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InterviewSolution
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How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium ? Take the fusion reaction as " "_(1)^(2)H+ " "_(1)^(2)H to " "_(2)^(3)He+n+3.27 MeV. |
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Answer» Solution :The amount of energy released per ACT of fusion = 3.27 MeV and a fusion required 2 nuclei of `" "_(1)^(2)H`, hence Energy released in 2.0 kg of `" "_(1)^(2)H = (("Number of atoms in 2 kg"))/2 xx ("Energy per act of fusion")=(6.023xx10^(26))/2xx 3.27 MeV =9.85 xx 10^(26) MeV = 15.76 xx 10^(13) J` LET the 100 W bulb run for t s, then total energy consumed is given by 100`xx t = 15.76 xx 10^(13) implies t =15.76xx10^(11)s= (15.76xx10^(11))/(3.154xx10^(7))year = 4.9 xx 10^(4)`year. |
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