The isotopic masses of `._(1)^(2)H` and `._(2)^(4)He` are `2.0141` and `4.0026` amu respectively and the velocity of light in vacuum is `2.998xx10^(8)

Question

The isotopic masses of . 2 1 H . and . 4 2 H e . are 2.0141 2.0141 and 4.0026 4.0026 amu respectively and the velocity of light in vacuum is 2.998 × 10 8 m / s 2.998 . Calculate the quantity of energy (in J J ) liberated when two mole of . 2 1 H . undergo fusion to form one mole of . 4 2 H e . Select the correct answer from above options

Answer 1

Correct Answer - 2.3 × 10 12 J 2.3 Δ E = Δ m × c 2 Δ . 1 H 2 + . _ ( 1 ) H 2 → _ . 2 H e 4 . Mass defect = Δ m = = = Mass of 2 ( . 1 H 2 ) − 2 Mass of H e H = 2 × 2.0141 − 4.0026 = = 4.0282 − 0026 = = 0.0256 g = = 0.256 × 10 − 3 k g = c = c speed of light = 2.998 × 10 8 m s − 1 = Δ E = Δ m × c 20 Δ = 0.0256 × 10 − 3 × ( 2.998 × 10 8 ) 2 = = 0.23 × 10 13 J = 2.3 × 10 12 J =