13.4 Mass-Energy and Nuclear Binding Energy

13.4.1 Mass – Energy

Einstein showed from his theory of special relativity that it is necessary to treat mass as another form of energy. Before the advent of this theory of special relativity it was presumed that mass and energy were conserved separately in a reaction. However, Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa.

Einstein gave the famous mass-energy equivalence relation

\[ E=m c^{2} \]

Here the energy equivalent of mass \(m\) is related by the above equation and \(c\) is the velocity of light in vacuum and is approximately equal to \(3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)

Experimental verification of the Einstein’s mass-energy relation has been achieved in the study of nuclear reactions amongst nucleons, nuclei, electrons and other more recently discovered particles. In a reaction the conservation law of energy states that the initial energy and the final energy are equal provided the energy associated with mass is also included. This concept is important in understanding nuclear masses and the interaction of nuclei with one another. They form the subject matter of the next few sections.

13.4.2 Nuclear binding energy

In Section 13.2 we have seen that the nucleus is made up of neutrons and protons. Therefore it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons. However, the nuclear mass \(M\) is found to be always less than this. For example, let us consider \({ }_{8}^{16} \mathrm{O}\); a nucleus which has 8 neutrons and 8 protons. We have

\begin{array}{l} \text { Mass of } 8 \text { neutrons }=8 \times 1.00866 \mathrm{u} \\ \text { Mass of } 8 \text { protons }=8 \times 1.00727 \mathrm{u} \\ \text { Mass of } 8 \text { electrons }=8 \times 0.00055 \mathrm{u} \end{array} \begin{aligned} &\text { Therefore the expected mass of }{ }_{8}^{16} \mathrm{O} \text { nucleus }\\ &=8 \times 2.01593 \mathrm{u}=16.12744 \mathrm{u} \end{aligned}

The atomic mass of \({ }_{8}^{16} \mathrm{O}\) found from mass spectroscopy experiments is seen to be \(15.99493 \mathrm{u}\). Substracting the mass of 8 electrons \((8 \times 0.00055 \mathrm{u})\) from this, we get the experimental mass of \({ }_{8}^{16} \mathrm{O}\) nucleus to be \(15.99053 \mathrm{u}\).

Thus, we find that the mass of the \({ }_{8}^{16} \mathrm{O}\) nucleus is less than the total mass of its constituents by \(0.13691 \mathrm{u}\). The difference in mass of a nucleus and its constituents, \(\Delta M\), is called the mass defect, and is given by

\[ \Delta M=\left[Z m_{p}+(A-Z) m_{n}\right]-M \]

What is the meaning of the mass defect? It is here that Einstein's equivalence of mass and energy plays a role. Since the mass of the oxygen nucleus is less that the sum of the masses of its constituents (8 protons and 8 neutrons, in the unbound state), the equivalent energy of the oxygen nucleus is less than that of the sum of the equivalent energies of its constituents. If one wants to break the oxygen nucleus into 8 protons and 8 neutrons, this extra energy \(\Delta M c^{2},\) has to supplied. This energy required \(E_{\mathrm{b}}\) is related to the mass defect by

\[ E_{b}=\Delta M c^{2} \]

If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy \(E_{b}\) will be released in the process. The energy \(E_{b}\) is called the binding energy of the nucleus. If we separate a nucleus into its nucleons, we would have to supply a total energy equal to \(E_{b},\) to those particles. Although we cannot tear apart a nucleus in this way, the nuclear binding energy is still a convenient measure of how well a nucleus is held together. A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon, \(E_{b n}\), which is the ratio of the binding energy \(E_{b}\) of a nucleus to the number of the nucleons, A, in that nucleus:

\[ E_{b n}=E_{b} / A \]

We can think of binding energy per nucleon as the average energy per nucleon needed to separate a nucleus into its individual nucleons.

Figure 13.1 is a plot of the binding energy per nucleon \(E_{bn}\) versus the mass number A for a large number of nuclei. We notice the following main features of the plot:

the binding energy per nucleon, \(E_{bn}\) , is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number ( 30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238
E bn is lower for both light nuclei (A<30) and heavy nuclei (A>170).

We can draw some conclusions from these two observations:

The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon
The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. Consider a particular nucleon inside a sufficiently large nucleus. It will be under the influence of only some of its neighbours, which come within the range of the nuclear force. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration. If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Let the binding energy of the nucleus be pk, where k is a constant having the dimensions of energy. If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk. The property that a given nucleon influences only nucleons close to it is also referred to as saturation property of the nuclear force.
A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. It has very important implications for energy production through fission, to be discussed later in Section 13.7.1.
Consider two very light nuclei (A ≤ 10) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion. This is the energy source of sun, to be discussed later in Section 13.7.3.