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: Use binding energy curve concept

The variation of binding energy per nucleon (BE/A) with mass number A is represented by the well-known binding energy curve.

This curve rises sharply for light nuclei, reaches a maximum near iron (A ≈ 56), and then slowly decreases for heavier nuclei.

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Step 2: Evaluate Statement I using the curve

Although BE/A decreases for very heavy nuclei, the total binding energy (BE) is given by:

BE = (BE/A) × A

Since A itself is large for heavy nuclei, the product (BE/A) × A continues to increase with mass number.

Hence, heavier nuclei generally possess larger total binding energy.

Therefore, Statement I is true.

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Step 3: Evaluate Statement II using nuclear stability argument

The stability of a nucleus depends on how strongly its nucleons are bound together. This is measured by binding energy per nucleon (BE/A).

Nuclei with higher BE/A are more stable, while nuclei with lower BE/A tend to undergo nuclear reactions such as fusion or fission to reach a more stable configuration.

Thus, BE/A is a direct indicator of nuclear stability.

Therefore, Statement II is true.

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Final Answer:

Both Statement I and Statement II are correct.