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How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reactions as ""_(1)^(2)H + ""_(1)^(2)H to ""_(2)^(3)He + n + 3.27 MeV. |
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Answer» Solution :MASS of deuterium = `2.0141 xx 2 = 4.0282u = 4.0282 xx 1.66 xx 10^(-27) KG` No. of fusion in kg of deuterium `= (2)/(4.0282 xx 1.66 xx 10^(-27)) = 2.991 xx 10^(26)kg` ENERGY released per fusion = `3.2 MeV = 3.2 xx 10^6 xx 1.6 xx 10^(-19)J` `:.` Energy released from 2 kg of `""_(1)^(2)H = E = 3.2 xx 10^(6) xx 1.6 xx 10^(-19) xx 2.991 xx 10^(26)J` `= 1.5314 xx 10^(14)J` `E = P xx t` `t = E/P = (1.5314 xx 10^(14))/(100) s = (1.5314 xx 10^(14))/(100 xx 365 xx 24 xx 60 xx 60) = 4.85 xx 10^4` years . |
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