Question

For given nuclear reactions:
\[ ^4_2\mathrm{He} + ^1_0\mathrm{n} \rightarrow ^3_2\mathrm{He} + 20\,\text{MeV} \] \[ ^4_2\mathrm{He} + ^1_0\mathrm{n} \rightarrow ^4_2\mathrm{He} - 0.9\,\text{MeV} \] $X_3$ represents stability of $^3_2\mathrm{He}$, $X_4$ represents stability of $^4_2\mathrm{He}$ and $X_5$ represents stability of $^5_2\mathrm{He}$. Compare the stabilities.

• A. $X_4>X_3>X_5$
• B. $X_4<X_3>X_5$
• C. $X_3>X_4>X_5$
• D. $X_4>X_5>X_3$

Hint:

Higher binding energy per nucleon always indicates higher nuclear stability.

Answer 1

The Correct Option is A

Step 1: Analyze the given nuclear reactions

The stability of a nucleus can be judged from whether energy is released (exothermic) or absorbed (endothermic) during its formation.

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Step 2: Reaction involving ³₂He

⁴₂He + ¹₀n → ³₂He + 20 MeV

This reaction is exothermic and releases 20 MeV of energy.

Release of energy indicates that the product nucleus ³₂He is relatively stable with respect to the reactants.

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Step 3: Reaction involving ⁵₂He

⁴₂He + ¹₀n → ⁴₂He − 0.9 MeV

This process is endothermic, requiring 0.9 MeV of energy.

Hence, the combined system corresponding to ⁵₂He is less stable than ⁴₂He.

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Step 4: Compare the stabilities

• ⁴₂He is highly stable, as it does not favor combination with an extra neutron.
• ³₂He is stable, but less stable than ⁴₂He.
• ⁵₂He is the least stable due to energy absorption.

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Final Answer:

The correct order of stability is:
X₄ > X₃ > X₅