, since they have the largest B.E./A values. The maximum stability is reached in the neighborhood of mass number 50. The importance of B.E./A curve in order to understand nuclear fission, fusion ^#* is immense and will be discussed in the respective modules of this lecture series.


Nucleons are bound inside the nucleus by a mutual interacting force known as nuclear force. Therefore, nuclear binding energy is proportional to the volume of the nucleus or to the total number of nucleons A, which contributes the volume energy term B₁ and can be expressed as

	B ₁ ∝ V ∝ A
	∴ B₁ = a₁A where a₁ > 0	—————– (m1.27)


This can be better understood on the basis of liquid drop picture. For a liquid drop to evaporate, a certain amount of heat must be supplied, which is the product of heat of vaporization, mass of the material that is number of molecules times the mass of each molecule. The reason for this simple relationship is that binding energy is the sum of all interactions between molecules, and each molecule interacts only with its neighbors. This consideration is correct for a system in which the range of interactions between particles is small compared to the dimensions of the system.


————————————————————————————————————– *Fission is a physical process by which a heavy nucleus is fragmented into two relatively less massive nuclei.
^# By fusion process two lighter nuclei fuse together to make a relatively massive nucleus. Both fission and fusion process has tremendous potential to produce energy for electricity generation and you will learn it in module III. ————————————————————————————————————–

However, the nucleons those are situated in the surface region of the nucleus are necessarily more weakly bound than those residing inside the nuclear interior because they have fewer immediate neighbors. The number of such nucleons is proportional to the surface area of the nucleus and therefore to R ² and so is proportional to A ^2/3 (because R ∝ A ^1/3). Thus we have

	B ₂ = -α ₂ A^2/3	——————(m1.28)

Where	α ₂ >0


The sign is negative because it weakens the binding energy as it corresponds to the surface tension of a liquid drop. This effect is pronounced for lower A nuclei and hence the B.E./A decreases for lower A (Fig. m1.9). B₂ is also referred to as surface energy term.


Now let us address the issue why at higher A, B.E./A again decreases quite significantly. Definitely, there is some sort of interaction which is trying to destabilize nuclei having higher A. The most probable reason is enhanced Coulomb repulsion between positively charged protons. The number of protons and hence the Coulomb repulsion energy increases with increase in A value. The contribution due to Coulomb energy can be calculated in the following manner: Approximately, a nucleus may be considered to have a constant charge density ρ , and for r > R (nuclear radius) there is no charge.

Therefore,



The work dW required to bring a thin spherical shell (thickness dr) of this charge upto its radius r is obtained from the coulomb’s law:


Integrating this from r = 0 to r = R, which gives total work done in assembling the sphere



Even if there were only one proton in the nucleus, we would obtain



Now considering Z protons and from the above equations the coulomb energy term can be written as



or



where a³ includes all the constant terms of Eqn. m1.20.

Nuclear Engineering

Nuclear Engineering

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Liquid Drop Model: Nuclear Stability

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