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How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as `._1H^2+._1H^2to ._1H^3+n+3.17MeV`A. `2.4xx10^6` yearsB. `7.4xx10^4` yearsC. `1.6xx10^6` yearsD. `4.9xx10^4` years |
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Answer» Correct Answer - d Number of atoms present in 2 g of deuterium `=(6.023xx10^23xx2000)/2=6.023xx10^26` Energy released in the fusion of 2 deuterium atoms =3.27 MeV Total energy released in the fusion of 2.0 kg of deuterium atoms `E=3.27/2xx6.023xx10^26=9.81xx10^26` MeV `=15.696xx10^13` J Energy consumed by the bulb per second = 100 J Time for which the bulb will glow `t=(15.69xx10^13)/100 s` or `t=(15.69xx10^11)/(3.15xx10^7)` years `=4.9xx10^4` years . |
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