How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
Solution
The given fusion reaction is:
\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.
∴2.0 kg of deuterium contains = `(6.023 xx 10^23)/2 xx 2000 = 6.023 xx 10^26` atoms
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴ Total energy per nucleus released in the fusion reaction:
`"E" = 3.27/2 xx 6.023 xx 10^26 "MeV"`
`= 3.27/2 xx 6.023 xx 10^26 xx 1.6 xx 10^(-19) xx 10^6`
`= 1.576 xx 10^14 " J"`
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
`(1.576 xx 10^14)/100`s
`(1.576 xx 10^14)/(100 xx 60 xx 60 xx 24 xx 365) ~~ 4.9 xx 10^4` year
