# Calculate for how many years will the fusion of 2.0 kg deuterium keep 800 W electric lamp glowing. Take the fusion reaction as

The energy released per atom

$=\frac{3.27}{2}\times1.6\times10^{-13}J$

Number of atoms in 2Kg deuterium

$=\frac{6.023\times10^{23}\times2000}{2}=6.023\times10^{26}$

Total energy released

$=6.023\times10^{26}\times\frac{3.27}{2}\times1.6\times10^{-13}=15.76\times10^{13}J$

Time required=total energy / power of lamp

$=\frac{15.76\times10^{13}}{800\times(365\times24\times3600)}=2.28\times10^{8}years$

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