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 × 10^{23} 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