How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
`""_(1)^(2)H + ""_(1)^(2)H to ""_(2)^(3)He + n + 3.27 MeV`.

Explanation

2g of `._(1)^(2)H` consists of `6.023xx10^(23)` atoms
`2000g "of" ._(1)^(2)H` consists of `(cancel2xx10^(3)xx6.023xx10^(23))/(cancel2)` atoms
i.e., `N=6.023xx10^(26)` atoms
Number of possible reactions =`(N)/(2)=3.012xx10^(26)`
Energy released in the fusion `=3.012xx10^(26)xx3.27` MeV
`=9.849xx10^(26)`MeV
`=9.849xx10^(26)xx1.6xx10^(-19)xx10^(6)J`
`=15.76xx10^(13)J`
Given power of lamp =100W
`=100 J s^(-1)`
i.e., to spend 100 J It takes 1s, for `15.76xx10^(13)J` of energy to spent, it takes 'x' s `{:(x=(15.76xx10^(13))/(100)15.76xx10^(11)s," "1 year=365xx8.64xx10^(4)s),("or "x=(15.76xx10^(11))/(3.15xx10^(7))years ," "=3.15xx10^(7)s):}`
`x=5.0xx10^(4)` years