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How long can an electric lamp of 100 W be kept glowing by fusion of 2 kg of deuterium? Take the fusion reaction as\(^2_1\)H + \(^2_1\)H → \(^3_2\)He + n + 3.27 MeV |
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Answer» Number of atoms present in 2 g of deuterium = 6 × 1023 Number of atoms present in 2.0 Kg of deuterium = 6 × 1026 Energy released in fusion of 2 deuterium atoms = 3.27 MeV Energy released in fusion of 2.0 Kg of deuterium atoms = \(\frac{3.27}{2}\) × 6 × 1026 MeV = 9.81 × 1026 MeV = 9.81 × 1026 MeV = 15.696 × 1013 J Energy consumed by bulb per sec = 100 J Time for which bulb will glow = \(\frac{15.696\times10^{13}}{100}\) s = 4.97 × 104 year |
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