1.

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`

Answer» Number of deuterium atoms in `2.0kg=(6.023xx10^(23)xx2000)/2=6.023xx10^(26)`
Energy released when 2 atoms fuse `=3.27MeV`
`:.` Total energy released `=3.27/2xx6.023xx10^(26)MeV=1.635xx6.023xx10^(26)xx1.6xx10^(-13)J`
`=15.75xx10^(13)J`
Energy consumed by the bulb/sec`=100J`
`:.` Time for which bulb will glow `=(15.75xx10^(13))/100s=(15.75xx10^(11))/(60xx60xx24xx365)years=4.99xx10^4 "years"`


Discussion

No Comment Found

Related InterviewSolutions