2He4 + 0n1 → 2He5 − 0.9 MeV If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5, respectively, then the correct order is
Q. ₂He⁴ + ₀n¹ → ₂He⁵ − 0.9 MeV 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
(C)  X₄ > X₅ < X₃
(D)  X₄ = X₅ = X₃

Correct Answer: X₄ > X₅ > X₃

Explanation

Stability of a nucleus depends on its binding energy per nucleon. Higher binding energy per nucleon implies greater stability.

The given nuclear reaction is:

₂He⁴ + ₀n¹ → ₂He⁵ − 0.9 MeV

Negative energy (−0.9 MeV) indicates that energy is absorbed, meaning ₂He⁵ is less stable than ₂He⁴.

Now compare the three isotopes:

₂He⁴ is a highly stable nucleus with very high binding energy per nucleon.

₂He⁵ is unstable but still more stable than ₂He³, since it has one more neutron providing additional nuclear force.

₂He³ has fewer nucleons and lower binding energy per nucleon, hence it is the least stable among the three.

Therefore, the correct order of stability is:

X₄ > X₅ > X₃

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