Given: mass(⁷₃Li) = 7.0160 u, mass(⁴₂He) = 4.0026 u and mass(¹₁H) = 1.0079 u. When 20 g of ⁷₃Li is converted into ⁴₂He by proton capture, find the energy liberated in kWh. Use 1 u c² ≈ 1 GeV.
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Question

Given: mass(⁷₃Li) = 7.0160 u, mass(⁴₂He) = 4.0026 u and mass(¹₁H) = 1.0079 u. When 20 g of ⁷₃Li is converted into ⁴₂He by proton capture, find the energy liberated in kWh. Use 1 u c² ≈ 1 GeV.

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G1 Physics · Accepted Answer

For ⁷₃Li + ¹₁H → 2(⁴₂He), the mass defect is Δm = 7.0160 + 1.0079 − 2(4.0026) = 0.0187 u. Thus Q ≈ 0.0187 × 931 ≈ 17.4 MeV per ⁷Li nucleus. The number of ⁷Li nuclei in 20 g is (20/7.016) times Avogadro's constant. Multiplying by Q and converting to kWh gives approximately 1.33 × 10⁶ kWh. Correct answer: B.