The binding energy for the following nuclear reactions are expressed in MeV.
³₂He + ¹₀n → ⁴₂He + 20 MeV
⁴₂He + ¹₀n → ⁵₂He - 0.9 MeV
If X₃, X₄, X₅ denote the stability of ³₂He, ⁴₂He and ⁵₂He, respectively, then the correct order is :

Question

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

³₂He + ¹₀n → ⁴₂He + 20 MeV

⁴₂He + ¹₀n → ⁵₂He - 0.9 MeV

If X₃, X₄, X₅ denote the stability of ³₂He, ⁴₂He and ⁵₂He, respectively, then the correct order is :

G1 Physics · Accepted Answer

For the first reaction, ³₂He + ¹₀n → ⁴₂He + 20 MeV. The released energy is Q₁ = +20 MeV. Since the free neutron has zero binding energy, BE(⁴₂He) - BE(³₂He) = 20 MeV. Therefore, BE(⁴₂He) > BE(³₂He). For the second reaction, ⁴₂He + ¹₀n → ⁵₂He - 0.9 MeV. This means the reaction is endothermic by 0.9 MeV: Q₂ = -0.9 MeV. Hence, BE(⁵₂He) - BE(⁴₂He) = -0.9 MeV. Therefore, BE(⁴₂He) > BE(⁵₂He). Also, BE(⁵₂He) = BE(⁴₂He)-0.9 MeV and BE(³₂He) = BE(⁴₂He)-20 MeV. Thus, BE(⁴₂He) > BE(⁵₂He) > BE(…