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: Understanding stability and binding energy.
Greater binding energy implies greater nuclear stability.
Step 2: Analysis of first reaction.
\[ BE_{^4\mathrm{He}} - BE_{^3\mathrm{He}} = 20\,\text{MeV} \]
This implies $^4_2\mathrm{He}$ is more stable than $^3_2\mathrm{He}$.
Step 3: Analysis of second reaction.
\[ BE_{^5\mathrm{He}} - BE_{^4\mathrm{He}} = -0.9\,\text{MeV} \]
Negative value indicates $^5_2\mathrm{He}$ is less stable than $^4_2\mathrm{He}$.
Step 4: Comparing stabilities.
\[ BE_{^4\mathrm{He}}>BE_{^3\mathrm{He}}>BE_{^5\mathrm{He}} \]
Step 5: Final order.
\[ X_4>X_3>X_5 \]