A star initially has 10 40 deuterons. It produces energy via the processes 1 H 2 + 1 H 2 ⟶ 1 H 3 + p & 1 H 2 + 1 H 3 ⟶ 2 He 4 + n . If the average power radiated by the star is 10 16 W , the deuteron supply of the star is exhausted in a time of the order of (mass of H 2 1 = 2 . 014 amu , mass of He 4 2 = 4 . 001 amu , m p = 1 . 007 amu , m n = 1 . 008 amu )

Step-by-step solution

The given reactions are: 1 H 2 + 1 H 2 → 1 H 3 + p 1 H 2 + 1 H 3 → 1 He 4 + n 3 1 H 2 → 2 He 4 + n + p Mass defect Δm = ( 3 × 2 . 014 - 4 . 001 - 1 . 007 - 1 . 008 ) amu = 0 . 026 amu Energy released = 0 . 026 × 931 MeV = 0 . 026 × 931 × 1 . 6 × 10 - 12 J = 3 . 87 × 10 - 13 J This is the energy produced by the consumption of three deuteron atoms. Therefore Total energy released by 10 40 deuterons = 1 0 4 0 3 × 3 . 87 × 10 - 12 J = 1 . 29 × 1028 J The average power radiated is P = 10 16 W or 10 16 J s . Therefore, t