The binding energy for the following nuclear reactions are expressed in MeV. ₂He³ + ₀n¹ → ₂He⁴ + 20MeV ₂He⁴ + ₀n¹ → ₂He⁵ - 0.9MeV If X₃, X₄, X₅ denote the stability of ₂He³, ₂He⁴ and ₂He⁵, respectively, then the correct order is: (a) X₄ > X₅ > X₃ (b) X₄ X₅ < X₃ (d) X₄ = X₅ = X₃

Answer: (a). Stability is judged by binding energy per nucleon. From the reactions: B₄ = B₃ + 20 MeV and B₅ = B₄ - 0.9 = B₃ + 19.1 MeV. With B₃ ≈ 7.7 MeV: B₃/3 ≈ 2.6, B₄/4 ≈ 6.9, B₅/5 ≈ 5.4 MeV So X₄ > X₅ > X₃, answer (a).

The binding energy for the following nuclear reactions are expressed… | JEE Main 2026 (24 Jan, Shift 2)

Q4JEE Main 2026 · 24 JAN · SHIFT 2Medium

Mass Defect & Binding Energy

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 $\mathrm{X_3, X_4, X_5}$ denote the stability of ${}_{2}\text{He}^{3}, {}_{2}\text{He}^{4}$ and ${}_{2}\text{He}^{5}$ , respectively, then the correct order is:

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Answer: (a)

Stability is judged by binding energy per nucleon. From the reactions: $B_4 = B_3 + 20$ MeV and $B_5 = B_4 - 0.9 = B_3 + 19.1$ MeV.
With $B_3 \approx 7.7$ MeV: $\dfrac{B_3}{3} \approx 2.6$ , $\dfrac{B_4}{4} \approx 6.9$ , $\dfrac{B_5}{5} \approx 5.4$ MeV
So $X_4 > X_5 > X_3$ , answer (a).

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Mass Defect & Binding Energy question from JEE Main 2026 (24 Jan, Shift 2)

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