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 &#160