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`

Explanation

the given fusion reaction is :
`""_(1)^(2)H+_(2)^(3)H to _(2)^(3)He +n +3.27 MeV`
Amount of deutrium ,m=2 kg
1 mole .i.e ., 2 g deuteriium contains `6.023xx10^(23)` atoms .
`therefore ` 2.0 kg of deuterium contains `=(6.023xx10^(23))/(2)xx2000=6.023xx10^(26)` atoms
it can be infrerred from the given reaction that when two atoms of seuterium fuse , 3.27 MeV energy is released .
`therefore` total energy per nucles released in the fusion reaction :
`E=(3.27)/(2)xx6.023 xx10^(26)MeV`
`=(3.27)/(2)xx6.023 xx10^(26) xx1.6 xx10^(-19)xx10^(6)`
`=1.576 xx 10^(14) j`
Power of the electric , p = 100 W = 100 j/s
Hence the energy consumed by the lamp per lamp per second = 100 j
the total time for which the electric lamp will glow is calculataed as:
`(1.576 xx10^(14))/(100)s`
`(1.576xx10^(14))/(100xx60xx60xx24xx365)~~4.9xx10^(4)` years