4.0 Nuclear binding energy

4.1 Binding energy per nucleon
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability

A short range force is known as a strong nuclear force which holds the nucleons tightly against the long range electrostatic force of repulsion among the protons.

Thus, the nucleus is itself a stable entity with an overall negative potential energy.

So, some energy must be supplied to separate all the nucleons.

More stable the nucleus is, the greater energy will be required to break it. This required energy is called the binding energy of the nucleus.

A simple nuclear reaction is given below where $Z$ number of protons binds with $\left( {A - Z} \right)$ number of neutrons having masses $\left( {{m_p}} \right)$ & $\left( {{m_n}} \right)$ respectively to form a nucleus ${}_Z^AX$.

$$Z{m_p} + \left( {A - Z} \right){m_n} \to {M_X} + Energy\left( {\Delta m{c^2}} \right)$$

So, whenever nucleons binds to form a nucleus some of the mass of the reacting species is lost and is converted into energy.

So, the nuclear binding energy holds the nucleons together and hence a stable nucleus is formed.

This difference in masses of the independent nucleons and mass of the nucleus is called mass defect(${\Delta m}$) of the nuclear reaction. It can be calculated as

$$\begin{equation} \begin{aligned} {\text{ }}\Delta m = Z{m_p} + \left( {A - Z} \right){m_n} - {M_X} \\ {\text{or }}\Delta m = \left[ {Z{m_p} + \left( {A - Z} \right){m_n}} \right] - \left[ {{M_{atom}} - Z{m_e}} \right] \\\end{aligned} \end{equation} $$

Using Einstein's mass-energy relationship if in a reaction 1 $amu$ mass is converted into energy, energy released

$$\begin{equation} \begin{aligned} \quad \quad \quad \;\Delta E = \Delta m{c^2} = \left( {1.656 \times {{10}^{ - 27}}{\text{ Kg}}} \right){\left( {3 \times {{10}^8}{\text{ m/s}}} \right)^2} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{{\left( {1.656 \times {{10}^{ - 27}}{\text{ Kg}}} \right){{\left( {3 \times {{10}^8}{\text{ m/s}}} \right)}^2}}}{{1.60 \times {{10}^{ - 19}}}}{\text{ ev}} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = 931.5 \times {10^6}{\text{ ev}} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = 931.5{\text{ Mev}} \\ \quad \quad \quad {\text{or }}1{\text{ amu = }}931.5{\text{ }}\frac{{{\text{Mev}}}}{{{c^2}}} \\\end{aligned} \end{equation} $$