Assuming that about 20 MeV of energy is released per fusion reaction 1 H 2 + 1 H 3 → 0 n 1 + 2 He 4 , then the mass of 1 H 2 consumed per day in a fusion reactor of power 1 MW will approximately be

Step-by-step solution

Energy produced U = P t = 10 6 × 24 × 36 × 10 2 = 24 × 36 × 10 8   J The energy released per fusion reaction = 20   MeV = 20 × 10 6 × 1.6 × 10 - 19 = 32 × 10 - 13   J The energy released per atom of 1 H 2 = 32 × 10 - 13   J Number of 1 H 2 atoms used = 24 × 36 × 10 8 32 × 10 - 13 = 27 × 10 21 Mass of 6 × 10 23 atom = 2   g ∴ Mass of 27 × 10 21 atoms = 2 6 × 10 23 × 27 × 10 21 ≈ 0 .1 g