Question

Statement - I : Greater is the mass of nucleus, more will be its binding energy.
Statement - II : Nucleus with less $\frac{BE}{A}$ (Binding energy/nucleon) breaks into nucleus with higher $\frac{BE}{A}$.
Choose the correct option :

• A. Statement I is true & statement II is false
• B. Statement I is false & statement II is true
• C. Both are true
• D. Both are false

Hint:

Do not confuse Total Binding Energy (BE) with Binding Energy per Nucleon ($BE/A$). Total BE almost always increases with mass number A. However, $BE/A$ peaks around Iron (Fe-56) and then decreases. It's the $BE/A$ that determines nuclear stability, not total BE.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is a statement-based question dealing with the concepts of nuclear binding energy (BE) and binding energy per nucleon ($BE/A$). We need to evaluate the truthfulness of both statements. 
Step 3: Detailed Explanation: 
Statement - I: Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. Since binding energy roughly scales with the number of nucleons (A), a nucleus with a greater mass (higher A) generally has a greater total binding energy. Even though the binding energy *per nucleon* drops for heavy nuclei (A $>$ 60), the *total* binding energy continues to increase with mass. Therefore, Statement I is generally True. 
Statement - II: Nuclear stability is determined by the binding energy per nucleon ($BE/A$). A higher $BE/A$ means the nucleus is more tightly bound and therefore more stable. Nuclei with low $BE/A$ (very light or very heavy nuclei) tend to undergo nuclear reactions (fusion or fission) to form nuclei with higher $BE/A$, releasing energy in the process. Therefore, Statement II is True. 
Since both Statement I and Statement II are correct in the context of nuclear physics, the correct option is (C). 
Step 4: Final Answer: 
Both Statement I and Statement II are true.