Number of atoms present in 2 g of deuterium = 6.023 × 1023
Total number of atoms present in 2000 g of deuterium
=

6.023×1023×2000

2

= 6.023 × 1026
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 =

3.27

2

× 6.023 × 1026 = 9.81 × 1026 MeV = 15.696 × 1013 J
Energy consumed by the bulb per second = 100 J
Time for which the bulb will glow
t =

15.69×1013

100

s or t =

15.69×1011

3.15×107

yearsz = 4.9 × 104 years