Number of atoms present in 2 g of deuterium = 6.023 × $10^{23}$ Total number of atoms present in 2000 g of deuterium = $\frac{6.023\times10^{23}\times2000}{2}$ = 6.023 × $10^{26}$ Energy released in the fusion of 2 deuterium atoms = 3.27 MeV Total energy released in the fusion of 2.0 kg of deuterium atoms E = $\frac{3.27}{2}$ × 6.023 × $10^{26}$ = 9.81 × $10^{26}$ MeV = 15.696 × $10^{13}$ J Energy consumed by the bulb per second = 100 J Time for which the bulb will glow t = $\frac{15.69\times10^{13}}{100}$ s or t = $\frac{15.69\times10^{11}}{3.15\times10^{7}}$ yearsz = 4.9 × $10^{4}$ years