You are given that Mass of ₃⁷Li = 7.0160u,
Mass of ₂⁴He = 4.0026u
and Mass of ₁¹H = 1.0079u.
When 20 g of ₃⁷Li is converted into ₂⁴He by proton capture, the energy liberated, (in kWh), is :
[Mass of nucleon = 1 GeV/c²]

Solution

Physics sir · 4 steps

Find the mass lost in one capture, turn it into MeV, then multiply by the number of Li nuclei in 20 g.

1. Write the reaction. One proton is captured by one lithium nucleus, and two helium nuclei come out: ⁷₃Li + ¹₁H → 2 ⁴₂He. The mass of the products is less than the mass of the reactants.

   Δ m = (7.0160 + 1.0079) - 2 × 4.0026 = 8.0239 - 8.0052 = 0.0187 u

2. Energy per capture. Use 1 u = 931.5 MeV/c². Each capture releases 0.0187 × 931.5 ≈ 17.42 MeV. If you round 1 u to the given 1 GeV/c², you get 18.7 MeV and about 1.43 × 10⁶ kWh. That is still nearest to (d), but the options were built with 931.5.
3. Count the nuclei. 20 g of lithium-7 contains 20/7 × 6.022 × 10²³ = 1.72 × 10²⁴ nuclei. Total energy = 1.72 × 10²⁴ × 17.42 MeV × 1.6 × 10⁻¹³ J MeV⁻¹ ≈ 4.8 × 10¹² J.
4. So. Divide by 1 kWh = 3.6 × 10⁶ J: (4.8 × 10¹²)/(3.6 × 10⁶) ≈ 1.33 × 10⁶ kWh, which is option (d).

The idea to keep. For energy released by a bulk sample: mass defect per reaction, times 931.5 MeV per u, times the number of nuclei (moles times N_A). Convert to joules last, then to kWh by dividing by 3.6 × 10⁶.

Answer from the NTA final answer key. Solution by PrepWiser.