Question

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
$^{2}_{1}H + ^{2}_{1}H →^{3}_{1}He + n + 3.27 \space MeV$

Answer 1

The given fusion reaction is:
$^{2}_{1}H + ^{2}_{1}H →^{3}_{1}He + n + 3.27 \space MeV$
Amount of deuterium, m = 2 kg
mole, i.e., 2 g of deuterium contains. $6.023 × 10^{23}$ atoms.
2.0 kg of deuterium contains = $\frac{6.023 \times 10^{23}}{2} \times 2000 = 6.023 \times 10^{26} atoms$
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:
$E = \frac{3.27}{2} \times 6.023 \times 10^{26} MeV$
$E = \frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19 }\times 10^{6}$
$E = 1.576 \times10^{14} J$
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:
$\frac{1.576 \times 10^{14}}{100} s$
$= \frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365}$
$≃ 4.9 \times 10^{4}$ Years