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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

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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