NTA Abhyas JEE Main2020PhysicsNuclear PhysicsPractice
The atomic masses of   1 2 H and   2 4 H   are 2 . 0141   u and 4 . 0026   u respectively. Taking the velocity of light in vacuum to be 2.998 × 10 8   m   s - 1 , calculate the amount of energy (in J ) liberated when two moles of   1 2 H undergo a fusion to form one mole of   2 4 H . [ u = 1 . 66057 × 10 - 27   kg ]
Options
- A2.3 × 10 12 J
- B3.3 × 10 12 J
- C5.3 × 10 12 J
- D2.9 × 10 12 J
Correct answer
A. 2.3 × 10 12 J
Step-by-step solution
2 1 2 H → 2 4 H + E n e r g y Mass defect = 2 × 2.0141 - 4.0026 = 0.0256 a m u ∴ ∆ E = ∆ m × c 2 = 0.0256 × 1.66057 × 10 - 27 × 6.02 × 10 23 × 2.998 × 10 8 2 = 2.3 × 10 12 J