Question

Write down the semi-empirical mass formula for the liquid drop model? Explain all the terms?

Answer 1

George Gamow was the first to propose a liquid drop model and later niels bohr and john Archibald wheeler worked and developed a liquid drop model. Nucleus is treated as a drop of incompressible fluid which has very high density, held together by a residual effect of strong force (nuclear force).

Complete answer:
Semi-empirical mass formula which gives expression for binding energy
The semi-empirical mass formula for the liquid drop model is given by:
${{E}_{B}}={{a}_{v}}A-{{a}_{S}}{{A}^{\dfrac{2}{3}}}-{{a}_{c}}\dfrac{Z(Z-1)}{{{A}^{\dfrac{1}{3}}}}-{{a}_{a}}\dfrac{{{(A-2Z)}^{2}}}{A}+\partial (A,Z)$
Where ${{E}_{B}}$= Binding energy
Volume energy term is the first term in the formula .When the number of nuclei with the same size packed into a small volume then nuclear energy is proportional to volume.

1. Volume energy is given as${{E}_{V}}\propto V$ and $v\propto {{R}^{2}}$
   Where $R={{R}_{0}}{{A}^{\dfrac{1}{3}}}$
   ${{E}_{V}}\propto A$
   ${{E}_{V}}={{a}_{v}}A$
2. The second term is the surface energy term: in volume term the interior nucleons are attracted equally from all sides but the nucleons which are on the surface of nucleus interact only with nucleus on one particular side this correction are mode in surface energy term.so surface term is a correction term
   Surface energy term ${{E}_{S}}\propto S$ where S is proportional to square of radius $({{R}^{2}})$
   $R\propto {{A}^{\dfrac{1}{3}}}$
   ${{E}_{S}}=-{{a}_{s}}{{A}^{\dfrac{2}{3}}}$
3. Coulomb energy is the third term
   Protons coulombic interaction destabilizes the nucleus. Protons are replied by the other (Z-1) protons so there are $\dfrac{Z(Z-1)}{2}$ repelling pairs of protons.
   Coulombic potential energy is inversely proportional to $R\propto {{A}^{\dfrac{1}{3}}}$
   $\therefore {{E}_{C}}\propto \dfrac{Z(Z-1)}{{{A}^{\dfrac{1}{3}}}}$
   ${{E}_{C}}=-{{a}_{c}}\dfrac{Z(Z-1)}{{{A}^{\dfrac{1}{3}}}}$
4. Asymmetry term is the fourth term
   When Z=N maximum stability is obtained and symmetry energy is directly proportional to excess neutron and the fraction of nuclear volume.
   $\begin{aligned} & {{E}_{A}}\propto (N-Z) \\\ & {{E}_{A}}\propto \dfrac{N-Z}{A} \\\ \end{aligned}$
   Symmetry term can be written as
   ${{E}_{A}}=-{{a}_{a}}\dfrac{{{(A-2Z)}^{2}}}{A}$
   Where N-Z=A-2Z
5. Pairing energy among the nucleus is the fifth term
   Mass of atomic nucleus is estimated as follows
   $m=Z{{m}_{p}}+N{{m}_{n}}-\dfrac{{{E}_{B}}}{{{C}^{2}}}$

Note:
Students atomic number is also called as proton number and mass number (A) of an atom is the addition of atomic number (Z) and number of neutrons (N).Atomic number is equal to number of electrons in uncharged atoms. Liquid drop model predicts a rough binding energy due to spherical shape of nuclei.