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How long can an electric lamp of 100W be kwpt glowing by fusion of 2.0 kg of deuterium ? Take fusion reaction as `._(1)^(2)H+._(1)^(2)Hto ._(2)^(3)He+n+3.27 MeV` |
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Answer» Number of deuterium atoms in 2.0 kg `= (6.023xx10^(23)xx2000)/(2)=6.023xx10^(26)` Energy released when 2 atoms fuse = 3.27 MeV `therefore` Total energy released `=(3.27)/(2)xx6.023xx10^(26)MeV` `= 1.635xx6.023xx10^(26)xx1.6xx10^(-13)J` `= 15.75xx10^(3)J` Energy consumed by the bulb/sec = 100 J `therefore` Time for which bulb will glow `= (15.75xx10^(13))/(100)S` `= (15.75xx10^(11))/(60xx60xx24xx365)=4.99xx10^(7)` years |
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