For a nucleus _A^A X having mass number A and atomic number Z
A. The surface energy per nucleon (bₛ)=-a₁ A^(2 / 3).
B. The Coulomb contribution to the binding energy b_c=-a₂ Z(Z-1)/A^(4 / 3)
C. The volume energy bᵥ=a₃ A
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. ( a₁, a₂ and a₃ are constants)
Choose the most appropriate answer from the options given below:

Solution

Physics sir · 3 steps

Check each term against the liquid drop formula and watch which ones are per nucleon and which are totals.

1. The model. The liquid drop model builds the binding energy from a few terms. The volume term adds to it. The surface and Coulomb terms subtract from it.

   B = a₃ A - a₁ A^(2/3) - a₂ Z(Z-1)/A^(1/3)

2. Where each comes from. Volume: every nucleon bonds only with its near neighbours, so B ∝ A. Surface: nucleons on the surface have fewer neighbours, and this loss scales with the area 4π R² ∝ A^(2/3). Coulomb: Z(Z-1)/2 proton pairs repel, with energy ∝ 1/R ∝ A^(-1/3).

   Statement	Correct form	Verdict
   A: surface energy per nucleon = -a₁ A^(2/3)	-a₁ A^(2/3) is the total. Per nucleon it is -a₁ A^(-1/3)	false
   B: Coulomb term = -a₂ Z(Z-1)/A^(4/3)	-a₂ Z(Z-1)/A^(1/3), because R ∝ A^(1/3)	false
   C: volume energy = a₃ A	a₃ A	true
   D: decrease ∝ surface area	∝ 4π R² ∝ A^(2/3)	true
   E: each nucleon has 12 partners	Surface nucleons have fewer neighbours than interior ones	false

3. So. If every nucleon had the full set of neighbours, there would be no surface term at all, so E fails. Only C and D are correct, which is option (a).

The idea to keep. Write each term from geometry: volume ∝ A, surface ∝ R² ∝ A^(2/3), Coulomb ∝ Z(Z-1)/R ∝ Z(Z-1)/A^(1/3). Then divide by A only if the question says per nucleon.

Answer from the NTA final answer key. Solution by PrepWiser.