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

Number of deuterium atoms in 2 kg.
`=(6.023 xx 10^(23))/2 xx 2,000 = 6.023 xx 10^(26)`
Energy released when two atoms fuse = 3.27 MeV Average energy released when`6.023 xx 10^(26)` nuclei of deuterium fuse together.
`E = 3.27/2 xx 6.023 xx 10^(26)` MeV
`=(3.27 xx 6.023 xx 10^(26))/2 xx 1.6 xx 10^(-12) J`
`=15.75 xx 10^(13)` J
Power of lamp, P = 100 W
If the lamp glows for time t, then electric energy consumed = P.t = 100t.
`therefore 100t =E = 15.75 xx 10^(13)`
`t = 15.75 xx 10^(11) = (15.75 xx 10^(11))/(365 xx 86,400)` years.
or `t=4.99 xx 10^(4)` years.