Obtain a semi empirical binding energy per nucleon formula. Using this as a guide, explain why the Coulomb force, which is only about$\frac{\mathbf{1}}{\mathbf{100}}$as strong as the inter nucleon attraction for two protons "in contact" (cf. Table 11.2), would eventually have to become a dominant factor in large nuclei. Assume that Z,N and A increase in rough proportion to one another.

$c_1=15.8,c_2=17.8,c_3=0.71$and$c_4=23.7$.First term in sum is volume term, second term is surface term, third term is Coulomb term and fourth term is asymmetry term. Obtained value of binding energy is in MeV.

Semi empirical binding Energy is given by the expression,

$\mathit{B}\mathit{E}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{A}\mathbf{-}{\mathit{c}}_{\mathbf{2}}{\mathit{A}}^{\mathbf{2}\mathbf{/}\mathbf{3}}\mathbf{-}{\mathit{c}}_{\mathbf{3}}\frac{\mathbf{Z}\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{1}\mathbf{)}}{{\mathbf{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}}\mathbf{-}{\mathit{c}}_{\mathbf{4}}\frac{{\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{Z}\mathbf{)}}^{\mathbf{2}}}{\mathbf{A}}$

Coulomb force is about $\frac{1}{100}$as strong as the inter nucleon attraction for two protons in contact.

The semi empirical binding energy formula is $BE=c_{1}A-c_2{A}^{2/3}-c_3\frac{Z(Z-1)}{A^{2/3}}-c_4\frac{{(N-Z)}^2}{A}$.

So, this formula for binding energy per nucleon would be:

$$\begin{gathered} BE=\frac{1}{A}\left(c_{1}A-c_2{A}^{2/3}-c_3\frac{Z(Z-1)}{A^{2/3}}-c_4\frac{{(N-Z)}^2}{A}\right) \\ BE=c_1-\frac{c_2}{A^{1/3}}-c_3\frac{Z(Z-1)}{{\left(A^{1/3}\right)}^4}-c_4\frac{{(N-Z)}^2}{A} \end{gathered}$$

If the atomic number is proportional to number of nucleons, the third term will increase like $A^{2/3}$ while all the other terms decrease as number of nucleons increase.

Thus, the Coulomb term will dominate eventually.