Question

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`

Answer 1

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