The Question : How long can an electric lamp of 100 W be kept glowing by fusion of 2 kg of deuterium? Take the fusion reaction as
Solution for the question :
Number of atoms present in 2 g of deuterium = 6 × 1023Number of atoms present in 2.0 Kg of deuterium = 6 × 1026Energy released in fusion of 2 deuterium atoms= 3.27 MeVEnergy released in fusion of 2.0 Kg of deuterium atoms, = 3.27/2 × 6 × 1026 MeV= 9.81 × 1026 MeV= 15.696 × 1013 JEnergy consumed by bulb per sec = 100 J∴ Time for which bulb will glow= 15.696 × 1013/100s = 4.97 × 104 year
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