JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

The binding energy for the following nuclear reactions are expressed in MeV.

$_{2} He^{3} +_{0} n^{1} \to_{2} He^{4} + 20 MeV$

$_{2} He^{4} +_{0} n^{1} \to_{2} He^{5} - 0.9 MeV$

If $X_{3}, X_{4}, X_{5}$ denote the stability of $_{2} He^{3}$ , $_{2} He^{4}$ and $_{2} He^{5}$ , respectively, then the correct order is :

Solution

The energy released in a nuclear reaction equals the difference in binding energies (Q-value form):

From reaction (1): $BE (He^{4}) - BE (He^{3}) = + 20$ MeV (energy released).

From reaction (2): $BE (He^{5}) - BE (He^{4}) = - 0.9$ MeV (energy absorbed).

So $BE (He^{4}) > BE (He^{5}) > BE (He^{3})$ . Stability mirrors binding energy.

$∴ X_{4} > X_{5} > X_{3}$ .

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