Question

The binding energy for the following nuclear reactions are expressed in MeV.
\[ {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \text{ MeV} \] \[ {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \text{ MeV} \] If \( X_3, X_4, X_5 \) denote the stability of \( {}^{3}_{2}\text{He}, {}^{4}_{2}\text{He} \) and \( {}^{5}_{2}\text{He} \), respectively, then the correct order is:

• A. \( X_4>X_5>X_3 \)
• B. \( X_4 = X_5 = X_3 \)
• C. \( X_4>X_5<X_3 \)
• D. \( X_4<X_5<X_3 \)

Hint:

Greater binding energy implies higher nuclear stability. A negative binding energy indicates an unstable nucleus.

Answer 1

The Correct Option is A

Step 1: Relation between binding energy and stability.
In nuclear physics, higher binding energy corresponds to greater nuclear stability. A nucleus that releases more energy during its formation is more tightly bound and hence more stable.
Step 2: Analysis of the first reaction.
\[ {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \text{ MeV} \] The release of a large amount of energy (20 MeV) indicates that \( {}^{4}_{2}\text{He} \) is highly stable compared to \( {}^{3}_{2}\text{He} \). Hence, \[ X_4>X_3 \]
Step 3: Analysis of the second reaction.
\[ {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \text{ MeV} \] The negative energy value shows that energy is required to form \( {}^{5}_{2}\text{He} \), making it less stable than \( {}^{4}_{2}\text{He} \). However, it is still more stable than \( {}^{3}_{2}\text{He} \).
Step 4: Final conclusion.
Therefore, the correct stability order is: \[ X_4>X_5>X_3 \]