JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2023
- Hard
- MCQ
For a nucleus \({ }_Z^A X\) having mass number \(A\) and atomic number \(Z\) \(A.\) The surface energy per nucleon \(\left(b_s\right)=a_1 A^{2 / 3}\) \(B.\) The Coulomb contribution to the binding energy \(b_c=-a_2 \frac{Z(Z-1)}{A^{4 / 3}}\) \(C.\) The volume energy \(b_v=a_3 A\) \(D.\) 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, \(\left(a_1, a_2\right.\) and \(a _3\) are constants) Choose the most appropriate answer from the options given below:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(C, D\) only
Step-by-step Solution
Detailed explanation
Surface energy per nucleon \(\propto \frac{r^2}{A} \propto \frac{A^{2 / 3}}{A} \propto \frac{1}{A^{1 / 3}}\) \(\left(\right.\) Mass number \(\left.A \propto r^3 \Rightarrow r \propto A^{1 / 3}\right)\).
A is incorrect
Contribution to binding energy by columbic forces is
\(=\frac{-a_2 Z(Z-1)}{A^{1 / 3}}\)
\(B\) is incorrect
Volume energy \(\propto A\)
\(C\) is correct
For \((D)\) ,if we consider only surface energy contribution then option is correct.
For \((E)\) only \(3\) interactions contribute to surface energy.
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