The binding energy for the following nuclear reactions are expressed in MeV. $_{2}He^{3} + _{0}n^{1} \rightarrow _{2}He^{4} + 20$ MeV $_{2}He^{4} + _{0}n^{1} \rightarrow _{2}He^{5} - 0.9$ MeV If X₃, X₄, X₅ denote the stability of $_{2}He^{3}$, $_{2}He^{4}$ and $_{2}He^{5}$, respectively, then the correct order is:

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Modern Physics

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

_{2} H e^{3} +_{0} n^{1} \to_{2} H e^{4} + 20

MeV

_{2} H e^{4} +_{0} n^{1} \to_{2} H e^{5} - 0.9

MeV
If X₃, X₄, X₅ denote the stability of

_{2} H e^{3}

_{2} H e^{4}

and

_{2} H e^{5}

, respectively, then the correct order is:

Answer: (A)

The energy released in

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

MeV equals the gain in binding energy, so

B (He^{4}) - B (He^{3}) = 20

MeV and

B (He^{4}) > B (He^{3})

. The second reaction absorbs

0.9

MeV, so

B (He^{5}) - B (He^{4}) = - 0.9

MeV, giving

B (He^{4}) > B (He^{5})

while

B (He^{5}) = B (He^{3}) + 19.1

MeV still exceeds

B (He^{3})

. Larger binding energy means greater stability, so the order is

X_{4} > X_{5} > X_{3}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option