How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:
21H + 21H → 32He + n + 3.27 MeV
The reaction is given As:
Amount of Deuterium fuel = 2.0 kg
∵ 2 gm of deuterium contains 6.023 × 1023 atoms
Hence, 2 kg of deuterium contains 6.023 × 1023 × 103 atoms
From the reaction we can infer 2 g of deuterium gives 3.27 MeV energy
∴ Hence, Total energy released in this reaction = 
MeV
= 9.847 × 1026 × 106 × 1.6 × 10-19J
= 1.576 × 1014 J.
∵ The power of the lamp is 100 W = 100J/s
Hence, time of glow by this much energy is = 
= 1.576 × 1012 sec
= (1.576 × 1012)/(60 × 60 × 24)
= 4.9 × 104 years.
19