Question

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as 21H + 21H -→ 31He + n + 3.27 MeV

Answer 1

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

so, total number of atoms present in (2kg = 2000g) of deuterium = 6.023 × 10²³ × 2000/2 = 6.023 × 10^26

given, energy released in the fusion of 2 deuterium atoms = 3.27MeV

so, total energy released in the fusion of 2kg of deuterium atoms = 3.27/2 × 6.023 × 10^26 = 9.81 × 10^26 MeV

= 15.96 × 10¹³ J [ as we know, 1eV = 1.6 × 10^-19 J ]

given, energy consumed by bulb per second = 100J

so, time taken for which the bulb will glow = (15.69 × 10¹³)/100 = 15.69 × 10¹¹ sec = 15.69 × 10¹¹/(3600 × 24 × 365) = 4.9 × 10⁴ years .