The binding energy for the following nuclear reactions are expressed in MeV .
${}_2\text{He}^3 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^4 + 20\text{MeV}$
${}_2\text{He}^4 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^5 - 0.9\text{MeV}$
If $\text{X}_3, \text{X}_4, \text{X}_5$ denote the stability of ${}_2\text{He}^3$, ${}_2\text{He}^4$ and ${}_2\text{He}^5$, respectively, then the correct order is :
Answer: (A) $X_4 > X_5 > X_3$
Energy released in a reaction = increase in binding energy (a free neutron has zero binding energy).
First reaction: $B_4 = B_3 + 20$ MeV.
Second reaction absorbs $0.9$ MeV: $B_5 = B_4 - 0.9$ MeV.
So $B_4 > B_5 > B_3$. He-4 gains a lot of binding energy by adding a neutron to He-3, while adding one more neutron to He-4 actually lowers the binding energy, so He-5 is less stable than He-4 but (having $19.1$ MeV more binding than He-3) more stable than He-3. The binding energy per nucleon follows the same order (with $B_3 \approx 7.7$ MeV: $2.6$, $6.9$ and $5.4$ MeV for He-3, He-4, He-5).
$$X_4 > X_5 > X_3$$
Solution by Sreeraj P, M.Sc Physics