Q36 (2026)
The binding energy for the following nuclear reactions are expressed in MeV.
$_{2}\mathrm{He}^{3} + _{0}\mathrm{n}^{1} \rightarrow _{2}\mathrm{He}^{4} + 20 \text{ MeV}$
$_{2}\mathrm{He}^{4} + _{0}\mathrm{n}^{1} \rightarrow _{2}\mathrm{He}^{5} - 0.9 \text{ MeV}$
If $X_{3}, X_{4}, X_{5}$ denote the stability of $_{2}\mathrm{He}^{3}, _{2}\mathrm{He}^{4}$ and $_{2}\mathrm{He}^{5}$ , respectively, then the correct order is:
- $X_{4} > X_{5} > X_{3}$
- $X_{4} = X_{5} = X_{3}$
- $X_{4} > X_{5} < X_{3}$
- $X_{4} < X_{5} < X_{3}$
### Related Formula
$$\Delta Q = BE_{\text{products}} - BE_{\text{reactants}}$$
### Core Logic
Reaction 1 releases energy ($+20 \text{ MeV}$), meaning the product is more stable:
$$BE_{He^{4}} - BE_{He^{3}} = 20 \text{ MeV} \quad \dots (1)$$
Reaction 2 absorbs energy ($-0.9 \text{ MeV}$), meaning the product has less total binding energy change but is relatively close:
$$BE_{He^{5}} - BE_{He^{4}} = -0.9 \text{ MeV} \quad \dots (2)$$
### Step 1: Compare Stabilities
From (1): $BE_{He^{4}}$ is significantly larger than $BE_{He^{3}}$.
From (2): $BE_{He^{5}} = BE_{He^{4}} - 0.9 \text{ MeV}$, which implies $BE_{He^{5}}$ is slightly less than $BE_{He^{4}}$.
Thus, $BE_{He^{4}} > BE_{He^{5}} > BE_{He^{3}}$.
The stability directly corresponds to the total binding energy here (and BE per nucleon leads to a similar ranking given small mass differences), so:
$X_4 > X_5 > X_3$
### Pattern Recognition
Positive Q-value means the products are structurally tighter (more stable). Negative Q-value means you lost stability. Simply tracking algebraic differences gives the hierarchical ranking.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
