Calculate for How many years will the fusion of 2.0 Kg deuterium keep the 800 W electric lamp glowing. Take the fusion reaction as
21H + 21H → 32He + 10n +3.27 MeV
[2020 | Delhi | Section - B | SET - 3 Q.24]
Solution :
Energy released by 1 atom of deuterium
= (3.27×106×1.6×10-19)/2
= 2.616×10-13 J
No. of moles in 2 Kg of deuterium
= given mass / molar mass
= 2000g / 2g
= 1000
No. of atoms in 2 kg (2000 g) of deuterium
= Avogadro Number × No. of moles
= 6.023×1023×1000
= 6.023×1026
Total energy released by the fusion of 2 kg of deuterium
= No. of atoms × Energy released by 1 atom of deuterium
= 6.023×1026×2.616×10-13
= 15.756168 × 1013 J
Power = Energy / time
Time = Energy/Power
Time = (Total Energy released by fusion of 2 kg of deuterium) / Power of lamp
Time = {[15.756168 × 1013]/800} sec
Time = {[15.756168 × 1013]/[800×60×60×24×365}
Time = 2.28×106 Years.
