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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
\[\ce{^2_1H + ^2_1H -> ^3_2He + n + 3.27 MeV}\]
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Solution
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 = `(3.27)/2 xx 6 xx 10^26` MeV
= 9.81 × 1026 MeV
= 15.696 × 1013 J
Energy consumed by bulb per sec = 100 J
Time for which bulb will glow = `(15.696 xx 10^13)/100` s = 4.97 × 104 year
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