13.19 How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:

\({ }_{1}^{2} \mathrm{H}+{ }_{2}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\)

Hint: 2 atoms of deuterium produces 3.27 MeV energy.
Step 1: Find the total number of atoms in 2 kg of deuterium.
The given fusion reaction is :
\({ }_{1}^{2} \mathrm{H}+{ }_{2}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\)
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 6.023×10232×2000=6.023×1026atoms
Step 2: Find the energy generated by 2 kg of deuterium.
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
Total energy released in the fusion reaction:
E=3.272×6.023×1026 MeV
=3.272×6.023×1026×1.6×10-19×106
=1.576×1014 J
Step 3: Find the time for which the lamp glows.
Power of the eletric lamp, P=100 W=100 J/s
Hene, the energy consumed by the lamp per second=100 J
The total time for which the electric lamp will glow = 1.576×1014100sec=1.576×1014100×60×60×24×3654.9×104 years