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
$_{1}^{2} H+_{1}^{2} H \rightarrow_{2}^{3} H e+n+3.27 \;M e V$
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
$_{1}^{2} H+_{1}^{2} H \rightarrow_{2}^{3} H e+n+3.27 \;M e V$
The given fusion reaction is:
$_{1}^{2} H+_{1}^{2} H \rightarrow_{2}^{3} H e+n+3.27 M e V$
Amount of deuterium, $m=2 kg$
1 mole, i.e., 2 g of deuterium contains $6.023 \times 10^{23}$ atoms.
$\therefore 2.0 kg$ 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.
$\therefore$ Total energy per nucleus released in the fusion reaction:
$E=\frac{3.27}{2} \times 6.023 \times 10^{26} MeV$
$=\frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^{6}$
$=1.576 \times 10^{14} J$
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
$\frac{=1.576 \times 10^{14}}{100} s$
$\frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^{4}\; years$
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