How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?

Solution:

The given fusion reaction is:

\${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{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.

$\therefore 2.0 \mathrm{~kg}$ 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:

$E=\frac{3.27}{2} \times 6.023 \times 10^{26} \mathrm{MeV}$

$=\frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^{6}$

$=1.576 \times 10^{14} \mathrm{~J}$

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:

$\frac{1.576 \times 10^{14}}{100} \mathrm{~s}$

$\frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^{4}$ years