If the binding energy per nucleon in Li7 and He4 nuclei are 5.60 MeV and 7.06 MeV, then energy of the reaction 3Li7 + 1H1 → 2 2He4 are

Q: If the binding energy per nucleon in Li⁷ and He⁴ nuclei are 5.60 MeV and 7.06 MeV, then energy of the reaction ₃Li⁷ + ₁H¹ → 2 ₂He⁴ are

(a) 19.6 MeV

(b) 2.4 MeV

(d) 17.3 MeV

Ans: (d)

Sol: ₃Li⁷ + ₁H¹ → 2 2 ₂He⁴

Energy of reaction =8 × 7.06-7 × 5.60 = 17.28 Me V