How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium ? The fusion reaction can be taken as `._1^2H + ._1^2H to ._2^3 He + n + 3.2 MeV `

Explanation

`"" _(1) H^(2) + "" _(1) H^(2) to "" _(2)_ ^(3) He + n + 3.27 MeV`
f No. of atoms in 2 kg of
` "" _(1) H ^(2)=2 //2 x 6.023 x 10 ^(26)` br> `= 6. 023 xx 10 ^(26)` atoms
In the above reaction two deuterium nuclei are combined
power (p) = w x rate of fusion.
`3.27 MeV x ("Number of atoms")/("Time exp ended")`
` 100 = 3.27 x 10 ^(6) x 1.6 x 10 ^(-19) x (6. 023 x 10 ^(26))/(2x)`
` therefore x = ( 3.27 xx 1.6 xx 6.023 xx 10 ^(+ 11))/(2)`
` = 15. 756 x 10 ^(11) S`
` = (15.756 xx 10 ^(11))/(365 xx 24 xx 60 xx 60) = (15.756 xx 10 ^(11))/(3.15 xx 10 ^(7))`
`= 5x 10 ^(4)` years