For a nucleus $^A_ZX$ having mass number $A$ and atomic number $Z$
A. The surface energy per nucleon $(b_s)=-a_1A^{2/3}$.
B. The Coulomb contribution to the binding energy $b_c=-a_2\dfrac{Z(Z-1)}{A^{4/3}}$.
C. The volume energy $b_v=a_3A$.
D. Decrease in the binding energy is proportional to surface area.
E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ($a_1$, $a_2$ and $a_3$ are constants)
Choose the most appropriate answer from the options given below:
Answer: (B) C, D only
In the liquid drop model the total binding energy is $a_3A-a_1A^{2/3}-a_2\dfrac{Z(Z-1)}{A^{1/3}}$.
- A: the surface term per nucleon goes as $A^{-1/3}$, not $A^{2/3}$ — incorrect.
- B: the Coulomb term goes as $A^{-1/3}$ (total), not $A^{-4/3}$ — incorrect.
- C: volume energy $\propto A$ — correct.
- D: nucleons at the surface have fewer neighbours, so the loss of binding energy is proportional to the surface area — correct.
- E: not an assumption of the model — incorrect.
So C and D only.
Solution by Sreeraj P, M.Sc Physics